3.572 \(\int \frac{\sqrt [3]{a+b x^3}}{x^4 (a d-b d x^3)} \, dx\)

Optimal. Leaf size=268 \[ -\frac{\left (a+b x^3\right )^{4/3}}{3 a^2 d x^3}+\frac{b \sqrt [3]{a+b x^3}}{3 a^2 d}+\frac{b \log \left (a-b x^3\right )}{3\ 2^{2/3} a^{5/3} d}+\frac{2 b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{3 a^{5/3} d}-\frac{b \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2^{2/3} a^{5/3} d}-\frac{4 b \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{5/3} d}+\frac{\sqrt [3]{2} b \tan ^{-1}\left (\frac{2^{2/3} \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} a^{5/3} d}-\frac{2 b \log (x)}{3 a^{5/3} d} \]

[Out]

(b*(a + b*x^3)^(1/3))/(3*a^2*d) - (a + b*x^3)^(4/3)/(3*a^2*d*x^3) - (4*b*ArcTan[(a^(1/3) + 2*(a + b*x^3)^(1/3)
)/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(5/3)*d) + (2^(1/3)*b*ArcTan[(a^(1/3) + 2^(2/3)*(a + b*x^3)^(1/3))/(Sqrt[3]
*a^(1/3))])/(Sqrt[3]*a^(5/3)*d) - (2*b*Log[x])/(3*a^(5/3)*d) + (b*Log[a - b*x^3])/(3*2^(2/3)*a^(5/3)*d) + (2*b
*Log[a^(1/3) - (a + b*x^3)^(1/3)])/(3*a^(5/3)*d) - (b*Log[2^(1/3)*a^(1/3) - (a + b*x^3)^(1/3)])/(2^(2/3)*a^(5/
3)*d)

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Rubi [A]  time = 0.247268, antiderivative size = 268, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {446, 103, 156, 50, 57, 617, 204, 31} \[ -\frac{\left (a+b x^3\right )^{4/3}}{3 a^2 d x^3}+\frac{b \sqrt [3]{a+b x^3}}{3 a^2 d}+\frac{b \log \left (a-b x^3\right )}{3\ 2^{2/3} a^{5/3} d}+\frac{2 b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{3 a^{5/3} d}-\frac{b \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2^{2/3} a^{5/3} d}-\frac{4 b \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{5/3} d}+\frac{\sqrt [3]{2} b \tan ^{-1}\left (\frac{2^{2/3} \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} a^{5/3} d}-\frac{2 b \log (x)}{3 a^{5/3} d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(1/3)/(x^4*(a*d - b*d*x^3)),x]

[Out]

(b*(a + b*x^3)^(1/3))/(3*a^2*d) - (a + b*x^3)^(4/3)/(3*a^2*d*x^3) - (4*b*ArcTan[(a^(1/3) + 2*(a + b*x^3)^(1/3)
)/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(5/3)*d) + (2^(1/3)*b*ArcTan[(a^(1/3) + 2^(2/3)*(a + b*x^3)^(1/3))/(Sqrt[3]
*a^(1/3))])/(Sqrt[3]*a^(5/3)*d) - (2*b*Log[x])/(3*a^(5/3)*d) + (b*Log[a - b*x^3])/(3*2^(2/3)*a^(5/3)*d) + (2*b
*Log[a^(1/3) - (a + b*x^3)^(1/3)])/(3*a^(5/3)*d) - (b*Log[2^(1/3)*a^(1/3) - (a + b*x^3)^(1/3)])/(2^(2/3)*a^(5/
3)*d)

