\(\int x^7 (A+B x^2) \sqrt {b x^2+c x^4} \, dx\) [154]

Optimal result

Integrand size = 26, antiderivative size = 218 \[ \int x^7 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {7 b^3 (3 b B-4 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{1024 c^5}-\frac {7 b^2 (3 b B-4 A c) \left (b x^2+c x^4\right )^{3/2}}{384 c^4}+\frac {7 b (3 b B-4 A c) x^2 \left (b x^2+c x^4\right )^{3/2}}{320 c^3}-\frac {(3 b B-4 A c) x^4 \left (b x^2+c x^4\right )^{3/2}}{40 c^2}+\frac {B x^6 \left (b x^2+c x^4\right )^{3/2}}{12 c}-\frac {7 b^5 (3 b B-4 A c) \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{1024 c^{11/2}} \] Output:

7/1024*b^3*(-4*A*c+3*B*b)*(2*c*x^2+b)*(c*x^4+b*x^2)^(1/2)/c^5-7/384*b^2*(- 
4*A*c+3*B*b)*(c*x^4+b*x^2)^(3/2)/c^4+7/320*b*(-4*A*c+3*B*b)*x^2*(c*x^4+b*x 
^2)^(3/2)/c^3-1/40*(-4*A*c+3*B*b)*x^4*(c*x^4+b*x^2)^(3/2)/c^2+1/12*B*x^6*( 
c*x^4+b*x^2)^(3/2)/c-7/1024*b^5*(-4*A*c+3*B*b)*arctanh(c^(1/2)*x^2/(c*x^4+ 
b*x^2)^(1/2))/c^(11/2)
 

Mathematica [A] (verified)

Time = 0.91 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.87 \[ \int x^7 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (\sqrt {c} \left (315 b^5 B-210 b^4 c \left (2 A+B x^2\right )+64 b c^4 x^6 \left (3 A+2 B x^2\right )+56 b^3 c^2 x^2 \left (5 A+3 B x^2\right )+256 c^5 x^8 \left (6 A+5 B x^2\right )-16 b^2 c^3 x^4 \left (14 A+9 B x^2\right )\right )+\frac {210 b^5 (3 b B-4 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b}-\sqrt {b+c x^2}}\right )}{x \sqrt {b+c x^2}}\right )}{15360 c^{11/2}} \] Input:

Integrate[x^7*(A + B*x^2)*Sqrt[b*x^2 + c*x^4],x]
 

Output:

(Sqrt[x^2*(b + c*x^2)]*(Sqrt[c]*(315*b^5*B - 210*b^4*c*(2*A + B*x^2) + 64* 
b*c^4*x^6*(3*A + 2*B*x^2) + 56*b^3*c^2*x^2*(5*A + 3*B*x^2) + 256*c^5*x^8*( 
6*A + 5*B*x^2) - 16*b^2*c^3*x^4*(14*A + 9*B*x^2)) + (210*b^5*(3*b*B - 4*A* 
c)*ArcTanh[(Sqrt[c]*x)/(Sqrt[b] - Sqrt[b + c*x^2])])/(x*Sqrt[b + c*x^2]))) 
/(15360*c^(11/2))
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.98, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {1940, 1221, 1134, 1134, 1160, 1087, 1091, 219}

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.

\(\displaystyle \int x^7 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx\)

\(\Big \downarrow \) 1940

\(\displaystyle \frac {1}{2} \int x^6 \left (B x^2+A\right ) \sqrt {c x^4+b x^2}dx^2\)

\(\Big \downarrow \) 1221

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

\(\Big \downarrow \) 1134

\(\displaystyle \frac {1}{2} \left (\frac {B x^6 \left (b x^2+c x^4\right )^{3/2}}{6 c}-\frac {(3 b B-4 A c) \left (\frac {x^4 \left (b x^2+c x^4\right )^{3/2}}{5 c}-\frac {7 b \int x^4 \sqrt {c x^4+b x^2}dx^2}{10 c}\right )}{4 c}\right )\)

