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

Optimal. Leaf size=168 \[ -\frac {c \log \left (c+d x^3\right )}{6 d^{5/3} \sqrt [3]{b c-a d}}+\frac {c \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{5/3} \sqrt [3]{b c-a d}}+\frac {c \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{5/3} \sqrt [3]{b c-a d}}+\frac {\left (a+b x^3\right )^{2/3}}{2 b d} \]

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Rubi [A]  time = 0.16, antiderivative size = 168, 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 = {446, 80, 56, 617, 204, 31} \begin {gather*} -\frac {c \log \left (c+d x^3\right )}{6 d^{5/3} \sqrt [3]{b c-a d}}+\frac {c \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{5/3} \sqrt [3]{b c-a d}}+\frac {c \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^{5/3} \sqrt [3]{b c-a d}}+\frac {\left (a+b x^3\right )^{2/3}}{2 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a + b*x^3)^(2/3)/(2*b*d) + (c*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]])/(Sqrt[3]
*d^(5/3)*(b*c - a*d)^(1/3)) - (c*Log[c + d*x^3])/(6*d^(5/3)*(b*c - a*d)^(1/3)) + (c*Log[(b*c - a*d)^(1/3) + d^
(1/3)*(a + b*x^3)^(1/3)])/(2*d^(5/3)*(b*c - a*d)^(1/3))

Rule 31

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

Rule 56

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

Rule 80

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 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 446

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

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Mathematica [C]  time = 0.03, size = 69, normalized size = 0.41 \begin {gather*} -\frac {\left (a+b x^3\right )^{2/3} \left (b c \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {d \left (b x^3+a\right )}{a d-b c}\right )+a d-b c\right )}{2 b d (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.21, size = 224, normalized size = 1.33 \begin {gather*} \frac {c \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{3 d^{5/3} \sqrt [3]{b c-a d}}-\frac {c \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^{5/3} \sqrt [3]{b c-a d}}+\frac {c \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^{5/3} \sqrt [3]{b c-a d}}+\frac {\left (a+b x^3\right )^{2/3}}{2 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.73, size = 667, normalized size = 3.97 \begin {gather*} \left [-\frac {{\left (b c d^{2} - a d^{3}\right )}^{\frac {2}{3}} b c \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} d^{2} - {\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d + {\left (b c d^{2} - a d^{3}\right )}^{\frac {2}{3}}\right ) - 2 \, {\left (b c d^{2} - a d^{3}\right )}^{\frac {2}{3}} b c \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} d + {\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}}\right ) - 3 \, \sqrt {\frac {1}{3}} {\left (b^{2} c^{2} d - a b c d^{2}\right )} \sqrt {-\frac {{\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}}}{b c - a d}} \log \left (\frac {2 \, b d^{2} x^{3} - b c d + 3 \, a d^{2} - 3 \, \sqrt {\frac {1}{3}} {\left (2 \, {\left (b c d^{2} - a d^{3}\right )}^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c d - a d^{2}\right )} - {\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}} {\left (b c - a d\right )}\right )} \sqrt {-\frac {{\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}}}{b c - a d}} - 3 \, {\left (b c d^{2} - a d^{3}\right )}^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{d x^{3} + c}\right ) - 3 \, {\left (b c d^{2} - a d^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{6 \, {\left (b^{2} c d^{3} - a b d^{4}\right )}}, -\frac {{\left (b c d^{2} - a d^{3}\right )}^{\frac {2}{3}} b c \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} d^{2} - {\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d + {\left (b c d^{2} - a d^{3}\right )}^{\frac {2}{3}}\right ) - 2 \, {\left (b c d^{2} - a d^{3}\right )}^{\frac {2}{3}} b c \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} d + {\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}}\right ) + 6 \, \sqrt {\frac {1}{3}} {\left (b^{2} c^{2} d - a b c d^{2}\right )} \sqrt {\frac {{\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}}}{b c - a d}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} d - {\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}}\right )} \sqrt {\frac {{\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}}}{b c - a d}}}{d}\right ) - 3 \, {\left (b c d^{2} - a d^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{6 \, {\left (b^{2} c d^{3} - a b d^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*((b*c*d^2 - a*d^3)^(2/3)*b*c*log((b*x^3 + a)^(2/3)*d^2 - (b*c*d^2 - a*d^3)^(1/3)*(b*x^3 + a)^(1/3)*d + (
b*c*d^2 - a*d^3)^(2/3)) - 2*(b*c*d^2 - a*d^3)^(2/3)*b*c*log((b*x^3 + a)^(1/3)*d + (b*c*d^2 - a*d^3)^(1/3)) - 3
*sqrt(1/3)*(b^2*c^2*d - a*b*c*d^2)*sqrt(-(b*c*d^2 - a*d^3)^(1/3)/(b*c - a*d))*log((2*b*d^2*x^3 - b*c*d + 3*a*d
^2 - 3*sqrt(1/3)*(2*(b*c*d^2 - a*d^3)^(2/3)*(b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*(b*c*d - a*d^2) - (b*c*d^2 -
a*d^3)^(1/3)*(b*c - a*d))*sqrt(-(b*c*d^2 - a*d^3)^(1/3)/(b*c - a*d)) - 3*(b*c*d^2 - a*d^3)^(2/3)*(b*x^3 + a)^
(1/3))/(d*x^3 + c)) - 3*(b*c*d^2 - a*d^3)*(b*x^3 + a)^(2/3))/(b^2*c*d^3 - a*b*d^4), -1/6*((b*c*d^2 - a*d^3)^(2
/3)*b*c*log((b*x^3 + a)^(2/3)*d^2 - (b*c*d^2 - a*d^3)^(1/3)*(b*x^3 + a)^(1/3)*d + (b*c*d^2 - a*d^3)^(2/3)) - 2
*(b*c*d^2 - a*d^3)^(2/3)*b*c*log((b*x^3 + a)^(1/3)*d + (b*c*d^2 - a*d^3)^(1/3)) + 6*sqrt(1/3)*(b^2*c^2*d - a*b
*c*d^2)*sqrt((b*c*d^2 - a*d^3)^(1/3)/(b*c - a*d))*arctan(sqrt(1/3)*(2*(b*x^3 + a)^(1/3)*d - (b*c*d^2 - a*d^3)^
(1/3))*sqrt((b*c*d^2 - a*d^3)^(1/3)/(b*c - a*d))/d) - 3*(b*c*d^2 - a*d^3)*(b*x^3 + a)^(2/3))/(b^2*c*d^3 - a*b*
d^4)]

