\(\int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx\) [202]

Optimal result

Integrand size = 28, antiderivative size = 213 \[ \int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=-\frac {(A-i B) x}{4 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\sqrt {3} (i A+B) \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {(i A+B) \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {3 (i A+B) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {3 (i A-B)}{2 d \sqrt [3]{a+i a \tan (c+d x)}} \]

[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] (verified)

Time = 0.27 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3607, 3562, 57, 631, 210, 31} \[ \int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {3} (B+i A) \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {3 (-B+i A)}{2 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 (B+i A) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {(B+i A) \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {x (A-i B)}{4 \sqrt [3]{2} \sqrt [3]{a}} \]

[In]

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

[Out]

-1/4*((A - I*B)*x)/(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*
Tan[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 57

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 210

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

Rule 631

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 3562

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 3607

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*} \text {integral}& = \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) \text {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)) \text {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)) \text {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)) \text {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) \arctan \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 [A] (verified)

Time = 0.89 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.70 \[ \int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {\frac {2^{2/3} (i A+B) \left (2 \sqrt {3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )-\log (i+\tan (c+d x))+3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )\right )}{\sqrt [3]{a}}+\frac {12 i (A+i B)}{\sqrt [3]{a+i a \tan (c+d x)}}}{8 d} \]

[In]

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

[Out]

((2^(2/3)*(I*A + B)*(2*Sqrt[3]*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3))/(Sqrt[3]*a^(1/3))] - Lo
g[I + Tan[c + d*x]] + 3*Log[2^(1/3)*a^(1/3) - (a + I*a*Tan[c + d*x])^(1/3)]))/a^(1/3) + ((12*I)*(A + I*B))/(a
+ I*a*Tan[c + d*x])^(1/3))/(8*d)

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.78

method	result	size
derivativedivides	\(\frac {3 i \left (-\frac {-\frac {A}{2}-\frac {i B}{2}}{\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}+\left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right ) \left (\frac {A}{2}-\frac {i B}{2}\right )\right )}{d}\)	\(166\)
default	\(\frac {3 i \left (-\frac {-\frac {A}{2}-\frac {i B}{2}}{\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}+\left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right ) \left (\frac {A}{2}-\frac {i B}{2}\right )\right )}{d}\)	\(166\)
parts	\(\frac {3 i A a \left (\frac {\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}}{2 a}+\frac {1}{2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}\right )}{d}+\frac {B \left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{4 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{8 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{4 a^{\frac {1}{3}}}-\frac {3}{2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}\right )}{d}\)	\(306\)

[In]

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

[Out]

3*I/d*(-(-1/2*A-1/2*I*B)/(a+I*a*tan(d*x+c))^(1/3)+(1/6*2^(2/3)/a^(1/3)*ln((a+I*a*tan(d*x+c))^(1/3)-2^(1/3)*a^(
1/3))-1/12*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
))+1/6*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)))*(1/2*A-1/2*I*
B))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 547 vs. \(2 (156) = 312\).

Time = 0.26 (sec) , antiderivative size = 547, normalized size of antiderivative = 2.57 \[ \int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\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 ) - 3 \cdot 2^{\frac {2}{3}} {\left ({\left (-i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, A + 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} \]

[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)) - 3*2^(2/3)*((-I*A + B)*e^(2*I*d*x + 2*I*c) - I*A + 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)

Sympy [F]

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

[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)

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.81 \[ \int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {i \, {\left (2 \, \sqrt {3} 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) - 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) + 2 \cdot 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) + \frac {12 \, {\left (A + i \, B\right )} a}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}\right )}}{8 \, a d} \]

[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)

Giac [F]

\[ \int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\int { \frac {B \tan \left (d x + c\right ) + A}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]

[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)

Mupad [B] (verification not implemented)

Time = 8.52 (sec) , antiderivative size = 383, normalized size of antiderivative = 1.80 \[ \int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\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} \]

[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)