3.746 \(\int (c x)^{10/3} \sqrt [3]{a+b x^2} \, dx\)

Optimal. Leaf size=451 \[ \frac{7 a^2 c^{7/3} \sqrt [3]{c x} \sqrt [3]{a+b x^2} \left (c^{2/3}-\frac{\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right ) \sqrt{\frac{\frac{b^{2/3} (c x)^{4/3}}{\left (a+b x^2\right )^{2/3}}+\frac{\sqrt [3]{b} c^{2/3} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}+c^{4/3}}{\left (c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}} \text{EllipticF}\left (\cos ^{-1}\left (\frac{c^{2/3}-\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}{c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}\right ),\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{135 \sqrt [4]{3} b^2 \sqrt{-\frac{\sqrt [3]{b} (c x)^{2/3} \left (c^{2/3}-\frac{\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{\sqrt [3]{a+b x^2} \left (c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}}}-\frac{14 a^2 c^3 \sqrt [3]{c x} \sqrt [3]{a+b x^2}}{135 b^2}+\frac{(c x)^{13/3} \sqrt [3]{a+b x^2}}{5 c}+\frac{2 a c (c x)^{7/3} \sqrt [3]{a+b x^2}}{45 b} \]

[Out]

(-14*a^2*c^3*(c*x)^(1/3)*(a + b*x^2)^(1/3))/(135*b^2) + (2*a*c*(c*x)^(7/3)*(a + b*x^2)^(1/3))/(45*b) + ((c*x)^
(13/3)*(a + b*x^2)^(1/3))/(5*c) + (7*a^2*c^(7/3)*(c*x)^(1/3)*(a + b*x^2)^(1/3)*(c^(2/3) - (b^(1/3)*(c*x)^(2/3)
)/(a + b*x^2)^(1/3))*Sqrt[(c^(4/3) + (b^(2/3)*(c*x)^(4/3))/(a + b*x^2)^(2/3) + (b^(1/3)*c^(2/3)*(c*x)^(2/3))/(
a + b*x^2)^(1/3))/(c^(2/3) - ((1 + Sqrt[3])*b^(1/3)*(c*x)^(2/3))/(a + b*x^2)^(1/3))^2]*EllipticF[ArcCos[(c^(2/
3) - ((1 - Sqrt[3])*b^(1/3)*(c*x)^(2/3))/(a + b*x^2)^(1/3))/(c^(2/3) - ((1 + Sqrt[3])*b^(1/3)*(c*x)^(2/3))/(a
+ b*x^2)^(1/3))], (2 + Sqrt[3])/4])/(135*3^(1/4)*b^2*Sqrt[-((b^(1/3)*(c*x)^(2/3)*(c^(2/3) - (b^(1/3)*(c*x)^(2/
3))/(a + b*x^2)^(1/3)))/((a + b*x^2)^(1/3)*(c^(2/3) - ((1 + Sqrt[3])*b^(1/3)*(c*x)^(2/3))/(a + b*x^2)^(1/3))^2
))])

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Rubi [A]  time = 0.970588, antiderivative size = 451, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {279, 321, 329, 241, 225} \[ -\frac{14 a^2 c^3 \sqrt [3]{c x} \sqrt [3]{a+b x^2}}{135 b^2}+\frac{7 a^2 c^{7/3} \sqrt [3]{c x} \sqrt [3]{a+b x^2} \left (c^{2/3}-\frac{\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right ) \sqrt{\frac{\frac{b^{2/3} (c x)^{4/3}}{\left (a+b x^2\right )^{2/3}}+\frac{\sqrt [3]{b} c^{2/3} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}+c^{4/3}}{\left (c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}} F\left (\cos ^{-1}\left (\frac{c^{2/3}-\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{b x^2+a}}}{c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{b x^2+a}}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{135 \sqrt [4]{3} b^2 \sqrt{-\frac{\sqrt [3]{b} (c x)^{2/3} \left (c^{2/3}-\frac{\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{\sqrt [3]{a+b x^2} \left (c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}}}+\frac{(c x)^{13/3} \sqrt [3]{a+b x^2}}{5 c}+\frac{2 a c (c x)^{7/3} \sqrt [3]{a+b x^2}}{45 b} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(10/3)*(a + b*x^2)^(1/3),x]

