\(\int x^5 (A+B x^2) \sqrt {b x^2+c x^4} \, dx\) [155]

Optimal result

Integrand size = 26, antiderivative size = 181 \[ \int x^5 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=-\frac {b^2 (7 b B-10 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{256 c^4}+\frac {b (7 b B-10 A c) \left (b x^2+c x^4\right )^{3/2}}{96 c^3}-\frac {(7 b B-10 A c) x^2 \left (b x^2+c x^4\right )^{3/2}}{80 c^2}+\frac {B x^4 \left (b x^2+c x^4\right )^{3/2}}{10 c}+\frac {b^4 (7 b B-10 A c) \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{256 c^{9/2}} \] Output:

-1/256*b^2*(-10*A*c+7*B*b)*(2*c*x^2+b)*(c*x^4+b*x^2)^(1/2)/c^4+1/96*b*(-10 
*A*c+7*B*b)*(c*x^4+b*x^2)^(3/2)/c^3-1/80*(-10*A*c+7*B*b)*x^2*(c*x^4+b*x^2) 
^(3/2)/c^2+1/10*B*x^4*(c*x^4+b*x^2)^(3/2)/c+1/256*b^4*(-10*A*c+7*B*b)*arct 
anh(c^(1/2)*x^2/(c*x^4+b*x^2)^(1/2))/c^(9/2)
 

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.94 \[ \int x^5 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (\sqrt {c} x \left (-105 b^4 B+16 b c^3 x^4 \left (5 A+3 B x^2\right )+96 c^4 x^6 \left (5 A+4 B x^2\right )+10 b^3 c \left (15 A+7 B x^2\right )-4 b^2 c^2 x^2 \left (25 A+14 B x^2\right )\right )+\frac {30 b^4 (7 b B-10 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {b}+\sqrt {b+c x^2}}\right )}{\sqrt {b+c x^2}}\right )}{3840 c^{9/2} x} \] Input:

Integrate[x^5*(A + B*x^2)*Sqrt[b*x^2 + c*x^4],x]
 

Output:

(Sqrt[x^2*(b + c*x^2)]*(Sqrt[c]*x*(-105*b^4*B + 16*b*c^3*x^4*(5*A + 3*B*x^ 
2) + 96*c^4*x^6*(5*A + 4*B*x^2) + 10*b^3*c*(15*A + 7*B*x^2) - 4*b^2*c^2*x^ 
2*(25*A + 14*B*x^2)) + (30*b^4*(7*b*B - 10*A*c)*ArcTanh[(Sqrt[c]*x)/(-Sqrt 
[b] + Sqrt[b + c*x^2])])/Sqrt[b + c*x^2]))/(3840*c^(9/2)*x)
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {1940, 1221, 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.

\(\displaystyle \int x^5 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx\)

\(\Big \downarrow \) 1940

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

\(\Big \downarrow \) 1221

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

\(\Big \downarrow \) 1134

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

\(\Big \downarrow \) 1160

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

\(\Big \downarrow \) 1087

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

\(\Big \downarrow \) 1091

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

\(\Big \downarrow \) 219

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

Input:

Int[x^5*(A + B*x^2)*Sqrt[b*x^2 + c*x^4],x]
 

Output:

((B*x^4*(b*x^2 + c*x^4)^(3/2))/(5*c) - ((7*b*B - 10*A*c)*((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))/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 = 0.48 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.88

