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

Optimal. Leaf size=274 \[ \frac {35 (B+3 i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{256 \sqrt {2} a^3 c^{3/2} f}+\frac {-B+i A}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}-\frac {35 (B+3 i A)}{256 a^3 c f \sqrt {c-i c \tan (e+f x)}}-\frac {35 (B+3 i A)}{384 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac {7 (B+3 i A)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac {B+3 i A}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}} \]

[Out]

35/512*(3*I*A+B)*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)-35/256*(3*I*A+B)/
a^3/c/f/(c-I*c*tan(f*x+e))^(1/2)-35/384*(3*I*A+B)/a^3/f/(c-I*c*tan(f*x+e))^(3/2)+1/6*(I*A-B)/a^3/f/(1+I*tan(f*
x+e))^3/(c-I*c*tan(f*x+e))^(3/2)+1/16*(3*I*A+B)/a^3/f/(1+I*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(3/2)+7/64*(3*I*A+
B)/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 = 0.31, antiderivative size = 274, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ \frac {35 (B+3 i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{256 \sqrt {2} a^3 c^{3/2} f}+\frac {-B+i A}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}-\frac {35 (B+3 i A)}{256 a^3 c f \sqrt {c-i c \tan (e+f x)}}-\frac {35 (B+3 i A)}{384 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac {7 (B+3 i A)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac {B+3 i A}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(35*((3*I)*A + B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(256*Sqrt[2]*a^3*c^(3/2)*f) - (35*((3
*I)*A + B))/(384*a^3*f*(c - I*c*Tan[e + f*x])^(3/2)) + (I*A - B)/(6*a^3*f*(1 + I*Tan[e + f*x])^3*(c - I*c*Tan[
e + f*x])^(3/2)) + ((3*I)*A + B)/(16*a^3*f*(1 + I*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^(3/2)) + (7*((3*I)*A
+ B))/(64*a^3*f*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2)) - (35*((3*I)*A + B))/(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 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)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {A+B x}{(a+i a x)^4 (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {((3 A-i B) c) \operatorname {Subst}\left (\int \frac {1}{(a+i a x)^3 (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=\frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i A+B}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {(7 (3 A-i B) c) \operatorname {Subst}\left (\int \frac {1}{(a+i a x)^2 (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{32 a f}\\ &=\frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i A+B}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {7 (3 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac {(35 (3 A-i B) c) \operatorname {Subst}\left (\int \frac {1}{(a+i a x) (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{128 a^2 f}\\ &=-\frac {35 (3 i A+B)}{384 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i A+B}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {7 (3 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac {(35 (3 A-i B)) \operatorname {Subst}\left (\int \frac {1}{(a+i a x) (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{256 a^2 f}\\ &=-\frac {35 (3 i A+B)}{384 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i A+B}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {7 (3 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {35 (3 i A+B)}{256 a^3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {(35 (3 A-i B)) \operatorname {Subst}\left (\int \frac {1}{(a+i a x) \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{512 a^2 c f}\\ &=-\frac {35 (3 i A+B)}{384 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i A+B}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {7 (3 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {35 (3 i A+B)}{256 a^3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {(35 (3 i A+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 )}{256 a^2 c^2 f}\\ &=\frac {35 (3 i A+B) \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 (3 i A+B)}{384 