3.666 \(\int \frac {x^4 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx\)

Optimal. Leaf size=276 \[ \frac {(3 b c-a d) \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{6 b^{2/3} d^2}+\frac {(3 b c-a d) \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} b^{2/3} d^2}+\frac {c^{2/3} \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^2}-\frac {c^{2/3} \sqrt [3]{b c-a d} \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 d^2}-\frac {c^{2/3} \sqrt [3]{b c-a d} \tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} d^2}+\frac {x^2 \sqrt [3]{a+b x^3}}{3 d} \]

[Out]

1/3*x^2*(b*x^3+a)^(1/3)/d+1/6*c^(2/3)*(-a*d+b*c)^(1/3)*ln(d*x^3+c)/d^2+1/6*(-a*d+3*b*c)*ln(b^(1/3)*x-(b*x^3+a)
^(1/3))/b^(2/3)/d^2-1/2*c^(2/3)*(-a*d+b*c)^(1/3)*ln((-a*d+b*c)^(1/3)*x/c^(1/3)-(b*x^3+a)^(1/3))/d^2+1/9*(-a*d+
3*b*c)*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))/b^(2/3)/d^2*3^(1/2)-1/3*c^(2/3)*(-a*d+b*c)^(1/3)*ar
ctan(1/3*(1+2*(-a*d+b*c)^(1/3)*x/c^(1/3)/(b*x^3+a)^(1/3))*3^(1/2))/d^2*3^(1/2)

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Rubi [C]  time = 0.05, antiderivative size = 64, normalized size of antiderivative = 0.23, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {511, 510} \[ \frac {x^5 \sqrt [3]{a+b x^3} F_1\left (\frac {5}{3};-\frac {1}{3},1;\frac {8}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{5 c \sqrt [3]{\frac {b x^3}{a}+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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^4 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx &=\frac {\sqrt [3]{a+b x^3} \int \frac {x^4 \sqrt [3]{1+\frac {b x^3}{a}}}{c+d x^3} \, dx}{\sqrt [3]{1+\frac {b x^3}{a}}}\\ &=\frac {x^5 \sqrt [3]{a+b x^3} F_1\left (\frac {5}{3};-\frac {1}{3},1;\frac {8}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{5 c \sqrt [3]{1+\frac {b x^3}{a}}}\\ \end {align*}

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Mathematica [C]  time = 0.19, size = 185, normalized size = 0.67 \[ \frac {x^5 \left (\frac {b x^3}{a}+1\right )^{2/3} \left (\frac {d x^3}{c}+1\right )^{2/3} (a d-3 b c) F_1\left (\frac {5}{3};\frac {2}{3},1;\frac {8}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+5 c x^2 \left (\left (a+b x^3\right ) \left (\frac {d x^3}{c}+1\right )^{2/3}-a \left (\frac {b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {2}{3};\frac {5}{3};\frac {(a d-b c) x^3}{a \left (d x^3+c\right )}\right )\right )}{15 c d \left (a+b x^3\right )^{2/3} \left (\frac {d x^3}{c}+1\right )^{2/3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-3*b*c + a*d)*x^5*(1 + (b*x^3)/a)^(2/3)*(1 + (d*x^3)/c)^(2/3)*AppellF1[5/3, 2/3, 1, 8/3, -((b*x^3)/a), -((d*
x^3)/c)] + 5*c*x^2*((a + b*x^3)*(1 + (d*x^3)/c)^(2/3) - a*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[2/3, 2/3, 5/
3, ((-(b*c) + a*d)*x^3)/(a*(c + d*x^3))]))/(15*c*d*(a + b*x^3)^(2/3)*(1 + (d*x^3)/c)^(2/3))

