\(\int \frac {1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^3} \, dx\) [376]

Optimal result

Integrand size = 25, antiderivative size = 189 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^3} \, dx=-\frac {31 \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 d^{3/2} f}-\frac {\text {arctanh}\left (\frac {\sqrt {d}+\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{2 \sqrt {2} a^3 d^{3/2} f}-\frac {27}{8 a^3 d f \sqrt {d \tan (e+f x)}}+\frac {9}{8 a^3 d f \sqrt {d \tan (e+f x)} (1+\tan (e+f x))}+\frac {1}{4 a d f \sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^2} \] Output:

-31/8*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/d^(3/2)/f-1/4*arctanh(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^(3 
/2)/f-27/8/a^3/d/f/(d*tan(f*x+e))^(1/2)+9/8/a^3/d/f/(d*tan(f*x+e))^(1/2)/( 
1+tan(f*x+e))+1/4/a/d/f/(d*tan(f*x+e))^(1/2)/(a+a*tan(f*x+e))^2
 

Mathematica [A] (verified)

Time = 2.21 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.96 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^3} \, dx=\frac {-\frac {31 \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {d}}+\frac {\sqrt {2} \left (\log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )-\log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )\right )}{\sqrt {d}}+\frac {9 d}{(1+\cot (e+f x)) (d \tan (e+f x))^{3/2}}+\frac {-27+\frac {2}{(1+\tan (e+f x))^2}}{\sqrt {d \tan (e+f x)}}}{8 a^3 d f} \] Input:

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

Output:

((-31*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/Sqrt[d] + (Sqrt[2]*(Log[Sqrt[d 
] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]] - Log[Sqrt[d] + S 
qrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]]))/Sqrt[d] + (9*d)/((1 
+ Cot[e + f*x])*(d*Tan[e + f*x])^(3/2)) + (-27 + 2/(1 + Tan[e + f*x])^2)/S 
qrt[d*Tan[e + f*x]])/(8*a^3*d*f)
 

Rubi [A] (verified)

Time = 1.52 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.14, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.720, Rules used = {3042, 4052, 27, 3042, 4132, 3042, 4132, 27, 3042, 4137, 27, 3042, 4015, 221, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4052

\(\displaystyle \frac {\int \frac {5 d \tan ^2(e+f x) a^2+9 d a^2-4 d \tan (e+f x) a^2}{2 (d \tan (e+f x))^{3/2} (\tan (e+f x) a+a)^2}dx}{4 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {5 d \tan ^2(e+f x) a^2+9 d a^2-4 d \tan (e+f x) a^2}{(d \tan (e+f x))^{3/2} (\tan (e+f x) a+a)^2}dx}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {5 d \tan (e+f x)^2 a^2+9 d a^2-4 d \tan (e+f x) a^2}{(d \tan (e+f x))^{3/2} (\tan (e+f x) a+a)^2}dx}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 4132

\(\displaystyle \frac {\frac {\int \frac {27 d^2 a^4+27 d^2 \tan ^2(e+f x) a^4-8 d^2 \tan (e+f x) a^4}{(d \tan (e+f x))^{3/2} (\tan (e+f x) a+a)}dx}{2 a^3 d}+\frac {9}{f (\tan (e+f x)+1) \sqrt {d \tan (e+f x)}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {27 d^2 a^4+27 d^2 \tan (e+f x)^2 a^4-8 d^2 \tan (e+f x) a^4}{(d \tan (e+f x))^{3/2} (\tan (e+f x) a+a)}dx}{2 a^3 d}+\frac {9}{f (\tan (e+f x)+1) \sqrt {d \tan (e+f x)}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 4132

\(\displaystyle \frac {\frac {-\frac {2 \int \frac {35 d^4 a^5+27 d^4 \tan ^2(e+f x) a^5}{2 \sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx}{a d^3}-\frac {54 a^3 d}{f \sqrt {d \tan (e+f x)}}}{2 a^3 d}+\frac {9}{f (\tan (e+f x)+1) \sqrt {d \tan (e+f x)}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-\frac {\int \frac {35 d^4 a^5+27 d^4 \tan ^2(e+f x) a^5}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx}{a d^3}-\frac {54 a^3 d}{f \sqrt {d \tan (e+f x)}}}{2 a^3 d}+\frac {9}{f (\tan (e+f x)+1) \sqrt {d \tan (e+f x)}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {-\frac {\int \frac {35 d^4 a^5+27 d^4 \tan (e+f x)^2 a^5}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx}{a d^3}-\frac {54 a^3 d}{f \sqrt {d \tan (e+f x)}}}{2 a^3 d}+\frac {9}{f (\tan (e+f x)+1) \sqrt {d \tan (e+f x)}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 4137

