3.679 \(\int \frac {x^5 (a+b x^3)^{2/3}}{c+d x^3} \, dx\)

Optimal. Leaf size=188 \[ \frac {c (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^{8/3}}-\frac {c (b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{8/3}}-\frac {c (b c-a d)^{2/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{8/3}}-\frac {c \left (a+b x^3\right )^{2/3}}{2 d^2}+\frac {\left (a+b x^3\right )^{5/3}}{5 b d} \]

[Out]

-1/2*c*(b*x^3+a)^(2/3)/d^2+1/5*(b*x^3+a)^(5/3)/b/d+1/6*c*(-a*d+b*c)^(2/3)*ln(d*x^3+c)/d^(8/3)-1/2*c*(-a*d+b*c)
^(2/3)*ln((-a*d+b*c)^(1/3)+d^(1/3)*(b*x^3+a)^(1/3))/d^(8/3)-1/3*c*(-a*d+b*c)^(2/3)*arctan(1/3*(1-2*d^(1/3)*(b*
x^3+a)^(1/3)/(-a*d+b*c)^(1/3))*3^(1/2))/d^(8/3)*3^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.20, antiderivative size = 188, 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, 56, 617, 204, 31} \[ -\frac {c \left (a+b x^3\right )^{2/3}}{2 d^2}+\frac {c (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^{8/3}}-\frac {c (b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{8/3}}-\frac {c (b c-a d)^{2/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{8/3}}+\frac {\left (a+b x^3\right )^{5/3}}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*(a + b*x^3)^(2/3))/(2*d^2) + (a + b*x^3)^(5/3)/(5*b*d) - (c*(b*c - a*d)^(2/3)*ArcTan[(1 - (2*d^(1/3)*(a +
b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]])/(Sqrt[3]*d^(8/3)) + (c*(b*c - a*d)^(2/3)*Log[c + d*x^3])/(6*d^(8/3)
) - (c*(b*c - a*d)^(2/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*d^(8/3))

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

________________________________________________________________________________________

Mathematica [C]  time = 0.05, size = 68, normalized size = 0.36 \[ \frac {\left (a+b x^3\right )^{2/3} \left (5 b c \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {d \left (b x^3+a\right )}{a d-b c}\right )+2 a d-5 b c+2 b d x^3\right )}{10 b d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^3)^(2/3)*(-5*b*c + 2*a*d + 2*b*d*x^3 + 5*b*c*Hypergeometric2F1[2/3, 1, 5/3, (d*(a + b*x^3))/(-(b*c)
+ a*d)]))/(10*b*d^2)

________________________________________________________________________________________

fricas [B]  time = 0.83, size = 353, normalized size = 1.88 \[ -\frac {10 \, \sqrt {3} b c \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}} \arctan \left (-\frac {2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}} + \sqrt {3} {\left (b c - a d\right )}}{3 \, {\left (b c - a d\right )}}\right ) + 5 \, b c \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} d \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {2}{3}} - {\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (b c - a d\right )} + {\left (b c - a d\right )} \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}}\right ) - 10 \, b c \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}} \log \left (-d \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {2}{3}} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c - a d\right )}\right ) - 3 \, {\left (2 \, b d x^{3} - 5 \, b c + 2 \, a d\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{30 \, b d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

-1/30*(10*sqrt(3)*b*c*(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(1/3)*arctan(-1/3*(2*sqrt(3)*(b*x^3 + a)^(1/3)*d*
(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(1/3) + sqrt(3)*(b*c - a*d))/(b*c - a*d)) + 5*b*c*(-(b^2*c^2 - 2*a*b*c*
d + a^2*d^2)/d^2)^(1/3)*log((b*x^3 + a)^(1/3)*d*(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(2/3) - (b*x^3 + a)^(2/
3)*(b*c - a*d) + (b*c - a*d)*(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(1/3)) - 10*b*c*(-(b^2*c^2 - 2*a*b*c*d + a
^2*d^2)/d^2)^(1/3)*log(-d*(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(2/3) - (b*x^3 + a)^(1/3)*(b*c - a*d)) - 3*(2
*b*d*x^3 - 5*b*c + 2*a*d)*(b*x^3 + a)^(2/3))/(b*d^2)

