\(\int \frac {x^{19}}{(a+b x^8)^2 \sqrt {c+d x^8}} \, dx\) [118]

Optimal result

Integrand size = 24, antiderivative size = 141 \[ \int \frac {x^{19}}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\frac {a x^4 \sqrt {c+d x^8}}{8 b (b c-a d) \left (a+b x^8\right )}-\frac {\sqrt {a} (3 b c-2 a d) \arctan \left (\frac {\sqrt {b c-a d} x^4}{\sqrt {a} \sqrt {c+d x^8}}\right )}{8 b^2 (b c-a d)^{3/2}}+\frac {\text {arctanh}\left (\frac {\sqrt {d} x^4}{\sqrt {c+d x^8}}\right )}{4 b^2 \sqrt {d}} \] Output:

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

Mathematica [A] (verified)

Time = 2.91 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.09 \[ \int \frac {x^{19}}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\frac {\frac {a b x^4 \sqrt {c+d x^8}}{(b c-a d) \left (a+b x^8\right )}+\frac {\sqrt {a} (-3 b c+2 a d) \arctan \left (\frac {a \sqrt {d}+b x^4 \left (\sqrt {d} x^4+\sqrt {c+d x^8}\right )}{\sqrt {a} \sqrt {b c-a d}}\right )}{(b c-a d)^{3/2}}+\frac {2 \log \left (\sqrt {d} x^4+\sqrt {c+d x^8}\right )}{\sqrt {d}}}{8 b^2} \] Input:

Integrate[x^19/((a + b*x^8)^2*Sqrt[c + d*x^8]),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.18, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {965, 372, 398, 224, 219, 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.

\(\displaystyle \int \frac {x^{19}}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx\)

\(\Big \downarrow \) 965

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

\(\Big \downarrow \) 372

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

\(\Big \downarrow \) 398

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 218

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

Input:

Int[x^19/((a + b*x^8)^2*Sqrt[c + d*x^8]),x]
 

Output:

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

Defintions of rubi rules used

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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

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

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 372

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 
)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 
))   Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + 
 (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, 
e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a 
, b, c, d, e, m, 2, p, q, x]
 

rule 398

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) 
, x_Symbol] :> Simp[f/b   Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ 
b   Int[1/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f} 
, 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]
 

Maple [A] (verified)

Time = 20.28 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.83

method	result	size
pseudoelliptic	\(-\frac {-\frac {a \left (-\frac {\sqrt {d \,x^{8}+c}\, b \,x^{4}}{b \,x^{8}+a}-\frac {\left (2 a d -3 c b \right ) \operatorname {arctanh}\left (\frac {a \sqrt {d \,x^{8}+c}}{x^{4} \sqrt {a \left (a d -c b \right )}}\right )}{\sqrt {a \left (a d -c b \right )}}\right )}{a d -c b}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{8}+c}}{x^{4} \sqrt {d}}\right )}{\sqrt {d}}}{8 b^{2}}\)	\(117\)

Input:

int(x^19/(b*x^8+a)^2/(d*x^8+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 1083, normalized size of antiderivative = 7.68 \[ \int \frac {x^{19}}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx =\text {Too large to display} \] Input:

integrate(x^19/(b*x^8+a)^2/(d*x^8+c)^(1/2),x, algorithm="fricas")
 

Output:

