\(\int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx\) [1118]

Optimal result

Integrand size = 30, antiderivative size = 285 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {i (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{8 a^3 f}+\frac {\left (2 i c^3+4 c^2 d-i c d^2+2 d^3\right ) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{16 a^3 \sqrt {c+i d} f}+\frac {(c+i d) (i c+2 d) \sqrt {c+d \tan (e+f x)}}{8 a f (a+i a \tan (e+f x))^2}+\frac {\left (2 i c^2+5 c d-4 i d^2\right ) \sqrt {c+d \tan (e+f x)}}{16 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {(i c-d) (c+d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3} \] Output:

-1/8*I*(c-I*d)^(5/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/a^3/f+1 
/16*(2*I*c^3+4*c^2*d-I*c*d^2+2*d^3)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d) 
^(1/2))/a^3/(c+I*d)^(1/2)/f+1/8*(c+I*d)*(I*c+2*d)*(c+d*tan(f*x+e))^(1/2)/a 
/f/(a+I*a*tan(f*x+e))^2+1/16*(2*I*c^2+5*c*d-4*I*d^2)*(c+d*tan(f*x+e))^(1/2 
)/f/(a^3+I*a^3*tan(f*x+e))+1/6*(I*c-d)*(c+d*tan(f*x+e))^(3/2)/f/(a+I*a*tan 
(f*x+e))^3
 

Mathematica [A] (verified)

Time = 3.85 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.14 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {\sec ^3(e+f x) (\cos (f x)+i \sin (f x))^3 \left (\frac {2 \left (-i \sqrt {-c+i d} \left (2 c^3-4 i c^2 d-c d^2-2 i d^3\right ) \arctan \left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {-c-i d}}\right )-2 \sqrt {-c-i d} (i c+d)^3 \arctan \left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {-c+i d}}\right )\right ) (\cos (3 e)+i \sin (3 e))}{\sqrt {-c-i d} \sqrt {-c+i d}}+\frac {2}{3} \cos (e+f x) (i \cos (3 f x)+\sin (3 f x)) \left (7 c^2+i c d+6 d^2+\left (13 c^2-14 i c d-6 d^2\right ) \cos (2 (e+f x))+\left (9 i c^2+22 c d-2 i d^2\right ) \sin (2 (e+f x))\right ) \sqrt {c+d \tan (e+f x)}\right )}{32 f (a+i a \tan (e+f x))^3} \] Input:

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

Output:

(Sec[e + f*x]^3*(Cos[f*x] + I*Sin[f*x])^3*((2*((-I)*Sqrt[-c + I*d]*(2*c^3 
- (4*I)*c^2*d - c*d^2 - (2*I)*d^3)*ArcTan[Sqrt[c + d*Tan[e + f*x]]/Sqrt[-c 
 - I*d]] - 2*Sqrt[-c - I*d]*(I*c + d)^3*ArcTan[Sqrt[c + d*Tan[e + f*x]]/Sq 
rt[-c + I*d]])*(Cos[3*e] + I*Sin[3*e]))/(Sqrt[-c - I*d]*Sqrt[-c + I*d]) + 
(2*Cos[e + f*x]*(I*Cos[3*f*x] + Sin[3*f*x])*(7*c^2 + I*c*d + 6*d^2 + (13*c 
^2 - (14*I)*c*d - 6*d^2)*Cos[2*(e + f*x)] + ((9*I)*c^2 + 22*c*d - (2*I)*d^ 
2)*Sin[2*(e + f*x)])*Sqrt[c + d*Tan[e + f*x]])/3))/(32*f*(a + I*a*Tan[e + 
f*x])^3)
 

Rubi [A] (warning: unable to verify)

Time = 1.76 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.08, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {3042, 4041, 27, 3042, 4078, 25, 3042, 4079, 25, 3042, 4022, 3042, 4020, 25, 73, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4041

