3.762 \(\int (c x)^{10/3} (a+b x^2)^{4/3} \, dx\)

Optimal. Leaf size=479 \[ \frac{8 a^3 c^{7/3} \sqrt [3]{c x} \sqrt [3]{a+b x^2} \left (c^{2/3}-\frac{\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right ) \sqrt{\frac{\frac{b^{2/3} (c x)^{4/3}}{\left (a+b x^2\right )^{2/3}}+\frac{\sqrt [3]{b} c^{2/3} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}+c^{4/3}}{\left (c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}} \text{EllipticF}\left (\cos ^{-1}\left (\frac{c^{2/3}-\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}{c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}\right ),\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{405 \sqrt [4]{3} b^2 \sqrt{-\frac{\sqrt [3]{b} (c x)^{2/3} \left (c^{2/3}-\frac{\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{\sqrt [3]{a+b x^2} \left (c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}}}-\frac{16 a^3 c^3 \sqrt [3]{c x} \sqrt [3]{a+b x^2}}{405 b^2}+\frac{16 a^2 c (c x)^{7/3} \sqrt [3]{a+b x^2}}{945 b}+\frac{(c x)^{13/3} \left (a+b x^2\right )^{4/3}}{7 c}+\frac{8 a (c x)^{13/3} \sqrt [3]{a+b x^2}}{105 c} \]

[Out]

(-16*a^3*c^3*(c*x)^(1/3)*(a + b*x^2)^(1/3))/(405*b^2) + (16*a^2*c*(c*x)^(7/3)*(a + b*x^2)^(1/3))/(945*b) + (8*
a*(c*x)^(13/3)*(a + b*x^2)^(1/3))/(105*c) + ((c*x)^(13/3)*(a + b*x^2)^(4/3))/(7*c) + (8*a^3*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])/(405*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 = 0.829501, antiderivative size = 479, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ -\frac{16 a^3 c^3 \sqrt [3]{c x} \sqrt [3]{a+b x^2}}{405 b^2}+\frac{8 a^3 c^{7/3} \sqrt [3]{c x} \sqrt [3]{a+b x^2} \left (c^{2/3}-\frac{\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right ) \sqrt{\frac{\frac{b^{2/3} (c x)^{4/3}}{\left (a+b x^2\right )^{2/3}}+\frac{\sqrt [3]{b} c^{2/3} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}+c^{4/3}}{\left (c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}} F\left (\cos ^{-1}\left (\frac{c^{2/3}-\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{b x^2+a}}}{c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{b x^2+a}}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{405 \sqrt [4]{3} b^2 \sqrt{-\frac{\sqrt [3]{b} (c x)^{2/3} \left (c^{2/3}-\frac{\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{\sqrt [3]{a+b x^2} \left (c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}}}+\frac{16 a^2 c (c x)^{7/3} \sqrt [3]{a+b x^2}}{945 b}+\frac{(c x)^{13/3} \left (a+b x^2\right )^{4/3}}{7 c}+\frac{8 a (c x)^{13/3} \sqrt [3]{a+b x^2}}{105 c} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(10/3)*(a + b*x^2)^(4/3),x]

[Out]

