3.696 \(\int \frac{1}{x^{4/3} (a+b x)^3} \, dx\)

Optimal. Leaf size=152 \[ \frac{7}{6 a^2 \sqrt [3]{x} (a+b x)}+\frac{7 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{3 a^{10/3}}-\frac{7 \sqrt [3]{b} \log (a+b x)}{9 a^{10/3}}+\frac{14 \sqrt [3]{b} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{10/3}}-\frac{14}{3 a^3 \sqrt [3]{x}}+\frac{1}{2 a \sqrt [3]{x} (a+b x)^2} \]

[Out]

-14/(3*a^3*x^(1/3)) + 1/(2*a*x^(1/3)*(a + b*x)^2) + 7/(6*a^2*x^(1/3)*(a + b*x)) + (14*b^(1/3)*ArcTan[(a^(1/3)
- 2*b^(1/3)*x^(1/3))/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(10/3)) + (7*b^(1/3)*Log[a^(1/3) + b^(1/3)*x^(1/3)])/(3*
a^(10/3)) - (7*b^(1/3)*Log[a + b*x])/(9*a^(10/3))

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Rubi [A]  time = 0.0596014, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {51, 56, 617, 204, 31} \[ \frac{7}{6 a^2 \sqrt [3]{x} (a+b x)}+\frac{7 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{3 a^{10/3}}-\frac{7 \sqrt [3]{b} \log (a+b x)}{9 a^{10/3}}+\frac{14 \sqrt [3]{b} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{10/3}}-\frac{14}{3 a^3 \sqrt [3]{x}}+\frac{1}{2 a \sqrt [3]{x} (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(4/3)*(a + b*x)^3),x]

[Out]

-14/(3*a^3*x^(1/3)) + 1/(2*a*x^(1/3)*(a + b*x)^2) + 7/(6*a^2*x^(1/3)*(a + b*x)) + (14*b^(1/3)*ArcTan[(a^(1/3)
- 2*b^(1/3)*x^(1/3))/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(10/3)) + (7*b^(1/3)*Log[a^(1/3) + b^(1/3)*x^(1/3)])/(3*
a^(10/3)) - (7*b^(1/3)*Log[a + b*x])/(9*a^(10/3))

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 56

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, Simp
[Log[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] && Ne
gQ[(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}{x^{4/3} (a+b x)^3} \, dx &=\frac{1}{2 a \sqrt [3]{x} (a+b x)^2}+\frac{7 \int \frac{1}{x^{4/3} (a+b x)^2} \, dx}{6 a}\\ &=\frac{1}{2 a \sqrt [3]{x} (a+b x)^2}+\frac{7}{6 a^2 \sqrt [3]{x} (a+b x)}+\frac{14 \int \frac{1}{x^{4/3} (a+b x)} \, dx}{9 a^2}\\ &=-\frac{14}{3 a^3 \sqrt [3]{x}}+\frac{1}{2 a \sqrt [3]{x} (a+b x)^2}+\frac{7}{6 a^2 \sqrt [3]{x} (a+b x)}-\frac{(14 b) \int \frac{1}{\sqrt [3]{x} (a+b x)} \, dx}{9 a^3}\\ &=-\frac{14}{3 a^3 \sqrt [3]{x}}+\frac{1}{2 a \sqrt [3]{x} (a+b x)^2}+\frac{7}{6 a^2 \sqrt [3]{x} (a+b x)}-\frac{7 \sqrt [3]{b} \log (a+b x)}{9 a^{10/3}}-\frac{7 \operatorname{Subst}\left (\int \frac{1}{\frac{a^{2/3}}{b^{2/3}}-\frac{\sqrt [3]{a} x}{\sqrt [3]{b}}+x^2} \, dx,x,\sqrt [3]{x}\right )}{3 a^3}+\frac{\left (7 \sqrt [3]{b}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt [3]{a}}{\sqrt [3]{b}}+x} \, dx,x,\sqrt [3]{x}\right )}{3 a^{10/3}}\\ &=-\frac{14}{3 a^3 \sqrt [3]{x}}+\frac{1}{2 a \sqrt [3]{x} (a+b x)^2}+\frac{7}{6 a^2 \sqrt [3]{x} (a+b x)}+\frac{7 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{3 a^{10/3}}-\frac{7 \sqrt [3]{b} \log (a+b x)}{9 a^{10/3}}-\frac{\left (14 \sqrt [3]{b}\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{a}}\right )}{3 a^{10/3}}\\ &=-\frac{14}{3 a^3 \sqrt [3]{x}}+\frac{1}{2 a \sqrt [3]{x} (a+b x)^2}+\frac{7}{6 a^2 \sqrt [3]{x} (a+b x)}+\frac{14 \sqrt [3]{b} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{3 \sqrt{3} a^{10/3}}+\frac{7 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{3 a^{10/3}}-\frac{7 \sqrt [3]{b} \log (a+b x)}{9 a^{10/3}}\\ \end{align*}