\(\Big \downarrow \) 1134

\(\displaystyle \frac {1}{2} \left (\frac {B x^6 \left (b x^2+c x^4\right )^{3/2}}{6 c}-\frac {(3 b B-4 A c) \left (\frac {x^4 \left (b x^2+c x^4\right )^{3/2}}{5 c}-\frac {7 b \left (\frac {x^2 \left (b x^2+c x^4\right )^{3/2}}{4 c}-\frac {5 b \int x^2 \sqrt {c x^4+b x^2}dx^2}{8 c}\right )}{10 c}\right )}{4 c}\right )\)

\(\Big \downarrow \) 1160

\(\displaystyle \frac {1}{2} \left (\frac {B x^6 \left (b x^2+c x^4\right )^{3/2}}{6 c}-\frac {(3 b B-4 A c) \left (\frac {x^4 \left (b x^2+c x^4\right )^{3/2}}{5 c}-\frac {7 b \left (\frac {x^2 \left (b x^2+c x^4\right )^{3/2}}{4 c}-\frac {5 b \left (\frac {\left (b x^2+c x^4\right )^{3/2}}{3 c}-\frac {b \int \sqrt {c x^4+b x^2}dx^2}{2 c}\right )}{8 c}\right )}{10 c}\right )}{4 c}\right )\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {1}{2} \left (\frac {B x^6 \left (b x^2+c x^4\right )^{3/2}}{6 c}-\frac {(3 b B-4 A c) \left (\frac {x^4 \left (b x^2+c x^4\right )^{3/2}}{5 c}-\frac {7 b \left (\frac {x^2 \left (b x^2+c x^4\right )^{3/2}}{4 c}-\frac {5 b \left (\frac {\left (b x^2+c x^4\right )^{3/2}}{3 c}-\frac {b \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2}{8 c}\right )}{2 c}\right )}{8 c}\right )}{10 c}\right )}{4 c}\right )\)

\(\Big \downarrow \) 1091

\(\displaystyle \frac {1}{2} \left (\frac {B x^6 \left (b x^2+c x^4\right )^{3/2}}{6 c}-\frac {(3 b B-4 A c) \left (\frac {x^4 \left (b x^2+c x^4\right )^{3/2}}{5 c}-\frac {7 b \left (\frac {x^2 \left (b x^2+c x^4\right )^{3/2}}{4 c}-\frac {5 b \left (\frac {\left (b x^2+c x^4\right )^{3/2}}{3 c}-\frac {b \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}}{4 c}\right )}{2 c}\right )}{8 c}\right )}{10 c}\right )}{4 c}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\frac {B x^6 \left (b x^2+c x^4\right )^{3/2}}{6 c}-\frac {(3 b B-4 A c) \left (\frac {x^4 \left (b x^2+c x^4\right )^{3/2}}{5 c}-\frac {7 b \left (\frac {x^2 \left (b x^2+c x^4\right )^{3/2}}{4 c}-\frac {5 b \left (\frac {\left (b x^2+c x^4\right )^{3/2}}{3 c}-\frac {b \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{4 c^{3/2}}\right )}{2 c}\right )}{8 c}\right )}{10 c}\right )}{4 c}\right )\)

Input:

Int[x^7*(A + B*x^2)*Sqrt[b*x^2 + c*x^4],x]
 

Output:

((B*x^6*(b*x^2 + c*x^4)^(3/2))/(6*c) - ((3*b*B - 4*A*c)*((x^4*(b*x^2 + c*x 
^4)^(3/2))/(5*c) - (7*b*((x^2*(b*x^2 + c*x^4)^(3/2))/(4*c) - (5*b*((b*x^2 
+ c*x^4)^(3/2)/(3*c) - (b*(((b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(4*c) - (b^ 
2*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(4*c^(3/2))))/(2*c)))/(8*c)) 
)/(10*c)))/(4*c))/2
 

Defintions of rubi rules used

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 1087

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) 
*((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* 
p + 1)))   Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && 
GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
 

rule 1091

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

rule 1134

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[(m + p)*((2*c*d - b*e)/(c*(m + 2*p + 1)))   Int[(d + e*x)^ 
(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[ 
c*d^2 - b*d*e + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2 
*p]
 

rule 1160

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

rule 1221

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1 
)/(c*(m + 2*p + 2))), x] + Simp[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)*(2*c 
*f - b*g))/(c*e*(m + 2*p + 2))   Int[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x 
] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] 
 && NeQ[m + 2*p + 2, 0]
 

rule 1940

Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_) 
^(n_))^(q_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1) 
*(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x] /; 
FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] &&  !IntegerQ[p] && NeQ[k, j] && I 
ntegerQ[Simplify[j/n]] && IntegerQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 
 1)/n]] && NeQ[n^2, 1]
 

