\(\int \frac {x^{14}}{(a+b x^6)^2 \sqrt {c+d x^6}} \, dx\) [79]

Optimal result

Integrand size = 24, antiderivative size = 141 \[ \int \frac {x^{14}}{\left (a+b x^6\right )^2 \sqrt {c+d x^6}} \, dx=\frac {a x^3 \sqrt {c+d x^6}}{6 b (b c-a d) \left (a+b x^6\right )}-\frac {\sqrt {a} (3 b c-2 a d) \arctan \left (\frac {\sqrt {b c-a d} x^3}{\sqrt {a} \sqrt {c+d x^6}}\right )}{6 b^2 (b c-a d)^{3/2}}+\frac {\text {arctanh}\left (\frac {\sqrt {d} x^3}{\sqrt {c+d x^6}}\right )}{3 b^2 \sqrt {d}} \] Output:

1/6*a*x^3*(d*x^6+c)^(1/2)/b/(-a*d+b*c)/(b*x^6+a)-1/6*a^(1/2)*(-2*a*d+3*b*c 
)*arctan((-a*d+b*c)^(1/2)*x^3/a^(1/2)/(d*x^6+c)^(1/2))/b^2/(-a*d+b*c)^(3/2 
)+1/3*arctanh(d^(1/2)*x^3/(d*x^6+c)^(1/2))/b^2/d^(1/2)
 

Mathematica [A] (verified)

Time = 2.20 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.09 \[ \int \frac {x^{14}}{\left (a+b x^6\right )^2 \sqrt {c+d x^6}} \, dx=\frac {\frac {a b x^3 \sqrt {c+d x^6}}{(b c-a d) \left (a+b x^6\right )}+\frac {\sqrt {a} (-3 b c+2 a d) \arctan \left (\frac {a \sqrt {d}+b x^3 \left (\sqrt {d} x^3+\sqrt {c+d x^6}\right )}{\sqrt {a} \sqrt {b c-a d}}\right )}{(b c-a d)^{3/2}}+\frac {2 \log \left (\sqrt {d} x^3+\sqrt {c+d x^6}\right )}{\sqrt {d}}}{6 b^2} \] Input:

Integrate[x^14/((a + b*x^6)^2*Sqrt[c + d*x^6]),x]
 

Output:

((a*b*x^3*Sqrt[c + d*x^6])/((b*c - a*d)*(a + b*x^6)) + (Sqrt[a]*(-3*b*c + 
2*a*d)*ArcTan[(a*Sqrt[d] + b*x^3*(Sqrt[d]*x^3 + Sqrt[c + d*x^6]))/(Sqrt[a] 
*Sqrt[b*c - a*d])])/(b*c - a*d)^(3/2) + (2*Log[Sqrt[d]*x^3 + Sqrt[c + d*x^ 
6]])/Sqrt[d])/(6*b^2)
 

Rubi [A] (verified)

Time = 0.55 (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.

\(\displaystyle \int \frac {x^{14}}{\left (a+b x^6\right )^2 \sqrt {c+d x^6}} \, dx\)

\(\Big \downarrow \) 965

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

\(\Big \downarrow \) 372

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

\(\Big \downarrow \) 398

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{3} \left (\frac {a x^3 \sqrt {c+d x^6}}{2 b \left (a+b x^6\right ) (b c-a d)}-\frac {\frac {a (3 b c-2 a d) \int \frac {1}{\left (b x^6+a\right ) \sqrt {d x^6+c}}dx^3}{b}-\frac {2 (b c-a d) \text {arctanh}\left (\frac {\sqrt {d} x^3}{\sqrt {c+d x^6}}\right )}{b \sqrt {d}}}{2 b (b c-a d)}\right )\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {1}{3} \left (\frac {a x^3 \sqrt {c+d x^6}}{2 b \left (a+b x^6\right ) (b c-a d)}-\frac {\frac {a (3 b c-2 a d) \int \frac {1}{a-(a d-b c) x^6}d\frac {x^3}{\sqrt {d x^6+c}}}{b}-\frac {2 (b c-a d) \text {arctanh}\left (\frac {\sqrt {d} x^3}{\sqrt {c+d x^6}}\right )}{b \sqrt {d}}}{2 b (b c-a d)}\right )\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {1}{3} \left (\frac {a x^3 \sqrt {c+d x^6}}{2 b \left (a+b x^6\right ) (b c-a d)}-\frac {\frac {\sqrt {a} (3 b c-2 a d) \arctan \left (\frac {x^3 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^6}}\right )}{b \sqrt {b c-a d}}-\frac {2 (b c-a d) \text {arctanh}\left (\frac {\sqrt {d} x^3}{\sqrt {c+d x^6}}\right )}{b \sqrt {d}}}{2 b (b c-a d)}\right )\)

