\(\int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [129]

Optimal result

Integrand size = 26, antiderivative size = 133 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}-\frac {i}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i}{2 a d (a+i a \tan (c+d x))^{3/2}}-\frac {i}{4 a^2 d \sqrt {a+i a \tan (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.09 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {i \sec ^2(c+d x) \left (2 \sqrt {a} (-3+2 \cos (2 (c+d x)))-5 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x))) \sqrt {a+i a \tan (c+d x)}\right )}{40 a^{5/2} d (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \] Input:

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

Output:

((I/40)*Sec[c + d*x]^2*(2*Sqrt[a]*(-3 + 2*Cos[2*(c + d*x)]) - 5*Sqrt[2]*Ar 
cTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])]*(Cos[2*(c + d*x)] + I* 
Sin[2*(c + d*x)])*Sqrt[a + I*a*Tan[c + d*x]]))/(a^(5/2)*d*(-I + Tan[c + d* 
x])^2*Sqrt[a + I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3042, 4023, 3042, 4009, 3042, 3960, 3042, 3961, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4023

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4009

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3960

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3961

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{2} \left (\frac {i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}-\frac {i}{d \sqrt {a+i a \tan (c+d x)}}\right )+\frac {i a}{d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}-\frac {i}{5 d (a+i a \tan (c+d x))^{5/2}}\)

Input:

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

Output:

(-1/5*I)/(d*(a + I*a*Tan[c + d*x])^(5/2)) + ((I*a)/(d*(a + I*a*Tan[c + d*x 
])^(3/2)) + ((I*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sq 
rt[2]*Sqrt[a]*d) - I/(d*Sqrt[a + I*a*Tan[c + d*x]]))/2)/(2*a^2)
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 3960

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a*((a + 
b*Tan[c + d*x])^n/(2*b*d*n)), x] + Simp[1/(2*a)   Int[(a + b*Tan[c + d*x])^ 
(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n, 0]
 

rule 3961

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

rule 4009

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

rule 4023

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^2, x_Symbol] :> Simp[(-b)*(a*c + b*d)^2*((a + b*Tan[e + f*x])^ 
m/(2*a^3*f*m)), x] + Simp[1/(2*a^2)   Int[(a + b*Tan[e + f*x])^(m + 1)*Simp 
[a*c^2 - 2*b*c*d + a*d^2 - 2*b*d^2*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, 
 c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LeQ[m, -1] && EqQ[a^2 + b^2, 0]
 

Maple [A] (verified)

Time = 1.36 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.71

method	result	size
derivativedivides	\(\frac {2 i \left (\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {3}{2}}}+\frac {1}{4 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {1}{8 a \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {a}{10 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}\right )}{d a}\)	\(94\)
default	\(\frac {2 i \left (\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {3}{2}}}+\frac {1}{4 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {1}{8 a \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {a}{10 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}\right )}{d a}\)	\(94\)

Input:

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

Output:

2*I/d/a*(1/16/a^(3/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2) 
/a^(1/2))+1/4/(a+I*a*tan(d*x+c))^(3/2)-1/8/a/(a+I*a*tan(d*x+c))^(1/2)-1/10 
*a/(a+I*a*tan(d*x+c))^(5/2))
 

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. 283 vs. \(2 (94) = 188\).

Time = 0.09 (sec) , antiderivative size = 283, normalized size of antiderivative = 2.13 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {{\left (5 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 5 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 2 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{40 \, a^{3} d} \] Input:

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

Output:

1/40*(5*I*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(4*(sqr 
t(2)*sqrt(1/2)*(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 2* 
I*c) + 1))*sqrt(1/(a^5*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 5*I* 
sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-4*(sqrt(2)*sqrt 
(1/2)*(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1) 
)*sqrt(1/(a^5*d^2)) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + sqrt(2)*sqrt( 
a/(e^(2*I*d*x + 2*I*c) + 1))*(-I*e^(6*I*d*x + 6*I*c) + 2*I*e^(4*I*d*x + 4* 
I*c) + 2*I*e^(2*I*d*x + 2*I*c) - I))*e^(-5*I*d*x - 5*I*c)/(a^3*d)
 

Sympy [F]

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

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

Output:

Integral(tan(c + d*x)**2/(I*a*(tan(c + d*x) - I))**(5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.89 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {i \, {\left (5 \, \sqrt {2} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (5 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a - 10 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} + 4 \, a^{3}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\right )}}{80 \, a^{3} d} \] Input:

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

Output:

-1/80*I*(5*sqrt(2)*sqrt(a)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + 
 a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a))) + 4*(5*(I*a*tan(d*x + 
 c) + a)^2*a - 10*(I*a*tan(d*x + c) + a)*a^2 + 4*a^3)/(I*a*tan(d*x + c) + 
a)^(5/2))/(a^3*d)
 

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeDone
 

Mupad [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.66 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {1{}\mathrm {i}}{20\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{4\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{8\,{\left (-a\right )}^{5/2}\,d} \] Input:

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

Output:

1i/(20*d*(a + a*tan(c + d*x)*1i)^(5/2)) + (tan(c + d*x)^2*1i)/(4*d*(a + a* 
tan(c + d*x)*1i)^(5/2)) - (2^(1/2)*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^( 
1/2))/(2*(-a)^(1/2)))*1i)/(8*(-a)^(5/2)*d)
 

Reduce [F]

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

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

Output:

( - int(tan(c + d*x)**2/(sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2 - 2*sqrt 
(tan(c + d*x)*i + 1)*tan(c + d*x)*i - sqrt(tan(c + d*x)*i + 1)),x))/(sqrt( 
a)*a**2)