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

Optimal. Leaf size=255 \[ \frac {2 a^{9/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 c^{3/4} \sqrt {a+c x^4}}-\frac {4 a^{9/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 c^{3/4} \sqrt {a+c x^4}}+\frac {4 a^2 x \sqrt {a+c x^4}}{15 \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {1}{9} x^3 \left (a+c x^4\right )^{3/2}+\frac {2}{15} a x^3 \sqrt {a+c x^4} \]

[Out]

1/9*x^3*(c*x^4+a)^(3/2)+2/15*a*x^3*(c*x^4+a)^(1/2)+4/15*a^2*x*(c*x^4+a)^(1/2)/c^(1/2)/(a^(1/2)+x^2*c^(1/2))-4/
15*a^(9/4)*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticE(sin(2*arctan(
c^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*c^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/c^(3/4)/(c*x^
4+a)^(1/2)+2/15*a^(9/4)*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticF(
sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*c^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/
c^(3/4)/(c*x^4+a)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {279, 305, 220, 1196} \[ \frac {2 a^{9/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 c^{3/4} \sqrt {a+c x^4}}-\frac {4 a^{9/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 c^{3/4} \sqrt {a+c x^4}}+\frac {4 a^2 x \sqrt {a+c x^4}}{15 \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {1}{9} x^3 \left (a+c x^4\right )^{3/2}+\frac {2}{15} a x^3 \sqrt {a+c x^4} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + c*x^4)^(3/2),x]

[Out]

(2*a*x^3*Sqrt[a + c*x^4])/15 + (4*a^2*x*Sqrt[a + c*x^4])/(15*Sqrt[c]*(Sqrt[a] + Sqrt[c]*x^2)) + (x^3*(a + c*x^
4)^(3/2))/9 - (4*a^(9/4)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticE[2*ArcTa
n[(c^(1/4)*x)/a^(1/4)], 1/2])/(15*c^(3/4)*Sqrt[a + c*x^4]) + (2*a^(9/4)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^
4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(15*c^(3/4)*Sqrt[a + c*x^4])

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

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 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin {align*} \int x^2 \left (a+c x^4\right )^{3/2} \, dx &=\frac {1}{9} x^3 \left (a+c x^4\right )^{3/2}+\frac {1}{3} (2 a) \int x^2 \sqrt {a+c x^4} \, dx\\ &=\frac {2}{15} a x^3 \sqrt {a+c x^4}+\frac {1}{9} x^3 \left (a+c x^4\right )^{3/2}+\frac {1}{15} \left (4 a^2\right ) \int \frac {x^2}{\sqrt {a+c x^4}} \, dx\\ &=\frac {2}{15} a x^3 \sqrt {a+c x^4}+\frac {1}{9} x^3 \left (a+c x^4\right )^{3/2}+\frac {\left (4 a^{5/2}\right ) \int \frac {1}{\sqrt {a+c x^4}} \, dx}{15 \sqrt {c}}-\frac {\left (4 a^{5/2}\right ) \int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {a+c x^4}} \, dx}{15 \sqrt {c}}\\ &=\frac {2}{15} a x^3 \sqrt {a+c x^4}+\frac {4 a^2 x \sqrt {a+c x^4}}{15 \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {1}{9} x^3 \left (a+c x^4\right )^{3/2}-\frac {4 a^{9/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 c^{3/4} \sqrt {a+c x^4}}+\frac {2 a^{9/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 c^{3/4} \sqrt {a+c x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 52, normalized size = 0.20 \[ \frac {a x^3 \sqrt {a+c x^4} \, _2F_1\left (-\frac {3}{2},\frac {3}{4};\frac {7}{4};-\frac {c x^4}{a}\right )}{3 \sqrt {\frac {c x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + c*x^4)^(3/2),x]

[Out]

(a*x^3*Sqrt[a + c*x^4]*Hypergeometric2F1[-3/2, 3/4, 7/4, -((c*x^4)/a)])/(3*Sqrt[1 + (c*x^4)/a])

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (c x^{6} + a x^{2}\right )} \sqrt {c x^{4} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(c*x^4+a)^(3/2),x, algorithm="fricas")

[Out]

integral((c*x^6 + a*x^2)*sqrt(c*x^4 + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(c*x^4+a)^(3/2),x, algorithm="giac")

[Out]

integrate((c*x^4 + a)^(3/2)*x^2, x)

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maple [C]  time = 0.01, size = 128, normalized size = 0.50 \[ \frac {\sqrt {c \,x^{4}+a}\, c \,x^{7}}{9}+\frac {11 \sqrt {c \,x^{4}+a}\, a \,x^{3}}{45}+\frac {4 i \sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \left (-\EllipticE \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )+\EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )\right ) a^{\frac {5}{2}}}{15 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*c*x^7*(c*x^4+a)^(1/2)+11/45*a*x^3*(c*x^4+a)^(1/2)+4/15*I*a^(5/2)/(I/a^(1/2)*c^(1/2))^(1/2)*(-I/a^(1/2)*c^(
1/2)*x^2+1)^(1/2)*(I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)/(c*x^4+a)^(1/2)/c^(1/2)*(EllipticF((I/a^(1/2)*c^(1/2))^(1/2)
*x,I)-EllipticE((I/a^(1/2)*c^(1/2))^(1/2)*x,I))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(c*x^4+a)^(3/2),x, algorithm="maxima")

[Out]

integrate((c*x^4 + a)^(3/2)*x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + c*x^4)^(3/2),x)

[Out]

int(x^2*(a + c*x^4)^(3/2), x)

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sympy [C]  time = 1.87, size = 39, normalized size = 0.15 \[ \frac {a^{\frac {3}{2}} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(3/2)*x**3*gamma(3/4)*hyper((-3/2, 3/4), (7/4,), c*x**4*exp_polar(I*pi)/a)/(4*gamma(7/4))

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