\(\int x \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx\) [72]

Optimal result

Integrand size = 38, antiderivative size = 309 \[ \int x \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {e \left (12 b c d-16 a d^2-7 b e^2\right ) (e+2 d x) \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{128 d^4 \left (a+b x^2\right )}+\frac {b x^2 \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d \left (a+b x^2\right )}-\frac {\left (32 b c d-80 a d^2-35 b e^2+42 b d e x\right ) \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{240 d^3 \left (a+b x^2\right )}+\frac {e \left (4 c d-e^2\right ) \left (12 b c d-16 a d^2-7 b e^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4} \text {arctanh}\left (\frac {e+2 d x}{2 \sqrt {d} \sqrt {c+e x+d x^2}}\right )}{256 d^{9/2} \left (a+b x^2\right )} \] Output:

1/128*e*(-16*a*d^2+12*b*c*d-7*b*e^2)*(2*d*x+e)*(d*x^2+e*x+c)^(1/2)*((b*x^2 
+a)^2)^(1/2)/d^4/(b*x^2+a)+1/5*b*x^2*(d*x^2+e*x+c)^(3/2)*((b*x^2+a)^2)^(1/ 
2)/d/(b*x^2+a)-1/240*(42*b*d*e*x-80*a*d^2+32*b*c*d-35*b*e^2)*(d*x^2+e*x+c) 
^(3/2)*((b*x^2+a)^2)^(1/2)/d^3/(b*x^2+a)+1/256*e*(4*c*d-e^2)*(-16*a*d^2+12 
*b*c*d-7*b*e^2)*((b*x^2+a)^2)^(1/2)*arctanh(1/2*(2*d*x+e)/d^(1/2)/(d*x^2+e 
*x+c)^(1/2))/d^(9/2)/(b*x^2+a)
 

Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.69 \[ \int x \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {\sqrt {\left (a+b x^2\right )^2} \left (2 \sqrt {d} \sqrt {c+x (e+d x)} \left (80 a d^2 \left (8 c d-3 e^2+2 d e x+8 d^2 x^2\right )+b \left (-256 c^2 d^2-105 e^4+70 d e^3 x-56 d^2 e^2 x^2+48 d^3 e x^3+384 d^4 x^4+4 c d \left (115 e^2-58 d e x+32 d^2 x^2\right )\right )\right )-15 e \left (-4 c d+e^2\right ) \left (-12 b c d+16 a d^2+7 b e^2\right ) \log \left (e+2 d x-2 \sqrt {d} \sqrt {c+x (e+d x)}\right )\right )}{3840 d^{9/2} \left (a+b x^2\right )} \] Input:

Integrate[x*Sqrt[c + e*x + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]
 

Output:

(Sqrt[(a + b*x^2)^2]*(2*Sqrt[d]*Sqrt[c + x*(e + d*x)]*(80*a*d^2*(8*c*d - 3 
*e^2 + 2*d*e*x + 8*d^2*x^2) + b*(-256*c^2*d^2 - 105*e^4 + 70*d*e^3*x - 56* 
d^2*e^2*x^2 + 48*d^3*e*x^3 + 384*d^4*x^4 + 4*c*d*(115*e^2 - 58*d*e*x + 32* 
d^2*x^2))) - 15*e*(-4*c*d + e^2)*(-12*b*c*d + 16*a*d^2 + 7*b*e^2)*Log[e + 
2*d*x - 2*Sqrt[d]*Sqrt[c + x*(e + d*x)]]))/(3840*d^(9/2)*(a + b*x^2))
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.70, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {1384, 27, 2184, 27, 1225, 1087, 1092, 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.

