3.985 \(\int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=242 \[ \frac {105 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{256 \sqrt {2} a^3 c^{3/2} f}-\frac {105 i}{256 a^3 c f \sqrt {c-i c \tan (e+f x)}}-\frac {35 i}{128 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac {21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac {3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \]

[Out]

105/512*I*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a^3/c^(3/2)/f*2^(1/2)-105/256*I/a^3/c/f/(c-I*c
*tan(f*x+e))^(1/2)-35/128*I/a^3/f/(c-I*c*tan(f*x+e))^(3/2)+1/6*I/a^3/f/(1+I*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(
3/2)+3/16*I/a^3/f/(1+I*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(3/2)+21/64*I/a^3/f/(1+I*tan(f*x+e))/(c-I*c*tan(f*x+e)
)^(3/2)

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Rubi [A]  time = 0.24, antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3522, 3487, 51, 63, 206} \[ \frac {105 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{256 \sqrt {2} a^3 c^{3/2} f}-\frac {105 i}{256 a^3 c f \sqrt {c-i c \tan (e+f x)}}-\frac {35 i}{128 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac {21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac {3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^(3/2)),x]

[Out]

(((105*I)/256)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a^3*c^(3/2)*f) - ((35*I)/128)/(
a^3*f*(c - I*c*Tan[e + f*x])^(3/2)) + (I/6)/(a^3*f*(1 + I*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^(3/2)) + ((3*
I)/16)/(a^3*f*(1 + I*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^(3/2)) + ((21*I)/64)/(a^3*f*(1 + I*Tan[e + f*x])*(
c - I*c*Tan[e + f*x])^(3/2)) - ((105*I)/256)/(a^3*c*f*Sqrt[c - I*c*Tan[e + f*x]])

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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 {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx &=\frac {\int \cos ^6(e+f x) (c-i c \tan (e+f x))^{3/2} \, dx}{a^3 c^3}\\ &=\frac {\left (i c^4\right ) \operatorname {Subst}\left (\int \frac {1}{(c-x)^4 (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{a^3 f}\\ &=\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {\left (3 i c^3\right ) \operatorname {Subst}\left (\int \frac {1}{(c-x)^3 (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{4 a^3 f}\\ &=\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {\left (21 i c^2\right ) \operatorname {Subst}\left (\int \frac {1}{(c-x)^2 (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{32 a^3 f}\\ &=\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac {(105 i c) \operatorname {Subst}\left (\int \frac {1}{(c-x) (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{128 a^3 f}\\ &=-\frac {35 i}{128 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac {(105 i) \operatorname {Subst}\left (\int \frac {1}{(c-x) (c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{256 a^3 f}\\ &=-\frac {35 i}{128 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {105 i}{256 a^3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {(105 i) \operatorname {Subst}\left (\int \frac {1}{(c-x) \sqrt {c+x}} \, dx,x,-i c \tan (e+f x)\right )}{512 a^3 c f}\\ &=-\frac {35 i}{128 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {105 i}{256 a^3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {(105 i) \operatorname {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{256 a^3 c f}\\ &=\frac {105 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{256 \sqrt {2} a^3 c^{3/2} f}-\frac {35 i}{128 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {105 i}{256 a^3 c f \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 4.19, size = 160, normalized size = 0.66 \[ \frac {\sqrt {c-i c \tan (e+f x)} (\cos (e+f x)-i \sin (e+f x)) \left (258 \sin (e+f x)+282 \sin (3 (e+f x))+24 \sin (5 (e+f x))+172 i \cos (e+f x)-166 i \cos (3 (e+f x))-8 i \cos (5 (e+f x))+315 i e^{i (e+f x)} \sqrt {1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt {1+e^{2 i (e+f x)}}\right )\right )}{1536 a^3 c^2 f} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^(3/2)),x]

[Out]

((Cos[e + f*x] - I*Sin[e + f*x])*((315*I)*E^(I*(e + f*x))*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcTanh[Sqrt[1 + E^((2
*I)*(e + f*x))]] + (172*I)*Cos[e + f*x] - (166*I)*Cos[3*(e + f*x)] - (8*I)*Cos[5*(e + f*x)] + 258*Sin[e + f*x]
+ 282*Sin[3*(e + f*x)] + 24*Sin[5*(e + f*x)])*Sqrt[c - I*c*Tan[e + f*x]])/(1536*a^3*c^2*f)

