3.105 \(\int \frac{\sqrt [3]{a+b x^3}}{(c+d x^3)^2} \, dx\)

Optimal. Leaf size=59 \[ \frac{x \sqrt [3]{a+b x^3} F_1\left (\frac{1}{3};-\frac{1}{3},2;\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{c^2 \sqrt [3]{\frac{b x^3}{a}+1}} \]

[Out]

(x*(a + b*x^3)^(1/3)*AppellF1[1/3, -1/3, 2, 4/3, -((b*x^3)/a), -((d*x^3)/c)])/(c^2*(1 + (b*x^3)/a)^(1/3))

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Rubi [A]  time = 0.026517, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {430, 429} \[ \frac{x \sqrt [3]{a+b x^3} F_1\left (\frac{1}{3};-\frac{1}{3},2;\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{c^2 \sqrt [3]{\frac{b x^3}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(1/3)/(c + d*x^3)^2,x]

[Out]

(x*(a + b*x^3)^(1/3)*AppellF1[1/3, -1/3, 2, 4/3, -((b*x^3)/a), -((d*x^3)/c)])/(c^2*(1 + (b*x^3)/a)^(1/3))

Rule 430

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

Rule 429

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

Rubi steps

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

Mathematica [B]  time = 0.202273, size = 232, normalized size = 3.93 \[ \frac{x \left (\frac{4 \left (\frac{a+b x^3}{c}-\frac{8 a^2 F_1\left (\frac{1}{3};\frac{2}{3},1;\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{x^3 \left (3 a d F_1\left (\frac{4}{3};\frac{2}{3},2;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )+2 b c F_1\left (\frac{4}{3};\frac{5}{3},1;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )\right )-4 a c F_1\left (\frac{1}{3};\frac{2}{3},1;\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}\right )}{c+d x^3}+\frac{b x^3 \left (\frac{b x^3}{a}+1\right )^{2/3} F_1\left (\frac{4}{3};\frac{2}{3},1;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{c^2}\right )}{12 \left (a+b x^3\right )^{2/3}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*x^3)^(1/3)/(c + d*x^3)^2,x]

[Out]

(x*((b*x^3*(1 + (b*x^3)/a)^(2/3)*AppellF1[4/3, 2/3, 1, 7/3, -((b*x^3)/a), -((d*x^3)/c)])/c^2 + (4*((a + b*x^3)
/c - (8*a^2*AppellF1[1/3, 2/3, 1, 4/3, -((b*x^3)/a), -((d*x^3)/c)])/(-4*a*c*AppellF1[1/3, 2/3, 1, 4/3, -((b*x^
3)/a), -((d*x^3)/c)] + x^3*(3*a*d*AppellF1[4/3, 2/3, 2, 7/3, -((b*x^3)/a), -((d*x^3)/c)] + 2*b*c*AppellF1[4/3,
5/3, 1, 7/3, -((b*x^3)/a), -((d*x^3)/c)]))))/(c + d*x^3)))/(12*(a + b*x^3)^(2/3))

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Maple [F]  time = 0.417, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( d{x}^{3}+c \right ) ^{2}}\sqrt [3]{b{x}^{3}+a}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{3} + a\right )}^{\frac{1}{3}}}{{\left (d x^{3} + c\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*x**3)**(1/3)/(c + d*x**3)**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{3} + a\right )}^{\frac{1}{3}}}{{\left (d x^{3} + c\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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