3.1458 \(\int \frac{1}{(a+b x) \sqrt [3]{c+d x}} \, dx\)

Optimal. Leaf size=139 \[ -\frac{\log (a+b x)}{2 b^{2/3} \sqrt [3]{b c-a d}}+\frac{3 \log \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{2 b^{2/3} \sqrt [3]{b c-a d}}+\frac{\sqrt{3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}}+1}{\sqrt{3}}\right )}{b^{2/3} \sqrt [3]{b c-a d}} \]

[Out]

(Sqrt[3]*ArcTan[(1 + (2*b^(1/3)*(c + d*x)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]])/(b^(2/3)*(b*c - a*d)^(1/3)) - Lo
g[a + b*x]/(2*b^(2/3)*(b*c - a*d)^(1/3)) + (3*Log[(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3)])/(2*b^(2/3)*(b*
c - a*d)^(1/3))

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Rubi [A]  time = 0.108445, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {55, 617, 204, 31} \[ -\frac{\log (a+b x)}{2 b^{2/3} \sqrt [3]{b c-a d}}+\frac{3 \log \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{2 b^{2/3} \sqrt [3]{b c-a d}}+\frac{\sqrt{3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}}+1}{\sqrt{3}}\right )}{b^{2/3} \sqrt [3]{b c-a d}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)*(c + d*x)^(1/3)),x]

[Out]

(Sqrt[3]*ArcTan[(1 + (2*b^(1/3)*(c + d*x)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]])/(b^(2/3)*(b*c - a*d)^(1/3)) - Lo
g[a + b*x]/(2*b^(2/3)*(b*c - a*d)^(1/3)) + (3*Log[(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3)])/(2*b^(2/3)*(b*
c - a*d)^(1/3))

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

Mathematica [A]  time = 0.0726313, size = 106, normalized size = 0.76 \[ \frac{3 \log \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )+2 \sqrt{3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}}+1}{\sqrt{3}}\right )-\log (a+b x)}{2 b^{2/3} \sqrt [3]{b c-a d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[3]*ArcTan[(1 + (2*b^(1/3)*(c + d*x)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]] - Log[a + b*x] + 3*Log[(b*c - a
*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3)])/(2*b^(2/3)*(b*c - a*d)^(1/3))

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Maple [A]  time = 0.008, size = 161, normalized size = 1.2 \begin{align*} -{\frac{1}{b}\ln \left ( \sqrt [3]{dx+c}+\sqrt [3]{{\frac{ad-bc}{b}}} \right ){\frac{1}{\sqrt [3]{{\frac{ad-bc}{b}}}}}}+{\frac{1}{2\,b}\ln \left ( \left ( dx+c \right ) ^{{\frac{2}{3}}}-\sqrt [3]{{\frac{ad-bc}{b}}}\sqrt [3]{dx+c}+ \left ({\frac{ad-bc}{b}} \right ) ^{{\frac{2}{3}}} \right ){\frac{1}{\sqrt [3]{{\frac{ad-bc}{b}}}}}}+{\frac{\sqrt{3}}{b}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\sqrt [3]{dx+c}{\frac{1}{\sqrt [3]{{\frac{ad-bc}{b}}}}}}-1 \right ) } \right ){\frac{1}{\sqrt [3]{{\frac{ad-bc}{b}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/b/((a*d-b*c)/b)^(1/3)*ln((d*x+c)^(1/3)+((a*d-b*c)/b)^(1/3))+1/2/b/((a*d-b*c)/b)^(1/3)*ln((d*x+c)^(2/3)-((a*
d-b*c)/b)^(1/3)*(d*x+c)^(1/3)+((a*d-b*c)/b)^(2/3))+3^(1/2)/b/((a*d-b*c)/b)^(1/3)*arctan(1/3*3^(1/2)*(2/((a*d-b
*c)/b)^(1/3)*(d*x+c)^(1/3)-1))

