3.5.79 \(\int \frac {x^{14}}{(a+b x^3)^{4/3} (c+d x^3)} \, dx\)

Optimal. Leaf size=347 \[ -\frac {a^4}{b^4 \sqrt [3]{a+b x^3} (b c-a d)}+\frac {a^2 \left (a+b x^3\right )^{2/3}}{2 b^4 d}+\frac {\left (a+b x^3\right )^{2/3} \left (a^2 d^2+a b c d+b^2 c^2\right )}{2 b^4 d^3}+\frac {a \left (a+b x^3\right )^{2/3} (a d+b c)}{2 b^4 d^2}-\frac {\left (a+b x^3\right )^{5/3} (a d+b c)}{5 b^4 d^2}-\frac {2 a \left (a+b x^3\right )^{5/3}}{5 b^4 d}+\frac {\left (a+b x^3\right )^{8/3}}{8 b^4 d}-\frac {c^4 \log \left (c+d x^3\right )}{6 d^{11/3} (b c-a d)^{4/3}}+\frac {c^4 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{11/3} (b c-a d)^{4/3}}+\frac {c^4 \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{11/3} (b c-a d)^{4/3}} \]

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Rubi [A]  time = 0.44, antiderivative size = 347, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {446, 87, 43, 56, 617, 204, 31} \begin {gather*} \frac {\left (a+b x^3\right )^{2/3} \left (a^2 d^2+a b c d+b^2 c^2\right )}{2 b^4 d^3}-\frac {a^4}{b^4 \sqrt [3]{a+b x^3} (b c-a d)}+\frac {a^2 \left (a+b x^3\right )^{2/3}}{2 b^4 d}+\frac {a \left (a+b x^3\right )^{2/3} (a d+b c)}{2 b^4 d^2}-\frac {\left (a+b x^3\right )^{5/3} (a d+b c)}{5 b^4 d^2}-\frac {2 a \left (a+b x^3\right )^{5/3}}{5 b^4 d}+\frac {\left (a+b x^3\right )^{8/3}}{8 b^4 d}-\frac {c^4 \log \left (c+d x^3\right )}{6 d^{11/3} (b c-a d)^{4/3}}+\frac {c^4 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{11/3} (b c-a d)^{4/3}}+\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^{11/3} (b c-a d)^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^14/((a + b*x^3)^(4/3)*(c + d*x^3)),x]

[Out]

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

Rule 31

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 87

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], ((c + d*x)^n*(e + f*x)^IntegerPart[p])/(a + b*x), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

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

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Mathematica [C]  time = 0.27, size = 157, normalized size = 0.45 \begin {gather*} \frac {\frac {81 a^3 d^3+9 a^2 b d^2 \left (8 c+3 d x^3\right )+3 a b^2 d \left (20 c^2+8 c d x^3-3 d^2 x^6\right )+b^3 \left (40 c^3+20 c^2 d x^3-8 c d^2 x^6+5 d^3 x^9\right )}{b^4}-\frac {40 c^4 \, _2F_1\left (-\frac {1}{3},1;\frac {2}{3};\frac {d \left (b x^3+a\right )}{a d-b c}\right )}{b c-a d}}{40 d^4 \sqrt [3]{a+b x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^14/((a + b*x^3)^(4/3)*(c + d*x^3)),x]

[Out]

((81*a^3*d^3 + 9*a^2*b*d^2*(8*c + 3*d*x^3) + 3*a*b^2*d*(20*c^2 + 8*c*d*x^3 - 3*d^2*x^6) + b^3*(40*c^3 + 20*c^2
*d*x^3 - 8*c*d^2*x^6 + 5*d^3*x^9))/b^4 - (40*c^4*Hypergeometric2F1[-1/3, 1, 2/3, (d*(a + b*x^3))/(-(b*c) + a*d
)])/(b*c - a*d))/(40*d^4*(a + b*x^3)^(1/3))

