\(\int \frac {1}{x^3 (a+b x^8)^2 \sqrt {c+d x^8}} \, dx\) [130]

Optimal result

Integrand size = 24, antiderivative size = 1266 \[ \int \frac {1}{x^3 \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx =\text {Too large to display} \] Output:

-1/8*(-4*a*d+5*b*c)*(d*x^8+c)^(1/2)/a^2/c/(-a*d+b*c)/x^2+1/8*d^(1/2)*(-4*a 
*d+5*b*c)*x^2*(d*x^8+c)^(1/2)/a^2/c/(-a*d+b*c)/(c^(1/2)+d^(1/2)*x^4)+1/8*b 
*(d*x^8+c)^(1/2)/a/(-a*d+b*c)/x^2/(b*x^8+a)-1/32*b^(3/4)*(-7*a*d+5*b*c)*ar 
ctan((-a*d+b*c)^(1/2)*x^2/(-a)^(1/4)/b^(1/4)/(d*x^8+c)^(1/2))/(-a)^(9/4)/( 
-a*d+b*c)^(3/2)+1/32*b^(3/4)*(-7*a*d+5*b*c)*arctanh((-a*d+b*c)^(1/2)*x^2/( 
-a)^(1/4)/b^(1/4)/(d*x^8+c)^(1/2))/(-a)^(9/4)/(-a*d+b*c)^(3/2)-1/8*d^(1/4) 
*(-4*a*d+5*b*c)*(c^(1/2)+d^(1/2)*x^4)*((d*x^8+c)/(c^(1/2)+d^(1/2)*x^4)^2)^ 
(1/2)*EllipticE(sin(2*arctan(d^(1/4)*x^2/c^(1/4))),1/2*2^(1/2))/a^2/c^(3/4 
)/(-a*d+b*c)/(d*x^8+c)^(1/2)+1/32*b^(1/2)*d^(1/4)*(-7*a*d+5*b*c)*(c^(1/2)+ 
d^(1/2)*x^4)*((d*x^8+c)/(c^(1/2)+d^(1/2)*x^4)^2)^(1/2)*InverseJacobiAM(2*a 
rctan(d^(1/4)*x^2/c^(1/4)),1/2*2^(1/2))/a^2/c^(1/4)/(b^(1/2)*c^(1/2)-(-a)^ 
(1/2)*d^(1/2))/(-a*d+b*c)/(d*x^8+c)^(1/2)+1/32*b^(1/2)*d^(1/4)*(-7*a*d+5*b 
*c)*(c^(1/2)+d^(1/2)*x^4)*((d*x^8+c)/(c^(1/2)+d^(1/2)*x^4)^2)^(1/2)*Invers 
eJacobiAM(2*arctan(d^(1/4)*x^2/c^(1/4)),1/2*2^(1/2))/a^2/c^(1/4)/(b^(1/2)* 
c^(1/2)+(-a)^(1/2)*d^(1/2))/(-a*d+b*c)/(d*x^8+c)^(1/2)+1/16*d^(1/4)*(-4*a* 
d+5*b*c)*(c^(1/2)+d^(1/2)*x^4)*((d*x^8+c)/(c^(1/2)+d^(1/2)*x^4)^2)^(1/2)*I 
nverseJacobiAM(2*arctan(d^(1/4)*x^2/c^(1/4)),1/2*2^(1/2))/a^2/c^(3/4)/(-a* 
d+b*c)/(d*x^8+c)^(1/2)-1/64*b^(1/2)*(b^(1/2)*c^(1/2)+(-a)^(1/2)*d^(1/2))*( 
-7*a*d+5*b*c)*(c^(1/2)+d^(1/2)*x^4)*((d*x^8+c)/(c^(1/2)+d^(1/2)*x^4)^2)^(1 
/2)*EllipticPi(sin(2*arctan(d^(1/4)*x^2/c^(1/4))),-1/4*(b^(1/2)*c^(1/2)...
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 10.23 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.18 \[ \int \frac {1}{x^3 \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\frac {21 a \left (c+d x^8\right ) \left (4 a^2 d-5 b^2 c x^8-4 a b \left (c-d x^8\right )\right )-7 \left (5 b^2 c^2-12 a b c d+4 a^2 d^2\right ) x^8 \left (a+b x^8\right ) \sqrt {1+\frac {d x^8}{c}} \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},-\frac {d x^8}{c},-\frac {b x^8}{a}\right )+3 b d (5 b c-4 a d) x^{16} \left (a+b x^8\right ) \sqrt {1+\frac {d x^8}{c}} \operatorname {AppellF1}\left (\frac {7}{4},\frac {1}{2},1,\frac {11}{4},-\frac {d x^8}{c},-\frac {b x^8}{a}\right )}{168 a^3 c (b c-a d) x^2 \left (a+b x^8\right ) \sqrt {c+d x^8}} \] Input:

