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]

-(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

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Rubi [A]  time = 0.165943, antiderivative size = 201, normalized size of antiderivative = 1., 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 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]

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 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 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 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 31

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

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 = 180.004, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

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

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Maple [A]  time = 0.019, size = 297, normalized size = 1.5 \begin{align*} 3\,{\frac{B\sqrt [3]{a+ia\tan \left ( dx+c \right ) }}{d}}+{\frac{\sqrt [3]{2}B}{2\,d}\sqrt [3]{a}\ln \left ( \sqrt [3]{a+ia\tan \left ( dx+c \right ) }-\sqrt [3]{2}\sqrt [3]{a} \right ) }+{\frac{{\frac{i}{2}}\sqrt [3]{2}A}{d}\sqrt [3]{a}\ln \left ( \sqrt [3]{a+ia\tan \left ( dx+c \right ) }-\sqrt [3]{2}\sqrt [3]{a} \right ) }-{\frac{\sqrt [3]{2}B}{4\,d}\sqrt [3]{a}\ln \left ( \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}+\sqrt [3]{2}\sqrt [3]{a}\sqrt [3]{a+ia\tan \left ( dx+c \right ) }+{2}^{{\frac{2}{3}}}{a}^{{\frac{2}{3}}} \right ) }-{\frac{{\frac{i}{4}}\sqrt [3]{2}A}{d}\sqrt [3]{a}\ln \left ( \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}+\sqrt [3]{2}\sqrt [3]{a}\sqrt [3]{a+ia\tan \left ( dx+c \right ) }+{2}^{{\frac{2}{3}}}{a}^{{\frac{2}{3}}} \right ) }-{\frac{\sqrt [3]{2}\sqrt{3}B}{2\,d}\sqrt [3]{a}\arctan \left ({\frac{\sqrt{3}}{3} \left ({{2}^{{\frac{2}{3}}}\sqrt [3]{a+ia\tan \left ( dx+c \right ) }{\frac{1}{\sqrt [3]{a}}}}+1 \right ) } \right ) }-{\frac{{\frac{i}{2}}\sqrt [3]{2}\sqrt{3}A}{d}\sqrt [3]{a}\arctan \left ({\frac{\sqrt{3}}{3} \left ({{2}^{{\frac{2}{3}}}\sqrt [3]{a+ia\tan \left ( dx+c \right ) }{\frac{1}{\sqrt [3]{a}}}}+1 \right ) } \right ) } \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [B]  time = 1.76749, size = 1131, normalized size = 5.63 \begin{align*} \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} \end{align*}

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt [3]{a \left (i \tan{\left (c + d x \right )} + 1\right )} \left (A + B \tan{\left (c + d x \right )}\right )\, dx \end{align*}

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)