Mathematica [C]  time = 0.0052801, size = 25, normalized size = 0.16 \[ -\frac{3 \, _2F_1\left (-\frac{1}{3},3;\frac{2}{3};-\frac{b x}{a}\right )}{a^3 \sqrt [3]{x}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(4/3)*(a + b*x)^3),x]

[Out]

(-3*Hypergeometric2F1[-1/3, 3, 2/3, -((b*x)/a)])/(a^3*x^(1/3))

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Maple [A]  time = 0.013, size = 139, normalized size = 0.9 \begin{align*} -3\,{\frac{1}{{a}^{3}\sqrt [3]{x}}}-{\frac{5\,{b}^{2}}{3\,{a}^{3} \left ( bx+a \right ) ^{2}}{x}^{{\frac{5}{3}}}}-{\frac{13\,b}{6\,{a}^{2} \left ( bx+a \right ) ^{2}}{x}^{{\frac{2}{3}}}}+{\frac{14}{9\,{a}^{3}}\ln \left ( \sqrt [3]{x}+\sqrt [3]{{\frac{a}{b}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-{\frac{7}{9\,{a}^{3}}\ln \left ({x}^{{\frac{2}{3}}}-\sqrt [3]{{\frac{a}{b}}}\sqrt [3]{x}+ \left ({\frac{a}{b}} \right ) ^{{\frac{2}{3}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-{\frac{14\,\sqrt{3}}{9\,{a}^{3}}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\sqrt [3]{x}{\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-1 \right ) } \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(4/3)/(b*x+a)^3,x)

[Out]

