\(\int \frac {x^7}{\sqrt {a+b x^2-c x^4}} \, dx\) [995]

Optimal result

Integrand size = 21, antiderivative size = 124 \[ \int \frac {x^7}{\sqrt {a+b x^2-c x^4}} \, dx=-\frac {x^4 \sqrt {a+b x^2-c x^4}}{6 c}-\frac {\left (15 b^2+16 a c+10 b c x^2\right ) \sqrt {a+b x^2-c x^4}}{48 c^3}-\frac {b \left (5 b^2+12 a c\right ) \arctan \left (\frac {b-2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2-c x^4}}\right )}{32 c^{7/2}} \] Output:

-1/6*x^4*(-c*x^4+b*x^2+a)^(1/2)/c-1/48*(10*b*c*x^2+16*a*c+15*b^2)*(-c*x^4+ 
b*x^2+a)^(1/2)/c^3-1/32*b*(12*a*c+5*b^2)*arctan(1/2*(-2*c*x^2+b)/c^(1/2)/( 
-c*x^4+b*x^2+a)^(1/2))/c^(7/2)
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.86 \[ \int \frac {x^7}{\sqrt {a+b x^2-c x^4}} \, dx=\frac {\sqrt {a+b x^2-c x^4} \left (-15 b^2-16 a c-10 b c x^2-8 c^2 x^4\right )}{48 c^3}+\frac {\left (5 b^3+12 a b c\right ) \arctan \left (\frac {\sqrt {c} x^2}{-\sqrt {a}+\sqrt {a+b x^2-c x^4}}\right )}{16 c^{7/2}} \] Input:

Integrate[x^7/Sqrt[a + b*x^2 - c*x^4],x]
 

Output:

(Sqrt[a + b*x^2 - c*x^4]*(-15*b^2 - 16*a*c - 10*b*c*x^2 - 8*c^2*x^4))/(48* 
c^3) + ((5*b^3 + 12*a*b*c)*ArcTan[(Sqrt[c]*x^2)/(-Sqrt[a] + Sqrt[a + b*x^2 
 - c*x^4])])/(16*c^(7/2))
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1434, 1166, 27, 1225, 1092, 217}

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.

\(\displaystyle \int \frac {x^7}{\sqrt {a+b x^2-c x^4}} \, dx\)

\(\Big \downarrow \) 1434

\(\displaystyle \frac {1}{2} \int \frac {x^6}{\sqrt {-c x^4+b x^2+a}}dx^2\)

\(\Big \downarrow \) 1166

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1225

\(\displaystyle \frac {1}{2} \left (\frac {\frac {3 b \left (12 a c+5 b^2\right ) \int \frac {1}{\sqrt {-c x^4+b x^2+a}}dx^2}{8 c^2}-\frac {\left (16 a c+15 b^2+10 b c x^2\right ) \sqrt {a+b x^2-c x^4}}{4 c^2}}{6 c}-\frac {x^4 \sqrt {a+b x^2-c x^4}}{3 c}\right )\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {1}{2} \left (\frac {\frac {3 b \left (12 a c+5 b^2\right ) \int \frac {1}{-x^4-4 c}d\frac {b-2 c x^2}{\sqrt {-c x^4+b x^2+a}}}{4 c^2}-\frac {\left (16 a c+15 b^2+10 b c x^2\right ) \sqrt {a+b x^2-c x^4}}{4 c^2}}{6 c}-\frac {x^4 \sqrt {a+b x^2-c x^4}}{3 c}\right )\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{2} \left (\frac {-\frac {3 b \left (12 a c+5 b^2\right ) \arctan \left (\frac {b-2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2-c x^4}}\right )}{8 c^{5/2}}-\frac {\sqrt {a+b x^2-c x^4} \left (16 a c+15 b^2+10 b c x^2\right )}{4 c^2}}{6 c}-\frac {x^4 \sqrt {a+b x^2-c x^4}}{3 c}\right )\)

Input:

Int[x^7/Sqrt[a + b*x^2 - c*x^4],x]
 

Output:

(-1/3*(x^4*Sqrt[a + b*x^2 - c*x^4])/c + (-1/4*((15*b^2 + 16*a*c + 10*b*c*x 
^2)*Sqrt[a + b*x^2 - c*x^4])/c^2 - (3*b*(5*b^2 + 12*a*c)*ArcTan[(b - 2*c*x 
^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 - c*x^4])])/(8*c^(5/2)))/(6*c))/2
 

