\(\int \frac {1}{(d \tan (e+f x))^{5/2} (a+a \tan (e+f x))^3} \, dx\) [377]

Optimal result

Integrand size = 25, antiderivative size = 215 \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+a \tan (e+f x))^3} \, dx=\frac {59 \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 d^{5/2} f}-\frac {\arctan \left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{2 \sqrt {2} a^3 d^{5/2} f}-\frac {55}{24 a^3 d f (d \tan (e+f x))^{3/2}}+\frac {63}{8 a^3 d^2 f \sqrt {d \tan (e+f x)}}+\frac {11}{8 a^3 d f (d \tan (e+f x))^{3/2} (1+\tan (e+f x))}+\frac {1}{4 a d f (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2} \] Output:

59/8*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/d^(5/2)/f-1/4*arctan(1/2*(d^ 
(1/2)-d^(1/2)*tan(f*x+e))*2^(1/2)/(d*tan(f*x+e))^(1/2))*2^(1/2)/a^3/d^(5/2 
)/f-55/24/a^3/d/f/(d*tan(f*x+e))^(3/2)+63/8/a^3/d^2/f/(d*tan(f*x+e))^(1/2) 
+11/8/a^3/d/f/(d*tan(f*x+e))^(3/2)/(1+tan(f*x+e))+1/4/a/d/f/(d*tan(f*x+e)) 
^(3/2)/(a+a*tan(f*x+e))^2
 

Mathematica [A] (verified)

Time = 6.47 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.14 \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+a \tan (e+f x))^3} \, dx=\frac {1}{4 a d f (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2}+\frac {\frac {11 a}{2 f (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))}+\frac {-\frac {55 a^3 d}{3 f (d \tan (e+f x))^{3/2}}+\frac {a^3 \left (59 d^{5/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )-2 \sqrt {2} d^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )+2 \sqrt {2} d^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )+\frac {63 d^3}{\sqrt {d \tan (e+f x)}}\right )}{d^3 f}}{2 a^3 d}}{4 a^3 d} \] Input:

Integrate[1/((d*Tan[e + f*x])^(5/2)*(a + a*Tan[e + f*x])^3),x]
 

Output:

1/(4*a*d*f*(d*Tan[e + f*x])^(3/2)*(a + a*Tan[e + f*x])^2) + ((11*a)/(2*f*( 
d*Tan[e + f*x])^(3/2)*(a + a*Tan[e + f*x])) + ((-55*a^3*d)/(3*f*(d*Tan[e + 
 f*x])^(3/2)) + (a^3*(59*d^(5/2)*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]] - 2* 
Sqrt[2]*d^(5/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]] + 2*Sqr 
t[2]*d^(5/2)*ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]] + (63*d^3) 
/Sqrt[d*Tan[e + f*x]]))/(d^3*f))/(2*a^3*d))/(4*a^3*d)
 

Rubi [A] (verified)

Time = 1.93 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.16, number of steps used = 22, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.840, Rules used = {3042, 4052, 27, 3042, 4132, 3042, 4132, 27, 3042, 4133, 27, 3042, 4136, 27, 3042, 4015, 218, 4117, 27, 73, 216}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{(a \tan (e+f x)+a)^3 (d \tan (e+f x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a \tan (e+f x)+a)^3 (d \tan (e+f x))^{5/2}}dx\)

\(\Big \downarrow \) 4052

\(\displaystyle \frac {\int \frac {7 d \tan ^2(e+f x) a^2+11 d a^2-4 d \tan (e+f x) a^2}{2 (d \tan (e+f x))^{5/2} (\tan (e+f x) a+a)^2}dx}{4 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 (d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {7 d \tan ^2(e+f x) a^2+11 d a^2-4 d \tan (e+f x) a^2}{(d \tan (e+f x))^{5/2} (\tan (e+f x) a+a)^2}dx}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 (d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {7 d \tan (e+f x)^2 a^2+11 d a^2-4 d \tan (e+f x) a^2}{(d \tan (e+f x))^{5/2} (\tan (e+f x) a+a)^2}dx}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 (d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 4132

