3.203 \(\int \frac{A+B \tan (c+d x)}{(a+i a \tan (c+d x))^{2/3}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.159795, antiderivative size = 213, 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 = {3526, 3481, 57, 617, 204, 31} \[ -\frac{\sqrt{3} (B+i A) \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{2\ 2^{2/3} a^{2/3} d}+\frac{3 (B+i A) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4\ 2^{2/3} a^{2/3} d}+\frac{(B+i A) \log (\cos (c+d x))}{4\ 2^{2/3} a^{2/3} d}-\frac{x (A-i B)}{4\ 2^{2/3} a^{2/3}}+\frac{3 (-B+i A)}{4 d (a+i a \tan (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[c + d*x])/(a + I*a*Tan[c + d*x])^(2/3),x]

[Out]

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

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]

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

Mathematica [F]  time = 0.628965, size = 0, normalized size = 0. \[ \int \frac{A+B \tan (c+d x)}{(a+i a \tan (c+d x))^{2/3}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(A + B*Tan[c + d*x])/(a + I*a*Tan[c + d*x])^(2/3),x]

[Out]

Integrate[(A + B*Tan[c + d*x])/(a + I*a*Tan[c + d*x])^(2/3), x]

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Maple [A]  time = 0.02, size = 318, normalized size = 1.5 \begin{align*}{\frac{\sqrt [3]{2}B}{4\,d}\ln \left ( \sqrt [3]{a+ia\tan \left ( dx+c \right ) }-\sqrt [3]{2}\sqrt [3]{a} \right ){a}^{-{\frac{2}{3}}}}+{\frac{{\frac{i}{4}}\sqrt [3]{2}A}{d}\ln \left ( \sqrt [3]{a+ia\tan \left ( dx+c \right ) }-\sqrt [3]{2}\sqrt [3]{a} \right ){a}^{-{\frac{2}{3}}}}-{\frac{\sqrt [3]{2}B}{8\,d}\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 ){a}^{-{\frac{2}{3}}}}-{\frac{{\frac{i}{8}}\sqrt [3]{2}A}{d}\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 ){a}^{-{\frac{2}{3}}}}-{\frac{\sqrt [3]{2}\sqrt{3}B}{4\,d}\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 ){a}^{-{\frac{2}{3}}}}-{\frac{{\frac{i}{4}}\sqrt [3]{2}\sqrt{3}A}{d}\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 ){a}^{-{\frac{2}{3}}}}+{\frac{{\frac{3\,i}{4}}A}{d} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{-{\frac{2}{3}}}}-{\frac{3\,B}{4\,d} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{-{\frac{2}{3}}}} \end{align*}

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))^(2/3),x)

[Out]

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

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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+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(2/3),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.73408, size = 1411, normalized size = 6.62 \begin{align*} \text{result too large to display} \end{align*}

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))^(2/3),x, algorithm="fricas")

[Out]

1/16*(8*(1/4)^(1/3)*a*d*((-I*A^3 - 3*A^2*B + 3*I*A*B^2 + B^3)/(a^2*d^3))^(1/3)*e^(2*I*d*x + 2*I*c)*log(-(2*(1/
4)^(1/3)*a*d*((-I*A^3 - 3*A^2*B + 3*I*A*B^2 + B^3)/(a^2*d^3))^(1/3) - 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))/(I*A + B)) + (1/4)^(1/3)*(-4*I*sqrt(3)*a*d - 4*a*d)*((-I*A^3 - 3*A^2*B
+ 3*I*A*B^2 + B^3)/(a^2*d^3))^(1/3)*e^(2*I*d*x + 2*I*c)*log(1/2*(2*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)*(2*I*sqrt(3)*a*d + 2*a*d)*((-I*A^3 - 3*A^2*B + 3*I*A*B^2 + B^
3)/(a^2*d^3))^(1/3))/(I*A + B)) + (1/4)^(1/3)*(4*I*sqrt(3)*a*d - 4*a*d)*((-I*A^3 - 3*A^2*B + 3*I*A*B^2 + B^3)/
(a^2*d^3))^(1/3)*e^(2*I*d*x + 2*I*c)*log(1/2*(2*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)*(-2*I*sqrt(3)*a*d + 2*a*d)*((-I*A^3 - 3*A^2*B + 3*I*A*B^2 + B^3)/(a^2*d^3))^(1/3
))/(I*A + B)) + 2*2^(1/3)*((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))^(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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B \tan{\left (c + d x \right )}}{\left (a \left (i \tan{\left (c + d x \right )} + 1\right )\right )^{\frac{2}{3}}}\, dx \end{align*}

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))**(2/3),x)

[Out]

Integral((A + B*tan(c + d*x))/(a*(I*tan(c + d*x) + 1))**(2/3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \tan \left (d x + c\right ) + A}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{2}{3}}}\,{d x} \end{align*}

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))^(2/3),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)/(I*a*tan(d*x + c) + a)^(2/3), x)