3.267 \(\int \frac{x^3 \sqrt{c+d x^3}}{4 c+d x^3} \, dx\)
Optimal. Leaf size=66 \[ \frac{x^4 \sqrt{c+d x^3} F_1\left (\frac{4}{3};1,-\frac{1}{2};\frac{7}{3};-\frac{d x^3}{4 c},-\frac{d x^3}{c}\right )}{16 c \sqrt{\frac{d x^3}{c}+1}} \]
[Out]
(x^4*Sqrt[c + d*x^3]*AppellF1[4/3, 1, -1/2, 7/3, -(d*x^3)/(4*c), -((d*x^3)/c)])/(16*c*Sqrt[1 + (d*x^3)/c])
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Rubi [A] time = 0.0537595, 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} \[ \frac{x^4 \sqrt{c+d x^3} F_1\left (\frac{4}{3};1,-\frac{1}{2};\frac{7}{3};-\frac{d x^3}{4 c},-\frac{d x^3}{c}\right )}{16 c \sqrt{\frac{d x^3}{c}+1}} \]
Antiderivative was successfully verified.
[In]
Int[(x^3*Sqrt[c + d*x^3])/(4*c + d*x^3),x]
[Out]
(x^4*Sqrt[c + d*x^3]*AppellF1[4/3, 1, -1/2, 7/3, -(d*x^3)/(4*c), -((d*x^3)/c)])/(16*c*Sqrt[1 + (d*x^3)/c])
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}}{4 c+d x^3} \, dx &=\frac{\sqrt{c+d x^3} \int \frac{x^3 \sqrt{1+\frac{d x^3}{c}}}{4 c+d x^3} \, dx}{\sqrt{1+\frac{d x^3}{c}}}\\ &=\frac{x^4 \sqrt{c+d x^3} 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{1+\frac{d x^3}{c}}}\\ \end{align*}
Mathematica [B] time = 0.274762, size = 236, normalized size = 3.58 \[ \frac{x \left (32 \left (\frac{64 c^3 F_1\left (\frac{1}{3};\frac{1}{2},1;\frac{4}{3};-\frac{d x^3}{c},-\frac{d x^3}{4 c}\right )}{d \left (4 c+d x^3\right ) \left (3 d x^3 \left (F_1\left (\frac{4}{3};\frac{1}{2},2;\frac{7}{3};-\frac{d x^3}{c},-\frac{d x^3}{4 c}\right )+2 F_1\left (\frac{4}{3};\frac{3}{2},1;\frac{7}{3};-\frac{d x^3}{c},-\frac{d x^3}{4 c}\right )\right )-16 c F_1\left (\frac{1}{3};\frac{1}{2},1;\frac{4}{3};-\frac{d x^3}{c},-\frac{d x^3}{4 c}\right )\right )}+\frac{c}{d}+x^3\right )-17 x^3 \sqrt{\frac{d x^3}{c}+1} F_1\left (\frac{4}{3};\frac{1}{2},1;\frac{7}{3};-\frac{d x^3}{c},-\frac{d x^3}{4 c}\right )\right )}{80 \sqrt{c+d x^3}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(x^3*Sqrt[c + d*x^3])/(4*c + d*x^3),x]
[Out]
(x*(-17*x^3*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 1/2, 1, 7/3, -((d*x^3)/c), -(d*x^3)/(4*c)] + 32*(c/d + x^3 + (64
*c^3*AppellF1[1/3, 1/2, 1, 4/3, -((d*x^3)/c), -(d*x^3)/(4*c)])/(d*(4*c + d*x^3)*(-16*c*AppellF1[1/3, 1/2, 1, 4
/3, -((d*x^3)/c), -(d*x^3)/(4*c)] + 3*d*x^3*(AppellF1[4/3, 1/2, 2, 7/3, -((d*x^3)/c), -(d*x^3)/(4*c)] + 2*Appe
llF1[4/3, 3/2, 1, 7/3, -((d*x^3)/c), -(d*x^3)/(4*c)]))))))/(80*Sqrt[c + d*x^3])
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Maple [C] time = 0.033, size = 1003, normalized size = 15.2 \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+c)^(1/2)/(d*x^3+4*c),x)
[Out]
1/d*(2/5*x*(d*x^3+c)^(1/2)-2/5*I*c*3^(1/2)/d*(-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/d*c*(-2/3*I*3^(1/2)/d*(-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)*Ellipt
icF(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))+1/3*I/d^3*2^(1/2)*sum(
1/_alpha^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)
*_alpha^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{\sqrt{d x^{3} + c} x^{3}}{d x^{3} + 4 \, c}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(d*x^3+c)^(1/2)/(d*x^3+4*c),x, algorithm="maxima")
[Out]
integrate(sqrt(d*x^3 + c)*x^3/(d*x^3 + 4*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 x^{3} + 4 \, c}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(d*x^3+c)^(1/2)/(d*x^3+4*c),x, algorithm="fricas")
[Out]
integral(sqrt(d*x^3 + c)*x^3/(d*x^3 + 4*c), 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}}}{4 c + d x^{3}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**3*(d*x**3+c)**(1/2)/(d*x**3+4*c),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{\sqrt{d x^{3} + c} x^{3}}{d x^{3} + 4 \, c}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(d*x^3+c)^(1/2)/(d*x^3+4*c),x, algorithm="giac")
[Out]
integrate(sqrt(d*x^3 + c)*x^3/(d*x^3 + 4*c), x)