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

Optimal. Leaf size=165 \[ \frac{c \log \left (c+d x^3\right )}{6 d^{4/3} (b c-a d)^{2/3}}-\frac{c \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{4/3} (b c-a d)^{2/3}}+\frac{c \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt{3}}\right )}{\sqrt{3} d^{4/3} (b c-a d)^{2/3}}+\frac{\sqrt [3]{a+b x^3}}{b d} \]

[Out]

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

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Rubi [A]  time = 0.155348, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {446, 80, 58, 617, 204, 31} \[ \frac{c \log \left (c+d x^3\right )}{6 d^{4/3} (b c-a d)^{2/3}}-\frac{c \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{4/3} (b c-a d)^{2/3}}+\frac{c \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt{3}}\right )}{\sqrt{3} d^{4/3} (b c-a d)^{2/3}}+\frac{\sqrt [3]{a+b x^3}}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 58

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, -Sim
p[Log[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
] && NegQ[(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} \left (c+d x^3\right )} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{x}{(a+b x)^{2/3} (c+d x)} \, dx,x,x^3\right )\\ &=\frac{\sqrt [3]{a+b x^3}}{b d}-\frac{c \operatorname{Subst}\left (\int \frac{1}{(a+b x)^{2/3} (c+d x)} \, dx,x,x^3\right )}{3 d}\\ &=\frac{\sqrt [3]{a+b x^3}}{b d}+\frac{c \log \left (c+d x^3\right )}{6 d^{4/3} (b c-a d)^{2/3}}-\frac{c \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^{4/3} (b c-a d)^{2/3}}-\frac{c \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^{5/3} \sqrt [3]{b c-a d}}\\ &=\frac{\sqrt [3]{a+b x^3}}{b d}+\frac{c \log \left (c+d x^3\right )}{6 d^{4/3} (b c-a d)^{2/3}}-\frac{c \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{4/3} (b c-a d)^{2/3}}-\frac{c \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^{4/3} (b c-a d)^{2/3}}\\ &=\frac{\sqrt [3]{a+b x^3}}{b d}+\frac{c \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^{4/3} (b c-a d)^{2/3}}+\frac{c \log \left (c+d x^3\right )}{6 d^{4/3} (b c-a d)^{2/3}}-\frac{c \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{4/3} (b c-a d)^{2/3}}\\ \end{align*}

Mathematica [A]  time = 0.103148, size = 202, normalized size = 1.22 \[ \frac{b c \log \left (-\sqrt [3]{d} \sqrt [3]{a+b x^3} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )+6 \sqrt [3]{d} \sqrt [3]{a+b x^3} (b c-a d)^{2/3}-2 b c \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )-2 \sqrt{3} b c \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}-1}{\sqrt{3}}\right )}{6 b d^{4/3} (b c-a d)^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.04, 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 = 1.52908, size = 2276, normalized size = 13.79 \begin{align*} \text{result too large to display} \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]

[1/6*((-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(2/3)*b*c*log(-(b*x^3 + a)^(2/3)*(b*c*d - a*d^2) + (-b^2*c^2*d + 2*
a*b*c*d^2 - a^2*d^3)^(1/3)*(b*c - a*d) + (-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(2/3)*(b*x^3 + a)^(1/3)) - 2*(-b
^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(2/3)*b*c*log(-(b*x^3 + a)^(1/3)*(b*c*d - a*d^2) - (-b^2*c^2*d + 2*a*b*c*d^2
- a^2*d^3)^(2/3)) - 3*sqrt(1/3)*(b^2*c^2*d - a*b*c*d^2)*sqrt((-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(1/3)/d)*lo
g((b^2*c^2 - 4*a*b*c*d + 3*a^2*d^2 - 2*(b^2*c*d - a*b*d^2)*x^3 - 3*sqrt(1/3)*(2*(b*x^3 + a)^(2/3)*(b*c*d - a*d
^2) + (-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(1/3)*(b*c - a*d) + (-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(2/3)*(b*x
^3 + a)^(1/3))*sqrt((-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(1/3)/d) - 3*(-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(1/
3)*(b*x^3 + a)^(1/3)*(b*c - a*d))/(d*x^3 + c)) + 6*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*(b*x^3 + a)^(1/3))/(b^3
*c^2*d^2 - 2*a*b^2*c*d^3 + a^2*b*d^4), 1/6*((-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(2/3)*b*c*log(-(b*x^3 + a)^(2
/3)*(b*c*d - a*d^2) + (-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(1/3)*(b*c - a*d) + (-b^2*c^2*d + 2*a*b*c*d^2 - a^2
*d^3)^(2/3)*(b*x^3 + a)^(1/3)) - 2*(-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(2/3)*b*c*log(-(b*x^3 + a)^(1/3)*(b*c*
d - a*d^2) - (-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(2/3)) - 6*sqrt(1/3)*(b^2*c^2*d - a*b*c*d^2)*sqrt(-(-b^2*c^2
*d + 2*a*b*c*d^2 - a^2*d^3)^(1/3)/d)*arctan(sqrt(1/3)*((-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(1/3)*(b*c - a*d)
+ 2*(-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(2/3)*(b*x^3 + a)^(1/3))*sqrt(-(-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(
1/3)/d)/(b^2*c^2 - 2*a*b*c*d + a^2*d^2)) + 6*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*(b*x^3 + a)^(1/3))/(b^3*c^2*d
^2 - 2*a*b^2*c*d^3 + a^2*b*d^4)]

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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}} \left (c + d x^{3}\right )}\, 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 = 1.18109, size = 342, normalized size = 2.07 \begin{align*} -\frac{\frac{6 \,{\left (-b c d^{2} + a d^{3}\right )}^{\frac{1}{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 )}{\sqrt{3} b c d^{2} - \sqrt{3} a d^{3}} + \frac{{\left (-b c d^{2} + a d^{3}\right )}^{\frac{1}{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 )}{b c d^{2} - a d^{3}} - \frac{2 \, b c \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 - a d^{2}} - \frac{6 \,{\left (b x^{3} + a\right )}^{\frac{1}{3}}}{d}}{6 \, 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]

-1/6*(6*(-b*c*d^2 + a*d^3)^(1/3)*b*c*arctan(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))/(sqrt(3)*b*c*d^2 - sqrt(3)*a*d^3) + (-b*c*d^2 + a*d^3)^(1/3)*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))/(b*c*d^2 - a*d^3) - 2*b*c*(-(b*c - a*d)/d)^(1/3
)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b*c*d - a*d^2) - 6*(b*x^3 + a)^(1/3)/d)/b