\(\displaystyle \frac {\frac {-\frac {31 a^5 d^4 \int \frac {\tan ^2(e+f x)+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {\int \frac {8 \left (a^6 d^4-a^6 d^4 \tan (e+f x)\right )}{\sqrt {d \tan (e+f x)}}dx}{2 a^2}}{a d^3}-\frac {54 a^3 d}{f \sqrt {d \tan (e+f x)}}}{2 a^3 d}+\frac {9}{f (\tan (e+f x)+1) \sqrt {d \tan (e+f x)}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-\frac {31 a^5 d^4 \int \frac {\tan ^2(e+f x)+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {4 \int \frac {a^6 d^4-a^6 d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx}{a^2}}{a d^3}-\frac {54 a^3 d}{f \sqrt {d \tan (e+f x)}}}{2 a^3 d}+\frac {9}{f (\tan (e+f x)+1) \sqrt {d \tan (e+f x)}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {-\frac {31 a^5 d^4 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {4 \int \frac {a^6 d^4-a^6 d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx}{a^2}}{a d^3}-\frac {54 a^3 d}{f \sqrt {d \tan (e+f x)}}}{2 a^3 d}+\frac {9}{f (\tan (e+f x)+1) \sqrt {d \tan (e+f x)}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 4015

\(\displaystyle \frac {\frac {-\frac {31 a^5 d^4 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx-\frac {8 a^{10} d^8 \int \frac {1}{\cot (e+f x) \left (d^4 a^6+d^4 \tan (e+f x) a^6\right )^2-2 a^{12} d^8}d\frac {d^4 a^6+d^4 \tan (e+f x) a^6}{\sqrt {d \tan (e+f x)}}}{f}}{a d^3}-\frac {54 a^3 d}{f \sqrt {d \tan (e+f x)}}}{2 a^3 d}+\frac {9}{f (\tan (e+f x)+1) \sqrt {d \tan (e+f x)}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {-\frac {31 a^5 d^4 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {4 \sqrt {2} a^4 d^{7/2} \text {arctanh}\left (\frac {a^6 d^4 \tan (e+f x)+a^6 d^4}{\sqrt {2} a^6 d^{7/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{a d^3}-\frac {54 a^3 d}{f \sqrt {d \tan (e+f x)}}}{2 a^3 d}+\frac {9}{f (\tan (e+f x)+1) \sqrt {d \tan (e+f x)}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 4117

\(\displaystyle \frac {\frac {-\frac {\frac {31 a^5 d^4 \int \frac {1}{a \sqrt {d \tan (e+f x)} (\tan (e+f x)+1)}d\tan (e+f x)}{f}+\frac {4 \sqrt {2} a^4 d^{7/2} \text {arctanh}\left (\frac {a^6 d^4 \tan (e+f x)+a^6 d^4}{\sqrt {2} a^6 d^{7/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{a d^3}-\frac {54 a^3 d}{f \sqrt {d \tan (e+f x)}}}{2 a^3 d}+\frac {9}{f (\tan (e+f x)+1) \sqrt {d \tan (e+f x)}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-\frac {\frac {31 a^4 d^4 \int \frac {1}{\sqrt {d \tan (e+f x)} (\tan (e+f x)+1)}d\tan (e+f x)}{f}+\frac {4 \sqrt {2} a^4 d^{7/2} \text {arctanh}\left (\frac {a^6 d^4 \tan (e+f x)+a^6 d^4}{\sqrt {2} a^6 d^{7/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{a d^3}-\frac {54 a^3 d}{f \sqrt {d \tan (e+f x)}}}{2 a^3 d}+\frac {9}{f (\tan (e+f x)+1) \sqrt {d \tan (e+f x)}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {-\frac {\frac {62 a^4 d^3 \int \frac {1}{\tan (e+f x)+1}d\sqrt {d \tan (e+f x)}}{f}+\frac {4 \sqrt {2} a^4 d^{7/2} \text {arctanh}\left (\frac {a^6 d^4 \tan (e+f x)+a^6 d^4}{\sqrt {2} a^6 d^{7/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{a d^3}-\frac {54 a^3 d}{f \sqrt {d \tan (e+f x)}}}{2 a^3 d}+\frac {9}{f (\tan (e+f x)+1) \sqrt {d \tan (e+f x)}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {-\frac {54 a^3 d}{f \sqrt {d \tan (e+f x)}}-\frac {\frac {62 a^4 d^{7/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}+\frac {4 \sqrt {2} a^4 d^{7/2} \text {arctanh}\left (\frac {a^6 d^4 \tan (e+f x)+a^6 d^4}{\sqrt {2} a^6 d^{7/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{a d^3}}{2 a^3 d}+\frac {9}{f (\tan (e+f x)+1) \sqrt {d \tan (e+f x)}}}{8 a^3 d}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 \sqrt {d \tan (e+f x)}}\)

