3.278 \(\int \frac{x^3}{\sqrt{c+d x^3} (4 c+d x^3)} \, dx\)

Optimal. Leaf size=66 \[ \frac{x^4 \sqrt{\frac{d x^3}{c}+1} F_1\left (\frac{4}{3};1,\frac{1}{2};\frac{7}{3};-\frac{d x^3}{4 c},-\frac{d x^3}{c}\right )}{16 c \sqrt{c+d x^3}} \]

[Out]

(x^4*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 1, 1/2, 7/3, -(d*x^3)/(4*c), -((d*x^3)/c)])/(16*c*Sqrt[c + d*x^3])

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Rubi [A]  time = 0.0525761, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {511, 510} \[ \frac{x^4 \sqrt{\frac{d x^3}{c}+1} F_1\left (\frac{4}{3};1,\frac{1}{2};\frac{7}{3};-\frac{d x^3}{4 c},-\frac{d x^3}{c}\right )}{16 c \sqrt{c+d x^3}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(Sqrt[c + d*x^3]*(4*c + d*x^3)),x]

[Out]

(x^4*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 1, 1/2, 7/3, -(d*x^3)/(4*c), -((d*x^3)/c)])/(16*c*Sqrt[c + d*x^3])

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^3}{\sqrt{c+d x^3} \left (4 c+d x^3\right )} \, dx &=\frac{\sqrt{1+\frac{d x^3}{c}} \int \frac{x^3}{\left (4 c+d x^3\right ) \sqrt{1+\frac{d x^3}{c}}} \, dx}{\sqrt{c+d x^3}}\\ &=\frac{x^4 \sqrt{1+\frac{d x^3}{c}} F_1\left (\frac{4}{3};1,\frac{1}{2};\frac{7}{3};-\frac{d x^3}{4 c},-\frac{d x^3}{c}\right )}{16 c \sqrt{c+d x^3}}\\ \end{align*}

Mathematica [A]  time = 0.0317753, size = 67, normalized size = 1.02 \[ \frac{x^4 \sqrt{\frac{c+d x^3}{c}} F_1\left (\frac{4}{3};\frac{1}{2},1;\frac{7}{3};-\frac{d x^3}{c},-\frac{d x^3}{4 c}\right )}{16 c \sqrt{c+d x^3}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^3/(Sqrt[c + d*x^3]*(4*c + d*x^3)),x]

[Out]

(x^4*Sqrt[(c + d*x^3)/c]*AppellF1[4/3, 1/2, 1, 7/3, -((d*x^3)/c), -(d*x^3)/(4*c)])/(16*c*Sqrt[c + d*x^3])

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Maple [C]  time = 0.032, size = 696, normalized size = 10.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(d*x^3+4*c)/(d*x^3+c)^(1/2),x)

[Out]

-2/3*I/d^2*3^(1/2)*(-d^2*c)^(1/3)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c
)^(1/3))^(1/2)*((x-1/d*(-d^2*c)^(1/3))/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2)*(-I*(x+1/
2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*EllipticF(1
/3*3^(1/2)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),(I*3^(1/
2)/d*(-d^2*c)^(1/3)/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2))+4/9*I/d^4*2^(1/2)*sum(1/_al
pha^2*(-d^2*c)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)*(d*(x
-1/d*(-d^2*c)^(1/3))/(-3*(-d^2*c)^(1/3)+I*3^(1/2)*(-d^2*c)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-d^2*c
)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-d^2*c)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-
d^2*c)^(2/3)+2*_alpha^2*d^2-(-d^2*c)^(1/3)*_alpha*d-(-d^2*c)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-d^2*c
)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),1/6/d*(2*I*(-d^2*c)^(1/3)*3^(1/2)*_alp
ha^2*d-I*(-d^2*c)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-d^2*c)^(2/3)*_alpha-3*c*d)/c,(I*3^(1/2)/d*(-d^2*c)^(1
/3)/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*d+4*c))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{{\left (d x^{3} + 4 \, c\right )} \sqrt{d x^{3} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(d*x^3+4*c)/(d*x^3+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^3/((d*x^3 + 4*c)*sqrt(d*x^3 + c)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{d x^{3} + c} x^{3}}{d^{2} x^{6} + 5 \, c d x^{3} + 4 \, c^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(d*x^3+4*c)/(d*x^3+c)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*x^3 + c)*x^3/(d^2*x^6 + 5*c*d*x^3 + 4*c^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\sqrt{c + d x^{3}} \left (4 c + d x^{3}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(d*x**3+4*c)/(d*x**3+c)**(1/2),x)

[Out]

Integral(x**3/(sqrt(c + d*x**3)*(4*c + d*x**3)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{{\left (d x^{3} + 4 \, c\right )} \sqrt{d x^{3} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(d*x^3+4*c)/(d*x^3+c)^(1/2),x, algorithm="giac")

[Out]

integrate(x^3/((d*x^3 + 4*c)*sqrt(d*x^3 + c)), x)