3.5.55 \(\int \frac {x^{14}}{\sqrt [3]{a+b x^3} (c+d x^3)} \, dx\)

Optimal. Leaf size=290 \[ -\frac {\left (a+b x^3\right )^{2/3} (a d+b c) \left (a^2 d^2+b^2 c^2\right )}{2 b^4 d^4}+\frac {\left (a+b x^3\right )^{5/3} \left (3 a^2 d^2+2 a b c d+b^2 c^2\right )}{5 b^4 d^3}-\frac {\left (a+b x^3\right )^{8/3} (3 a d+b c)}{8 b^4 d^2}+\frac {\left (a+b x^3\right )^{11/3}}{11 b^4 d}+\frac {c^4 \log \left (c+d x^3\right )}{6 d^{14/3} \sqrt [3]{b c-a d}}-\frac {c^4 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{14/3} \sqrt [3]{b c-a d}}-\frac {c^4 \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{14/3} \sqrt [3]{b c-a d}} \]

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Rubi [A]  time = 0.32, antiderivative size = 290, 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 = {446, 88, 56, 617, 204, 31} \begin {gather*} \frac {\left (a+b x^3\right )^{5/3} \left (3 a^2 d^2+2 a b c d+b^2 c^2\right )}{5 b^4 d^3}-\frac {\left (a+b x^3\right )^{2/3} (a d+b c) \left (a^2 d^2+b^2 c^2\right )}{2 b^4 d^4}-\frac {\left (a+b x^3\right )^{8/3} (3 a d+b c)}{8 b^4 d^2}+\frac {\left (a+b x^3\right )^{11/3}}{11 b^4 d}+\frac {c^4 \log \left (c+d x^3\right )}{6 d^{14/3} \sqrt [3]{b c-a d}}-\frac {c^4 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{14/3} \sqrt [3]{b c-a d}}-\frac {c^4 \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^{14/3} \sqrt [3]{b c-a d}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^14/((a + b*x^3)^(1/3)*(c + d*x^3)),x]

[Out]

-((b*c + a*d)*(b^2*c^2 + a^2*d^2)*(a + b*x^3)^(2/3))/(2*b^4*d^4) + ((b^2*c^2 + 2*a*b*c*d + 3*a^2*d^2)*(a + b*x
^3)^(5/3))/(5*b^4*d^3) - ((b*c + 3*a*d)*(a + b*x^3)^(8/3))/(8*b^4*d^2) + (a + b*x^3)^(11/3)/(11*b^4*d) - (c^4*
ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]])/(Sqrt[3]*d^(14/3)*(b*c - a*d)^(1/3)) +
(c^4*Log[c + d*x^3])/(6*d^(14/3)*(b*c - a*d)^(1/3)) - (c^4*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])
/(2*d^(14/3)*(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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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^{14}}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^4}{\sqrt [3]{a+b x} (c+d x)} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (\frac {(b c+a d) \left (-b^2 c^2-a^2 d^2\right )}{b^3 d^4 \sqrt [3]{a+b x}}+\frac {\left (b^2 c^2+2 a b c d+3 a^2 d^2\right ) (a+b x)^{2/3}}{b^3 d^3}+\frac {(-b c-3 a d) (a+b x)^{5/3}}{b^3 d^2}+\frac {(a+b x)^{8/3}}{b^3 d}+\frac {c^4}{d^4 \sqrt [3]{a+b x} (c+d x)}\right ) \, dx,x,x^3\right )\\ &=-\frac {(b c+a d) \left (b^2 c^2+a^2 d^2\right ) \left (a+b x^3\right )^{2/3}}{2 b^4 d^4}+\frac {\left (b^2 c^2+2 a b c d+3 a^2 d^2\right ) \left (a+b x^3\right )^{5/3}}{5 b^4 d^3}-\frac {(b c+3 a d) \left (a+b x^3\right )^{8/3}}{8 b^4 d^2}+\frac {\left (a+b x^3\right )^{11/3}}{11 b^4 d}+\frac {c^4 \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a+b x} (c+d x)} \, dx,x,x^3\right )}{3 d^4}\\ &=-\frac {(b c+a d) \left (b^2 c^2+a^2 d^2\right ) \left (a+b x^3\right )^{2/3}}{2 b^4 d^4}+\frac {\left (b^2 c^2+2 a b c d+3 a^2 d^2\right ) \left (a+b x^3\right )^{5/3}}{5 b^4 d^3}-\frac {(b c+3 a d) \left (a+b x^3\right )^{8/3}}{8 b^4 d^2}+\frac {\left (a+b x^3\right )^{11/3}}{11 b^4 d}+\frac {c^4 \log \left (c+d x^3\right )}{6 d^{14/3} \sqrt [3]{b c-a d}}+\frac {c^4 \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^5}-\frac {c^4 \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^{14/3} \sqrt [3]{b c-a d}}\\ &=-\frac {(b c+a d) \left (b^2 c^2+a^2 d^2\right ) \left (a+b x^3\right )^{2/3}}{2 b^4 d^4}+\frac {\left (b^2 c^2+2 a b c d+3 a^2 d^2\right ) \left (a+b x^3\right )^{5/3}}{5 b^4 d^3}-\frac {(b c+3 a d) \left (a+b x^3\right )^{8/3}}{8 b^4 d^2}+\frac {\left (a+b x^3\right )^{11/3}}{11 b^4 d}+\frac {c^4 \log \left (c+d x^3\right )}{6 d^{14/3} \sqrt [3]{b c-a d}}-\frac {c^4 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{14/3} \sqrt [3]{b c-a d}}+\frac {c^4 \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^{14/3} \sqrt [3]{b c-a d}}\\ &=-\frac {(b c+a d) \left (b^2 c^2+a^2 d^2\right ) \left (a+b x^3\right )^{2/3}}{2 b^4 d^4}+\frac {\left (b^2 c^2+2 a b c d+3 a^2 d^2\right ) \left (a+b x^3\right )^{5/3}}{5 b^4 d^3}-\frac {(b c+3 a d) \left (a+b x^3\right )^{8/3}}{8 b^4 d^2}+\frac {\left (a+b x^3\right )^{11/3}}{11 b^4 d}-\frac {c^4 \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^{14/3} \sqrt [3]{b c-a d}}+\frac {c^4 \log \left (c+d x^3\right )}{6 d^{14/3} \sqrt [3]{b c-a d}}-\frac {c^4 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{14/3} \sqrt [3]{b c-a d}}\\ \end {align*}