Defintions of rubi rules used

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 217

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

rule 1092

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[I 
nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a 
, b, c}, x]
 

rule 1166

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 
 1))), x] + Simp[1/(c*(m + 2*p + 1))   Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m 
+ 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]* 
(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && If[Ration 
alQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadrat 
icQ[a, b, c, d, e, m, p, x]
 

rule 1225

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*( 
x_)^2)^(p_), x_Symbol] :> Simp[(-(b*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 
 2*c*e*g*(p + 1)*x))*((a + b*x + c*x^2)^(p + 1)/(2*c^2*(p + 1)*(2*p + 3))), 
 x] + Simp[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p 
+ 3))/(2*c^2*(2*p + 3))   Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c 
, d, e, f, g, p}, x] &&  !LeQ[p, -1]
 

rule 1434

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp 
[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x + c*x^2)^p, x], x, x^2], x] /; Free 
Q[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]
 

Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.75

method	result	size
risch	\(-\frac {\left (8 c^{2} x^{4}+10 b c \,x^{2}+16 a c +15 b^{2}\right ) \sqrt {-c \,x^{4}+b \,x^{2}+a}}{48 c^{3}}+\frac {b \left (12 a c +5 b^{2}\right ) \arctan \left (\frac {\sqrt {c}\, \left (x^{2}-\frac {b}{2 c}\right )}{\sqrt {-c \,x^{4}+b \,x^{2}+a}}\right )}{32 c^{\frac {7}{2}}}\)	\(93\)
pseudoelliptic	\(\frac {-16 x^{4} \sqrt {-c \,x^{4}+b \,x^{2}+a}\, c^{\frac {5}{2}}-20 b \,x^{2} \sqrt {-c \,x^{4}+b \,x^{2}+a}\, c^{\frac {3}{2}}-36 b a \arctan \left (\frac {-2 c \,x^{2}+b}{2 \sqrt {c}\, \sqrt {-c \,x^{4}+b \,x^{2}+a}}\right ) c -15 b^{3} \arctan \left (\frac {-2 c \,x^{2}+b}{2 \sqrt {c}\, \sqrt {-c \,x^{4}+b \,x^{2}+a}}\right )-32 a \sqrt {-c \,x^{4}+b \,x^{2}+a}\, c^{\frac {3}{2}}-30 b^{2} \sqrt {-c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}}{96 c^{\frac {7}{2}}}\)	\(166\)
default	\(-\frac {x^{4} \sqrt {-c \,x^{4}+b \,x^{2}+a}}{6 c}-\frac {5 b \,x^{2} \sqrt {-c \,x^{4}+b \,x^{2}+a}}{24 c^{2}}-\frac {5 b^{2} \sqrt {-c \,x^{4}+b \,x^{2}+a}}{16 c^{3}}+\frac {5 b^{3} \arctan \left (\frac {\sqrt {c}\, \left (x^{2}-\frac {b}{2 c}\right )}{\sqrt {-c \,x^{4}+b \,x^{2}+a}}\right )}{32 c^{\frac {7}{2}}}+\frac {3 b a \arctan \left (\frac {\sqrt {c}\, \left (x^{2}-\frac {b}{2 c}\right )}{\sqrt {-c \,x^{4}+b \,x^{2}+a}}\right )}{8 c^{\frac {5}{2}}}-\frac {a \sqrt {-c \,x^{4}+b \,x^{2}+a}}{3 c^{2}}\)	\(168\)
elliptic	\(-\frac {x^{4} \sqrt {-c \,x^{4}+b \,x^{2}+a}}{6 c}-\frac {5 b \,x^{2} \sqrt {-c \,x^{4}+b \,x^{2}+a}}{24 c^{2}}-\frac {5 b^{2} \sqrt {-c \,x^{4}+b \,x^{2}+a}}{16 c^{3}}+\frac {5 b^{3} \arctan \left (\frac {\sqrt {c}\, \left (x^{2}-\frac {b}{2 c}\right )}{\sqrt {-c \,x^{4}+b \,x^{2}+a}}\right )}{32 c^{\frac {7}{2}}}+\frac {3 b a \arctan \left (\frac {\sqrt {c}\, \left (x^{2}-\frac {b}{2 c}\right )}{\sqrt {-c \,x^{4}+b \,x^{2}+a}}\right )}{8 c^{\frac {5}{2}}}-\frac {a \sqrt {-c \,x^{4}+b \,x^{2}+a}}{3 c^{2}}\)	\(168\)

