3.203 \(\int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^{2/3}} \, dx\)

Optimal. Leaf size=213 \[ -\frac {\sqrt {3} (B+i A) \tan ^{-1}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2\ 2^{2/3} a^{2/3} d}+\frac {3 (B+i A) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4\ 2^{2/3} a^{2/3} d}+\frac {(B+i A) \log (\cos (c+d x))}{4\ 2^{2/3} a^{2/3} d}-\frac {x (A-i B)}{4\ 2^{2/3} a^{2/3}}+\frac {3 (-B+i A)}{4 d (a+i a \tan (c+d x))^{2/3}} \]

[Out]

-1/8*(A-I*B)*x*2^(1/3)/a^(2/3)+1/8*(I*A+B)*ln(cos(d*x+c))*2^(1/3)/a^(2/3)/d+3/8*(I*A+B)*ln(2^(1/3)*a^(1/3)-(a+
I*a*tan(d*x+c))^(1/3))*2^(1/3)/a^(2/3)/d-1/4*(I*A+B)*arctan(1/3*(a^(1/3)+2^(2/3)*(a+I*a*tan(d*x+c))^(1/3))/a^(
1/3)*3^(1/2))*3^(1/2)*2^(1/3)/a^(2/3)/d+3/4*(I*A-B)/d/(a+I*a*tan(d*x+c))^(2/3)

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Rubi [A]  time = 0.16, antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3526, 3481, 57, 617, 204, 31} \[ -\frac {\sqrt {3} (B+i A) \tan ^{-1}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2\ 2^{2/3} a^{2/3} d}+\frac {3 (B+i A) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4\ 2^{2/3} a^{2/3} d}+\frac {(B+i A) \log (\cos (c+d x))}{4\ 2^{2/3} a^{2/3} d}-\frac {x (A-i B)}{4\ 2^{2/3} a^{2/3}}+\frac {3 (-B+i A)}{4 d (a+i a \tan (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((A - I*B)*x)/(4*2^(2/3)*a^(2/3)) - (Sqrt[3]*(I*A + B)*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3)
)/(Sqrt[3]*a^(1/3))])/(2*2^(2/3)*a^(2/3)*d) + ((I*A + B)*Log[Cos[c + d*x]])/(4*2^(2/3)*a^(2/3)*d) + (3*(I*A +
B)*Log[2^(1/3)*a^(1/3) - (a + I*a*Tan[c + d*x])^(1/3)])/(4*2^(2/3)*a^(2/3)*d) + (3*(I*A - B))/(4*d*(a + I*a*Ta
n[c + d*x])^(2/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 57

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Dist[3/(2*b*q), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x
)^(1/3)], x] - Dist[3/(2*b*q^2), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x]
&& PosQ[(b*c - a*d)/b]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 3481

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

Rule 3526

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

Rubi steps

\begin {align*} \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^{2/3}} \, dx &=\frac {3 (i A-B)}{4 d (a+i a \tan (c+d x))^{2/3}}+\frac {(A-i B) \int \sqrt [3]{a+i a \tan (c+d x)} \, dx}{2 a}\\ &=\frac {3 (i A-B)}{4 d (a+i a \tan (c+d x))^{2/3}}-\frac {(i A+B) \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{2/3}} \, dx,x,i a \tan (c+d x)\right )}{2 d}\\ &=-\frac {(A-i B) x}{4\ 2^{2/3} a^{2/3}}+\frac {(i A+B) \log (\cos (c+d x))}{4\ 2^{2/3} a^{2/3} d}+\frac {3 (i A-B)}{4 d (a+i a \tan (c+d x))^{2/3}}-\frac {(3 (i A+B)) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{2} \sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+i a \tan (c+d x)}\right )}{4\ 2^{2/3} a^{2/3} d}-\frac {(3 (i A+B)) \operatorname {Subst}\left (\int \frac {1}{2^{2/3} a^{2/3}+\sqrt [3]{2} \sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}\\ &=-\frac {(A-i B) x}{4\ 2^{2/3} a^{2/3}}+\frac {(i A+B) \log (\cos (c+d x))}{4\ 2^{2/3} a^{2/3} d}+\frac {3 (i A+B) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4\ 2^{2/3} a^{2/3} d}+\frac {3 (i A-B)}{4 d (a+i a \tan (c+d x))^{2/3}}+\frac {(3 (i A+B)) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}\right )}{2\ 2^{2/3} a^{2/3} d}\\ &=-\frac {(A-i B) x}{4\ 2^{2/3} a^{2/3}}-\frac {\sqrt {3} (i A+B) \tan ^{-1}\left (\frac {1+\frac {2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{2\ 2^{2/3} a^{2/3} d}+\frac {(i A+B) \log (\cos (c+d x))}{4\ 2^{2/3} a^{2/3} d}+\frac {3 (i A+B) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4\ 2^{2/3} a^{2/3} d}+\frac {3 (i A-B)}{4 d (a+i a \tan (c+d x))^{2/3}}\\ \end {align*}

