\(\int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^{5/2}} \, dx\) [986]

Optimal result

Integrand size = 33, antiderivative size = 92 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {8 i a^3}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {8 i a^3}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 i a^3}{c^2 f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

-2*I*a^3/c^2/f/(c-I*c*tan(f*x+e))^(1/2)-8/5*I*a^3/f/(c-I*c*tan(f*x+e))^(5/2)+8/3*I*a^3/c/f/(c-I*c*tan(f*x+e))^
(3/2)

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 92, 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} \[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {2 i a^3}{c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {8 i a^3}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {8 i a^3}{5 f (c-i c \tan (e+f x))^{5/2}} \]

[In]

Int[(a + I*a*Tan[e + f*x])^3/(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

(((-8*I)/5)*a^3)/(f*(c - I*c*Tan[e + f*x])^(5/2)) + (((8*I)/3)*a^3)/(c*f*(c - I*c*Tan[e + f*x])^(3/2)) - ((2*I
)*a^3)/(c^2*f*Sqrt[c - I*c*Tan[e + f*x]])

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))^{11/2}} \, dx \\ & = \frac {\left (i a^3\right ) \text {Subst}\left (\int \frac {(c-x)^2}{(c+x)^{7/2}} \, 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)^{7/2}}-\frac {4 c}{(c+x)^{5/2}}+\frac {1}{(c+x)^{3/2}}\right ) \, dx,x,-i c \tan (e+f x)\right )}{c^2 f} \\ & = -\frac {8 i a^3}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {8 i a^3}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 i a^3}{c^2 f \sqrt {c-i c \tan (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.44 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.72 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {2 a^3 \left (7 i+10 \tan (e+f x)-15 i \tan ^2(e+f x)\right )}{15 c^2 f (i+\tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}} \]

[In]

Integrate[(a + I*a*Tan[e + f*x])^3/(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

(2*a^3*(7*I + 10*Tan[e + f*x] - (15*I)*Tan[e + f*x]^2))/(15*c^2*f*(I + Tan[e + f*x])^2*Sqrt[c - I*c*Tan[e + f*
x]])

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.62

method	result	size
risch	\(-\frac {i a^{3} \left (3 \,{\mathrm e}^{4 i \left (f x +e \right )}-4 \,{\mathrm e}^{2 i \left (f x +e \right )}+8\right ) \sqrt {2}}{15 c^{2} \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(57\)
derivativedivides	\(\frac {2 i a^{3} \left (-\frac {1}{\sqrt {c -i c \tan \left (f x +e \right )}}-\frac {4 c^{2}}{5 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {4 c}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f \,c^{2}}\)	\(66\)
default	\(\frac {2 i a^{3} \left (-\frac {1}{\sqrt {c -i c \tan \left (f x +e \right )}}-\frac {4 c^{2}}{5 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {4 c}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f \,c^{2}}\)	\(66\)
parts	\(\frac {2 i a^{3} c \left (\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{16 c^{\frac {7}{2}}}-\frac {1}{8 c^{3} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{12 c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {1}{10 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f}-\frac {2 i a^{3} \left (\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{16 \sqrt {c}}+\frac {7}{8 \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {5 c}{12 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {c^{2}}{10 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f \,c^{2}}+\frac {3 i a^{3} \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 c^{\frac {5}{2}}}-\frac {1}{5 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {1}{4 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {1}{6 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f}-\frac {6 i a^{3} \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{16 c^{\frac {3}{2}}}-\frac {1}{4 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {1}{8 c \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {c}{10 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f c}\)	\(388\)

[In]

int((a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/15*I*a^3/c^2/(c/(exp(2*I*(f*x+e))+1))^(1/2)*(3*exp(4*I*(f*x+e))-4*exp(2*I*(f*x+e))+8)/f*2^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.83 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {\sqrt {2} {\left (-3 i \, a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 4 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 8 i \, a^{3}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{15 \, c^{3} f} \]

[In]

integrate((a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/15*sqrt(2)*(-3*I*a^3*e^(6*I*f*x + 6*I*e) + 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^3*f)

Sympy [F]

\[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^{5/2}} \, dx=- i a^{3} \left (\int \frac {i}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {3 \tan {\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {\tan ^{3}{\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {3 i \tan ^{2}{\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx\right ) \]

[In]

integrate((a+I*a*tan(f*x+e))**3/(c-I*c*tan(f*x+e))**(5/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.71 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {2 i \, {\left (15 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} a^{3} - 20 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{3} c + 12 \, a^{3} c^{2}\right )}}{15 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} c^{2} f} \]

[In]

integrate((a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

-2/15*I*(15*(-I*c*tan(f*x + e) + c)^2*a^3 - 20*(-I*c*tan(f*x + e) + c)*a^3*c + 12*a^3*c^2)/((-I*c*tan(f*x + e)
+ c)^(5/2)*c^2*f)

Giac [F]

\[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^{5/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^3/(-I*c*tan(f*x + e) + c)^(5/2), x)

Mupad [B] (verification not implemented)

Time = 7.29 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.32 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {a^3\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (\cos \left (2\,e+2\,f\,x\right )\,4{}\mathrm {i}-\cos \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+\cos \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}-4\,\sin \left (2\,e+2\,f\,x\right )+\sin \left (4\,e+4\,f\,x\right )-3\,\sin \left (6\,e+6\,f\,x\right )+8{}\mathrm {i}\right )}{15\,c^3\,f} \]

[In]

int((a + a*tan(e + f*x)*1i)^3/(c - c*tan(e + f*x)*1i)^(5/2),x)

[Out]

-(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 - c
os(4*e + 4*f*x)*1i + cos(6*e + 6*f*x)*3i - 4*sin(2*e + 2*f*x) + sin(4*e + 4*f*x) - 3*sin(6*e + 6*f*x) + 8i))/(
15*c^3*f)