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.28262, size = 1289, normalized size = 9.27 \begin{align*} \left [\frac{\sqrt{3}{\left (b^{2} c - a b d\right )} \sqrt{-\frac{{\left (b^{3} c - a b^{2} d\right )}^{\frac{1}{3}}}{b c - a d}} \log \left (\frac{2 \, b^{2} d x + 3 \, b^{2} c - a b d - \sqrt{3}{\left ({\left (b^{3} c - a b^{2} d\right )}^{\frac{1}{3}}{\left (b c - a d\right )} +{\left (b^{2} c - a b d\right )}{\left (d x + c\right )}^{\frac{1}{3}} - 2 \,{\left (b^{3} c - a b^{2} d\right )}^{\frac{2}{3}}{\left (d x + c\right )}^{\frac{2}{3}}\right )} \sqrt{-\frac{{\left (b^{3} c - a b^{2} d\right )}^{\frac{1}{3}}}{b c - a d}} - 3 \,{\left (b^{3} c - a b^{2} d\right )}^{\frac{2}{3}}{\left (d x + c\right )}^{\frac{1}{3}}}{b x + a}\right ) -{\left (b^{3} c - a b^{2} d\right )}^{\frac{2}{3}} \log \left ({\left (d x + c\right )}^{\frac{2}{3}} b^{2} +{\left (b^{3} c - a b^{2} d\right )}^{\frac{1}{3}}{\left (d x + c\right )}^{\frac{1}{3}} b +{\left (b^{3} c - a b^{2} d\right )}^{\frac{2}{3}}\right ) + 2 \,{\left (b^{3} c - a b^{2} d\right )}^{\frac{2}{3}} \log \left ({\left (d x + c\right )}^{\frac{1}{3}} b -{\left (b^{3} c - a b^{2} d\right )}^{\frac{1}{3}}\right )}{2 \,{\left (b^{3} c - a b^{2} d\right )}}, \frac{2 \, \sqrt{3}{\left (b^{2} c - a b d\right )} \sqrt{\frac{{\left (b^{3} c - a b^{2} d\right )}^{\frac{1}{3}}}{b c - a d}} \arctan \left (\frac{\sqrt{3}{\left (2 \,{\left (d x + c\right )}^{\frac{1}{3}} b +{\left (b^{3} c - a b^{2} d\right )}^{\frac{1}{3}}\right )} \sqrt{\frac{{\left (b^{3} c - a b^{2} d\right )}^{\frac{1}{3}}}{b c - a d}}}{3 \, b}\right ) -{\left (b^{3} c - a b^{2} d\right )}^{\frac{2}{3}} \log \left ({\left (d x + c\right )}^{\frac{2}{3}} b^{2} +{\left (b^{3} c - a b^{2} d\right )}^{\frac{1}{3}}{\left (d x + c\right )}^{\frac{1}{3}} b +{\left (b^{3} c - a b^{2} d\right )}^{\frac{2}{3}}\right ) + 2 \,{\left (b^{3} c - a b^{2} d\right )}^{\frac{2}{3}} \log \left ({\left (d x + c\right )}^{\frac{1}{3}} b -{\left (b^{3} c - a b^{2} d\right )}^{\frac{1}{3}}\right )}{2 \,{\left (b^{3} c - a b^{2} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(3)*(b^2*c - a*b*d)*sqrt(-(b^3*c - a*b^2*d)^(1/3)/(b*c - a*d))*log((2*b^2*d*x + 3*b^2*c - a*b*d - sq
rt(3)*((b^3*c - a*b^2*d)^(1/3)*(b*c - a*d) + (b^2*c - a*b*d)*(d*x + c)^(1/3) - 2*(b^3*c - a*b^2*d)^(2/3)*(d*x
+ c)^(2/3))*sqrt(-(b^3*c - a*b^2*d)^(1/3)/(b*c - a*d)) - 3*(b^3*c - a*b^2*d)^(2/3)*(d*x + c)^(1/3))/(b*x + a))
- (b^3*c - a*b^2*d)^(2/3)*log((d*x + c)^(2/3)*b^2 + (b^3*c - a*b^2*d)^(1/3)*(d*x + c)^(1/3)*b + (b^3*c - a*b^
2*d)^(2/3)) + 2*(b^3*c - a*b^2*d)^(2/3)*log((d*x + c)^(1/3)*b - (b^3*c - a*b^2*d)^(1/3)))/(b^3*c - a*b^2*d), 1
/2*(2*sqrt(3)*(b^2*c - a*b*d)*sqrt((b^3*c - a*b^2*d)^(1/3)/(b*c - a*d))*arctan(1/3*sqrt(3)*(2*(d*x + c)^(1/3)*
b + (b^3*c - a*b^2*d)^(1/3))*sqrt((b^3*c - a*b^2*d)^(1/3)/(b*c - a*d))/b) - (b^3*c - a*b^2*d)^(2/3)*log((d*x +
c)^(2/3)*b^2 + (b^3*c - a*b^2*d)^(1/3)*(d*x + c)^(1/3)*b + (b^3*c - a*b^2*d)^(2/3)) + 2*(b^3*c - a*b^2*d)^(2/
3)*log((d*x + c)^(1/3)*b - (b^3*c - a*b^2*d)^(1/3)))/(b^3*c - a*b^2*d)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.11485, size = 265, normalized size = 1.91 \begin{align*} \frac{3 \,{\left (b^{3} c - a b^{2} d\right )}^{\frac{2}{3}} \arctan \left (\frac{\sqrt{3}{\left (2 \,{\left (d x + c\right )}^{\frac{1}{3}} + \left (\frac{b c - a d}{b}\right )^{\frac{1}{3}}\right )}}{3 \, \left (\frac{b c - a d}{b}\right )^{\frac{1}{3}}}\right )}{\sqrt{3} b^{3} c - \sqrt{3} a b^{2} d} - \frac{\log \left ({\left (d x + c\right )}^{\frac{2}{3}} +{\left (d x + c\right )}^{\frac{1}{3}} \left (\frac{b c - a d}{b}\right )^{\frac{1}{3}} + \left (\frac{b c - a d}{b}\right )^{\frac{2}{3}}\right )}{2 \,{\left (b^{3} c - a b^{2} d\right )}^{\frac{1}{3}}} + \frac{\left (\frac{b c - a d}{b}\right )^{\frac{2}{3}} \log \left ({\left |{\left (d x + c\right )}^{\frac{1}{3}} - \left (\frac{b c - a d}{b}\right )^{\frac{1}{3}} \right |}\right )}{b c - a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*(b^3*c - a*b^2*d)^(2/3)*arctan(1/3*sqrt(3)*(2*(d*x + c)^(1/3) + ((b*c - a*d)/b)^(1/3))/((b*c - a*d)/b)^(1/3)
)/(sqrt(3)*b^3*c - sqrt(3)*a*b^2*d) - 1/2*log((d*x + c)^(2/3) + (d*x + c)^(1/3)*((b*c - a*d)/b)^(1/3) + ((b*c
- a*d)/b)^(2/3))/(b^3*c - a*b^2*d)^(1/3) + ((b*c - a*d)/b)^(2/3)*log(abs((d*x + c)^(1/3) - ((b*c - a*d)/b)^(1/
3)))/(b*c - a*d)