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IntegrateAlgebraic [A]  time = 0.71, size = 393, normalized size = 1.13 \begin {gather*} -\frac {81 a^4 d^3-9 a^3 b c d^2+27 a^3 b d^3 x^3-12 a^2 b^2 c^2 d-3 a^2 b^2 c d^2 x^3-9 a^2 b^2 d^3 x^6-20 a b^3 c^3-4 a b^3 c^2 d x^3+a b^3 c d^2 x^6+5 a b^3 d^3 x^9-20 b^4 c^3 x^3+8 b^4 c^2 d x^6-5 b^4 c d^2 x^9}{40 b^4 d^3 \sqrt [3]{a+b x^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 )}{3 d^{11/3} (b c-a d)^{4/3}}-\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^{11/3} (b c-a d)^{4/3}}+\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^{11/3} (b c-a d)^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^14/((a + b*x^3)^(4/3)*(c + d*x^3)),x]

[Out]

-1/40*(-20*a*b^3*c^3 - 12*a^2*b^2*c^2*d - 9*a^3*b*c*d^2 + 81*a^4*d^3 - 20*b^4*c^3*x^3 - 4*a*b^3*c^2*d*x^3 - 3*
a^2*b^2*c*d^2*x^3 + 27*a^3*b*d^3*x^3 + 8*b^4*c^2*d*x^6 + a*b^3*c*d^2*x^6 - 9*a^2*b^2*d^3*x^6 - 5*b^4*c*d^2*x^9
+ 5*a*b^3*d^3*x^9)/(b^4*d^3*(b*c - a*d)*(a + b*x^3)^(1/3)) + (c^4*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^(11/3)*(b*c - a*d)^(4/3)) + (c^4*Log[(b*c - a*d)^(1/3) + d^(1/3
)*(a + b*x^3)^(1/3)])/(3*d^(11/3)*(b*c - a*d)^(4/3)) - (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^(11/3)*(b*c - a*d)^(4/3))

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fricas [B]  time = 0.84, size = 1300, normalized size = 3.75

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

[-1/120*(60*sqrt(1/3)*(a*b^5*c^5*d - a^2*b^4*c^4*d^2 + (b^6*c^5*d - a*b^5*c^4*d^2)*x^3)*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)) + 20*(b^5*c^4*x^3 + a*b^4*c^4)*(-b
*c*d^2 + a*d^3)^(2/3)*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)) - 40*(b^5*c^4*x^3 + a*b^4*c^4)*(-b*c*d^2 + a*d^3)^(2/3)*log((b*x^3 + a)^(1/3)*d - (-b*c*d^2 + a*d
^3)^(1/3)) - 3*(20*a*b^4*c^4*d^2 - 8*a^2*b^3*c^3*d^3 - 3*a^3*b^2*c^2*d^4 - 90*a^4*b*c*d^5 + 81*a^5*d^6 + 5*(b^
5*c^2*d^4 - 2*a*b^4*c*d^5 + a^2*b^3*d^6)*x^9 - (8*b^5*c^3*d^3 - 7*a*b^4*c^2*d^4 - 10*a^2*b^3*c*d^5 + 9*a^3*b^2
*d^6)*x^6 + (20*b^5*c^4*d^2 - 16*a*b^4*c^3*d^3 - a^2*b^3*c^2*d^4 - 30*a^3*b^2*c*d^5 + 27*a^4*b*d^6)*x^3)*(b*x^
3 + a)^(2/3))/(a*b^6*c^2*d^5 - 2*a^2*b^5*c*d^6 + a^3*b^4*d^7 + (b^7*c^2*d^5 - 2*a*b^6*c*d^6 + a^2*b^5*d^7)*x^3
), -1/120*(120*sqrt(1/3)*(a*b^5*c^5*d - a^2*b^4*c^4*d^2 + (b^6*c^5*d - a*b^5*c^4*d^2)*x^3)*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) + 20*(b^5*c^4*x^3 + a*b^4*c^4)*(-b*c*d^2 + a*d^3)^(2/3)*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)) - 40*(b^5*c^4*x^3 + a*b^4*c^4)*(-
b*c*d^2 + a*d^3)^(2/3)*log((b*x^3 + a)^(1/3)*d - (-b*c*d^2 + a*d^3)^(1/3)) - 3*(20*a*b^4*c^4*d^2 - 8*a^2*b^3*c
^3*d^3 - 3*a^3*b^2*c^2*d^4 - 90*a^4*b*c*d^5 + 81*a^5*d^6 + 5*(b^5*c^2*d^4 - 2*a*b^4*c*d^5 + a^2*b^3*d^6)*x^9 -
(8*b^5*c^3*d^3 - 7*a*b^4*c^2*d^4 - 10*a^2*b^3*c*d^5 + 9*a^3*b^2*d^6)*x^6 + (20*b^5*c^4*d^2 - 16*a*b^4*c^3*d^3
- a^2*b^3*c^2*d^4 - 30*a^3*b^2*c*d^5 + 27*a^4*b*d^6)*x^3)*(b*x^3 + a)^(2/3))/(a*b^6*c^2*d^5 - 2*a^2*b^5*c*d^6
+ a^3*b^4*d^7 + (b^7*c^2*d^5 - 2*a*b^6*c*d^6 + a^2*b^5*d^7)*x^3)]