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i A+B}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {7 (3 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {35 (3 i A+B)}{256 a^3 c f \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 8.32, size = 206, normalized size = 0.75 \[ \frac {\sqrt {c-i c \tan (e+f x)} (\sin (e+f x)+i \cos (e+f x)) \left (105 (3 A-i B) e^{i (e+f x)} \sqrt {1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt {1+e^{2 i (e+f x)}}\right )-2 \cos (e+f x) (2 (79 A-69 i B) \cos (2 (e+f x))+8 (A-3 i B) \cos (4 (e+f x))+258 i A \sin (2 (e+f x))+24 i A \sin (4 (e+f x))-165 A+86 B \sin (2 (e+f x))+8 B \sin (4 (e+f x))-9 i B)\right )}{1536 a^3 c^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I*Cos[e + f*x] + Sin[e + f*x])*(105*(3*A - I*B)*E^(I*(e + f*x))*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcTanh[Sqrt[1
+ E^((2*I)*(e + f*x))]] - 2*Cos[e + f*x]*(-165*A - (9*I)*B + 2*(79*A - (69*I)*B)*Cos[2*(e + f*x)] + 8*(A - (3
*I)*B)*Cos[4*(e + f*x)] + (258*I)*A*Sin[2*(e + f*x)] + 86*B*Sin[2*(e + f*x)] + (24*I)*A*Sin[4*(e + f*x)] + 8*B
*Sin[4*(e + f*x)]))*Sqrt[c - I*c*Tan[e + f*x]])/(1536*a^3*c^2*f)

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fricas [A]  time = 0.73, size = 436, normalized size = 1.59 \[ \frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{3} c^{2} f \sqrt {-\frac {11025 \, A^{2} - 7350 i \, A B - 1225 \, B^{2}}{a^{6} c^{3} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} c f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} c f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {11025 \, A^{2} - 7350 i \, A B - 1225 \, B^{2}}{a^{6} c^{3} f^{2}}} + 105 i \, A + 35 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{128 \, a^{3} c f}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{3} c^{2} f \sqrt {-\frac {11025 \, A^{2} - 7350 i \, A B - 1225 \, B^{2}}{a^{6} c^{3} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} c f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} c f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {11025 \, A^{2} - 7350 i \, A B - 1225 \, B^{2}}{a^{6} c^{3} f^{2}}} - 105 i \, A - 35 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{128 \, a^{3} c f}\right ) + \sqrt {2} {\left ({\left (-16 i \, A - 16 \, B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} + {\left (-224 i \, A - 128 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + {\left (-43 i \, A - 121 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (215 i \, A - 35 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (58 i \, A - 34 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 i \, A - 8 \, B\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{1536 \, a^{3} c^{2} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1536*(3*sqrt(1/2)*a^3*c^2*f*sqrt(-(11025*A^2 - 7350*I*A*B - 1225*B^2)/(a^6*c^3*f^2))*e^(6*I*f*x + 6*I*e)*log
(1/128*(sqrt(2)*sqrt(1/2)*(a^3*c*f*e^(2*I*f*x + 2*I*e) + a^3*c*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(110
25*A^2 - 7350*I*A*B - 1225*B^2)/(a^6*c^3*f^2)) + 105*I*A + 35*B)*e^(-I*f*x - I*e)/(a^3*c*f)) - 3*sqrt(1/2)*a^3
*c^2*f*sqrt(-(11025*A^2 - 7350*I*A*B - 1225*B^2)/(a^6*c^3*f^2))*e^(6*I*f*x + 6*I*e)*log(-1/128*(sqrt(2)*sqrt(1
/2)*(a^3*c*f*e^(2*I*f*x + 2*I*e) + a^3*c*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(11025*A^2 - 7350*I*A*B -
1225*B^2)/(a^6*c^3*f^2)) - 105*I*A - 35*B)*e^(-I*f*x - I*e)/(a^3*c*f)) + sqrt(2)*((-16*I*A - 16*B)*e^(10*I*f*x
+ 10*I*e) + (-224*I*A - 128*B)*e^(8*I*f*x + 8*I*e) + (-43*I*A - 121*B)*e^(6*I*f*x + 6*I*e) + (215*I*A - 35*B)
