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

Optimal. Leaf size=186 \[ -\frac {c \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^{7/3}}+\frac {c \sqrt [3]{b c-a d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{7/3}}-\frac {c \sqrt [3]{b c-a d} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{7/3}}-\frac {c \sqrt [3]{a+b x^3}}{d^2}+\frac {\left (a+b x^3\right )^{4/3}}{4 b d} \]

________________________________________________________________________________________

Rubi [A]  time = 0.20, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {446, 80, 50, 58, 617, 204, 31} \begin {gather*} -\frac {c \sqrt [3]{a+b x^3}}{d^2}-\frac {c \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^{7/3}}+\frac {c \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^{7/3}}-\frac {c \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^{7/3}}+\frac {\left (a+b x^3\right )^{4/3}}{4 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.47, size = 204, normalized size = 1.10 \begin {gather*} \frac {c \sqrt [3]{b c-a d} \left (-\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 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )+2 \sqrt {3} \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 )\right )}{6 d^{7/3}}-\frac {c \sqrt [3]{a+b x^3}}{d^2}+\frac {\left (a+b x^3\right )^{4/3}}{4 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.22, size = 239, normalized size = 1.28 \begin {gather*} \frac {c \sqrt [3]{b c-a d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{3 d^{7/3}}-\frac {c \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 )}{6 d^{7/3}}-\frac {c \sqrt [3]{b c-a d} \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 (a d-4 b c+b d x^3\right )}{4 b d^2} \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)^(1/3)*(-4*b*c + a*d + b*d*x^3))/(4*b*d^2) - (c*(b*c - a*d)^(1/3)*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)) + (c*(b*c - a*d)^(1/3)*Log[(b*c - a*d)^(1/3)
+ d^(1/3)*(a + b*x^3)^(1/3)])/(3*d^(7/3)) - (c*(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))

________________________________________________________________________________________

fricas [A]  time = 0.45, size = 222, normalized size = 1.19 \begin {gather*} -\frac {4 \, \sqrt {3} b c \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 \, b c \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 \, b c \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 + a d\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{12 \, b d^{2}} \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]

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

________________________________________________________________________________________

giac [A]  time = 0.25, size = 276, normalized size = 1.48 \begin {gather*} -\frac {{\left (b^{6} c^{2} d^{2} - a b^{5} c d^{3}\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^{6} c d^{4} - a b^{5} d^{5}\right )}} + \frac {\sqrt {3} {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} 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 )}{3 \, d^{3}} + \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} 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 )}{6 \, d^{3}} - \frac {4 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{4} c d^{2} - {\left (b x^{3} + a\right )}^{\frac {4}{3}} b^{3} d^{3}}{4 \, b^{4} d^{4}} \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]

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

________________________________________________________________________________________

maple [F]  time = 0.62, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{5}}{d \,x^{3}+c}\, 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)

________________________________________________________________________________________

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?

________________________________________________________________________________________

mupad [B]  time = 4.62, size = 298, normalized size = 1.60 \begin {gather*} \frac {{\left (b\,x^3+a\right )}^{4/3}}{4\,b\,d}-{\left (b\,x^3+a\right )}^{1/3}\,\left (\frac {a}{b\,d}+\frac {b^2\,c-a\,b\,d}{b^2\,d^2}\right )-\frac {c\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (3\,b\,c^2-3\,a\,c\,d\right )+\frac {c\,{\left (a\,d-b\,c\right )}^{1/3}\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{3\,d^{7/3}}\right )\,{\left (a\,d-b\,c\right )}^{1/3}}{3\,d^{7/3}}-\frac {c\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (3\,b\,c^2-3\,a\,c\,d\right )+\frac {c\,\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^{7/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{1/3}}{3\,d^{7/3}}+\frac {c\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (3\,b\,c^2-3\,a\,c\,d\right )-\frac {c\,\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^{7/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{1/3}}{3\,d^{7/3}} \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)^(4/3)/(4*b*d) - (a + b*x^3)^(1/3)*(a/(b*d) + (b^2*c - a*b*d)/(b^2*d^2)) - (c*log((a + b*x^3)^(1/3)
*(3*b*c^2 - 3*a*c*d) + (c*(a*d - b*c)^(1/3)*(9*a*d^3 - 9*b*c*d^2))/(3*d^(7/3)))*(a*d - b*c)^(1/3))/(3*d^(7/3))
- (c*log((a + b*x^3)^(1/3)*(3*b*c^2 - 3*a*c*d) + (c*((3^(1/2)*1i)/2 - 1/2)*(a*d - b*c)^(1/3)*(9*a*d^3 - 9*b*c
*d^2))/(3*d^(7/3)))*((3^(1/2)*1i)/2 - 1/2)*(a*d - b*c)^(1/3))/(3*d^(7/3)) + (c*log((a + b*x^3)^(1/3)*(3*b*c^2
- 3*a*c*d) - (c*((3^(1/2)*1i)/2 + 1/2)*(a*d - b*c)^(1/3)*(9*a*d^3 - 9*b*c*d^2))/(3*d^(7/3)))*((3^(1/2)*1i)/2 +
1/2)*(a*d - b*c)^(1/3))/(3*d^(7/3))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{5} \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**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)

________________________________________________________________________________________