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4078

\(\displaystyle \frac {\frac {a (-d+i c) (c-2 i d) \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))^2}-\frac {\int -\frac {\left (4 c^3-9 i d c^2-5 d^2 c-2 i d^3\right ) a^2+d \left (3 c^2-7 i d c-6 d^2\right ) \tan (e+f x) a^2}{(i \tan (e+f x) a+a) \sqrt {c+d \tan (e+f x)}}dx}{4 a^2}}{4 a^2}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {\left (4 c^3-9 i d c^2-5 d^2 c-2 i d^3\right ) a^2+d \left (3 c^2-7 i d c-6 d^2\right ) \tan (e+f x) a^2}{(i \tan (e+f x) a+a) \sqrt {c+d \tan (e+f x)}}dx}{4 a^2}+\frac {a (-d+i c) (c-2 i d) \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))^2}}{4 a^2}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\left (4 c^3-9 i d c^2-5 d^2 c-2 i d^3\right ) a^2+d \left (3 c^2-7 i d c-6 d^2\right ) \tan (e+f x) a^2}{(i \tan (e+f x) a+a) \sqrt {c+d \tan (e+f x)}}dx}{4 a^2}+\frac {a (-d+i c) (c-2 i d) \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))^2}}{4 a^2}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\frac {\frac {a^2 \left (5 c d+i \left (2 c^2-4 d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f (a+i a \tan (e+f x))}-\frac {\int -\frac {c (i c-d) \left (4 c^2-10 i d c-7 d^2\right ) a^3+(c+i d) d \left (2 i c^2+5 d c-4 i d^2\right ) \tan (e+f x) a^3}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}}{4 a^2}+\frac {a (-d+i c) (c-2 i d) \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))^2}}{4 a^2}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {\int \frac {c (c+i d) \left (4 i c^2+10 d c-7 i d^2\right ) a^3+(c+i d) d \left (2 i c^2+5 d c-4 i d^2\right ) \tan (e+f x) a^3}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}+\frac {a^2 \left (5 c d+i \left (2 c^2-4 d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f (a+i a \tan (e+f x))}}{4 a^2}+\frac {a (-d+i c) (c-2 i d) \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))^2}}{4 a^2}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int \frac {c (c+i d) \left (4 i c^2+10 d c-7 i d^2\right ) a^3+(c+i d) d \left (2 i c^2+5 d c-4 i d^2\right ) \tan (e+f x) a^3}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}+\frac {a^2 \left (5 c d+i \left (2 c^2-4 d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f (a+i a \tan (e+f x))}}{4 a^2}+\frac {a (-d+i c) (c-2 i d) \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))^2}}{4 a^2}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {\frac {\frac {a^3 (c+i d) \left (2 i c^3+4 c^2 d-i c d^2+2 d^3\right ) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-2 a^3 (c+i d) (d+i c)^3 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}+\frac {a^2 \left (5 c d+i \left (2 c^2-4 d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f (a+i a \tan (e+f x))}}{4 a^2}+\frac {a (-d+i c) (c-2 i d) \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))^2}}{4 a^2}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {a^3 (c+i d) \left (2 i c^3+4 c^2 d-i c d^2+2 d^3\right ) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-2 a^3 (c+i d) (d+i c)^3 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}+\frac {a^2 \left (5 c d+i \left (2 c^2-4 d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f (a+i a \tan (e+f x))}}{4 a^2}+\frac {a (-d+i c) (c-2 i d) \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))^2}}{4 a^2}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {\frac {\frac {-\frac {i a^3 (c+i d) \left (2 i c^3+4 c^2 d-i c d^2+2 d^3\right ) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{f}-\frac {2 i a^3 (c+i d) (d+i c)^3 \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{f}}{2 a^2 (-d+i c)}+\frac {a^2 \left (5 c d+i \left (2 c^2-4 d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f (a+i a \tan (e+f x))}}{4 a^2}+\frac {a (-d+i c) (c-2 i d) \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))^2}}{4 a^2}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {\frac {i a^3 (c+i d) \left (2 i c^3+4 c^2 d-i c d^2+2 d^3\right ) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{f}+\frac {2 i a^3 (c+i d) (d+i c)^3 \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{f}}{2 a^2 (-d+i c)}+\frac {a^2 \left (5 c d+i \left (2 c^2-4 d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f (a+i a \tan (e+f x))}}{4 a^2}+\frac {a (-d+i c) (c-2 i d) \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))^2}}{4 a^2}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {\frac {\frac {2 a^3 (c+i d) \left (2 i c^3+4 c^2 d-i c d^2+2 d^3\right ) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}-\frac {4 a^3 (c+i d) (d+i c)^3 \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}}{2 a^2 (-d+i c)}+\frac {a^2 \left (5 c d+i \left (2 c^2-4 d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f (a+i a \tan (e+f x))}}{4 a^2}+\frac {a (-d+i c) (c-2 i d) \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))^2}}{4 a^2}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\frac {a^2 \left (5 c d+i \left (2 c^2-4 d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f (a+i a \tan (e+f x))}+\frac {\frac {2 a^3 \sqrt {c+i d} \left (2 i c^3+4 c^2 d-i c d^2+2 d^3\right ) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}-\frac {4 a^3 (c+i d) (d+i c)^3 \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}}{2 a^2 (-d+i c)}}{4 a^2}+\frac {a (-d+i c) (c-2 i d) \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))^2}}{4 a^2}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\)