\(\displaystyle \int x \sqrt {a^2+2 a b x^2+b^2 x^4} \sqrt {c+d x^2+e x} \, dx\)

\(\Big \downarrow \) 1384

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2184

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1225

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {b x^2 \left (c+d x^2+e x\right )^{3/2}}{5 d}-\frac {\frac {\left (c+d x^2+e x\right )^{3/2} \left (16 d (2 b c-5 a d)+42 b d e x-35 b e^2\right )}{24 d^2}-\frac {5 e \left (-16 a d^2+12 b c d-7 b e^2\right ) \int \sqrt {d x^2+e x+c}dx}{16 d^2}}{10 d}\right )}{a+b x^2}\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {b x^2 \left (c+d x^2+e x\right )^{3/2}}{5 d}-\frac {\frac {\left (c+d x^2+e x\right )^{3/2} \left (16 d (2 b c-5 a d)+42 b d e x-35 b e^2\right )}{24 d^2}-\frac {5 e \left (-16 a d^2+12 b c d-7 b e^2\right ) \left (\frac {\left (4 c d-e^2\right ) \int \frac {1}{\sqrt {d x^2+e x+c}}dx}{8 d}+\frac {(2 d x+e) \sqrt {c+d x^2+e x}}{4 d}\right )}{16 d^2}}{10 d}\right )}{a+b x^2}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {b x^2 \left (c+d x^2+e x\right )^{3/2}}{5 d}-\frac {\frac {\left (c+d x^2+e x\right )^{3/2} \left (16 d (2 b c-5 a d)+42 b d e x-35 b e^2\right )}{24 d^2}-\frac {5 e \left (-16 a d^2+12 b c d-7 b e^2\right ) \left (\frac {\left (4 c d-e^2\right ) \int \frac {1}{4 d-\frac {(e+2 d x)^2}{d x^2+e x+c}}d\frac {e+2 d x}{\sqrt {d x^2+e x+c}}}{4 d}+\frac {(2 d x+e) \sqrt {c+d x^2+e x}}{4 d}\right )}{16 d^2}}{10 d}\right )}{a+b x^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {b x^2 \left (c+d x^2+e x\right )^{3/2}}{5 d}-\frac {\frac {\left (c+d x^2+e x\right )^{3/2} \left (16 d (2 b c-5 a d)+42 b d e x-35 b e^2\right )}{24 d^2}-\frac {5 e \left (-16 a d^2+12 b c d-7 b e^2\right ) \left (\frac {\left (4 c d-e^2\right ) \text {arctanh}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{8 d^{3/2}}+\frac {(2 d x+e) \sqrt {c+d x^2+e x}}{4 d}\right )}{16 d^2}}{10 d}\right )}{a+b x^2}\)

Input:

Int[x*Sqrt[c + e*x + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]
 

Output:

(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*((b*x^2*(c + e*x + d*x^2)^(3/2))/(5*d) - 
(((16*d*(2*b*c - 5*a*d) - 35*b*e^2 + 42*b*d*e*x)*(c + e*x + d*x^2)^(3/2))/ 
(24*d^2) - (5*e*(12*b*c*d - 16*a*d^2 - 7*b*e^2)*(((e + 2*d*x)*Sqrt[c + e*x 
 + d*x^2])/(4*d) + ((4*c*d - e^2)*ArcTanh[(e + 2*d*x)/(2*Sqrt[d]*Sqrt[c + 
e*x + d*x^2])])/(8*d^(3/2))))/(16*d^2))/(10*d)))/(a + b*x^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 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 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 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 1384

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S 
imp[(a + b*x^n + c*x^(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*Frac 
Part[p]))   Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] 
&& EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && NeQ[u, x^(n 
- 1)] && NeQ[u, x^(2*n - 1)] &&  !(EqQ[p, 1/2] && EqQ[u, x^(-2*n - 1)])
 

rule 2184

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p 
_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, S 
imp[f*(d + e*x)^(m + q - 1)*((a + b*x + c*x^2)^(p + 1)/(c*e^(q - 1)*(m + q 
+ 2*p + 1))), x] + Simp[1/(c*e^q*(m + q + 2*p + 1))   Int[(d + e*x)^m*(a + 
b*x + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 
1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q - 1) - c 
*d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[ 
q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && Pol 
yQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &&  !(IGt 
Q[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))
 

