3.783 \(\int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=209 \[ \frac {(-B+i A) \sqrt {c-i c \tan (e+f x)}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac {(7 B+5 i A) \sqrt {c-i c \tan (e+f x)}}{64 a^3 f (1+i \tan (e+f x))}+\frac {(7 B+5 i A) \sqrt {c-i c \tan (e+f x)}}{48 a^3 f (1+i \tan (e+f x))^2}+\frac {\sqrt {c} (7 B+5 i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{64 \sqrt {2} a^3 f} \]

[Out]

1/128*(5*I*A+7*B)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^(1/2)/a^3/f*2^(1/2)+1/6*(I*A-B)*(c-I
*c*tan(f*x+e))^(1/2)/a^3/f/(1+I*tan(f*x+e))^3+1/48*(5*I*A+7*B)*(c-I*c*tan(f*x+e))^(1/2)/a^3/f/(1+I*tan(f*x+e))
^2+1/64*(5*I*A+7*B)*(c-I*c*tan(f*x+e))^(1/2)/a^3/f/(1+I*tan(f*x+e))

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Rubi [A]  time = 0.24, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {3588, 78, 51, 63, 208} \[ \frac {(-B+i A) \sqrt {c-i c \tan (e+f x)}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac {(7 B+5 i A) \sqrt {c-i c \tan (e+f x)}}{64 a^3 f (1+i \tan (e+f x))}+\frac {(7 B+5 i A) \sqrt {c-i c \tan (e+f x)}}{48 a^3 f (1+i \tan (e+f x))^2}+\frac {\sqrt {c} (7 B+5 i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{64 \sqrt {2} a^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((5*I)*A + 7*B)*Sqrt[c]*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(64*Sqrt[2]*a^3*f) + ((I*A - B
)*Sqrt[c - I*c*Tan[e + f*x]])/(6*a^3*f*(1 + I*Tan[e + f*x])^3) + (((5*I)*A + 7*B)*Sqrt[c - I*c*Tan[e + f*x]])/
(48*a^3*f*(1 + I*Tan[e + f*x])^2) + (((5*I)*A + 7*B)*Sqrt[c - I*c*Tan[e + f*x]])/(64*a^3*f*(1 + I*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 78

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

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 3588

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(a*c)/f, Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1)*(A + B*x), x
], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rubi steps

\begin {align*} \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^3} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {A+B x}{(a+i a x)^4 \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(i A-B) \sqrt {c-i c \tan (e+f x)}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac {((5 A-7 i B) c) \operatorname {Subst}\left (\int \frac {1}{(a+i a x)^3 \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{12 f}\\ &=\frac {(i A-B) \sqrt {c-i c \tan (e+f x)}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac {(5 i A+7 B) \sqrt {c-i c \tan (e+f x)}}{48 a^3 f (1+i \tan (e+f x))^2}+\frac {((5 A-7 i B) c) \operatorname {Subst}\left (\int \frac {1}{(a+i a x)^2 \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{32 a f}\\ &=\frac {(i A-B) \sqrt {c-i c \tan (e+f x)}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac {(5 i A+7 B) \sqrt {c-i c \tan (e+f x)}}{48 a^3 f (1+i \tan (e+f x))^2}+\frac {(5 i A+7 B) \sqrt {c-i c \tan (e+f x)}}{64 a^3 f (1+i \tan (e+f x))}+\frac {((5 A-7 i B) c) \operatorname {Subst}\left (\int \frac {1}{(a+i a x) \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{128 a^2 f}\\ &=\frac {(i A-B) \sqrt {c-i c \tan (e+f x)}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac {(5 i A+7 B) \sqrt {c-i c \tan (e+f x)}}{48 a^3 f (1+i \tan (e+f x))^2}+\frac {(5 i A+7 B) \sqrt {c-i c \tan (e+f x)}}{64 a^3 f (1+i \tan (e+f x))}+\frac {(5 i A+7 B) \operatorname {Subst}\left (\int \frac {1}{2 a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{64 a^2 f}\\ &=\frac {(5 i A+7 B) \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{64 \sqrt {2} a^3 f}+\frac {(i A-B) \sqrt {c-i c \tan (e+f x)}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac {(5 i A+7 B) \sqrt {c-i c \tan (e+f x)}}{48 a^3 f (1+i \tan (e+f x))^2}+\frac {(5 i A+7 B) \sqrt {c-i c \tan (e+f x)}}{64 a^3 f (1+i \tan (e+f x))}\\ \end {align*}