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Mathematica [F]  time = 0.79, size = 0, normalized size = 0.00 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^{2/3}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.70, size = 496, normalized size = 2.33 \[ \frac {{\left (4 \, \left (\frac {1}{4}\right )^{\frac {1}{3}} a d \left (\frac {-i \, A^{3} - 3 \, A^{2} B + 3 i \, A B^{2} + B^{3}}{a^{2} d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\frac {2 \, \left (\frac {1}{4}\right )^{\frac {1}{3}} a d \left (\frac {-i \, A^{3} - 3 \, A^{2} B + 3 i \, A B^{2} + B^{3}}{a^{2} d^{3}}\right )^{\frac {1}{3}} - 2^{\frac {1}{3}} {\left (i \, A + B\right )} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}}{i \, A + B}\right ) - 2 \, \left (\frac {1}{4}\right )^{\frac {1}{3}} {\left (-i \, \sqrt {3} a d + a d\right )} \left (\frac {-i \, A^{3} - 3 \, A^{2} B + 3 i \, A B^{2} + B^{3}}{a^{2} d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac {2^{\frac {1}{3}} {\left (i \, A + B\right )} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )} - \left (\frac {1}{4}\right )^{\frac {1}{3}} {\left (i \, \sqrt {3} a d - a d\right )} \left (\frac {-i \, A^{3} - 3 \, A^{2} B + 3 i \, A B^{2} + B^{3}}{a^{2} d^{3}}\right )^{\frac {1}{3}}}{i \, A + B}\right ) - 2 \, \left (\frac {1}{4}\right )^{\frac {1}{3}} {\left (i \, \sqrt {3} a d + a d\right )} \left (\frac {-i \, A^{3} - 3 \, A^{2} B + 3 i \, A B^{2} + B^{3}}{a^{2} d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac {2^{\frac {1}{3}} {\left (i \, A + B\right )} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )} - \left (\frac {1}{4}\right )^{\frac {1}{3}} {\left (-i \, \sqrt {3} a d - a d\right )} \left (\frac {-i \, A^{3} - 3 \, A^{2} B + 3 i \, A B^{2} + B^{3}}{a^{2} d^{3}}\right )^{\frac {1}{3}}}{i \, A + B}\right ) + 2^{\frac {1}{3}} {\left ({\left (3 i \, A - 3 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, A - 3 \, B\right )} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*(1/4)^(1/3)*a*d*((-I*A^3 - 3*A^2*B + 3*I*A*B^2 + B^3)/(a^2*d^3))^(1/3)*e^(2*I*d*x + 2*I*c)*log(-(2*(1/4
)^(1/3)*a*d*((-I*A^3 - 3*A^2*B + 3*I*A*B^2 + B^3)/(a^2*d^3))^(1/3) - 2^(1/3)*(I*A + B)*(a/(e^(2*I*d*x + 2*I*c)