Input:

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

Output:

1/(4*a*d*f*Sqrt[d*Tan[e + f*x]]*(a + a*Tan[e + f*x])^2) + (9/(f*Sqrt[d*Tan 
[e + f*x]]*(1 + Tan[e + f*x])) + (-(((62*a^4*d^(7/2)*ArcTan[Sqrt[d*Tan[e + 
 f*x]]/Sqrt[d]])/f + (4*Sqrt[2]*a^4*d^(7/2)*ArcTanh[(a^6*d^4 + a^6*d^4*Tan 
[e + f*x])/(Sqrt[2]*a^6*d^(7/2)*Sqrt[d*Tan[e + f*x]])])/f)/(a*d^3)) - (54* 
a^3*d)/(f*Sqrt[d*Tan[e + f*x]]))/(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 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[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 4137

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*tan[(e_.) 
+ (f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Sim 
p[1/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^n*Simp[a*(A - C) - (A*b - b*C)*T 
an[e + f*x], x], x], x] + Simp[(A*b^2 + 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, 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. \(363\) vs. \(2(156)=312\).

Time = 1.48 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.93

method	result	size
derivativedivides	\(\frac {2 d^{4} \left (\frac {-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}}{4 d^{5}}-\frac {\frac {\frac {11 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{4}+\frac {13 d \sqrt {d \tan \left (f x +e \right )}}{4}}{\left (d \tan \left (f x +e \right )+d \right )^{2}}+\frac {31 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{4 \sqrt {d}}}{4 d^{5}}-\frac {1}{d^{5} \sqrt {d \tan \left (f x +e \right )}}\right )}{f \,a^{3}}\)	\(364\)
default	\(\frac {2 d^{4} \left (\frac {-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}}{4 d^{5}}-\frac {\frac {\frac {11 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{4}+\frac {13 d \sqrt {d \tan \left (f x +e \right )}}{4}}{\left (d \tan \left (f x +e \right )+d \right )^{2}}+\frac {31 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{4 \sqrt {d}}}{4 d^{5}}-\frac {1}{d^{5} \sqrt {d \tan \left (f x +e \right )}}\right )}{f \,a^{3}}\)	\(364\)

Input:

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

Output:

2/f/a^3*d^4*(1/4/d^5*(-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* 
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)))-1/4/d^5*((11/4*(d*ta 
n(f*x+e))^(3/2)+13/4*d*(d*tan(f*x+e))^(1/2))/(d*tan(f*x+e)+d)^2+31/4/d^(1/ 
2)*arctan((d*tan(f*x+e))^(1/2)/d^(1/2)))-1/d^5/(d*tan(f*x+e))^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 465, normalized size of antiderivative = 2.46 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^3} \, dx=\left [\frac {4 \, \sqrt {2} {\left (\tan \left (f x + e\right )^{3} + 2 \, \tan \left (f x + e\right )^{2} + \tan \left (f x + e\right )\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} {\left (\tan \left (f x + e\right ) + 1\right )}}{2 \, d \tan \left (f x + e\right )}\right ) - 31 \, {\left (\tan \left (f x + e\right )^{3} + 2 \, \tan \left (f x + e\right )^{2} + \tan \left (f x + e\right )\right )} \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right ) + 2 \, \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} - d}{\tan \left (f x + e\right ) + 1}\right ) - 2 \, \sqrt {d \tan \left (f x + e\right )} {\left (27 \, \tan \left (f x + e\right )^{2} + 45 \, \tan \left (f x + e\right ) + 16\right )}}{16 \, {\left (a^{3} d^{2} f \tan \left (f x + e\right )^{3} + 2 \, a^{3} d^{2} f \tan \left (f x + e\right )^{2} + a^{3} d^{2} f \tan \left (f x + e\right )\right )}}, \frac {\sqrt {2} {\left (\tan \left (f x + e\right )^{3} + 2 \, \tan \left (f x + e\right )^{2} + \tan \left (f x + e\right )\right )} \sqrt {d} \log \left (\frac {d \tan \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} {\left (\tan \left (f x + e\right ) + 1\right )} + 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) + 31 \, {\left (\tan \left (f x + e\right )^{3} + 2 \, \tan \left (f x + e\right )^{2} + \tan \left (f x + e\right )\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d} \tan \left (f x + e\right )}\right ) - \sqrt {d \tan \left (f x + e\right )} {\left (27 \, \tan \left (f x + e\right )^{2} + 45 \, \tan \left (f x + e\right ) + 16\right )}}{8 \, {\left (a^{3} d^{2} f \tan \left (f x + e\right )^{3} + 2 \, a^{3} d^{2} f \tan \left (f x + e\right )^{2} + a^{3} d^{2} f \tan \left (f x + e\right )\right )}}\right ] \] Input:

