\(\int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c-i c \tan (e+f x)}} \, dx\) [974]

Optimal result

Integrand size = 33, antiderivative size = 90 \[ \int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {8 i a^3}{f \sqrt {c-i c \tan (e+f x)}}-\frac {8 i a^3 \sqrt {c-i c \tan (e+f x)}}{c f}+\frac {2 i a^3 (c-i c \tan (e+f x))^{3/2}}{3 c^2 f} \] Output:

-8*I*a^3/f/(c-I*c*tan(f*x+e))^(1/2)-8*I*a^3*(c-I*c*tan(f*x+e))^(1/2)/c/f+2 
/3*I*a^3*(c-I*c*tan(f*x+e))^(3/2)/c^2/f
 

Mathematica [A] (verified)

Time = 1.36 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.54 \[ \int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {2 i a^3 \left (23-10 i \tan (e+f x)+\tan ^2(e+f x)\right )}{3 f \sqrt {c-i c \tan (e+f x)}} \] Input:

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

Output:

(((-2*I)/3)*a^3*(23 - (10*I)*Tan[e + f*x] + Tan[e + f*x]^2))/(f*Sqrt[c - I 
*c*Tan[e + f*x]])
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.86, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4005

\(\displaystyle a^3 c^3 \int \frac {\sec ^6(e+f x)}{(c-i c \tan (e+f x))^{7/2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a^3 c^3 \int \frac {\sec (e+f x)^6}{(c-i c \tan (e+f x))^{7/2}}dx\)

\(\Big \downarrow \) 3968

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

\(\Big \downarrow \) 53

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {i a^3 \left (-\frac {8 c^2}{\sqrt {c-i c \tan (e+f x)}}-8 c \sqrt {c-i c \tan (e+f x)}+\frac {2}{3} (c-i c \tan (e+f x))^{3/2}\right )}{c^2 f}\)

Input:

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

Output:

(I*a^3*((-8*c^2)/Sqrt[c - I*c*Tan[e + f*x]] - 8*c*Sqrt[c - I*c*Tan[e + f*x 
]] + (2*(c - I*c*Tan[e + f*x])^(3/2))/3))/(c^2*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 = 0.42 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(\frac {2 i a^{3} \left (\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-4 \sqrt {c -i c \tan \left (f x +e \right )}\, c -\frac {4 c^{2}}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f \,c^{2}}\)	\(66\)
default	\(\frac {2 i a^{3} \left (\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-4 \sqrt {c -i c \tan \left (f x +e \right )}\, c -\frac {4 c^{2}}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f \,c^{2}}\)	\(66\)
parts	\(\frac {2 i a^{3} c \left (\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 c^{\frac {3}{2}}}-\frac {1}{2 c \sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f}-\frac {2 i a^{3} \left (-\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+\sqrt {c -i c \tan \left (f x +e \right )}\, c +\frac {c^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4}+\frac {c^{2}}{2 \sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f \,c^{2}}+\frac {3 i a^{3} \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{2 \sqrt {c}}-\frac {1}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f}-\frac {6 i a^{3} \left (\sqrt {c -i c \tan \left (f x +e \right )}-\frac {\sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4}+\frac {c}{2 \sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f c}\)	\(290\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.82 \[ \int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {4 \, \sqrt {2} {\left (3 i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 12 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 i \, a^{3}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{3 \, {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )}} \] Input:

integrate((a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="fric 
as")
 

Output:

-4/3*sqrt(2)*(3*I*a^3*e^(4*I*f*x + 4*I*e) + 12*I*a^3*e^(2*I*f*x + 2*I*e) + 
 8*I*a^3)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(c*f*e^(2*I*f*x + 2*I*e) + c*f 
)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.78 \[ \int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {2 i \, {\left (\frac {12 \, a^{3} c}{\sqrt {-i \, c \tan \left (f x + e\right ) + c}} - \frac {{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{3} - 12 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} a^{3} c}{c}\right )}}{3 \, c f} \] Input:

integrate((a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="maxi 
ma")
 

Output:

-2/3*I*(12*a^3*c/sqrt(-I*c*tan(f*x + e) + c) - ((-I*c*tan(f*x + e) + c)^(3 
/2)*a^3 - 12*sqrt(-I*c*tan(f*x + e) + c)*a^3*c)/c)/(c*f)
 

Giac [F(-2)]

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

integrate((a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(1/2),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 = 2.45 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.26 \[ \int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {2\,a^3\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (\cos \left (2\,e+2\,f\,x\right )\,23{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}-7\,\sin \left (2\,e+2\,f\,x\right )-3\,\sin \left (4\,e+4\,f\,x\right )+20{}\mathrm {i}\right )}{3\,c\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \] Input:

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

Output:

-(2*a^3*((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)*23i + cos(4*e + 4*f*x)*3i - 7*sin(2*e + 2* 
f*x) - 3*sin(4*e + 4*f*x) + 20i))/(3*c*f*(cos(2*e + 2*f*x) + 1))
 

Reduce [F]

\[ \int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c-i c \tan (e+f x)}} \, dx=\frac {\sqrt {c}\, a^{3} \left (2 \sqrt {-\tan \left (f x +e \right ) i +1}\, i +\left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{4}}{\tan \left (f x +e \right )^{2}+1}d x \right ) \tan \left (f x +e \right )^{2} f +\left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{4}}{\tan \left (f x +e \right )^{2}+1}d x \right ) f -4 \left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{2}+1}d x \right ) \tan \left (f x +e \right )^{2} f i -4 \left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{2}+1}d x \right ) f i -6 \left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2}+1}d x \right ) \tan \left (f x +e \right )^{2} f -6 \left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2}+1}d x \right ) f +7 \left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{2}+1}d x \right ) \tan \left (f x +e \right )^{2} f i +7 \left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{2}+1}d x \right ) f i \right )}{c f \left (\tan \left (f x +e \right )^{2}+1\right )} \] Input:

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

Output:

(sqrt(c)*a**3*(2*sqrt( - tan(e + f*x)*i + 1)*i + int((sqrt( - tan(e + f*x) 
*i + 1)*tan(e + f*x)**4)/(tan(e + f*x)**2 + 1),x)*tan(e + f*x)**2*f + int( 
(sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**4)/(tan(e + f*x)**2 + 1),x)*f - 
 4*int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**3)/(tan(e + f*x)**2 + 1) 
,x)*tan(e + f*x)**2*f*i - 4*int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)* 
*3)/(tan(e + f*x)**2 + 1),x)*f*i - 6*int((sqrt( - tan(e + f*x)*i + 1)*tan( 
e + f*x)**2)/(tan(e + f*x)**2 + 1),x)*tan(e + f*x)**2*f - 6*int((sqrt( - t 
an(e + f*x)*i + 1)*tan(e + f*x)**2)/(tan(e + f*x)**2 + 1),x)*f + 7*int((sq 
rt( - tan(e + f*x)*i + 1)*tan(e + f*x))/(tan(e + f*x)**2 + 1),x)*tan(e + f 
*x)**2*f*i + 7*int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x))/(tan(e + f*x 
)**2 + 1),x)*f*i))/(c*f*(tan(e + f*x)**2 + 1))