\(\int \frac {1}{x^{13} (a+b x^8)^2 \sqrt {c+d x^8}} \, dx\) [122]

Optimal result

Integrand size = 24, antiderivative size = 208 \[ \int \frac {1}{x^{13} \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=-\frac {(5 b c-2 a d) \sqrt {c+d x^8}}{24 a^2 c (b c-a d) x^{12}}+\frac {\left (15 b^2 c^2-8 a b c d-4 a^2 d^2\right ) \sqrt {c+d x^8}}{24 a^3 c^2 (b c-a d) x^4}+\frac {b \sqrt {c+d x^8}}{8 a (b c-a d) x^{12} \left (a+b x^8\right )}+\frac {b^2 (5 b c-6 a d) \arctan \left (\frac {\sqrt {b c-a d} x^4}{\sqrt {a} \sqrt {c+d x^8}}\right )}{8 a^{7/2} (b c-a d)^{3/2}} \] Output:

-1/24*(-2*a*d+5*b*c)*(d*x^8+c)^(1/2)/a^2/c/(-a*d+b*c)/x^12+1/24*(-4*a^2*d^ 
2-8*a*b*c*d+15*b^2*c^2)*(d*x^8+c)^(1/2)/a^3/c^2/(-a*d+b*c)/x^4+1/8*b*(d*x^ 
8+c)^(1/2)/a/(-a*d+b*c)/x^12/(b*x^8+a)+1/8*b^2*(-6*a*d+5*b*c)*arctan((-a*d 
+b*c)^(1/2)*x^4/a^(1/2)/(d*x^8+c)^(1/2))/a^(7/2)/(-a*d+b*c)^(3/2)
 

Mathematica [A] (verified)

Time = 3.57 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.97 \[ \int \frac {1}{x^{13} \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=-\frac {\sqrt {c+d x^8} \left (15 b^3 c^2 x^{16}+2 a b^2 c x^8 \left (5 c-4 d x^8\right )+2 a^3 d \left (c-2 d x^8\right )-2 a^2 b \left (c^2+3 c d x^8+2 d^2 x^{16}\right )\right )}{24 a^3 c^2 (-b c+a d) x^{12} \left (a+b x^8\right )}+\frac {b^2 (5 b c-6 a d) \arctan \left (\frac {a \sqrt {d}+b x^4 \left (\sqrt {d} x^4+\sqrt {c+d x^8}\right )}{\sqrt {a} \sqrt {b c-a d}}\right )}{8 a^{7/2} (b c-a d)^{3/2}} \] Input:

Integrate[1/(x^13*(a + b*x^8)^2*Sqrt[c + d*x^8]),x]
 

Output:

-1/24*(Sqrt[c + d*x^8]*(15*b^3*c^2*x^16 + 2*a*b^2*c*x^8*(5*c - 4*d*x^8) + 
2*a^3*d*(c - 2*d*x^8) - 2*a^2*b*(c^2 + 3*c*d*x^8 + 2*d^2*x^16)))/(a^3*c^2* 
(-(b*c) + a*d)*x^12*(a + b*x^8)) + (b^2*(5*b*c - 6*a*d)*ArcTan[(a*Sqrt[d] 
+ b*x^4*(Sqrt[d]*x^4 + Sqrt[c + d*x^8]))/(Sqrt[a]*Sqrt[b*c - a*d])])/(8*a^ 
(7/2)*(b*c - a*d)^(3/2))
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {965, 374, 25, 445, 445, 27, 291, 218}

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.