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 50

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

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{\sqrt [3]{a+b x^3}}{x^4 \left (a d-b d x^3\right )} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{\sqrt [3]{a+b x}}{x^2 (a d-b d x)} \, dx,x,x^3\right )\\ &=-\frac{\left (a+b x^3\right )^{4/3}}{3 a^2 d x^3}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt [3]{a+b x} \left (-\frac{4}{3} a b d+\frac{1}{3} b^2 d x\right )}{x (a d-b d x)} \, dx,x,x^3\right )}{3 a^2 d}\\ &=-\frac{\left (a+b x^3\right )^{4/3}}{3 a^2 d x^3}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\sqrt [3]{a+b x}}{a d-b d x} \, dx,x,x^3\right )}{3 a^2}+\frac{(4 b) \operatorname{Subst}\left (\int \frac{\sqrt [3]{a+b x}}{x} \, dx,x,x^3\right )}{9 a^2 d}\\ &=\frac{b \sqrt [3]{a+b x^3}}{3 a^2 d}-\frac{\left (a+b x^3\right )^{4/3}}{3 a^2 d x^3}+\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{(a+b x)^{2/3} (a d-b d x)} \, dx,x,x^3\right )}{3 a}+\frac{(4 b) \operatorname{Subst}\left (\int \frac{1}{x (a+b x)^{2/3}} \, dx,x,x^3\right )}{9 a d}\\ &=\frac{b \sqrt [3]{a+b x^3}}{3 a^2 d}-\frac{\left (a+b x^3\right )^{4/3}}{3 a^2 d x^3}-\frac{2 b \log (x)}{3 a^{5/3} d}+\frac{b \log \left (a-b x^3\right )}{3\ 2^{2/3} a^{5/3} d}-\frac{(2 b) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x^3}\right )}{3 a^{5/3} d}+\frac{b \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{2} \sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x^3}\right )}{2^{2/3} a^{5/3} d}-\frac{(2 b) \operatorname{Subst}\left (\int \frac{1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+b x^3}\right )}{3 a^{4/3} d}+\frac{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+b x^3}\right )}{\sqrt [3]{2} a^{4/3} d}\\ &=\frac{b \sqrt [3]{a+b x^3}}{3 a^2 d}-\frac{\left (a+b x^3\right )^{4/3}}{3 a^2 d x^3}-\frac{2 b \log (x)}{3 a^{5/3} d}+\frac{b \log \left (a-b x^3\right )}{3\ 2^{2/3} a^{5/3} d}+\frac{2 b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{3 a^{5/3} d}-\frac{b \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2^{2/3} a^{5/3} d}+\frac{(4 b) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}\right )}{3 a^{5/3} d}-\frac{\left (\sqrt [3]{2} b\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2^{2/3} \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}\right )}{a^{5/3} d}\\ &=\frac{b \sqrt [3]{a+b x^3}}{3 a^2 d}-\frac{\left (a+b x^3\right )^{4/3}}{3 a^2 d x^3}-\frac{4 b \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{3 \sqrt{3} a^{5/3} d}+\frac{\sqrt [3]{2} b \tan ^{-1}\left (\frac{1+\frac{2^{2/3} \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{\sqrt{3} a^{5/3} d}-\frac{2 b \log (x)}{3 a^{5/3} d}+\frac{b \log \left (a-b x^3\right )}{3\ 2^{2/3} a^{5/3} d}+\frac{2 b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{3 a^{5/3} d}-\frac{b \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2^{2/3} a^{5/3} d}\\ \end{align*}

Mathematica [A]  time = 0.108945, size = 280, normalized size = 1.04 \[ -\frac{6 a^{2/3} \sqrt [3]{a+b x^3}+4 b x^3 \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )-3 \sqrt [3]{2} b x^3 \log \left (2^{2/3} a^{2/3}+\sqrt [3]{2} \sqrt [3]{a} \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )-8 b x^3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )+6 \sqrt [3]{2} b x^3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )+8 \sqrt{3} b x^3 \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}+1}{\sqrt{3}}\right )-6 \sqrt [3]{2} \sqrt{3} b x^3 \tan ^{-1}\left (\frac{\frac{2^{2/3} \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}+1}{\sqrt{3}}\right )}{18 a^{5/3} d x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(1/3)/(x^4*(a*d - b*d*x^3)),x]

[Out]

-(6*a^(2/3)*(a + b*x^3)^(1/3) + 8*Sqrt[3]*b*x^3*ArcTan[(1 + (2*(a + b*x^3)^(1/3))/a^(1/3))/Sqrt[3]] - 6*2^(1/3
)*Sqrt[3]*b*x^3*ArcTan[(1 + (2^(2/3)*(a + b*x^3)^(1/3))/a^(1/3))/Sqrt[3]] - 8*b*x^3*Log[a^(1/3) - (a + b*x^3)^
(1/3)] + 6*2^(1/3)*b*x^3*Log[2^(1/3)*a^(1/3) - (a + b*x^3)^(1/3)] + 4*b*x^3*Log[a^(2/3) + a^(1/3)*(a + b*x^3)^
(1/3) + (a + b*x^3)^(2/3)] - 3*2^(1/3)*b*x^3*Log[2^(2/3)*a^(2/3) + 2^(1/3)*a^(1/3)*(a + b*x^3)^(1/3) + (a + b*
x^3)^(2/3)])/(18*a^(5/3)*d*x^3)

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Maple [F]  time = 0.055, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{4} \left ( -bd{x}^{3}+ad \right ) }\sqrt [3]{b{x}^{3}+a}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{{\left (b x^{3} + a\right )}^{\frac{1}{3}}}{{\left (b d x^{3} - a d\right )} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-integrate((b*x^3 + a)^(1/3)/((b*d*x^3 - a*d)*x^4), x)

