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

Optimal. Leaf size=156 \[ -\frac {i \sqrt {3} \sqrt [3]{a} \tan ^{-1}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2^{2/3} d}+\frac {3 i \sqrt [3]{a} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2\ 2^{2/3} d}+\frac {i \sqrt [3]{a} \log (\cos (c+d x))}{2\ 2^{2/3} d}-\frac {\sqrt [3]{a} x}{2\ 2^{2/3}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {3481, 57, 617, 204, 31} \[ -\frac {i \sqrt {3} \sqrt [3]{a} \tan ^{-1}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2^{2/3} d}+\frac {3 i \sqrt [3]{a} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2\ 2^{2/3} d}+\frac {i \sqrt [3]{a} \log (\cos (c+d x))}{2\ 2^{2/3} d}-\frac {\sqrt [3]{a} x}{2\ 2^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(1/3),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.95, size = 223, normalized size = 1.43 \[ -\frac {i e^{-\frac {2}{3} i (c+d x)} \sqrt [3]{1+e^{2 i (c+d x)}} \sqrt [3]{a+i a \tan (c+d x)} \left (-2 \log \left (1-\frac {e^{\frac {2}{3} i (c+d x)}}{\sqrt [3]{1+e^{2 i (c+d x)}}}\right )+\log \left (\frac {\left (1+e^{2 i (c+d x)}\right )^{2/3}+e^{\frac {2}{3} i (c+d x)} \sqrt [3]{1+e^{2 i (c+d x)}}+e^{\frac {4}{3} i (c+d x)}}{\left (1+e^{2 i (c+d x)}\right )^{2/3}}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 e^{\frac {2}{3} i (c+d x)}}{\sqrt [3]{1+e^{2 i (c+d x)}}}}{\sqrt {3}}\right )\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.48, size = 188, normalized size = 1.21 \[ \frac {1}{2} \, {\left (i \, \sqrt {3} - 1\right )} \left (-\frac {i \, a}{4 \, d^{3}}\right )^{\frac {1}{3}} \log \left (2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )} - {\left (\sqrt {3} d + i \, d\right )} \left (-\frac {i \, a}{4 \, d^{3}}\right )^{\frac {1}{3}}\right ) + \frac {1}{2} \, {\left (-i \, \sqrt {3} - 1\right )} \left (-\frac {i \, a}{4 \, d^{3}}\right )^{\frac {1}{3}} \log \left (2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )} + {\left (\sqrt {3} d - i \, d\right )} \left (-\frac {i \, a}{4 \, d^{3}}\right )^{\frac {1}{3}}\right ) + \left (-\frac {i \, a}{4 \, d^{3}}\right )^{\frac {1}{3}} \log \left (2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )} + 2 i \, d \left (-\frac {i \, a}{4 \, d^{3}}\right )^{\frac {1}{3}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(I*sqrt(3) - 1)*(-1/4*I*a/d^3)^(1/3)*log(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) - (sqrt(3)*d + I*d)*(-1/4*I*a/d^3)^(1/3)) + 1/2*(-I*sqrt(3) - 1)*(-1/4*I*a/d^3)^(1/3)*log(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) + (sqrt(3)*d - I*d)*(-1/4*I*a/d^3)^(1/3)) + (-1/4*I*a/d^3)^
(1/3)*log(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*d*(-1/4*I*a/d^3)^(1/3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.12, size = 138, normalized size = 0.88 \[ \frac {i a^{\frac {1}{3}} 2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{2 d}-\frac {i a^{\frac {1}{3}} 2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{4 d}-\frac {i a^{\frac {1}{3}} 2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))-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))-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))

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maxima [A]  time = 0.63, size = 135, normalized size = 0.87 \[ -\frac {i \, {\left (2 \, \sqrt {3} 2^{\frac {1}{3}} a^{\frac {4}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) + 2^{\frac {1}{3}} a^{\frac {4}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) - 2 \cdot 2^{\frac {1}{3}} a^{\frac {4}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )\right )}}{4 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.97, size = 184, normalized size = 1.18 \[ \frac {{\left (\frac {1}{4}{}\mathrm {i}\right )}^{1/3}\,{\left (-a\right )}^{1/3}\,\ln \left (18\,{\left (\frac {1}{4}{}\mathrm {i}\right )}^{1/3}\,{\left (-a\right )}^{4/3}\,d^2+a\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,9{}\mathrm {i}\right )}{d}+\frac {{\left (\frac {1}{4}{}\mathrm {i}\right )}^{1/3}\,{\left (-a\right )}^{1/3}\,\ln \left (a\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,9{}\mathrm {i}+18\,{\left (\frac {1}{4}{}\mathrm {i}\right )}^{1/3}\,{\left (-a\right )}^{4/3}\,d^2\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{d}-\frac {{\left (\frac {1}{4}{}\mathrm {i}\right )}^{1/3}\,{\left (-a\right )}^{1/3}\,\ln \left (a\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,9{}\mathrm {i}-18\,{\left (\frac {1}{4}{}\mathrm {i}\right )}^{1/3}\,{\left (-a\right )}^{4/3}\,d^2\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^(1/3),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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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