3.374 \(\int \frac {x (c+d x^3)^{3/2}}{a+b x^3} \, dx\)

Optimal. Leaf size=65 \[ \frac {c x^2 \sqrt {c+d x^3} F_1\left (\frac {2}{3};1,-\frac {3}{2};\frac {5}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a \sqrt {\frac {d x^3}{c}+1}} \]

[Out]

1/2*c*x^2*AppellF1(2/3,1,-3/2,5/3,-b*x^3/a,-d*x^3/c)*(d*x^3+c)^(1/2)/a/(1+d*x^3/c)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {511, 510} \[ \frac {c x^2 \sqrt {c+d x^3} F_1\left (\frac {2}{3};1,-\frac {3}{2};\frac {5}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a \sqrt {\frac {d x^3}{c}+1}} \]

Antiderivative was successfully verified.

[In]

Int[(x*(c + d*x^3)^(3/2))/(a + b*x^3),x]

[Out]

(c*x^2*Sqrt[c + d*x^3]*AppellF1[2/3, 1, -3/2, 5/3, -((b*x^3)/a), -((d*x^3)/c)])/(2*a*Sqrt[1 + (d*x^3)/c])

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])

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])

Rubi steps

\begin {align*} \int \frac {x \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx &=\frac {\left (c \sqrt {c+d x^3}\right ) \int \frac {x \left (1+\frac {d x^3}{c}\right )^{3/2}}{a+b x^3} \, dx}{\sqrt {1+\frac {d x^3}{c}}}\\ &=\frac {c x^2 \sqrt {c+d x^3} F_1\left (\frac {2}{3};1,-\frac {3}{2};\frac {5}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a \sqrt {1+\frac {d x^3}{c}}}\\ \end {align*}

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Mathematica [B]  time = 0.21, size = 149, normalized size = 2.29 \[ \frac {x^2 \left (2 d x^3 \sqrt {\frac {d x^3}{c}+1} (10 b c-7 a d) F_1\left (\frac {5}{3};\frac {1}{2},1;\frac {8}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+5 c \sqrt {\frac {d x^3}{c}+1} (7 b c-4 a d) F_1\left (\frac {2}{3};\frac {1}{2},1;\frac {5}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+20 a d \left (c+d x^3\right )\right )}{70 a b \sqrt {c+d x^3}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x*(c + d*x^3)^(3/2))/(a + b*x^3),x]

[Out]

(x^2*(20*a*d*(c + d*x^3) + 5*c*(7*b*c - 4*a*d)*Sqrt[1 + (d*x^3)/c]*AppellF1[2/3, 1/2, 1, 5/3, -((d*x^3)/c), -(
(b*x^3)/a)] + 2*d*(10*b*c - 7*a*d)*x^3*Sqrt[1 + (d*x^3)/c]*AppellF1[5/3, 1/2, 1, 8/3, -((d*x^3)/c), -((b*x^3)/
a)]))/(70*a*b*Sqrt[c + d*x^3])

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fricas [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(x*(d*x^3+c)^(3/2)/(b*x^3+a),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x^3 + c)^(3/2)*x/(b*x^3 + a), x)

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maple [C]  time = 0.37, size = 930, normalized size = 14.31 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*d/b*x^2*(d*x^3+c)^(1/2)-2/3*I*(-(a*d-2*b*c)/b^2*d-4/7/b*c*d)*3^(1/2)*(-c*d^2)^(1/3)/d*(I*(x+1/2*(-c*d^2)^(
1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2)*((x-(-c*d^2)^(1/3)/d)/(-3/2*(-c*d^2)^(1
/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d))^(1/2)*(-I*(x+1/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/
2)/(-c*d^2)^(1/3)*d)^(1/2)/(d*x^3+c)^(1/2)*((-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*EllipticE(1
/3*3^(1/2)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2),(I*3^(1/
2)*(-c*d^2)^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)/d)^(1/2))+(-c*d^2)^(1/3)/d*EllipticF(
1/3*3^(1/2)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2),(I*3^(1
/2)*(-c*d^2)^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)/d)^(1/2)))+1/3*I/b^2/d^2*2^(1/2)*sum
((-a^2*d^2+2*a*b*c*d-b^2*c^2)/_alpha/(a*d-b*c)*(-c*d^2)^(1/3)*(1/2*I*(2*x+(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^
(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)*((x-(-c*d^2)^(1/3)/d)/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3))*d)^(1/2)*
(-1/2*I*(2*x+(I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)/(d*x^3+c)^(1/2)*(2*_alpha^2*
d^2+I*(-c*d^2)^(1/3)*3^(1/2)*_alpha*d-(-c*d^2)^(1/3)*_alpha*d-I*3^(1/2)*(-c*d^2)^(2/3)-(-c*d^2)^(2/3))*Ellipti
cPi(1/3*3^(1/2)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2),1/2
*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_alpha^2*d+I*3^(1/2)*c*d-3*c*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha-3*(-c*d^2)^(2/3)*_
alpha)/(a*d-b*c)*b/d,(I*3^(1/2)*(-c*d^2)^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)/d)^(1/2)
),_alpha=RootOf(_Z^3*b+a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x^3 + c)^(3/2)*x/(b*x^3 + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x\,{\left (d\,x^3+c\right )}^{3/2}}{b\,x^3+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*(c + d*x**3)**(3/2)/(a + b*x**3), x)

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