3.6.69 \(\int \frac {x^5 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx\) [569]

Optimal. Leaf size=172 \[ -\frac {a \sqrt [3]{a+b x^3}}{b^2 d}-\frac {\left (a+b x^3\right )^{4/3}}{4 b^2 d}+\frac {\sqrt [3]{2} a^{4/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+b x^3}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^2 d}+\frac {a^{4/3} \log \left (a-b x^3\right )}{3\ 2^{2/3} b^2 d}-\frac {a^{4/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2^{2/3} b^2 d} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {457, 81, 52, 59, 631, 210, 31} \begin {gather*} \frac {\sqrt [3]{2} a^{4/3} \text {ArcTan}\left (\frac {2^{2/3} \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^2 d}+\frac {a^{4/3} \log \left (a-b x^3\right )}{3\ 2^{2/3} b^2 d}-\frac {a^{4/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2^{2/3} b^2 d}-\frac {a \sqrt [3]{a+b x^3}}{b^2 d}-\frac {\left (a+b x^3\right )^{4/3}}{4 b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*(a + b*x^3)^(1/3))/(b^2*d)) - (a + b*x^3)^(4/3)/(4*b^2*d) + (2^(1/3)*a^(4/3)*ArcTan[(a^(1/3) + 2^(2/3)*(a
+ b*x^3)^(1/3))/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*b^2*d) + (a^(4/3)*Log[a - b*x^3])/(3*2^(2/3)*b^2*d) - (a^(4/3)*L
og[2^(1/3)*a^(1/3) - (a + b*x^3)^(1/3)])/(2^(2/3)*b^2*d)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 59

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, Simp[-L
og[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Dist[3/(2*b*q), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x
)^(1/3)], x] - Dist[3/(2*b*q^2), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x]
&& PosQ[(b*c - a*d)/b]

Rule 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 457

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

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {x^5 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {x \sqrt [3]{a+b x}}{a d-b d x} \, dx,x,x^3\right )\\ &=-\frac {\left (a+b x^3\right )^{4/3}}{4 b^2 d}+\frac {a \text {Subst}\left (\int \frac {\sqrt [3]{a+b x}}{a d-b d x} \, dx,x,x^3\right )}{3 b}\\ &=-\frac {a \sqrt [3]{a+b x^3}}{b^2 d}-\frac {\left (a+b x^3\right )^{4/3}}{4 b^2 d}+\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {1}{(a+b x)^{2/3} (a d-b d x)} \, dx,x,x^3\right )}{3 b}\\ &=-\frac {a \sqrt [3]{a+b x^3}}{b^2 d}-\frac {\left (a+b x^3\right )^{4/3}}{4 b^2 d}+\frac {a^{4/3} \log \left (a-b x^3\right )}{3\ 2^{2/3} b^2 d}+\frac {a^{4/3} \text {Subst}\left (\int \frac {1}{\sqrt [3]{2} \sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x^3}\right )}{2^{2/3} b^2 d}+\frac {a^{5/3} \text {Subst}\left (\int \frac {1}{2^{2/3} a^{2/3}+\sqrt [3]{2} \sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+b x^3}\right )}{\sqrt [3]{2} b^2 d}\\ &=-\frac {a \sqrt [3]{a+b x^3}}{b^2 d}-\frac {\left (a+b x^3\right )^{4/3}}{4 b^2 d}+\frac {a^{4/3} \log \left (a-b x^3\right )}{3\ 2^{2/3} b^2 d}-\frac {a^{4/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2^{2/3} b^2 d}-\frac {\left (\sqrt [3]{2} a^{4/3}\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2^{2/3} \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}\right )}{b^2 d}\\ &=-\frac {a \sqrt [3]{a+b x^3}}{b^2 d}-\frac {\left (a+b x^3\right )^{4/3}}{4 b^2 d}+\frac {\sqrt [3]{2} a^{4/3} \tan ^{-1}\left (\frac {1+\frac {2^{2/3} \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt {3} b^2 d}+\frac {a^{4/3} \log \left (a-b x^3\right )}{3\ 2^{2/3} b^2 d}-\frac {a^{4/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2^{2/3} b^2 d}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 188, normalized size = 1.09 \begin {gather*} -\frac {15 a \sqrt [3]{a+b x^3}+3 b x^3 \sqrt [3]{a+b x^3}-4 \sqrt [3]{2} \sqrt {3} a^{4/3} \tan ^{-1}\left (\frac {1+\frac {2^{2/3} \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}}{\sqrt {3}}\right )+4 \sqrt [3]{2} a^{4/3} \log \left (-2 \sqrt [3]{a}+2^{2/3} \sqrt [3]{a+b x^3}\right )-2 \sqrt [3]{2} a^{4/3} \log \left (2 a^{2/3}+2^{2/3} \sqrt [3]{a} \sqrt [3]{a+b x^3}+\sqrt [3]{2} \left (a+b x^3\right )^{2/3}\right )}{12 b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(15*a*(a + b*x^3)^(1/3) + 3*b*x^3*(a + b*x^3)^(1/3) - 4*2^(1/3)*Sqrt[3]*a^(4/3)*ArcTan[(1 + (2^(2/3)*(a
+ b*x^3)^(1/3))/a^(1/3))/Sqrt[3]] + 4*2^(1/3)*a^(4/3)*Log[-2*a^(1/3) + 2^(2/3)*(a + b*x^3)^(1/3)] - 2*2^(1/3)*
a^(4/3)*Log[2*a^(2/3) + 2^(2/3)*a^(1/3)*(a + b*x^3)^(1/3) + 2^(1/3)*(a + b*x^3)^(2/3)])/(b^2*d)

