∫ x(c+dx3)3∕2 __________________

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]

(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])

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Rubi [A] time = 0.0399799, antiderivative size = 65, normalized size of antiderivative = 1., 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 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 \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*}

Mathematica [B] time = 0.152379, 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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Maple [C] time = 0.027, size = 930, normalized size = 14.3 \begin{align*} \text{result too large to display} \end{align*}

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*(-d*(a*d-2*b*c)/b^2-4/7*d/b*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)*((-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*EllipticE

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

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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{3} + c\right )}^{\frac{3}{2}} x}{b x^{3} + a}\,{d x} \end{align*}

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

[Out]

Timed out

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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \left (c + d x^{3}\right )^{\frac{3}{2}}}{a + b x^{3}}\, dx \end{align*}

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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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{3} + c\right )}^{\frac{3}{2}} x}{b x^{3} + a}\,{d x} \end{align*}

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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