Input:

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

Output:

((I*c - d)*(c + d*Tan[e + f*x])^(3/2))/(6*f*(a + I*a*Tan[e + f*x])^3) + (( 
a*(I*c - d)*(c - (2*I)*d)*Sqrt[c + d*Tan[e + f*x]])/(2*f*(a + I*a*Tan[e + 
f*x])^2) + (((-4*a^3*(c + I*d)*(I*c + d)^3*ArcTan[Tan[e + f*x]/Sqrt[c - I* 
d]])/(Sqrt[c - I*d]*f) + (2*a^3*Sqrt[c + I*d]*((2*I)*c^3 + 4*c^2*d - I*c*d 
^2 + 2*d^3)*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/f)/(2*a^2*(I*c - d)) + (a^ 
2*(5*c*d + I*(2*c^2 - 4*d^2))*Sqrt[c + d*Tan[e + f*x]])/(f*(a + I*a*Tan[e 
+ f*x])))/(4*a^2))/(4*a^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 4020

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

rule 4022

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

rule 4041

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

rule 4078

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

rule 4079

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

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 452, normalized size of antiderivative = 1.59

method	result	size
derivativedivides	\(\frac {2 d^{4} \left (\frac {\frac {-\frac {d \left (i c^{4} d +i c^{2} d^{3}+4 i d^{5}+2 c^{5}+5 c^{3} d^{2}+7 c \,d^{4}\right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{2 \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right )}+\frac {2 d \left (6 i c^{5} d +20 i c^{3} d^{3}-2 i c \,d^{5}+3 c^{6}+5 c^{4} d^{2}-15 c^{2} d^{4}-d^{6}\right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3 \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right )}-\frac {d c \left (7 i c^{5} d +10 i c^{3} d^{3}-13 i c \,d^{5}+2 c^{6}-5 c^{4} d^{2}-20 c^{2} d^{4}+3 d^{6}\right ) \sqrt {c +d \tan \left (f x +e \right )}}{2 \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right )}}{\left (-d \tan \left (f x +e \right )+i d \right )^{3}}-\frac {\left (2 i c^{6}+5 i c^{4} d^{2}+5 i c^{2} d^{4}-2 i d^{6}-2 c^{5} d -5 c^{3} d^{3}-7 c \,d^{5}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{2 \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right ) \sqrt {-i d -c}}}{16 d^{4}}-\frac {i \left (i d -c \right )^{\frac {5}{2}} \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{16 d^{4}}\right )}{f \,a^{3}}\)	\(452\)
default	\(\frac {2 d^{4} \left (\frac {\frac {-\frac {d \left (i c^{4} d +i c^{2} d^{3}+4 i d^{5}+2 c^{5}+5 c^{3} d^{2}+7 c \,d^{4}\right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{2 \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right )}+\frac {2 d \left (6 i c^{5} d +20 i c^{3} d^{3}-2 i c \,d^{5}+3 c^{6}+5 c^{4} d^{2}-15 c^{2} d^{4}-d^{6}\right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3 \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right )}-\frac {d c \left (7 i c^{5} d +10 i c^{3} d^{3}-13 i c \,d^{5}+2 c^{6}-5 c^{4} d^{2}-20 c^{2} d^{4}+3 d^{6}\right ) \sqrt {c +d \tan \left (f x +e \right )}}{2 \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right )}}{\left (-d \tan \left (f x +e \right )+i d \right )^{3}}-\frac {\left (2 i c^{6}+5 i c^{4} d^{2}+5 i c^{2} d^{4}-2 i d^{6}-2 c^{5} d -5 c^{3} d^{3}-7 c \,d^{5}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{2 \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right ) \sqrt {-i d -c}}}{16 d^{4}}-\frac {i \left (i d -c \right )^{\frac {5}{2}} \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{16 d^{4}}\right )}{f \,a^{3}}\)	\(452\)

