\(\int \frac {(d \tan (e+f x))^{5/2}}{a+a \tan (e+f x)} \, dx\) [358]

Optimal result

Integrand size = 25, antiderivative size = 111 \[ \int \frac {(d \tan (e+f x))^{5/2}}{a+a \tan (e+f x)} \, dx=-\frac {d^{5/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{a f}+\frac {d^{5/2} \arctan \left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a f}+\frac {2 d^2 \sqrt {d \tan (e+f x)}}{a f} \] Output:

-d^(5/2)*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))/a/f+1/2*d^(5/2)*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/f+2*d^ 
2*(d*tan(f*x+e))^(1/2)/a/f
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.17 \[ \int \frac {(d \tan (e+f x))^{5/2}}{a+a \tan (e+f x)} \, dx=\frac {-2 d^{5/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )-(1+i) (-1)^{3/4} d^{5/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )+(1+i) \sqrt [4]{-1} d^{5/2} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )+4 d^2 \sqrt {d \tan (e+f x)}}{2 a f} \] Input:

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

Output:

(-2*d^(5/2)*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]] - (1 + I)*(-1)^(3/4)*d^(5 
/2)*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]] + (1 + I)*(-1)^(1/4) 
*d^(5/2)*ArcTanh[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]] + 4*d^2*Sqrt[d 
*Tan[e + f*x]])/(2*a*f)
 

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.10, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 4049, 27, 3042, 4136, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4049

\(\displaystyle \frac {2 \int -\frac {a \tan ^2(e+f x) d^3+a d^3+a \tan (e+f x) d^3}{2 \sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx}{a}+\frac {2 d^2 \sqrt {d \tan (e+f x)}}{a f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 d^2 \sqrt {d \tan (e+f x)}}{a f}-\frac {\int \frac {a \tan ^2(e+f x) d^3+a d^3+a \tan (e+f x) d^3}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 d^2 \sqrt {d \tan (e+f x)}}{a f}-\frac {\int \frac {a \tan (e+f x)^2 d^3+a d^3+a \tan (e+f x) d^3}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx}{a}\)

\(\Big \downarrow \) 4136

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4015

\(\displaystyle \frac {2 d^2 \sqrt {d \tan (e+f x)}}{a f}-\frac {\frac {1}{2} a d^3 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx-\frac {a^2 d^6 \int \frac {1}{2 a^4 d^6+\cot (e+f x) \left (a^2 d^3-a^2 d^3 \tan (e+f x)\right )^2}d\frac {a^2 d^3-a^2 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}}{f}}{a}\)

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 4117

\(\displaystyle \frac {2 d^2 \sqrt {d \tan (e+f x)}}{a f}-\frac {\frac {a d^3 \int \frac {1}{a \sqrt {d \tan (e+f x)} (\tan (e+f x)+1)}d\tan (e+f x)}{2 f}-\frac {d^{5/2} \arctan \left (\frac {a^2 d^3-a^2 d^3 \tan (e+f x)}{\sqrt {2} a^2 d^{5/2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} f}}{a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 d^2 \sqrt {d \tan (e+f x)}}{a f}-\frac {\frac {d^3 \int \frac {1}{\sqrt {d \tan (e+f x)} (\tan (e+f x)+1)}d\tan (e+f x)}{2 f}-\frac {d^{5/2} \arctan \left (\frac {a^2 d^3-a^2 d^3 \tan (e+f x)}{\sqrt {2} a^2 d^{5/2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} f}}{a}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 d^2 \sqrt {d \tan (e+f x)}}{a f}-\frac {\frac {d^2 \int \frac {1}{\tan (e+f x)+1}d\sqrt {d \tan (e+f x)}}{f}-\frac {d^{5/2} \arctan \left (\frac {a^2 d^3-a^2 d^3 \tan (e+f x)}{\sqrt {2} a^2 d^{5/2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} f}}{a}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {2 d^2 \sqrt {d \tan (e+f x)}}{a f}-\frac {\frac {d^{5/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {d^{5/2} \arctan \left (\frac {a^2 d^3-a^2 d^3 \tan (e+f x)}{\sqrt {2} a^2 d^{5/2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} f}}{a}\)

Input:

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

Output:

-(((d^(5/2)*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/f - (d^(5/2)*ArcTan[(a^2 
*d^3 - a^2*d^3*Tan[e + f*x])/(Sqrt[2]*a^2*d^(5/2)*Sqrt[d*Tan[e + f*x]])])/ 
(Sqrt[2]*f))/a) + (2*d^2*Sqrt[d*Tan[e + f*x]])/(a*f)
 

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 4049

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 - 2)*((c 
+ d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[1/(d*(m + n - 1)) 
 Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n 
- 1) - b^2*(b*c*(m - 2) + a*d*(1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[ 
e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*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] && GtQ[m, 2] && (GeQ[n, -1] || I 
ntegerQ[m]) &&  !(IGtQ[n, 2] && ( !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 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. \(311\) vs. \(2(93)=186\).