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giac [A]  time = 0.28, size = 257, normalized size = 1.53 \begin {gather*} \frac {\frac {2 \, b c d \left (-\frac {b c - a d}{d}\right )^{\frac {2}{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 )}{b c d^{2} - a d^{3}} + \frac {6 \, {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} b c \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 )}{\sqrt {3} b c d^{3} - \sqrt {3} a d^{4}} - \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} b c \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 )}{b c d^{3} - a d^{4}} + \frac {3 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{d}}{6 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.61, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{5}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (d \,x^{3}+c \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^5/(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^5/(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 details)Is a*d-b*c positive or negative?

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mupad [B]  time = 5.10, size = 219, normalized size = 1.30 \begin {gather*} \frac {{\left (b\,x^3+a\right )}^{2/3}}{2\,b\,d}+\frac {\ln \left (\frac {c^2\,{\left (b\,x^3+a\right )}^{1/3}}{d}-\frac {c^2\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (a\,d-b\,c\right )}^{1/3}}{4\,d^{4/3}}\right )\,\left (c-\sqrt {3}\,c\,1{}\mathrm {i}\right )}{6\,d^{5/3}\,{\left (a\,d-b\,c\right )}^{1/3}}+\frac {\ln \left (\frac {c^2\,{\left (b\,x^3+a\right )}^{1/3}}{d}-\frac {c^2\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (a\,d-b\,c\right )}^{1/3}}{4\,d^{4/3}}\right )\,\left (c+\sqrt {3}\,c\,1{}\mathrm {i}\right )}{6\,d^{5/3}\,{\left (a\,d-b\,c\right )}^{1/3}}-\frac {c\,\ln \left (\frac {c^2\,{\left (b\,x^3+a\right )}^{1/3}}{d}+\frac {b\,c^3-a\,c^2\,d}{d^{4/3}\,{\left (a\,d-b\,c\right )}^{2/3}}\right )}{3\,d^{5/3}\,{\left (a\,d-b\,c\right )}^{1/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{5}}{\sqrt [3]{a + b x^{3}} \left (c + d x^{3}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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