[Out]

(-14*a^2*c^3*(c*x)^(1/3)*(a + b*x^2)^(1/3))/(135*b^2) + (2*a*c*(c*x)^(7/3)*(a + b*x^2)^(1/3))/(45*b) + ((c*x)^
(13/3)*(a + b*x^2)^(1/3))/(5*c) + (7*a^2*c^(7/3)*(c*x)^(1/3)*(a + b*x^2)^(1/3)*(c^(2/3) - (b^(1/3)*(c*x)^(2/3)
)/(a + b*x^2)^(1/3))*Sqrt[(c^(4/3) + (b^(2/3)*(c*x)^(4/3))/(a + b*x^2)^(2/3) + (b^(1/3)*c^(2/3)*(c*x)^(2/3))/(
a + b*x^2)^(1/3))/(c^(2/3) - ((1 + Sqrt[3])*b^(1/3)*(c*x)^(2/3))/(a + b*x^2)^(1/3))^2]*EllipticF[ArcCos[(c^(2/
3) - ((1 - Sqrt[3])*b^(1/3)*(c*x)^(2/3))/(a + b*x^2)^(1/3))/(c^(2/3) - ((1 + Sqrt[3])*b^(1/3)*(c*x)^(2/3))/(a
+ b*x^2)^(1/3))], (2 + Sqrt[3])/4])/(135*3^(1/4)*b^2*Sqrt[-((b^(1/3)*(c*x)^(2/3)*(c^(2/3) - (b^(1/3)*(c*x)^(2/
3))/(a + b*x^2)^(1/3)))/((a + b*x^2)^(1/3)*(c^(2/3) - ((1 + Sqrt[3])*b^(1/3)*(c*x)^(2/3))/(a + b*x^2)^(1/3))^2
))])

Rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 241

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a/(a + b*x^n))^(p + 1/n)*(a + b*x^n)^(p + 1/n), Subst[In
t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p,
0] && NeQ[p, -2^(-1)] && LtQ[Denominator[p + 1/n], Denominator[p]]

