3.592 \(\int \frac{(a+b x^3)^{2/3}}{x^7 (a d-b d x^3)} \, dx\)

Optimal. Leaf size=284 \[ \frac{b^2 \log \left (a-b x^3\right )}{3 \sqrt [3]{2} a^{7/3} d}+\frac{7 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{9 a^{7/3} d}-\frac{b^2 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{\sqrt [3]{2} a^{7/3} d}+\frac{14 b^2 \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{9 \sqrt{3} a^{7/3} d}-\frac{2^{2/3} b^2 \tan ^{-1}\left (\frac{2^{2/3} \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} a^{7/3} d}-\frac{7 b^2 \log (x)}{9 a^{7/3} d}-\frac{5 b \left (a+b x^3\right )^{2/3}}{18 a^2 d x^3}-\frac{\left (a+b x^3\right )^{5/3}}{6 a^2 d x^6} \]

[Out]

(-5*b*(a + b*x^3)^(2/3))/(18*a^2*d*x^3) - (a + b*x^3)^(5/3)/(6*a^2*d*x^6) + (14*b^2*ArcTan[(a^(1/3) + 2*(a + b
*x^3)^(1/3))/(Sqrt[3]*a^(1/3))])/(9*Sqrt[3]*a^(7/3)*d) - (2^(2/3)*b^2*ArcTan[(a^(1/3) + 2^(2/3)*(a + b*x^3)^(1
/3))/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(7/3)*d) - (7*b^2*Log[x])/(9*a^(7/3)*d) + (b^2*Log[a - b*x^3])/(3*2^(1/3)*
a^(7/3)*d) + (7*b^2*Log[a^(1/3) - (a + b*x^3)^(1/3)])/(9*a^(7/3)*d) - (b^2*Log[2^(1/3)*a^(1/3) - (a + b*x^3)^(
1/3)])/(2^(1/3)*a^(7/3)*d)

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Rubi [A]  time = 0.279263, antiderivative size = 284, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {446, 103, 149, 156, 55, 617, 204, 31} \[ \frac{b^2 \log \left (a-b x^3\right )}{3 \sqrt [3]{2} a^{7/3} d}+\frac{7 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{9 a^{7/3} d}-\frac{b^2 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{\sqrt [3]{2} a^{7/3} d}+\frac{14 b^2 \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{9 \sqrt{3} a^{7/3} d}-\frac{2^{2/3} b^2 \tan ^{-1}\left (\frac{2^{2/3} \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} a^{7/3} d}-\frac{7 b^2 \log (x)}{9 a^{7/3} d}-\frac{5 b \left (a+b x^3\right )^{2/3}}{18 a^2 d x^3}-\frac{\left (a+b x^3\right )^{5/3}}{6 a^2 d x^6} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(2/3)/(x^7*(a*d - b*d*x^3)),x]

[Out]

(-5*b*(a + b*x^3)^(2/3))/(18*a^2*d*x^3) - (a + b*x^3)^(5/3)/(6*a^2*d*x^6) + (14*b^2*ArcTan[(a^(1/3) + 2*(a + b
*x^3)^(1/3))/(Sqrt[3]*a^(1/3))])/(9*Sqrt[3]*a^(7/3)*d) - (2^(2/3)*b^2*ArcTan[(a^(1/3) + 2^(2/3)*(a + b*x^3)^(1
/3))/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(7/3)*d) - (7*b^2*Log[x])/(9*a^(7/3)*d) + (b^2*Log[a - b*x^3])/(3*2^(1/3)*
a^(7/3)*d) + (7*b^2*Log[a^(1/3) - (a + b*x^3)^(1/3)])/(9*a^(7/3)*d) - (b^2*Log[2^(1/3)*a^(1/3) - (a + b*x^3)^(
1/3)])/(2^(1/3)*a^(7/3)*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 149

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegerQ[m]

