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

Optimal. Leaf size=211 \[ -\frac {c^{3/2} (3 B+i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{32 \sqrt {2} a^3 f}-\frac {c (3 B+i A) \sqrt {c-i c \tan (e+f x)}}{32 a^3 f (1+i \tan (e+f x))}+\frac {c (3 B+i A) \sqrt {c-i c \tan (e+f x)}}{8 a^3 f (1+i \tan (e+f x))^2}+\frac {(-B+i A) (c-i c \tan (e+f x))^{3/2}}{6 a^3 f (1+i \tan (e+f x))^3} \]

[Out]

-1/64*(I*A+3*B)*c^(3/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a^3/f*2^(1/2)+1/8*(I*A+3*B)*c*(c
-I*c*tan(f*x+e))^(1/2)/a^3/f/(1+I*tan(f*x+e))^2-1/32*(I*A+3*B)*c*(c-I*c*tan(f*x+e))^(1/2)/a^3/f/(1+I*tan(f*x+e
))+1/6*(I*A-B)*(c-I*c*tan(f*x+e))^(3/2)/a^3/f/(1+I*tan(f*x+e))^3

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Rubi [A]  time = 0.25, antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {3588, 78, 47, 51, 63, 208} \[ -\frac {c^{3/2} (3 B+i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{32 \sqrt {2} a^3 f}-\frac {c (3 B+i A) \sqrt {c-i c \tan (e+f x)}}{32 a^3 f (1+i \tan (e+f x))}+\frac {c (3 B+i A) \sqrt {c-i c \tan (e+f x)}}{8 a^3 f (1+i \tan (e+f x))^2}+\frac {(-B+i A) (c-i c \tan (e+f x))^{3/2}}{6 a^3 f (1+i \tan (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((I*A + 3*B)*c^(3/2)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(32*Sqrt[2]*a^3*f) + ((I*A + 3*B)
*c*Sqrt[c - I*c*Tan[e + f*x]])/(8*a^3*f*(1 + I*Tan[e + f*x])^2) - ((I*A + 3*B)*c*Sqrt[c - I*c*Tan[e + f*x]])/(
32*a^3*f*(1 + I*Tan[e + f*x])) + ((I*A - B)*(c - I*c*Tan[e + f*x])^(3/2))/(6*a^3*f*(1 + I*Tan[e + f*x])^3)

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, 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)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^3} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(A+B x) \sqrt {c-i c x}}{(a+i a x)^4} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(i A-B) (c-i c \tan (e+f x))^{3/2}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac {((A-3 i B) c) \operatorname {Subst}\left (\int \frac {\sqrt {c-i c x}}{(a+i a x)^3} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=\frac {(i A+3 B) c \sqrt {c-i c \tan (e+f x)}}{8 a^3 f (1+i \tan (e+f x))^2}+\frac {(i A-B) (c-i c \tan (e+f x))^{3/2}}{6 a^3 f (1+i \tan (e+f x))^3}-\frac {\left ((A-3 i B) c^2\right ) \operatorname {Subst}\left (\int \frac {1}{(a+i a x)^2 \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{16 a f}\\ &=\frac {(i A+3 B) c \sqrt {c-i c \tan (e+f x)}}{8 a^3 f (1+i \tan (e+f x))^2}-\frac {(i A+3 B) c \sqrt {c-i c \tan (e+f x)}}{32 a^3 f (1+i \tan (e+f x))}+\frac {(i A-B) (c-i c \tan (e+f x))^{3/2}}{6 a^3 f (1+i \tan (e+f x))^3}-\frac {\left ((A-3 i B) c^2\right ) \operatorname {Subst}\left (\int \frac {1}{(a+i a x) \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{64 a^2 f}\\ &=\frac {(i A+3 B) c \sqrt {c-i c \tan (e+f x)}}{8 a^3 f (1+i \tan (e+f x))^2}-\frac {(i A+3 B) c \sqrt {c-i c \tan (e+f x)}}{32 a^3 f (1+i \tan (e+f x))}+\frac {(i A-B) (c-i c \tan (e+f x))^{3/2}}{6 a^3 f (1+i \tan (e+f x))^3}-\frac {((i A+3 B) c) \operatorname {Subst}\left (\int \frac {1}{2 a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{32 a^2 f}\\ &=-\frac {(i A+3 B) c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{32 \sqrt {2} a^3 f}+\frac {(i A+3 B) c \sqrt {c-i c \tan (e+f x)}}{8 a^3 f (1+i \tan (e+f x))^2}-\frac {(i A+3 B) c \sqrt {c-i c \tan (e+f x)}}{32 a^3 f (1+i \tan (e+f x))}+\frac {(i A-B) (c-i c \tan (e+f x))^{3/2}}{6 a^3 f (1+i \tan (e+f x))^3}\\ \end {align*}