Input:

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

Output:

2/f/a^3*d^4*(1/16/d^4*((-1/2*d*(I*c^4*d+I*c^2*d^3+4*I*d^5+2*c^5+5*c^3*d^2+ 
7*c*d^4)/(3*I*c^2*d-I*d^3+c^3-3*c*d^2)*(c+d*tan(f*x+e))^(5/2)+2/3*d*(5*c^4 
*d^2-15*c^2*d^4-d^6+6*I*c^5*d+20*I*c^3*d^3-2*I*c*d^5+3*c^6)/(3*I*c^2*d-I*d 
^3+c^3-3*c*d^2)*(c+d*tan(f*x+e))^(3/2)-1/2*d*c*(-5*c^4*d^2-20*c^2*d^4+3*d^ 
6+7*I*c^5*d+10*I*c^3*d^3-13*I*c*d^5+2*c^6)/(3*I*c^2*d-I*d^3+c^3-3*c*d^2)*( 
c+d*tan(f*x+e))^(1/2))/(-d*tan(f*x+e)+I*d)^3-1/2*(-2*c^5*d-5*c^3*d^3-7*c*d 
^5+2*I*c^6+5*I*c^4*d^2+5*I*c^2*d^4-2*I*d^6)/(3*I*c^2*d-I*d^3+c^3-3*c*d^2)/ 
(-c-I*d)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(-c-I*d)^(1/2)))-1/16*I*(I*d- 
c)^(5/2)/d^4*arctan((c+d*tan(f*x+e))^(1/2)/(I*d-c)^(1/2)))
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1252 vs. \(2 (225) = 450\).

Time = 0.44 (sec) , antiderivative size = 1252, normalized size of antiderivative = 4.39 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

1/192*(6*a^3*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^ 
4 - I*d^5)/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(2*(c^3 - 2*I*c^2*d - c*d^2 + 
 (I*a^3*f*e^(2*I*f*x + 2*I*e) + I*a^3*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I* 
e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d 
^2 + 10*I*c^2*d^3 + 5*c*d^4 - I*d^5)/(a^6*f^2)) + (c^3 - 3*I*c^2*d - 3*c*d 
^2 + I*d^3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(c^2 - 2*I*c*d - d^2 
)) - 6*a^3*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 
- I*d^5)/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(2*(c^3 - 2*I*c^2*d - c*d^2 + ( 
-I*a^3*f*e^(2*I*f*x + 2*I*e) - I*a^3*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e 
) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^ 
2 + 10*I*c^2*d^3 + 5*c*d^4 - I*d^5)/(a^6*f^2)) + (c^3 - 3*I*c^2*d - 3*c*d^ 
2 + I*d^3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(c^2 - 2*I*c*d - d^2) 
) + 3*a^3*f*sqrt(-(4*I*c^6 + 16*c^5*d - 20*I*c^4*d^2 - 15*I*c^2*d^4 - 4*c* 
d^5 - 4*I*d^6)/((I*a^6*c - a^6*d)*f^2))*e^(6*I*f*x + 6*I*e)*log(-1/16*(2*c 
^4 - 2*I*c^3*d + 3*c^2*d^2 - 3*I*c*d^3 + 2*d^4 - ((I*a^3*c - a^3*d)*f*e^(2 
*I*f*x + 2*I*e) + (I*a^3*c - a^3*d)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) 
 + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(4*I*c^6 + 16*c^5*d - 20*I*c^ 
4*d^2 - 15*I*c^2*d^4 - 4*c*d^5 - 4*I*d^6)/((I*a^6*c - a^6*d)*f^2)) + (2*c^ 
4 - 4*I*c^3*d - c^2*d^2 - 2*I*c*d^3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2* 
I*e)/((I*a^3*c - a^3*d)*f)) - 3*a^3*f*sqrt(-(4*I*c^6 + 16*c^5*d - 20*I*...
 

