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

Optimal. Leaf size=90 \[ \frac {2 i a^3 \sqrt {c-i c \tan (e+f x)}}{c^2 f}+\frac {8 i a^3}{c f \sqrt {c-i c \tan (e+f x)}}-\frac {8 i a^3}{3 f (c-i c \tan (e+f x))^{3/2}} \]

[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)

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Rubi [A]  time = 0.16, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3522, 3487, 43} \[ \frac {2 i a^3 \sqrt {c-i c \tan (e+f x)}}{c^2 f}+\frac {8 i a^3}{c f \sqrt {c-i c \tan (e+f x)}}-\frac {8 i a^3}{3 f (c-i c \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[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 43

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 3487

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 3522

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

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Mathematica [A]  time = 5.36, size = 94, normalized size = 1.04 \[ \frac {2 a^3 \sqrt {c-i c \tan (e+f x)} (9 \sin (2 (e+f x))+7 i \cos (2 (e+f x))+4 i) (\cos (2 e+5 f x)+i \sin (2 e+5 f x))}{3 c^2 f (\cos (f x)+i \sin (f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*(4*I + (7*I)*Cos[2*(e + f*x)] + 9*Sin[2*(e + f*x)])*(Cos[2*e + 5*f*x] + I*Sin[2*e + 5*f*x])*Sqrt[c - I*
c*Tan[e + f*x]])/(3*c^2*f*(Cos[f*x] + I*Sin[f*x])^3)

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fricas [A]  time = 0.48, size = 62, normalized size = 0.69 \[ \frac {\sqrt {2} {\left (-2 i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 8 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 16 i \, a^{3}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{3 \, c^{2} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(2)*(-2*I*a^3*e^(4*I*f*x + 4*I*e) + 8*I*a^3*e^(2*I*f*x + 2*I*e) + 16*I*a^3)*sqrt(c/(e^(2*I*f*x + 2*I*e
) + 1))/(c^2*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.22, size = 64, normalized size = 0.71 \[ \frac {2 i a^{3} \left (\sqrt {c -i c \tan \left (f x +e \right )}-\frac {4 c^{2}}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {4 c}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f \,c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.65, size = 68, normalized size = 0.76 \[ \frac {2 i \, {\left (\frac {3 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} a^{3}}{c} + \frac {4 \, {\left (3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{3} - a^{3} c\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\right )}}{3 \, c f} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 5.15, size = 98, normalized size = 1.09 \[ \frac {2\,a^3\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (\cos \left (2\,e+2\,f\,x\right )\,4{}\mathrm {i}-\cos \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}-4\,\sin \left (2\,e+2\,f\,x\right )+\sin \left (4\,e+4\,f\,x\right )+8{}\mathrm {i}\right )}{3\,c^2\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - i a^{3} \left (\int \frac {i}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {3 \tan {\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {\tan ^{3}{\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {3 i \tan ^{2}{\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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
))

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