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 245, normalized size of antiderivative = 0.79

method	result	size
risch	\(\frac {\left (384 b \,x^{4} d^{4}+48 b e \,x^{3} d^{3}+640 a \,d^{4} x^{2}+128 b c \,d^{3} x^{2}-56 b \,d^{2} e^{2} x^{2}+160 a \,d^{3} e x -232 b c \,d^{2} e x +70 x b d \,e^{3}+640 a c \,d^{3}-240 d^{2} e^{2} a -256 b \,c^{2} d^{2}+460 b c d \,e^{2}-105 b \,e^{4}\right ) \sqrt {d \,x^{2}+e x +c}\, \sqrt {\left (b \,x^{2}+a \right )^{2}}}{1920 d^{4} \left (b \,x^{2}+a \right )}-\frac {e \left (64 a c \,d^{3}-16 d^{2} e^{2} a -48 b \,c^{2} d^{2}+40 b c d \,e^{2}-7 b \,e^{4}\right ) \ln \left (\frac {\frac {e}{2}+d x}{\sqrt {d}}+\sqrt {d \,x^{2}+e x +c}\right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{256 d^{\frac {9}{2}} \left (b \,x^{2}+a \right )}\)	\(245\)
default	\(-\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-768 d^{\frac {9}{2}} \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} b \,x^{2}+672 d^{\frac {7}{2}} \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} b e x -1280 d^{\frac {9}{2}} \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} a +512 d^{\frac {7}{2}} \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} b c -560 d^{\frac {5}{2}} \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} b \,e^{2}+960 d^{\frac {9}{2}} \sqrt {d \,x^{2}+e x +c}\, a e x -720 d^{\frac {7}{2}} \sqrt {d \,x^{2}+e x +c}\, b c e x +420 d^{\frac {5}{2}} \sqrt {d \,x^{2}+e x +c}\, b \,e^{3} x +480 d^{\frac {7}{2}} \sqrt {d \,x^{2}+e x +c}\, a \,e^{2}-360 d^{\frac {5}{2}} \sqrt {d \,x^{2}+e x +c}\, b c \,e^{2}+210 d^{\frac {3}{2}} \sqrt {d \,x^{2}+e x +c}\, b \,e^{4}+960 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) a c \,d^{4} e -240 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) a \,d^{3} e^{3}-720 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) b \,c^{2} d^{3} e +600 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) b c \,d^{2} e^{3}-105 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) b d \,e^{5}\right )}{3840 \left (b \,x^{2}+a \right ) d^{\frac {11}{2}}}\)	\(442\)

Input:

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

Output:

1/1920*(384*b*d^4*x^4+48*b*d^3*e*x^3+640*a*d^4*x^2+128*b*c*d^3*x^2-56*b*d^ 
2*e^2*x^2+160*a*d^3*e*x-232*b*c*d^2*e*x+70*b*d*e^3*x+640*a*c*d^3-240*a*d^2 
*e^2-256*b*c^2*d^2+460*b*c*d*e^2-105*b*e^4)*(d*x^2+e*x+c)^(1/2)/d^4*((b*x^ 
2+a)^2)^(1/2)/(b*x^2+a)-1/256*e*(64*a*c*d^3-16*a*d^2*e^2-48*b*c^2*d^2+40*b 
*c*d*e^2-7*b*e^4)/d^(9/2)*ln((1/2*e+d*x)/d^(1/2)+(d*x^2+e*x+c)^(1/2))*((b* 
x^2+a)^2)^(1/2)/(b*x^2+a)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 469, normalized size of antiderivative = 1.52 \[ \int x \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\left [\frac {15 \, {\left (7 \, b e^{5} - 8 \, {\left (5 \, b c d - 2 \, a d^{2}\right )} e^{3} + 16 \, {\left (3 \, b c^{2} d^{2} - 4 \, a c d^{3}\right )} e\right )} \sqrt {d} \log \left (8 \, d^{2} x^{2} + 8 \, d e x + 4 \, \sqrt {d x^{2} + e x + c} {\left (2 \, d x + e\right )} \sqrt {d} + 4 \, c d + e^{2}\right ) + 4 \, {\left (384 \, b d^{5} x^{4} + 48 \, b d^{4} e x^{3} - 256 \, b c^{2} d^{3} + 640 \, a c d^{4} - 105 \, b d e^{4} + 20 \, {\left (23 \, b c d^{2} - 12 \, a d^{3}\right )} e^{2} + 8 \, {\left (16 \, b c d^{4} + 80 \, a d^{5} - 7 \, b d^{3} e^{2}\right )} x^{2} + 2 \, {\left (35 \, b d^{2} e^{3} - 4 \, {\left (29 \, b c d^{3} - 20 \, a d^{4}\right )} e\right )} x\right )} \sqrt {d x^{2} + e x + c}}{7680 \, d^{5}}, -\frac {15 \, {\left (7 \, b e^{5} - 8 \, {\left (5 \, b c d - 2 \, a d^{2}\right )} e^{3} + 16 \, {\left (3 \, b c^{2} d^{2} - 4 \, a c d^{3}\right )} e\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {d x^{2} + e x + c} {\left (2 \, d x + e\right )} \sqrt {-d}}{2 \, {\left (d^{2} x^{2} + d e x + c d\right )}}\right ) - 2 \, {\left (384 \, b d^{5} x^{4} + 48 \, b d^{4} e x^{3} - 256 \, b c^{2} d^{3} + 640 \, a c d^{4} - 105 \, b d e^{4} + 20 \, {\left (23 \, b c d^{2} - 12 \, a d^{3}\right )} e^{2} + 8 \, {\left (16 \, b c d^{4} + 80 \, a d^{5} - 7 \, b d^{3} e^{2}\right )} x^{2} + 2 \, {\left (35 \, b d^{2} e^{3} - 4 \, {\left (29 \, b c d^{3} - 20 \, a d^{4}\right )} e\right )} x\right )} \sqrt {d x^{2} + e x + c}}{3840 \, d^{5}}\right ] \] Input:

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

Output:

[1/7680*(15*(7*b*e^5 - 8*(5*b*c*d - 2*a*d^2)*e^3 + 16*(3*b*c^2*d^2 - 4*a*c 
*d^3)*e)*sqrt(d)*log(8*d^2*x^2 + 8*d*e*x + 4*sqrt(d*x^2 + e*x + c)*(2*d*x 
+ e)*sqrt(d) + 4*c*d + e^2) + 4*(384*b*d^5*x^4 + 48*b*d^4*e*x^3 - 256*b*c^ 
2*d^3 + 640*a*c*d^4 - 105*b*d*e^4 + 20*(23*b*c*d^2 - 12*a*d^3)*e^2 + 8*(16 
*b*c*d^4 + 80*a*d^5 - 7*b*d^3*e^2)*x^2 + 2*(35*b*d^2*e^3 - 4*(29*b*c*d^3 - 
 20*a*d^4)*e)*x)*sqrt(d*x^2 + e*x + c))/d^5, -1/3840*(15*(7*b*e^5 - 8*(5*b 
*c*d - 2*a*d^2)*e^3 + 16*(3*b*c^2*d^2 - 4*a*c*d^3)*e)*sqrt(-d)*arctan(1/2* 
sqrt(d*x^2 + e*x + c)*(2*d*x + e)*sqrt(-d)/(d^2*x^2 + d*e*x + c*d)) - 2*(3 
84*b*d^5*x^4 + 48*b*d^4*e*x^3 - 256*b*c^2*d^3 + 640*a*c*d^4 - 105*b*d*e^4 
+ 20*(23*b*c*d^2 - 12*a*d^3)*e^2 + 8*(16*b*c*d^4 + 80*a*d^5 - 7*b*d^3*e^2) 
*x^2 + 2*(35*b*d^2*e^3 - 4*(29*b*c*d^3 - 20*a*d^4)*e)*x)*sqrt(d*x^2 + e*x 
+ c))/d^5]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int x \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(e>0)', see `assume?` for more de 
tails)Is e
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.16 \[ \int x \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {1}{1920} \, \sqrt {d x^{2} + e x + c} {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, b x \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {b e \mathrm {sgn}\left (b x^{2} + a\right )}{d}\right )} x + \frac {16 \, b c d^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 80 \, a d^{4} \mathrm {sgn}\left (b x^{2} + a\right ) - 7 \, b d^{2} e^{2} \mathrm {sgn}\left (b x^{2} + a\right )}{d^{4}}\right )} x - \frac {116 \, b c d^{2} e \mathrm {sgn}\left (b x^{2} + a\right ) - 80 \, a d^{3} e \mathrm {sgn}\left (b x^{2} + a\right ) - 35 \, b d e^{3} \mathrm {sgn}\left (b x^{2} + a\right )}{d^{4}}\right )} x - \frac {256 \, b c^{2} d^{2} \mathrm {sgn}\left (b x^{2} + a\right ) - 640 \, a c d^{3} \mathrm {sgn}\left (b x^{2} + a\right ) - 460 \, b c d e^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 240 \, a d^{2} e^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 105 \, b e^{4} \mathrm {sgn}\left (b x^{2} + a\right )}{d^{4}}\right )} - \frac {{\left (48 \, b c^{2} d^{2} e \mathrm {sgn}\left (b x^{2} + a\right ) - 64 \, a c d^{3} e \mathrm {sgn}\left (b x^{2} + a\right ) - 40 \, b c d e^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 16 \, a d^{2} e^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 7 \, b e^{5} \mathrm {sgn}\left (b x^{2} + a\right )\right )} \log \left ({\left | 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + e x + c}\right )} \sqrt {d} + e \right |}\right )}{256 \, d^{\frac {9}{2}}} \] Input:

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

Output:

1/1920*sqrt(d*x^2 + e*x + c)*(2*(4*(6*(8*b*x*sgn(b*x^2 + a) + b*e*sgn(b*x^ 
2 + a)/d)*x + (16*b*c*d^3*sgn(b*x^2 + a) + 80*a*d^4*sgn(b*x^2 + a) - 7*b*d 
^2*e^2*sgn(b*x^2 + a))/d^4)*x - (116*b*c*d^2*e*sgn(b*x^2 + a) - 80*a*d^3*e 
*sgn(b*x^2 + a) - 35*b*d*e^3*sgn(b*x^2 + a))/d^4)*x - (256*b*c^2*d^2*sgn(b 
*x^2 + a) - 640*a*c*d^3*sgn(b*x^2 + a) - 460*b*c*d*e^2*sgn(b*x^2 + a) + 24 
0*a*d^2*e^2*sgn(b*x^2 + a) + 105*b*e^4*sgn(b*x^2 + a))/d^4) - 1/256*(48*b* 
c^2*d^2*e*sgn(b*x^2 + a) - 64*a*c*d^3*e*sgn(b*x^2 + a) - 40*b*c*d*e^3*sgn( 
b*x^2 + a) + 16*a*d^2*e^3*sgn(b*x^2 + a) + 7*b*e^5*sgn(b*x^2 + a))*log(abs 
(2*(sqrt(d)*x - sqrt(d*x^2 + e*x + c))*sqrt(d) + e))/d^(9/2)
 

Mupad [F(-1)]