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Fricas [A]  time = 1.39113, size = 905, normalized size = 3.38 \begin{align*} -\frac{6 \, \sqrt{3} 2^{\frac{1}{3}} a^{2} b x^{3} \left (-\frac{1}{a^{2}}\right )^{\frac{1}{3}} \arctan \left (\frac{1}{3} \, \sqrt{3} 2^{\frac{2}{3}}{\left (b x^{3} + a\right )}^{\frac{1}{3}} a \left (-\frac{1}{a^{2}}\right )^{\frac{2}{3}} + \frac{1}{3} \, \sqrt{3}\right ) + 3 \cdot 2^{\frac{1}{3}} a^{2} b x^{3} \left (-\frac{1}{a^{2}}\right )^{\frac{1}{3}} \log \left (2^{\frac{2}{3}} a^{2} \left (-\frac{1}{a^{2}}\right )^{\frac{2}{3}} - 2^{\frac{1}{3}}{\left (b x^{3} + a\right )}^{\frac{1}{3}} a \left (-\frac{1}{a^{2}}\right )^{\frac{1}{3}} +{\left (b x^{3} + a\right )}^{\frac{2}{3}}\right ) - 6 \cdot 2^{\frac{1}{3}} a^{2} b x^{3} \left (-\frac{1}{a^{2}}\right )^{\frac{1}{3}} \log \left (2^{\frac{1}{3}} a \left (-\frac{1}{a^{2}}\right )^{\frac{1}{3}} +{\left (b x^{3} + a\right )}^{\frac{1}{3}}\right ) + 8 \, \sqrt{3}{\left (a^{2}\right )}^{\frac{1}{6}} a b x^{3} \arctan \left (\frac{{\left (a^{2}\right )}^{\frac{1}{6}}{\left (\sqrt{3}{\left (a^{2}\right )}^{\frac{1}{3}} a + 2 \, \sqrt{3}{\left (b x^{3} + a\right )}^{\frac{1}{3}}{\left (a^{2}\right )}^{\frac{2}{3}}\right )}}{3 \, a^{2}}\right ) + 4 \,{\left (a^{2}\right )}^{\frac{2}{3}} b x^{3} \log \left ({\left (b x^{3} + a\right )}^{\frac{2}{3}} a +{\left (a^{2}\right )}^{\frac{1}{3}} a +{\left (b x^{3} + a\right )}^{\frac{1}{3}}{\left (a^{2}\right )}^{\frac{2}{3}}\right ) - 8 \,{\left (a^{2}\right )}^{\frac{2}{3}} b x^{3} \log \left ({\left (b x^{3} + a\right )}^{\frac{1}{3}} a -{\left (a^{2}\right )}^{\frac{2}{3}}\right ) + 6 \,{\left (b x^{3} + a\right )}^{\frac{1}{3}} a^{2}}{18 \, a^{3} d x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18*(6*sqrt(3)*2^(1/3)*a^2*b*x^3*(-1/a^2)^(1/3)*arctan(1/3*sqrt(3)*2^(2/3)*(b*x^3 + a)^(1/3)*a*(-1/a^2)^(2/3
) + 1/3*sqrt(3)) + 3*2^(1/3)*a^2*b*x^3*(-1/a^2)^(1/3)*log(2^(2/3)*a^2*(-1/a^2)^(2/3) - 2^(1/3)*(b*x^3 + a)^(1/
3)*a*(-1/a^2)^(1/3) + (b*x^3 + a)^(2/3)) - 6*2^(1/3)*a^2*b*x^3*(-1/a^2)^(1/3)*log(2^(1/3)*a*(-1/a^2)^(1/3) + (
b*x^3 + a)^(1/3)) + 8*sqrt(3)*(a^2)^(1/6)*a*b*x^3*arctan(1/3*(a^2)^(1/6)*(sqrt(3)*(a^2)^(1/3)*a + 2*sqrt(3)*(b
*x^3 + a)^(1/3)*(a^2)^(2/3))/a^2) + 4*(a^2)^(2/3)*b*x^3*log((b*x^3 + a)^(2/3)*a + (a^2)^(1/3)*a + (b*x^3 + a)^
(1/3)*(a^2)^(2/3)) - 8*(a^2)^(2/3)*b*x^3*log((b*x^3 + a)^(1/3)*a - (a^2)^(2/3)) + 6*(b*x^3 + a)^(1/3)*a^2)/(a^
3*d*x^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{\sqrt [3]{a + b x^{3}}}{- a x^{4} + b x^{7}}\, dx}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(1/3)/x**4/(-b*d*x**3+a*d),x)

[Out]

-Integral((a + b*x**3)**(1/3)/(-a*x**4 + b*x**7), x)/d

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x