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fricas [A]  time = 0.48, size = 331, normalized size = 1.37 \[ \frac {{\left (315 i \, \sqrt {\frac {1}{2}} a^{3} c^{2} f \sqrt {\frac {1}{a^{6} c^{3} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (13440 i \, a^{3} c f e^{\left (2 i \, f x + 2 i \, e\right )} + 13440 i \, a^{3} c f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{6} c^{3} f^{2}}} + 13440 i\right )} e^{\left (-i \, f x - i \, e\right )}}{16384 \, a^{3} c f}\right ) - 315 i \, \sqrt {\frac {1}{2}} a^{3} c^{2} f \sqrt {\frac {1}{a^{6} c^{3} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-13440 i \, a^{3} c f e^{\left (2 i \, f x + 2 i \, e\right )} - 13440 i \, a^{3} c f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{6} c^{3} f^{2}}} + 13440 i\right )} e^{\left (-i \, f x - i \, e\right )}}{16384 \, a^{3} c f}\right ) + \sqrt {2} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-16 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 224 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 43 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 215 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 58 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 8 i\right )}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{1536 \, a^{3} c^{2} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1536*(315*I*sqrt(1/2)*a^3*c^2*f*sqrt(1/(a^6*c^3*f^2))*e^(6*I*f*x + 6*I*e)*log(1/16384*(sqrt(2)*sqrt(1/2)*(13
440*I*a^3*c*f*e^(2*I*f*x + 2*I*e) + 13440*I*a^3*c*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^6*c^3*f^2)) +
13440*I)*e^(-I*f*x - I*e)/(a^3*c*f)) - 315*I*sqrt(1/2)*a^3*c^2*f*sqrt(1/(a^6*c^3*f^2))*e^(6*I*f*x + 6*I*e)*lo
g(1/16384*(sqrt(2)*sqrt(1/2)*(-13440*I*a^3*c*f*e^(2*I*f*x + 2*I*e) - 13440*I*a^3*c*f)*sqrt(c/(e^(2*I*f*x + 2*I
*e) + 1))*sqrt(1/(a^6*c^3*f^2)) + 13440*I)*e^(-I*f*x - I*e)/(a^3*c*f)) + sqrt(2)*sqrt(c/(e^(2*I*f*x + 2*I*e) +
1))*(-16*I*e^(10*I*f*x + 10*I*e) - 224*I*e^(8*I*f*x + 8*I*e) - 43*I*e^(6*I*f*x + 6*I*e) + 215*I*e^(4*I*f*x +
4*I*e) + 58*I*e^(2*I*f*x + 2*I*e) + 8*I))*e^(-6*I*f*x - 6*I*e)/(a^3*c^2*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((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 = 0.40, size = 159, normalized size = 0.66 \[ \frac {2 i c^{4} \left (-\frac {\frac {\frac {41 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{32}-\frac {35 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}} c}{6}+\frac {55 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}{8}}{\left (-c -i c \tan \left (f x +e \right )\right )^{3}}-\frac {105 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{64 \sqrt {c}}}{16 c^{5}}-\frac {1}{8 c^{5} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{48 c^{4} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(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^4*(-1/16/c^5*((41/32*(c-I*c*tan(f*x+e))^(5/2)-35/6*(c-I*c*tan(f*x+e))^(3/2)*c+55/8*c^2*(c-I*c*tan(
f*x+e))^(1/2))/(-c-I*c*tan(f*x+e))^3-105/64*2^(1/2)/c^(1/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/
2)))-1/8/c^5/(c-I*c*tan(f*x+e))^(1/2)-1/48/c^4/(c-I*c*tan(f*x+e))^(3/2))

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maxima [A]  time = 0.69, size = 223, normalized size = 0.92 \[ -\frac {i \, {\left (\frac {4 \, {\left (315 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{4} - 1680 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} c + 2772 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} c^{2} - 1152 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} c^{3} - 256 \, c^{4}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}} a^{3} - 6 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{3} c + 12 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{3} c^{2} - 8 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{3} c^{3}} + \frac {315 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{3} \sqrt {c}}\right )}}{3072 \, c f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3072*I*(4*(315*(-I*c*tan(f*x + e) + c)^4 - 1680*(-I*c*tan(f*x + e) + c)^3*c + 2772*(-I*c*tan(f*x + e) + c)^
2*c^2 - 1152*(-I*c*tan(f*x + e) + c)*c^3 - 256*c^4)/((-I*c*tan(f*x + e) + c)^(9/2)*a^3 - 6*(-I*c*tan(f*x + e)
+ c)^(7/2)*a^3*c + 12*(-I*c*tan(f*x + e) + c)^(5/2)*a^3*c^2 - 8*(-I*c*tan(f*x + e) + c)^(3/2)*a^3*c^3) + 315*s
qrt(2)*log(-(sqrt(2)*sqrt(c) - sqrt(-I*c*tan(f*x + e) + c))/(sqrt(2)*sqrt(c) + sqrt(-I*c*tan(f*x + e) + c)))/(
a^3*sqrt(c)))/(c*f)

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mupad [B]  time = 5.07, size = 229, normalized size = 0.95 \[ -\frac {\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,35{}\mathrm {i}}{16\,a^3\,f}+\frac {c^3\,1{}\mathrm {i}}{3\,a^3\,f}-\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4\,105{}\mathrm {i}}{256\,a^3\,c\,f}-\frac {c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,231{}\mathrm {i}}{64\,a^3\,f}+\frac {c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{2\,a^3\,f}}{6\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{9/2}+8\,c^3\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}-12\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,105{}\mathrm {i}}{512\,a^3\,{\left (-c\right )}^{3/2}\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2^(1/2)*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*105i)/(512*a^3*(-c)^(3/2)*f) - (((c - c*
tan(e + f*x)*1i)^3*35i)/(16*a^3*f) + (c^3*1i)/(3*a^3*f) - ((c - c*tan(e + f*x)*1i)^4*105i)/(256*a^3*c*f) - (c*
(c - c*tan(e + f*x)*1i)^2*231i)/(64*a^3*f) + (c^2*(c - c*tan(e + f*x)*1i)*3i)/(2*a^3*f))/(6*c*(c - c*tan(e + f
*x)*1i)^(7/2) - (c - c*tan(e + f*x)*1i)^(9/2) + 8*c^3*(c - c*tan(e + f*x)*1i)^(3/2) - 12*c^2*(c - c*tan(e + f*
x)*1i)^(5/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {i \int \frac {1}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )} - 2 c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 2 c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + i c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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