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.82

method	result	size
pseudoelliptic	\(\frac {\frac {7 \left (A \,b^{5} c -\frac {3}{4} B \,b^{6}\right ) \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right )}{512}+\frac {7 \left (\frac {192 x^{8} \left (\frac {5 B \,x^{2}}{6}+A \right ) c^{\frac {11}{2}}}{35}+\left (-\frac {3 \left (\frac {B \,x^{2}}{2}+A \right ) b^{3} c^{\frac {3}{2}}}{2}+b^{2} x^{2} \left (\frac {3 B \,x^{2}}{5}+A \right ) c^{\frac {5}{2}}-\frac {4 x^{4} \left (\frac {9 B \,x^{2}}{14}+A \right ) b \,c^{\frac {7}{2}}}{5}+\frac {24 x^{6} \left (\frac {2 B \,x^{2}}{3}+A \right ) c^{\frac {9}{2}}}{35}+\frac {9 B \sqrt {c}\, b^{4}}{8}\right ) b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{384}-\frac {7 \ln \left (2\right ) \left (A c -\frac {3 B b}{4}\right ) b^{5}}{512}}{c^{\frac {11}{2}}}\)	\(178\)
risch	\(-\frac {\left (-1280 B \,c^{5} x^{10}-1536 A \,c^{5} x^{8}-128 B b \,c^{4} x^{8}-192 A b \,c^{4} x^{6}+144 B \,b^{2} c^{3} x^{6}+224 A \,b^{2} c^{3} x^{4}-168 B \,b^{3} c^{2} x^{4}-280 A \,b^{3} c^{2} x^{2}+210 B \,b^{4} c \,x^{2}+420 A \,b^{4} c -315 b^{5} B \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{15360 c^{5}}+\frac {7 b^{5} \left (4 A c -3 B b \right ) \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{1024 c^{\frac {11}{2}} x \sqrt {c \,x^{2}+b}}\)	\(188\)
default	\(\frac {\sqrt {c \,x^{4}+b \,x^{2}}\, \left (1280 B \,c^{\frac {9}{2}} \left (c \,x^{2}+b \right )^{\frac {3}{2}} x^{9}+1536 A \,c^{\frac {9}{2}} \left (c \,x^{2}+b \right )^{\frac {3}{2}} x^{7}-1152 B \,c^{\frac {7}{2}} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b \,x^{7}-1344 A \,c^{\frac {7}{2}} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b \,x^{5}+1008 B \,c^{\frac {5}{2}} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{2} x^{5}+1120 A \,c^{\frac {5}{2}} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{2} x^{3}-840 B \,c^{\frac {3}{2}} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{3} x^{3}-840 A \,c^{\frac {3}{2}} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{3} x +420 A \,c^{\frac {3}{2}} \sqrt {c \,x^{2}+b}\, b^{4} x +630 B \sqrt {c}\, \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{4} x -315 B \sqrt {c}\, \sqrt {c \,x^{2}+b}\, b^{5} x +420 A \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{5} c -315 B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{6}\right )}{15360 x \sqrt {c \,x^{2}+b}\, c^{\frac {11}{2}}}\)	\(290\)

Input:

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

Output:

7/384/c^(11/2)*(3/4*(A*b^5*c-3/4*B*b^6)*ln((2*c*x^2+2*(x^2*(c*x^2+b))^(1/2 
)*c^(1/2)+b)/c^(1/2))+(192/35*x^8*(5/6*B*x^2+A)*c^(11/2)+(-3/2*(1/2*B*x^2+ 
A)*b^3*c^(3/2)+b^2*x^2*(3/5*B*x^2+A)*c^(5/2)-4/5*x^4*(9/14*B*x^2+A)*b*c^(7 
/2)+24/35*x^6*(2/3*B*x^2+A)*c^(9/2)+9/8*B*c^(1/2)*b^4)*b)*(x^2*(c*x^2+b))^ 
(1/2)-3/4*ln(2)*(A*c-3/4*B*b)*b^5)
 