\(\displaystyle \frac {\frac {\int \frac {55 d^2 a^4+55 d^2 \tan ^2(e+f x) a^4-8 d^2 \tan (e+f x) a^4}{(d \tan (e+f x))^{5/2} (\tan (e+f x) a+a)}dx}{2 a^3 d}+\frac {11}{f (\tan (e+f x)+1) (d \tan (e+f x))^{3/2}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 (d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {55 d^2 a^4+55 d^2 \tan (e+f x)^2 a^4-8 d^2 \tan (e+f x) a^4}{(d \tan (e+f x))^{5/2} (\tan (e+f x) a+a)}dx}{2 a^3 d}+\frac {11}{f (\tan (e+f x)+1) (d \tan (e+f x))^{3/2}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 (d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 4132

\(\displaystyle \frac {\frac {-\frac {2 \int \frac {3 \left (63 d^4 a^5+55 d^4 \tan ^2(e+f x) a^5\right )}{2 (d \tan (e+f x))^{3/2} (\tan (e+f x) a+a)}dx}{3 a d^3}-\frac {110 a^3 d}{3 f (d \tan (e+f x))^{3/2}}}{2 a^3 d}+\frac {11}{f (\tan (e+f x)+1) (d \tan (e+f x))^{3/2}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 (d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-\frac {\int \frac {63 d^4 a^5+55 d^4 \tan ^2(e+f x) a^5}{(d \tan (e+f x))^{3/2} (\tan (e+f x) a+a)}dx}{a d^3}-\frac {110 a^3 d}{3 f (d \tan (e+f x))^{3/2}}}{2 a^3 d}+\frac {11}{f (\tan (e+f x)+1) (d \tan (e+f x))^{3/2}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 (d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {-\frac {\int \frac {63 d^4 a^5+55 d^4 \tan (e+f x)^2 a^5}{(d \tan (e+f x))^{3/2} (\tan (e+f x) a+a)}dx}{a d^3}-\frac {110 a^3 d}{3 f (d \tan (e+f x))^{3/2}}}{2 a^3 d}+\frac {11}{f (\tan (e+f x)+1) (d \tan (e+f x))^{3/2}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 (d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 4133

\(\displaystyle \frac {\frac {-\frac {-\frac {2 \int \frac {63 a^6 d^6+63 a^6 \tan ^2(e+f x) d^6+8 a^6 \tan (e+f x) d^6}{2 \sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx}{a d^3}-\frac {126 a^4 d^3}{f \sqrt {d \tan (e+f x)}}}{a d^3}-\frac {110 a^3 d}{3 f (d \tan (e+f x))^{3/2}}}{2 a^3 d}+\frac {11}{f (\tan (e+f x)+1) (d \tan (e+f x))^{3/2}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 (d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-\frac {-\frac {\int \frac {63 a^6 d^6+63 a^6 \tan ^2(e+f x) d^6+8 a^6 \tan (e+f x) d^6}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx}{a d^3}-\frac {126 a^4 d^3}{f \sqrt {d \tan (e+f x)}}}{a d^3}-\frac {110 a^3 d}{3 f (d \tan (e+f x))^{3/2}}}{2 a^3 d}+\frac {11}{f (\tan (e+f x)+1) (d \tan (e+f x))^{3/2}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 (d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {-\frac {-\frac {\int \frac {63 a^6 d^6+63 a^6 \tan (e+f x)^2 d^6+8 a^6 \tan (e+f x) d^6}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx}{a d^3}-\frac {126 a^4 d^3}{f \sqrt {d \tan (e+f x)}}}{a d^3}-\frac {110 a^3 d}{3 f (d \tan (e+f x))^{3/2}}}{2 a^3 d}+\frac {11}{f (\tan (e+f x)+1) (d \tan (e+f x))^{3/2}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 (d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 4136

