3.6.58 \(\int \frac {x^{14}}{(a+b x^6)^2 \sqrt {c+d x^6}} \, dx\)

Optimal. Leaf size=141 \[ -\frac {\sqrt {a} (3 b c-2 a d) \tan ^{-1}\left (\frac {x^3 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^6}}\right )}{6 b^2 (b c-a d)^{3/2}}+\frac {a x^3 \sqrt {c+d x^6}}{6 b \left (a+b x^6\right ) (b c-a d)}+\frac {\tanh ^{-1}\left (\frac {\sqrt {d} x^3}{\sqrt {c+d x^6}}\right )}{3 b^2 \sqrt {d}} \]

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Rubi [A]  time = 0.16, antiderivative size = 141, 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 = {465, 470, 523, 217, 206, 377, 205} \begin {gather*} -\frac {\sqrt {a} (3 b c-2 a d) \tan ^{-1}\left (\frac {x^3 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^6}}\right )}{6 b^2 (b c-a d)^{3/2}}+\frac {a x^3 \sqrt {c+d x^6}}{6 b \left (a+b x^6\right ) (b c-a d)}+\frac {\tanh ^{-1}\left (\frac {\sqrt {d} x^3}{\sqrt {c+d x^6}}\right )}{3 b^2 \sqrt {d}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^14/((a + b*x^6)^2*Sqrt[c + d*x^6]),x]

[Out]

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 465

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = GCD[m + 1,
n]}, Dist[1/k, Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^q, x], x, x^k], x] /; k != 1] /;
FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rubi steps

\begin {align*} \int \frac {x^{14}}{\left (a+b x^6\right )^2 \sqrt {c+d x^6}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^4}{\left (a+b x^2\right )^2 \sqrt {c+d x^2}} \, dx,x,x^3\right )\\ &=\frac {a x^3 \sqrt {c+d x^6}}{6 b (b c-a d) \left (a+b x^6\right )}-\frac {\operatorname {Subst}\left (\int \frac {a c-2 (b c-a d) x^2}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx,x,x^3\right )}{6 b (b c-a d)}\\ &=\frac {a x^3 \sqrt {c+d x^6}}{6 b (b c-a d) \left (a+b x^6\right )}+\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {c+d x^2}} \, dx,x,x^3\right )}{3 b^2}-\frac {(a (3 b c-2 a d)) \operatorname {Subst}\left (\int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx,x,x^3\right )}{6 b^2 (b c-a d)}\\ &=\frac {a x^3 \sqrt {c+d x^6}}{6 b (b c-a d) \left (a+b x^6\right )}+\frac {\operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x^3}{\sqrt {c+d x^6}}\right )}{3 b^2}-\frac {(a (3 b c-2 a d)) \operatorname {Subst}\left (\int \frac {1}{a-(-b c+a d) x^2} \, dx,x,\frac {x^3}{\sqrt {c+d x^6}}\right )}{6 b^2 (b c-a d)}\\ &=\frac {a x^3 \sqrt {c+d x^6}}{6 b (b c-a d) \left (a+b x^6\right )}-\frac {\sqrt {a} (3 b c-2 a d) \tan ^{-1}\left (\frac {\sqrt {b c-a d} x^3}{\sqrt {a} \sqrt {c+d x^6}}\right )}{6 b^2 (b c-a d)^{3/2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {d} x^3}{\sqrt {c+d x^6}}\right )}{3 b^2 \sqrt {d}}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 135, normalized size = 0.96 \begin {gather*} \frac {\frac {a b x^3 \sqrt {c+d x^6}}{\left (a+b x^6\right ) (b c-a d)}+\frac {\sqrt {a} (2 a d-3 b c) \tan ^{-1}\left (\frac {x^3 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^6}}\right )}{(b c-a d)^{3/2}}+\frac {2 \log \left (\sqrt {d} \sqrt {c+d x^6}+d x^3\right )}{\sqrt {d}}}{6 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^14/((a + b*x^6)^2*Sqrt[c + d*x^6]),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 2.62, size = 166, normalized size = 1.18 \begin {gather*} \frac {\left (2 a^{3/2} d-3 \sqrt {a} b c\right ) \tan ^{-1}\left (\frac {a \sqrt {d}+b x^3 \sqrt {c+d x^6}+b \sqrt {d} x^6}{\sqrt {a} \sqrt {b c-a d}}\right )}{6 b^2 (b c-a d)^{3/2}}+\frac {a x^3 \sqrt {c+d x^6}}{6 b \left (a+b x^6\right ) (b c-a d)}+\frac {\log \left (\sqrt {c+d x^6}+\sqrt {d} x^3\right )}{3 b^2 \sqrt {d}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^14/((a + b*x^6)^2*Sqrt[c + d*x^6]),x]