Sympy [F]

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

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

Output:

I*(Integral(c**2*sqrt(c + d*tan(e + f*x))/(tan(e + f*x)**3 - 3*I*tan(e + f 
*x)**2 - 3*tan(e + f*x) + I), x) + Integral(d**2*sqrt(c + d*tan(e + f*x))* 
tan(e + f*x)**2/(tan(e + f*x)**3 - 3*I*tan(e + f*x)**2 - 3*tan(e + f*x) + 
I), x) + Integral(2*c*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)/(tan(e + f*x 
)**3 - 3*I*tan(e + f*x)**2 - 3*tan(e + f*x) + I), x))/a**3
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 583 vs. \(2 (225) = 450\).

Time = 0.79 (sec) , antiderivative size = 583, normalized size of antiderivative = 2.05 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {\frac {3 \, \sqrt {2} {\left (-2 i \, c^{3} - 4 \, c^{2} d + i \, c d^{2} - 2 \, d^{3}\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} + i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{\sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {6 \, \sqrt {2} {\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} - i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{\sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {6 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} c^{2} d - 12 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c^{3} d + 6 \, \sqrt {d \tan \left (f x + e\right ) + c} c^{4} d - 15 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} c d^{2} + 12 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c^{2} d^{2} + 3 i \, \sqrt {d \tan \left (f x + e\right ) + c} c^{3} d^{2} - 12 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} d^{3} - 20 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c d^{3} + 12 \, \sqrt {d \tan \left (f x + e\right ) + c} c^{2} d^{3} + 4 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} d^{4} + 9 i \, \sqrt {d \tan \left (f x + e\right ) + c} c d^{4}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3}}}{48 \, a^{3} f} \] Input:

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

Output:

1/48*(3*sqrt(2)*(-2*I*c^3 - 4*c^2*d + I*c*d^2 - 2*d^3)*arctan(2*(sqrt(d*ta 
n(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(sqrt(2)*c*s 
qrt(-c + sqrt(c^2 + d^2)) + I*sqrt(2)*sqrt(-c + sqrt(c^2 + d^2))*d - sqrt( 
2)*sqrt(c^2 + d^2)*sqrt(-c + sqrt(c^2 + d^2))))/(sqrt(-c + sqrt(c^2 + d^2) 
)*(I*d/(c - sqrt(c^2 + d^2)) + 1)) - 6*sqrt(2)*(-I*c^3 - 3*c^2*d + 3*I*c*d 
^2 + d^3)*arctan(2*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*ta 
n(f*x + e) + c))/(sqrt(2)*c*sqrt(-c + sqrt(c^2 + d^2)) - I*sqrt(2)*sqrt(-c 
 + sqrt(c^2 + d^2))*d - sqrt(2)*sqrt(c^2 + d^2)*sqrt(-c + sqrt(c^2 + d^2)) 
))/(sqrt(-c + sqrt(c^2 + d^2))*(-I*d/(c - sqrt(c^2 + d^2)) + 1)) + (6*(d*t 
an(f*x + e) + c)^(5/2)*c^2*d - 12*(d*tan(f*x + e) + c)^(3/2)*c^3*d + 6*sqr 
t(d*tan(f*x + e) + c)*c^4*d - 15*I*(d*tan(f*x + e) + c)^(5/2)*c*d^2 + 12*I 
*(d*tan(f*x + e) + c)^(3/2)*c^2*d^2 + 3*I*sqrt(d*tan(f*x + e) + c)*c^3*d^2 
 - 12*(d*tan(f*x + e) + c)^(5/2)*d^3 - 20*(d*tan(f*x + e) + c)^(3/2)*c*d^3 
 + 12*sqrt(d*tan(f*x + e) + c)*c^2*d^3 + 4*I*(d*tan(f*x + e) + c)^(3/2)*d^ 
4 + 9*I*sqrt(d*tan(f*x + e) + c)*c*d^4)/(d*tan(f*x + e) - I*d)^3)/(a^3*f)
 