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

Output:

[1/16*(4*sqrt(2)*(tan(f*x + e)^3 + 2*tan(f*x + e)^2 + tan(f*x + e))*sqrt(- 
d)*arctan(1/2*sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(-d)*(tan(f*x + e) + 1)/(d* 
tan(f*x + e))) - 31*(tan(f*x + e)^3 + 2*tan(f*x + e)^2 + tan(f*x + e))*sqr 
t(-d)*log((d*tan(f*x + e) + 2*sqrt(d*tan(f*x + e))*sqrt(-d) - d)/(tan(f*x 
+ e) + 1)) - 2*sqrt(d*tan(f*x + e))*(27*tan(f*x + e)^2 + 45*tan(f*x + e) + 
 16))/(a^3*d^2*f*tan(f*x + e)^3 + 2*a^3*d^2*f*tan(f*x + e)^2 + a^3*d^2*f*t 
an(f*x + e)), 1/8*(sqrt(2)*(tan(f*x + e)^3 + 2*tan(f*x + e)^2 + tan(f*x + 
e))*sqrt(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)) + 31*(ta 
n(f*x + e)^3 + 2*tan(f*x + e)^2 + tan(f*x + e))*sqrt(d)*arctan(sqrt(d*tan( 
f*x + e))/(sqrt(d)*tan(f*x + e))) - sqrt(d*tan(f*x + e))*(27*tan(f*x + e)^ 
2 + 45*tan(f*x + e) + 16))/(a^3*d^2*f*tan(f*x + e)^3 + 2*a^3*d^2*f*tan(f*x 
 + e)^2 + a^3*d^2*f*tan(f*x + e))]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.01 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^3} \, dx=-\frac {\frac {27 \, d^{2} \tan \left (f x + e\right )^{2} + 45 \, d^{2} \tan \left (f x + e\right ) + 16 \, d^{2}}{\left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} a^{3} + 2 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a^{3} d + \sqrt {d \tan \left (f x + e\right )} a^{3} d^{2}} + \frac {\frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}}{a^{3}} + \frac {31 \, \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{3} \sqrt {d}}}{8 \, d f} \] Input:

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

Output:

-1/8*((27*d^2*tan(f*x + e)^2 + 45*d^2*tan(f*x + e) + 16*d^2)/((d*tan(f*x + 
 e))^(5/2)*a^3 + 2*(d*tan(f*x + e))^(3/2)*a^3*d + sqrt(d*tan(f*x + e))*a^3 
*d^2) + (sqrt(2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) 
 + d)/sqrt(d) - sqrt(2)*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))* 
sqrt(d) + d)/sqrt(d))/a^3 + 31*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a^3*s 
qrt(d)))/(d*f)
 

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 1.93 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.93 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^3} \, dx=-\frac {\frac {27\,d\,{\mathrm {tan}\left (e+f\,x\right )}^2}{8}+\frac {45\,d\,\mathrm {tan}\left (e+f\,x\right )}{8}+2\,d}{a^3\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}+2\,a^3\,d\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}+a^3\,d^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}-\frac {31\,\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,d^{3/2}\,f}-\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {63504384\,\sqrt {2}\,a^9\,d^{15/2}\,f^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{63504384\,a^9\,d^8\,f^3+63504384\,a^9\,d^8\,f^3\,\mathrm {tan}\left (e+f\,x\right )}\right )}{4\,a^3\,d^{3/2}\,f} \] Input:

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

Output:

- (2*d + (45*d*tan(e + f*x))/8 + (27*d*tan(e + f*x)^2)/8)/(a^3*f*(d*tan(e 
+ f*x))^(5/2) + 2*a^3*d*f*(d*tan(e + f*x))^(3/2) + a^3*d^2*f*(d*tan(e + f* 
x))^(1/2)) - (31*atan((d*tan(e + f*x))^(1/2)/d^(1/2)))/(8*a^3*d^(3/2)*f) - 
 (2^(1/2)*atanh((63504384*2^(1/2)*a^9*d^(15/2)*f^3*(d*tan(e + f*x))^(1/2)) 
/(63504384*a^9*d^8*f^3 + 63504384*a^9*d^8*f^3*tan(e + f*x))))/(4*a^3*d^(3/ 
2)*f)
 

Reduce [F]

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

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

Output:

(sqrt(d)*int(sqrt(tan(e + f*x))/(tan(e + f*x)**5 + 3*tan(e + f*x)**4 + 3*t 
an(e + f*x)**3 + tan(e + f*x)**2),x))/(a**3*d**2)