Rule 225

Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(x*(s
+ r*x^2)*Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2
)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4])/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[(r*x^2*(s + r*x^2))/(s + (1
+ Sqrt[3])*r*x^2)^2]), x]] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int (c x)^{10/3} \sqrt [3]{a+b x^2} \, dx &=\frac{(c x)^{13/3} \sqrt [3]{a+b x^2}}{5 c}+\frac{1}{15} (2 a) \int \frac{(c x)^{10/3}}{\left (a+b x^2\right )^{2/3}} \, dx\\ &=\frac{2 a c (c x)^{7/3} \sqrt [3]{a+b x^2}}{45 b}+\frac{(c x)^{13/3} \sqrt [3]{a+b x^2}}{5 c}-\frac{\left (14 a^2 c^2\right ) \int \frac{(c x)^{4/3}}{\left (a+b x^2\right )^{2/3}} \, dx}{135 b}\\ &=-\frac{14 a^2 c^3 \sqrt [3]{c x} \sqrt [3]{a+b x^2}}{135 b^2}+\frac{2 a c (c x)^{7/3} \sqrt [3]{a+b x^2}}{45 b}+\frac{(c x)^{13/3} \sqrt [3]{a+b x^2}}{5 c}+\frac{\left (14 a^3 c^4\right ) \int \frac{1}{(c x)^{2/3} \left (a+b x^2\right )^{2/3}} \, dx}{405 b^2}\\ &=-\frac{14 a^2 c^3 \sqrt [3]{c x} \sqrt [3]{a+b x^2}}{135 b^2}+\frac{2 a c (c x)^{7/3} \sqrt [3]{a+b x^2}}{45 b}+\frac{(c x)^{13/3} \sqrt [3]{a+b x^2}}{5 c}+\frac{\left (14 a^3 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a+\frac{b x^6}{c^2}\right )^{2/3}} \, dx,x,\sqrt [3]{c x}\right )}{135 b^2}\\ &=-\frac{14 a^2 c^3 \sqrt [3]{c x} \sqrt [3]{a+b x^2}}{135 b^2}+\frac{2 a c (c x)^{7/3} \sqrt [3]{a+b x^2}}{45 b}+\frac{(c x)^{13/3} \sqrt [3]{a+b x^2}}{5 c}+\frac{\left (14 a^3 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{b x^6}{c^2}}} \, dx,x,\frac{\sqrt [3]{c x}}{\sqrt [6]{a+b x^2}}\right )}{135 b^2 \sqrt{\frac{a}{a+b x^2}} \sqrt{a+b x^2}}\\ &=-\frac{14 a^2 c^3 \sqrt [3]{c x} \sqrt [3]{a+b x^2}}{135 b^2}+\frac{2 a c (c x)^{7/3} \sqrt [3]{a+b x^2}}{45 b}+\frac{(c x)^{13/3} \sqrt [3]{a+b x^2}}{5 c}+\frac{7 a^2 c^{7/3} \sqrt [3]{c x} \sqrt [3]{a+b x^2} \left (c^{2/3}-\frac{\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right ) \sqrt{\frac{c^{4/3}+\frac{b^{2/3} (c x)^{4/3}}{\left (a+b x^2\right )^{2/3}}+\frac{\sqrt [3]{b} c^{2/3} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}{\left (c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}} F\left (\cos ^{-1}\left (\frac{c^{2/3}-\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}{c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{135 \sqrt [4]{3} b^2 \sqrt{-\frac{\sqrt [3]{b} (c x)^{2/3} \left (c^{2/3}-\frac{\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{\sqrt [3]{a+b x^2} \left (c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}}}\\ \end{align*}

Mathematica [C]  time = 0.0577519, size = 103, normalized size = 0.23 \[ \frac{c^3 \sqrt [3]{c x} \sqrt [3]{a+b x^2} \left (\sqrt [3]{\frac{b x^2}{a}+1} \left (-7 a^2+2 a b x^2+9 b^2 x^4\right )+7 a^2 \, _2F_1\left (-\frac{1}{3},\frac{1}{6};\frac{7}{6};-\frac{b x^2}{a}\right )\right )}{45 b^2 \sqrt [3]{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^(10/3)*(a + b*x^2)^(1/3),x]

[Out]

(c^3*(c*x)^(1/3)*(a + b*x^2)^(1/3)*((1 + (b*x^2)/a)^(1/3)*(-7*a^2 + 2*a*b*x^2 + 9*b^2*x^4) + 7*a^2*Hypergeomet
ric2F1[-1/3, 1/6, 7/6, -((b*x^2)/a)]))/(45*b^2*(1 + (b*x^2)/a)^(1/3))

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Maple [F]  time = 0.02, size = 0, normalized size = 0. \begin{align*} \int \left ( cx \right ) ^{{\frac{10}{3}}}\sqrt [3]{b{x}^{2}+a}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(10/3)*(b*x^2+a)^(1/3),x)

[Out]

int((c*x)^(10/3)*(b*x^2+a)^(1/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{2} + a\right )}^{\frac{1}{3}} \left (c x\right )^{\frac{10}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(10/3)*(b*x^2+a)^(1/3),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(1/3)*(c*x)^(10/3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b x^{2} + a\right )}^{\frac{1}{3}} \left (c x\right )^{\frac{1}{3}} c^{3} x^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(10/3)*(b*x^2+a)^(1/3),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(1/3)*(c*x)^(1/3)*c^3*x^3, x)

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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((c*x)**(10/3)*(b*x**2+a)**(1/3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{2} + a\right )}^{\frac{1}{3}} \left (c x\right )^{\frac{10}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(10/3)*(b*x^2+a)^(1/3),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(1/3)*(c*x)^(10/3), x)