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giac [A]  time = 0.26, size = 431, normalized size = 1.24 \begin {gather*} \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^{2} c^{2} d^{5} - 2 \, \sqrt {3} a b c d^{6} + \sqrt {3} a^{2} 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^{2} c^{2} d^{5} - 2 \, a b c d^{6} + a^{2} d^{7}\right )}} + \frac {c^{4} \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^{2} c^{2} d^{3} - 2 \, a b c d^{4} + a^{2} d^{5}\right )}} - \frac {a^{4}}{{\left (b^{5} c - a b^{4} d\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}} + \frac {20 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{30} c^{2} d^{5} - 8 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} b^{29} c d^{6} + 40 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a b^{29} c d^{6} + 5 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} b^{28} d^{7} - 24 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a b^{28} d^{7} + 60 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{2} b^{28} d^{7}}{40 \, b^{32} d^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="giac")

[Out]

(-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^2*c^2*d^5 - 2*sqrt(3)*a*b*c*d^6 + sqrt(3)*a^2*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^2*c^2*d^5 - 2*a*b*c*
d^6 + a^2*d^7) + 1/3*c^4*(-(b*c - a*d)/d)^(2/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b^2*c^2*
d^3 - 2*a*b*c*d^4 + a^2*d^5) - a^4/((b^5*c - a*b^4*d)*(b*x^3 + a)^(1/3)) + 1/40*(20*(b*x^3 + a)^(2/3)*b^30*c^2
*d^5 - 8*(b*x^3 + a)^(5/3)*b^29*c*d^6 + 40*(b*x^3 + a)^(2/3)*a*b^29*c*d^6 + 5*(b*x^3 + a)^(8/3)*b^28*d^7 - 24*
(b*x^3 + a)^(5/3)*a*b^28*d^7 + 60*(b*x^3 + a)^(2/3)*a^2*b^28*d^7)/(b^32*d^8)

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maple [F]  time = 0.46, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{14}}{\left (b \,x^{3}+a \right )^{\frac {4}{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)^(4/3)/(d*x^3+c),x)

[Out]

