3.588 \(\int \frac{x^5 (a+b x^3)^{2/3}}{a d-b d x^3} \, dx\)

Optimal. Leaf size=175 \[ \frac{a^{5/3} \log \left (a-b x^3\right )}{3 \sqrt [3]{2} b^2 d}-\frac{a^{5/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{\sqrt [3]{2} b^2 d}-\frac{2^{2/3} a^{5/3} \tan ^{-1}\left (\frac{2^{2/3} \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} b^2 d}-\frac{a \left (a+b x^3\right )^{2/3}}{2 b^2 d}-\frac{\left (a+b x^3\right )^{5/3}}{5 b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.156465, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {446, 80, 50, 55, 617, 204, 31} \[ \frac{a^{5/3} \log \left (a-b x^3\right )}{3 \sqrt [3]{2} b^2 d}-\frac{a^{5/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{\sqrt [3]{2} b^2 d}-\frac{2^{2/3} a^{5/3} \tan ^{-1}\left (\frac{2^{2/3} \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} b^2 d}-\frac{a \left (a+b x^3\right )^{2/3}}{2 b^2 d}-\frac{\left (a+b x^3\right )^{5/3}}{5 b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(a + b*x^3)^(2/3))/(a*d - b*d*x^3),x]

[Out]

-(a*(a + b*x^3)^(2/3))/(2*b^2*d) - (a + b*x^3)^(5/3)/(5*b^2*d) - (2^(2/3)*a^(5/3)*ArcTan[(a^(1/3) + 2^(2/3)*(a
+ b*x^3)^(1/3))/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*b^2*d) + (a^(5/3)*Log[a - b*x^3])/(3*2^(1/3)*b^2*d) - (a^(5/3)*L
og[2^(1/3)*a^(1/3) - (a + b*x^3)^(1/3)])/(2^(1/3)*b^2*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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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

Mathematica [A]  time = 0.129372, size = 143, normalized size = 0.82 \[ \frac{5\ 2^{2/3} a^{5/3} \log \left (a-b x^3\right )-3 \left (5\ 2^{2/3} a^{5/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )+\left (a+b x^3\right )^{2/3} \left (7 a+2 b x^3\right )\right )-10\ 2^{2/3} \sqrt{3} a^{5/3} \tan ^{-1}\left (\frac{\frac{2^{2/3} \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}+1}{\sqrt{3}}\right )}{30 b^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(a + b*x^3)^(2/3))/(a*d - b*d*x^3),x]

[Out]

(-10*2^(2/3)*Sqrt[3]*a^(5/3)*ArcTan[(1 + (2^(2/3)*(a + b*x^3)^(1/3))/a^(1/3))/Sqrt[3]] + 5*2^(2/3)*a^(5/3)*Log
[a - b*x^3] - 3*((a + b*x^3)^(2/3)*(7*a + 2*b*x^3) + 5*2^(2/3)*a^(5/3)*Log[2^(1/3)*a^(1/3) - (a + b*x^3)^(1/3)
]))/(30*b^2*d)

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Maple [F]  time = 0.04, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{5}}{-bd{x}^{3}+ad} \left ( b{x}^{3}+a \right ) ^{{\frac{2}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(b*x^3+a)^(2/3)/(-b*d*x^3+a*d),x)

[Out]

int(x^5*(b*x^3+a)^(2/3)/(-b*d*x^3+a*d),x)

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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(x^5*(b*x^3+a)^(2/3)/(-b*d*x^3+a*d),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.84128, size = 493, normalized size = 2.82 \begin{align*} -\frac{10 \cdot 4^{\frac{1}{3}} \sqrt{3} \left (-a^{2}\right )^{\frac{1}{3}} a \arctan \left (\frac{4^{\frac{1}{3}} \sqrt{3}{\left (b x^{3} + a\right )}^{\frac{1}{3}} \left (-a^{2}\right )^{\frac{1}{3}} - \sqrt{3} a}{3 \, a}\right ) + 5 \cdot 4^{\frac{1}{3}} \left (-a^{2}\right )^{\frac{1}{3}} a \log \left (4^{\frac{2}{3}}{\left (b x^{3} + a\right )}^{\frac{1}{3}} \left (-a^{2}\right )^{\frac{2}{3}} + 2 \,{\left (b x^{3} + a\right )}^{\frac{2}{3}} a - 2 \cdot 4^{\frac{1}{3}} \left (-a^{2}\right )^{\frac{1}{3}} a\right ) - 10 \cdot 4^{\frac{1}{3}} \left (-a^{2}\right )^{\frac{1}{3}} a \log \left (-4^{\frac{2}{3}} \left (-a^{2}\right )^{\frac{2}{3}} + 2 \,{\left (b x^{3} + a\right )}^{\frac{1}{3}} a\right ) + 3 \,{\left (2 \, b x^{3} + 7 \, a\right )}{\left (b x^{3} + a\right )}^{\frac{2}{3}}}{30 \, b^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^3+a)^(2/3)/(-b*d*x^3+a*d),x, algorithm="fricas")

[Out]

-1/30*(10*4^(1/3)*sqrt(3)*(-a^2)^(1/3)*a*arctan(1/3*(4^(1/3)*sqrt(3)*(b*x^3 + a)^(1/3)*(-a^2)^(1/3) - sqrt(3)*
a)/a) + 5*4^(1/3)*(-a^2)^(1/3)*a*log(4^(2/3)*(b*x^3 + a)^(1/3)*(-a^2)^(2/3) + 2*(b*x^3 + a)^(2/3)*a - 2*4^(1/3
)*(-a^2)^(1/3)*a) - 10*4^(1/3)*(-a^2)^(1/3)*a*log(-4^(2/3)*(-a^2)^(2/3) + 2*(b*x^3 + a)^(1/3)*a) + 3*(2*b*x^3
+ 7*a)*(b*x^3 + a)^(2/3))/(b^2*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{x^{5} \left (a + b x^{3}\right )^{\frac{2}{3}}}{- a + b x^{3}}\, dx}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(b*x**3+a)**(2/3)/(-b*d*x**3+a*d),x)

[Out]

-Integral(x**5*(a + b*x**3)**(2/3)/(-a + b*x**3), 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(x^5*(b*x^3+a)^(2/3)/(-b*d*x^3+a*d),x, algorithm="giac")

[Out]

sage0*x