3.9.99 \(\int \frac {1}{x^7 (a+b x^8) \sqrt {c+d x^8}} \, dx\) [899]

Optimal. Leaf size=878 \[ -\frac {\sqrt {c+d x^8}}{6 a c x^6}-\frac {b^{5/4} \tan ^{-1}\left (\frac {\sqrt {b c-a d} x^2}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {c+d x^8}}\right )}{8 (-a)^{7/4} \sqrt {b c-a d}}-\frac {b^{5/4} \tan ^{-1}\left (\frac {\sqrt {-b c+a d} x^2}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {c+d x^8}}\right )}{8 (-a)^{7/4} \sqrt {-b c+a d}}-\frac {d^{3/4} \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 )}{12 a c^{5/4} \sqrt {c+d x^8}}-\frac {b \left (\frac {\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 a \sqrt [4]{c} (b c+a d) \sqrt {c+d x^8}}-\frac {b \left (\sqrt {-a} \sqrt {b} \sqrt {c}+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 a^2 \sqrt [4]{c} (b c+a d) \sqrt {c+d x^8}}-\frac {b \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 a^2 \sqrt [4]{c} \sqrt [4]{d} (b c+a d) \sqrt {c+d x^8}}-\frac {b \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 a^2 \sqrt [4]{c} \sqrt [4]{d} (b c+a d) \sqrt {c+d x^8}} \]

[Out]