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 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^7 \left (a d-b d x^3\right )} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{(a+b x)^{2/3}}{x^3 (a d-b d x)} \, dx,x,x^3\right )\\ &=-\frac{\left (a+b x^3\right )^{5/3}}{6 a^2 d x^6}-\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^{2/3} \left (-\frac{5}{3} a b d-\frac{1}{3} b^2 d x\right )}{x^2 (a d-b d x)} \, dx,x,x^3\right )}{6 a^2 d}\\ &=-\frac{5 b \left (a+b x^3\right )^{2/3}}{18 a^2 d x^3}-\frac{\left (a+b x^3\right )^{5/3}}{6 a^2 d x^6}-\frac{\operatorname{Subst}\left (\int \frac{-\frac{28}{9} a^2 b^2 d^2-\frac{8}{9} a b^3 d^2 x}{x \sqrt [3]{a+b x} (a d-b d x)} \, dx,x,x^3\right )}{6 a^3 d^2}\\ &=-\frac{5 b \left (a+b x^3\right )^{2/3}}{18 a^2 d x^3}-\frac{\left (a+b x^3\right )^{5/3}}{6 a^2 d x^6}+\frac{\left (2 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a+b x} (a d-b d x)} \, dx,x,x^3\right )}{3 a^2}+\frac{\left (14 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt [3]{a+b x}} \, dx,x,x^3\right )}{27 a^2 d}\\ &=-\frac{5 b \left (a+b x^3\right )^{2/3}}{18 a^2 d x^3}-\frac{\left (a+b x^3\right )^{5/3}}{6 a^2 d x^6}-\frac{7 b^2 \log (x)}{9 a^{7/3} d}+\frac{b^2 \log \left (a-b x^3\right )}{3 \sqrt [3]{2} a^{7/3} d}-\frac{\left (7 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x^3}\right )}{9 a^{7/3} d}+\frac{b^2 \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^{7/3} d}+\frac{\left (7 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+b x^3}\right )}{9 a^2 d}-\frac{b^2 \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^2 d}\\ &=-\frac{5 b \left (a+b x^3\right )^{2/3}}{18 a^2 d x^3}-\frac{\left (a+b x^3\right )^{5/3}}{6 a^2 d x^6}-\frac{7 b^2 \log (x)}{9 a^{7/3} d}+\frac{b^2 \log \left (a-b x^3\right )}{3 \sqrt [3]{2} a^{7/3} d}+\frac{7 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{9 a^{7/3} d}-\frac{b^2 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{\sqrt [3]{2} a^{7/3} d}-\frac{\left (14 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}\right )}{9 a^{7/3} d}+\frac{\left (2^{2/3} b^2\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^{7/3} d}\\ &=-\frac{5 b \left (a+b x^3\right )^{2/3}}{18 a^2 d x^3}-\frac{\left (a+b x^3\right )^{5/3}}{6 a^2 d x^6}+\frac{14 b^2 \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{9 \sqrt{3} a^{7/3} d}-\frac{2^{2/3} b^2 \tan ^{-1}\left (\frac{1+\frac{2^{2/3} \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{\sqrt{3} a^{7/3} d}-\frac{7 b^2 \log (x)}{9 a^{7/3} d}+\frac{b^2 \log \left (a-b x^3\right )}{3 \sqrt [3]{2} a^{7/3} d}+\frac{7 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{9 a^{7/3} d}-\frac{b^2 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{\sqrt [3]{2} a^{7/3} d}\\ \end{align*}

Mathematica [A]  time = 0.14117, size = 247, normalized size = 0.87 \[ \frac{28 \sqrt{3} b^2 x^6 \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}+1}{\sqrt{3}}\right )-3 \left (3 a^{4/3} \left (a+b x^3\right )^{2/3}-3\ 2^{2/3} b^2 x^6 \log \left (a-b x^3\right )-14 b^2 x^6 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )+9\ 2^{2/3} b^2 x^6 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )+6\ 2^{2/3} \sqrt{3} b^2 x^6 \tan ^{-1}\left (\frac{\frac{2^{2/3} \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}+1}{\sqrt{3}}\right )+8 \sqrt [3]{a} b x^3 \left (a+b x^3\right )^{2/3}+14 b^2 x^6 \log (x)\right )}{54 a^{7/3} d x^6} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(2/3)/(x^7*(a*d - b*d*x^3)),x]

