3.591 \(\int \frac{(a+b x^3)^{2/3}}{x^4 (a d-b d x^3)} \, dx\)

Optimal. Leaf size=269 \[ -\frac{\left (a+b x^3\right )^{5/3}}{3 a^2 d x^3}+\frac{b \left (a+b x^3\right )^{2/3}}{3 a^2 d}+\frac{b \log \left (a-b x^3\right )}{3 \sqrt [3]{2} a^{4/3} d}+\frac{5 b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{6 a^{4/3} d}-\frac{b \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{\sqrt [3]{2} a^{4/3} d}+\frac{5 b \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{4/3} d}-\frac{2^{2/3} b \tan ^{-1}\left (\frac{2^{2/3} \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} a^{4/3} d}-\frac{5 b \log (x)}{6 a^{4/3} d} \]

[Out]

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

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Rubi [A]  time = 0.263398, antiderivative size = 269, 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, 55, 617, 204, 31} \[ -\frac{\left (a+b x^3\right )^{5/3}}{3 a^2 d x^3}+\frac{b \left (a+b x^3\right )^{2/3}}{3 a^2 d}+\frac{b \log \left (a-b x^3\right )}{3 \sqrt [3]{2} a^{4/3} d}+\frac{5 b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{6 a^{4/3} d}-\frac{b \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{\sqrt [3]{2} a^{4/3} d}+\frac{5 b \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{4/3} d}-\frac{2^{2/3} b \tan ^{-1}\left (\frac{2^{2/3} \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} a^{4/3} d}-\frac{5 b \log (x)}{6 a^{4/3} d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(2/3)/(x^4*(a*d - b*d*x^3)),x]

[Out]

(b*(a + b*x^3)^(2/3))/(3*a^2*d) - (a + b*x^3)^(5/3)/(3*a^2*d*x^3) + (5*b*ArcTan[(a^(1/3) + 2*(a + b*x^3)^(1/3)
)/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(4/3)*d) - (2^(2/3)*b*ArcTan[(a^(1/3) + 2^(2/3)*(a + b*x^3)^(1/3))/(Sqrt[3]
*a^(1/3))])/(Sqrt[3]*a^(4/3)*d) - (5*b*Log[x])/(6*a^(4/3)*d) + (b*Log[a - b*x^3])/(3*2^(1/3)*a^(4/3)*d) + (5*b
*Log[a^(1/3) - (a + b*x^3)^(1/3)])/(6*a^(4/3)*d) - (b*Log[2^(1/3)*a^(1/3) - (a + b*x^3)^(1/3)])/(2^(1/3)*a^(4/
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 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{\left (a+b x^3\right )^{2/3}}{x^4 \left (a d-b d x^3\right )} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{(a+b x)^{2/3}}{x^2 (a d-b d x)} \, dx,x,x^3\right )\\ &=-\frac{\left (a+b x^3\right )^{5/3}}{3 a^2 d x^3}-\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^{2/3} \left (-\frac{5}{3} a b d+\frac{2}{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 )^{5/3}}{3 a^2 d x^3}+\frac{b^2 \operatorname{Subst}\left (\int \frac{(a+b x)^{2/3}}{a d-b d x} \, dx,x,x^3\right )}{3 a^2}+\frac{(5 b) \operatorname{Subst}\left (\int \frac{(a+b x)^{2/3}}{x} \, dx,x,x^3\right )}{9 a^2 d}\\ &=\frac{b \left (a+b x^3\right )^{2/3}}{3 a^2 d}-\frac{\left (a+b x^3\right )^{5/3}}{3 a^2 d x^3}+\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a+b x} (a d-b d x)} \, dx,x,x^3\right )}{3 a}+\frac{(5 b) \operatorname{Subst}\left (\int \frac{1}{x \sqrt [3]{a+b x}} \, dx,x,x^3\right )}{9 a d}\\ &=\frac{b \left (a+b x^3\right )^{2/3}}{3 a^2 d}-\frac{\left (a+b x^3\right )^{5/3}}{3 a^2 d x^3}-\frac{5 b \log (x)}{6 a^{4/3} d}+\frac{b \log \left (a-b x^3\right )}{3 \sqrt [3]{2} a^{4/3} d}-\frac{(5 b) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x^3}\right )}{6 a^{4/3} 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 )}{\sqrt [3]{2} a^{4/3} d}+\frac{(5 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 )}{6 a 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 )}{a d}\\ &=\frac{b \left (a+b x^3\right )^{2/3}}{3 a^2 d}-\frac{\left (a+b x^3\right )^{5/3}}{3 a^2 d x^3}-\frac{5 b \log (x)}{6 a^{4/3} d}+\frac{b \log \left (a-b x^3\right )}{3 \sqrt [3]{2} a^{4/3} d}+\frac{5 b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{6 a^{4/3} d}-\frac{b \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{\sqrt [3]{2} a^{4/3} d}-\frac{(5 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^{4/3} d}+\frac{\left (2^{2/3} 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^{4/3} d}\\ &=\frac{b \left (a+b x^3\right )^{2/3}}{3 a^2 d}-\frac{\left (a+b x^3\right )^{5/3}}{3 a^2 d x^3}+\frac{5 b \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{3 \sqrt{3} a^{4/3} d}-\frac{2^{2/3} 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^{4/3} d}-\frac{5 b \log (x)}{6 a^{4/3} d}+\frac{b \log \left (a-b x^3\right )}{3 \sqrt [3]{2} a^{4/3} d}+\frac{5 b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{6 a^{4/3} d}-\frac{b \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{\sqrt [3]{2} a^{4/3} d}\\ \end{align*}