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fricas [B]  time = 0.94, size = 452, normalized size = 1.64 \[ \frac {6 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{2} d x^{2} + 6 \, \sqrt {3} {\left (-b c^{3} + a c^{2} d\right )}^{\frac {1}{3}} b^{2} \arctan \left (-\frac {\sqrt {3} {\left (b c^{2} - a c d\right )} x + 2 \, \sqrt {3} {\left (-b c^{3} + a c^{2} d\right )}^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{3 \, {\left (b c^{2} - a c d\right )} x}\right ) + 6 \, {\left (-b c^{3} + a c^{2} d\right )}^{\frac {1}{3}} b^{2} \log \left (\frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} c + {\left (-b c^{3} + a c^{2} d\right )}^{\frac {1}{3}} x}{x}\right ) - 3 \, {\left (-b c^{3} + a c^{2} d\right )}^{\frac {1}{3}} b^{2} \log \left (\frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}} c^{2} - {\left (-b c^{3} + a c^{2} d\right )}^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} c x + {\left (-b c^{3} + a c^{2} d\right )}^{\frac {2}{3}} x^{2}}{x^{2}}\right ) - 2 \, \sqrt {3} {\left (3 \, b^{2} c - a b d\right )} \sqrt {-\left (-b^{2}\right )^{\frac {1}{3}}} \arctan \left (-\frac {{\left (\sqrt {3} \left (-b^{2}\right )^{\frac {1}{3}} b x - 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {2}{3}}\right )} \sqrt {-\left (-b^{2}\right )^{\frac {1}{3}}}}{3 \, b^{2} x}\right ) + 2 \, \left (-b^{2}\right )^{\frac {2}{3}} {\left (3 \, b c - a d\right )} \log \left (-\frac {\left (-b^{2}\right )^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{x}\right ) - \left (-b^{2}\right )^{\frac {2}{3}} {\left (3 \, b c - a d\right )} \log \left (-\frac {\left (-b^{2}\right )^{\frac {1}{3}} b x^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {2}{3}} b}{x^{2}}\right )}{18 \, b^{2} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

1/18*(6*(b*x^3 + a)^(1/3)*b^2*d*x^2 + 6*sqrt(3)*(-b*c^3 + a*c^2*d)^(1/3)*b^2*arctan(-1/3*(sqrt(3)*(b*c^2 - a*c
*d)*x + 2*sqrt(3)*(-b*c^3 + a*c^2*d)^(2/3)*(b*x^3 + a)^(1/3))/((b*c^2 - a*c*d)*x)) + 6*(-b*c^3 + a*c^2*d)^(1/3
)*b^2*log(((b*x^3 + a)^(1/3)*c + (-b*c^3 + a*c^2*d)^(1/3)*x)/x) - 3*(-b*c^3 + a*c^2*d)^(1/3)*b^2*log(((b*x^3 +
a)^(2/3)*c^2 - (-b*c^3 + a*c^2*d)^(1/3)*(b*x^3 + a)^(1/3)*c*x + (-b*c^3 + a*c^2*d)^(2/3)*x^2)/x^2) - 2*sqrt(3
)*(3*b^2*c - a*b*d)*sqrt(-(-b^2)^(1/3))*arctan(-1/3*(sqrt(3)*(-b^2)^(1/3)*b*x - 2*sqrt(3)*(b*x^3 + a)^(1/3)*(-
b^2)^(2/3))*sqrt(-(-b^2)^(1/3))/(b^2*x)) + 2*(-b^2)^(2/3)*(3*b*c - a*d)*log(-((-b^2)^(2/3)*x - (b*x^3 + a)^(1/
3)*b)/x) - (-b^2)^(2/3)*(3*b*c - a*d)*log(-((-b^2)^(1/3)*b*x^2 - (b*x^3 + a)^(1/3)*(-b^2)^(2/3)*x - (b*x^3 + a
)^(2/3)*b)/x^2))/(b^2*d^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} x^{4}}{d x^{3} + c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.60, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{4}}{d \,x^{3}+c}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} x^{4}}{d x^{3} + c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4\,{\left (b\,x^3+a\right )}^{1/3}}{d\,x^3+c} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4} \sqrt [3]{a + b x^{3}}}{c + d x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(b*x**3+a)**(1/3)/(d*x**3+c),x)

[Out]

Integral(x**4*(a + b*x**3)**(1/3)/(c + d*x**3), x)

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