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Mathematica [A]  time = 6.07, size = 224, normalized size = 1.06 \[ \frac {\sec ^2(e+f x) (\cos (f x)+i \sin (f x))^3 (A+B \tan (e+f x)) \left (\sqrt {2} c^{3/2} (A-3 i B) (\sin (3 e)-i \cos (3 e)) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )+\frac {2}{3} c \cos (e+f x) (\cos (3 f x)-i \sin (3 f x)) \sqrt {c-i c \tan (e+f x)} ((5 A+17 i B) \sin (2 (e+f x))+(B+11 i A) \cos (2 (e+f x))+2 (5 B+7 i A))\right )}{64 f (a+i a \tan (e+f x))^3 (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2))/(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]*(A - (3*I)*B)*c^(3/2)*ArcTanh[Sqrt[c -
I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])]*((-I)*Cos[3*e] + Sin[3*e]) + (2*c*Cos[e + f*x]*(Cos[3*f*x] - I*Sin[3*f*x
])*(2*((7*I)*A + 5*B) + ((11*I)*A + B)*Cos[2*(e + f*x)] + (5*A + (17*I)*B)*Sin[2*(e + f*x)])*Sqrt[c - I*c*Tan[
e + f*x]])/3))/(64*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 = 0.93, size = 392, normalized size = 1.86 \[ \frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {{\left (A^{2} - 6 i \, A B - 9 \, B^{2}\right )} c^{3}}{a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left ({\left (-i \, A - 3 \, B\right )} c^{2} + \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {{\left (A^{2} - 6 i \, A B - 9 \, B^{2}\right )} c^{3}}{a^{6} f^{2}}}\right )} e^{\left (-i \, f x - i \, e\right )}}{16 \, a^{3} f}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {{\left (A^{2} - 6 i \, A B - 9 \, B^{2}\right )} c^{3}}{a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left ({\left (-i \, A - 3 \, B\right )} c^{2} - \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {{\left (A^{2} - 6 i \, A B - 9 \, B^{2}\right )} c^{3}}{a^{6} f^{2}}}\right )} e^{\left (-i \, f x - i \, e\right )}}{16 \, a^{3} f}\right ) + \sqrt {2} {\left ({\left (3 i \, A + 9 \, B\right )} c e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (17 i \, A + 19 \, B\right )} c e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (22 i \, A + 2 \, B\right )} c e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (8 i \, A - 8 \, B\right )} c\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{192 \, a^{3} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(3*sqrt(1/2)*a^3*f*sqrt(-(A^2 - 6*I*A*B - 9*B^2)*c^3/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/16*((-I*A - 3*
B)*c^2 + 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(-(A^2 -
6*I*A*B - 9*B^2)*c^3/(a^6*f^2)))*e^(-I*f*x - I*e)/(a^3*f)) - 3*sqrt(1/2)*a^3*f*sqrt(-(A^2 - 6*I*A*B - 9*B^2)*c
^3/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/16*((-I*A - 3*B)*c^2 - 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(-(A^2 - 6*I*A*B - 9*B^2)*c^3/(a^6*f^2)))*e^(-I*f*x - I*e)/(a^3*f
)) + sqrt(2)*((3*I*A + 9*B)*c*e^(6*I*f*x + 6*I*e) + (17*I*A + 19*B)*c*e^(4*I*f*x + 4*I*e) + (22*I*A + 2*B)*c*e
^(2*I*f*x + 2*I*e) + (8*I*A - 8*B)*c)*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 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.66, size = 140, normalized size = 0.66 \[ \frac {2 i c^{3} \left (\frac {\frac {\left (-3 i B +A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{64 c}+\left (-\frac {A}{12}-\frac {i B}{12}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}-\frac {c \left (-3 i B +A \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{16}}{\left (-c -i c \tan \left (f x +e \right )\right )^{3}}-\frac {\left (-3 i B +A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{128 c^{\frac {3}{2}}}\right )}{f \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I/f/a^3*c^3*((1/64/c*(A-3*I*B)*(c-I*c*tan(f*x+e))^(5/2)+(-1/12*A-1/12*I*B)*(c-I*c*tan(f*x+e))^(3/2)-1/16*c*(
A-3*I*B)*(c-I*c*tan(f*x+e))^(1/2))/(-c-I*c*tan(f*x+e))^3-1/128/c^(3/2)*(A-3*I*B)*2^(1/2)*arctanh(1/2*(c-I*c*ta
n(f*x+e))^(1/2)*2^(1/2)/c^(1/2)))