\(\displaystyle \int \frac {1}{x^{13} \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx\)

\(\Big \downarrow \) 965

\(\displaystyle \frac {1}{4} \int \frac {1}{x^{16} \left (b x^8+a\right )^2 \sqrt {d x^8+c}}dx^4\)

\(\Big \downarrow \) 374

\(\displaystyle \frac {1}{4} \left (\frac {b \sqrt {c+d x^8}}{2 a x^{12} \left (a+b x^8\right ) (b c-a d)}-\frac {\int -\frac {4 b d x^8+5 b c-2 a d}{x^{16} \left (b x^8+a\right ) \sqrt {d x^8+c}}dx^4}{2 a (b c-a d)}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {4 b d x^8+5 b c-2 a d}{x^{16} \left (b x^8+a\right ) \sqrt {d x^8+c}}dx^4}{2 a (b c-a d)}+\frac {b \sqrt {c+d x^8}}{2 a x^{12} \left (a+b x^8\right ) (b c-a d)}\right )\)

\(\Big \downarrow \) 445

\(\displaystyle \frac {1}{4} \left (\frac {-\frac {\int \frac {2 b d (5 b c-2 a d) x^8+15 b^2 c^2-4 a^2 d^2-8 a b c d}{x^8 \left (b x^8+a\right ) \sqrt {d x^8+c}}dx^4}{3 a c}-\frac {\sqrt {c+d x^8} (5 b c-2 a d)}{3 a c x^{12}}}{2 a (b c-a d)}+\frac {b \sqrt {c+d x^8}}{2 a x^{12} \left (a+b x^8\right ) (b c-a d)}\right )\)

\(\Big \downarrow \) 445

\(\displaystyle \frac {1}{4} \left (\frac {-\frac {-\frac {\int \frac {3 b^2 c^2 (5 b c-6 a d)}{\left (b x^8+a\right ) \sqrt {d x^8+c}}dx^4}{a c}-\frac {\sqrt {c+d x^8} \left (\frac {15 b^2 c}{a}-\frac {4 a d^2}{c}-8 b d\right )}{x^4}}{3 a c}-\frac {\sqrt {c+d x^8} (5 b c-2 a d)}{3 a c x^{12}}}{2 a (b c-a d)}+\frac {b \sqrt {c+d x^8}}{2 a x^{12} \left (a+b x^8\right ) (b c-a d)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \left (\frac {-\frac {-\frac {3 b^2 c (5 b c-6 a d) \int \frac {1}{\left (b x^8+a\right ) \sqrt {d x^8+c}}dx^4}{a}-\frac {\sqrt {c+d x^8} \left (\frac {15 b^2 c}{a}-\frac {4 a d^2}{c}-8 b d\right )}{x^4}}{3 a c}-\frac {\sqrt {c+d x^8} (5 b c-2 a d)}{3 a c x^{12}}}{2 a (b c-a d)}+\frac {b \sqrt {c+d x^8}}{2 a x^{12} \left (a+b x^8\right ) (b c-a d)}\right )\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {1}{4} \left (\frac {-\frac {-\frac {3 b^2 c (5 b c-6 a d) \int \frac {1}{a-(a d-b c) x^8}d\frac {x^4}{\sqrt {d x^8+c}}}{a}-\frac {\sqrt {c+d x^8} \left (\frac {15 b^2 c}{a}-\frac {4 a d^2}{c}-8 b d\right )}{x^4}}{3 a c}-\frac {\sqrt {c+d x^8} (5 b c-2 a d)}{3 a c x^{12}}}{2 a (b c-a d)}+\frac {b \sqrt {c+d x^8}}{2 a x^{12} \left (a+b x^8\right ) (b c-a d)}\right )\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {1}{4} \left (\frac {-\frac {-\frac {3 b^2 c (5 b c-6 a d) \arctan \left (\frac {x^4 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^8}}\right )}{a^{3/2} \sqrt {b c-a d}}-\frac {\sqrt {c+d x^8} \left (\frac {15 b^2 c}{a}-\frac {4 a d^2}{c}-8 b d\right )}{x^4}}{3 a c}-\frac {\sqrt {c+d x^8} (5 b c-2 a d)}{3 a c x^{12}}}{2 a (b c-a d)}+\frac {b \sqrt {c+d x^8}}{2 a x^{12} \left (a+b x^8\right ) (b c-a d)}\right )\)

Input:

Int[1/(x^13*(a + b*x^8)^2*Sqrt[c + d*x^8]),x]
 

Output:

((b*Sqrt[c + d*x^8])/(2*a*(b*c - a*d)*x^12*(a + b*x^8)) + (-1/3*((5*b*c - 
2*a*d)*Sqrt[c + d*x^8])/(a*c*x^12) - (-((((15*b^2*c)/a - 8*b*d - (4*a*d^2) 
/c)*Sqrt[c + d*x^8])/x^4) - (3*b^2*c*(5*b*c - 6*a*d)*ArcTan[(Sqrt[b*c - a* 
d]*x^4)/(Sqrt[a]*Sqrt[c + d*x^8])])/(a^(3/2)*Sqrt[b*c - a*d]))/(3*a*c))/(2 
*a*(b*c - a*d)))/4
 

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

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

rule 291

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

rule 374

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

rule 445

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

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]
 

Maple [A] (verified)

Time = 45.94 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.64

method	result	size
pseudoelliptic	\(\frac {-\frac {\sqrt {d \,x^{8}+c}\, \left (-2 a d \,x^{8}-6 b c \,x^{8}+a c \right )}{3 x^{12}}-\frac {b^{2} c^{2} \left (\frac {\sqrt {d \,x^{8}+c}\, b \,x^{4}}{b \,x^{8}+a}-\frac {\left (6 a d -5 c b \right ) \operatorname {arctanh}\left (\frac {a \sqrt {d \,x^{8}+c}}{x^{4} \sqrt {a \left (a d -c b \right )}}\right )}{\sqrt {a \left (a d -c b \right )}}\right )}{2 \left (a d -c b \right )}}{4 a^{3} c^{2}}\)	\(134\)

Input:

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

Output:

1/4/a^3*(-1/3*(d*x^8+c)^(1/2)*(-2*a*d*x^8-6*b*c*x^8+a*c)/x^12-1/2*b^2*c^2/ 
(a*d-b*c)*((d*x^8+c)^(1/2)*b*x^4/(b*x^8+a)-(6*a*d-5*b*c)/(a*(a*d-b*c))^(1/ 
2)*arctanh(a*(d*x^8+c)^(1/2)/x^4/(a*(a*d-b*c))^(1/2))))/c^2
 

Fricas [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 760, normalized size of antiderivative = 3.65 \[ \int \frac {1}{x^{13} \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/96*(3*((5*b^4*c^3 - 6*a*b^3*c^2*d)*x^20 + (5*a*b^3*c^3 - 6*a^2*b^2*c^2 
*d)*x^12)*sqrt(-a*b*c + a^2*d)*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^16 
 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^8 + a^2*c^2 - 4*((b*c - 2*a*d)*x^12 - a*c*x 
^4)*sqrt(d*x^8 + c)*sqrt(-a*b*c + a^2*d))/(b^2*x^16 + 2*a*b*x^8 + a^2)) - 
4*((15*a*b^4*c^3 - 23*a^2*b^3*c^2*d + 4*a^3*b^2*c*d^2 + 4*a^4*b*d^3)*x^16 
+ 2*(5*a^2*b^3*c^3 - 8*a^3*b^2*c^2*d + a^4*b*c*d^2 + 2*a^5*d^3)*x^8 - 2*a^ 
3*b^2*c^3 + 4*a^4*b*c^2*d - 2*a^5*c*d^2)*sqrt(d*x^8 + c))/((a^4*b^3*c^4 - 
2*a^5*b^2*c^3*d + a^6*b*c^2*d^2)*x^20 + (a^5*b^2*c^4 - 2*a^6*b*c^3*d + a^7 
*c^2*d^2)*x^12), 1/48*(3*((5*b^4*c^3 - 6*a*b^3*c^2*d)*x^20 + (5*a*b^3*c^3 
- 6*a^2*b^2*c^2*d)*x^12)*sqrt(a*b*c - a^2*d)*arctan(1/2*((b*c - 2*a*d)*x^8 
 - a*c)*sqrt(d*x^8 + c)*sqrt(a*b*c - a^2*d)/((a*b*c*d - a^2*d^2)*x^12 + (a 
*b*c^2 - a^2*c*d)*x^4)) + 2*((15*a*b^4*c^3 - 23*a^2*b^3*c^2*d + 4*a^3*b^2* 
c*d^2 + 4*a^4*b*d^3)*x^16 + 2*(5*a^2*b^3*c^3 - 8*a^3*b^2*c^2*d + a^4*b*c*d 
^2 + 2*a^5*d^3)*x^8 - 2*a^3*b^2*c^3 + 4*a^4*b*c^2*d - 2*a^5*c*d^2)*sqrt(d* 
x^8 + c))/((a^4*b^3*c^4 - 2*a^5*b^2*c^3*d + a^6*b*c^2*d^2)*x^20 + (a^5*b^2 
*c^4 - 2*a^6*b*c^3*d + a^7*c^2*d^2)*x^12)]
 

