\(\int \frac {1}{x^{13} (a+b x^8) \sqrt {c+d x^8}} \, dx\) [100]

Optimal result

Integrand size = 24, antiderivative size = 115 \[ \int \frac {1}{x^{13} \left (a+b x^8\right ) \sqrt {c+d x^8}} \, dx=-\frac {\sqrt {c+d x^8}}{12 a c x^{12}}+\frac {(3 b c+2 a d) \sqrt {c+d x^8}}{12 a^2 c^2 x^4}+\frac {b^2 \arctan \left (\frac {\sqrt {b c-a d} x^4}{\sqrt {a} \sqrt {c+d x^8}}\right )}{4 a^{5/2} \sqrt {b c-a d}} \] Output:

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

Mathematica [A] (verified)

Time = 1.69 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.05 \[ \int \frac {1}{x^{13} \left (a+b x^8\right ) \sqrt {c+d x^8}} \, dx=\frac {\sqrt {c+d x^8} \left (-a c+3 b c x^8+2 a d x^8\right )}{12 a^2 c^2 x^{12}}+\frac {b^2 \arctan \left (\frac {a \sqrt {d}+b x^4 \left (\sqrt {d} x^4+\sqrt {c+d x^8}\right )}{\sqrt {a} \sqrt {b c-a d}}\right )}{4 a^{5/2} \sqrt {b c-a d}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {965, 382, 25, 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.

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

\(\Big \downarrow \) 965

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

\(\Big \downarrow \) 382

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 445

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

(-1/3*Sqrt[c + d*x^8]/(a*c*x^12) - (-(((3*b*c + 2*a*d)*Sqrt[c + d*x^8])/(a 
*c*x^4)) - (3*b^2*c*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))/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 382

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) 
, x_Symbol] :> Simp[(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/ 
(a*c*e*(m + 1))), x] - Simp[1/(a*c*e^2*(m + 1))   Int[(e*x)^(m + 2)*(a + b* 
x^2)^p*(c + d*x^2)^q*Simp[(b*c + a*d)*(m + 3) + 2*(b*c*p + a*d*q) + b*d*(m 
+ 2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[ 
b*c - a*d, 0] && LtQ[m, -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 = 20.18 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.77

method	result	size
pseudoelliptic	\(\frac {-\frac {\sqrt {d \,x^{8}+c}\, \left (-2 a d \,x^{8}-3 b c \,x^{8}+a c \right )}{3 x^{12}}+\frac {b^{2} c^{2} \operatorname {arctanh}\left (\frac {a \sqrt {d \,x^{8}+c}}{x^{4} \sqrt {a \left (a d -c b \right )}}\right )}{\sqrt {a \left (a d -c b \right )}}}{4 a^{2} c^{2}}\)	\(88\)

Input:

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

Output:

1/4/a^2*(-1/3*(d*x^8+c)^(1/2)*(-2*a*d*x^8-3*b*c*x^8+a*c)/x^12+b^2*c^2/(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 = 0.16 (sec) , antiderivative size = 416, normalized size of antiderivative = 3.62 \[ \int \frac {1}{x^{13} \left (a+b x^8\right ) \sqrt {c+d x^8}} \, dx=\left [-\frac {3 \, \sqrt {-a b c + a^{2} d} b^{2} c^{2} x^{12} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{16} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{8} + a^{2} c^{2} - 4 \, {\left ({\left (b c - 2 \, a d\right )} x^{12} - a c x^{4}\right )} \sqrt {d x^{8} + c} \sqrt {-a b c + a^{2} d}}{b^{2} x^{16} + 2 \, a b x^{8} + a^{2}}\right ) - 4 \, {\left ({\left (3 \, a b^{2} c^{2} - a^{2} b c d - 2 \, a^{3} d^{2}\right )} x^{8} - a^{2} b c^{2} + a^{3} c d\right )} \sqrt {d x^{8} + c}}{48 \, {\left (a^{3} b c^{3} - a^{4} c^{2} d\right )} x^{12}}, \frac {3 \, \sqrt {a b c - a^{2} d} b^{2} c^{2} x^{12} \arctan \left (\frac {{\left ({\left (b c - 2 \, a d\right )} x^{8} - a c\right )} \sqrt {d x^{8} + c} \sqrt {a b c - a^{2} d}}{2 \, {\left ({\left (a b c d - a^{2} d^{2}\right )} x^{12} + {\left (a b c^{2} - a^{2} c d\right )} x^{4}\right )}}\right ) + 2 \, {\left ({\left (3 \, a b^{2} c^{2} - a^{2} b c d - 2 \, a^{3} d^{2}\right )} x^{8} - a^{2} b c^{2} + a^{3} c d\right )} \sqrt {d x^{8} + c}}{24 \, {\left (a^{3} b c^{3} - a^{4} c^{2} d\right )} x^{12}}\right ] \] Input:

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

Output:

[-1/48*(3*sqrt(-a*b*c + a^2*d)*b^2*c^2*x^12*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*((3*a*b^2*c^2 - a^2*b*c*d - 2*a^3*d^2)*x^8 - a^2*b*c^2 + a^ 
3*c*d)*sqrt(d*x^8 + c))/((a^3*b*c^3 - a^4*c^2*d)*x^12), 1/24*(3*sqrt(a*b*c 
 - a^2*d)*b^2*c^2*x^12*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*((3*a*b^2*c^2 - a^2*b*c*d - 2*a^3*d^2)*x^8 - a^2*b*c^2 + a^3*c*d)*sq 
rt(d*x^8 + c))/((a^3*b*c^3 - a^4*c^2*d)*x^12)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate(1/((b*x^8 + a)*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. 205 vs. \(2 (95) = 190\).

Time = 0.28 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.78 \[ \int \frac {1}{x^{13} \left (a+b x^8\right ) \sqrt {c+d x^8}} \, dx=-\frac {1}{12} \, d^{\frac {5}{2}} {\left (\frac {3 \, b^{2} \arctan \left (\frac {{\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{\sqrt {a b c d - a^{2} d^{2}} a^{2} d^{2}} + \frac {2 \, {\left (3 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{4} b - 6 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} b c - 6 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} a d + 3 \, b c^{2} + 2 \, a c d\right )}}{{\left ({\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} - c\right )}^{3} a^{2} d^{2}}\right )} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{x^{13} \left (a+b x^8\right ) \sqrt {c+d x^8}} \, dx=\frac {-\sqrt {d \,x^{8}+c}\, c +2 \sqrt {d \,x^{8}+c}\, d \,x^{8}-12 \left (\int \frac {\sqrt {d \,x^{8}+c}}{b d \,x^{21}+a d \,x^{13}+b c \,x^{13}+a c \,x^{5}}d x \right ) b \,c^{2} x^{12}}{12 a \,c^{2} x^{12}} \] Input:

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

Output:

( - sqrt(c + d*x**8)*c + 2*sqrt(c + d*x**8)*d*x**8 - 12*int(sqrt(c + d*x** 
8)/(a*c*x**5 + a*d*x**13 + b*c*x**13 + b*d*x**21),x)*b*c**2*x**12)/(12*a*c 
**2*x**12)