Input:

int(x^7/(-c*x^4+b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/48*(8*c^2*x^4+10*b*c*x^2+16*a*c+15*b^2)/c^3*(-c*x^4+b*x^2+a)^(1/2)+1/32 
*b*(12*a*c+5*b^2)/c^(7/2)*arctan(c^(1/2)*(x^2-1/2/c*b)/(-c*x^4+b*x^2+a)^(1 
/2))
 

Fricas [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.01 \[ \int \frac {x^7}{\sqrt {a+b x^2-c x^4}} \, dx=\left [-\frac {3 \, {\left (5 \, b^{3} + 12 \, a b c\right )} \sqrt {-c} \log \left (8 \, c^{2} x^{4} - 8 \, b c x^{2} + b^{2} - 4 \, \sqrt {-c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} - b\right )} \sqrt {-c} - 4 \, a c\right ) + 4 \, {\left (8 \, c^{3} x^{4} + 10 \, b c^{2} x^{2} + 15 \, b^{2} c + 16 \, a c^{2}\right )} \sqrt {-c x^{4} + b x^{2} + a}}{192 \, c^{4}}, -\frac {3 \, {\left (5 \, b^{3} + 12 \, a b c\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} - b\right )} \sqrt {c}}{2 \, {\left (c^{2} x^{4} - b c x^{2} - a c\right )}}\right ) + 2 \, {\left (8 \, c^{3} x^{4} + 10 \, b c^{2} x^{2} + 15 \, b^{2} c + 16 \, a c^{2}\right )} \sqrt {-c x^{4} + b x^{2} + a}}{96 \, c^{4}}\right ] \] Input:

integrate(x^7/(-c*x^4+b*x^2+a)^(1/2),x, algorithm="fricas")
 

Output:

[-1/192*(3*(5*b^3 + 12*a*b*c)*sqrt(-c)*log(8*c^2*x^4 - 8*b*c*x^2 + b^2 - 4 
*sqrt(-c*x^4 + b*x^2 + a)*(2*c*x^2 - b)*sqrt(-c) - 4*a*c) + 4*(8*c^3*x^4 + 
 10*b*c^2*x^2 + 15*b^2*c + 16*a*c^2)*sqrt(-c*x^4 + b*x^2 + a))/c^4, -1/96* 
(3*(5*b^3 + 12*a*b*c)*sqrt(c)*arctan(1/2*sqrt(-c*x^4 + b*x^2 + a)*(2*c*x^2 
 - b)*sqrt(c)/(c^2*x^4 - b*c*x^2 - a*c)) + 2*(8*c^3*x^4 + 10*b*c^2*x^2 + 1 
5*b^2*c + 16*a*c^2)*sqrt(-c*x^4 + b*x^2 + a))/c^4]
 

Sympy [F]

\[ \int \frac {x^7}{\sqrt {a+b x^2-c x^4}} \, dx=\int \frac {x^{7}}{\sqrt {a + b x^{2} - c x^{4}}}\, dx \] Input:

integrate(x**7/(-c*x**4+b*x**2+a)**(1/2),x)
 

Output:

Integral(x**7/sqrt(a + b*x**2 - c*x**4), x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.23 \[ \int \frac {x^7}{\sqrt {a+b x^2-c x^4}} \, dx=-\frac {\sqrt {-c x^{4} + b x^{2} + a} x^{4}}{6 \, c} - \frac {5 \, \sqrt {-c x^{4} + b x^{2} + a} b x^{2}}{24 \, c^{2}} - \frac {5 \, b^{3} \arcsin \left (-\frac {2 \, c x^{2} - b}{\sqrt {b^{2} + 4 \, a c}}\right )}{32 \, c^{\frac {7}{2}}} - \frac {3 \, a b \arcsin \left (-\frac {2 \, c x^{2} - b}{\sqrt {b^{2} + 4 \, a c}}\right )}{8 \, c^{\frac {5}{2}}} - \frac {5 \, \sqrt {-c x^{4} + b x^{2} + a} b^{2}}{16 \, c^{3}} - \frac {\sqrt {-c x^{4} + b x^{2} + a} a}{3 \, c^{2}} \] Input:

integrate(x^7/(-c*x^4+b*x^2+a)^(1/2),x, algorithm="maxima")
 