-3/a^3/x^(1/3)-5/3*b^2/a^3/(b*x+a)^2*x^(5/3)-13/6*b/a^2/(b*x+a)^2*x^(2/3)+14/9/a^3/(1/b*a)^(1/3)*ln(x^(1/3)+(1
/b*a)^(1/3))-7/9/a^3/(1/b*a)^(1/3)*ln(x^(2/3)-(1/b*a)^(1/3)*x^(1/3)+(1/b*a)^(2/3))-14/9/a^3*3^(1/2)/(1/b*a)^(1
/3)*arctan(1/3*3^(1/2)*(2/(1/b*a)^(1/3)*x^(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/x^(4/3)/(b*x+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.7421, size = 510, normalized size = 3.36 \begin{align*} -\frac{28 \, \sqrt{3}{\left (b^{2} x^{3} + 2 \, a b x^{2} + a^{2} x\right )} \left (\frac{b}{a}\right )^{\frac{1}{3}} \arctan \left (\frac{2}{3} \, \sqrt{3} x^{\frac{1}{3}} \left (\frac{b}{a}\right )^{\frac{1}{3}} - \frac{1}{3} \, \sqrt{3}\right ) + 14 \,{\left (b^{2} x^{3} + 2 \, a b x^{2} + a^{2} x\right )} \left (\frac{b}{a}\right )^{\frac{1}{3}} \log \left (-a x^{\frac{1}{3}} \left (\frac{b}{a}\right )^{\frac{2}{3}} + b x^{\frac{2}{3}} + a \left (\frac{b}{a}\right )^{\frac{1}{3}}\right ) - 28 \,{\left (b^{2} x^{3} + 2 \, a b x^{2} + a^{2} x\right )} \left (\frac{b}{a}\right )^{\frac{1}{3}} \log \left (a \left (\frac{b}{a}\right )^{\frac{2}{3}} + b x^{\frac{1}{3}}\right ) + 3 \,{\left (28 \, b^{2} x^{2} + 49 \, a b x + 18 \, a^{2}\right )} x^{\frac{2}{3}}}{18 \,{\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(4/3)/(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/18*(28*sqrt(3)*(b^2*x^3 + 2*a*b*x^2 + a^2*x)*(b/a)^(1/3)*arctan(2/3*sqrt(3)*x^(1/3)*(b/a)^(1/3) - 1/3*sqrt(
3)) + 14*(b^2*x^3 + 2*a*b*x^2 + a^2*x)*(b/a)^(1/3)*log(-a*x^(1/3)*(b/a)^(2/3) + b*x^(2/3) + a*(b/a)^(1/3)) - 2
8*(b^2*x^3 + 2*a*b*x^2 + a^2*x)*(b/a)^(1/3)*log(a*(b/a)^(2/3) + b*x^(1/3)) + 3*(28*b^2*x^2 + 49*a*b*x + 18*a^2
)*x^(2/3))/(a^3*b^2*x^3 + 2*a^4*b*x^2 + a^5*x)

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Sympy [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(1/x**(4/3)/(b*x+a)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.08608, size = 209, normalized size = 1.38 \begin{align*} \frac{14 \, b \left (-\frac{a}{b}\right )^{\frac{2}{3}} \log \left ({\left | x^{\frac{1}{3}} - \left (-\frac{a}{b}\right )^{\frac{1}{3}} \right |}\right )}{9 \, a^{4}} + \frac{14 \, \sqrt{3} \left (-a b^{2}\right )^{\frac{2}{3}} \arctan \left (\frac{\sqrt{3}{\left (2 \, x^{\frac{1}{3}} + \left (-\frac{a}{b}\right )^{\frac{1}{3}}\right )}}{3 \, \left (-\frac{a}{b}\right )^{\frac{1}{3}}}\right )}{9 \, a^{4} b} - \frac{3}{a^{3} x^{\frac{1}{3}}} - \frac{7 \, \left (-a b^{2}\right )^{\frac{2}{3}} \log \left (x^{\frac{2}{3}} + x^{\frac{1}{3}} \left (-\frac{a}{b}\right )^{\frac{1}{3}} + \left (-\frac{a}{b}\right )^{\frac{2}{3}}\right )}{9 \, a^{4} b} - \frac{10 \, b^{2} x^{\frac{5}{3}} + 13 \, a b x^{\frac{2}{3}}}{6 \,{\left (b x + a\right )}^{2} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(4/3)/(b*x+a)^3,x, algorithm="giac")

[Out]

14/9*b*(-a/b)^(2/3)*log(abs(x^(1/3) - (-a/b)^(1/3)))/a^4 + 14/9*sqrt(3)*(-a*b^2)^(2/3)*arctan(1/3*sqrt(3)*(2*x
^(1/3) + (-a/b)^(1/3))/(-a/b)^(1/3))/(a^4*b) - 3/(a^3*x^(1/3)) - 7/9*(-a*b^2)^(2/3)*log(x^(2/3) + x^(1/3)*(-a/
b)^(1/3) + (-a/b)^(2/3))/(a^4*b) - 1/6*(10*b^2*x^(5/3) + 13*a*b*x^(2/3))/((b*x + a)^2*a^3)