\(\int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^{3/2}} \, dx\) [980]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.22 (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} \[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {2 i a^3 \sqrt {c-i c \tan (e+f x)}}{c^2 f}+\frac {8 i a^3}{c f \sqrt {c-i c \tan (e+f x)}}-\frac {8 i a^3}{3 f (c-i c \tan (e+f x))^{3/2}} \]

[In]

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

[Out]

(((-8*I)/3)*a^3)/(f*(c - I*c*Tan[e + f*x])^(3/2)) + ((8*I)*a^3)/(c*f*Sqrt[c - I*c*Tan[e + f*x]]) + ((2*I)*a^3*
Sqrt[c - I*c*Tan[e + f*x]])/(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))^{9/2}} \, dx \\ & = \frac {\left (i a^3\right ) \text {Subst}\left (\int \frac {(c-x)^2}{(c+x)^{5/2}} \, 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)^{5/2}}-\frac {4 c}{(c+x)^{3/2}}+\frac {1}{\sqrt {c+x}}\right ) \, dx,x,-i c \tan (e+f x)\right )}{c^2 f} \\ & = -\frac {8 i a^3}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {8 i a^3}{c f \sqrt {c-i c \tan (e+f x)}}+\frac {2 i a^3 \sqrt {c-i c \tan (e+f x)}}{c^2 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.81 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.71 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {2 a^3 \left (-11+18 i \tan (e+f x)+3 \tan ^2(e+f x)\right )}{3 c f (i+\tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \]

[In]

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

[Out]

(2*a^3*(-11 + (18*I)*Tan[e + f*x] + 3*Tan[e + f*x]^2))/(3*c*f*(I + Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.71

method	result	size
derivativedivides	\(\frac {2 i a^{3} \left (\sqrt {c -i c \tan \left (f x +e \right )}+\frac {4 c}{\sqrt {c -i c \tan \left (f x +e \right )}}-\frac {4 c^{2}}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f \,c^{2}}\)	\(64\)
default	\(\frac {2 i a^{3} \left (\sqrt {c -i c \tan \left (f x +e \right )}+\frac {4 c}{\sqrt {c -i c \tan \left (f x +e \right )}}-\frac {4 c^{2}}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f \,c^{2}}\)	\(64\)
parts	\(\frac {2 i a^{3} c \left (\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 c^{\frac {5}{2}}}-\frac {1}{4 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{6 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f}-\frac {2 i a^{3} \left (-\sqrt {c -i c \tan \left (f x +e \right )}+\frac {\sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8}-\frac {5 c}{4 \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {c^{2}}{6 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f \,c^{2}}+\frac {3 i a^{3} \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 c^{\frac {3}{2}}}-\frac {1}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {1}{2 c \sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f}-\frac {6 i a^{3} \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 \sqrt {c}}-\frac {3}{4 \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {c}{6 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f c}\)	\(331\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.69 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {2 \, \sqrt {2} {\left (i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 4 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 8 i \, a^{3}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{3 \, c^{2} f} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

-I*a**3*(Integral(I/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sqrt(-I*c*tan(e + f*x) + c)), x) + Inte
gral(-3*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) + Int
egral(tan(e + f*x)**3/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sqrt(-I*c*tan(e + f*x) + c)), x) + In
tegral(-3*I*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
))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.76 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {2 i \, {\left (\frac {3 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} a^{3}}{c} + \frac {4 \, {\left (3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{3} - a^{3} c\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\right )}}{3 \, c f} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.35 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.09 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {2\,a^3\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (\cos \left (2\,e+2\,f\,x\right )\,4{}\mathrm {i}-\cos \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}-4\,\sin \left (2\,e+2\,f\,x\right )+\sin \left (4\,e+4\,f\,x\right )+8{}\mathrm {i}\right )}{3\,c^2\,f} \]

[In]

int((a + a*tan(e + f*x)*1i)^3/(c - c*tan(e + f*x)*1i)^(3/2),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)*4i -
cos(4*e + 4*f*x)*1i - 4*sin(2*e + 2*f*x) + sin(4*e + 4*f*x) + 8i))/(3*c^2*f)