method	result	size
pseudoelliptic	\(\frac {\frac {5 \left (-\frac {1}{2} A \,b^{4} c +\frac {7}{20} b^{5} B \right ) \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right )}{128}+\frac {5 \left (b^{3} \left (\frac {7 B \,x^{2}}{15}+A \right ) c^{\frac {3}{2}}-\frac {2 x^{2} \left (\frac {14 B \,x^{2}}{25}+A \right ) b^{2} c^{\frac {5}{2}}}{3}+\frac {8 \left (\frac {3 B \,x^{2}}{5}+A \right ) x^{4} b \,c^{\frac {7}{2}}}{15}+\frac {16 \left (\frac {4 B \,x^{2}}{5}+A \right ) x^{6} c^{\frac {9}{2}}}{5}-\frac {7 B \sqrt {c}\, b^{4}}{10}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{128}+\frac {5 \ln \left (2\right ) b^{4} \left (A c -\frac {7 B b}{10}\right )}{256}}{c^{\frac {9}{2}}}\)	\(159\)
risch	\(\frac {\left (384 B \,x^{8} c^{4}+480 A \,x^{6} c^{4}+48 B \,x^{6} b \,c^{3}+80 A \,x^{4} b \,c^{3}-56 B \,x^{4} b^{2} c^{2}-100 A \,b^{2} c^{2} x^{2}+70 B \,b^{3} c \,x^{2}+150 A \,b^{3} c -105 B \,b^{4}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{3840 c^{4}}-\frac {b^{4} \left (10 A c -7 B b \right ) \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{256 c^{\frac {9}{2}} x \sqrt {c \,x^{2}+b}}\)	\(164\)
default	\(\frac {\sqrt {c \,x^{4}+b \,x^{2}}\, \left (384 B \,c^{\frac {7}{2}} \left (c \,x^{2}+b \right )^{\frac {3}{2}} x^{7}+480 A \,c^{\frac {7}{2}} \left (c \,x^{2}+b \right )^{\frac {3}{2}} x^{5}-336 B \,c^{\frac {5}{2}} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b \,x^{5}-400 A \,c^{\frac {5}{2}} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b \,x^{3}+280 B \,c^{\frac {3}{2}} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{2} x^{3}+300 A \,c^{\frac {3}{2}} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{2} x -210 B \sqrt {c}\, \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{3} x -150 A \,c^{\frac {3}{2}} \sqrt {c \,x^{2}+b}\, b^{3} x +105 B \sqrt {c}\, \sqrt {c \,x^{2}+b}\, b^{4} x -150 A \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{4} c +105 B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{5}\right )}{3840 x \sqrt {c \,x^{2}+b}\, c^{\frac {9}{2}}}\)	\(248\)

Input:

int(x^5*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

5/128/c^(9/2)*((-1/2*A*b^4*c+7/20*b^5*B)*ln((2*c*x^2+2*(x^2*(c*x^2+b))^(1/ 
2)*c^(1/2)+b)/c^(1/2))+(b^3*(7/15*B*x^2+A)*c^(3/2)-2/3*x^2*(14/25*B*x^2+A) 
*b^2*c^(5/2)+8/15*(3/5*B*x^2+A)*x^4*b*c^(7/2)+16/5*(4/5*B*x^2+A)*x^6*c^(9/ 
2)-7/10*B*c^(1/2)*b^4)*(x^2*(c*x^2+b))^(1/2)+1/2*ln(2)*b^4*(A*c-7/10*B*b))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.77 \[ \int x^5 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\left [-\frac {15 \, {\left (7 \, B b^{5} - 10 \, A b^{4} c\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left (384 \, B c^{5} x^{8} + 48 \, {\left (B b c^{4} + 10 \, A c^{5}\right )} x^{6} - 105 \, B b^{4} c + 150 \, A b^{3} c^{2} - 8 \, {\left (7 \, B b^{2} c^{3} - 10 \, A b c^{4}\right )} x^{4} + 10 \, {\left (7 \, B b^{3} c^{2} - 10 \, A b^{2} c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{7680 \, c^{5}}, -\frac {15 \, {\left (7 \, B b^{5} - 10 \, A b^{4} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) - {\left (384 \, B c^{5} x^{8} + 48 \, {\left (B b c^{4} + 10 \, A c^{5}\right )} x^{6} - 105 \, B b^{4} c + 150 \, A b^{3} c^{2} - 8 \, {\left (7 \, B b^{2} c^{3} - 10 \, A b c^{4}\right )} x^{4} + 10 \, {\left (7 \, B b^{3} c^{2} - 10 \, A b^{2} c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{3840 \, c^{5}}\right ] \] Input:

integrate(x^5*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
 

Output:

[-1/7680*(15*(7*B*b^5 - 10*A*b^4*c)*sqrt(c)*log(-2*c*x^2 - b + 2*sqrt(c*x^ 
4 + b*x^2)*sqrt(c)) - 2*(384*B*c^5*x^8 + 48*(B*b*c^4 + 10*A*c^5)*x^6 - 105 
*B*b^4*c + 150*A*b^3*c^2 - 8*(7*B*b^2*c^3 - 10*A*b*c^4)*x^4 + 10*(7*B*b^3* 
c^2 - 10*A*b^2*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/c^5, -1/3840*(15*(7*B*b^5 - 
10*A*b^4*c)*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) - (3 
84*B*c^5*x^8 + 48*(B*b*c^4 + 10*A*c^5)*x^6 - 105*B*b^4*c + 150*A*b^3*c^2 - 
 8*(7*B*b^2*c^3 - 10*A*b*c^4)*x^4 + 10*(7*B*b^3*c^2 - 10*A*b^2*c^3)*x^2)*s 
qrt(c*x^4 + b*x^2))/c^5]
 

Sympy [A] (verification not implemented)