\(\displaystyle \frac {\frac {-\frac {-\frac {59 a^6 d^6 \int \frac {\tan ^2(e+f x)+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {\int \frac {8 \left (d^6 a^7+d^6 \tan (e+f x) a^7\right )}{\sqrt {d \tan (e+f x)}}dx}{2 a^2}}{a d^3}-\frac {126 a^4 d^3}{f \sqrt {d \tan (e+f x)}}}{a d^3}-\frac {110 a^3 d}{3 f (d \tan (e+f x))^{3/2}}}{2 a^3 d}+\frac {11}{f (\tan (e+f x)+1) (d \tan (e+f x))^{3/2}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 (d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-\frac {-\frac {59 a^6 d^6 \int \frac {\tan ^2(e+f x)+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {4 \int \frac {d^6 a^7+d^6 \tan (e+f x) a^7}{\sqrt {d \tan (e+f x)}}dx}{a^2}}{a d^3}-\frac {126 a^4 d^3}{f \sqrt {d \tan (e+f x)}}}{a d^3}-\frac {110 a^3 d}{3 f (d \tan (e+f x))^{3/2}}}{2 a^3 d}+\frac {11}{f (\tan (e+f x)+1) (d \tan (e+f x))^{3/2}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 (d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {-\frac {-\frac {59 a^6 d^6 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {4 \int \frac {d^6 a^7+d^6 \tan (e+f x) a^7}{\sqrt {d \tan (e+f x)}}dx}{a^2}}{a d^3}-\frac {126 a^4 d^3}{f \sqrt {d \tan (e+f x)}}}{a d^3}-\frac {110 a^3 d}{3 f (d \tan (e+f x))^{3/2}}}{2 a^3 d}+\frac {11}{f (\tan (e+f x)+1) (d \tan (e+f x))^{3/2}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 (d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 4015

\(\displaystyle \frac {\frac {-\frac {-\frac {59 a^6 d^6 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx-\frac {8 a^{12} d^{12} \int \frac {1}{2 d^{12} a^{14}+\cot (e+f x) \left (a^7 d^6-a^7 d^6 \tan (e+f x)\right )^2}d\frac {a^7 d^6-a^7 d^6 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}}{f}}{a d^3}-\frac {126 a^4 d^3}{f \sqrt {d \tan (e+f x)}}}{a d^3}-\frac {110 a^3 d}{3 f (d \tan (e+f x))^{3/2}}}{2 a^3 d}+\frac {11}{f (\tan (e+f x)+1) (d \tan (e+f x))^{3/2}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 (d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {-\frac {-\frac {59 a^6 d^6 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx-\frac {4 \sqrt {2} a^5 d^{11/2} \arctan \left (\frac {a^7 d^6-a^7 d^6 \tan (e+f x)}{\sqrt {2} a^7 d^{11/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{a d^3}-\frac {126 a^4 d^3}{f \sqrt {d \tan (e+f x)}}}{a d^3}-\frac {110 a^3 d}{3 f (d \tan (e+f x))^{3/2}}}{2 a^3 d}+\frac {11}{f (\tan (e+f x)+1) (d \tan (e+f x))^{3/2}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 (d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 4117

