3.900 \(\int \frac{x^{13}}{(a+b x^8) \sqrt{c+d x^8}} \, dx\)

Optimal. Leaf size=1005 \[ \text{result too large to display} \]

[Out]

(x^2*Sqrt[c + d*x^8])/(2*b*Sqrt[d]*(Sqrt[c] + Sqrt[d]*x^4)) + ((-a)^(3/4)*ArcTan[(Sqrt[b*c - a*d]*x^2)/((-a)^(
1/4)*b^(1/4)*Sqrt[c + d*x^8])])/(8*b^(5/4)*Sqrt[b*c - a*d]) - ((-a)^(3/4)*ArcTan[(Sqrt[-(b*c) + a*d]*x^2)/((-a
)^(1/4)*b^(1/4)*Sqrt[c + d*x^8])])/(8*b^(5/4)*Sqrt[-(b*c) + a*d]) - (c^(1/4)*(Sqrt[c] + Sqrt[d]*x^4)*Sqrt[(c +
d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticE[2*ArcTan[(d^(1/4)*x^2)/c^(1/4)], 1/2])/(2*b*d^(3/4)*Sqrt[c + d*x^
8]) + (c^(1/4)*(Sqrt[c] + Sqrt[d]*x^4)*Sqrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticF[2*ArcTan[(d^(1/4)
*x^2)/c^(1/4)], 1/2])/(4*b*d^(3/4)*Sqrt[c + d*x^8]) + (a*(Sqrt[c] - (Sqrt[-a]*Sqrt[d])/Sqrt[b])*d^(1/4)*(Sqrt[
c] + Sqrt[d]*x^4)*Sqrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticF[2*ArcTan[(d^(1/4)*x^2)/c^(1/4)], 1/2])
/(8*b*c^(1/4)*(b*c + a*d)*Sqrt[c + d*x^8]) + (a*(Sqrt[c] + (Sqrt[-a]*Sqrt[d])/Sqrt[b])*d^(1/4)*(Sqrt[c] + Sqrt
[d]*x^4)*Sqrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticF[2*ArcTan[(d^(1/4)*x^2)/c^(1/4)], 1/2])/(8*b*c^(
1/4)*(b*c + a*d)*Sqrt[c + d*x^8]) + (Sqrt[-a]*(Sqrt[b]*Sqrt[c] + Sqrt[-a]*Sqrt[d])^2*(Sqrt[c] + Sqrt[d]*x^4)*S
qrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticPi[-(Sqrt[b]*Sqrt[c] - Sqrt[-a]*Sqrt[d])^2/(4*Sqrt[-a]*Sqrt
[b]*Sqrt[c]*Sqrt[d]), 2*ArcTan[(d^(1/4)*x^2)/c^(1/4)], 1/2])/(16*b^(3/2)*c^(1/4)*d^(1/4)*(b*c + a*d)*Sqrt[c +
d*x^8]) - (Sqrt[-a]*(Sqrt[b]*Sqrt[c] - Sqrt[-a]*Sqrt[d])^2*(Sqrt[c] + Sqrt[d]*x^4)*Sqrt[(c + d*x^8)/(Sqrt[c] +
Sqrt[d]*x^4)^2]*EllipticPi[(Sqrt[b]*Sqrt[c] + Sqrt[-a]*Sqrt[d])^2/(4*Sqrt[-a]*Sqrt[b]*Sqrt[c]*Sqrt[d]), 2*Arc
Tan[(d^(1/4)*x^2)/c^(1/4)], 1/2])/(16*b^(3/2)*c^(1/4)*d^(1/4)*(b*c + a*d)*Sqrt[c + d*x^8])

