\(\int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [120]

Optimal result

Integrand size = 26, antiderivative size = 104 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {i}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {3 i}{2 a d \sqrt {a+i a \tan (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.21 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {2 \sqrt {a} (7+9 i \tan (c+d x))+3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right ) (1+i \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{12 a^{3/2} d (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}} \] Input:

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

Output:

(2*Sqrt[a]*(7 + (9*I)*Tan[c + d*x]) + 3*Sqrt[2]*ArcTanh[Sqrt[a + I*a*Tan[c 
 + d*x]]/(Sqrt[2]*Sqrt[a])]*(1 + I*Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x] 
])/(12*a^(3/2)*d*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {3042, 4023, 3042, 4009, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4023

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4009

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3961

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(-1/3*I)/(d*(a + I*a*Tan[c + d*x])^(3/2)) + ((I*Sqrt[a]*ArcTanh[Sqrt[a + I 
*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*d) + ((3*I)*a)/(d*Sqrt[a + I 
*a*Tan[c + d*x]]))/(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 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 = 75, normalized size of antiderivative = 0.72

method	result	size
derivativedivides	\(\frac {2 i \left (\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 \sqrt {a}}+\frac {3}{4 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {a}{6 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}\right )}{d a}\)	\(75\)
default	\(\frac {2 i \left (\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 \sqrt {a}}+\frac {3}{4 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {a}{6 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}\right )}{d a}\)	\(75\)

Input:

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

Output:

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

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. 272 vs. \(2 (73) = 146\).

Time = 0.09 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.62 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {{\left (3 i \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 3 i \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (8 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 7 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{12 \, a^{2} d} \] Input:

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

Output:

1/12*(3*I*sqrt(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log(4*(sqr 
t(2)*sqrt(1/2)*(a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)*sqrt(a/(e^(2*I*d*x + 2* 
I*c) + 1))*sqrt(1/(a^3*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 3*I* 
sqrt(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log(-4*(sqrt(2)*sqrt 
(1/2)*(a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1) 
)*sqrt(1/(a^3*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))*(8*I*e^(4*I*d*x + 4*I*c) + 7*I*e^(2*I*d*x + 2 
*I*c) - I))*e^(-3*I*d*x - 3*I*c)/(a^2*d)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.99 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {i \, {\left (3 \, \sqrt {2} a^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - \frac {4 \, {\left (9 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} - 2 \, a^{3}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\right )}}{24 \, a^{3} d} \] Input:

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

Output:

-1/24*I*(3*sqrt(2)*a^(3/2)*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*(9*(I*a*tan(d*x + 
 c) + a)*a^2 - 2*a^3)/(I*a*tan(d*x + c) + a)^(3/2))/(a^3*d)
 

Giac [F(-2)]

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

integrate(tan(d*x+c)^2/(a+I*a*tan(d*x+c))^(3/2),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 = 1.03 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.81 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {\frac {1{}\mathrm {i}}{3\,d}-\frac {\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{2\,a\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{4\,{\left (-a\right )}^{3/2}\,d} \] Input:

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

Output:

(2^(1/2)*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*(-a)^(1/2)))*1i)/ 
(4*(-a)^(3/2)*d) - (1i/(3*d) - ((a + a*tan(c + d*x)*1i)*3i)/(2*a*d))/(a + 
a*tan(c + d*x)*1i)^(3/2)
 

Reduce [F]

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

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

Output:

(sqrt(a)*( - 4*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x) + 2*sqrt(tan(c + d*x) 
*i + 1)*i + 5*int(sqrt(tan(c + d*x)*i + 1)/(tan(c + d*x)**3*i + tan(c + d* 
x)**2 + tan(c + d*x)*i + 1),x)*tan(c + d*x)**2*d + 5*int(sqrt(tan(c + d*x) 
*i + 1)/(tan(c + d*x)**3*i + tan(c + d*x)**2 + tan(c + d*x)*i + 1),x)*d + 
5*int((sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2)/(tan(c + d*x)**3*i + tan( 
c + d*x)**2 + tan(c + d*x)*i + 1),x)*tan(c + d*x)**2*d + 5*int((sqrt(tan(c 
 + d*x)*i + 1)*tan(c + d*x)**2)/(tan(c + d*x)**3*i + tan(c + d*x)**2 + tan 
(c + d*x)*i + 1),x)*d))/(2*a**2*d*(tan(c + d*x)**2 + 1))