Mathematica [A]  time = 0.0839781, size = 213, normalized size = 0.79 \[ \frac{10 \sqrt{3} b x^3 \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}+1}{\sqrt{3}}\right )-3 \left (2 \sqrt [3]{a} \left (a+b x^3\right )^{2/3}-2^{2/3} b x^3 \log \left (a-b x^3\right )-5 b x^3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )+3\ 2^{2/3} b x^3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )+2\ 2^{2/3} \sqrt{3} b x^3 \tan ^{-1}\left (\frac{\frac{2^{2/3} \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}+1}{\sqrt{3}}\right )+5 b x^3 \log (x)\right )}{18 a^{4/3} d x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(2/3)/(x^4*(a*d - b*d*x^3)),x]

[Out]

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

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Maple [F]  time = 0.058, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{4} \left ( -bd{x}^{3}+ad \right ) } \left ( b{x}^{3}+a \right ) ^{{\frac{2}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(2/3)/x^4/(-b*d*x^3+a*d),x)

[Out]

int((b*x^3+a)^(2/3)/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{2}{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)^(2/3)/x^4/(-b*d*x^3+a*d),x, algorithm="maxima")

[Out]

-integrate((b*x^3 + a)^(2/3)/((b*d*x^3 - a*d)*x^4), x)

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Fricas [A]  time = 1.69358, size = 1727, normalized size = 6.42 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(2/3)/x^4/(-b*d*x^3+a*d),x, algorithm="fricas")

[Out]

[-1/18*(6*4^(1/3)*sqrt(3)*a*b*x^3*(-1/a)^(1/3)*arctan(1/3*4^(1/3)*sqrt(3)*(b*x^3 + a)^(1/3)*(-1/a)^(1/3) - 1/3
*sqrt(3)) - 15*sqrt(1/3)*a*b*x^3*sqrt(-1/a^(2/3))*log((2*b*x^3 + 3*sqrt(1/3)*(2*(b*x^3 + a)^(2/3)*a^(2/3) - (b
*x^3 + a)^(1/3)*a - a^(4/3))*sqrt(-1/a^(2/3)) - 3*(b*x^3 + a)^(1/3)*a^(2/3) + 3*a)/x^3) + 3*4^(1/3)*a*b*x^3*(-
1/a)^(1/3)*log(4^(2/3)*(b*x^3 + a)^(1/3)*a*(-1/a)^(2/3) - 2*4^(1/3)*a*(-1/a)^(1/3) + 2*(b*x^3 + a)^(2/3)) - 6*
4^(1/3)*a*b*x^3*(-1/a)^(1/3)*log(-4^(2/3)*a*(-1/a)^(2/3) + 2*(b*x^3 + a)^(1/3)) + 5*a^(2/3)*b*x^3*log((b*x^3 +
a)^(2/3) + (b*x^3 + a)^(1/3)*a^(1/3) + a^(2/3)) - 10*a^(2/3)*b*x^3*log((b*x^3 + a)^(1/3) - a^(1/3)) + 6*(b*x^
3 + a)^(2/3)*a)/(a^2*d*x^3), -1/18*(6*4^(1/3)*sqrt(3)*a*b*x^3*(-1/a)^(1/3)*arctan(1/3*4^(1/3)*sqrt(3)*(b*x^3 +
a)^(1/3)*(-1/a)^(1/3) - 1/3*sqrt(3)) + 3*4^(1/3)*a*b*x^3*(-1/a)^(1/3)*log(4^(2/3)*(b*x^3 + a)^(1/3)*a*(-1/a)^
(2/3) - 2*4^(1/3)*a*(-1/a)^(1/3) + 2*(b*x^3 + a)^(2/3)) - 6*4^(1/3)*a*b*x^3*(-1/a)^(1/3)*log(-4^(2/3)*a*(-1/a)
^(2/3) + 2*(b*x^3 + a)^(1/3)) - 30*sqrt(1/3)*a^(2/3)*b*x^3*arctan(sqrt(1/3)*(2*(b*x^3 + a)^(1/3) + a^(1/3))/a^
(1/3)) + 5*a^(2/3)*b*x^3*log((b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*a^(1/3) + a^(2/3)) - 10*a^(2/3)*b*x^3*log((
b*x^3 + a)^(1/3) - a^(1/3)) + 6*(b*x^3 + a)^(2/3)*a)/(a^2*d*x^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{\left (a + b x^{3}\right )^{\frac{2}{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)**(2/3)/x**4/(-b*d*x**3+a*d),x)

[Out]

-Integral((a + b*x**3)**(2/3)/(-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)^(2/3)/x^4/(-b*d*x^3+a*d),x, algorithm="giac")

[Out]

sage0*x