Sympy [F]

\[ \int \frac {1}{x^{13} \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int \frac {1}{x^{13} \left (a + b x^{8}\right )^{2} \sqrt {c + d x^{8}}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {1}{x^{13} \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int { \frac {1}{{\left (b x^{8} + a\right )}^{2} \sqrt {d x^{8} + c} x^{13}} \,d x } \] Input:

integrate(1/x^13/(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^13), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 395 vs. \(2 (184) = 368\).

Time = 0.54 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.90 \[ \int \frac {1}{x^{13} \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\frac {1}{24} \, d^{\frac {7}{2}} {\left (\frac {3 \, {\left (5 \, b^{3} c - 6 \, a b^{2} d\right )} \arctan \left (-\frac {{\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{{\left (a^{3} b c d^{3} - a^{4} d^{4}\right )} \sqrt {a b c d - a^{2} d^{2}}} - \frac {6 \, {\left ({\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} b^{3} c - 2 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} a b^{2} d - b^{3} c^{2}\right )}}{{\left (a^{3} b c d^{3} - a^{4} d^{4}\right )} {\left ({\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} a d + b c^{2}\right )}} - \frac {8 \, {\left (3 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{4} b - 6 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} b c - 3 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} a d + 3 \, b c^{2} + a c d\right )}}{{\left ({\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} - c\right )}^{3} a^{3} d^{3}}\right )} \] Input:

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

Output:

1/24*d^(7/2)*(3*(5*b^3*c - 6*a*b^2*d)*arctan(-1/2*((sqrt(d)*x^4 - sqrt(d*x 
^8 + c))^2*b - b*c + 2*a*d)/sqrt(a*b*c*d - a^2*d^2))/((a^3*b*c*d^3 - a^4*d 
^4)*sqrt(a*b*c*d - a^2*d^2)) - 6*((sqrt(d)*x^4 - sqrt(d*x^8 + c))^2*b^3*c 
- 2*(sqrt(d)*x^4 - sqrt(d*x^8 + c))^2*a*b^2*d - b^3*c^2)/((a^3*b*c*d^3 - a 
^4*d^4)*((sqrt(d)*x^4 - sqrt(d*x^8 + c))^4*b - 2*(sqrt(d)*x^4 - sqrt(d*x^8 
 + c))^2*b*c + 4*(sqrt(d)*x^4 - sqrt(d*x^8 + c))^2*a*d + b*c^2)) - 8*(3*(s 
qrt(d)*x^4 - sqrt(d*x^8 + c))^4*b - 6*(sqrt(d)*x^4 - sqrt(d*x^8 + c))^2*b* 
c - 3*(sqrt(d)*x^4 - sqrt(d*x^8 + c))^2*a*d + 3*b*c^2 + a*c*d)/(((sqrt(d)* 
x^4 - sqrt(d*x^8 + c))^2 - c)^3*a^3*d^3))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^{13} \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int \frac {1}{x^{13}\,{\left (b\,x^8+a\right )}^2\,\sqrt {d\,x^8+c}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{x^{13} \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int \frac {1}{x^{13} \left (b \,x^{8}+a \right )^{2} \sqrt {d \,x^{8}+c}}d x \] Input:

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

Output:

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