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Mathematica [A]  time = 4.03, size = 225, normalized size = 1.08 \[ \frac {\sec ^2(e+f x) (\cos (f x)+i \sin (f x))^3 (A+B \tan (e+f x)) \left (\frac {2}{3} \cos (e+f x) (\sin (3 f x)+i \cos (3 f x)) \sqrt {c-i c \tan (e+f x)} (5 (7 B+5 i A) \sin (2 (e+f x))+(41 A-19 i B) \cos (2 (e+f x))+26 A+2 i B)+\sqrt {2} \sqrt {c} (7 B+5 i A) (\cos (3 e)+i \sin (3 e)) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )\right )}{128 f (a+i a \tan (e+f x))^3 (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]^2*(Cos[f*x] + I*Sin[f*x])^3*(A + B*Tan[e + f*x])*(Sqrt[2]*((5*I)*A + 7*B)*Sqrt[c]*ArcTanh[Sqrt[c
- I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])]*(Cos[3*e] + I*Sin[3*e]) + (2*Cos[e + f*x]*(I*Cos[3*f*x] + Sin[3*f*x])*
(26*A + (2*I)*B + (41*A - (19*I)*B)*Cos[2*(e + f*x)] + 5*((5*I)*A + 7*B)*Sin[2*(e + f*x)])*Sqrt[c - I*c*Tan[e
+ f*x]])/3))/(128*f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^3)

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fricas [B]  time = 0.60, size = 382, normalized size = 1.83 \[ \frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {{\left (25 \, A^{2} - 70 i \, A B - 49 \, B^{2}\right )} c}{a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {{\left (25 \, A^{2} - 70 i \, A B - 49 \, B^{2}\right )} c}{a^{6} f^{2}}} + {\left (5 i \, A + 7 \, B\right )} c\right )} e^{\left (-i \, f x - i \, e\right )}}{32 \, a^{3} f}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {{\left (25 \, A^{2} - 70 i \, A B - 49 \, B^{2}\right )} c}{a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {{\left (25 \, A^{2} - 70 i \, A B - 49 \, B^{2}\right )} c}{a^{6} f^{2}}} - {\left (5 i \, A + 7 \, B\right )} c\right )} e^{\left (-i \, f x - i \, e\right )}}{32 \, a^{3} f}\right ) + \sqrt {2} {\left ({\left (33 i \, A + 27 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (59 i \, A + 25 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (34 i \, A - 10 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 i \, A - 8 \, B\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{384 \, a^{3} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(3*sqrt(1/2)*a^3*f*sqrt(-(25*A^2 - 70*I*A*B - 49*B^2)*c/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/32*(sqrt(2)
*sqrt(1/2)*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(25*A^2 - 70*I*A*B - 49
*B^2)*c/(a^6*f^2)) + (5*I*A + 7*B)*c)*e^(-I*f*x - I*e)/(a^3*f)) - 3*sqrt(1/2)*a^3*f*sqrt(-(25*A^2 - 70*I*A*B -
49*B^2)*c/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(-1/32*(sqrt(2)*sqrt(1/2)*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqr
t(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(25*A^2 - 70*I*A*B - 49*B^2)*c/(a^6*f^2)) - (5*I*A + 7*B)*c)*e^(-I*f*x -
I*e)/(a^3*f)) + sqrt(2)*((33*I*A + 27*B)*e^(6*I*f*x + 6*I*e) + (59*I*A + 25*B)*e^(4*I*f*x + 4*I*e) + (34*I*A -
10*B)*e^(2*I*f*x + 2*I*e) + 8*I*A - 8*B)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-6*I*f*x - 6*I*e)/(a^3*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.67, size = 148, normalized size = 0.71 \[ \frac {2 i c^{3} \left (\frac {-\frac {\left (-7 i B +5 A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{128 c^{2}}+\frac {\left (-7 i B +5 A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{24 c}+\left (-\frac {11 A}{32}+\frac {9 i B}{32}\right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (-c -i c \tan \left (f x +e \right )\right )^{3}}+\frac {\left (-7 i B +5 A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{256 c^{\frac {5}{2}}}\right )}{f \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I/f/a^3*c^3*((-1/128/c^2*(5*A-7*I*B)*(c-I*c*tan(f*x+e))^(5/2)+1/24/c*(5*A-7*I*B)*(c-I*c*tan(f*x+e))^(3/2)+(-
11/32*A+9/32*I*B)*(c-I*c*tan(f*x+e))^(1/2))/(-c-I*c*tan(f*x+e))^3+1/256/c^(5/2)*(5*A-7*I*B)*2^(1/2)*arctanh(1/
2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2)))