Integrate[1/(x^3*(a + b*x^8)^2*Sqrt[c + d*x^8]),x]
 

Output:

(21*a*(c + d*x^8)*(4*a^2*d - 5*b^2*c*x^8 - 4*a*b*(c - d*x^8)) - 7*(5*b^2*c 
^2 - 12*a*b*c*d + 4*a^2*d^2)*x^8*(a + b*x^8)*Sqrt[1 + (d*x^8)/c]*AppellF1[ 
3/4, 1/2, 1, 7/4, -((d*x^8)/c), -((b*x^8)/a)] + 3*b*d*(5*b*c - 4*a*d)*x^16 
*(a + b*x^8)*Sqrt[1 + (d*x^8)/c]*AppellF1[7/4, 1/2, 1, 11/4, -((d*x^8)/c), 
 -((b*x^8)/a)])/(168*a^3*c*(b*c - a*d)*x^2*(a + b*x^8)*Sqrt[c + d*x^8])
 

Rubi [A] (verified)

Time = 2.38 (sec) , antiderivative size = 1170, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {965, 972, 25, 1053, 1054, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{x^3 \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx\)

\(\Big \downarrow \) 965

\(\displaystyle \frac {1}{2} \int \frac {1}{x^4 \left (b x^8+a\right )^2 \sqrt {d x^8+c}}dx^2\)

\(\Big \downarrow \) 972

\(\displaystyle \frac {1}{2} \left (\frac {b \sqrt {c+d x^8}}{4 a x^2 \left (a+b x^8\right ) (b c-a d)}-\frac {\int -\frac {3 b d x^8+5 b c-4 a d}{x^4 \left (b x^8+a\right ) \sqrt {d x^8+c}}dx^2}{4 a (b c-a d)}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {3 b d x^8+5 b c-4 a d}{x^4 \left (b x^8+a\right ) \sqrt {d x^8+c}}dx^2}{4 a (b c-a d)}+\frac {b \sqrt {c+d x^8}}{4 a x^2 \left (a+b x^8\right ) (b c-a d)}\right )\)

\(\Big \downarrow \) 1053

\(\displaystyle \frac {1}{2} \left (\frac {-\frac {\int \frac {x^4 \left ((b c-2 a d) (5 b c-2 a d)-b d (5 b c-4 a d) x^8\right )}{\left (b x^8+a\right ) \sqrt {d x^8+c}}dx^2}{a c}-\frac {\sqrt {c+d x^8} (5 b c-4 a d)}{a c x^2}}{4 a (b c-a d)}+\frac {b \sqrt {c+d x^8}}{4 a x^2 \left (a+b x^8\right ) (b c-a d)}\right )\)