[1/32*(4*sqrt(d*x^8 + c)*a*b*d*x^4 + 4*((b^2*c - a*b*d)*x^8 + a*b*c - a^2* 
d)*sqrt(d)*log(-2*d*x^8 - 2*sqrt(d*x^8 + c)*sqrt(d)*x^4 - c) + ((3*b^2*c*d 
 - 2*a*b*d^2)*x^8 + 3*a*b*c*d - 2*a^2*d^2)*sqrt(-a/(b*c - a*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^2*c^2 - 3*a*b*c*d + 2*a^2*d^2)*x^12 - (a*b*c^2 - a^2*c*d)*x^4)*s 
qrt(d*x^8 + c)*sqrt(-a/(b*c - a*d)))/(b^2*x^16 + 2*a*b*x^8 + a^2)))/((b^4* 
c*d - a*b^3*d^2)*x^8 + a*b^3*c*d - a^2*b^2*d^2), 1/32*(4*sqrt(d*x^8 + c)*a 
*b*d*x^4 - 8*((b^2*c - a*b*d)*x^8 + a*b*c - a^2*d)*sqrt(-d)*arctan(sqrt(d* 
x^8 + c)*sqrt(-d)/(d*x^4)) + ((3*b^2*c*d - 2*a*b*d^2)*x^8 + 3*a*b*c*d - 2* 
a^2*d^2)*sqrt(-a/(b*c - a*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^2*c^2 - 3*a*b*c*d + 2*a^ 
2*d^2)*x^12 - (a*b*c^2 - a^2*c*d)*x^4)*sqrt(d*x^8 + c)*sqrt(-a/(b*c - a*d) 
))/(b^2*x^16 + 2*a*b*x^8 + a^2)))/((b^4*c*d - a*b^3*d^2)*x^8 + a*b^3*c*d - 
 a^2*b^2*d^2), 1/16*(2*sqrt(d*x^8 + c)*a*b*d*x^4 + ((3*b^2*c*d - 2*a*b*d^2 
)*x^8 + 3*a*b*c*d - 2*a^2*d^2)*sqrt(a/(b*c - a*d))*arctan(-1/2*((b*c - 2*a 
*d)*x^8 - a*c)*sqrt(d*x^8 + c)*sqrt(a/(b*c - a*d))/(a*d*x^12 + a*c*x^4)) + 
 2*((b^2*c - a*b*d)*x^8 + a*b*c - a^2*d)*sqrt(d)*log(-2*d*x^8 - 2*sqrt(d*x 
^8 + c)*sqrt(d)*x^4 - c))/((b^4*c*d - a*b^3*d^2)*x^8 + a*b^3*c*d - a^2*b^2 
*d^2), 1/16*(2*sqrt(d*x^8 + c)*a*b*d*x^4 + ((3*b^2*c*d - 2*a*b*d^2)*x^8 + 
3*a*b*c*d - 2*a^2*d^2)*sqrt(a/(b*c - a*d))*arctan(-1/2*((b*c - 2*a*d)*x...
 

Sympy [F]

\[ \int \frac {x^{19}}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int \frac {x^{19}}{\left (a + b x^{8}\right )^{2} \sqrt {c + d x^{8}}}\, dx \] Input:

integrate(x**19/(b*x**8+a)**2/(d*x**8+c)**(1/2),x)
 

Output:

Integral(x**19/((a + b*x**8)**2*sqrt(c + d*x**8)), x)
 

Maxima [F]

\[ \int \frac {x^{19}}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int { \frac {x^{19}}{{\left (b x^{8} + a\right )}^{2} \sqrt {d x^{8} + c}} \,d x } \] Input:

integrate(x^19/(b*x^8+a)^2/(d*x^8+c)^(1/2),x, algorithm="maxima")
 

Output:

integrate(x^19/((b*x^8 + a)^2*sqrt(d*x^8 + c)), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 298 vs. \(2 (117) = 234\).

Time = 0.27 (sec) , antiderivative size = 298, normalized size of antiderivative = 2.11 \[ \int \frac {x^{19}}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=-\frac {{\left (3 \, a b c \sqrt {d} - 2 \, a^{2} d^{\frac {3}{2}}\right )} \arctan \left (-\frac {{\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{8 \, {\left (b^{3} c - a b^{2} d\right )} \sqrt {a b c d - a^{2} d^{2}}} - \frac {{\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} a b c \sqrt {d} - 2 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} a^{2} d^{\frac {3}{2}} - a b c^{2} \sqrt {d}}{4 \, {\left ({\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} a d + b c^{2}\right )} {\left (b^{3} c - a b^{2} d\right )}} - \frac {\log \left ({\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2}\right )}{8 \, b^{2} \sqrt {d}} \] Input:

integrate(x^19/(b*x^8+a)^2/(d*x^8+c)^(1/2),x, algorithm="giac")
 

Output:

-1/8*(3*a*b*c*sqrt(d) - 2*a^2*d^(3/2))*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))/((b^3*c - a*b^2*d)*s 
qrt(a*b*c*d - a^2*d^2)) - 1/4*((sqrt(d)*x^4 - sqrt(d*x^8 + c))^2*a*b*c*sqr 
t(d) - 2*(sqrt(d)*x^4 - sqrt(d*x^8 + c))^2*a^2*d^(3/2) - a*b*c^2*sqrt(d))/ 
(((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)*(b^3*c - a*b^2*d) 
) - 1/8*log((sqrt(d)*x^4 - sqrt(d*x^8 + c))^2)/(b^2*sqrt(d))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{19}}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int \frac {x^{19}}{{\left (b\,x^8+a\right )}^2\,\sqrt {d\,x^8+c}} \,d x \] Input:

int(x^19/((a + b*x^8)^2*(c + d*x^8)^(1/2)),x)
 

Output:

int(x^19/((a + b*x^8)^2*(c + d*x^8)^(1/2)), x)
 

Reduce [F]

\[ \int \frac {x^{19}}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx =\text {Too large to display} \] Input:

int(x^19/(b*x^8+a)^2/(d*x^8+c)^(1/2),x)
 

Output:

( - 4*sqrt(c + d*x**8)*a*b*d*x**4 - 2*sqrt(d)*log(sqrt(c + d*x**8) - sqrt( 
d)*x**4)*a**2*d + sqrt(d)*log(sqrt(c + d*x**8) - sqrt(d)*x**4)*a*b*c - 2*s 
qrt(d)*log(sqrt(c + d*x**8) - sqrt(d)*x**4)*a*b*d*x**8 + sqrt(d)*log(sqrt( 
c + d*x**8) - sqrt(d)*x**4)*b**2*c*x**8 + 2*sqrt(d)*log(sqrt(c + d*x**8) + 
 sqrt(d)*x**4)*a**2*d - sqrt(d)*log(sqrt(c + d*x**8) + sqrt(d)*x**4)*a*b*c 
 + 2*sqrt(d)*log(sqrt(c + d*x**8) + sqrt(d)*x**4)*a*b*d*x**8 - sqrt(d)*log 
(sqrt(c + d*x**8) + sqrt(d)*x**4)*b**2*c*x**8 - 32*int((sqrt(c + d*x**8)*x 
**3)/(2*a**3*c*d + 2*a**3*d**2*x**8 - a**2*b*c**2 + 3*a**2*b*c*d*x**8 + 4* 
a**2*b*d**2*x**16 - 2*a*b**2*c**2*x**8 + 2*a*b**2*d**2*x**24 - b**3*c**2*x 
**16 - b**3*c*d*x**24),x)*a**5*d**3 + 64*int((sqrt(c + d*x**8)*x**3)/(2*a* 
*3*c*d + 2*a**3*d**2*x**8 - a**2*b*c**2 + 3*a**2*b*c*d*x**8 + 4*a**2*b*d** 
2*x**16 - 2*a*b**2*c**2*x**8 + 2*a*b**2*d**2*x**24 - b**3*c**2*x**16 - b** 
3*c*d*x**24),x)*a**4*b*c*d**2 - 32*int((sqrt(c + d*x**8)*x**3)/(2*a**3*c*d 
 + 2*a**3*d**2*x**8 - a**2*b*c**2 + 3*a**2*b*c*d*x**8 + 4*a**2*b*d**2*x**1 
6 - 2*a*b**2*c**2*x**8 + 2*a*b**2*d**2*x**24 - b**3*c**2*x**16 - b**3*c*d* 
x**24),x)*a**4*b*d**3*x**8 - 24*int((sqrt(c + d*x**8)*x**3)/(2*a**3*c*d + 
2*a**3*d**2*x**8 - a**2*b*c**2 + 3*a**2*b*c*d*x**8 + 4*a**2*b*d**2*x**16 - 
 2*a*b**2*c**2*x**8 + 2*a*b**2*d**2*x**24 - b**3*c**2*x**16 - b**3*c*d*x** 
24),x)*a**3*b**2*c**2*d + 64*int((sqrt(c + d*x**8)*x**3)/(2*a**3*c*d + 2*a 
**3*d**2*x**8 - a**2*b*c**2 + 3*a**2*b*c*d*x**8 + 4*a**2*b*d**2*x**16 -...