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maxima [A]  time = 0.62, size = 220, normalized size = 1.05 \[ -\frac {i \, {\left (\frac {3 \, \sqrt {2} {\left (5 \, A - 7 i \, B\right )} c^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{3}} + \frac {4 \, {\left (3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (5 \, A - 7 i \, B\right )} c^{2} - 16 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (5 \, A - 7 i \, B\right )} c^{3} + 12 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (11 \, A - 9 i \, B\right )} c^{4}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} a^{3} - 6 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} a^{3} c + 12 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{3} c^{2} - 8 \, a^{3} c^{3}}\right )}}{768 \, c f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/768*I*(3*sqrt(2)*(5*A - 7*I*B)*c^(3/2)*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 + 4*(3*(-I*c*tan(f*x + e) + c)^(5/2)*(5*A - 7*I*B)*c^2 - 16*(-I*c*tan(f*
x + e) + c)^(3/2)*(5*A - 7*I*B)*c^3 + 12*sqrt(-I*c*tan(f*x + e) + c)*(11*A - 9*I*B)*c^4)/((-I*c*tan(f*x + e) +
c)^3*a^3 - 6*(-I*c*tan(f*x + e) + c)^2*a^3*c + 12*(-I*c*tan(f*x + e) + c)*a^3*c^2 - 8*a^3*c^3))/(c*f)

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mupad [B]  time = 9.36, size = 355, normalized size = 1.70 \[ \frac {\frac {7\,B\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{64}+\frac {9\,B\,c^3\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{16}-\frac {7\,B\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{12}}{8\,a^3\,c^3\,f-a^3\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3+6\,a^3\,c\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-12\,a^3\,c^2\,f\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}+\frac {\frac {A\,c^3\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,11{}\mathrm {i}}{16\,a^3\,f}-\frac {A\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,5{}\mathrm {i}}{12\,a^3\,f}+\frac {A\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,5{}\mathrm {i}}{64\,a^3\,f}}{6\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-12\,c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3+8\,c^3}+\frac {\sqrt {2}\,A\,\sqrt {-c}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,5{}\mathrm {i}}{128\,a^3\,f}+\frac {7\,\sqrt {2}\,B\,\sqrt {c}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{128\,a^3\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((7*B*c*(c - c*tan(e + f*x)*1i)^(5/2))/64 + (9*B*c^3*(c - c*tan(e + f*x)*1i)^(1/2))/16 - (7*B*c^2*(c - c*tan(e
+ f*x)*1i)^(3/2))/12)/(8*a^3*c^3*f - a^3*f*(c - c*tan(e + f*x)*1i)^3 + 6*a^3*c*f*(c - c*tan(e + f*x)*1i)^2 -
12*a^3*c^2*f*(c - c*tan(e + f*x)*1i)) + ((A*c^3*(c - c*tan(e + f*x)*1i)^(1/2)*11i)/(16*a^3*f) - (A*c^2*(c - c*
tan(e + f*x)*1i)^(3/2)*5i)/(12*a^3*f) + (A*c*(c - c*tan(e + f*x)*1i)^(5/2)*5i)/(64*a^3*f))/(6*c*(c - c*tan(e +
f*x)*1i)^2 - 12*c^2*(c - c*tan(e + f*x)*1i) - (c - c*tan(e + f*x)*1i)^3 + 8*c^3) + (2^(1/2)*A*(-c)^(1/2)*atan
((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*5i)/(128*a^3*f) + (7*2^(1/2)*B*c^(1/2)*atanh((2^(1/2)
*(c - c*tan(e + f*x)*1i)^(1/2))/(2*c^(1/2))))/(128*a^3*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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