-1/8*b^(5/4)*arctan(x^2*(-a*d+b*c)^(1/2)/(-a)^(1/4)/b^(1/4)/(d*x^8+c)^(1/2))/(-a)^(7/4)/(-a*d+b*c)^(1/2)-1/8*b
^(5/4)*arctan(x^2*(a*d-b*c)^(1/2)/(-a)^(1/4)/b^(1/4)/(d*x^8+c)^(1/2))/(-a)^(7/4)/(a*d-b*c)^(1/2)-1/6*(d*x^8+c)
^(1/2)/a/c/x^6-1/12*d^(3/4)*(cos(2*arctan(d^(1/4)*x^2/c^(1/4)))^2)^(1/2)/cos(2*arctan(d^(1/4)*x^2/c^(1/4)))*El
lipticF(sin(2*arctan(d^(1/4)*x^2/c^(1/4))),1/2*2^(1/2))*(c^(1/2)+x^4*d^(1/2))*((d*x^8+c)/(c^(1/2)+x^4*d^(1/2))
^2)^(1/2)/a/c^(5/4)/(d*x^8+c)^(1/2)-1/8*b*d^(1/4)*(cos(2*arctan(d^(1/4)*x^2/c^(1/4)))^2)^(1/2)/cos(2*arctan(d^
(1/4)*x^2/c^(1/4)))*EllipticF(sin(2*arctan(d^(1/4)*x^2/c^(1/4))),1/2*2^(1/2))*(b^(1/2)*c^(1/2)/(-a)^(1/2)+d^(1
/2))*(c^(1/2)+x^4*d^(1/2))*((d*x^8+c)/(c^(1/2)+x^4*d^(1/2))^2)^(1/2)/a/c^(1/4)/(a*d+b*c)/(d*x^8+c)^(1/2)-1/8*b
*d^(1/4)*(cos(2*arctan(d^(1/4)*x^2/c^(1/4)))^2)^(1/2)/cos(2*arctan(d^(1/4)*x^2/c^(1/4)))*EllipticF(sin(2*arcta
n(d^(1/4)*x^2/c^(1/4))),1/2*2^(1/2))*((-a)^(1/2)*b^(1/2)*c^(1/2)+a*d^(1/2))*(c^(1/2)+x^4*d^(1/2))*((d*x^8+c)/(
c^(1/2)+x^4*d^(1/2))^2)^(1/2)/a^2/c^(1/4)/(a*d+b*c)/(d*x^8+c)^(1/2)-1/16*b*(cos(2*arctan(d^(1/4)*x^2/c^(1/4)))
^2)^(1/2)/cos(2*arctan(d^(1/4)*x^2/c^(1/4)))*EllipticPi(sin(2*arctan(d^(1/4)*x^2/c^(1/4))),1/4*(b^(1/2)*c^(1/2
)+(-a)^(1/2)*d^(1/2))^2/(-a)^(1/2)/b^(1/2)/c^(1/2)/d^(1/2),1/2*2^(1/2))*(c^(1/2)+x^4*d^(1/2))*(b^(1/2)*c^(1/2)
-(-a)^(1/2)*d^(1/2))^2*((d*x^8+c)/(c^(1/2)+x^4*d^(1/2))^2)^(1/2)/a^2/c^(1/4)/d^(1/4)/(a*d+b*c)/(d*x^8+c)^(1/2)
-1/16*b*(cos(2*arctan(d^(1/4)*x^2/c^(1/4)))^2)^(1/2)/cos(2*arctan(d^(1/4)*x^2/c^(1/4)))*EllipticPi(sin(2*arcta
n(d^(1/4)*x^2/c^(1/4))),-1/4*(b^(1/2)*c^(1/2)-(-a)^(1/2)*d^(1/2))^2/(-a)^(1/2)/b^(1/2)/c^(1/2)/d^(1/2),1/2*2^(
1/2))*(c^(1/2)+x^4*d^(1/2))*(b^(1/2)*c^(1/2)+(-a)^(1/2)*d^(1/2))^2*((d*x^8+c)/(c^(1/2)+x^4*d^(1/2))^2)^(1/2)/a
^2/c^(1/4)/d^(1/4)/(a*d+b*c)/(d*x^8+c)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 1.15, antiderivative size = 878, 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 = {476, 491, 537, 226, 418, 1231, 1721} \begin {gather*} -\frac {b \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 \text {ArcTan}\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 a^2 \sqrt [4]{c} \sqrt [4]{d} (b c+a d) \sqrt {d x^8+c}}-\frac {b^{5/4} \text {ArcTan}\left (\frac {\sqrt {b c-a d} x^2}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {d x^8+c}}\right )}{8 (-a)^{7/4} \sqrt {b c-a d}}-\frac {b^{5/4} \text {ArcTan}\left (\frac {\sqrt {a d-b c} x^2}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {d x^8+c}}\right )}{8 (-a)^{7/4} \sqrt {a d-b c}}-\frac {d^{3/4} \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 \text {ArcTan}\left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{12 a c^{5/4} \sqrt {d x^8+c}}-\frac {b \left (\frac {\sqrt {b} \sqrt {c}}{\sqrt {-a}}+\sqrt {d}\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 \text {ArcTan}\left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{8 a \sqrt [4]{c} (b c+a d) \sqrt {d x^8+c}}-\frac {b \left (\sqrt {d} a+\sqrt {-a} \sqrt {b} \sqrt {c}\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 \text {ArcTan}\left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{8 a^2 \sqrt [4]{c} (b c+a d) \sqrt {d x^8+c}}-\frac {b \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 \text {ArcTan}\left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{16 a^2 \sqrt [4]{c} \sqrt [4]{d} (b c+a d) \sqrt {d x^8+c}}-\frac {\sqrt {d x^8+c}}{6 a c x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^7*(a + b*x^8)*Sqrt[c + d*x^8]),x]

[Out]