Fricas [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 368, normalized size of antiderivative = 1.69 \[ \int x^7 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\left [-\frac {105 \, {\left (3 \, B b^{6} - 4 \, A b^{5} c\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left (1280 \, B c^{6} x^{10} + 128 \, {\left (B b c^{5} + 12 \, A c^{6}\right )} x^{8} + 315 \, B b^{5} c - 420 \, A b^{4} c^{2} - 48 \, {\left (3 \, B b^{2} c^{4} - 4 \, A b c^{5}\right )} x^{6} + 56 \, {\left (3 \, B b^{3} c^{3} - 4 \, A b^{2} c^{4}\right )} x^{4} - 70 \, {\left (3 \, B b^{4} c^{2} - 4 \, A b^{3} c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{30720 \, c^{6}}, \frac {105 \, {\left (3 \, B b^{6} - 4 \, A b^{5} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + {\left (1280 \, B c^{6} x^{10} + 128 \, {\left (B b c^{5} + 12 \, A c^{6}\right )} x^{8} + 315 \, B b^{5} c - 420 \, A b^{4} c^{2} - 48 \, {\left (3 \, B b^{2} c^{4} - 4 \, A b c^{5}\right )} x^{6} + 56 \, {\left (3 \, B b^{3} c^{3} - 4 \, A b^{2} c^{4}\right )} x^{4} - 70 \, {\left (3 \, B b^{4} c^{2} - 4 \, A b^{3} c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{15360 \, c^{6}}\right ] \] Input:

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

Output:

[-1/30720*(105*(3*B*b^6 - 4*A*b^5*c)*sqrt(c)*log(-2*c*x^2 - b - 2*sqrt(c*x 
^4 + b*x^2)*sqrt(c)) - 2*(1280*B*c^6*x^10 + 128*(B*b*c^5 + 12*A*c^6)*x^8 + 
 315*B*b^5*c - 420*A*b^4*c^2 - 48*(3*B*b^2*c^4 - 4*A*b*c^5)*x^6 + 56*(3*B* 
b^3*c^3 - 4*A*b^2*c^4)*x^4 - 70*(3*B*b^4*c^2 - 4*A*b^3*c^3)*x^2)*sqrt(c*x^ 
4 + b*x^2))/c^6, 1/15360*(105*(3*B*b^6 - 4*A*b^5*c)*sqrt(-c)*arctan(sqrt(c 
*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) + (1280*B*c^6*x^10 + 128*(B*b*c^5 + 12 
*A*c^6)*x^8 + 315*B*b^5*c - 420*A*b^4*c^2 - 48*(3*B*b^2*c^4 - 4*A*b*c^5)*x 
^6 + 56*(3*B*b^3*c^3 - 4*A*b^2*c^4)*x^4 - 70*(3*B*b^4*c^2 - 4*A*b^3*c^3)*x 
^2)*sqrt(c*x^4 + b*x^2))/c^6]
 

Sympy [A] (verification not implemented)

Time = 0.73 (sec) , antiderivative size = 345, normalized size of antiderivative = 1.58 \[ \int x^7 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {A \left (\begin {cases} \frac {7 b^{5} \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x^{2} + c x^{4}} + 2 c x^{2} \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x^{2}\right ) \log {\left (\frac {b}{2 c} + x^{2} \right )}}{\sqrt {c \left (\frac {b}{2 c} + x^{2}\right )^{2}}} & \text {otherwise} \end {cases}\right )}{256 c^{4}} + \sqrt {b x^{2} + c x^{4}} \left (- \frac {7 b^{4}}{128 c^{4}} + \frac {7 b^{3} x^{2}}{192 c^{3}} - \frac {7 b^{2} x^{4}}{240 c^{2}} + \frac {b x^{6}}{40 c} + \frac {x^{8}}{5}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (b x^{2}\right )^{\frac {9}{2}}}{9 b^{4}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases}\right )}{2} + \frac {B \left (\begin {cases} - \frac {21 b^{6} \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x^{2} + c x^{4}} + 2 c x^{2} \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x^{2}\right ) \log {\left (\frac {b}{2 c} + x^{2} \right )}}{\sqrt {c \left (\frac {b}{2 c} + x^{2}\right )^{2}}} & \text {otherwise} \end {cases}\right )}{1024 c^{5}} + \sqrt {b x^{2} + c x^{4}} \cdot \left (\frac {21 b^{5}}{512 c^{5}} - \frac {7 b^{4} x^{2}}{256 c^{4}} + \frac {7 b^{3} x^{4}}{320 c^{3}} - \frac {3 b^{2} x^{6}}{160 c^{2}} + \frac {b x^{8}}{60 c} + \frac {x^{10}}{6}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (b x^{2}\right )^{\frac {11}{2}}}{11 b^{5}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases}\right )}{2} \] Input:

integrate(x**7*(B*x**2+A)*(c*x**4+b*x**2)**(1/2),x)
 

Output:

A*Piecewise((7*b**5*Piecewise((log(b + 2*sqrt(c)*sqrt(b*x**2 + c*x**4) + 2 
*c*x**2)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x**2)*log(b/(2*c) + x**2)/sq 
rt(c*(b/(2*c) + x**2)**2), True))/(256*c**4) + sqrt(b*x**2 + c*x**4)*(-7*b 
**4/(128*c**4) + 7*b**3*x**2/(192*c**3) - 7*b**2*x**4/(240*c**2) + b*x**6/ 
(40*c) + x**8/5), Ne(c, 0)), (2*(b*x**2)**(9/2)/(9*b**4), Ne(b, 0)), (0, T 
rue))/2 + B*Piecewise((-21*b**6*Piecewise((log(b + 2*sqrt(c)*sqrt(b*x**2 + 
 c*x**4) + 2*c*x**2)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x**2)*log(b/(2*c 
) + x**2)/sqrt(c*(b/(2*c) + x**2)**2), True))/(1024*c**5) + sqrt(b*x**2 + 
c*x**4)*(21*b**5/(512*c**5) - 7*b**4*x**2/(256*c**4) + 7*b**3*x**4/(320*c* 
*3) - 3*b**2*x**6/(160*c**2) + b*x**8/(60*c) + x**10/6), Ne(c, 0)), (2*(b* 
x**2)**(11/2)/(11*b**5), Ne(b, 0)), (0, True))/2
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.47 \[ \int x^7 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {1}{7680} \, {\left (\frac {768 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{4}}{c} - \frac {420 \, \sqrt {c x^{4} + b x^{2}} b^{3} x^{2}}{c^{3}} - \frac {672 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b x^{2}}{c^{2}} + \frac {105 \, b^{5} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {9}{2}}} - \frac {210 \, \sqrt {c x^{4} + b x^{2}} b^{4}}{c^{4}} + \frac {560 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b^{2}}{c^{3}}\right )} A + \frac {1}{30720} \, {\left (\frac {2560 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{6}}{c} - \frac {2304 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b x^{4}}{c^{2}} + \frac {1260 \, \sqrt {c x^{4} + b x^{2}} b^{4} x^{2}}{c^{4}} + \frac {2016 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b^{2} x^{2}}{c^{3}} - \frac {315 \, b^{6} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {11}{2}}} + \frac {630 \, \sqrt {c x^{4} + b x^{2}} b^{5}}{c^{5}} - \frac {1680 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b^{3}}{c^{4}}\right )} B \] Input:

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

Output:

1/7680*(768*(c*x^4 + b*x^2)^(3/2)*x^4/c - 420*sqrt(c*x^4 + b*x^2)*b^3*x^2/ 
c^3 - 672*(c*x^4 + b*x^2)^(3/2)*b*x^2/c^2 + 105*b^5*log(2*c*x^2 + b + 2*sq 
rt(c*x^4 + b*x^2)*sqrt(c))/c^(9/2) - 210*sqrt(c*x^4 + b*x^2)*b^4/c^4 + 560 
*(c*x^4 + b*x^2)^(3/2)*b^2/c^3)*A + 1/30720*(2560*(c*x^4 + b*x^2)^(3/2)*x^ 
6/c - 2304*(c*x^4 + b*x^2)^(3/2)*b*x^4/c^2 + 1260*sqrt(c*x^4 + b*x^2)*b^4* 
x^2/c^4 + 2016*(c*x^4 + b*x^2)^(3/2)*b^2*x^2/c^3 - 315*b^6*log(2*c*x^2 + b 
 + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(11/2) + 630*sqrt(c*x^4 + b*x^2)*b^5/c 
^5 - 1680*(c*x^4 + b*x^2)^(3/2)*b^3/c^4)*B
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.12 \[ \int x^7 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {1}{15360} \, {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (10 \, B x^{2} \mathrm {sgn}\left (x\right ) + \frac {B b c^{9} \mathrm {sgn}\left (x\right ) + 12 \, A c^{10} \mathrm {sgn}\left (x\right )}{c^{10}}\right )} x^{2} - \frac {3 \, {\left (3 \, B b^{2} c^{8} \mathrm {sgn}\left (x\right ) - 4 \, A b c^{9} \mathrm {sgn}\left (x\right )\right )}}{c^{10}}\right )} x^{2} + \frac {7 \, {\left (3 \, B b^{3} c^{7} \mathrm {sgn}\left (x\right ) - 4 \, A b^{2} c^{8} \mathrm {sgn}\left (x\right )\right )}}{c^{10}}\right )} x^{2} - \frac {35 \, {\left (3 \, B b^{4} c^{6} \mathrm {sgn}\left (x\right ) - 4 \, A b^{3} c^{7} \mathrm {sgn}\left (x\right )\right )}}{c^{10}}\right )} x^{2} + \frac {105 \, {\left (3 \, B b^{5} c^{5} \mathrm {sgn}\left (x\right ) - 4 \, A b^{4} c^{6} \mathrm {sgn}\left (x\right )\right )}}{c^{10}}\right )} \sqrt {c x^{2} + b} x + \frac {7 \, {\left (3 \, B b^{6} \mathrm {sgn}\left (x\right ) - 4 \, A b^{5} c \mathrm {sgn}\left (x\right )\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{1024 \, c^{\frac {11}{2}}} - \frac {7 \, {\left (3 \, B b^{6} \log \left ({\left | b \right |}\right ) - 4 \, A b^{5} c \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\left (x\right )}{2048 \, c^{\frac {11}{2}}} \] Input:

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

Output:

1/15360*(2*(4*(2*(8*(10*B*x^2*sgn(x) + (B*b*c^9*sgn(x) + 12*A*c^10*sgn(x)) 
/c^10)*x^2 - 3*(3*B*b^2*c^8*sgn(x) - 4*A*b*c^9*sgn(x))/c^10)*x^2 + 7*(3*B* 
b^3*c^7*sgn(x) - 4*A*b^2*c^8*sgn(x))/c^10)*x^2 - 35*(3*B*b^4*c^6*sgn(x) - 
4*A*b^3*c^7*sgn(x))/c^10)*x^2 + 105*(3*B*b^5*c^5*sgn(x) - 4*A*b^4*c^6*sgn( 
x))/c^10)*sqrt(c*x^2 + b)*x + 7/1024*(3*B*b^6*sgn(x) - 4*A*b^5*c*sgn(x))*l 
og(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/c^(11/2) - 7/2048*(3*B*b^6*log(abs(b 
)) - 4*A*b^5*c*log(abs(b)))*sgn(x)/c^(11/2)
 

Mupad [B] (verification not implemented)

Time = 10.10 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.33 \[ \int x^7 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {A\,x^4\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{10\,c}+\frac {B\,x^6\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{12\,c}-\frac {3\,B\,b\,\left (\frac {7\,b\,\left (\frac {5\,b\,\left (\frac {b^3\,\ln \left (b+2\,c\,x^2+2\,\sqrt {c}\,\left |x\right |\,\sqrt {c\,x^2+b}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^4+b\,x^2}\,\left (-3\,b^2+2\,b\,c\,x^2+8\,c^2\,x^4\right )}{24\,c^2}\right )}{8\,c}-\frac {x^2\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{4\,c}\right )}{10\,c}+\frac {x^4\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{5\,c}\right )}{8\,c}+\frac {7\,A\,b\,\left (\frac {5\,b\,\left (\frac {b^3\,\ln \left (b+2\,c\,x^2+2\,\sqrt {c}\,\left |x\right |\,\sqrt {c\,x^2+b}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^4+b\,x^2}\,\left (-3\,b^2+2\,b\,c\,x^2+8\,c^2\,x^4\right )}{24\,c^2}\right )}{8\,c}-\frac {x^2\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{4\,c}\right )}{20\,c} \] Input:

int(x^7*(A + B*x^2)*(b*x^2 + c*x^4)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