Time = 1.36 (sec) , antiderivative size = 312, normalized size of antiderivative = 2.81

method	result	size
derivativedivides	\(\frac {2 d^{2} \left (\sqrt {d \tan \left (f x +e \right )}-\frac {\sqrt {d}\, \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2}-\frac {d \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{2}\right )}{f a}\)	\(312\)
default	\(\frac {2 d^{2} \left (\sqrt {d \tan \left (f x +e \right )}-\frac {\sqrt {d}\, \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2}-\frac {d \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{2}\right )}{f a}\)	\(312\)

Input:

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

Output:

2/f/a*d^2*((d*tan(f*x+e))^(1/2)-1/2*d^(1/2)*arctan((d*tan(f*x+e))^(1/2)/d^ 
(1/2))-1/2*d*(1/8/d*(d^2)^(1/4)*2^(1/2)*(ln((d*tan(f*x+e)+(d^2)^(1/4)*(d*t 
an(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*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*arctan(-2 
^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1))))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 243, normalized size of antiderivative = 2.19 \[ \int \frac {(d \tan (e+f x))^{5/2}}{a+a \tan (e+f x)} \, dx=\left [\frac {\sqrt {\frac {1}{2}} \sqrt {-d} d^{2} \log \left (\frac {d \tan \left (f x + e\right )^{2} - 4 \, \sqrt {\frac {1}{2}} \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} {\left (\tan \left (f x + e\right ) - 1\right )} - 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) + \sqrt {-d} d^{2} \log \left (\frac {d \tan \left (f x + e\right ) - 2 \, \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} - d}{\tan \left (f x + e\right ) + 1}\right ) + 4 \, \sqrt {d \tan \left (f x + e\right )} d^{2}}{2 \, a f}, -\frac {\sqrt {\frac {1}{2}} d^{\frac {5}{2}} \arctan \left (\frac {\sqrt {\frac {1}{2}} \sqrt {d \tan \left (f x + e\right )} {\left (\tan \left (f x + e\right ) - 1\right )}}{\sqrt {d} \tan \left (f x + e\right )}\right ) - d^{\frac {5}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d} \tan \left (f x + e\right )}\right ) - 2 \, \sqrt {d \tan \left (f x + e\right )} d^{2}}{a f}\right ] \] Input:

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

Output:

[1/2*(sqrt(1/2)*sqrt(-d)*d^2*log((d*tan(f*x + e)^2 - 4*sqrt(1/2)*sqrt(d*ta 
n(f*x + e))*sqrt(-d)*(tan(f*x + e) - 1) - 4*d*tan(f*x + e) + d)/(tan(f*x + 
 e)^2 + 1)) + sqrt(-d)*d^2*log((d*tan(f*x + e) - 2*sqrt(d*tan(f*x + e))*sq 
rt(-d) - d)/(tan(f*x + e) + 1)) + 4*sqrt(d*tan(f*x + e))*d^2)/(a*f), -(sqr 
t(1/2)*d^(5/2)*arctan(sqrt(1/2)*sqrt(d*tan(f*x + e))*(tan(f*x + e) - 1)/(s 
qrt(d)*tan(f*x + e))) - d^(5/2)*arctan(sqrt(d*tan(f*x + e))/(sqrt(d)*tan(f 
*x + e))) - 2*sqrt(d*tan(f*x + e))*d^2)/(a*f)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.17 \[ \int \frac {(d \tan (e+f x))^{5/2}}{a+a \tan (e+f x)} \, dx=-\frac {\frac {d^{4} {\left (\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}}\right )}}{a} + \frac {2 \, d^{\frac {7}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a} - \frac {4 \, \sqrt {d \tan \left (f x + e\right )} d^{3}}{a}}{2 \, d f} \] Input:

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

Output:

-1/2*(d^4*(sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(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 + 2*d^(7/2)*arctan(sqrt(d*tan(f 
*x + e))/sqrt(d))/a - 4*sqrt(d*tan(f*x + e))*d^3/a)/(d*f)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(d \tan (e+f x))^{5/2}}{a+a \tan (e+f x)} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{1,[4,13]%%%}+%%%{6,[4,11]%%%}+%%%{15,[4,9]%%%}+%%%{20, 
[4,7]%%%}
 

Mupad [B] (verification not implemented)

Time = 1.57 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.12 \[ \int \frac {(d \tan (e+f x))^{5/2}}{a+a \tan (e+f x)} \, dx=\frac {2\,d^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{a\,f}-\frac {d^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{a\,f}-\frac {\sqrt {2}\,d^{5/2}\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}+\frac {\sqrt {2}\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{2\,d^{3/2}}\right )\right )}{4\,a\,f} \] Input:

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

Output:

(2*d^2*(d*tan(e + f*x))^(1/2))/(a*f) - (d^(5/2)*atan((d*tan(e + f*x))^(1/2 
)/d^(1/2)))/(a*f) - (2^(1/2)*d^(5/2)*(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)))))/(4*a*f)
 

Reduce [F]

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

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

Output:

(sqrt(d)*int((sqrt(tan(e + f*x))*tan(e + f*x)**2)/(tan(e + f*x) + 1),x)*d* 
*2)/a