________________________________________________________________________________________

giac [B]  time = 0.55, size = 306, normalized size = 1.63 \[ -\frac {{\left (b^{7} c^{2} d^{3} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} - a b^{6} c d^{4} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b^{7} c d^{5} - a b^{6} d^{6}\right )}} - \frac {\sqrt {3} {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} c \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{3 \, d^{4}} + \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} c \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{6 \, d^{4}} - \frac {5 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{5} c d^{3} - 2 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} b^{4} d^{4}}{10 \, b^{5} d^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="giac")

[Out]

-1/3*(b^7*c^2*d^3*(-(b*c - a*d)/d)^(1/3) - a*b^6*c*d^4*(-(b*c - a*d)/d)^(1/3))*(-(b*c - a*d)/d)^(1/3)*log(abs(
(b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b^7*c*d^5 - a*b^6*d^6) - 1/3*sqrt(3)*(-b*c*d^2 + a*d^3)^(2/3)*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^4 + 1/6*(-b*c*d^2
+ a*d^3)^(2/3)*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^
4 - 1/10*(5*(b*x^3 + a)^(2/3)*b^5*c*d^3 - 2*(b*x^3 + a)^(5/3)*b^4*d^4)/(b^5*d^5)

________________________________________________________________________________________

maple [F]  time = 0.68, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} x^{5}}{d \,x^{3}+c}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(b*x^3+a)^(2/3)/(d*x^3+c),x)

[Out]

int(x^5*(b*x^3+a)^(2/3)/(d*x^3+c),x)

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^3+a)^(2/3)/(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 = 5.06, size = 302, normalized size = 1.61 \[ \frac {{\left (b\,x^3+a\right )}^{5/3}}{5\,b\,d}-{\left (b\,x^3+a\right )}^{2/3}\,\left (\frac {a}{2\,b\,d}+\frac {b^2\,c-a\,b\,d}{2\,b^2\,d^2}\right )-\frac {c\,\ln \left (\frac {{\left (b\,x^3+a\right )}^{1/3}\,\left (a^2\,c^2\,d^2-2\,a\,b\,c^3\,d+b^2\,c^4\right )}{d^3}-\frac {c^2\,{\left (a\,d-b\,c\right )}^{4/3}\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{9\,d^{16/3}}\right )\,{\left (a\,d-b\,c\right )}^{2/3}}{3\,d^{8/3}}-\frac {c\,\ln \left (\frac {c^2\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{7/3}}{d^{10/3}}+\frac {c^2\,{\left (b\,x^3+a\right )}^{1/3}\,{\left (a\,d-b\,c\right )}^2}{d^3}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{2/3}}{3\,d^{8/3}}+\frac {c\,\ln \left (\frac {c^2\,{\left (b\,x^3+a\right )}^{1/3}\,{\left (a\,d-b\,c\right )}^2}{d^3}-\frac {c^2\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{7/3}}{d^{10/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{2/3}}{3\,d^{8/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*(a + b*x^3)^(2/3))/(c + d*x^3),x)

[Out]

(a + b*x^3)^(5/3)/(5*b*d) - (a + b*x^3)^(2/3)*(a/(2*b*d) + (b^2*c - a*b*d)/(2*b^2*d^2)) - (c*log(((a + b*x^3)^
(1/3)*(b^2*c^4 + a^2*c^2*d^2 - 2*a*b*c^3*d))/d^3 - (c^2*(a*d - b*c)^(4/3)*(9*a*d^3 - 9*b*c*d^2))/(9*d^(16/3)))
*(a*d - b*c)^(2/3))/(3*d^(8/3)) - (c*log((c^2*((3^(1/2)*1i)/2 + 1/2)*(a*d - b*c)^(7/3))/d^(10/3) + (c^2*(a + b
*x^3)^(1/3)*(a*d - b*c)^2)/d^3)*((3^(1/2)*1i)/2 - 1/2)*(a*d - b*c)^(2/3))/(3*d^(8/3)) + (c*log((c^2*(a + b*x^3
)^(1/3)*(a*d - b*c)^2)/d^3 - (c^2*((3^(1/2)*1i)/2 - 1/2)*(a*d - b*c)^(7/3))/d^(10/3))*((3^(1/2)*1i)/2 + 1/2)*(
a*d - b*c)^(2/3))/(3*d^(8/3))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5} \left (a + b x^{3}\right )^{\frac {2}{3}}}{c + d x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(b*x**3+a)**(2/3)/(d*x**3+c),x)

[Out]

Integral(x**5*(a + b*x**3)**(2/3)/(c + d*x**3), x)

________________________________________________________________________________________