Mupad [B] (verification not implemented)

Time = 6.42 (sec) , antiderivative size = 9472, normalized size of antiderivative = 33.24 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

- atan((((2*a^3*f*(1536*a^6*c*d^5*f^2 + a^6*c^2*d^4*f^2*2560i - 1024*a^6*c 
^3*d^3*f^2) - 65536*a^12*c*d^2*f^4*(c + d*tan(e + f*x))^(1/2)*(-(20*c*d^10 
 + c^2*d^9*55i - 35*c^3*d^8 + c^4*d^7*40i - 72*c^5*d^6 - c^6*d^5*40i + 8*c 
^7*d^4 - 4*a^6*f^2*(((5*c*d^10 - (35*c^3*d^8)/4 - 18*c^5*d^6 + 2*c^7*d^4)/ 
(a^6*f^2) + (((55*c^2*d^9)/4 + 10*c^4*d^7 - 10*c^6*d^5)*1i)/(a^6*f^2))^2 - 
 4*(256*d^6 + 256*c^2*d^4)*((((5*c*d^15)/256 + (35*c^3*d^13)/512 - (25*c^5 
*d^11)/128 - (285*c^7*d^9)/512 + (55*c^9*d^7)/128 - (5*c^11*d^5)/128)*1i)/ 
(a^12*f^4) - ((21*c^2*d^14)/1024 - d^16/256 + (225*c^4*d^12)/1024 + (155*c 
^6*d^10)/1024 - (665*c^8*d^8)/1024 + (11*c^10*d^6)/64 - (c^12*d^4)/256)/(a 
^12*f^4)))^(1/2))/(2048*a^6*f^2*(d^6 + c^2*d^4)))^(1/2))*(-(20*c*d^10 + c^ 
2*d^9*55i - 35*c^3*d^8 + c^4*d^7*40i - 72*c^5*d^6 - c^6*d^5*40i + 8*c^7*d^ 
4 - 4*a^6*f^2*(((5*c*d^10 - (35*c^3*d^8)/4 - 18*c^5*d^6 + 2*c^7*d^4)/(a^6* 
f^2) + (((55*c^2*d^9)/4 + 10*c^4*d^7 - 10*c^6*d^5)*1i)/(a^6*f^2))^2 - 4*(2 
56*d^6 + 256*c^2*d^4)*((((5*c*d^15)/256 + (35*c^3*d^13)/512 - (25*c^5*d^11 
)/128 - (285*c^7*d^9)/512 + (55*c^9*d^7)/128 - (5*c^11*d^5)/128)*1i)/(a^12 
*f^4) - ((21*c^2*d^14)/1024 - d^16/256 + (225*c^4*d^12)/1024 + (155*c^6*d^ 
10)/1024 - (665*c^8*d^8)/1024 + (11*c^10*d^6)/64 - (c^12*d^4)/256)/(a^12*f 
^4)))^(1/2))/(2048*a^6*f^2*(d^6 + c^2*d^4)))^(1/2) + 32*a^6*f^2*(c + d*tan 
(e + f*x))^(1/2)*(c*d^7*20i + 8*d^8 - 45*c^2*d^6 - c^3*d^5*80i + 80*c^4*d^ 
4 + c^5*d^3*40i - 8*c^6*d^2))*(-(20*c*d^10 + c^2*d^9*55i - 35*c^3*d^8 +...
 

Reduce [F]

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

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

Output:

( - int(sqrt(tan(e + f*x)*d + c)/(tan(e + f*x)**3*i + 3*tan(e + f*x)**2 - 
3*tan(e + f*x)*i - 1),x)*c**2 - int((sqrt(tan(e + f*x)*d + c)*tan(e + f*x) 
**2)/(tan(e + f*x)**3*i + 3*tan(e + f*x)**2 - 3*tan(e + f*x)*i - 1),x)*d** 
2 - 2*int((sqrt(tan(e + f*x)*d + c)*tan(e + f*x))/(tan(e + f*x)**3*i + 3*t 
an(e + f*x)**2 - 3*tan(e + f*x)*i - 1),x)*c*d)/a**3