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

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

[Out]

-16/405*a^3*c^3*(c*x)^(1/3)*(b*x^2+a)^(1/3)/b^2+16/945*a^2*c*(c*x)^(7/3)*(b*x^2+a)^(1/3)/b+8/105*a*(c*x)^(13/3
)*(b*x^2+a)^(1/3)/c+1/7*(c*x)^(13/3)*(b*x^2+a)^(4/3)/c+8/1215*a^3*c^(7/3)*(c*x)^(1/3)*(b*x^2+a)^(1/3)*(c^(2/3)
-b^(1/3)*(c*x)^(2/3)/(b*x^2+a)^(1/3))*((c^(2/3)-b^(1/3)*(c*x)^(2/3)*(1-3^(1/2))/(b*x^2+a)^(1/3))^2/(c^(2/3)-b^
(1/3)*(c*x)^(2/3)*(1+3^(1/2))/(b*x^2+a)^(1/3))^2)^(1/2)/(c^(2/3)-b^(1/3)*(c*x)^(2/3)*(1-3^(1/2))/(b*x^2+a)^(1/
3))*(c^(2/3)-b^(1/3)*(c*x)^(2/3)*(1+3^(1/2))/(b*x^2+a)^(1/3))*EllipticF((1-(c^(2/3)-b^(1/3)*(c*x)^(2/3)*(1-3^(
1/2))/(b*x^2+a)^(1/3))^2/(c^(2/3)-b^(1/3)*(c*x)^(2/3)*(1+3^(1/2))/(b*x^2+a)^(1/3))^2)^(1/2),1/4*6^(1/2)+1/4*2^
(1/2))*((c^(4/3)+b^(2/3)*(c*x)^(4/3)/(b*x^2+a)^(2/3)+b^(1/3)*c^(2/3)*(c*x)^(2/3)/(b*x^2+a)^(1/3))/(c^(2/3)-b^(
1/3)*(c*x)^(2/3)*(1+3^(1/2))/(b*x^2+a)^(1/3))^2)^(1/2)*3^(3/4)/b^2/(-b^(1/3)*(c*x)^(2/3)*(c^(2/3)-b^(1/3)*(c*x
)^(2/3)/(b*x^2+a)^(1/3))/(b*x^2+a)^(1/3)/(c^(2/3)-b^(1/3)*(c*x)^(2/3)*(1+3^(1/2))/(b*x^2+a)^(1/3))^2)^(1/2)

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Rubi [A]  time = 0.83, antiderivative size = 479, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [C]  time = 0.07, 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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fricas [F]  time = 1.36, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b c^{3} x^{5} + a c^{3} x^{3}\right )} {\left (b x^{2} + a\right )}^{\frac {1}{3}} \left (c x\right )^{\frac {1}{3}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{2} + a\right )}^{\frac {4}{3}} \left (c x\right )^{\frac {10}{3}}\,{d x} \]

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)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \left (c x \right )^{\frac {10}{3}} \left (b \,x^{2}+a \right )^{\frac {4}{3}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{2} + a\right )}^{\frac {4}{3}} \left (c x\right )^{\frac {10}{3}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (c\,x\right )}^{10/3}\,{\left (b\,x^2+a\right )}^{4/3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(10/3)*(a + b*x^2)^(4/3),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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