int(x^14/(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^14/(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 = 5.19, size = 564, normalized size = 1.63 \begin {gather*} \left (\frac {3\,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 )}{2\,b^4\,d}\right )\,{\left (b\,x^3+a\right )}^{2/3}-\left (\frac {4\,a}{5\,b^4\,d}+\frac {b^5\,c-a\,b^4\,d}{5\,b^8\,d^2}\right )\,{\left (b\,x^3+a\right )}^{5/3}+\frac {{\left (b\,x^3+a\right )}^{8/3}}{8\,b^4\,d}+\frac {a^4}{b^4\,{\left (b\,x^3+a\right )}^{1/3}\,\left (a\,d-b\,c\right )}+\frac {c^4\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,c^8\,d^5-b\,c^9\,d^4\right )-\frac {c^8\,\left (9\,a^4\,d^{15}-36\,a^3\,b\,c\,d^{14}+54\,a^2\,b^2\,c^2\,d^{13}-36\,a\,b^3\,c^3\,d^{12}+9\,b^4\,c^4\,d^{11}\right )}{9\,d^{22/3}\,{\left (a\,d-b\,c\right )}^{8/3}}\right )}{3\,d^{11/3}\,{\left (a\,d-b\,c\right )}^{4/3}}-\frac {\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,c^8\,d^5-b\,c^9\,d^4\right )-\frac {{\left (c^4+\sqrt {3}\,c^4\,1{}\mathrm {i}\right )}^2\,\left (9\,a^4\,d^{15}-36\,a^3\,b\,c\,d^{14}+54\,a^2\,b^2\,c^2\,d^{13}-36\,a\,b^3\,c^3\,d^{12}+9\,b^4\,c^4\,d^{11}\right )}{36\,d^{22/3}\,{\left (a\,d-b\,c\right )}^{8/3}}\right )\,\left (c^4+\sqrt {3}\,c^4\,1{}\mathrm {i}\right )}{6\,d^{11/3}\,{\left (a\,d-b\,c\right )}^{4/3}}+\frac {c^4\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,c^8\,d^5-b\,c^9\,d^4\right )-\frac {c^8\,{\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2\,\left (9\,a^4\,d^{15}-36\,a^3\,b\,c\,d^{14}+54\,a^2\,b^2\,c^2\,d^{13}-36\,a\,b^3\,c^3\,d^{12}+9\,b^4\,c^4\,d^{11}\right )}{d^{22/3}\,{\left (a\,d-b\,c\right )}^{8/3}}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}{d^{11/3}\,{\left (a\,d-b\,c\right )}^{4/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^14/((a + b*x^3)^(4/3)*(c + d*x^3)),x)

[Out]

((3*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))/(2*b^4*d))*(a + b*x^3)^(2
/3) - ((4*a)/(5*b^4*d) + (b^5*c - a*b^4*d)/(5*b^8*d^2))*(a + b*x^3)^(5/3) + (a + b*x^3)^(8/3)/(8*b^4*d) + a^4/
(b^4*(a + b*x^3)^(1/3)*(a*d - b*c)) + (c^4*log((a + b*x^3)^(1/3)*(a*c^8*d^5 - b*c^9*d^4) - (c^8*(9*a^4*d^15 +
9*b^4*c^4*d^11 - 36*a*b^3*c^3*d^12 + 54*a^2*b^2*c^2*d^13 - 36*a^3*b*c*d^14))/(9*d^(22/3)*(a*d - b*c)^(8/3))))/
(3*d^(11/3)*(a*d - b*c)^(4/3)) - (log((a + b*x^3)^(1/3)*(a*c^8*d^5 - b*c^9*d^4) - ((3^(1/2)*c^4*1i + c^4)^2*(9
*a^4*d^15 + 9*b^4*c^4*d^11 - 36*a*b^3*c^3*d^12 + 54*a^2*b^2*c^2*d^13 - 36*a^3*b*c*d^14))/(36*d^(22/3)*(a*d - b
*c)^(8/3)))*(3^(1/2)*c^4*1i + c^4))/(6*d^(11/3)*(a*d - b*c)^(4/3)) + (c^4*log((a + b*x^3)^(1/3)*(a*c^8*d^5 - b
*c^9*d^4) - (c^8*((3^(1/2)*1i)/6 - 1/6)^2*(9*a^4*d^15 + 9*b^4*c^4*d^11 - 36*a*b^3*c^3*d^12 + 54*a^2*b^2*c^2*d^
13 - 36*a^3*b*c*d^14))/(d^(22/3)*(a*d - b*c)^(8/3)))*((3^(1/2)*1i)/6 - 1/6))/(d^(11/3)*(a*d - b*c)^(4/3))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{14}}{\left (a + b x^{3}\right )^{\frac {4}{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)**(4/3)/(d*x**3+c),x)

[Out]

Integral(x**14/((a + b*x**3)**(4/3)*(c + d*x**3)), x)

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