\(\int \frac {(a+i a \tan (e+f x))^2}{(c-i c \tan (e+f x))^{3/2}} \, dx\) [981]

Optimal result

Integrand size = 33, antiderivative size = 60 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {4 i a^2}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {2 i a^2}{c f \sqrt {c-i c \tan (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 1.90 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.92 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {i a^2 \left (-\frac {4 c}{3 (c-i c \tan (e+f x))^{3/2}}+\frac {2}{\sqrt {c-i c \tan (e+f x)}}\right )}{c f} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3042, 4005, 3042, 3968, 53, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4005

\(\displaystyle a^2 c^2 \int \frac {\sec ^4(e+f x)}{(c-i c \tan (e+f x))^{7/2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 c^2 \int \frac {\sec (e+f x)^4}{(c-i c \tan (e+f x))^{7/2}}dx\)

\(\Big \downarrow \) 3968

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

\(\Big \downarrow \) 53

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 53

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, 
x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) 
|| LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3968

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ 
), x_Symbol] :> Simp[1/(a^(m - 2)*b*f)   Subst[Int[(a - x)^(m/2 - 1)*(a + x 
)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && 
 EqQ[a^2 + b^2, 0] && IntegerQ[m/2]
 

rule 4005

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

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.73

method	result	size
risch	\(-\frac {i a^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}-2\right ) \sqrt {2}}{3 c \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(44\)
derivativedivides	\(\frac {2 i a^{2} \left (\frac {1}{\sqrt {c -i c \tan \left (f x +e \right )}}-\frac {2 c}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f c}\)	\(45\)
default	\(\frac {2 i a^{2} \left (\frac {1}{\sqrt {c -i c \tan \left (f x +e \right )}}-\frac {2 c}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f c}\)	\(45\)
parts	\(\frac {2 i a^{2} c \left (\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 c^{\frac {5}{2}}}-\frac {1}{4 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{6 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f}+\frac {2 i a^{2} \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 c^{\frac {3}{2}}}-\frac {1}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {1}{2 c \sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f}-\frac {2 i a^{2} \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 \sqrt {c}}-\frac {3}{4 \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {c}{6 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f c}\)	\(235\)

Input:

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

Output:

-1/3*I*a^2/c/(c/(exp(2*I*(f*x+e))+1))^(1/2)*(exp(2*I*(f*x+e))-2)/f*2^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.03 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {\sqrt {2} {\left (-i \, a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + i \, a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, a^{2}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{3 \, c^{2} f} \] Input:

integrate((a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fric 
as")
 

Output:

1/3*sqrt(2)*(-I*a^2*e^(4*I*f*x + 4*I*e) + I*a^2*e^(2*I*f*x + 2*I*e) + 2*I* 
a^2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(c^2*f)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

-a**2*(Integral(tan(e + f*x)**2/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + 
f*x) + c*sqrt(-I*c*tan(e + f*x) + c)), x) + Integral(-2*I*tan(e + f*x)/(-I 
*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sqrt(-I*c*tan(e + f*x) + c 
)), x) + Integral(-1/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sq 
rt(-I*c*tan(e + f*x) + c)), x))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.73 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {2 i \, {\left (3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{2} - 2 \, a^{2} c\right )}}{3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c f} \] Input:

integrate((a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="maxi 
ma")
 

Output:

2/3*I*(3*(-I*c*tan(f*x + e) + c)*a^2 - 2*a^2*c)/((-I*c*tan(f*x + e) + c)^( 
3/2)*c*f)
 

Giac [F(-2)]

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

integrate((a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(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 = 2.21 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.63 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {a^2\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (\cos \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}-\cos \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}-\sin \left (2\,e+2\,f\,x\right )+\sin \left (4\,e+4\,f\,x\right )+2{}\mathrm {i}\right )}{3\,c^2\,f} \] Input:

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

Output:

(a^2*((c*(cos(2*e + 2*f*x) - sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 
 1))^(1/2)*(cos(2*e + 2*f*x)*1i - cos(4*e + 4*f*x)*1i - sin(2*e + 2*f*x) + 
 sin(4*e + 4*f*x) + 2i))/(3*c^2*f)
 

Reduce [F]

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

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

Output:

(sqrt(c)*a**2*(2*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x) + int(sqrt( - ta 
n(e + f*x)*i + 1)/(tan(e + f*x)**3*i - tan(e + f*x)**2 + tan(e + f*x)*i - 
1),x)*tan(e + f*x)**2*f + int(sqrt( - tan(e + f*x)*i + 1)/(tan(e + f*x)**3 
*i - tan(e + f*x)**2 + tan(e + f*x)*i - 1),x)*f - 3*int((sqrt( - tan(e + f 
*x)*i + 1)*tan(e + f*x)**2)/(tan(e + f*x)**3*i - tan(e + f*x)**2 + tan(e + 
 f*x)*i - 1),x)*tan(e + f*x)**2*f - 3*int((sqrt( - tan(e + f*x)*i + 1)*tan 
(e + f*x)**2)/(tan(e + f*x)**3*i - tan(e + f*x)**2 + tan(e + f*x)*i - 1),x 
)*f - 4*int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x))/(tan(e + f*x)**3*i 
- tan(e + f*x)**2 + tan(e + f*x)*i - 1),x)*tan(e + f*x)**2*f*i - 4*int((sq 
rt( - tan(e + f*x)*i + 1)*tan(e + f*x))/(tan(e + f*x)**3*i - tan(e + f*x)* 
*2 + tan(e + f*x)*i - 1),x)*f*i))/(c**2*f*(tan(e + f*x)**2 + 1))