(A*x^4*(b*x^2 + c*x^4)^(3/2))/(10*c) + (B*x^6*(b*x^2 + c*x^4)^(3/2))/(12*c 
) - (3*B*b*((7*b*((5*b*((b^3*log(b + 2*c*x^2 + 2*c^(1/2)*abs(x)*(b + c*x^2 
)^(1/2)))/(16*c^(5/2)) + ((b*x^2 + c*x^4)^(1/2)*(8*c^2*x^4 - 3*b^2 + 2*b*c 
*x^2))/(24*c^2)))/(8*c) - (x^2*(b*x^2 + c*x^4)^(3/2))/(4*c)))/(10*c) + (x^ 
4*(b*x^2 + c*x^4)^(3/2))/(5*c)))/(8*c) + (7*A*b*((5*b*((b^3*log(b + 2*c*x^ 
2 + 2*c^(1/2)*abs(x)*(b + c*x^2)^(1/2)))/(16*c^(5/2)) + ((b*x^2 + c*x^4)^( 
1/2)*(8*c^2*x^4 - 3*b^2 + 2*b*c*x^2))/(24*c^2)))/(8*c) - (x^2*(b*x^2 + c*x 
^4)^(3/2))/(4*c)))/(20*c)
 

Reduce [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.20 \[ \int x^7 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {-420 \sqrt {c \,x^{2}+b}\, a \,b^{4} c^{2} x +280 \sqrt {c \,x^{2}+b}\, a \,b^{3} c^{3} x^{3}-224 \sqrt {c \,x^{2}+b}\, a \,b^{2} c^{4} x^{5}+192 \sqrt {c \,x^{2}+b}\, a b \,c^{5} x^{7}+1536 \sqrt {c \,x^{2}+b}\, a \,c^{6} x^{9}+315 \sqrt {c \,x^{2}+b}\, b^{6} c x -210 \sqrt {c \,x^{2}+b}\, b^{5} c^{2} x^{3}+168 \sqrt {c \,x^{2}+b}\, b^{4} c^{3} x^{5}-144 \sqrt {c \,x^{2}+b}\, b^{3} c^{4} x^{7}+128 \sqrt {c \,x^{2}+b}\, b^{2} c^{5} x^{9}+1280 \sqrt {c \,x^{2}+b}\, b \,c^{6} x^{11}+420 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a \,b^{5} c -315 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{7}}{15360 c^{6}} \] Input:

int(x^7*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x)
 

Output:

( - 420*sqrt(b + c*x**2)*a*b**4*c**2*x + 280*sqrt(b + c*x**2)*a*b**3*c**3* 
x**3 - 224*sqrt(b + c*x**2)*a*b**2*c**4*x**5 + 192*sqrt(b + c*x**2)*a*b*c* 
*5*x**7 + 1536*sqrt(b + c*x**2)*a*c**6*x**9 + 315*sqrt(b + c*x**2)*b**6*c* 
x - 210*sqrt(b + c*x**2)*b**5*c**2*x**3 + 168*sqrt(b + c*x**2)*b**4*c**3*x 
**5 - 144*sqrt(b + c*x**2)*b**3*c**4*x**7 + 128*sqrt(b + c*x**2)*b**2*c**5 
*x**9 + 1280*sqrt(b + c*x**2)*b*c**6*x**11 + 420*sqrt(c)*log((sqrt(b + c*x 
**2) + sqrt(c)*x)/sqrt(b))*a*b**5*c - 315*sqrt(c)*log((sqrt(b + c*x**2) + 
sqrt(c)*x)/sqrt(b))*b**7)/(15360*c**6)