*e^(4*I*f*x + 4*I*e) + (58*I*A - 34*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*c^2*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*tan(f*x + e) + A)/((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 = 0.58, size = 206, normalized size = 0.75 \[ \frac {2 i c^{3} \left (-\frac {\frac {\left (-\frac {3 i B}{32}+\frac {41 A}{32}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}+\left (\frac {1}{6} i B c -\frac {35}{6} c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+\left (\frac {55}{8} A \,c^{2}+\frac {3}{8} i B \,c^{2}\right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (-c -i c \tan \left (f x +e \right )\right )^{3}}-\frac {35 \left (-i B +3 A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{64 \sqrt {c}}}{16 c^{4}}-\frac {-i B +2 A}{16 c^{4} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {-i B +A}{48 c^{3} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [A]  time = 0.68, size = 263, normalized size = 0.96 \[ -\frac {i \, {\left (\frac {4 \, {\left (105 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{4} {\left (3 \, A - i \, B\right )} - 560 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} {\left (3 \, A - i \, B\right )} c + 924 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} {\left (3 \, A - i \, B\right )} c^{2} - 384 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (3 \, A - i \, B\right )} c^{3} - 256 \, {\left (A - i \, B\right )} c^{4}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}} a^{3} - 6 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{3} c + 12 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{3} c^{2} - 8 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{3} c^{3}} + \frac {105 \, \sqrt {2} {\left (3 \, A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{3} \sqrt {c}}\right )}}{3072 \, c f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))/(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*(105*(-I*c*tan(f*x + e) + c)^4*(3*A - I*B) - 560*(-I*c*tan(f*x + e) + c)^3*(3*A - I*B)*c + 924*(-
I*c*tan(f*x + e) + c)^2*(3*A - I*B)*c^2 - 384*(-I*c*tan(f*x + e) + c)*(3*A - I*B)*c^3 - 256*(A - I*B)*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) + 105*sqrt(2)*(3*A - I*B)*log(-(sqrt(2)*sqrt(c) - sqrt(-I*c*ta
n(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 = 9.98, size = 443, normalized size = 1.62 \[ -\frac {\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,35{}\mathrm {i}}{16\,a^3\,f}+\frac {A\,c^3\,1{}\mathrm {i}}{3\,a^3\,f}-\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4\,105{}\mathrm {i}}{256\,a^3\,c\,f}-\frac {A\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,231{}\mathrm {i}}{64\,a^3\,f}+\frac {A\,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 {\frac {B\,c^3}{3}+\frac {35\,B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3}{48}-\frac {77\,B\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2}{64}+\frac {B\,c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}{2}-\frac {35\,B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4}{256\,c}}{a^3\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{9/2}-6\,a^3\,c\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}-8\,a^3\,c^3\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}+12\,a^3\,c^2\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\sqrt {2}\,A\,\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}+\frac {35\,\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{512\,a^3\,c^{3/2}\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((B*c^3)/3 + (35*B*(c - c*tan(e + f*x)*1i)^3)/48 - (77*B*c*(c - c*tan(e + f*x)*1i)^2)/64 + (B*c^2*(c - c*tan(e
+ f*x)*1i))/2 - (35*B*(c - c*tan(e + f*x)*1i)^4)/(256*c))/(a^3*f*(c - c*tan(e + f*x)*1i)^(9/2) - 6*a^3*c*f*(c
- c*tan(e + f*x)*1i)^(7/2) - 8*a^3*c^3*f*(c - c*tan(e + f*x)*1i)^(3/2) + 12*a^3*c^2*f*(c - c*tan(e + f*x)*1i)
^(5/2)) - ((A*(c - c*tan(e + f*x)*1i)^3*35i)/(16*a^3*f) + (A*c^3*1i)/(3*a^3*f) - (A*(c - c*tan(e + f*x)*1i)^4*
105i)/(256*a^3*c*f) - (A*c*(c - c*tan(e + f*x)*1i)^2*231i)/(64*a^3*f) + (A*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)) + (2^(1/2)*A*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) + (35*2^(1/2)*B*atanh((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*c^(1/2)))
)/(512*a^3*c^(3/2)*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*(Integral(A/(-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) + Integral(B*tan(e +
f*x)/(-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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