3.196 \(\int \sqrt [3]{a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=201 \[ -\frac {\sqrt {3} \sqrt [3]{a} (B+i A) \tan ^{-1}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2^{2/3} d}+\frac {3 \sqrt [3]{a} (B+i A) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2\ 2^{2/3} d}+\frac {\sqrt [3]{a} (B+i A) \log (\cos (c+d x))}{2\ 2^{2/3} d}-\frac {\sqrt [3]{a} x (A-i B)}{2\ 2^{2/3}}+\frac {3 B \sqrt [3]{a+i a \tan (c+d x)}}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.17, antiderivative size = 201, 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 = {3527, 3481, 57, 617, 204, 31} \[ -\frac {\sqrt {3} \sqrt [3]{a} (B+i A) \tan ^{-1}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2^{2/3} d}+\frac {3 \sqrt [3]{a} (B+i A) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2\ 2^{2/3} d}+\frac {\sqrt [3]{a} (B+i A) \log (\cos (c+d x))}{2\ 2^{2/3} d}-\frac {\sqrt [3]{a} x (A-i B)}{2\ 2^{2/3}}+\frac {3 B \sqrt [3]{a+i a \tan (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(1/3)*(A + B*Tan[c + d*x]),x]

[Out]

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

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 3527

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [F]  time = 180.00, size = 0, normalized size = 0.00 \[ \text {\$Aborted} \]

Verification is Not applicable to the result.

[In]

Integrate[(a + I*a*Tan[c + d*x])^(1/3)*(A + B*Tan[c + d*x]),x]

[Out]

$Aborted

________________________________________________________________________________________

fricas [B]  time = 0.73, size = 405, normalized size = 2.01 \[ \frac {6 \cdot 2^{\frac {1}{3}} B \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} d - d\right )} \left (\frac {{\left (-i \, A^{3} - 3 \, A^{2} B + 3 i \, A B^{2} + B^{3}\right )} a}{d^{3}}\right )^{\frac {1}{3}} \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} d + d\right )} \left (\frac {{\left (-i \, A^{3} - 3 \, A^{2} B + 3 i \, A B^{2} + B^{3}\right )} a}{d^{3}}\right )^{\frac {1}{3}}}{i \, A + B}\right ) + \left (\frac {1}{4}\right )^{\frac {1}{3}} {\left (i \, \sqrt {3} d - d\right )} \left (\frac {{\left (-i \, A^{3} - 3 \, A^{2} B + 3 i \, A B^{2} + B^{3}\right )} a}{d^{3}}\right )^{\frac {1}{3}} \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} d + d\right )} \left (\frac {{\left (-i \, A^{3} - 3 \, A^{2} B + 3 i \, A B^{2} + B^{3}\right )} a}{d^{3}}\right )^{\frac {1}{3}}}{i \, A + B}\right ) + 2 \, \left (\frac {1}{4}\right )^{\frac {1}{3}} d \left (\frac {{\left (-i \, A^{3} - 3 \, A^{2} B + 3 i \, A B^{2} + B^{3}\right )} a}{d^{3}}\right )^{\frac {1}{3}} \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 )} - 2 \, \left (\frac {1}{4}\right )^{\frac {1}{3}} d \left (\frac {{\left (-i \, A^{3} - 3 \, A^{2} B + 3 i \, A B^{2} + B^{3}\right )} a}{d^{3}}\right )^{\frac {1}{3}}}{i \, A + B}\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(6*2^(1/3)*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)*d - d)*
((-I*A^3 - 3*A^2*B + 3*I*A*B^2 + B^3)*a/d^3)^(1/3)*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)*d + d)*((-I*A^3 - 3*A^2*B + 3*I*A*B^2 + B^3)*a/d^3)^(1/3))/(I
*A + B)) + (1/4)^(1/3)*(I*sqrt(3)*d - d)*((-I*A^3 - 3*A^2*B + 3*I*A*B^2 + B^3)*a/d^3)^(1/3)*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)*d + d)*((-I*A^3 - 3
*A^2*B + 3*I*A*B^2 + B^3)*a/d^3)^(1/3))/(I*A + B)) + 2*(1/4)^(1/3)*d*((-I*A^3 - 3*A^2*B + 3*I*A*B^2 + B^3)*a/d
^3)^(1/3)*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) - 2*(1/4)^(1/3)*d
*((-I*A^3 - 3*A^2*B + 3*I*A*B^2 + B^3)*a/d^3)^(1/3))/(I*A + B)))/d

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^(1/3), x)