[Out]

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

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fricas [A]  time = 1.41, size = 1077, normalized size = 7.64 \begin {gather*} \left [\frac {4 \, \sqrt {d x^{6} + c} a b d x^{3} + 4 \, {\left ({\left (b^{2} c - a b d\right )} x^{6} + a b c - a^{2} d\right )} \sqrt {d} \log \left (-2 \, d x^{6} - 2 \, \sqrt {d x^{6} + c} \sqrt {d} x^{3} - c\right ) + {\left ({\left (3 \, b^{2} c d - 2 \, a b d^{2}\right )} x^{6} + 3 \, a b c d - 2 \, a^{2} d^{2}\right )} \sqrt {-\frac {a}{b c - a d}} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{12} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{6} + a^{2} c^{2} - 4 \, {\left ({\left (b^{2} c^{2} - 3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{9} - {\left (a b c^{2} - a^{2} c d\right )} x^{3}\right )} \sqrt {d x^{6} + c} \sqrt {-\frac {a}{b c - a d}}}{b^{2} x^{12} + 2 \, a b x^{6} + a^{2}}\right )}{24 \, {\left ({\left (b^{4} c d - a b^{3} d^{2}\right )} x^{6} + a b^{3} c d - a^{2} b^{2} d^{2}\right )}}, \frac {4 \, \sqrt {d x^{6} + c} a b d x^{3} - 8 \, {\left ({\left (b^{2} c - a b d\right )} x^{6} + a b c - a^{2} d\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x^{3}}{\sqrt {d x^{6} + c}}\right ) + {\left ({\left (3 \, b^{2} c d - 2 \, a b d^{2}\right )} x^{6} + 3 \, a b c d - 2 \, a^{2} d^{2}\right )} \sqrt {-\frac {a}{b c - a d}} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{12} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{6} + a^{2} c^{2} - 4 \, {\left ({\left (b^{2} c^{2} - 3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{9} - {\left (a b c^{2} - a^{2} c d\right )} x^{3}\right )} \sqrt {d x^{6} + c} \sqrt {-\frac {a}{b c - a d}}}{b^{2} x^{12} + 2 \, a b x^{6} + a^{2}}\right )}{24 \, {\left ({\left (b^{4} c d - a b^{3} d^{2}\right )} x^{6} + a b^{3} c d - a^{2} b^{2} d^{2}\right )}}, \frac {2 \, \sqrt {d x^{6} + c} a b d x^{3} + {\left ({\left (3 \, b^{2} c d - 2 \, a b d^{2}\right )} x^{6} + 3 \, a b c d - 2 \, a^{2} d^{2}\right )} \sqrt {\frac {a}{b c - a d}} \arctan \left (-\frac {{\left ({\left (b c - 2 \, a d\right )} x^{6} - a c\right )} \sqrt {d x^{6} + c} \sqrt {\frac {a}{b c - a d}}}{2 \, {\left (a d x^{9} + a c x^{3}\right )}}\right ) + 2 \, {\left ({\left (b^{2} c - a b d\right )} x^{6} + a b c - a^{2} d\right )} \sqrt {d} \log \left (-2 \, d x^{6} - 2 \, \sqrt {d x^{6} + c} \sqrt {d} x^{3} - c\right )}{12 \, {\left ({\left (b^{4} c d - a b^{3} d^{2}\right )} x^{6} + a b^{3} c d - a^{2} b^{2} d^{2}\right )}}, \frac {2 \, \sqrt {d x^{6} + c} a b d x^{3} - 4 \, {\left ({\left (b^{2} c - a b d\right )} x^{6} + a b c - a^{2} d\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x^{3}}{\sqrt {d x^{6} + c}}\right ) + {\left ({\left (3 \, b^{2} c d - 2 \, a b d^{2}\right )} x^{6} + 3 \, a b c d - 2 \, a^{2} d^{2}\right )} \sqrt {\frac {a}{b c - a d}} \arctan \left (-\frac {{\left ({\left (b c - 2 \, a d\right )} x^{6} - a c\right )} \sqrt {d x^{6} + c} \sqrt {\frac {a}{b c - a d}}}{2 \, {\left (a d x^{9} + a c x^{3}\right )}}\right )}{12 \, {\left ({\left (b^{4} c d - a b^{3} d^{2}\right )} x^{6} + a b^{3} c d - a^{2} b^{2} d^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14/(b*x^6+a)^2/(d*x^6+c)^(1/2),x, algorithm="fricas")