-1/6*Sqrt[c + d*x^8]/(a*c*x^6) - (b^(5/4)*ArcTan[(Sqrt[b*c - a*d]*x^2)/((-a)^(1/4)*b^(1/4)*Sqrt[c + d*x^8])])/
(8*(-a)^(7/4)*Sqrt[b*c - a*d]) - (b^(5/4)*ArcTan[(Sqrt[-(b*c) + a*d]*x^2)/((-a)^(1/4)*b^(1/4)*Sqrt[c + d*x^8])
])/(8*(-a)^(7/4)*Sqrt[-(b*c) + a*d]) - (d^(3/4)*(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])/(12*a*c^(5/4)*Sqrt[c + d*x^8]) - (b*((Sqrt[b]*Sqrt[c])/
Sqrt[-a] + Sqrt[d])*d^(1/4)*(Sqrt[c] + Sqrt[d]*x^4)*Sqrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticF[2*Ar
cTan[(d^(1/4)*x^2)/c^(1/4)], 1/2])/(8*a*c^(1/4)*(b*c + a*d)*Sqrt[c + d*x^8]) - (b*(Sqrt[-a]*Sqrt[b]*Sqrt[c] +
a*Sqrt[d])*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*a^2*c^(1/4)*(b*c + a*d)*Sqrt[c + d*x^8]) - (b*(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[-1/4*(Sqrt[b]*Sqrt[c] - Sqr
t[-a]*Sqrt[d])^2/(Sqrt[-a]*Sqrt[b]*Sqrt[c]*Sqrt[d]), 2*ArcTan[(d^(1/4)*x^2)/c^(1/4)], 1/2])/(16*a^2*c^(1/4)*d^
(1/4)*(b*c + a*d)*Sqrt[c + d*x^8]) - (b*(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]*Sqr
t[c]*Sqrt[d]), 2*ArcTan[(d^(1/4)*x^2)/c^(1/4)], 1/2])/(16*a^2*c^(1/4)*d^(1/4)*(b*c + a*d)*Sqrt[c + d*x^8])

Rule 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 418

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

Rule 476

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 491

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

Rule 537

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]

Rule 1231

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 1721

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))]/(4*d*e*A*q*Sqrt[a + c*x^4]))*El
lipticPi[Cancel[-(B*d - A*e)^2/(4*d*e*A*B)], 2*ArcTan[q*x], 1/2], 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 {1}{x^7 \left (a+b x^8\right ) \sqrt {c+d x^8}} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {1}{x^4 \left (a+b x^4\right ) \sqrt {c+d x^4}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {c+d x^8}}{6 a c x^6}+\frac {\text {Subst}\left (\int \frac {-3 b c-a d-b d x^4}{\left (a+b x^4\right ) \sqrt {c+d x^4}} \, dx,x,x^2\right )}{6 a c}\\ &=-\frac {\sqrt {c+d x^8}}{6 a c x^6}-\frac {b \text {Subst}\left (\int \frac {1}{\left (a+b x^4\right ) \sqrt {c+d x^4}} \, dx,x,x^2\right )}{2 a}-\frac {d \text {Subst}\left (\int \frac {1}{\sqrt {c+d x^4}} \, dx,x,x^2\right )}{6 a c}\\ &=-\frac {\sqrt {c+d x^8}}{6 a c x^6}-\frac {d^{3/4} \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 )}{12 a c^{5/4} \sqrt {c+d x^8}}-\frac {b \text {Subst}\left (\int \frac {1}{\left (1-\frac {\sqrt {b} x^2}{\sqrt {-a}}\right ) \sqrt {c+d x^4}} \, dx,x,x^2\right )}{4 a^2}-\frac {b \text {Subst}\left (\int \frac {1}{\left (1+\frac {\sqrt {b} x^2}{\sqrt {-a}}\right ) \sqrt {c+d x^4}} \, dx,x,x^2\right )}{4 a^2}\\ &=-\frac {\sqrt {c+d x^8}}{6 a c x^6}-\frac {d^{3/4} \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 )}{12 a c^{5/4} \sqrt {c+d x^8}}-\frac {\left (b^{3/2} \sqrt {c} \left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right )\right ) \text {Subst}\left (\int \frac {1+\frac {\sqrt {d} x^2}{\sqrt {c}}}{\left (1-\frac {\sqrt {b} x^2}{\sqrt {-a}}\right ) \sqrt {c+d x^4}} \, dx,x,x^2\right )}{4 a^2 (b c+a d)}-\frac {\left (b^{3/2} \sqrt {c} \left (\sqrt {b} \sqrt {c}+\sqrt {-a} \sqrt {d}\right )\right ) \text {Subst}\left (\int \frac {1+\frac {\sqrt {d} x^2}{\sqrt {c}}}{\left (1+\frac {\sqrt {b} x^2}{\sqrt {-a}}\right ) \sqrt {c+d x^4}} \, dx,x,x^2\right )}{4 a^2 (b c+a d)}-\frac {\left (b \left (\frac {\sqrt {b} \sqrt {c}}{\sqrt {-a}}+\sqrt {d}\right ) \sqrt {d}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c+d x^4}} \, dx,x,x^2\right )}{4 a (b c+a d)}-\frac {\left (b \left (\sqrt {-a} \sqrt {b} \sqrt {c}+a \sqrt {d}\right ) \sqrt {d}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c+d x^4}} \, dx,x,x^2\right )}{4 a^2 (b c+a d)}\\ &=-\frac {\sqrt {c+d x^8}}{6 a c x^6}-\frac {b^{5/4} \tan ^{-1}\left (\frac {\sqrt {b c-a d} x^2}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {c+d x^8}}\right )}{8 (-a)^{7/4} \sqrt {b c-a d}}-\frac {b^{5/4} \tan ^{-1}\left (\frac {\sqrt {-b c+a d} x^2}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {c+d x^8}}\right )}{8 (-a)^{7/4} \sqrt {-b c+a d}}-\frac {d^{3/4} \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 )}{12 a c^{5/4} \sqrt {c+d x^8}}-\frac {b \left (\frac {\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 a \sqrt [4]{c} (b c+a d) \sqrt {c+d x^8}}-\frac {b \left (\sqrt {-a} \sqrt {b} \sqrt {c}+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 a^2 \sqrt [4]{c} (b c+a d) \sqrt {c+d x^8}}-\frac {b \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 a^2 \sqrt [4]{c} \sqrt [4]{d} (b c+a d) \sqrt {c+d x^8}}-\frac {b \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 a^2 \sqrt [4]{c} \sqrt [4]{d} (b c+a d) \sqrt {c+d x^8}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
time = 10.12, size = 141, normalized size = 0.16 \begin {gather*} \frac {-5 a \left (c+d x^8\right )-5 (3 b c+a d) x^8 \sqrt {1+\frac {d x^8}{c}} F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};-\frac {d x^8}{c},-\frac {b x^8}{a}\right )-b d x^{16} \sqrt {1+\frac {d x^8}{c}} F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};-\frac {d x^8}{c},-\frac {b x^8}{a}\right )}{30 a^2 c x^6 \sqrt {c+d x^8}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^7*(a + b*x^8)*Sqrt[c + d*x^8]),x]

