3.904 \(\int \frac{x^2}{(a+b x^8) \sqrt{c+d x^8}} \, dx\)

Optimal. Leaf size=64 \[ \frac{x^3 \sqrt{\frac{d x^8}{c}+1} F_1\left (\frac{3}{8};1,\frac{1}{2};\frac{11}{8};-\frac{b x^8}{a},-\frac{d x^8}{c}\right )}{3 a \sqrt{c+d x^8}} \]

[Out]

(x^3*Sqrt[1 + (d*x^8)/c]*AppellF1[3/8, 1, 1/2, 11/8, -((b*x^8)/a), -((d*x^8)/c)])/(3*a*Sqrt[c + d*x^8])

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Rubi [A]  time = 0.053032, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {511, 510} \[ \frac{x^3 \sqrt{\frac{d x^8}{c}+1} F_1\left (\frac{3}{8};1,\frac{1}{2};\frac{11}{8};-\frac{b x^8}{a},-\frac{d x^8}{c}\right )}{3 a \sqrt{c+d x^8}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/((a + b*x^8)*Sqrt[c + d*x^8]),x]

[Out]

(x^3*Sqrt[1 + (d*x^8)/c]*AppellF1[3/8, 1, 1/2, 11/8, -((b*x^8)/a), -((d*x^8)/c)])/(3*a*Sqrt[c + d*x^8])

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rubi steps

\begin{align*} \int \frac{x^2}{\left (a+b x^8\right ) \sqrt{c+d x^8}} \, dx &=\frac{\sqrt{1+\frac{d x^8}{c}} \int \frac{x^2}{\left (a+b x^8\right ) \sqrt{1+\frac{d x^8}{c}}} \, dx}{\sqrt{c+d x^8}}\\ &=\frac{x^3 \sqrt{1+\frac{d x^8}{c}} F_1\left (\frac{3}{8};1,\frac{1}{2};\frac{11}{8};-\frac{b x^8}{a},-\frac{d x^8}{c}\right )}{3 a \sqrt{c+d x^8}}\\ \end{align*}

Mathematica [A]  time = 0.037005, size = 65, normalized size = 1.02 \[ \frac{x^3 \sqrt{\frac{c+d x^8}{c}} F_1\left (\frac{3}{8};\frac{1}{2},1;\frac{11}{8};-\frac{d x^8}{c},-\frac{b x^8}{a}\right )}{3 a \sqrt{c+d x^8}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^2/((a + b*x^8)*Sqrt[c + d*x^8]),x]

[Out]

(x^3*Sqrt[(c + d*x^8)/c]*AppellF1[3/8, 1/2, 1, 11/8, -((d*x^8)/c), -((b*x^8)/a)])/(3*a*Sqrt[c + d*x^8])

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Maple [F]  time = 0.036, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{2}}{b{x}^{8}+a}{\frac{1}{\sqrt{d{x}^{8}+c}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(b*x^8+a)/(d*x^8+c)^(1/2),x)

[Out]

int(x^2/(b*x^8+a)/(d*x^8+c)^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{{\left (b x^{8} + a\right )} \sqrt{d x^{8} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x^8+a)/(d*x^8+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^2/((b*x^8 + a)*sqrt(d*x^8 + c)), x)

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Fricas [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(x^2/(b*x^8+a)/(d*x^8+c)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\left (a + b x^{8}\right ) \sqrt{c + d x^{8}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(b*x**8+a)/(d*x**8+c)**(1/2),x)

[Out]

Integral(x**2/((a + b*x**8)*sqrt(c + d*x**8)), x)

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Giac [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(x^2/(b*x^8+a)/(d*x^8+c)^(1/2),x, algorithm="giac")

[Out]

Timed out