3.2.18 \(\int \frac {\sqrt [3]{a+b x^3}}{(c+d x^3)^3} \, dx\) [118]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 59, normalized size of antiderivative = 1.00, 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 = {441, 440} \begin {gather*} \frac {x \sqrt [3]{a+b x^3} F_1\left (\frac {1}{3};-\frac {1}{3},3;\frac {4}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c^3 \sqrt [3]{\frac {b x^3}{a}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 440

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])

Rule 441

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])

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(431\) vs. \(2(59)=118\).
time = 10.40, size = 431, normalized size = 7.31 \begin {gather*} \frac {-b (-4 b c+5 a d) x^4 \left (1+\frac {b x^3}{a}\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 )+\frac {c \left (16 a c x \left (-b^2 c x^3 \left (7 c+4 d x^3\right )+3 a^2 d \left (6 c+5 d x^3\right )+a b \left (-18 c^2-7 c d x^3+5 d^2 x^6\right )\right ) F_1\left (\frac {1}{3};\frac {2}{3},1;\frac {4}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )-4 x^4 \left (-b^2 c x^3 \left (7 c+4 d x^3\right )+a^2 d \left (8 c+5 d x^3\right )+a b \left (-7 c^2+4 c d x^3+5 d^2 x^6\right )\right ) \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 )\right )}{\left (c+d x^3\right )^2 \left (-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 )+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 )\right )}}{72 c^3 (b c-a d) \left (a+b x^3\right )^{2/3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-(b*(-4*b*c + 5*a*d)*x^4*(1 + (b*x^3)/a)^(2/3)*AppellF1[4/3, 2/3, 1, 7/3, -((b*x^3)/a), -((d*x^3)/c)]) + (c*(
16*a*c*x*(-(b^2*c*x^3*(7*c + 4*d*x^3)) + 3*a^2*d*(6*c + 5*d*x^3) + a*b*(-18*c^2 - 7*c*d*x^3 + 5*d^2*x^6))*Appe
llF1[1/3, 2/3, 1, 4/3, -((b*x^3)/a), -((d*x^3)/c)] - 4*x^4*(-(b^2*c*x^3*(7*c + 4*d*x^3)) + a^2*d*(8*c + 5*d*x^
3) + a*b*(-7*c^2 + 4*c*d*x^3 + 5*d^2*x^6))*(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)^2*(-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*Appel
lF1[4/3, 5/3, 1, 7/3, -((b*x^3)/a), -((d*x^3)/c)]))))/(72*c^3*(b*c - a*d)*(a + b*x^3)^(2/3))

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{\left (d \,x^{3}+c \right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [3]{a + b x^{3}}}{\left (c + d x^{3}\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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