[Out]

(28*Sqrt[3]*b^2*x^6*ArcTan[(1 + (2*(a + b*x^3)^(1/3))/a^(1/3))/Sqrt[3]] - 3*(3*a^(4/3)*(a + b*x^3)^(2/3) + 8*a
^(1/3)*b*x^3*(a + b*x^3)^(2/3) + 6*2^(2/3)*Sqrt[3]*b^2*x^6*ArcTan[(1 + (2^(2/3)*(a + b*x^3)^(1/3))/a^(1/3))/Sq
rt[3]] + 14*b^2*x^6*Log[x] - 3*2^(2/3)*b^2*x^6*Log[a - b*x^3] - 14*b^2*x^6*Log[a^(1/3) - (a + b*x^3)^(1/3)] +
9*2^(2/3)*b^2*x^6*Log[2^(1/3)*a^(1/3) - (a + b*x^3)^(1/3)]))/(54*a^(7/3)*d*x^6)

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Maple [F]  time = 0.063, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{7} \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^7/(-b*d*x^3+a*d),x)

[Out]

int((b*x^3+a)^(2/3)/x^7/(-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^{7}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(2/3)/x^7/(-b*d*x^3+a*d),x, algorithm="maxima")

[Out]

-integrate((b*x^3 + a)^(2/3)/((b*d*x^3 - a*d)*x^7), x)

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Fricas [A]  time = 1.72148, size = 1816, normalized size = 6.39 \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^7/(-b*d*x^3+a*d),x, algorithm="fricas")

[Out]

[-1/54*(18*4^(1/3)*sqrt(3)*a*b^2*x^6*(-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)) - 42*sqrt(1/3)*a*b^2*x^6*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) + 9*4^(1/3)*a*b^2
*x^6*(-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
)) - 18*4^(1/3)*a*b^2*x^6*(-1/a)^(1/3)*log(-4^(2/3)*a*(-1/a)^(2/3) + 2*(b*x^3 + a)^(1/3)) + 14*a^(2/3)*b^2*x^6
*log((b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*a^(1/3) + a^(2/3)) - 28*a^(2/3)*b^2*x^6*log((b*x^3 + a)^(1/3) - a^(
1/3)) + 3*(8*a*b*x^3 + 3*a^2)*(b*x^3 + a)^(2/3))/(a^3*d*x^6), -1/54*(18*4^(1/3)*sqrt(3)*a*b^2*x^6*(-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)) + 9*4^(1/3)*a*b^2*x^6*(-1/a)^(1/3)*l
og(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)) - 18*4^(1/3)*a*b
^2*x^6*(-1/a)^(1/3)*log(-4^(2/3)*a*(-1/a)^(2/3) + 2*(b*x^3 + a)^(1/3)) - 84*sqrt(1/3)*a^(2/3)*b^2*x^6*arctan(s
qrt(1/3)*(2*(b*x^3 + a)^(1/3) + a^(1/3))/a^(1/3)) + 14*a^(2/3)*b^2*x^6*log((b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/
3)*a^(1/3) + a^(2/3)) - 28*a^(2/3)*b^2*x^6*log((b*x^3 + a)^(1/3) - a^(1/3)) + 3*(8*a*b*x^3 + 3*a^2)*(b*x^3 + a
)^(2/3))/(a^3*d*x^6)]

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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((b*x**3+a)**(2/3)/x**7/(-b*d*x**3+a*d),x)

[Out]

Timed out

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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^7/(-b*d*x^3+a*d),x, algorithm="giac")

[Out]

sage0*x