Time = 0.67 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.76 \[ \int x^5 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {A \left (\begin {cases} - \frac {5 b^{4} \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x^{2} + c x^{4}} + 2 c x^{2} \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x^{2}\right ) \log {\left (\frac {b}{2 c} + x^{2} \right )}}{\sqrt {c \left (\frac {b}{2 c} + x^{2}\right )^{2}}} & \text {otherwise} \end {cases}\right )}{128 c^{3}} + \sqrt {b x^{2} + c x^{4}} \cdot \left (\frac {5 b^{3}}{64 c^{3}} - \frac {5 b^{2} x^{2}}{96 c^{2}} + \frac {b x^{4}}{24 c} + \frac {x^{6}}{4}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (b x^{2}\right )^{\frac {7}{2}}}{7 b^{3}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases}\right )}{2} + \frac {B \left (\begin {cases} \frac {7 b^{5} \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x^{2} + c x^{4}} + 2 c x^{2} \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x^{2}\right ) \log {\left (\frac {b}{2 c} + x^{2} \right )}}{\sqrt {c \left (\frac {b}{2 c} + x^{2}\right )^{2}}} & \text {otherwise} \end {cases}\right )}{256 c^{4}} + \sqrt {b x^{2} + c x^{4}} \left (- \frac {7 b^{4}}{128 c^{4}} + \frac {7 b^{3} x^{2}}{192 c^{3}} - \frac {7 b^{2} x^{4}}{240 c^{2}} + \frac {b x^{6}}{40 c} + \frac {x^{8}}{5}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (b x^{2}\right )^{\frac {9}{2}}}{9 b^{4}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases}\right )}{2} \] Input:

integrate(x**5*(B*x**2+A)*(c*x**4+b*x**2)**(1/2),x)
 

Output:

A*Piecewise((-5*b**4*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)/s 
qrt(c*(b/(2*c) + x**2)**2), True))/(128*c**3) + sqrt(b*x**2 + c*x**4)*(5*b 
**3/(64*c**3) - 5*b**2*x**2/(96*c**2) + b*x**4/(24*c) + x**6/4), Ne(c, 0)) 
, (2*(b*x**2)**(7/2)/(7*b**3), Ne(b, 0)), (0, True))/2 + B*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)/sqrt(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, True))/2
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.51 \[ \int x^5 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {1}{768} \, {\left (\frac {60 \, \sqrt {c x^{4} + b x^{2}} b^{2} x^{2}}{c^{2}} + \frac {96 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{2}}{c} - \frac {15 \, b^{4} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {7}{2}}} + \frac {30 \, \sqrt {c x^{4} + b x^{2}} b^{3}}{c^{3}} - \frac {80 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b}{c^{2}}\right )} A + \frac {1}{7680} \, {\left (\frac {768 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{4}}{c} - \frac {420 \, \sqrt {c x^{4} + b x^{2}} b^{3} x^{2}}{c^{3}} - \frac {672 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b x^{2}}{c^{2}} + \frac {105 \, b^{5} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {9}{2}}} - \frac {210 \, \sqrt {c x^{4} + b x^{2}} b^{4}}{c^{4}} + \frac {560 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b^{2}}{c^{3}}\right )} B \] Input:

integrate(x^5*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")
 

Output:

1/768*(60*sqrt(c*x^4 + b*x^2)*b^2*x^2/c^2 + 96*(c*x^4 + b*x^2)^(3/2)*x^2/c 
 - 15*b^4*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(7/2) + 30*sq 
rt(c*x^4 + b*x^2)*b^3/c^3 - 80*(c*x^4 + b*x^2)^(3/2)*b/c^2)*A + 1/7680*(76 
8*(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*sqrt(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)*B
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.17 \[ \int x^5 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {1}{3840} \, {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, B x^{2} \mathrm {sgn}\left (x\right ) + \frac {B b c^{7} \mathrm {sgn}\left (x\right ) + 10 \, A c^{8} \mathrm {sgn}\left (x\right )}{c^{8}}\right )} x^{2} - \frac {7 \, B b^{2} c^{6} \mathrm {sgn}\left (x\right ) - 10 \, A b c^{7} \mathrm {sgn}\left (x\right )}{c^{8}}\right )} x^{2} + \frac {5 \, {\left (7 \, B b^{3} c^{5} \mathrm {sgn}\left (x\right ) - 10 \, A b^{2} c^{6} \mathrm {sgn}\left (x\right )\right )}}{c^{8}}\right )} x^{2} - \frac {15 \, {\left (7 \, B b^{4} c^{4} \mathrm {sgn}\left (x\right ) - 10 \, A b^{3} c^{5} \mathrm {sgn}\left (x\right )\right )}}{c^{8}}\right )} \sqrt {c x^{2} + b} x - \frac {{\left (7 \, B b^{5} \mathrm {sgn}\left (x\right ) - 10 \, A b^{4} c \mathrm {sgn}\left (x\right )\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{256 \, c^{\frac {9}{2}}} + \frac {{\left (7 \, B b^{5} \log \left ({\left | b \right |}\right ) - 10 \, A b^{4} c \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\left (x\right )}{512 \, c^{\frac {9}{2}}} \] Input:

integrate(x^5*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x, algorithm="giac")
 