[Out]

[1/24*(4*sqrt(d*x^6 + c)*a*b*d*x^3 + 4*((b^2*c - a*b*d)*x^6 + a*b*c - a^2*d)*sqrt(d)*log(-2*d*x^6 - 2*sqrt(d*x
^6 + c)*sqrt(d)*x^3 - c) + ((3*b^2*c*d - 2*a*b*d^2)*x^6 + 3*a*b*c*d - 2*a^2*d^2)*sqrt(-a/(b*c - a*d))*log(((b^
2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^12 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^6 + a^2*c^2 - 4*((b^2*c^2 - 3*a*b*c*d + 2*a^
2*d^2)*x^9 - (a*b*c^2 - a^2*c*d)*x^3)*sqrt(d*x^6 + c)*sqrt(-a/(b*c - a*d)))/(b^2*x^12 + 2*a*b*x^6 + a^2)))/((b
^4*c*d - a*b^3*d^2)*x^6 + a*b^3*c*d - a^2*b^2*d^2), 1/24*(4*sqrt(d*x^6 + c)*a*b*d*x^3 - 8*((b^2*c - a*b*d)*x^6
+ a*b*c - a^2*d)*sqrt(-d)*arctan(sqrt(-d)*x^3/sqrt(d*x^6 + c)) + ((3*b^2*c*d - 2*a*b*d^2)*x^6 + 3*a*b*c*d - 2
*a^2*d^2)*sqrt(-a/(b*c - a*d))*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^12 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^6 + a
^2*c^2 - 4*((b^2*c^2 - 3*a*b*c*d + 2*a^2*d^2)*x^9 - (a*b*c^2 - a^2*c*d)*x^3)*sqrt(d*x^6 + c)*sqrt(-a/(b*c - a*
d)))/(b^2*x^12 + 2*a*b*x^6 + a^2)))/((b^4*c*d - a*b^3*d^2)*x^6 + a*b^3*c*d - a^2*b^2*d^2), 1/12*(2*sqrt(d*x^6
+ c)*a*b*d*x^3 + ((3*b^2*c*d - 2*a*b*d^2)*x^6 + 3*a*b*c*d - 2*a^2*d^2)*sqrt(a/(b*c - a*d))*arctan(-1/2*((b*c -
2*a*d)*x^6 - a*c)*sqrt(d*x^6 + c)*sqrt(a/(b*c - a*d))/(a*d*x^9 + a*c*x^3)) + 2*((b^2*c - a*b*d)*x^6 + a*b*c -
a^2*d)*sqrt(d)*log(-2*d*x^6 - 2*sqrt(d*x^6 + c)*sqrt(d)*x^3 - c))/((b^4*c*d - a*b^3*d^2)*x^6 + a*b^3*c*d - a^
2*b^2*d^2), 1/12*(2*sqrt(d*x^6 + c)*a*b*d*x^3 - 4*((b^2*c - a*b*d)*x^6 + a*b*c - a^2*d)*sqrt(-d)*arctan(sqrt(-
d)*x^3/sqrt(d*x^6 + c)) + ((3*b^2*c*d - 2*a*b*d^2)*x^6 + 3*a*b*c*d - 2*a^2*d^2)*sqrt(a/(b*c - a*d))*arctan(-1/
2*((b*c - 2*a*d)*x^6 - a*c)*sqrt(d*x^6 + c)*sqrt(a/(b*c - a*d))/(a*d*x^9 + a*c*x^3)))/((b^4*c*d - a*b^3*d^2)*x
^6 + a*b^3*c*d - a^2*b^2*d^2)]