________________________________________________________________________________________

Rubi [A]  time = 1.36333, antiderivative size = 1005, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {465, 483, 305, 220, 1196, 490, 1217, 1707} \[ -\frac{\sqrt{-a} \left (\sqrt{d} x^4+\sqrt{c}\right ) \sqrt{\frac{d x^8+c}{\left (\sqrt{d} x^4+\sqrt{c}\right )^2}} \Pi \left (\frac{\left (\sqrt{b} \sqrt{c}+\sqrt{-a} \sqrt{d}\right )^2}{4 \sqrt{-a} \sqrt{b} \sqrt{c} \sqrt{d}};2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac{1}{2}\right ) \left (\sqrt{b} \sqrt{c}-\sqrt{-a} \sqrt{d}\right )^2}{16 b^{3/2} \sqrt [4]{c} \sqrt [4]{d} (b c+a d) \sqrt{d x^8+c}}+\frac{(-a)^{3/4} \tan ^{-1}\left (\frac{\sqrt{b c-a d} x^2}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt{d x^8+c}}\right )}{8 b^{5/4} \sqrt{b c-a d}}-\frac{(-a)^{3/4} \tan ^{-1}\left (\frac{\sqrt{a d-b c} x^2}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt{d x^8+c}}\right )}{8 b^{5/4} \sqrt{a d-b c}}-\frac{\sqrt [4]{c} \left (\sqrt{d} x^4+\sqrt{c}\right ) \sqrt{\frac{d x^8+c}{\left (\sqrt{d} x^4+\sqrt{c}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{2 b d^{3/4} \sqrt{d x^8+c}}+\frac{\sqrt [4]{c} \left (\sqrt{d} x^4+\sqrt{c}\right ) \sqrt{\frac{d x^8+c}{\left (\sqrt{d} x^4+\sqrt{c}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{4 b d^{3/4} \sqrt{d x^8+c}}+\frac{a \left (\sqrt{c}-\frac{\sqrt{-a} \sqrt{d}}{\sqrt{b}}\right ) \sqrt [4]{d} \left (\sqrt{d} x^4+\sqrt{c}\right ) \sqrt{\frac{d x^8+c}{\left (\sqrt{d} x^4+\sqrt{c}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{8 b \sqrt [4]{c} (b c+a d) \sqrt{d x^8+c}}+\frac{a \left (\sqrt{c}+\frac{\sqrt{-a} \sqrt{d}}{\sqrt{b}}\right ) \sqrt [4]{d} \left (\sqrt{d} x^4+\sqrt{c}\right ) \sqrt{\frac{d x^8+c}{\left (\sqrt{d} x^4+\sqrt{c}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{8 b \sqrt [4]{c} (b c+a d) \sqrt{d x^8+c}}+\frac{\sqrt{-a} \left (\sqrt{b} \sqrt{c}+\sqrt{-a} \sqrt{d}\right )^2 \left (\sqrt{d} x^4+\sqrt{c}\right ) \sqrt{\frac{d x^8+c}{\left (\sqrt{d} x^4+\sqrt{c}\right )^2}} \Pi \left (-\frac{\left (\sqrt{b} \sqrt{c}-\sqrt{-a} \sqrt{d}\right )^2}{4 \sqrt{-a} \sqrt{b} \sqrt{c} \sqrt{d}};2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{16 b^{3/2} \sqrt [4]{c} \sqrt [4]{d} (b c+a d) \sqrt{d x^8+c}}+\frac{x^2 \sqrt{d x^8+c}}{2 b \sqrt{d} \left (\sqrt{d} x^4+\sqrt{c}\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^13/((a + b*x^8)*Sqrt[c + d*x^8]),x]

[Out]

