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

Optimal. Leaf size=188 \[ -\frac{c \left (a+b x^3\right )^{2/3}}{2 d^2}+\frac{c (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^{8/3}}-\frac{c (b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{8/3}}-\frac{c (b c-a d)^{2/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt{3}}\right )}{\sqrt{3} d^{8/3}}+\frac{\left (a+b x^3\right )^{5/3}}{5 b d} \]

[Out]

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

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Rubi [A]  time = 0.198692, antiderivative size = 188, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {446, 80, 50, 56, 617, 204, 31} \[ -\frac{c \left (a+b x^3\right )^{2/3}}{2 d^2}+\frac{c (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^{8/3}}-\frac{c (b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{8/3}}-\frac{c (b c-a d)^{2/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt{3}}\right )}{\sqrt{3} d^{8/3}}+\frac{\left (a+b x^3\right )^{5/3}}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 0.0247067, size = 68, normalized size = 0.36 \[ \frac{\left (a+b x^3\right )^{2/3} \left (5 b c \, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\frac{d \left (b x^3+a\right )}{a d-b c}\right )+2 a d-5 b c+2 b d x^3\right )}{10 b d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^3)^(2/3)*(-5*b*c + 2*a*d + 2*b*d*x^3 + 5*b*c*Hypergeometric2F1[2/3, 1, 5/3, (d*(a + b*x^3))/(-(b*c)
+ a*d)]))/(10*b*d^2)

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Maple [F]  time = 0.041, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{5}}{d{x}^{3}+c} \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)/(d*x^3+c),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.59756, size = 815, normalized size = 4.34 \begin{align*} -\frac{10 \, \sqrt{3} b c \left (-\frac{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac{1}{3}} \arctan \left (-\frac{2 \, \sqrt{3}{\left (b x^{3} + a\right )}^{\frac{1}{3}} d \left (-\frac{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac{1}{3}} + \sqrt{3}{\left (b c - a d\right )}}{3 \,{\left (b c - a d\right )}}\right ) + 5 \, b c \left (-\frac{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac{1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac{1}{3}} d \left (-\frac{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac{2}{3}} -{\left (b x^{3} + a\right )}^{\frac{2}{3}}{\left (b c - a d\right )} +{\left (b c - a d\right )} \left (-\frac{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac{1}{3}}\right ) - 10 \, b c \left (-\frac{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac{1}{3}} \log \left (-d \left (-\frac{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac{2}{3}} -{\left (b x^{3} + a\right )}^{\frac{1}{3}}{\left (b c - a d\right )}\right ) - 3 \,{\left (2 \, b d x^{3} - 5 \, b c + 2 \, a d\right )}{\left (b x^{3} + a\right )}^{\frac{2}{3}}}{30 \, b d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**5*(a + b*x**3)**(2/3)/(c + d*x**3), x)

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Giac [A]  time = 1.19439, size = 402, normalized size = 2.14 \begin{align*} -\frac{\frac{10 \,{\left (b^{2} c^{2} d^{3} \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}} - a b c d^{4} \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}}\right )} \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}} \log \left ({\left |{\left (b x^{3} + a\right )}^{\frac{1}{3}} - \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}} \right |}\right )}{b c d^{5} - a d^{6}} + \frac{10 \, \sqrt{3}{\left (-b c d^{2} + a d^{3}\right )}^{\frac{2}{3}} b c \arctan \left (\frac{\sqrt{3}{\left (2 \,{\left (b x^{3} + a\right )}^{\frac{1}{3}} + \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}}\right )}}{3 \, \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}}}\right )}{d^{4}} - \frac{5 \,{\left (-b c d^{2} + a d^{3}\right )}^{\frac{2}{3}} b c \log \left ({\left (b x^{3} + a\right )}^{\frac{2}{3}} +{\left (b x^{3} + a\right )}^{\frac{1}{3}} \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}} + \left (-\frac{b c - a d}{d}\right )^{\frac{2}{3}}\right )}{d^{4}} + \frac{3 \,{\left (5 \,{\left (b x^{3} + a\right )}^{\frac{2}{3}} b c d^{3} - 2 \,{\left (b x^{3} + a\right )}^{\frac{5}{3}} d^{4}\right )}}{d^{5}}}{30 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(10*(b^2*c^2*d^3*(-(b*c - a*d)/d)^(1/3) - a*b*c*d^4*(-(b*c - a*d)/d)^(1/3))*(-(b*c - a*d)/d)^(1/3)*log(a
bs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b*c*d^5 - a*d^6) + 10*sqrt(3)*(-b*c*d^2 + a*d^3)^(2/3)*b*c*ar
ctan(1/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) + (-(b*c - a*d)/d)^(1/3))/(-(b*c - a*d)/d)^(1/3))/d^4 - 5*(-b*c*d^2 + a*
d^3)^(2/3)*b*c*log((b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*(-(b*c - a*d)/d)^(1/3) + (-(b*c - a*d)/d)^(2/3))/d^4
+ 3*(5*(b*x^3 + a)^(2/3)*b*c*d^3 - 2*(b*x^3 + a)^(5/3)*d^4)/d^5)/b