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

Optimal. Leaf size=233 \[ \frac {a \log \left (a d-b d x^3\right )}{3\ 2^{2/3} b^{5/3} d}+\frac {2 a \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{3 b^{5/3} d}-\frac {a \log \left (\sqrt [3]{2} \sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2^{2/3} b^{5/3} d}+\frac {4 a \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} b^{5/3} d}-\frac {\sqrt [3]{2} a \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{2} \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} b^{5/3} d}-\frac {x^2 \sqrt [3]{a+b x^3}}{3 b d} \]

[Out]

-1/3*x^2*(b*x^3+a)^(1/3)/b/d+1/6*a*ln(-b*d*x^3+a*d)*2^(1/3)/b^(5/3)/d+2/3*a*ln(b^(1/3)*x-(b*x^3+a)^(1/3))/b^(5
/3)/d-1/2*a*ln(2^(1/3)*b^(1/3)*x-(b*x^3+a)^(1/3))*2^(1/3)/b^(5/3)/d+4/9*a*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^
(1/3))*3^(1/2))/b^(5/3)/d*3^(1/2)-1/3*2^(1/3)*a*arctan(1/3*(1+2*2^(1/3)*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))/b^
(5/3)/d*3^(1/2)

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Rubi [C]  time = 0.06, antiderivative size = 66, normalized size of antiderivative = 0.28, number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {511, 510} \[ \frac {x^5 \sqrt [3]{a+b x^3} F_1\left (\frac {5}{3};-\frac {1}{3},1;\frac {8}{3};-\frac {b x^3}{a},\frac {b x^3}{a}\right )}{5 a d \sqrt [3]{\frac {b x^3}{a}+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

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Mathematica [C]  time = 0.17, size = 160, normalized size = 0.69 \[ \frac {4 b x^5 \left (1-\frac {b^2 x^6}{a^2}\right )^{2/3} F_1\left (\frac {5}{3};\frac {2}{3},1;\frac {8}{3};-\frac {b x^3}{a},\frac {b x^3}{a}\right )-5 x^2 \left (\left (a+b x^3\right ) \left (1-\frac {b x^3}{a}\right )^{2/3}-a \left (\frac {b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {2}{3};\frac {5}{3};-\frac {2 b x^3}{a-b x^3}\right )\right )}{15 b d \left (a+b x^3\right )^{2/3} \left (1-\frac {b x^3}{a}\right )^{2/3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*b*x^5*(1 - (b^2*x^6)/a^2)^(2/3)*AppellF1[5/3, 2/3, 1, 8/3, -((b*x^3)/a), (b*x^3)/a] - 5*x^2*((a + b*x^3)*(1
- (b*x^3)/a)^(2/3) - a*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[2/3, 2/3, 5/3, (-2*b*x^3)/(a - b*x^3)]))/(15*b
*d*(a + b*x^3)^(2/3)*(1 - (b*x^3)/a)^(2/3))

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fricas [A]  time = 1.02, size = 338, normalized size = 1.45 \[ -\frac {6 \, \sqrt {3} 2^{\frac {1}{3}} a b^{2} \left (-\frac {1}{b^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} b \left (-\frac {1}{b^{2}}\right )^{\frac {2}{3}} + \sqrt {3} x}{3 \, x}\right ) - 6 \cdot 2^{\frac {1}{3}} a b^{2} \left (-\frac {1}{b^{2}}\right )^{\frac {1}{3}} \log \left (\frac {2^{\frac {1}{3}} b x \left (-\frac {1}{b^{2}}\right )^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) + 3 \cdot 2^{\frac {1}{3}} a b^{2} \left (-\frac {1}{b^{2}}\right )^{\frac {1}{3}} \log \left (\frac {2^{\frac {2}{3}} b^{2} x^{2} \left (-\frac {1}{b^{2}}\right )^{\frac {2}{3}} - 2^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} b x \left (-\frac {1}{b^{2}}\right )^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) + 6 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{2} x^{2} + 8 \, \sqrt {3} a {\left (b^{2}\right )}^{\frac {1}{6}} b \arctan \left (\frac {{\left (\sqrt {3} {\left (b^{2}\right )}^{\frac {1}{3}} b x + 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}}\right )} {\left (b^{2}\right )}^{\frac {1}{6}}}{3 \, b^{2} x}\right ) - 8 \, a {\left (b^{2}\right )}^{\frac {2}{3}} \log \left (-\frac {{\left (b^{2}\right )}^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{x}\right ) + 4 \, a {\left (b^{2}\right )}^{\frac {2}{3}} \log \left (\frac {{\left (b^{2}\right )}^{\frac {1}{3}} b x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}} b}{x^{2}}\right )}{18 \, b^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18*(6*sqrt(3)*2^(1/3)*a*b^2*(-1/b^2)^(1/3)*arctan(1/3*(sqrt(3)*2^(2/3)*(b*x^3 + a)^(1/3)*b*(-1/b^2)^(2/3) +
sqrt(3)*x)/x) - 6*2^(1/3)*a*b^2*(-1/b^2)^(1/3)*log((2^(1/3)*b*x*(-1/b^2)^(1/3) + (b*x^3 + a)^(1/3))/x) + 3*2^
(1/3)*a*b^2*(-1/b^2)^(1/3)*log((2^(2/3)*b^2*x^2*(-1/b^2)^(2/3) - 2^(1/3)*(b*x^3 + a)^(1/3)*b*x*(-1/b^2)^(1/3)
+ (b*x^3 + a)^(2/3))/x^2) + 6*(b*x^3 + a)^(1/3)*b^2*x^2 + 8*sqrt(3)*a*(b^2)^(1/6)*b*arctan(1/3*(sqrt(3)*(b^2)^
(1/3)*b*x + 2*sqrt(3)*(b*x^3 + a)^(1/3)*(b^2)^(2/3))*(b^2)^(1/6)/(b^2*x)) - 8*a*(b^2)^(2/3)*log(-((b^2)^(2/3)*
x - (b*x^3 + a)^(1/3)*b)/x) + 4*a*(b^2)^(2/3)*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^3*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.74, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{4}}{-b d \,x^{3}+a d}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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