\(\displaystyle \frac {\frac {-\frac {-\frac {\frac {59 a^6 d^6 \int \frac {1}{a \sqrt {d \tan (e+f x)} (\tan (e+f x)+1)}d\tan (e+f x)}{f}-\frac {4 \sqrt {2} a^5 d^{11/2} \arctan \left (\frac {a^7 d^6-a^7 d^6 \tan (e+f x)}{\sqrt {2} a^7 d^{11/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{a d^3}-\frac {126 a^4 d^3}{f \sqrt {d \tan (e+f x)}}}{a d^3}-\frac {110 a^3 d}{3 f (d \tan (e+f x))^{3/2}}}{2 a^3 d}+\frac {11}{f (\tan (e+f x)+1) (d \tan (e+f x))^{3/2}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 (d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-\frac {-\frac {\frac {59 a^5 d^6 \int \frac {1}{\sqrt {d \tan (e+f x)} (\tan (e+f x)+1)}d\tan (e+f x)}{f}-\frac {4 \sqrt {2} a^5 d^{11/2} \arctan \left (\frac {a^7 d^6-a^7 d^6 \tan (e+f x)}{\sqrt {2} a^7 d^{11/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{a d^3}-\frac {126 a^4 d^3}{f \sqrt {d \tan (e+f x)}}}{a d^3}-\frac {110 a^3 d}{3 f (d \tan (e+f x))^{3/2}}}{2 a^3 d}+\frac {11}{f (\tan (e+f x)+1) (d \tan (e+f x))^{3/2}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 (d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {-\frac {-\frac {\frac {118 a^5 d^5 \int \frac {1}{\tan (e+f x)+1}d\sqrt {d \tan (e+f x)}}{f}-\frac {4 \sqrt {2} a^5 d^{11/2} \arctan \left (\frac {a^7 d^6-a^7 d^6 \tan (e+f x)}{\sqrt {2} a^7 d^{11/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{a d^3}-\frac {126 a^4 d^3}{f \sqrt {d \tan (e+f x)}}}{a d^3}-\frac {110 a^3 d}{3 f (d \tan (e+f x))^{3/2}}}{2 a^3 d}+\frac {11}{f (\tan (e+f x)+1) (d \tan (e+f x))^{3/2}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 (d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {-\frac {110 a^3 d}{3 f (d \tan (e+f x))^{3/2}}-\frac {-\frac {126 a^4 d^3}{f \sqrt {d \tan (e+f x)}}-\frac {\frac {118 a^5 d^{11/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {4 \sqrt {2} a^5 d^{11/2} \arctan \left (\frac {a^7 d^6-a^7 d^6 \tan (e+f x)}{\sqrt {2} a^7 d^{11/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{a d^3}}{a d^3}}{2 a^3 d}+\frac {11}{f (\tan (e+f x)+1) (d \tan (e+f x))^{3/2}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 (d \tan (e+f x))^{3/2}}\)

Input:

Int[1/((d*Tan[e + f*x])^(5/2)*(a + a*Tan[e + f*x])^3),x]
 

Output:

1/(4*a*d*f*(d*Tan[e + f*x])^(3/2)*(a + a*Tan[e + f*x])^2) + (11/(f*(d*Tan[ 
e + f*x])^(3/2)*(1 + Tan[e + f*x])) + ((-110*a^3*d)/(3*f*(d*Tan[e + f*x])^ 
(3/2)) - (-(((118*a^5*d^(11/2)*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/f - ( 
4*Sqrt[2]*a^5*d^(11/2)*ArcTan[(a^7*d^6 - a^7*d^6*Tan[e + f*x])/(Sqrt[2]*a^ 
7*d^(11/2)*Sqrt[d*Tan[e + f*x]])])/f)/(a*d^3)) - (126*a^4*d^3)/(f*Sqrt[d*T 
an[e + f*x]]))/(a*d^3))/(2*a^3*d))/(8*a^3*d)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4015

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ 
)]], x_Symbol] :> Simp[-2*(d^2/f)   Subst[Int[1/(2*c*d + b*x^2), x], x, (c 
- d*Tan[e + f*x])/Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && 
 EqQ[c^2 - d^2, 0]
 

rule 4052

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m + 1)*((c 
+ d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d))), x] + Simp[1 
/((m + 1)*(a^2 + b^2)*(b*c - a*d))   Int[(a + b*Tan[e + f*x])^(m + 1)*(c + 
d*Tan[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - 
 a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x], x], x] / 
; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] 
 && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || Integ 
erQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
 

rule 4117

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
 Simp[A/f   Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; 
FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
 

rule 4132

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*(b*B - a*C))*(a + b*Tan[e + 
 f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + 
b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 + b^2))   Int[(a + b*Tan[e + 
f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1) - b^2*d* 
(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d 
)*(A*b - a*B - b*C)*Tan[e + f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Ta 
n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ 
[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] && 
!(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
 

rule 4133

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
Simp[(A*b^2 + a^2*C)*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n 
+ 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Simp[1/((m + 1)*(b*c - a*d) 
*(a^2 + b^2))   Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Sim 
p[A*(a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2)) - a*C*(b*c*(m + 1) + a*d*(n 
 + 1)) - (m + 1)*(b*c - a*d)*(A*b - b*C)*Tan[e + f*x] - d*(A*b^2 + a^2*C)*( 
m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, 
 x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m 
, -1] &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
 

