3.7.61 \(\int \frac {x^2 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx\) [661]

Optimal. Leaf size=159 \[ \frac {\sqrt [3]{a+b x^3}}{d}+\frac {\sqrt [3]{b c-a d} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{4/3}}+\frac {\sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^{4/3}}-\frac {\sqrt [3]{b c-a d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{4/3}} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {455, 52, 60, 631, 210, 31} \begin {gather*} \frac {\sqrt [3]{b c-a d} \text {ArcTan}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{4/3}}+\frac {\sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^{4/3}}-\frac {\sqrt [3]{b c-a d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{4/3}}+\frac {\sqrt [3]{a+b x^3}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a + b*x^3)^(1/3)/d + ((b*c - a*d)^(1/3)*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]]
)/(Sqrt[3]*d^(4/3)) + ((b*c - a*d)^(1/3)*Log[c + d*x^3])/(6*d^(4/3)) - ((b*c - a*d)^(1/3)*Log[(b*c - a*d)^(1/3
) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*d^(4/3))

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 60

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-(b*c - a*d)/b, 3]}, Simp[-
Log[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]
&& NegQ[(b*c - a*d)/b]

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 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

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^2 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {\sqrt [3]{a+b x}}{c+d x} \, dx,x,x^3\right )\\ &=\frac {\sqrt [3]{a+b x^3}}{d}-\frac {(b c-a d) \text {Subst}\left (\int \frac {1}{(a+b x)^{2/3} (c+d x)} \, dx,x,x^3\right )}{3 d}\\ &=\frac {\sqrt [3]{a+b x^3}}{d}+\frac {\sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^{4/3}}-\frac {\sqrt [3]{b c-a d} \text {Subst}\left (\int \frac {1}{\frac {\sqrt [3]{b c-a d}}{\sqrt [3]{d}}+x} \, dx,x,\sqrt [3]{a+b x^3}\right )}{2 d^{4/3}}-\frac {(b c-a d)^{2/3} \text {Subst}\left (\int \frac {1}{\frac {(b c-a d)^{2/3}}{d^{2/3}}-\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{d}}+x^2} \, dx,x,\sqrt [3]{a+b x^3}\right )}{2 d^{5/3}}\\ &=\frac {\sqrt [3]{a+b x^3}}{d}+\frac {\sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^{4/3}}-\frac {\sqrt [3]{b c-a d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{4/3}}-\frac {\sqrt [3]{b c-a d} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}\right )}{d^{4/3}}\\ &=\frac {\sqrt [3]{a+b x^3}}{d}+\frac {\sqrt [3]{b c-a d} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{4/3}}+\frac {\sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^{4/3}}-\frac {\sqrt [3]{b c-a d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{4/3}}\\ \end {align*}

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Mathematica [A]
time = 0.19, size = 205, normalized size = 1.29 \begin {gather*} \frac {6 \sqrt [3]{d} \sqrt [3]{a+b x^3}+2 \sqrt {3} \sqrt [3]{b c-a d} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )-2 \sqrt [3]{b c-a d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )+\sqrt [3]{b c-a d} \log \left ((b c-a d)^{2/3}-\sqrt [3]{d} \sqrt [3]{b c-a d} \sqrt [3]{a+b x^3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{6 d^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(6*d^(1/3)*(a + b*x^3)^(1/3) + 2*Sqrt[3]*(b*c - a*d)^(1/3)*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*
d)^(1/3))/Sqrt[3]] - 2*(b*c - a*d)^(1/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)] + (b*c - a*d)^(1/3
)*Log[(b*c - a*d)^(2/3) - d^(1/3)*(b*c - a*d)^(1/3)*(a + b*x^3)^(1/3) + d^(2/3)*(a + b*x^3)^(2/3)])/(6*d^(4/3)
)

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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int \frac {x^{2} \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{d \,x^{3}+c}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more detail

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Fricas [A]
time = 2.63, size = 206, normalized size = 1.30 \begin {gather*} -\frac {2 \, \sqrt {3} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \arctan \left (-\frac {2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}} - \sqrt {3} {\left (b c - a d\right )}}{3 \, {\left (b c - a d\right )}}\right ) + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right ) - 2 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right ) - 6 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \sqrt [3]{a + b x^{3}}}{c + d x^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.73, size = 223, normalized size = 1.40 \begin {gather*} \frac {{\left (b c - a d\right )} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b c d - a d^{2}\right )}} - \frac {\sqrt {3} {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{3 \, d^{2}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{d} - \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{6 \, d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 4.62, size = 249, normalized size = 1.57 \begin {gather*} \frac {{\left (b\,x^3+a\right )}^{1/3}}{d}+\frac {\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (3\,a\,d^2-3\,b\,c\,d\right )-\frac {{\left (a\,d-b\,c\right )}^{1/3}\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{3\,d^{4/3}}\right )\,{\left (a\,d-b\,c\right )}^{1/3}}{3\,d^{4/3}}-\frac {\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (3\,a\,d^2-3\,b\,c\,d\right )+\frac {\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{1/3}\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{3\,d^{4/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{1/3}}{3\,d^{4/3}}+\frac {\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (3\,a\,d^2-3\,b\,c\,d\right )-\frac {\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,{\left (a\,d-b\,c\right )}^{1/3}\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{d^{4/3}}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,{\left (a\,d-b\,c\right )}^{1/3}}{d^{4/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a + b*x^3)^(1/3)/d + (log((a + b*x^3)^(1/3)*(3*a*d^2 - 3*b*c*d) - ((a*d - b*c)^(1/3)*(9*a*d^3 - 9*b*c*d^2))/(
3*d^(4/3)))*(a*d - b*c)^(1/3))/(3*d^(4/3)) - (log((a + b*x^3)^(1/3)*(3*a*d^2 - 3*b*c*d) + (((3^(1/2)*1i)/2 + 1
/2)*(a*d - b*c)^(1/3)*(9*a*d^3 - 9*b*c*d^2))/(3*d^(4/3)))*((3^(1/2)*1i)/2 + 1/2)*(a*d - b*c)^(1/3))/(3*d^(4/3)
) + (log((a + b*x^3)^(1/3)*(3*a*d^2 - 3*b*c*d) - (((3^(1/2)*1i)/6 - 1/6)*(a*d - b*c)^(1/3)*(9*a*d^3 - 9*b*c*d^
2))/d^(4/3))*((3^(1/2)*1i)/6 - 1/6)*(a*d - b*c)^(1/3))/d^(4/3)

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