(x^2*Sqrt[c + d*x^8])/(2*b*Sqrt[d]*(Sqrt[c] + Sqrt[d]*x^4)) + ((-a)^(3/4)*ArcTan[(Sqrt[b*c - a*d]*x^2)/((-a)^(
1/4)*b^(1/4)*Sqrt[c + d*x^8])])/(8*b^(5/4)*Sqrt[b*c - a*d]) - ((-a)^(3/4)*ArcTan[(Sqrt[-(b*c) + a*d]*x^2)/((-a
)^(1/4)*b^(1/4)*Sqrt[c + d*x^8])])/(8*b^(5/4)*Sqrt[-(b*c) + a*d]) - (c^(1/4)*(Sqrt[c] + Sqrt[d]*x^4)*Sqrt[(c +
d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticE[2*ArcTan[(d^(1/4)*x^2)/c^(1/4)], 1/2])/(2*b*d^(3/4)*Sqrt[c + d*x^
8]) + (c^(1/4)*(Sqrt[c] + Sqrt[d]*x^4)*Sqrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticF[2*ArcTan[(d^(1/4)
*x^2)/c^(1/4)], 1/2])/(4*b*d^(3/4)*Sqrt[c + d*x^8]) + (a*(Sqrt[c] - (Sqrt[-a]*Sqrt[d])/Sqrt[b])*d^(1/4)*(Sqrt[
c] + Sqrt[d]*x^4)*Sqrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticF[2*ArcTan[(d^(1/4)*x^2)/c^(1/4)], 1/2])
/(8*b*c^(1/4)*(b*c + a*d)*Sqrt[c + d*x^8]) + (a*(Sqrt[c] + (Sqrt[-a]*Sqrt[d])/Sqrt[b])*d^(1/4)*(Sqrt[c] + Sqrt
[d]*x^4)*Sqrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticF[2*ArcTan[(d^(1/4)*x^2)/c^(1/4)], 1/2])/(8*b*c^(
1/4)*(b*c + a*d)*Sqrt[c + d*x^8]) + (Sqrt[-a]*(Sqrt[b]*Sqrt[c] + Sqrt[-a]*Sqrt[d])^2*(Sqrt[c] + Sqrt[d]*x^4)*S
qrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticPi[-(Sqrt[b]*Sqrt[c] - Sqrt[-a]*Sqrt[d])^2/(4*Sqrt[-a]*Sqrt
[b]*Sqrt[c]*Sqrt[d]), 2*ArcTan[(d^(1/4)*x^2)/c^(1/4)], 1/2])/(16*b^(3/2)*c^(1/4)*d^(1/4)*(b*c + a*d)*Sqrt[c +
d*x^8]) - (Sqrt[-a]*(Sqrt[b]*Sqrt[c] - Sqrt[-a]*Sqrt[d])^2*(Sqrt[c] + Sqrt[d]*x^4)*Sqrt[(c + d*x^8)/(Sqrt[c] +
Sqrt[d]*x^4)^2]*EllipticPi[(Sqrt[b]*Sqrt[c] + Sqrt[-a]*Sqrt[d])^2/(4*Sqrt[-a]*Sqrt[b]*Sqrt[c]*Sqrt[d]), 2*Arc
Tan[(d^(1/4)*x^2)/c^(1/4)], 1/2])/(16*b^(3/2)*c^(1/4)*d^(1/4)*(b*c + a*d)*Sqrt[c + d*x^8])

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 483

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

Rule 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 490

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(x_)^4]), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]],
s = Denominator[Rt[-(a/b), 2]]}, Dist[s/(2*b), Int[1/((r + s*x^2)*Sqrt[c + d*x^4]), x], x] - Dist[s/(2*b), In
t[1/((r - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1217

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(c*d + a*e*q
)/(c*d^2 - a*e^2), Int[1/Sqrt[a + c*x^4], x], x] - Dist[(a*e*(e + d*q))/(c*d^2 - a*e^2), Int[(1 + q*x^2)/((d +
e*x^2)*Sqrt[a + c*x^4]), x], x]] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0]
&& PosQ[c/a]

Rule 1707

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[B/A, 2]
}, -Simp[((B*d - A*e)*ArcTan[(Rt[(c*d)/e + (a*e)/d, 2]*x)/Sqrt[a + c*x^4]])/(2*d*e*Rt[(c*d)/e + (a*e)/d, 2]),
x] + Simp[((B*d + A*e)*(A + B*x^2)*Sqrt[(A^2*(a + c*x^4))/(a*(A + B*x^2)^2)]*EllipticPi[Cancel[-((B*d - A*e)^2
/(4*d*e*A*B))], 2*ArcTan[q*x], 1/2])/(4*d*e*A*q*Sqrt[a + c*x^4]), x]] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c
*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] && EqQ[c*A^2 - a*B^2, 0]