rule 4136

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) 
+ (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) 
+ (f_.)*(x_)]), x_Symbol] :> Simp[1/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^ 
n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Simp[ 
(A*b^2 - a*b*B + a^2*C)/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^n*((1 + Tan[ 
e + f*x]^2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, 
 C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] & 
&  !GtQ[n, 0] &&  !LeQ[n, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(378\) vs. \(2(178)=356\).

Time = 1.48 (sec) , antiderivative size = 379, normalized size of antiderivative = 1.76

method	result	size
derivativedivides	\(\frac {2 d^{4} \left (\frac {\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}}{4 d^{6}}-\frac {1}{3 d^{5} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {3}{d^{6} \sqrt {d \tan \left (f x +e \right )}}+\frac {\frac {\frac {15 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{4}+\frac {17 d \sqrt {d \tan \left (f x +e \right )}}{4}}{\left (d \tan \left (f x +e \right )+d \right )^{2}}+\frac {59 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{4 \sqrt {d}}}{4 d^{6}}\right )}{f \,a^{3}}\)	\(379\)
default	\(\frac {2 d^{4} \left (\frac {\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}}{4 d^{6}}-\frac {1}{3 d^{5} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {3}{d^{6} \sqrt {d \tan \left (f x +e \right )}}+\frac {\frac {\frac {15 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{4}+\frac {17 d \sqrt {d \tan \left (f x +e \right )}}{4}}{\left (d \tan \left (f x +e \right )+d \right )^{2}}+\frac {59 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{4 \sqrt {d}}}{4 d^{6}}\right )}{f \,a^{3}}\)	\(379\)

Input:

int(1/(d*tan(f*x+e))^(5/2)/(a+a*tan(f*x+e))^3,x,method=_RETURNVERBOSE)
 

Output:

2/f/a^3*d^4*(1/4/d^6*(1/8/d*(d^2)^(1/4)*2^(1/2)*(ln((d*tan(f*x+e)+(d^2)^(1 
/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)-(d^2)^(1/4)*(d 
*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*t 
an(f*x+e))^(1/2)+1)-2*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)) 
+1/8/(d^2)^(1/4)*2^(1/2)*(ln((d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2 
)*2^(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1 
/2)+(d^2)^(1/2)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-2*a 
rctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)))-1/3/d^5/(d*tan(f*x+e) 
)^(3/2)+3/d^6/(d*tan(f*x+e))^(1/2)+1/4/d^6*((15/4*(d*tan(f*x+e))^(3/2)+17/ 
4*d*(d*tan(f*x+e))^(1/2))/(d*tan(f*x+e)+d)^2+59/4/d^(1/2)*arctan((d*tan(f* 
x+e))^(1/2)/d^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 494, normalized size of antiderivative = 2.30 \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+a \tan (e+f x))^3} \, dx=\left [-\frac {6 \, \sqrt {2} {\left (\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{3} + \tan \left (f x + e\right )^{2}\right )} \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} {\left (\tan \left (f x + e\right ) - 1\right )} - 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) + 177 \, {\left (\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{3} + \tan \left (f x + e\right )^{2}\right )} \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right ) - 2 \, \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} - d}{\tan \left (f x + e\right ) + 1}\right ) - 2 \, {\left (189 \, \tan \left (f x + e\right )^{3} + 323 \, \tan \left (f x + e\right )^{2} + 112 \, \tan \left (f x + e\right ) - 16\right )} \sqrt {d \tan \left (f x + e\right )}}{48 \, {\left (a^{3} d^{3} f \tan \left (f x + e\right )^{4} + 2 \, a^{3} d^{3} f \tan \left (f x + e\right )^{3} + a^{3} d^{3} f \tan \left (f x + e\right )^{2}\right )}}, \frac {6 \, \sqrt {2} {\left (\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{3} + \tan \left (f x + e\right )^{2}\right )} \sqrt {d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} {\left (\tan \left (f x + e\right ) - 1\right )}}{2 \, \sqrt {d} \tan \left (f x + e\right )}\right ) - 177 \, {\left (\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{3} + \tan \left (f x + e\right )^{2}\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d} \tan \left (f x + e\right )}\right ) + {\left (189 \, \tan \left (f x + e\right )^{3} + 323 \, \tan \left (f x + e\right )^{2} + 112 \, \tan \left (f x + e\right ) - 16\right )} \sqrt {d \tan \left (f x + e\right )}}{24 \, {\left (a^{3} d^{3} f \tan \left (f x + e\right )^{4} + 2 \, a^{3} d^{3} f \tan \left (f x + e\right )^{3} + a^{3} d^{3} f \tan \left (f x + e\right )^{2}\right )}}\right ] \] Input:

integrate(1/(d*tan(f*x+e))^(5/2)/(a+a*tan(f*x+e))^3,x, algorithm="fricas")
 

Output:

[-1/48*(6*sqrt(2)*(tan(f*x + e)^4 + 2*tan(f*x + e)^3 + tan(f*x + e)^2)*sqr 
t(-d)*log((d*tan(f*x + e)^2 - 2*sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(-d)*(tan 
(f*x + e) - 1) - 4*d*tan(f*x + e) + d)/(tan(f*x + e)^2 + 1)) + 177*(tan(f* 
x + e)^4 + 2*tan(f*x + e)^3 + tan(f*x + e)^2)*sqrt(-d)*log((d*tan(f*x + e) 
 - 2*sqrt(d*tan(f*x + e))*sqrt(-d) - d)/(tan(f*x + e) + 1)) - 2*(189*tan(f 
*x + e)^3 + 323*tan(f*x + e)^2 + 112*tan(f*x + e) - 16)*sqrt(d*tan(f*x + e 
)))/(a^3*d^3*f*tan(f*x + e)^4 + 2*a^3*d^3*f*tan(f*x + e)^3 + a^3*d^3*f*tan 
(f*x + e)^2), 1/24*(6*sqrt(2)*(tan(f*x + e)^4 + 2*tan(f*x + e)^3 + tan(f*x 
 + e)^2)*sqrt(d)*arctan(1/2*sqrt(2)*sqrt(d*tan(f*x + e))*(tan(f*x + e) - 1 
)/(sqrt(d)*tan(f*x + e))) - 177*(tan(f*x + e)^4 + 2*tan(f*x + e)^3 + tan(f 
*x + e)^2)*sqrt(d)*arctan(sqrt(d*tan(f*x + e))/(sqrt(d)*tan(f*x + e))) + ( 
189*tan(f*x + e)^3 + 323*tan(f*x + e)^2 + 112*tan(f*x + e) - 16)*sqrt(d*ta 
n(f*x + e)))/(a^3*d^3*f*tan(f*x + e)^4 + 2*a^3*d^3*f*tan(f*x + e)^3 + a^3* 
d^3*f*tan(f*x + e)^2)]
 

Sympy [F]

\[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+a \tan (e+f x))^3} \, dx=\frac {\int \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \tan ^{3}{\left (e + f x \right )} + 3 \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \tan ^{2}{\left (e + f x \right )} + 3 \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \tan {\left (e + f x \right )} + \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx}{a^{3}} \] Input:

integrate(1/(d*tan(f*x+e))**(5/2)/(a+a*tan(f*x+e))**3,x)
 

Output:

Integral(1/((d*tan(e + f*x))**(5/2)*tan(e + f*x)**3 + 3*(d*tan(e + f*x))** 
(5/2)*tan(e + f*x)**2 + 3*(d*tan(e + f*x))**(5/2)*tan(e + f*x) + (d*tan(e 
+ f*x))**(5/2)), x)/a**3
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.98 \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+a \tan (e+f x))^3} \, dx=\frac {\frac {189 \, d^{3} \tan \left (f x + e\right )^{3} + 323 \, d^{3} \tan \left (f x + e\right )^{2} + 112 \, d^{3} \tan \left (f x + e\right ) - 16 \, d^{3}}{\left (d \tan \left (f x + e\right )\right )^{\frac {7}{2}} a^{3} d + 2 \, \left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} a^{3} d^{2} + \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a^{3} d^{3}} + \frac {6 \, {\left (\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}}\right )}}{a^{3} d} + \frac {177 \, \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{3} d^{\frac {3}{2}}}}{24 \, d f} \] Input:

integrate(1/(d*tan(f*x+e))^(5/2)/(a+a*tan(f*x+e))^3,x, algorithm="maxima")
 

Output:

1/24*((189*d^3*tan(f*x + e)^3 + 323*d^3*tan(f*x + e)^2 + 112*d^3*tan(f*x + 
 e) - 16*d^3)/((d*tan(f*x + e))^(7/2)*a^3*d + 2*(d*tan(f*x + e))^(5/2)*a^3 
*d^2 + (d*tan(f*x + e))^(3/2)*a^3*d^3) + 6*(sqrt(2)*arctan(1/2*sqrt(2)*(sq 
rt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) + sqrt(2)*arctan( 
-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d))/ 
(a^3*d) + 177*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a^3*d^(3/2)))/(d*f)
 

Giac [F(-1)]

Timed out. \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+a \tan (e+f x))^3} \, dx=\text {Timed out} \] Input:

integrate(1/(d*tan(f*x+e))^(5/2)/(a+a*tan(f*x+e))^3,x, algorithm="giac")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 2.17 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.89 \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+a \tan (e+f x))^3} \, dx=\frac {\frac {63\,d\,{\mathrm {tan}\left (e+f\,x\right )}^3}{8}+\frac {323\,d\,{\mathrm {tan}\left (e+f\,x\right )}^2}{24}+\frac {14\,d\,\mathrm {tan}\left (e+f\,x\right )}{3}-\frac {2\,d}{3}}{a^3\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{7/2}+2\,a^3\,d\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}+a^3\,d^2\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}+\frac {59\,\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,d^{5/2}\,f}+\frac {\sqrt {2}\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}+\frac {\sqrt {2}\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{2\,d^{3/2}}\right )\right )}{8\,a^3\,d^{5/2}\,f} \] Input:

int(1/((d*tan(e + f*x))^(5/2)*(a + a*tan(e + f*x))^3),x)
 

Output:

((14*d*tan(e + f*x))/3 - (2*d)/3 + (323*d*tan(e + f*x)^2)/24 + (63*d*tan(e 
 + f*x)^3)/8)/(a^3*f*(d*tan(e + f*x))^(7/2) + 2*a^3*d*f*(d*tan(e + f*x))^( 
5/2) + a^3*d^2*f*(d*tan(e + f*x))^(3/2)) + (59*atan((d*tan(e + f*x))^(1/2) 
/d^(1/2)))/(8*a^3*d^(5/2)*f) + (2^(1/2)*(2*atan((2^(1/2)*(d*tan(e + f*x))^ 
(1/2))/(2*d^(1/2))) + 2*atan((2^(1/2)*(d*tan(e + f*x))^(1/2))/(2*d^(1/2)) 
+ (2^(1/2)*(d*tan(e + f*x))^(3/2))/(2*d^(3/2)))))/(8*a^3*d^(5/2)*f)
 

Reduce [F]

\[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+a \tan (e+f x))^3} \, dx=\frac {\sqrt {d}\, \left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )^{6}+3 \tan \left (f x +e \right )^{5}+3 \tan \left (f x +e \right )^{4}+\tan \left (f x +e \right )^{3}}d x \right )}{a^{3} d^{3}} \] Input:

int(1/(d*tan(f*x+e))^(5/2)/(a+a*tan(f*x+e))^3,x)
 

Output:

(sqrt(d)*int(sqrt(tan(e + f*x))/(tan(e + f*x)**6 + 3*tan(e + f*x)**5 + 3*t 
an(e + f*x)**4 + tan(e + f*x)**3),x))/(a**3*d**3)