3.5.44 \(\int \frac {x^2 (a+b x^3)^{4/3}}{c+d x^3} \, dx\)

Optimal. Leaf size=187 \[ -\frac {(b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 d^{7/3}}+\frac {(b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{7/3}}-\frac {(b c-a d)^{4/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{7/3}}-\frac {\sqrt [3]{a+b x^3} (b c-a d)}{d^2}+\frac {\left (a+b x^3\right )^{4/3}}{4 d} \]

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Rubi [A]  time = 0.21, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {444, 50, 58, 617, 204, 31} \begin {gather*} -\frac {\sqrt [3]{a+b x^3} (b c-a d)}{d^2}-\frac {(b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 d^{7/3}}+\frac {(b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{7/3}}-\frac {(b c-a d)^{4/3} \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^{7/3}}+\frac {\left (a+b x^3\right )^{4/3}}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*x^3)^(4/3))/(c + d*x^3),x]

[Out]

-(((b*c - a*d)*(a + b*x^3)^(1/3))/d^2) + (a + b*x^3)^(4/3)/(4*d) - ((b*c - a*d)^(4/3)*ArcTan[(1 - (2*d^(1/3)*(
a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]])/(Sqrt[3]*d^(7/3)) - ((b*c - a*d)^(4/3)*Log[c + d*x^3])/(6*d^(7/
3)) + ((b*c - a*d)^(4/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*d^(7/3))

Rule 31

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

Rule 50

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 58

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, -Sim
p[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 204

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

Rule 444

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 617

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 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {(a+b x)^{4/3}}{c+d x} \, dx,x,x^3\right )\\ &=\frac {\left (a+b x^3\right )^{4/3}}{4 d}-\frac {(b c-a d) \operatorname {Subst}\left (\int \frac {\sqrt [3]{a+b x}}{c+d x} \, dx,x,x^3\right )}{3 d}\\ &=-\frac {(b c-a d) \sqrt [3]{a+b x^3}}{d^2}+\frac {\left (a+b x^3\right )^{4/3}}{4 d}+\frac {(b c-a d)^2 \operatorname {Subst}\left (\int \frac {1}{(a+b x)^{2/3} (c+d x)} \, dx,x,x^3\right )}{3 d^2}\\ &=-\frac {(b c-a d) \sqrt [3]{a+b x^3}}{d^2}+\frac {\left (a+b x^3\right )^{4/3}}{4 d}-\frac {(b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 d^{7/3}}+\frac {(b c-a d)^{4/3} \operatorname {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^{7/3}}+\frac {(b c-a d)^{5/3} \operatorname {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^{8/3}}\\ &=-\frac {(b c-a d) \sqrt [3]{a+b x^3}}{d^2}+\frac {\left (a+b x^3\right )^{4/3}}{4 d}-\frac {(b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 d^{7/3}}+\frac {(b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{7/3}}+\frac {(b c-a d)^{4/3} \operatorname {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^{7/3}}\\ &=-\frac {(b c-a d) \sqrt [3]{a+b x^3}}{d^2}+\frac {\left (a+b x^3\right )^{4/3}}{4 d}-\frac {(b c-a d)^{4/3} \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^{7/3}}-\frac {(b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 d^{7/3}}+\frac {(b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{7/3}}\\ \end {align*}

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Mathematica [A]  time = 0.32, size = 232, normalized size = 1.24 \begin {gather*} \frac {(a d-b c) \left (\sqrt [3]{b c-a d} \log \left (-\sqrt [3]{d} \sqrt [3]{a+b x^3} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+d^{2/3} \left (a+b x^3\right )^{2/3}\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 )-2 \sqrt {3} \sqrt [3]{b c-a d} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}-1}{\sqrt {3}}\right )+6 \sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{6 d^{7/3}}+\frac {\left (a+b x^3\right )^{4/3}}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(a + b*x^3)^(4/3))/(c + d*x^3),x]

[Out]

(a + b*x^3)^(4/3)/(4*d) + ((-(b*c) + a*d)*(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^(7/3))

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IntegrateAlgebraic [A]  time = 0.24, size = 234, normalized size = 1.25 \begin {gather*} \frac {(b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{3 d^{7/3}}-\frac {(b c-a d)^{4/3} \log \left (-\sqrt [3]{d} \sqrt [3]{a+b x^3} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{6 d^{7/3}}-\frac {(b c-a d)^{4/3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt {3} \sqrt [3]{b c-a d}}\right )}{\sqrt {3} d^{7/3}}+\frac {\sqrt [3]{a+b x^3} \left (5 a d-4 b c+b d x^3\right )}{4 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^2*(a + b*x^3)^(4/3))/(c + d*x^3),x]

[Out]