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {x^{5} \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{-b d \,x^{3}+a d}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(b*x^3+a)^(1/3)/(-b*d*x^3+a*d),x)

[Out]

int(x^5*(b*x^3+a)^(1/3)/(-b*d*x^3+a*d),x)

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Maxima [A]
time = 0.52, size = 153, normalized size = 0.89 \begin {gather*} \frac {\frac {4 \, \sqrt {3} 2^{\frac {1}{3}} a^{\frac {4}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right )}{d} + \frac {2 \cdot 2^{\frac {1}{3}} a^{\frac {4}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {2}{3}}\right )}{d} - \frac {4 \cdot 2^{\frac {1}{3}} a^{\frac {4}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )}{d} - \frac {3 \, {\left ({\left (b x^{3} + a\right )}^{\frac {4}{3}} + 4 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a\right )}}{d}}{12 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^3+a)^(1/3)/(-b*d*x^3+a*d),x, algorithm="maxima")

[Out]

1/12*(4*sqrt(3)*2^(1/3)*a^(4/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3)*a^(1/3) + 2*(b*x^3 + a)^(1/3))/a^(1/3))/d
+ 2*2^(1/3)*a^(4/3)*log(2^(2/3)*a^(2/3) + 2^(1/3)*(b*x^3 + a)^(1/3)*a^(1/3) + (b*x^3 + a)^(2/3))/d - 4*2^(1/3)
*a^(4/3)*log(-2^(1/3)*a^(1/3) + (b*x^3 + a)^(1/3))/d - 3*((b*x^3 + a)^(4/3) + 4*(b*x^3 + a)^(1/3)*a)/d)/b^2