Rubi steps

\begin{align*} \int \frac{x^{13}}{\left (a+b x^8\right ) \sqrt{c+d x^8}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^6}{\left (a+b x^4\right ) \sqrt{c+d x^4}} \, dx,x,x^2\right )\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{\sqrt{c+d x^4}} \, dx,x,x^2\right )}{2 b}-\frac{a \operatorname{Subst}\left (\int \frac{x^2}{\left (a+b x^4\right ) \sqrt{c+d x^4}} \, dx,x,x^2\right )}{2 b}\\ &=\frac{a \operatorname{Subst}\left (\int \frac{1}{\left (\sqrt{-a}-\sqrt{b} x^2\right ) \sqrt{c+d x^4}} \, dx,x,x^2\right )}{4 b^{3/2}}-\frac{a \operatorname{Subst}\left (\int \frac{1}{\left (\sqrt{-a}+\sqrt{b} x^2\right ) \sqrt{c+d x^4}} \, dx,x,x^2\right )}{4 b^{3/2}}+\frac{\sqrt{c} \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+d x^4}} \, dx,x,x^2\right )}{2 b \sqrt{d}}-\frac{\sqrt{c} \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{d} x^2}{\sqrt{c}}}{\sqrt{c+d x^4}} \, dx,x,x^2\right )}{2 b \sqrt{d}}\\ &=\frac{x^2 \sqrt{c+d x^8}}{2 b \sqrt{d} \left (\sqrt{c}+\sqrt{d} x^4\right )}-\frac{\sqrt [4]{c} \left (\sqrt{c}+\sqrt{d} x^4\right ) \sqrt{\frac{c+d x^8}{\left (\sqrt{c}+\sqrt{d} x^4\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{2 b d^{3/4} \sqrt{c+d x^8}}+\frac{\sqrt [4]{c} \left (\sqrt{c}+\sqrt{d} x^4\right ) \sqrt{\frac{c+d x^8}{\left (\sqrt{c}+\sqrt{d} x^4\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{4 b d^{3/4} \sqrt{c+d x^8}}+\frac{\left (a \sqrt{c} \left (\sqrt{b} \sqrt{c}-\sqrt{-a} \sqrt{d}\right )\right ) \operatorname{Subst}\left (\int \frac{1+\frac{\sqrt{d} x^2}{\sqrt{c}}}{\left (\sqrt{-a}-\sqrt{b} x^2\right ) \sqrt{c+d x^4}} \, dx,x,x^2\right )}{4 b (b c+a d)}-\frac{\left (a \sqrt{c} \left (\sqrt{b} \sqrt{c}+\sqrt{-a} \sqrt{d}\right )\right ) \operatorname{Subst}\left (\int \frac{1+\frac{\sqrt{d} x^2}{\sqrt{c}}}{\left (\sqrt{-a}+\sqrt{b} x^2\right ) \sqrt{c+d x^4}} \, dx,x,x^2\right )}{4 b (b c+a d)}+\frac{\left (a \left (\sqrt{b} \sqrt{c}-\sqrt{-a} \sqrt{d}\right ) \sqrt{d}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+d x^4}} \, dx,x,x^2\right )}{4 b^{3/2} (b c+a d)}+\frac{\left (a \left (\sqrt{b} \sqrt{c}+\sqrt{-a} \sqrt{d}\right ) \sqrt{d}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+d x^4}} \, dx,x,x^2\right )}{4 b^{3/2} (b c+a d)}\\ &=\frac{x^2 \sqrt{c+d x^8}}{2 b \sqrt{d} \left (\sqrt{c}+\sqrt{d} x^4\right )}+\frac{(-a)^{3/4} \tan ^{-1}\left (\frac{\sqrt{b c-a d} x^2}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt{c+d x^8}}\right )}{8 b^{5/4} \sqrt{b c-a d}}-\frac{(-a)^{3/4} \tan ^{-1}\left (\frac{\sqrt{-b c+a d} x^2}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt{c+d x^8}}\right )}{8 b^{5/4} \sqrt{-b c+a d}}-\frac{\sqrt [4]{c} \left (\sqrt{c}+\sqrt{d} x^4\right ) \sqrt{\frac{c+d x^8}{\left (\sqrt{c}+\sqrt{d} x^4\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{2 b d^{3/4} \sqrt{c+d x^8}}+\frac{\sqrt [4]{c} \left (\sqrt{c}+\sqrt{d} x^4\right ) \sqrt{\frac{c+d x^8}{\left (\sqrt{c}+\sqrt{d} x^4\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{4 b d^{3/4} \sqrt{c+d x^8}}+\frac{a \left (\sqrt{b} \sqrt{c}-\sqrt{-a} \sqrt{d}\right ) \sqrt [4]{d} \left (\sqrt{c}+\sqrt{d} x^4\right ) \sqrt{\frac{c+d x^8}{\left (\sqrt{c}+\sqrt{d} x^4\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{8 b^{3/2} \sqrt [4]{c} (b c+a d) \sqrt{c+d x^8}}+\frac{a \left (\sqrt{b} \sqrt{c}+\sqrt{-a} \sqrt{d}\right ) \sqrt [4]{d} \left (\sqrt{c}+\sqrt{d} x^4\right ) \sqrt{\frac{c+d x^8}{\left (\sqrt{c}+\sqrt{d} x^4\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{8 b^{3/2} \sqrt [4]{c} (b c+a d) \sqrt{c+d x^8}}+\frac{\sqrt{-a} \left (\sqrt{b} \sqrt{c}+\sqrt{-a} \sqrt{d}\right )^2 \left (\sqrt{c}+\sqrt{d} x^4\right ) \sqrt{\frac{c+d x^8}{\left (\sqrt{c}+\sqrt{d} x^4\right )^2}} \Pi \left (-\frac{\left (\sqrt{b} \sqrt{c}-\sqrt{-a} \sqrt{d}\right )^2}{4 \sqrt{-a} \sqrt{b} \sqrt{c} \sqrt{d}};2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{16 b^{3/2} \sqrt [4]{c} \sqrt [4]{d} (b c+a d) \sqrt{c+d x^8}}-\frac{\sqrt{-a} \left (\sqrt{b} \sqrt{c}-\sqrt{-a} \sqrt{d}\right )^2 \left (\sqrt{c}+\sqrt{d} x^4\right ) \sqrt{\frac{c+d x^8}{\left (\sqrt{c}+\sqrt{d} x^4\right )^2}} \Pi \left (\frac{\left (\sqrt{b} \sqrt{c}+\sqrt{-a} \sqrt{d}\right )^2}{4 \sqrt{-a} \sqrt{b} \sqrt{c} \sqrt{d}};2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{16 b^{3/2} \sqrt [4]{c} \sqrt [4]{d} (b c+a d) \sqrt{c+d x^8}}\\ \end{align*}