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giac [B]  time = 0.51, size = 343, normalized size = 2.43 \begin {gather*} -\frac {{\left (3 \, a b c \sqrt {-d} \arctan \left (\frac {a \sqrt {d}}{\sqrt {a b c - a^{2} d}}\right ) - 2 \, a^{2} \sqrt {-d} d \arctan \left (\frac {a \sqrt {d}}{\sqrt {a b c - a^{2} d}}\right ) - 2 \, \sqrt {a b c - a^{2} d} b c \arctan \left (\frac {\sqrt {d}}{\sqrt {-d}}\right ) + 2 \, \sqrt {a b c - a^{2} d} a d \arctan \left (\frac {\sqrt {d}}{\sqrt {-d}}\right ) + \sqrt {a b c - a^{2} d} a \sqrt {-d} \sqrt {d}\right )} \mathrm {sgn}\relax (x)}{6 \, {\left (\sqrt {a b c - a^{2} d} b^{3} c \sqrt {-d} - \sqrt {a b c - a^{2} d} a b^{2} \sqrt {-d} d\right )}} + \frac {a c \sqrt {d + \frac {c}{x^{6}}}}{6 \, {\left (b^{2} c \mathrm {sgn}\relax (x) - a b d \mathrm {sgn}\relax (x)\right )} {\left (b c + a {\left (d + \frac {c}{x^{6}}\right )} - a d\right )}} + \frac {{\left (3 \, a b c - 2 \, a^{2} d\right )} \arctan \left (\frac {a \sqrt {d + \frac {c}{x^{6}}}}{\sqrt {a b c - a^{2} d}}\right )}{6 \, {\left (b^{3} c \mathrm {sgn}\relax (x) - a b^{2} d \mathrm {sgn}\relax (x)\right )} \sqrt {a b c - a^{2} d}} - \frac {\arctan \left (\frac {\sqrt {d + \frac {c}{x^{6}}}}{\sqrt {-d}}\right )}{3 \, b^{2} \sqrt {-d} \mathrm {sgn}\relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14/(b*x^6+a)^2/(d*x^6+c)^(1/2),x, algorithm="giac")

[Out]

-1/6*(3*a*b*c*sqrt(-d)*arctan(a*sqrt(d)/sqrt(a*b*c - a^2*d)) - 2*a^2*sqrt(-d)*d*arctan(a*sqrt(d)/sqrt(a*b*c -
a^2*d)) - 2*sqrt(a*b*c - a^2*d)*b*c*arctan(sqrt(d)/sqrt(-d)) + 2*sqrt(a*b*c - a^2*d)*a*d*arctan(sqrt(d)/sqrt(-
d)) + sqrt(a*b*c - a^2*d)*a*sqrt(-d)*sqrt(d))*sgn(x)/(sqrt(a*b*c - a^2*d)*b^3*c*sqrt(-d) - sqrt(a*b*c - a^2*d)
*a*b^2*sqrt(-d)*d) + 1/6*a*c*sqrt(d + c/x^6)/((b^2*c*sgn(x) - a*b*d*sgn(x))*(b*c + a*(d + c/x^6) - a*d)) + 1/6
*(3*a*b*c - 2*a^2*d)*arctan(a*sqrt(d + c/x^6)/sqrt(a*b*c - a^2*d))/((b^3*c*sgn(x) - a*b^2*d*sgn(x))*sqrt(a*b*c
- a^2*d)) - 1/3*arctan(sqrt(d + c/x^6)/sqrt(-d))/(b^2*sqrt(-d)*sgn(x))

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maple [F]  time = 0.63, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{14}}{\left (b \,x^{6}+a \right )^{2} \sqrt {d \,x^{6}+c}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^14/(b*x^6+a)^2/(d*x^6+c)^(1/2),x)

[Out]

int(x^14/(b*x^6+a)^2/(d*x^6+c)^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{14}}{{\left (b x^{6} + a\right )}^{2} \sqrt {d x^{6} + c}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14/(b*x^6+a)^2/(d*x^6+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^14/((b*x^6 + a)^2*sqrt(d*x^6 + c)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{14}}{{\left (b\,x^6+a\right )}^2\,\sqrt {d\,x^6+c}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^14/((a + b*x^6)^2*(c + d*x^6)^(1/2)),x)

[Out]

int(x^14/((a + b*x^6)^2*(c + d*x^6)^(1/2)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**14/(b*x**6+a)**2/(d*x**6+c)**(1/2),x)

[Out]

Timed out

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