((a + b*x^3)^(1/3)*(-4*b*c + 5*a*d + b*d*x^3))/(4*d^2) - ((b*c - a*d)^(4/3)*ArcTan[1/Sqrt[3] - (2*d^(1/3)*(a +
b*x^3)^(1/3))/(Sqrt[3]*(b*c - a*d)^(1/3))])/(Sqrt[3]*d^(7/3)) + ((b*c - a*d)^(4/3)*Log[(b*c - a*d)^(1/3) + d^
(1/3)*(a + b*x^3)^(1/3)])/(3*d^(7/3)) - ((b*c - a*d)^(4/3)*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^(7/3))

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fricas [A]  time = 0.44, size = 246, normalized size = 1.32 \begin {gather*} \frac {4 \, \sqrt {3} {\left (b c - a d\right )} \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 ) + 2 \, {\left (b c - a d\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 ) - 4 \, {\left (b c - a d\right )} \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 ) + 3 \, {\left (b d x^{3} - 4 \, b c + 5 \, a d\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{12 \, d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.30, size = 297, normalized size = 1.59 \begin {gather*} -\frac {{\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\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^{4} - a d^{5}\right )}} + \frac {\sqrt {3} {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} {\left (b c - a d\right )} \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^{3}} + \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} {\left (b c - a d\right )} \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^{3}} - \frac {4 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b c d^{2} - {\left (b x^{3} + a\right )}^{\frac {4}{3}} d^{3} - 4 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a d^{3}}{4 \, d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="giac")

[Out]

-1/3*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*(-(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^4 - a*d^5) + 1/3*sqrt(3)*(-b*c*d^2 + a*d^3)^(1/3)*(b*c - a*d)*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^3 + 1/6*(-b*c*d^2 + a*d^3)^(1/3)*(b*c - a*d)*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^3 - 1/4*(4*(b*x^3 + a)
^(1/3)*b*c*d^2 - (b*x^3 + a)^(4/3)*d^3 - 4*(b*x^3 + a)^(1/3)*a*d^3)/d^4

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maple [F]  time = 0.65, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b \,x^{3}+a \right )^{\frac {4}{3}} x^{2}}{d \,x^{3}+c}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b*x^3+a)^(4/3)/(d*x^3+c),x)

[Out]

int(x^2*(b*x^3+a)^(4/3)/(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)^(4/3)/(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 details)Is a*d-b*c positive or negative?

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mupad [B]  time = 4.72, size = 304, normalized size = 1.63 \begin {gather*} \frac {{\left (b\,x^3+a\right )}^{4/3}}{4\,d}+\frac {\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (3\,a^2\,d^2-6\,a\,b\,c\,d+3\,b^2\,c^2\right )-\frac {{\left (a\,d-b\,c\right )}^{4/3}\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{3\,d^{7/3}}\right )\,{\left (a\,d-b\,c\right )}^{4/3}}{3\,d^{7/3}}+\frac {{\left (b\,x^3+a\right )}^{1/3}\,\left (a\,d-b\,c\right )}{d^2}-\frac {\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (3\,a^2\,d^2-6\,a\,b\,c\,d+3\,b^2\,c^2\right )+\frac {\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{4/3}\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{3\,d^{7/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{4/3}}{3\,d^{7/3}}+\frac {\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (3\,a^2\,d^2-6\,a\,b\,c\,d+3\,b^2\,c^2\right )-\frac {\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,{\left (a\,d-b\,c\right )}^{4/3}\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{d^{7/3}}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,{\left (a\,d-b\,c\right )}^{4/3}}{d^{7/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(a + b*x^3)^(4/3))/(c + d*x^3),x)

[Out]

(a + b*x^3)^(4/3)/(4*d) + (log((a + b*x^3)^(1/3)*(3*a^2*d^2 + 3*b^2*c^2 - 6*a*b*c*d) - ((a*d - b*c)^(4/3)*(9*a
*d^3 - 9*b*c*d^2))/(3*d^(7/3)))*(a*d - b*c)^(4/3))/(3*d^(7/3)) + ((a + b*x^3)^(1/3)*(a*d - b*c))/d^2 - (log((a
+ b*x^3)^(1/3)*(3*a^2*d^2 + 3*b^2*c^2 - 6*a*b*c*d) + (((3^(1/2)*1i)/2 + 1/2)*(a*d - b*c)^(4/3)*(9*a*d^3 - 9*b
*c*d^2))/(3*d^(7/3)))*((3^(1/2)*1i)/2 + 1/2)*(a*d - b*c)^(4/3))/(3*d^(7/3)) + (log((a + b*x^3)^(1/3)*(3*a^2*d^
2 + 3*b^2*c^2 - 6*a*b*c*d) - (((3^(1/2)*1i)/6 - 1/6)*(a*d - b*c)^(4/3)*(9*a*d^3 - 9*b*c*d^2))/d^(7/3))*((3^(1/
2)*1i)/6 - 1/6)*(a*d - b*c)^(4/3))/d^(7/3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \left (a + b x^{3}\right )^{\frac {4}{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)**(4/3)/(d*x**3+c),x)

[Out]

Integral(x**2*(a + b*x**3)**(4/3)/(c + d*x**3), x)

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