________________________________________________________________________________________

maple [A]  time = 0.18, size = 297, normalized size = 1.48 \[ \frac {3 B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{d}+\frac {a^{\frac {1}{3}} 2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right ) B}{2 d}+\frac {i a^{\frac {1}{3}} 2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right ) A}{2 d}-\frac {a^{\frac {1}{3}} 2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right ) B}{4 d}-\frac {i a^{\frac {1}{3}} 2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right ) A}{4 d}-\frac {a^{\frac {1}{3}} 2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right ) B}{2 d}-\frac {i a^{\frac {1}{3}} 2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right ) A}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.50, size = 167, normalized size = 0.83 \[ -\frac {i \, {\left (2 \, \sqrt {3} 2^{\frac {1}{3}} {\left (A - i \, B\right )} a^{\frac {4}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) + 2^{\frac {1}{3}} {\left (A - i \, B\right )} a^{\frac {4}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) - 2 \cdot 2^{\frac {1}{3}} {\left (A - i \, B\right )} a^{\frac {4}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) + 12 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} B a\right )}}{4 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*I*(2*sqrt(3)*2^(1/3)*(A - I*B)*a^(4/3)*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^(4/3)*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^(4/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) +
a)^(1/3)) + 12*I*(I*a*tan(d*x + c) + a)^(1/3)*B*a)/(a*d)

________________________________________________________________________________________

mupad [B]  time = 1.04, size = 365, normalized size = 1.82 \[ \frac {3\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d}+\frac {2^{1/3}\,B\,a^{1/3}\,\ln \left ({\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}-2^{1/3}\,a^{1/3}\right )}{2\,d}-\frac {{\left (\frac {1}{4}{}\mathrm {i}\right )}^{1/3}\,A\,a^{1/3}\,\ln \left (A\,a\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,9{}\mathrm {i}+18\,{\left (\frac {1}{4}{}\mathrm {i}\right )}^{1/3}\,A\,a^{4/3}\,d^2\right )}{d}-\frac {{\left (\frac {1}{4}{}\mathrm {i}\right )}^{1/3}\,A\,a^{1/3}\,\ln \left (A\,a\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,9{}\mathrm {i}+18\,{\left (\frac {1}{4}{}\mathrm {i}\right )}^{1/3}\,A\,a^{4/3}\,d^2\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{d}+\frac {{\left (\frac {1}{4}{}\mathrm {i}\right )}^{1/3}\,A\,a^{1/3}\,\ln \left (A\,a\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,9{}\mathrm {i}-18\,{\left (\frac {1}{4}{}\mathrm {i}\right )}^{1/3}\,A\,a^{4/3}\,d^2\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{d}+\frac {4^{2/3}\,B\,a^{1/3}\,\ln \left (\frac {9\,B\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d}-\frac {9\,2^{1/3}\,B\,a^{4/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2\,d}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{4\,d}-\frac {4^{2/3}\,B\,a^{1/3}\,\ln \left (\frac {9\,B\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d}+\frac {9\,2^{1/3}\,B\,a^{4/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2\,d}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{4\,d} \]

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)^(1/3),x)

[Out]

(3*B*(a + a*tan(c + d*x)*1i)^(1/3))/d + (2^(1/3)*B*a^(1/3)*log((a*(tan(c + d*x)*1i + 1))^(1/3) - 2^(1/3)*a^(1/
3)))/(2*d) - ((1i/4)^(1/3)*A*a^(1/3)*log(A*a*d^2*(a + a*tan(c + d*x)*1i)^(1/3)*9i + 18*(1i/4)^(1/3)*A*a^(4/3)*
d^2))/d - ((1i/4)^(1/3)*A*a^(1/3)*log(A*a*d^2*(a + a*tan(c + d*x)*1i)^(1/3)*9i + 18*(1i/4)^(1/3)*A*a^(4/3)*d^2
*((3^(1/2)*1i)/2 - 1/2))*((3^(1/2)*1i)/2 - 1/2))/d + ((1i/4)^(1/3)*A*a^(1/3)*log(A*a*d^2*(a + a*tan(c + d*x)*1
i)^(1/3)*9i - 18*(1i/4)^(1/3)*A*a^(4/3)*d^2*((3^(1/2)*1i)/2 + 1/2))*((3^(1/2)*1i)/2 + 1/2))/d + (4^(2/3)*B*a^(
1/3)*log((9*B*a*(a + a*tan(c + d*x)*1i)^(1/3))/d - (9*2^(1/3)*B*a^(4/3)*(3^(1/2)*1i - 1))/(2*d))*((3^(1/2)*1i)
/2 - 1/2))/(4*d) - (4^(2/3)*B*a^(1/3)*log((9*B*a*(a + a*tan(c + d*x)*1i)^(1/3))/d + (9*2^(1/3)*B*a^(4/3)*(3^(1
/2)*1i + 1))/(2*d))*((3^(1/2)*1i)/2 + 1/2))/(4*d)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**(1/3)*(A+B*tan(d*x+c)),x)

[Out]

Integral((I*a*(tan(c + d*x) - I))**(1/3)*(A + B*tan(c + d*x)), x)

________________________________________________________________________________________