Timed out. \[ \int x \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\int x\,\sqrt {{\left (b\,x^2+a\right )}^2}\,\sqrt {d\,x^2+e\,x+c} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.57 (sec) , antiderivative size = 499, normalized size of antiderivative = 1.61 \[ \int x \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {1280 \sqrt {d \,x^{2}+e x +c}\, a c \,d^{4}+1280 \sqrt {d \,x^{2}+e x +c}\, a \,d^{5} x^{2}+320 \sqrt {d \,x^{2}+e x +c}\, a \,d^{4} e x -480 \sqrt {d \,x^{2}+e x +c}\, a \,d^{3} e^{2}-512 \sqrt {d \,x^{2}+e x +c}\, b \,c^{2} d^{3}+256 \sqrt {d \,x^{2}+e x +c}\, b c \,d^{4} x^{2}-464 \sqrt {d \,x^{2}+e x +c}\, b c \,d^{3} e x +920 \sqrt {d \,x^{2}+e x +c}\, b c \,d^{2} e^{2}+768 \sqrt {d \,x^{2}+e x +c}\, b \,d^{5} x^{4}+96 \sqrt {d \,x^{2}+e x +c}\, b \,d^{4} e \,x^{3}-112 \sqrt {d \,x^{2}+e x +c}\, b \,d^{3} e^{2} x^{2}+140 \sqrt {d \,x^{2}+e x +c}\, b \,d^{2} e^{3} x -210 \sqrt {d \,x^{2}+e x +c}\, b d \,e^{4}-960 \sqrt {d}\, \mathrm {log}\left (\frac {2 \sqrt {d}\, \sqrt {d \,x^{2}+e x +c}+2 d x +e}{\sqrt {4 c d -e^{2}}}\right ) a c \,d^{3} e +240 \sqrt {d}\, \mathrm {log}\left (\frac {2 \sqrt {d}\, \sqrt {d \,x^{2}+e x +c}+2 d x +e}{\sqrt {4 c d -e^{2}}}\right ) a \,d^{2} e^{3}+720 \sqrt {d}\, \mathrm {log}\left (\frac {2 \sqrt {d}\, \sqrt {d \,x^{2}+e x +c}+2 d x +e}{\sqrt {4 c d -e^{2}}}\right ) b \,c^{2} d^{2} e -600 \sqrt {d}\, \mathrm {log}\left (\frac {2 \sqrt {d}\, \sqrt {d \,x^{2}+e x +c}+2 d x +e}{\sqrt {4 c d -e^{2}}}\right ) b c d \,e^{3}+105 \sqrt {d}\, \mathrm {log}\left (\frac {2 \sqrt {d}\, \sqrt {d \,x^{2}+e x +c}+2 d x +e}{\sqrt {4 c d -e^{2}}}\right ) b \,e^{5}}{3840 d^{5}} \] Input:

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

Output:

(1280*sqrt(c + d*x**2 + e*x)*a*c*d**4 + 1280*sqrt(c + d*x**2 + e*x)*a*d**5 
*x**2 + 320*sqrt(c + d*x**2 + e*x)*a*d**4*e*x - 480*sqrt(c + d*x**2 + e*x) 
*a*d**3*e**2 - 512*sqrt(c + d*x**2 + e*x)*b*c**2*d**3 + 256*sqrt(c + d*x** 
2 + e*x)*b*c*d**4*x**2 - 464*sqrt(c + d*x**2 + e*x)*b*c*d**3*e*x + 920*sqr 
t(c + d*x**2 + e*x)*b*c*d**2*e**2 + 768*sqrt(c + d*x**2 + e*x)*b*d**5*x**4 
 + 96*sqrt(c + d*x**2 + e*x)*b*d**4*e*x**3 - 112*sqrt(c + d*x**2 + e*x)*b* 
d**3*e**2*x**2 + 140*sqrt(c + d*x**2 + e*x)*b*d**2*e**3*x - 210*sqrt(c + d 
*x**2 + e*x)*b*d*e**4 - 960*sqrt(d)*log((2*sqrt(d)*sqrt(c + d*x**2 + e*x) 
+ 2*d*x + e)/sqrt(4*c*d - e**2))*a*c*d**3*e + 240*sqrt(d)*log((2*sqrt(d)*s 
qrt(c + d*x**2 + e*x) + 2*d*x + e)/sqrt(4*c*d - e**2))*a*d**2*e**3 + 720*s 
qrt(d)*log((2*sqrt(d)*sqrt(c + d*x**2 + e*x) + 2*d*x + e)/sqrt(4*c*d - e** 
2))*b*c**2*d**2*e - 600*sqrt(d)*log((2*sqrt(d)*sqrt(c + d*x**2 + e*x) + 2* 
d*x + e)/sqrt(4*c*d - e**2))*b*c*d*e**3 + 105*sqrt(d)*log((2*sqrt(d)*sqrt( 
c + d*x**2 + e*x) + 2*d*x + e)/sqrt(4*c*d - e**2))*b*e**5)/(3840*d**5)