3.294 \(\int \frac {1}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=184 \[ \frac {i \sqrt {3} \tan ^{-1}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {i \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {x}{4 \sqrt [3]{2} \sqrt [3]{a}} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {3479, 3481, 55, 617, 204, 31} \[ \frac {i \sqrt {3} \tan ^{-1}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {i \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {x}{4 \sqrt [3]{2} \sqrt [3]{a}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(-1/3),x]

[Out]

-x/(4*2^(1/3)*a^(1/3)) + ((I/2)*Sqrt[3]*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3))/(Sqrt[3]*a^(1/
3))])/(2^(1/3)*a^(1/3)*d) + ((I/4)*Log[Cos[c + d*x]])/(2^(1/3)*a^(1/3)*d) + (((3*I)/4)*Log[2^(1/3)*a^(1/3) - (
a + I*a*Tan[c + d*x])^(1/3)])/(2^(1/3)*a^(1/3)*d) + ((3*I)/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 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 3479

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

Rubi steps

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

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Mathematica [C]  time = 0.44, size = 141, normalized size = 0.77 \[ \frac {3 \left (e^{2 i d x} (\cos (c)+i \sin (c)) \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )-2 i \sin (c) \left (-1+e^{2 i d x}\right )-2 \cos (c) \left (1+e^{2 i d x}\right )\right )}{4 d \sqrt [3]{a+i a \tan (c+d x)} \left (i \cos (c) \left (1+e^{2 i d x}\right )-\sin (c) \left (-1+e^{2 i d x}\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.46, size = 314, normalized size = 1.71 \[ \frac {{\left (4 \, a d \left (-\frac {i}{16 \, a d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (8 \, a d^{2} \left (-\frac {i}{16 \, a d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) + 2^{\frac {2}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {2}{3}} {\left (3 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (\frac {4}{3} i \, d x + \frac {4}{3} i \, c\right )} - 2 \, {\left (-i \, \sqrt {3} a d + a d\right )} \left (-\frac {i}{16 \, a d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-4 \, {\left (i \, \sqrt {3} a d^{2} + a d^{2}\right )} \left (-\frac {i}{16 \, a d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - 2 \, {\left (i \, \sqrt {3} a d + a d\right )} \left (-\frac {i}{16 \, a d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-4 \, {\left (-i \, \sqrt {3} a d^{2} + a d^{2}\right )} \left (-\frac {i}{16 \, a d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right )\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*a*d*(-1/16*I/(a*d^3))^(1/3)*e^(2*I*d*x + 2*I*c)*log(8*a*d^2*(-1/16*I/(a*d^3))^(2/3) + 2^(1/3)*(a/(e^(2*
I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c)) + 2^(2/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*(3*I*e^(2*I*
d*x + 2*I*c) + 3*I)*e^(4/3*I*d*x + 4/3*I*c) - 2*(-I*sqrt(3)*a*d + a*d)*(-1/16*I/(a*d^3))^(1/3)*e^(2*I*d*x + 2*
I*c)*log(-4*(I*sqrt(3)*a*d^2 + a*d^2)*(-1/16*I/(a*d^3))^(2/3) + 2^(1/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^
(2/3*I*d*x + 2/3*I*c)) - 2*(I*sqrt(3)*a*d + a*d)*(-1/16*I/(a*d^3))^(1/3)*e^(2*I*d*x + 2*I*c)*log(-4*(-I*sqrt(3
)*a*d^2 + a*d^2)*(-1/16*I/(a*d^3))^(2/3) + 2^(1/3)*(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 \[ \int \frac {1}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.10, size = 158, normalized size = 0.86 \[ \frac {i 2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{4 d \,a^{\frac {1}{3}}}-\frac {i 2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{8 d \,a^{\frac {1}{3}}}+\frac {i \sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{4 d \,a^{\frac {1}{3}}}+\frac {3 i}{2 d \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.65, size = 152, normalized size = 0.83 \[ \frac {i \, {\left (2 \, \sqrt {3} 2^{\frac {2}{3}} a^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) - 2^{\frac {2}{3}} a^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) + 2 \cdot 2^{\frac {2}{3}} a^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) + \frac {12 \, a}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}\right )}}{8 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.39, size = 197, normalized size = 1.07 \[ \frac {3{}\mathrm {i}}{2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}+\frac {{\left (\frac {1}{16}{}\mathrm {i}\right )}^{1/3}\,\ln \left ({\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}-{\left (-1\right )}^{1/3}\,2^{1/3}\,{\left (-a\right )}^{1/3}\right )}{{\left (-a\right )}^{1/3}\,d}-\frac {{\left (\frac {1}{16}{}\mathrm {i}\right )}^{1/3}\,\ln \left (-\frac {9\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{4\,d^2}+\frac {9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,{\left (-a\right )}^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{8\,d^2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{{\left (-a\right )}^{1/3}\,d}+\frac {{\left (\frac {1}{16}{}\mathrm {i}\right )}^{1/3}\,\ln \left (-\frac {9\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{4\,d^2}-\frac {9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,{\left (-a\right )}^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{8\,d^2}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{{\left (-a\right )}^{1/3}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3i/(2*d*(a + a*tan(c + d*x)*1i)^(1/3)) + ((1i/16)^(1/3)*log((a*(tan(c + d*x)*1i + 1))^(1/3) - (-1)^(1/3)*2^(1/
3)*(-a)^(1/3)))/((-a)^(1/3)*d) - ((1i/16)^(1/3)*log((9*(-1)^(1/3)*2^(1/3)*(-a)^(1/3)*(3^(1/2)*1i - 1))/(8*d^2)
- (9*(a + a*tan(c + d*x)*1i)^(1/3))/(4*d^2))*((3^(1/2)*1i)/2 + 1/2))/((-a)^(1/3)*d) + ((1i/16)^(1/3)*log(- (9
*(a + a*tan(c + d*x)*1i)^(1/3))/(4*d^2) - (9*(-1)^(1/3)*2^(1/3)*(-a)^(1/3)*(3^(1/2)*1i + 1))/(8*d^2))*((3^(1/2
)*1i)/2 - 1/2))/((-a)^(1/3)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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