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

Optimal. Leaf size=357 \[ \frac{\sqrt{3} (B+i A) \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt{3}}\right )}{2 d \sqrt [3]{a-i b}}-\frac{\sqrt{3} (-B+i A) \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt{3}}\right )}{2 d \sqrt [3]{a+i b}}+\frac{3 (B+i A) \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )}{4 d \sqrt [3]{a-i b}}-\frac{3 (-B+i A) \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )}{4 d \sqrt [3]{a+i b}}-\frac{(-B+i A) \log (\cos (c+d x))}{4 d \sqrt [3]{a+i b}}+\frac{(B+i A) \log (\cos (c+d x))}{4 d \sqrt [3]{a-i b}}-\frac{x (A-i B)}{4 \sqrt [3]{a-i b}}-\frac{x (A+i B)}{4 \sqrt [3]{a+i b}} \]

[Out]

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

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Rubi [A]  time = 0.278317, antiderivative size = 357, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3539, 3537, 55, 617, 204, 31} \[ \frac{\sqrt{3} (B+i A) \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt{3}}\right )}{2 d \sqrt [3]{a-i b}}-\frac{\sqrt{3} (-B+i A) \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt{3}}\right )}{2 d \sqrt [3]{a+i b}}+\frac{3 (B+i A) \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )}{4 d \sqrt [3]{a-i b}}-\frac{3 (-B+i A) \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )}{4 d \sqrt [3]{a+i b}}-\frac{(-B+i A) \log (\cos (c+d x))}{4 d \sqrt [3]{a+i b}}+\frac{(B+i A) \log (\cos (c+d x))}{4 d \sqrt [3]{a-i b}}-\frac{x (A-i B)}{4 \sqrt [3]{a-i b}}-\frac{x (A+i B)}{4 \sqrt [3]{a+i b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3539

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c
+ I*d)/2, Int[(a + b*Tan[e + f*x])^m*(1 - I*Tan[e + f*x]), x], x] + Dist[(c - I*d)/2, Int[(a + b*Tan[e + f*x]
)^m*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0]
&& NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]

Rule 3537

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^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+b \tan (c+d x)}} \, dx &=\frac{1}{2} (A-i B) \int \frac{1+i \tan (c+d x)}{\sqrt [3]{a+b \tan (c+d x)}} \, dx+\frac{1}{2} (A+i B) \int \frac{1-i \tan (c+d x)}{\sqrt [3]{a+b \tan (c+d x)}} \, dx\\ &=-\frac{(i A-B) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt [3]{a+i b x}} \, dx,x,-i \tan (c+d x)\right )}{2 d}+\frac{(i A+B) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt [3]{a-i b x}} \, dx,x,i \tan (c+d x)\right )}{2 d}\\ &=-\frac{(A-i B) x}{4 \sqrt [3]{a-i b}}-\frac{(A+i B) x}{4 \sqrt [3]{a+i b}}-\frac{(i A-B) \log (\cos (c+d x))}{4 \sqrt [3]{a+i b} d}+\frac{(i A+B) \log (\cos (c+d x))}{4 \sqrt [3]{a-i b} d}-\frac{(3 (i A-B)) \operatorname{Subst}\left (\int \frac{1}{(a+i b)^{2/3}+\sqrt [3]{a+i b} x+x^2} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}+\frac{(3 (i A-B)) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a+i b}-x} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt [3]{a+i b} d}+\frac{(3 (i A+B)) \operatorname{Subst}\left (\int \frac{1}{(a-i b)^{2/3}+\sqrt [3]{a-i b} x+x^2} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}-\frac{(3 (i A+B)) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a-i b}-x} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt [3]{a-i b} d}\\ &=-\frac{(A-i B) x}{4 \sqrt [3]{a-i b}}-\frac{(A+i B) x}{4 \sqrt [3]{a+i b}}-\frac{(i A-B) \log (\cos (c+d x))}{4 \sqrt [3]{a+i b} d}+\frac{(i A+B) \log (\cos (c+d x))}{4 \sqrt [3]{a-i b} d}+\frac{3 (i A+B) \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt [3]{a-i b} d}-\frac{3 (i A-B) \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt [3]{a+i b} d}+\frac{(3 (i A-B)) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}\right )}{2 \sqrt [3]{a+i b} d}-\frac{(3 (i A+B)) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}\right )}{2 \sqrt [3]{a-i b} d}\\ &=-\frac{(A-i B) x}{4 \sqrt [3]{a-i b}}-\frac{(A+i B) x}{4 \sqrt [3]{a+i b}}+\frac{\sqrt{3} (i A+B) \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt{3}}\right )}{2 \sqrt [3]{a-i b} d}-\frac{\sqrt{3} (i A-B) \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt{3}}\right )}{2 \sqrt [3]{a+i b} d}-\frac{(i A-B) \log (\cos (c+d x))}{4 \sqrt [3]{a+i b} d}+\frac{(i A+B) \log (\cos (c+d x))}{4 \sqrt [3]{a-i b} d}+\frac{3 (i A+B) \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt [3]{a-i b} d}-\frac{3 (i A-B) \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt [3]{a+i b} d}\\ \end{align*}

Mathematica [A]  time = 0.433142, size = 227, normalized size = 0.64 \[ \frac{i \left (\frac{(A-i B) \left (2 \sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt{3}}\right )+3 \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )-\log (\tan (c+d x)+i)\right )}{\sqrt [3]{a-i b}}-\frac{(A+i B) \left (2 \sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt{3}}\right )+3 \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )-\log (-\tan (c+d x)+i)\right )}{\sqrt [3]{a+i b}}\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/4)*(((A - I*B)*(2*Sqrt[3]*ArcTan[(1 + (2*(a + b*Tan[c + d*x])^(1/3))/(a - I*b)^(1/3))/Sqrt[3]] - Log[I + T
an[c + d*x]] + 3*Log[(a - I*b)^(1/3) - (a + b*Tan[c + d*x])^(1/3)]))/(a - I*b)^(1/3) - ((A + I*B)*(2*Sqrt[3]*A
rcTan[(1 + (2*(a + b*Tan[c + d*x])^(1/3))/(a + I*b)^(1/3))/Sqrt[3]] - Log[I - Tan[c + d*x]] + 3*Log[(a + I*b)^
(1/3) - (a + b*Tan[c + d*x])^(1/3)]))/(a + I*b)^(1/3)))/d

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Maple [C]  time = 0.075, size = 72, normalized size = 0.2 \begin{align*}{\frac{1}{2\,d}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{6}-2\,{{\it \_Z}}^{3}a+{a}^{2}+{b}^{2} \right ) }{\frac{B{{\it \_R}}^{4}+ \left ( Ab-aB \right ){\it \_R}}{{{\it \_R}}^{5}-{{\it \_R}}^{2}a}\ln \left ( \sqrt [3]{a+b\tan \left ( dx+c \right ) }-{\it \_R} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*sum((B*_R^4+(A*b-B*a)*_R)/(_R^5-_R^2*a)*ln((a+b*tan(d*x+c))^(1/3)-_R),_R=RootOf(_Z^6-2*_Z^3*a+a^2+b^2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \tan \left (d x + c\right ) + A}{{\left (b \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+b*tan(d*x+c))^(1/3),x, algorithm="maxima")

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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 )}}{\sqrt [3]{a + b \tan{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 10.1553, size = 752, normalized size = 2.11 \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+b*tan(d*x+c))^(1/3),x, algorithm="giac")

[Out]

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