Output:

-1/6*sqrt(-c*x^4 + b*x^2 + a)*x^4/c - 5/24*sqrt(-c*x^4 + b*x^2 + a)*b*x^2/ 
c^2 - 5/32*b^3*arcsin(-(2*c*x^2 - b)/sqrt(b^2 + 4*a*c))/c^(7/2) - 3/8*a*b* 
arcsin(-(2*c*x^2 - b)/sqrt(b^2 + 4*a*c))/c^(5/2) - 5/16*sqrt(-c*x^4 + b*x^ 
2 + a)*b^2/c^3 - 1/3*sqrt(-c*x^4 + b*x^2 + a)*a/c^2
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.90 \[ \int \frac {x^7}{\sqrt {a+b x^2-c x^4}} \, dx=-\frac {1}{48} \, \sqrt {-c x^{4} + b x^{2} + a} {\left (2 \, x^{2} {\left (\frac {4 \, x^{2}}{c} + \frac {5 \, b}{c^{2}}\right )} + \frac {15 \, b^{2} + 16 \, a c}{c^{3}}\right )} - \frac {{\left (5 \, b^{3} + 12 \, a b c\right )} \log \left ({\left | 2 \, {\left (\sqrt {-c} x^{2} - \sqrt {-c x^{4} + b x^{2} + a}\right )} \sqrt {-c} + b \right |}\right )}{32 \, \sqrt {-c} c^{3}} \] Input:

integrate(x^7/(-c*x^4+b*x^2+a)^(1/2),x, algorithm="giac")
 

Output:

-1/48*sqrt(-c*x^4 + b*x^2 + a)*(2*x^2*(4*x^2/c + 5*b/c^2) + (15*b^2 + 16*a 
*c)/c^3) - 1/32*(5*b^3 + 12*a*b*c)*log(abs(2*(sqrt(-c)*x^2 - sqrt(-c*x^4 + 
 b*x^2 + a))*sqrt(-c) + b))/(sqrt(-c)*c^3)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^7}{\sqrt {a+b x^2-c x^4}} \, dx=\int \frac {x^7}{\sqrt {-c\,x^4+b\,x^2+a}} \,d x \] Input:

int(x^7/(a + b*x^2 - c*x^4)^(1/2),x)
 

Output:

int(x^7/(a + b*x^2 - c*x^4)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.44 \[ \int \frac {x^7}{\sqrt {a+b x^2-c x^4}} \, dx=\frac {-36 \sqrt {c}\, \mathit {asin} \left (\frac {-2 c \,x^{2}+b}{\sqrt {4 a c +b^{2}}}\right ) a b c -15 \sqrt {c}\, \mathit {asin} \left (\frac {-2 c \,x^{2}+b}{\sqrt {4 a c +b^{2}}}\right ) b^{3}+16 \sqrt {c}\, \sqrt {4 a c +b^{2}}\, a c +22 \sqrt {c}\, \sqrt {4 a c +b^{2}}\, b^{2}-32 \sqrt {-c \,x^{4}+b \,x^{2}+a}\, a \,c^{2}-30 \sqrt {-c \,x^{4}+b \,x^{2}+a}\, b^{2} c -20 \sqrt {-c \,x^{4}+b \,x^{2}+a}\, b \,c^{2} x^{2}-16 \sqrt {-c \,x^{4}+b \,x^{2}+a}\, c^{3} x^{4}}{96 c^{4}} \] Input:

int(x^7/(-c*x^4+b*x^2+a)^(1/2),x)
 

Output:

( - 36*sqrt(c)*asin((b - 2*c*x**2)/sqrt(4*a*c + b**2))*a*b*c - 15*sqrt(c)* 
asin((b - 2*c*x**2)/sqrt(4*a*c + b**2))*b**3 + 16*sqrt(c)*sqrt(4*a*c + b** 
2)*a*c + 22*sqrt(c)*sqrt(4*a*c + b**2)*b**2 - 32*sqrt(a + b*x**2 - c*x**4) 
*a*c**2 - 30*sqrt(a + b*x**2 - c*x**4)*b**2*c - 20*sqrt(a + b*x**2 - c*x** 
4)*b*c**2*x**2 - 16*sqrt(a + b*x**2 - c*x**4)*c**3*x**4)/(96*c**4)