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Mathematica [C]  time = 0.26, size = 157, normalized size = 0.54 \begin {gather*} \frac {\left (a+b x^3\right )^{2/3} \left (\frac {-81 a^3 d^3+9 a^2 b d^2 \left (6 d x^3-11 c\right )-3 a b^2 d \left (44 c^2-22 c d x^3+15 d^2 x^6\right )+b^3 \left (-220 c^3+88 c^2 d x^3-55 c d^2 x^6+40 d^3 x^9\right )}{b^4}+\frac {220 c^4 \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {d \left (b x^3+a\right )}{a d-b c}\right )}{b c-a d}\right )}{440 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^14/((a + b*x^3)^(1/3)*(c + d*x^3)),x]

[Out]

((a + b*x^3)^(2/3)*((-81*a^3*d^3 + 9*a^2*b*d^2*(-11*c + 6*d*x^3) - 3*a*b^2*d*(44*c^2 - 22*c*d*x^3 + 15*d^2*x^6
) + b^3*(-220*c^3 + 88*c^2*d*x^3 - 55*c*d^2*x^6 + 40*d^3*x^9))/b^4 + (220*c^4*Hypergeometric2F1[2/3, 1, 5/3, (
d*(a + b*x^3))/(-(b*c) + a*d)])/(b*c - a*d)))/(440*d^4)

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IntegrateAlgebraic [A]  time = 0.63, size = 340, normalized size = 1.17 \begin {gather*} \frac {\left (a+b x^3\right )^{2/3} \left (-81 a^3 d^3-99 a^2 b c d^2+54 a^2 b d^3 x^3-132 a b^2 c^2 d+66 a b^2 c d^2 x^3-45 a b^2 d^3 x^6-220 b^3 c^3+88 b^3 c^2 d x^3-55 b^3 c d^2 x^6+40 b^3 d^3 x^9\right )}{440 b^4 d^4}-\frac {c^4 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{3 d^{14/3} \sqrt [3]{b c-a d}}+\frac {c^4 \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^{14/3} \sqrt [3]{b c-a d}}-\frac {c^4 \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^{14/3} \sqrt [3]{b c-a d}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^14/((a + b*x^3)^(1/3)*(c + d*x^3)),x]

[Out]

((a + b*x^3)^(2/3)*(-220*b^3*c^3 - 132*a*b^2*c^2*d - 99*a^2*b*c*d^2 - 81*a^3*d^3 + 88*b^3*c^2*d*x^3 + 66*a*b^2
*c*d^2*x^3 + 54*a^2*b*d^3*x^3 - 55*b^3*c*d^2*x^6 - 45*a*b^2*d^3*x^6 + 40*b^3*d^3*x^9))/(440*b^4*d^4) - (c^4*Ar
cTan[1/Sqrt[3] - (2*d^(1/3)*(a + b*x^3)^(1/3))/(Sqrt[3]*(b*c - a*d)^(1/3))])/(Sqrt[3]*d^(14/3)*(b*c - a*d)^(1/
3)) - (c^4*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(3*d^(14/3)*(b*c - a*d)^(1/3)) + (c^4*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^(14/3)*(b*c - a*
d)^(1/3))

