3.1459 \(\int \frac{1}{(a+b x) (c+d x)^{2/3}} \, dx\)

Optimal. Leaf size=140 \[ -\frac{\log (a+b x)}{2 \sqrt [3]{b} (b c-a d)^{2/3}}+\frac{3 \log \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{2 \sqrt [3]{b} (b c-a d)^{2/3}}-\frac{\sqrt{3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}}+1}{\sqrt{3}}\right )}{\sqrt [3]{b} (b c-a d)^{2/3}} \]

[Out]

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

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Rubi [A]  time = 0.0744044, antiderivative size = 140, 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 = {57, 617, 204, 31} \[ -\frac{\log (a+b x)}{2 \sqrt [3]{b} (b c-a d)^{2/3}}+\frac{3 \log \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{2 \sqrt [3]{b} (b c-a d)^{2/3}}-\frac{\sqrt{3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}}+1}{\sqrt{3}}\right )}{\sqrt [3]{b} (b c-a d)^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)*(c + d*x)^(2/3)),x]

[Out]

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

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

Mathematica [A]  time = 0.0702584, size = 154, normalized size = 1.1 \[ -\frac{\log \left (\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}\right )-2 \log \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )+2 \sqrt{3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}}+1}{\sqrt{3}}\right )}{2 \sqrt [3]{b} (b c-a d)^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)*(c + d*x)^(2/3)),x]

[Out]

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

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Maple [A]  time = 0.005, size = 160, normalized size = 1.1 \begin{align*}{\frac{1}{b}\ln \left ( \sqrt [3]{dx+c}+\sqrt [3]{{\frac{ad-bc}{b}}} \right ) \left ({\frac{ad-bc}{b}} \right ) ^{-{\frac{2}{3}}}}-{\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 ) \left ({\frac{ad-bc}{b}} \right ) ^{-{\frac{2}{3}}}}+{\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 ) \left ({\frac{ad-bc}{b}} \right ) ^{-{\frac{2}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)/(d*x+c)^(2/3),x)

[Out]

1/b/((a*d-b*c)/b)^(2/3)*ln((d*x+c)^(1/3)+((a*d-b*c)/b)^(1/3))-1/2/b/((a*d-b*c)/b)^(2/3)*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))+1/b/((a*d-b*c)/b)^(2/3)*3^(1/2)*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)^(2/3),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.4266, size = 1995, normalized size = 14.25 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(d*x+c)^(2/3),x, algorithm="fricas")

[Out]

[-1/2*(sqrt(3)*(b^2*c - a*b*d)*sqrt(-(b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)^(1/3)/b)*log(-(3*b^2*c^2 - 4*a*b*c*d
+ a^2*d^2 + 2*(b^2*c*d - a*b*d^2)*x + sqrt(3)*(2*(b^2*c - a*b*d)*(d*x + c)^(2/3) - (b^3*c^2 - 2*a*b^2*c*d + a^
2*b*d^2)^(1/3)*(b*c - a*d) - (b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)^(2/3)*(d*x + c)^(1/3))*sqrt(-(b^3*c^2 - 2*a*b
^2*c*d + a^2*b*d^2)^(1/3)/b) - 3*(b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)^(1/3)*(b*c - a*d)*(d*x + c)^(1/3))/(b*x +
a)) + (b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)^(2/3)*log(-(b^2*c - a*b*d)*(d*x + c)^(2/3) - (b^3*c^2 - 2*a*b^2*c*d
+ a^2*b*d^2)^(1/3)*(b*c - a*d) - (b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)^(2/3)*(d*x + c)^(1/3)) - 2*(b^3*c^2 - 2*
a*b^2*c*d + a^2*b*d^2)^(2/3)*log(-(b^2*c - a*b*d)*(d*x + c)^(1/3) + (b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)^(2/3))
)/(b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2), -1/2*(2*sqrt(3)*(b^2*c - a*b*d)*sqrt((b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2
)^(1/3)/b)*arctan(1/3*sqrt(3)*((b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)^(1/3)*(b*c - a*d) + 2*(b^3*c^2 - 2*a*b^2*c*
d + a^2*b*d^2)^(2/3)*(d*x + c)^(1/3))*sqrt((b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)^(1/3)/b)/(b^2*c^2 - 2*a*b*c*d +
a^2*d^2)) + (b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)^(2/3)*log(-(b^2*c - a*b*d)*(d*x + c)^(2/3) - (b^3*c^2 - 2*a*b
^2*c*d + a^2*b*d^2)^(1/3)*(b*c - a*d) - (b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)^(2/3)*(d*x + c)^(1/3)) - 2*(b^3*c^
2 - 2*a*b^2*c*d + a^2*b*d^2)^(2/3)*log(-(b^2*c - a*b*d)*(d*x + c)^(1/3) + (b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)^
(2/3)))/(b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b x\right ) \left (c + d x\right )^{\frac{2}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(d*x+c)**(2/3),x)

[Out]

Integral(1/((a + b*x)*(c + d*x)**(2/3)), x)

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Giac [A]  time = 1.12254, size = 279, normalized size = 1.99 \begin{align*} -\frac{3 \,{\left (b^{3} c - a b^{2} d\right )}^{\frac{1}{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^{2} c - \sqrt{3} a b d} - \frac{{\left (b^{3} c - a b^{2} d\right )}^{\frac{1}{3}} \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^{2} c - a b d\right )}} + \frac{\left (\frac{b c - a d}{b}\right )^{\frac{1}{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)^(2/3),x, algorithm="giac")

[Out]

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