+ 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c))/(I*A + B)) - 2*(1/4)^(1/3)*(-I*sqrt(3)*a*d + a*d)*((-I*A^3 - 3*A^2*B + 3
*I*A*B^2 + B^3)/(a^2*d^3))^(1/3)*e^(2*I*d*x + 2*I*c)*log((2^(1/3)*(I*A + B)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3
)*e^(2/3*I*d*x + 2/3*I*c) - (1/4)^(1/3)*(I*sqrt(3)*a*d - a*d)*((-I*A^3 - 3*A^2*B + 3*I*A*B^2 + B^3)/(a^2*d^3))
^(1/3))/(I*A + B)) - 2*(1/4)^(1/3)*(I*sqrt(3)*a*d + a*d)*((-I*A^3 - 3*A^2*B + 3*I*A*B^2 + B^3)/(a^2*d^3))^(1/3
)*e^(2*I*d*x + 2*I*c)*log((2^(1/3)*(I*A + B)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) - (1/
4)^(1/3)*(-I*sqrt(3)*a*d - a*d)*((-I*A^3 - 3*A^2*B + 3*I*A*B^2 + B^3)/(a^2*d^3))^(1/3))/(I*A + B)) + 2^(1/3)*(
(3*I*A - 3*B)*e^(2*I*d*x + 2*I*c) + 3*I*A - 3*B)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c))*
e^(-2*I*d*x - 2*I*c)/(a*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.16, size = 318, normalized size = 1.49 \[ \frac {3 i A}{4 d \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}-\frac {3 B}{4 d \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}+\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right ) B}{4 d \,a^{\frac {2}{3}}}+\frac {i 2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right ) A}{4 d \,a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right ) B}{8 d \,a^{\frac {2}{3}}}-\frac {i 2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right ) A}{8 d \,a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right ) B}{4 d \,a^{\frac {2}{3}}}-\frac {i 2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right ) A}{4 d \,a^{\frac {2}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*I/d/(a+I*a*tan(d*x+c))^(2/3)*A-3/4/d/(a+I*a*tan(d*x+c))^(2/3)*B+1/4/d*2^(1/3)/a^(2/3)*ln((a+I*a*tan(d*x+c)
)^(1/3)-2^(1/3)*a^(1/3))*B+1/4*I/d*2^(1/3)/a^(2/3)*ln((a+I*a*tan(d*x+c))^(1/3)-2^(1/3)*a^(1/3))*A-1/8/d*2^(1/3
)/a^(2/3)*ln((a+I*a*tan(d*x+c))^(2/3)+2^(1/3)*a^(1/3)*(a+I*a*tan(d*x+c))^(1/3)+2^(2/3)*a^(2/3))*B-1/8*I/d*2^(1
/3)/a^(2/3)*ln((a+I*a*tan(d*x+c))^(2/3)+2^(1/3)*a^(1/3)*(a+I*a*tan(d*x+c))^(1/3)+2^(2/3)*a^(2/3))*A-1/4/d*2^(1
/3)/a^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2^(2/3)/a^(1/3)*(a+I*a*tan(d*x+c))^(1/3)+1))*B-1/4*I/d*2^(1/3)/a^(2/3)
*3^(1/2)*arctan(1/3*3^(1/2)*(2^(2/3)/a^(1/3)*(a+I*a*tan(d*x+c))^(1/3)+1))*A