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fricas [A]  time = 0.76, size = 1004, normalized size = 3.46

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14/(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

[1/1320*(220*(-b*c*d^2 + a*d^3)^(2/3)*b^4*c^4*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)) - 440*(-b*c*d^2 + a*d^3)^(2/3)*b^4*c^4*log((b*x^3 + a)^(1/3)*d - (-b*c*d^
2 + a*d^3)^(1/3)) + 660*sqrt(1/3)*(b^5*c^5*d - a*b^4*c^4*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*(220*b^4*c^4*d^2 - 88*a*b^3*c^3*d^3 - 33*a^2*b^2*c^2*d^4
- 18*a^3*b*c*d^5 - 81*a^4*d^6 - 40*(b^4*c*d^5 - a*b^3*d^6)*x^9 + 5*(11*b^4*c^2*d^4 - 2*a*b^3*c*d^5 - 9*a^2*b^
2*d^6)*x^6 - 2*(44*b^4*c^3*d^3 - 11*a*b^3*c^2*d^4 - 6*a^2*b^2*c*d^5 - 27*a^3*b*d^6)*x^3)*(b*x^3 + a)^(2/3))/(b
^5*c*d^6 - a*b^4*d^7), 1/1320*(220*(-b*c*d^2 + a*d^3)^(2/3)*b^4*c^4*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)) - 440*(-b*c*d^2 + a*d^3)^(2/3)*b^4*c^4*log((b*x^3 +
a)^(1/3)*d - (-b*c*d^2 + a*d^3)^(1/3)) + 1320*sqrt(1/3)*(b^5*c^5*d - a*b^4*c^4*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*(220*b^4*c^4*d^2 - 88*a*b^3*c^3*d^3 - 33*a^2*b^2*c^2*d^4 - 18*a^3*b*c*d^5 - 81*a^4
*d^6 - 40*(b^4*c*d^5 - a*b^3*d^6)*x^9 + 5*(11*b^4*c^2*d^4 - 2*a*b^3*c*d^5 - 9*a^2*b^2*d^6)*x^6 - 2*(44*b^4*c^3
*d^3 - 11*a*b^3*c^2*d^4 - 6*a^2*b^2*c*d^5 - 27*a^3*b*d^6)*x^3)*(b*x^3 + a)^(2/3))/(b^5*c*d^6 - a*b^4*d^7)]

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giac [A]  time = 0.33, size = 454, normalized size = 1.57 \begin {gather*} -\frac {b^{48} c^{4} d^{7} \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 )}{3 \, {\left (b^{49} c d^{11} - a b^{48} d^{12}\right )}} - \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} c^{4} \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^{6} - \sqrt {3} a d^{7}} + \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} c^{4} \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 \, {\left (b c d^{6} - a d^{7}\right )}} - \frac {220 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{43} c^{3} d^{7} - 88 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} b^{42} c^{2} d^{8} + 220 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a b^{42} c^{2} d^{8} + 55 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} b^{41} c d^{9} - 176 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a b^{41} c d^{9} + 220 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{2} b^{41} c d^{9} - 40 \, {\left (b x^{3} + a\right )}^{\frac {11}{3}} b^{40} d^{10} + 165 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} a b^{40} d^{10} - 264 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a^{2} b^{40} d^{10} + 220 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{3} b^{40} d^{10}}{440 \, b^{44} d^{11}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14/(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="giac")

[Out]

-1/3*b^48*c^4*d^7*(-(b*c - a*d)/d)^(2/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b^49*c*d^11 - a
*b^48*d^12) - (-b*c*d^2 + a*d^3)^(2/3)*c^4*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^6 - sqrt(3)*a*d^7) + 1/6*(-b*c*d^2 + a*d^3)^(2/3)*c^4*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^6 - a*d^7) - 1/440*(220*(b*x^3
+ a)^(2/3)*b^43*c^3*d^7 - 88*(b*x^3 + a)^(5/3)*b^42*c^2*d^8 + 220*(b*x^3 + a)^(2/3)*a*b^42*c^2*d^8 + 55*(b*x^
3 + a)^(8/3)*b^41*c*d^9 - 176*(b*x^3 + a)^(5/3)*a*b^41*c*d^9 + 220*(b*x^3 + a)^(2/3)*a^2*b^41*c*d^9 - 40*(b*x^
3 + a)^(11/3)*b^40*d^10 + 165*(b*x^3 + a)^(8/3)*a*b^40*d^10 - 264*(b*x^3 + a)^(5/3)*a^2*b^40*d^10 + 220*(b*x^3
+ a)^(2/3)*a^3*b^40*d^10)/(b^44*d^11)