(-16*a^3*c^3*(c*x)^(1/3)*(a + b*x^2)^(1/3))/(405*b^2) + (16*a^2*c*(c*x)^(7/3)*(a + b*x^2)^(1/3))/(945*b) + (8*
a*(c*x)^(13/3)*(a + b*x^2)^(1/3))/(105*c) + ((c*x)^(13/3)*(a + b*x^2)^(4/3))/(7*c) + (8*a^3*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])/(405*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} \left (a+b x^2\right )^{4/3} \, dx &=\frac{(c x)^{13/3} \left (a+b x^2\right )^{4/3}}{7 c}+\frac{1}{21} (8 a) \int (c x)^{10/3} \sqrt [3]{a+b x^2} \, dx\\ &=\frac{8 a (c x)^{13/3} \sqrt [3]{a+b x^2}}{105 c}+\frac{(c x)^{13/3} \left (a+b x^2\right )^{4/3}}{7 c}+\frac{1}{315} \left (16 a^2\right ) \int \frac{(c x)^{10/3}}{\left (a+b x^2\right )^{2/3}} \, dx\\ &=\frac{16 a^2 c (c x)^{7/3} \sqrt [3]{a+b x^2}}{945 b}+\frac{8 a (c x)^{13/3} \sqrt [3]{a+b x^2}}{105 c}+\frac{(c x)^{13/3} \left (a+b x^2\right )^{4/3}}{7 c}-\frac{\left (16 a^3 c^2\right ) \int \frac{(c x)^{4/3}}{\left (a+b x^2\right )^{2/3}} \, dx}{405 b}\\ &=-\frac{16 a^3 c^3 \sqrt [3]{c x} \sqrt [3]{a+b x^2}}{405 b^2}+\frac{16 a^2 c (c x)^{7/3} \sqrt [3]{a+b x^2}}{945 b}+\frac{8 a (c x)^{13/3} \sqrt [3]{a+b x^2}}{105 c}+\frac{(c x)^{13/3} \left (a+b x^2\right )^{4/3}}{7 c}+\frac{\left (16 a^4 c^4\right ) \int \frac{1}{(c x)^{2/3} \left (a+b x^2\right )^{2/3}} \, dx}{1215 b^2}\\ &=-\frac{16 a^3 c^3 \sqrt [3]{c x} \sqrt [3]{a+b x^2}}{405 b^2}+\frac{16 a^2 c (c x)^{7/3} \sqrt [3]{a+b x^2}}{945 b}+\frac{8 a (c x)^{13/3} \sqrt [3]{a+b x^2}}{105 c}+\frac{(c x)^{13/3} \left (a+b x^2\right )^{4/3}}{7 c}+\frac{\left (16 a^4 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 )}{405 b^2}\\ &=-\frac{16 a^3 c^3 \sqrt [3]{c x} \sqrt [3]{a+b x^2}}{405 b^2}+\frac{16 a^2 c (c x)^{7/3} \sqrt [3]{a+b x^2}}{945 b}+\frac{8 a (c x)^{13/3} \sqrt [3]{a+b x^2}}{105 c}+\frac{(c x)^{13/3} \left (a+b x^2\right )^{4/3}}{7 c}+\frac{\left (16 a^4 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 )}{405 b^2 \sqrt{\frac{a}{a+b x^2}} \sqrt{a+b x^2}}\\ &=-\frac{16 a^3 c^3 \sqrt [3]{c x} \sqrt [3]{a+b x^2}}{405 b^2}+\frac{16 a^2 c (c x)^{7/3} \sqrt [3]{a+b x^2}}{945 b}+\frac{8 a (c x)^{13/3} \sqrt [3]{a+b x^2}}{105 c}+\frac{(c x)^{13/3} \left (a+b x^2\right )^{4/3}}{7 c}+\frac{8 a^3 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 )}{405 \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 = 0.070091, size = 102, normalized size = 0.21 \[ \frac{c^3 \sqrt [3]{c x} \sqrt [3]{a+b x^2} \left (7 a^3 \, _2F_1\left (-\frac{4}{3},\frac{1}{6};\frac{7}{6};-\frac{b x^2}{a}\right )-\left (7 a-15 b x^2\right ) \left (a+b x^2\right )^2 \sqrt [3]{\frac{b x^2}{a}+1}\right )}{105 b^2 \sqrt [3]{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^(10/3)*(a + b*x^2)^(4/3),x]

[Out]

(c^3*(c*x)^(1/3)*(a + b*x^2)^(1/3)*(-((7*a - 15*b*x^2)*(a + b*x^2)^2*(1 + (b*x^2)/a)^(1/3)) + 7*a^3*Hypergeome
tric2F1[-4/3, 1/6, 7/6, -((b*x^2)/a)]))/(105*b^2*(1 + (b*x^2)/a)^(1/3))

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Maple [F]  time = 0.016, size = 0, normalized size = 0. \begin{align*} \int \left ( cx \right ) ^{{\frac{10}{3}}} \left ( b{x}^{2}+a \right ) ^{{\frac{4}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(10/3)*(b*x^2+a)^(4/3),x)

[Out]

int((c*x)^(10/3)*(b*x^2+a)^(4/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{2} + a\right )}^{\frac{4}{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)^(4/3),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(4/3)*(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 c^{3} x^{5} + a c^{3} x^{3}\right )}{\left (b x^{2} + a\right )}^{\frac{1}{3}} \left (c x\right )^{\frac{1}{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)^(4/3),x, algorithm="fricas")

[Out]

integral((b*c^3*x^5 + a*c^3*x^3)*(b*x^2 + a)^(1/3)*(c*x)^(1/3), 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)**(4/3),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{4}{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)^(4/3),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(4/3)*(c*x)^(10/3), x)