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maxima [A]  time = 0.43, size = 171, normalized size = 0.80 \[ -\frac {i \, {\left (2 \, \sqrt {3} 2^{\frac {1}{3}} {\left (A - i \, B\right )} a^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) + 2^{\frac {1}{3}} {\left (A - i \, B\right )} a^{\frac {1}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) - 2 \cdot 2^{\frac {1}{3}} {\left (A - i \, B\right )} a^{\frac {1}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) - \frac {6 \, {\left (A + i \, B\right )} a}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}}\right )}}{8 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*I*(2*sqrt(3)*2^(1/3)*(A - I*B)*a^(1/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3)*a^(1/3) + 2*(I*a*tan(d*x + c)
+ a)^(1/3))/a^(1/3)) + 2^(1/3)*(A - I*B)*a^(1/3)*log(2^(2/3)*a^(2/3) + 2^(1/3)*(I*a*tan(d*x + c) + a)^(1/3)*a^
(1/3) + (I*a*tan(d*x + c) + a)^(2/3)) - 2*2^(1/3)*(A - I*B)*a^(1/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) +
a)^(1/3)) - 6*(A + I*B)*a/(I*a*tan(d*x + c) + a)^(2/3))/(a*d)

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mupad [B]  time = 0.67, size = 390, normalized size = 1.83 \[ \frac {A\,3{}\mathrm {i}}{4\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{2/3}}-\frac {3\,B}{4\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{2/3}}-\frac {{\left (\frac {1}{32}{}\mathrm {i}\right )}^{1/3}\,A\,\ln \left (A\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,36{}\mathrm {i}+144\,{\left (\frac {1}{32}{}\mathrm {i}\right )}^{1/3}\,A\,a^{1/3}\,d^2\right )}{a^{2/3}\,d}+\frac {2^{1/3}\,B\,\ln \left (36\,B\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-36\,2^{1/3}\,B\,a^{1/3}\,d^2\right )}{4\,a^{2/3}\,d}-\frac {{\left (\frac {1}{32}{}\mathrm {i}\right )}^{1/3}\,A\,\ln \left (A\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,36{}\mathrm {i}+144\,{\left (\frac {1}{32}{}\mathrm {i}\right )}^{1/3}\,A\,a^{1/3}\,d^2\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{2/3}\,d}+\frac {{\left (\frac {1}{32}{}\mathrm {i}\right )}^{1/3}\,A\,\ln \left (A\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,36{}\mathrm {i}-144\,{\left (\frac {1}{32}{}\mathrm {i}\right )}^{1/3}\,A\,a^{1/3}\,d^2\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{2/3}\,d}+\frac {2^{1/3}\,B\,\ln \left (36\,B\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-36\,2^{1/3}\,B\,a^{1/3}\,d^2\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{4\,a^{2/3}\,d}-\frac {2^{1/3}\,B\,\ln \left (36\,B\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}+36\,2^{1/3}\,B\,a^{1/3}\,d^2\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{4\,a^{2/3}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*3i)/(4*d*(a + a*tan(c + d*x)*1i)^(2/3)) - (3*B)/(4*d*(a + a*tan(c + d*x)*1i)^(2/3)) - ((1i/32)^(1/3)*A*log(
A*d^2*(a + a*tan(c + d*x)*1i)^(1/3)*36i + 144*(1i/32)^(1/3)*A*a^(1/3)*d^2))/(a^(2/3)*d) + (2^(1/3)*B*log(36*B*
d^2*(a + a*tan(c + d*x)*1i)^(1/3) - 36*2^(1/3)*B*a^(1/3)*d^2))/(4*a^(2/3)*d) - ((1i/32)^(1/3)*A*log(A*d^2*(a +
a*tan(c + d*x)*1i)^(1/3)*36i + 144*(1i/32)^(1/3)*A*a^(1/3)*d^2*((3^(1/2)*1i)/2 - 1/2))*((3^(1/2)*1i)/2 - 1/2)
)/(a^(2/3)*d) + ((1i/32)^(1/3)*A*log(A*d^2*(a + a*tan(c + d*x)*1i)^(1/3)*36i - 144*(1i/32)^(1/3)*A*a^(1/3)*d^2
*((3^(1/2)*1i)/2 + 1/2))*((3^(1/2)*1i)/2 + 1/2))/(a^(2/3)*d) + (2^(1/3)*B*log(36*B*d^2*(a + a*tan(c + d*x)*1i)
^(1/3) - 36*2^(1/3)*B*a^(1/3)*d^2*((3^(1/2)*1i)/2 - 1/2))*((3^(1/2)*1i)/2 - 1/2))/(4*a^(2/3)*d) - (2^(1/3)*B*l
og(36*B*d^2*(a + a*tan(c + d*x)*1i)^(1/3) + 36*2^(1/3)*B*a^(1/3)*d^2*((3^(1/2)*1i)/2 + 1/2))*((3^(1/2)*1i)/2 +
1/2))/(4*a^(2/3)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*tan(c + d*x))/(I*a*(tan(c + d*x) - I))**(2/3), x)

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