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maple [F]  time = 0.60, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{14}}{\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^14/(b*x^3+a)^(1/3)/(d*x^3+c),x)

[Out]

int(x^14/(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^14/(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 = 5.11, size = 438, normalized size = 1.51 \begin {gather*} \left (\frac {6\,a^2}{5\,b^4\,d}+\frac {\left (\frac {4\,a}{b^4\,d}+\frac {b^5\,c-a\,b^4\,d}{b^8\,d^2}\right )\,\left (b^5\,c-a\,b^4\,d\right )}{5\,b^4\,d}\right )\,{\left (b\,x^3+a\right )}^{5/3}-\left (\frac {a}{2\,b^4\,d}+\frac {b^5\,c-a\,b^4\,d}{8\,b^8\,d^2}\right )\,{\left (b\,x^3+a\right )}^{8/3}-{\left (b\,x^3+a\right )}^{2/3}\,\left (\frac {2\,a^3}{b^4\,d}+\frac {\left (\frac {6\,a^2}{b^4\,d}+\frac {\left (\frac {4\,a}{b^4\,d}+\frac {b^5\,c-a\,b^4\,d}{b^8\,d^2}\right )\,\left (b^5\,c-a\,b^4\,d\right )}{b^4\,d}\right )\,\left (b^5\,c-a\,b^4\,d\right )}{2\,b^4\,d}\right )+\frac {{\left (b\,x^3+a\right )}^{11/3}}{11\,b^4\,d}+\frac {c^4\,\ln \left (\frac {c^8\,{\left (b\,x^3+a\right )}^{1/3}}{d^7}-\frac {c^8\,{\left (a\,d-b\,c\right )}^{1/3}}{d^{22/3}}\right )}{3\,d^{14/3}\,{\left (a\,d-b\,c\right )}^{1/3}}-\frac {\ln \left (\frac {c^8\,{\left (b\,x^3+a\right )}^{1/3}}{d^7}-\frac {c^8\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (a\,d-b\,c\right )}^{1/3}}{4\,d^{22/3}}\right )\,\left (c^4+\sqrt {3}\,c^4\,1{}\mathrm {i}\right )}{6\,d^{14/3}\,{\left (a\,d-b\,c\right )}^{1/3}}+\frac {c^4\,\ln \left (\frac {c^8\,{\left (b\,x^3+a\right )}^{1/3}}{d^7}-\frac {c^8\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (a\,d-b\,c\right )}^{1/3}}{4\,d^{22/3}}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}{d^{14/3}\,{\left (a\,d-b\,c\right )}^{1/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^14/((a + b*x^3)^(1/3)*(c + d*x^3)),x)

[Out]

((6*a^2)/(5*b^4*d) + (((4*a)/(b^4*d) + (b^5*c - a*b^4*d)/(b^8*d^2))*(b^5*c - a*b^4*d))/(5*b^4*d))*(a + b*x^3)^
(5/3) - (a/(2*b^4*d) + (b^5*c - a*b^4*d)/(8*b^8*d^2))*(a + b*x^3)^(8/3) - (a + b*x^3)^(2/3)*((2*a^3)/(b^4*d) +
(((6*a^2)/(b^4*d) + (((4*a)/(b^4*d) + (b^5*c - a*b^4*d)/(b^8*d^2))*(b^5*c - a*b^4*d))/(b^4*d))*(b^5*c - a*b^4
*d))/(2*b^4*d)) + (a + b*x^3)^(11/3)/(11*b^4*d) + (c^4*log((c^8*(a + b*x^3)^(1/3))/d^7 - (c^8*(a*d - b*c)^(1/3
))/d^(22/3)))/(3*d^(14/3)*(a*d - b*c)^(1/3)) - (log((c^8*(a + b*x^3)^(1/3))/d^7 - (c^8*(3^(1/2)*1i + 1)^2*(a*d
- b*c)^(1/3))/(4*d^(22/3)))*(3^(1/2)*c^4*1i + c^4))/(6*d^(14/3)*(a*d - b*c)^(1/3)) + (c^4*log((c^8*(a + b*x^3
)^(1/3))/d^7 - (c^8*(3^(1/2)*1i - 1)^2*(a*d - b*c)^(1/3))/(4*d^(22/3)))*((3^(1/2)*1i)/6 - 1/6))/(d^(14/3)*(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^{14}}{\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**14/(b*x**3+a)**(1/3)/(d*x**3+c),x)

[Out]

Integral(x**14/((a + b*x**3)**(1/3)*(c + d*x**3)), x)

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