[Out]

(-5*a*(c + d*x^8) - 5*(3*b*c + a*d)*x^8*Sqrt[1 + (d*x^8)/c]*AppellF1[1/4, 1/2, 1, 5/4, -((d*x^8)/c), -((b*x^8)
/a)] - b*d*x^16*Sqrt[1 + (d*x^8)/c]*AppellF1[5/4, 1/2, 1, 9/4, -((d*x^8)/c), -((b*x^8)/a)])/(30*a^2*c*x^6*Sqrt
[c + d*x^8])

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {1}{x^{7} \left (b \,x^{8}+a \right ) \sqrt {d \,x^{8}+c}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^7/(b*x^8+a)/(d*x^8+c)^(1/2),x)

[Out]

int(1/x^7/(b*x^8+a)/(d*x^8+c)^(1/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(b*x^8+a)/(d*x^8+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^8 + a)*sqrt(d*x^8 + c)*x^7), x)

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Fricas [F(-1)] Timed out
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(1/x^7/(b*x^8+a)/(d*x^8+c)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{7} \left (a + b x^{8}\right ) \sqrt {c + d x^{8}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**7/(b*x**8+a)/(d*x**8+c)**(1/2),x)

[Out]

Integral(1/(x**7*(a + b*x**8)*sqrt(c + d*x**8)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(b*x^8+a)/(d*x^8+c)^(1/2),x, algorithm="giac")

[Out]

integrate(1/((b*x^8 + a)*sqrt(d*x^8 + c)*x^7), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{x^7\,\left (b\,x^8+a\right )\,\sqrt {d\,x^8+c}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^7*(a + b*x^8)*(c + d*x^8)^(1/2)),x)

[Out]

int(1/(x^7*(a + b*x^8)*(c + d*x^8)^(1/2)), x)

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