Mathematica [C]  time = 0.0428873, size = 65, normalized size = 0.06 \[ \frac{x^{14} \sqrt{\frac{c+d x^8}{c}} F_1\left (\frac{7}{4};\frac{1}{2},1;\frac{11}{4};-\frac{d x^8}{c},-\frac{b x^8}{a}\right )}{14 a \sqrt{c+d x^8}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^13/((a + b*x^8)*Sqrt[c + d*x^8]),x]

[Out]

(x^14*Sqrt[(c + d*x^8)/c]*AppellF1[7/4, 1/2, 1, 11/4, -((d*x^8)/c), -((b*x^8)/a)])/(14*a*Sqrt[c + d*x^8])

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Maple [F]  time = 0.043, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{13}}{b{x}^{8}+a}{\frac{1}{\sqrt{d{x}^{8}+c}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^13/(b*x^8+a)/(d*x^8+c)^(1/2),x)

[Out]

int(x^13/(b*x^8+a)/(d*x^8+c)^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{13}}{{\left (b x^{8} + a\right )} \sqrt{d x^{8} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^13/(b*x^8+a)/(d*x^8+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^13/((b*x^8 + a)*sqrt(d*x^8 + c)), x)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^13/(b*x^8+a)/(d*x^8+c)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{13}}{\left (a + b x^{8}\right ) \sqrt{c + d x^{8}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**13/(b*x**8+a)/(d*x**8+c)**(1/2),x)

[Out]

Integral(x**13/((a + b*x**8)*sqrt(c + d*x**8)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{13}}{{\left (b x^{8} + a\right )} \sqrt{d x^{8} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^13/(b*x^8+a)/(d*x^8+c)^(1/2),x, algorithm="giac")

[Out]

integrate(x^13/((b*x^8 + a)*sqrt(d*x^8 + c)), x)