\(\Big \downarrow \) 1054

\(\displaystyle \frac {1}{2} \left (\frac {-\frac {\int \left (\frac {\left (5 b^2 c^2-7 a b c d\right ) x^4}{\left (b x^8+a\right ) \sqrt {d x^8+c}}-\frac {d (5 b c-4 a d) x^4}{\sqrt {d x^8+c}}\right )dx^2}{a c}-\frac {\sqrt {c+d x^8} (5 b c-4 a d)}{a c x^2}}{4 a (b c-a d)}+\frac {b \sqrt {c+d x^8}}{4 a x^2 \left (a+b x^8\right ) (b c-a d)}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (\frac {\sqrt {d x^8+c} b}{4 a (b c-a d) x^2 \left (b x^8+a\right )}+\frac {-\frac {\sqrt {d x^8+c} (5 b c-4 a d)}{a c x^2}-\frac {-\frac {\sqrt {b} c^{3/4} (5 b c-7 a d) \left (\sqrt {d} x^4+\sqrt {c}\right ) \sqrt {\frac {d x^8+c}{\left (\sqrt {d} x^4+\sqrt {c}\right )^2}} \operatorname {EllipticPi}\left (\frac {\left (\sqrt {b} \sqrt {c}+\sqrt {-a} \sqrt {d}\right )^2}{4 \sqrt {-a} \sqrt {b} \sqrt {c} \sqrt {d}},2 \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}{8 \sqrt {-a} \sqrt [4]{d} (b c+a d) \sqrt {d x^8+c}}+\frac {b^{3/4} c (5 b c-7 a d) \arctan \left (\frac {\sqrt {b c-a d} x^2}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {d x^8+c}}\right )}{4 \sqrt [4]{-a} \sqrt {b c-a d}}-\frac {b^{3/4} c (5 b c-7 a d) \text {arctanh}\left (\frac {\sqrt {b c-a d} x^2}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {d x^8+c}}\right )}{4 \sqrt [4]{-a} \sqrt {b c-a d}}+\frac {\sqrt [4]{c} \sqrt [4]{d} (5 b c-4 a d) \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 \arctan \left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{\sqrt {d x^8+c}}-\frac {\sqrt [4]{c} \sqrt [4]{d} (5 b c-4 a d) \left (\sqrt {d} x^4+\sqrt {c}\right ) \sqrt {\frac {d x^8+c}{\left (\sqrt {d} x^4+\sqrt {c}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right ),\frac {1}{2}\right )}{2 \sqrt {d x^8+c}}-\frac {b c^{3/4} \left (\sqrt {c}-\frac {\sqrt {-a} \sqrt {d}}{\sqrt {b}}\right ) \sqrt [4]{d} (5 b c-7 a d) \left (\sqrt {d} x^4+\sqrt {c}\right ) \sqrt {\frac {d x^8+c}{\left (\sqrt {d} x^4+\sqrt {c}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right ),\frac {1}{2}\right )}{4 (b c+a d) \sqrt {d x^8+c}}-\frac {b c^{3/4} \left (\sqrt {c}+\frac {\sqrt {-a} \sqrt {d}}{\sqrt {b}}\right ) \sqrt [4]{d} (5 b c-7 a d) \left (\sqrt {d} x^4+\sqrt {c}\right ) \sqrt {\frac {d x^8+c}{\left (\sqrt {d} x^4+\sqrt {c}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right ),\frac {1}{2}\right )}{4 (b c+a d) \sqrt {d x^8+c}}+\frac {\sqrt {b} c^{3/4} \left (\sqrt {b} \sqrt {c}+\sqrt {-a} \sqrt {d}\right )^2 (5 b c-7 a d) \left (\sqrt {d} x^4+\sqrt {c}\right ) \sqrt {\frac {d x^8+c}{\left (\sqrt {d} x^4+\sqrt {c}\right )^2}} \operatorname {EllipticPi}\left (-\frac {\sqrt {c} \left (\sqrt {b}-\frac {\sqrt {-a} \sqrt {d}}{\sqrt {c}}\right )^2}{4 \sqrt {-a} \sqrt {b} \sqrt {d}},2 \arctan \left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right ),\frac {1}{2}\right )}{8 \sqrt {-a} \sqrt [4]{d} (b c+a d) \sqrt {d x^8+c}}-\frac {\sqrt {d} (5 b c-4 a d) x^2 \sqrt {d x^8+c}}{\sqrt {d} x^4+\sqrt {c}}}{a c}}{4 a (b c-a d)}\right )\)

Input:

Int[1/(x^3*(a + b*x^8)^2*Sqrt[c + d*x^8]),x]
 

Output:

((b*Sqrt[c + d*x^8])/(4*a*(b*c - a*d)*x^2*(a + b*x^8)) + (-(((5*b*c - 4*a* 
d)*Sqrt[c + d*x^8])/(a*c*x^2)) - (-((Sqrt[d]*(5*b*c - 4*a*d)*x^2*Sqrt[c + 
d*x^8])/(Sqrt[c] + Sqrt[d]*x^4)) + (b^(3/4)*c*(5*b*c - 7*a*d)*ArcTan[(Sqrt 
[b*c - a*d]*x^2)/((-a)^(1/4)*b^(1/4)*Sqrt[c + d*x^8])])/(4*(-a)^(1/4)*Sqrt 
[b*c - a*d]) - (b^(3/4)*c*(5*b*c - 7*a*d)*ArcTanh[(Sqrt[b*c - a*d]*x^2)/(( 
-a)^(1/4)*b^(1/4)*Sqrt[c + d*x^8])])/(4*(-a)^(1/4)*Sqrt[b*c - a*d]) + (c^( 
1/4)*d^(1/4)*(5*b*c - 4*a*d)*(Sqrt[c] + Sqrt[d]*x^4)*Sqrt[(c + d*x^8)/(Sqr 
t[c] + Sqrt[d]*x^4)^2]*EllipticE[2*ArcTan[(d^(1/4)*x^2)/c^(1/4)], 1/2])/Sq 
rt[c + d*x^8] - (c^(1/4)*d^(1/4)*(5*b*c - 4*a*d)*(Sqrt[c] + Sqrt[d]*x^4)*S 
qrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticF[2*ArcTan[(d^(1/4)*x^2 
)/c^(1/4)], 1/2])/(2*Sqrt[c + d*x^8]) - (b*c^(3/4)*(Sqrt[c] - (Sqrt[-a]*Sq 
rt[d])/Sqrt[b])*d^(1/4)*(5*b*c - 7*a*d)*(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*c + a*d)*Sqrt[c + d*x^8]) - (b*c^(3/4)*(Sqrt[c] + (Sqrt[-a] 
*Sqrt[d])/Sqrt[b])*d^(1/4)*(5*b*c - 7*a*d)*(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*c + a*d)*Sqrt[c + d*x^8]) - (Sqrt[b]*c^(3/4)*(Sqrt[b]*Sq 
rt[c] - Sqrt[-a]*Sqrt[d])^2*(5*b*c - 7*a*d)*(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*ArcTan[(d^(1/4)*x...
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 965

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

rule 972

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

rule 1053

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

rule 1054

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n 
_)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a 
+ b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, 
 m, p}, x] && IGtQ[n, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [F]

\[\int \frac {1}{x^{3} \left (b \,x^{8}+a \right )^{2} \sqrt {d \,x^{8}+c}}d x\]

Input:

int(1/x^3/(b*x^8+a)^2/(d*x^8+c)^(1/2),x)
 

Output:

int(1/x^3/(b*x^8+a)^2/(d*x^8+c)^(1/2),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{x^3 \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\text {Timed out} \] Input:

integrate(1/x^3/(b*x^8+a)^2/(d*x^8+c)^(1/2),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {1}{x^3 \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int \frac {1}{x^{3} \left (a + b x^{8}\right )^{2} \sqrt {c + d x^{8}}}\, dx \] Input:

integrate(1/x**3/(b*x**8+a)**2/(d*x**8+c)**(1/2),x)
 

Output:

Integral(1/(x**3*(a + b*x**8)**2*sqrt(c + d*x**8)), x)
 

Maxima [F]

\[ \int \frac {1}{x^3 \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int { \frac {1}{{\left (b x^{8} + a\right )}^{2} \sqrt {d x^{8} + c} x^{3}} \,d x } \] Input:

integrate(1/x^3/(b*x^8+a)^2/(d*x^8+c)^(1/2),x, algorithm="maxima")
 

Output:

integrate(1/((b*x^8 + a)^2*sqrt(d*x^8 + c)*x^3), x)
 

Giac [F]

\[ \int \frac {1}{x^3 \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int { \frac {1}{{\left (b x^{8} + a\right )}^{2} \sqrt {d x^{8} + c} x^{3}} \,d x } \] Input:

integrate(1/x^3/(b*x^8+a)^2/(d*x^8+c)^(1/2),x, algorithm="giac")
 

Output:

integrate(1/((b*x^8 + a)^2*sqrt(d*x^8 + c)*x^3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^3 \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int \frac {1}{x^3\,{\left (b\,x^8+a\right )}^2\,\sqrt {d\,x^8+c}} \,d x \] Input:

int(1/(x^3*(a + b*x^8)^2*(c + d*x^8)^(1/2)),x)
 

Output:

int(1/(x^3*(a + b*x^8)^2*(c + d*x^8)^(1/2)), x)
 

Reduce [F]

\[ \int \frac {1}{x^3 \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int \frac {1}{x^{3} \left (b \,x^{8}+a \right )^{2} \sqrt {d \,x^{8}+c}}d x \] Input:

int(1/x^3/(b*x^8+a)^2/(d*x^8+c)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

int(1/x^3/(b*x^8+a)^2/(d*x^8+c)^(1/2),x)