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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.158228, 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, 55, 617, 204, 31} \[ \frac{\sqrt{3} (B+i A) \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} d}+\frac{3 (-B+i A)}{2 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac{3 (B+i A) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac{(B+i A) \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac{x (A-i B)}{4 \sqrt [3]{2} \sqrt [3]{a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx &=\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) \operatorname{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)) \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 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 \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)) \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{(A-i B) x}{4 \sqrt [3]{2} \sqrt [3]{a}}+\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 \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 [C]  time = 1.07252, size = 137, normalized size = 0.64 \[ -\frac{3 i e^{-2 i (c+d x)} \left (\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{2/3} \left ((A-i B) e^{2 i (c+d x)} \text{Hypergeometric2F1}\left (\frac{2}{3},1,\frac{5}{3},\frac{e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )-2 (A+i B) \left (1+e^{2 i (c+d x)}\right )\right )}{4 \sqrt [3]{2} a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.02, size = 318, normalized size = 1.5 \begin{align*}{\frac{{2}^{{\frac{2}{3}}}B}{4\,d}\ln \left ( \sqrt [3]{a+ia\tan \left ( dx+c \right ) }-\sqrt [3]{2}\sqrt [3]{a} \right ){\frac{1}{\sqrt [3]{a}}}}+{\frac{{\frac{i}{4}}{2}^{{\frac{2}{3}}}A}{d}\ln \left ( \sqrt [3]{a+ia\tan \left ( dx+c \right ) }-\sqrt [3]{2}\sqrt [3]{a} \right ){\frac{1}{\sqrt [3]{a}}}}-{\frac{{2}^{{\frac{2}{3}}}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 ){\frac{1}{\sqrt [3]{a}}}}-{\frac{{\frac{i}{8}}{2}^{{\frac{2}{3}}}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 ){\frac{1}{\sqrt [3]{a}}}}+{\frac{\sqrt{3}{2}^{{\frac{2}{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 ){\frac{1}{\sqrt [3]{a}}}}+{\frac{{\frac{i}{4}}\sqrt{3}{2}^{{\frac{2}{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 ){\frac{1}{\sqrt [3]{a}}}}-{\frac{3\,B}{2\,d}{\frac{1}{\sqrt [3]{a+ia\tan \left ( dx+c \right ) }}}}+{\frac{{\frac{3\,i}{2}}A}{d}{\frac{1}{\sqrt [3]{a+ia\tan \left ( dx+c \right ) }}}} \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))^(1/3),x)

[Out]

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

________________________________________________________________________________________

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))^(1/3),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 1.77855, size = 1503, normalized size = 7.06 \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))^(1/3),x, algorithm="fricas")

[Out]

1/8*(4*(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)*(2*I*sqrt(3)*a*d - 2*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(1/4*(4*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)*(4*I*sqrt(3)*a*d^2 + 4*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)*(-2*I*sqrt(3)*a*d - 2
*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(1/4*(4*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)*(-4*I*sqrt(3)*a*d^2 + 4*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)) + 2*2^(2/3)*((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))^(2/3)*e^(4/3*I*d*x + 4/3*I*c))*e^(-2*I*d*x - 2
*I*c)/(a*d)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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{1}{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))^(1/3),x, algorithm="giac")

[Out]

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