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

Optimal result

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

[Out]

-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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3603, 3568, 45} \[ \int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c-i c \tan (e+f x)}} \, dx=\frac {2 i a^3 (c-i c \tan (e+f x))^{3/2}}{3 c^2 f}-\frac {8 i a^3 \sqrt {c-i c \tan (e+f x)}}{c f}-\frac {8 i a^3}{f \sqrt {c-i c \tan (e+f x)}} \]

[In]

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

[Out]

((-8*I)*a^3)/(f*Sqrt[c - I*c*Tan[e + f*x]]) - ((8*I)*a^3*Sqrt[c - I*c*Tan[e + f*x]])/(c*f) + (((2*I)/3)*a^3*(c
- I*c*Tan[e + f*x])^(3/2))/(c^2*f)

Rule 45

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] && NeQ[b*c - a*d, 0] && 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 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[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 3603

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[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]))

Rubi steps \begin{align*} \text {integral}& = \left (a^3 c^3\right ) \int \frac {\sec ^6(e+f x)}{(c-i c \tan (e+f x))^{7/2}} \, dx \\ & = \frac {\left (i a^3\right ) \text {Subst}\left (\int \frac {(c-x)^2}{(c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{c^2 f} \\ & = \frac {\left (i a^3\right ) \text {Subst}\left (\int \left (\frac {4 c^2}{(c+x)^{3/2}}-\frac {4 c}{\sqrt {c+x}}+\sqrt {c+x}\right ) \, dx,x,-i c \tan (e+f x)\right )}{c^2 f} \\ & = -\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} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.20 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.54 \[ \int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {2 i a^3 \left (23-10 i \tan (e+f x)+\tan ^2(e+f x)\right )}{3 f \sqrt {c-i c \tan (e+f x)}} \]

[In]

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

[Out]

(((-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]])

Maple [A] (verified)

Time = 0.65 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.73

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.82 \[ \int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {4 \, \sqrt {2} {\left (3 i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 12 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 i \, a^{3}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{3 \, {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )}} \]

[In]

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

[Out]

-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]

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

[In]

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

[Out]

-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)

none

Time = 0.22 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.78 \[ \int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {2 i \, {\left (\frac {12 \, a^{3} c}{\sqrt {-i \, c \tan \left (f x + e\right ) + c}} - \frac {{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{3} - 12 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} a^{3} c}{c}\right )}}{3 \, c f} \]

[In]

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

[Out]

-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]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.94 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.26 \[ \int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {2\,a^3\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (\cos \left (2\,e+2\,f\,x\right )\,23{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}-7\,\sin \left (2\,e+2\,f\,x\right )-3\,\sin \left (4\,e+4\,f\,x\right )+20{}\mathrm {i}\right )}{3\,c\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

[In]

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

[Out]

-(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))