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Fricas [A]
time = 4.50, size = 157, normalized size = 0.91 \begin {gather*} -\frac {4 \, \sqrt {3} 2^{\frac {1}{3}} \left (-a\right )^{\frac {1}{3}} a \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-a\right )^{\frac {2}{3}} + \sqrt {3} a}{3 \, a}\right ) + 2 \cdot 2^{\frac {1}{3}} \left (-a\right )^{\frac {1}{3}} a \log \left (2^{\frac {2}{3}} \left (-a\right )^{\frac {2}{3}} - 2^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-a\right )^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {2}{3}}\right ) - 4 \cdot 2^{\frac {1}{3}} \left (-a\right )^{\frac {1}{3}} a \log \left (2^{\frac {1}{3}} \left (-a\right )^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right ) + 3 \, {\left (b x^{3} + 5 \, a\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{12 \, b^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^3+a)^(1/3)/(-b*d*x^3+a*d),x, algorithm="fricas")

[Out]

-1/12*(4*sqrt(3)*2^(1/3)*(-a)^(1/3)*a*arctan(1/3*(sqrt(3)*2^(2/3)*(b*x^3 + a)^(1/3)*(-a)^(2/3) + sqrt(3)*a)/a)
+ 2*2^(1/3)*(-a)^(1/3)*a*log(2^(2/3)*(-a)^(2/3) - 2^(1/3)*(b*x^3 + a)^(1/3)*(-a)^(1/3) + (b*x^3 + a)^(2/3)) -
4*2^(1/3)*(-a)^(1/3)*a*log(2^(1/3)*(-a)^(1/3) + (b*x^3 + a)^(1/3)) + 3*(b*x^3 + 5*a)*(b*x^3 + a)^(1/3))/(b^2*
d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {x^{5} \sqrt [3]{a + b x^{3}}}{- a + b x^{3}}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(b*x**3+a)**(1/3)/(-b*d*x**3+a*d),x)

[Out]

-Integral(x**5*(a + b*x**3)**(1/3)/(-a + b*x**3), x)/d

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^3+a)^(1/3)/(-b*d*x^3+a*d),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Algebraic extensions not allowed in a ro
otofAlgebraic extensions not allowed in a rootofAlgebraic extensions not allowed in a rootofAlgebraic extensio
ns not allowed in a r

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Mupad [B]
time = 4.66, size = 200, normalized size = 1.16 \begin {gather*} -\frac {{\left (b\,x^3+a\right )}^{4/3}}{4\,b^2\,d}-\frac {a\,{\left (b\,x^3+a\right )}^{1/3}}{b^2\,d}-\frac {2^{1/3}\,a^{4/3}\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}-2^{1/3}\,a^{1/3}\right )}{3\,b^2\,d}-\frac {2^{1/3}\,a^{4/3}\,\ln \left (\frac {6\,a^2\,{\left (b\,x^3+a\right )}^{1/3}}{b^2\,d}-\frac {6\,2^{1/3}\,a^{7/3}\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{b^2\,d}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,b^2\,d}+\frac {2^{1/3}\,a^{4/3}\,\ln \left (\frac {6\,a^2\,{\left (b\,x^3+a\right )}^{1/3}}{b^2\,d}+\frac {18\,2^{1/3}\,a^{7/3}\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}{b^2\,d}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}{b^2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*(a + b*x^3)^(1/3))/(a*d - b*d*x^3),x)

[Out]

(2^(1/3)*a^(4/3)*log((6*a^2*(a + b*x^3)^(1/3))/(b^2*d) + (18*2^(1/3)*a^(7/3)*((3^(1/2)*1i)/6 + 1/6))/(b^2*d))*
((3^(1/2)*1i)/6 + 1/6))/(b^2*d) - (a*(a + b*x^3)^(1/3))/(b^2*d) - (2^(1/3)*a^(4/3)*log((a + b*x^3)^(1/3) - 2^(
1/3)*a^(1/3)))/(3*b^2*d) - (2^(1/3)*a^(4/3)*log((6*a^2*(a + b*x^3)^(1/3))/(b^2*d) - (6*2^(1/3)*a^(7/3)*((3^(1/
2)*1i)/2 - 1/2))/(b^2*d))*((3^(1/2)*1i)/2 - 1/2))/(3*b^2*d) - (a + b*x^3)^(4/3)/(4*b^2*d)

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