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

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

[Out]

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

________________________________________________________________________________________

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, 55, 617, 204, 31} \[ \frac {\sqrt {3} (B+i A) \tan ^{-1}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {3 (-B+i A)}{2 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 (B+i A) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {(B+i A) \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {x (A-i B)}{4 \sqrt [3]{2} \sqrt [3]{a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

Rule 55

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), 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)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx &=\frac {3 (i A-B)}{2 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {(A-i B) \int (a+i a \tan (c+d x))^{2/3} \, dx}{2 a}\\ &=\frac {3 (i A-B)}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {(i A+B) \operatorname {Subst}\left (\int \frac {1}{(a-x) \sqrt [3]{a+x}} \, dx,x,i a \tan (c+d x)\right )}{2 d}\\ &=-\frac {(A-i B) x}{4 \sqrt [3]{2} \sqrt [3]{a}}+\frac {(i A+B) \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {3 (i A-B)}{2 d \sqrt [3]{a+i a \tan (c+d x)}}+\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 d}-\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 \sqrt [3]{2} \sqrt [3]{a} d}\\ &=-\frac {(A-i B) x}{4 \sqrt [3]{2} \sqrt [3]{a}}+\frac {(i A+B) \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} 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 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {3 (i A-B)}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\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 \sqrt [3]{2} \sqrt [3]{a} d}\\ &=-\frac {(A-i B) x}{4 \sqrt [3]{2} \sqrt [3]{a}}+\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 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {(i A+B) \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} 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 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {3 (i A-B)}{2 d \sqrt [3]{a+i a \tan (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 1.30, size = 137, normalized size = 0.64 \[ -\frac {3 i e^{-2 i (c+d x)} \left (\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{2/3} \left ((A-i B) e^{2 i (c+d x)} \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )-2 (A+i B) \left (1+e^{2 i (c+d x)}\right )\right )}{4 \sqrt [3]{2} a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-3*I)/4)*((a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x))))^(2/3)*(-2*(A + I*B)*(1 + E^((2*I)*(c + d*x)))
+ (A - I*B)*E^((2*I)*(c + d*x))*Hypergeometric2F1[2/3, 1, 5/3, E^((2*I)*(c + d*x))/(1 + E^((2*I)*(c + d*x)))])
)/(2^(1/3)*a*d*E^((2*I)*(c + d*x)))

________________________________________________________________________________________

fricas [B]  time = 0.67, size = 550, normalized size = 2.58 \[ \frac {{\left (2 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} a d \left (\frac {-i \, A^{3} - 3 \, A^{2} B + 3 i \, A B^{2} + B^{3}}{a d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac {2 \, \left (\frac {1}{2}\right )^{\frac {2}{3}} a d^{2} \left (\frac {-i \, A^{3} - 3 \, A^{2} B + 3 i \, A B^{2} + B^{3}}{a d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (A^{2} - 2 i \, A B - B^{2}\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 )}}{A^{2} - 2 i \, A B - B^{2}}\right ) - \left (\frac {1}{2}\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 d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac {2^{\frac {1}{3}} {\left (A^{2} - 2 i \, A B - B^{2}\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}{2}\right )^{\frac {2}{3}} {\left (i \, \sqrt {3} a d^{2} - a d^{2}\right )} \left (\frac {-i \, A^{3} - 3 \, A^{2} B + 3 i \, A B^{2} + B^{3}}{a d^{3}}\right )^{\frac {2}{3}}}{A^{2} - 2 i \, A B - B^{2}}\right ) - \left (\frac {1}{2}\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 d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac {2^{\frac {1}{3}} {\left (A^{2} - 2 i \, A B - B^{2}\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}{2}\right )^{\frac {2}{3}} {\left (-i \, \sqrt {3} a d^{2} - a d^{2}\right )} \left (\frac {-i \, A^{3} - 3 \, A^{2} B + 3 i \, A B^{2} + B^{3}}{a d^{3}}\right )^{\frac {2}{3}}}{A^{2} - 2 i \, A B - B^{2}}\right ) + 2^{\frac {2}{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 {2}{3}} e^{\left (\frac {4}{3} i \, d x + \frac {4}{3} i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, 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))^(1/3),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

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))^(1/3),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.17, size = 318, normalized size = 1.49 \[ \frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right ) B}{4 d \,a^{\frac {1}{3}}}+\frac {i 2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right ) A}{4 d \,a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right ) B}{8 d \,a^{\frac {1}{3}}}-\frac {i 2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right ) A}{8 d \,a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right ) B}{4 d \,a^{\frac {1}{3}}}+\frac {i \sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right ) A}{4 d \,a^{\frac {1}{3}}}-\frac {3 B}{2 d \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}+\frac {3 i A}{2 d \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}} \]