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maxima [A]  time = 1.12, size = 212, normalized size = 1.00 \[ \frac {i \, {\left (\frac {3 \, \sqrt {2} {\left (A - 3 i \, B\right )} c^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{3}} + \frac {4 \, {\left (3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (A - 3 i \, B\right )} c^{3} - 16 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (A + i \, B\right )} c^{4} - 12 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (A - 3 i \, B\right )} c^{5}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} a^{3} - 6 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} a^{3} c + 12 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{3} c^{2} - 8 \, a^{3} c^{3}}\right )}}{384 \, c f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*I*(3*sqrt(2)*(A - 3*I*B)*c^(5/2)*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)*(A - 3*I*B)*c^3 - 16*(-I*c*tan(f*x + e
) + c)^(3/2)*(A + I*B)*c^4 - 12*sqrt(-I*c*tan(f*x + e) + c)*(A - 3*I*B)*c^5)/((-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 = 9.41, size = 360, normalized size = 1.71 \[ \frac {\frac {A\,c^4\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{8\,a^3\,f}+\frac {A\,c^3\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,1{}\mathrm {i}}{6\,a^3\,f}-\frac {A\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,1{}\mathrm {i}}{32\,a^3\,f}}{6\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-12\,c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3+8\,c^3}-\frac {-\frac {3\,B\,c^4\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{8}+\frac {B\,c^3\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{6}+\frac {3\,B\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{32}}{8\,a^3\,c^3\,f-a^3\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3+6\,a^3\,c\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-12\,a^3\,c^2\,f\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}+\frac {\sqrt {2}\,A\,{\left (-c\right )}^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,1{}\mathrm {i}}{64\,a^3\,f}-\frac {3\,\sqrt {2}\,B\,c^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{64\,a^3\,f} \]

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)^(3/2))/(a + a*tan(e + f*x)*1i)^3,x)

[Out]

((A*c^4*(c - c*tan(e + f*x)*1i)^(1/2)*1i)/(8*a^3*f) + (A*c^3*(c - c*tan(e + f*x)*1i)^(3/2)*1i)/(6*a^3*f) - (A*
c^2*(c - c*tan(e + f*x)*1i)^(5/2)*1i)/(32*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) - ((B*c^3*(c - c*tan(e + f*x)*1i)^(3/2))/6 - (3*B*c^4*(c - c*tan(e +
f*x)*1i)^(1/2))/8 + (3*B*c^2*(c - c*tan(e + f*x)*1i)^(5/2))/32)/(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)) + (2^(1/2)*A*(-c)^(3/2)*atan((2
^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*1i)/(64*a^3*f) - (3*2^(1/2)*B*c^(3/2)*atanh((2^(1/2)*(c
- c*tan(e + f*x)*1i)^(1/2))/(2*c^(1/2))))/(64*a^3*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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