Input:

Int[x^14/((a + b*x^6)^2*Sqrt[c + d*x^6]),x]
 

Output:

((a*x^3*Sqrt[c + d*x^6])/(2*b*(b*c - a*d)*(a + b*x^6)) - ((Sqrt[a]*(3*b*c 
- 2*a*d)*ArcTan[(Sqrt[b*c - a*d]*x^3)/(Sqrt[a]*Sqrt[c + d*x^6])])/(b*Sqrt[ 
b*c - a*d]) - (2*(b*c - a*d)*ArcTanh[(Sqrt[d]*x^3)/Sqrt[c + d*x^6]])/(b*Sq 
rt[d]))/(2*b*(b*c - a*d)))/3
 

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 = 5.78 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.83

method	result	size
pseudoelliptic	\(-\frac {-\frac {a \left (-\frac {\sqrt {d \,x^{6}+c}\, b \,x^{3}}{b \,x^{6}+a}-\frac {\left (2 a d -3 c b \right ) \operatorname {arctanh}\left (\frac {a \sqrt {d \,x^{6}+c}}{x^{3} \sqrt {a \left (a d -c b \right )}}\right )}{\sqrt {a \left (a d -c b \right )}}\right )}{a d -c b}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{6}+c}}{x^{3} \sqrt {d}}\right )}{\sqrt {d}}}{6 b^{2}}\)	\(117\)

Input:

int(x^14/(b*x^6+a)^2/(d*x^6+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 1077, normalized size of antiderivative = 7.64 \[ \int \frac {x^{14}}{\left (a+b x^6\right )^2 \sqrt {c+d x^6}} \, dx =\text {Too large to display} \] Input:

integrate(x^14/(b*x^6+a)^2/(d*x^6+c)^(1/2),x, algorithm="fricas")
 

Output:

[1/24*(4*sqrt(d*x^6 + c)*a*b*d*x^3 + 4*((b^2*c - a*b*d)*x^6 + a*b*c - a^2* 
d)*sqrt(d)*log(-2*d*x^6 - 2*sqrt(d*x^6 + c)*sqrt(d)*x^3 - c) + ((3*b^2*c*d 
 - 2*a*b*d^2)*x^6 + 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^12 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^6 + a^2*c^ 
2 - 4*((b^2*c^2 - 3*a*b*c*d + 2*a^2*d^2)*x^9 - (a*b*c^2 - a^2*c*d)*x^3)*sq 
rt(d*x^6 + c)*sqrt(-a/(b*c - a*d)))/(b^2*x^12 + 2*a*b*x^6 + a^2)))/((b^4*c 
*d - a*b^3*d^2)*x^6 + a*b^3*c*d - a^2*b^2*d^2), 1/24*(4*sqrt(d*x^6 + c)*a* 
b*d*x^3 - 8*((b^2*c - a*b*d)*x^6 + a*b*c - a^2*d)*sqrt(-d)*arctan(sqrt(-d) 
*x^3/sqrt(d*x^6 + c)) + ((3*b^2*c*d - 2*a*b*d^2)*x^6 + 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^12 - 2*( 
3*a*b*c^2 - 4*a^2*c*d)*x^6 + a^2*c^2 - 4*((b^2*c^2 - 3*a*b*c*d + 2*a^2*d^2 
)*x^9 - (a*b*c^2 - a^2*c*d)*x^3)*sqrt(d*x^6 + c)*sqrt(-a/(b*c - a*d)))/(b^ 
2*x^12 + 2*a*b*x^6 + a^2)))/((b^4*c*d - a*b^3*d^2)*x^6 + a*b^3*c*d - a^2*b 
^2*d^2), 1/12*(2*sqrt(d*x^6 + c)*a*b*d*x^3 + ((3*b^2*c*d - 2*a*b*d^2)*x^6 
+ 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^ 
6 - a*c)*sqrt(d*x^6 + c)*sqrt(a/(b*c - a*d))/(a*d*x^9 + a*c*x^3)) + 2*((b^ 
2*c - a*b*d)*x^6 + a*b*c - a^2*d)*sqrt(d)*log(-2*d*x^6 - 2*sqrt(d*x^6 + c) 
*sqrt(d)*x^3 - c))/((b^4*c*d - a*b^3*d^2)*x^6 + a*b^3*c*d - a^2*b^2*d^2), 
1/12*(2*sqrt(d*x^6 + c)*a*b*d*x^3 - 4*((b^2*c - a*b*d)*x^6 + a*b*c - a^2*d 
)*sqrt(-d)*arctan(sqrt(-d)*x^3/sqrt(d*x^6 + c)) + ((3*b^2*c*d - 2*a*b*d...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{14}}{\left (a+b x^6\right )^2 \sqrt {c+d x^6}} \, dx=\text {Timed out} \] Input:

integrate(x**14/(b*x**6+a)**2/(d*x**6+c)**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {x^{14}}{\left (a+b x^6\right )^2 \sqrt {c+d x^6}} \, dx=\int { \frac {x^{14}}{{\left (b x^{6} + a\right )}^{2} \sqrt {d x^{6} + c}} \,d x } \] Input:

integrate(x^14/(b*x^6+a)^2/(d*x^6+c)^(1/2),x, algorithm="maxima")
 

Output:

integrate(x^14/((b*x^6 + a)^2*sqrt(d*x^6 + c)), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 343 vs. \(2 (117) = 234\).

Time = 0.14 (sec) , antiderivative size = 343, normalized size of antiderivative = 2.43 \[ \int \frac {x^{14}}{\left (a+b x^6\right )^2 \sqrt {c+d x^6}} \, dx=-\frac {{\left (3 \, a b c \sqrt {-d} \arctan \left (\frac {a \sqrt {d}}{\sqrt {a b c - a^{2} d}}\right ) - 2 \, a^{2} \sqrt {-d} d \arctan \left (\frac {a \sqrt {d}}{\sqrt {a b c - a^{2} d}}\right ) - 2 \, \sqrt {a b c - a^{2} d} b c \arctan \left (\frac {\sqrt {d}}{\sqrt {-d}}\right ) + 2 \, \sqrt {a b c - a^{2} d} a d \arctan \left (\frac {\sqrt {d}}{\sqrt {-d}}\right ) + \sqrt {a b c - a^{2} d} a \sqrt {-d} \sqrt {d}\right )} \mathrm {sgn}\left (x\right )}{6 \, {\left (\sqrt {a b c - a^{2} d} b^{3} c \sqrt {-d} - \sqrt {a b c - a^{2} d} a b^{2} \sqrt {-d} d\right )}} + \frac {a c \sqrt {d + \frac {c}{x^{6}}}}{6 \, {\left (b^{2} c \mathrm {sgn}\left (x\right ) - a b d \mathrm {sgn}\left (x\right )\right )} {\left (b c + a {\left (d + \frac {c}{x^{6}}\right )} - a d\right )}} + \frac {{\left (3 \, a b c - 2 \, a^{2} d\right )} \arctan \left (\frac {a \sqrt {d + \frac {c}{x^{6}}}}{\sqrt {a b c - a^{2} d}}\right )}{6 \, {\left (b^{3} c \mathrm {sgn}\left (x\right ) - a b^{2} d \mathrm {sgn}\left (x\right )\right )} \sqrt {a b c - a^{2} d}} - \frac {\arctan \left (\frac {\sqrt {d + \frac {c}{x^{6}}}}{\sqrt {-d}}\right )}{3 \, b^{2} \sqrt {-d} \mathrm {sgn}\left (x\right )} \] Input:

integrate(x^14/(b*x^6+a)^2/(d*x^6+c)^(1/2),x, algorithm="giac")
 

Output:

-1/6*(3*a*b*c*sqrt(-d)*arctan(a*sqrt(d)/sqrt(a*b*c - a^2*d)) - 2*a^2*sqrt( 
-d)*d*arctan(a*sqrt(d)/sqrt(a*b*c - a^2*d)) - 2*sqrt(a*b*c - a^2*d)*b*c*ar 
ctan(sqrt(d)/sqrt(-d)) + 2*sqrt(a*b*c - a^2*d)*a*d*arctan(sqrt(d)/sqrt(-d) 
) + sqrt(a*b*c - a^2*d)*a*sqrt(-d)*sqrt(d))*sgn(x)/(sqrt(a*b*c - a^2*d)*b^ 
3*c*sqrt(-d) - sqrt(a*b*c - a^2*d)*a*b^2*sqrt(-d)*d) + 1/6*a*c*sqrt(d + c/ 
x^6)/((b^2*c*sgn(x) - a*b*d*sgn(x))*(b*c + a*(d + c/x^6) - a*d)) + 1/6*(3* 
a*b*c - 2*a^2*d)*arctan(a*sqrt(d + c/x^6)/sqrt(a*b*c - a^2*d))/((b^3*c*sgn 
(x) - a*b^2*d*sgn(x))*sqrt(a*b*c - a^2*d)) - 1/3*arctan(sqrt(d + c/x^6)/sq 
rt(-d))/(b^2*sqrt(-d)*sgn(x))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{14}}{\left (a+b x^6\right )^2 \sqrt {c+d x^6}} \, dx=\int \frac {x^{14}}{{\left (b\,x^6+a\right )}^2\,\sqrt {d\,x^6+c}} \,d x \] Input:

int(x^14/((a + b*x^6)^2*(c + d*x^6)^(1/2)),x)
 

Output:

int(x^14/((a + b*x^6)^2*(c + d*x^6)^(1/2)), x)
 

Reduce [F]

\[ \int \frac {x^{14}}{\left (a+b x^6\right )^2 \sqrt {c+d x^6}} \, dx =\text {Too large to display} \] Input:

int(x^14/(b*x^6+a)^2/(d*x^6+c)^(1/2),x)
 

Output:

( - 4*sqrt(c + d*x**6)*a*b*d*x**3 - 2*sqrt(d)*log(sqrt(c + d*x**6) - sqrt( 
d)*x**3)*a**2*d + sqrt(d)*log(sqrt(c + d*x**6) - sqrt(d)*x**3)*a*b*c - 2*s 
qrt(d)*log(sqrt(c + d*x**6) - sqrt(d)*x**3)*a*b*d*x**6 + sqrt(d)*log(sqrt( 
c + d*x**6) - sqrt(d)*x**3)*b**2*c*x**6 + 2*sqrt(d)*log(sqrt(c + d*x**6) + 
 sqrt(d)*x**3)*a**2*d - sqrt(d)*log(sqrt(c + d*x**6) + sqrt(d)*x**3)*a*b*c 
 + 2*sqrt(d)*log(sqrt(c + d*x**6) + sqrt(d)*x**3)*a*b*d*x**6 - sqrt(d)*log 
(sqrt(c + d*x**6) + sqrt(d)*x**3)*b**2*c*x**6 - 24*int((sqrt(c + d*x**6)*x 
**2)/(2*a**3*c*d + 2*a**3*d**2*x**6 - a**2*b*c**2 + 3*a**2*b*c*d*x**6 + 4* 
a**2*b*d**2*x**12 - 2*a*b**2*c**2*x**6 + 2*a*b**2*d**2*x**18 - b**3*c**2*x 
**12 - b**3*c*d*x**18),x)*a**5*d**3 + 48*int((sqrt(c + d*x**6)*x**2)/(2*a* 
*3*c*d + 2*a**3*d**2*x**6 - a**2*b*c**2 + 3*a**2*b*c*d*x**6 + 4*a**2*b*d** 
2*x**12 - 2*a*b**2*c**2*x**6 + 2*a*b**2*d**2*x**18 - b**3*c**2*x**12 - b** 
3*c*d*x**18),x)*a**4*b*c*d**2 - 24*int((sqrt(c + d*x**6)*x**2)/(2*a**3*c*d 
 + 2*a**3*d**2*x**6 - a**2*b*c**2 + 3*a**2*b*c*d*x**6 + 4*a**2*b*d**2*x**1 
2 - 2*a*b**2*c**2*x**6 + 2*a*b**2*d**2*x**18 - b**3*c**2*x**12 - b**3*c*d* 
x**18),x)*a**4*b*d**3*x**6 - 18*int((sqrt(c + d*x**6)*x**2)/(2*a**3*c*d + 
2*a**3*d**2*x**6 - a**2*b*c**2 + 3*a**2*b*c*d*x**6 + 4*a**2*b*d**2*x**12 - 
 2*a*b**2*c**2*x**6 + 2*a*b**2*d**2*x**18 - b**3*c**2*x**12 - b**3*c*d*x** 
18),x)*a**3*b**2*c**2*d + 48*int((sqrt(c + d*x**6)*x**2)/(2*a**3*c*d + 2*a 
**3*d**2*x**6 - a**2*b*c**2 + 3*a**2*b*c*d*x**6 + 4*a**2*b*d**2*x**12 -...