Output:

1/3840*(2*(4*(6*(8*B*x^2*sgn(x) + (B*b*c^7*sgn(x) + 10*A*c^8*sgn(x))/c^8)* 
x^2 - (7*B*b^2*c^6*sgn(x) - 10*A*b*c^7*sgn(x))/c^8)*x^2 + 5*(7*B*b^3*c^5*s 
gn(x) - 10*A*b^2*c^6*sgn(x))/c^8)*x^2 - 15*(7*B*b^4*c^4*sgn(x) - 10*A*b^3* 
c^5*sgn(x))/c^8)*sqrt(c*x^2 + b)*x - 1/256*(7*B*b^5*sgn(x) - 10*A*b^4*c*sg 
n(x))*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/c^(9/2) + 1/512*(7*B*b^5*log( 
abs(b)) - 10*A*b^4*c*log(abs(b)))*sgn(x)/c^(9/2)
 

Mupad [B] (verification not implemented)

Time = 9.82 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.29 \[ \int x^5 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {A\,x^2\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{8\,c}-\frac {5\,A\,b\,\left (\frac {b^3\,\ln \left (b+2\,c\,x^2+2\,\sqrt {c}\,\left |x\right |\,\sqrt {c\,x^2+b}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^4+b\,x^2}\,\left (-3\,b^2+2\,b\,c\,x^2+8\,c^2\,x^4\right )}{24\,c^2}\right )}{16\,c}+\frac {B\,x^4\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{10\,c}+\frac {7\,B\,b\,\left (\frac {5\,b\,\left (\frac {b^3\,\ln \left (b+2\,c\,x^2+2\,\sqrt {c}\,\left |x\right |\,\sqrt {c\,x^2+b}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^4+b\,x^2}\,\left (-3\,b^2+2\,b\,c\,x^2+8\,c^2\,x^4\right )}{24\,c^2}\right )}{8\,c}-\frac {x^2\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{4\,c}\right )}{20\,c} \] Input:

int(x^5*(A + B*x^2)*(b*x^2 + c*x^4)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

(A*x^2*(b*x^2 + c*x^4)^(3/2))/(8*c) - (5*A*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)))/(16*c) + (B*x^4*(b*x^2 + c*x^4)^(3 
/2))/(10*c) + (7*B*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 = 0.32 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.23 \[ \int x^5 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {150 \sqrt {c \,x^{2}+b}\, a \,b^{3} c^{2} x -100 \sqrt {c \,x^{2}+b}\, a \,b^{2} c^{3} x^{3}+80 \sqrt {c \,x^{2}+b}\, a b \,c^{4} x^{5}+480 \sqrt {c \,x^{2}+b}\, a \,c^{5} x^{7}-105 \sqrt {c \,x^{2}+b}\, b^{5} c x +70 \sqrt {c \,x^{2}+b}\, b^{4} c^{2} x^{3}-56 \sqrt {c \,x^{2}+b}\, b^{3} c^{3} x^{5}+48 \sqrt {c \,x^{2}+b}\, b^{2} c^{4} x^{7}+384 \sqrt {c \,x^{2}+b}\, b \,c^{5} x^{9}-150 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a \,b^{4} c +105 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{6}}{3840 c^{5}} \] Input:

int(x^5*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x)
 

Output:

(150*sqrt(b + c*x**2)*a*b**3*c**2*x - 100*sqrt(b + c*x**2)*a*b**2*c**3*x** 
3 + 80*sqrt(b + c*x**2)*a*b*c**4*x**5 + 480*sqrt(b + c*x**2)*a*c**5*x**7 - 
 105*sqrt(b + c*x**2)*b**5*c*x + 70*sqrt(b + c*x**2)*b**4*c**2*x**3 - 56*s 
qrt(b + c*x**2)*b**3*c**3*x**5 + 48*sqrt(b + c*x**2)*b**2*c**4*x**7 + 384* 
sqrt(b + c*x**2)*b*c**5*x**9 - 150*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c) 
*x)/sqrt(b))*a*b**4*c + 105*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqr 
t(b))*b**6)/(3840*c**5)