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))^(1/3),x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.46, size = 172, normalized size = 0.81 \[ \frac {i \, {\left (2 \, \sqrt {3} 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) - 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) + 2 \cdot 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) + \frac {12 \, {\left (A + i \, B\right )} a}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}\right )}}{8 \, a d} \]

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))^(1/3),x, algorithm="maxima")

[Out]

1/8*I*(2*sqrt(3)*2^(2/3)*(A - I*B)*a^(2/3)*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^(2/3)*(A - I*B)*a^(2/3)*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^(2/3)*(A - I*B)*a^(2/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) +
a)^(1/3)) + 12*(A + I*B)*a/(I*a*tan(d*x + c) + a)^(1/3))/(a*d)

________________________________________________________________________________________

mupad [B]  time = 6.99, size = 383, normalized size = 1.80 \[ \frac {A\,3{}\mathrm {i}}{2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}-\frac {3\,B}{2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}-\frac {{\left (\frac {1}{16}{}\mathrm {i}\right )}^{1/3}\,A\,\ln \left ({\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}+{\left (-1\right )}^{1/3}\,2^{1/3}\,a^{1/3}\right )}{a^{1/3}\,d}+\frac {4^{1/3}\,B\,\ln \left (18\,B^2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-9\,4^{2/3}\,B^2\,a^{1/3}\,d\right )}{4\,a^{1/3}\,d}+\frac {4^{1/3}\,B\,\ln \left (18\,B^2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-9\,4^{2/3}\,B^2\,a^{1/3}\,d\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{4\,a^{1/3}\,d}-\frac {4^{1/3}\,B\,\ln \left (18\,B^2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-9\,4^{2/3}\,B^2\,a^{1/3}\,d\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{4\,a^{1/3}\,d}-\frac {{\left (\frac {1}{16}{}\mathrm {i}\right )}^{1/3}\,A\,\ln \left (\frac {{\left (-1\right )}^{1/3}\,2^{1/3}\,a^{1/3}}{2}-{\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}+\frac {{\left (-1\right )}^{5/6}\,2^{1/3}\,\sqrt {3}\,a^{1/3}}{2}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{1/3}\,d}+\frac {{\left (\frac {1}{16}{}\mathrm {i}\right )}^{1/3}\,A\,\ln \left ({\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}-\frac {{\left (-1\right )}^{1/3}\,2^{1/3}\,a^{1/3}}{2}+\frac {{\left (-1\right )}^{5/6}\,2^{1/3}\,\sqrt {3}\,a^{1/3}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{1/3}\,d} \]

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)^(1/3),x)

[Out]

(A*3i)/(2*d*(a + a*tan(c + d*x)*1i)^(1/3)) - (3*B)/(2*d*(a + a*tan(c + d*x)*1i)^(1/3)) - ((1i/16)^(1/3)*A*log(
(a*(tan(c + d*x)*1i + 1))^(1/3) + (-1)^(1/3)*2^(1/3)*a^(1/3)))/(a^(1/3)*d) + (4^(1/3)*B*log(18*B^2*d*(a + a*ta
n(c + d*x)*1i)^(1/3) - 9*4^(2/3)*B^2*a^(1/3)*d))/(4*a^(1/3)*d) + (4^(1/3)*B*log(18*B^2*d*(a + a*tan(c + d*x)*1
i)^(1/3) - 9*4^(2/3)*B^2*a^(1/3)*d*((3^(1/2)*1i)/2 - 1/2)^2)*((3^(1/2)*1i)/2 - 1/2))/(4*a^(1/3)*d) - (4^(1/3)*
B*log(18*B^2*d*(a + a*tan(c + d*x)*1i)^(1/3) - 9*4^(2/3)*B^2*a^(1/3)*d*((3^(1/2)*1i)/2 + 1/2)^2)*((3^(1/2)*1i)
/2 + 1/2))/(4*a^(1/3)*d) - ((1i/16)^(1/3)*A*log(((-1)^(1/3)*2^(1/3)*a^(1/3))/2 - (a*(tan(c + d*x)*1i + 1))^(1/
3) + ((-1)^(5/6)*2^(1/3)*3^(1/2)*a^(1/3))/2)*((3^(1/2)*1i)/2 - 1/2))/(a^(1/3)*d) + ((1i/16)^(1/3)*A*log((a*(ta
n(c + d*x)*1i + 1))^(1/3) - ((-1)^(1/3)*2^(1/3)*a^(1/3))/2 + ((-1)^(5/6)*2^(1/3)*3^(1/2)*a^(1/3))/2)*((3^(1/2)
*1i)/2 + 1/2))/(a^(1/3)*d)

________________________________________________________________________________________

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

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))**(1/3),x)

[Out]

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

________________________________________________________________________________________