\(\int \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx\) [73]

Optimal result

Integrand size = 37, antiderivative size = 283 \[ \int \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=-\frac {\left (4 b c d-16 a d^2-5 b e^2\right ) (e+2 d x) \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{64 d^3 \left (a+b x^2\right )}-\frac {5 b e \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{24 d^2 \left (a+b x^2\right )}+\frac {b x \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 d \left (a+b x^2\right )}-\frac {\left (4 c d-e^2\right ) \left (4 b c d-16 a d^2-5 b e^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4} \text {arctanh}\left (\frac {e+2 d x}{2 \sqrt {d} \sqrt {c+e x+d x^2}}\right )}{128 d^{7/2} \left (a+b x^2\right )} \] Output:

-1/64*(-16*a*d^2+4*b*c*d-5*b*e^2)*(2*d*x+e)*(d*x^2+e*x+c)^(1/2)*((b*x^2+a) 
^2)^(1/2)/d^3/(b*x^2+a)-5/24*b*e*(d*x^2+e*x+c)^(3/2)*((b*x^2+a)^2)^(1/2)/d 
^2/(b*x^2+a)+1/4*b*x*(d*x^2+e*x+c)^(3/2)*((b*x^2+a)^2)^(1/2)/d/(b*x^2+a)-1 
/128*(4*c*d-e^2)*(-16*a*d^2+4*b*c*d-5*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^(7/2)/(b*x^2+a)
 

Mathematica [A] (verified)

Time = 1.51 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.76 \[ \int \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {\sqrt {\left (a+b x^2\right )^2} \left (\sqrt {d} \sqrt {c+x (e+d x)} \left (48 a d^2 (e+2 d x)+b \left (15 e^3-10 d e^2 x+8 d^2 e x^2+48 d^3 x^3+4 c d (-13 e+6 d x)\right )\right )+3 \left (16 b c^2 d^2+16 a d^2 e^2+5 b e^4\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c}-\sqrt {c+x (e+d x)}}\right )+24 c d \left (8 a d^2+3 b e^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{-\sqrt {c}+\sqrt {c+x (e+d x)}}\right )\right )}{192 d^{7/2} \left (a+b x^2\right )} \] Input:

Integrate[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]*(Sqrt[d]*Sqrt[c + x*(e + d*x)]*(48*a*d^2*(e + 2*d*x) 
+ b*(15*e^3 - 10*d*e^2*x + 8*d^2*e*x^2 + 48*d^3*x^3 + 4*c*d*(-13*e + 6*d*x 
))) + 3*(16*b*c^2*d^2 + 16*a*d^2*e^2 + 5*b*e^4)*ArcTanh[(Sqrt[d]*x)/(Sqrt[ 
c] - Sqrt[c + x*(e + d*x)])] + 24*c*d*(8*a*d^2 + 3*b*e^2)*ArcTanh[(Sqrt[d] 
*x)/(-Sqrt[c] + Sqrt[c + x*(e + d*x)])]))/(192*d^(7/2)*(a + b*x^2))
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.67, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.216, Rules used = {1384, 27, 2192, 27, 1160, 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.

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

\(\Big \downarrow \) 1384

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2192

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {b x \left (c+d x^2+e x\right )^{3/2}}{4 d}-\frac {\int (2 (b c-4 a d)+5 b e x) \sqrt {d x^2+e x+c}dx}{8 d}\right )}{a+b x^2}\)

\(\Big \downarrow \) 1160

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {b x \left (c+d x^2+e x\right )^{3/2}}{4 d}-\frac {\frac {\left (4 d (b c-4 a d)-5 b e^2\right ) \int \sqrt {d x^2+e x+c}dx}{2 d}+\frac {5 b e \left (c+d x^2+e x\right )^{3/2}}{3 d}}{8 d}\right )}{a+b x^2}\)

\(\Big \downarrow \) 1087

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

\(\Big \downarrow \) 1092

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {b x \left (c+d x^2+e x\right )^{3/2}}{4 d}-\frac {\frac {\left (4 d (b c-4 a d)-5 b e^2\right ) \left (\frac {\left (4 c d-e^2\right ) \text {arctanh}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{8 d^{3/2}}+\frac {(2 d x+e) \sqrt {c+d x^2+e x}}{4 d}\right )}{2 d}+\frac {5 b e \left (c+d x^2+e x\right )^{3/2}}{3 d}}{8 d}\right )}{a+b x^2}\)

Input:

Int[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*(c + e*x + d*x^2)^(3/2))/(4*d) - (( 
5*b*e*(c + e*x + d*x^2)^(3/2))/(3*d) + ((4*d*(b*c - 4*a*d) - 5*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))))/(2*d))/(8*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 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 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 2192

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = 
Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x + 
 c*x^2)^(p + 1)/(c*(q + 2*p + 1))), x] + Simp[1/(c*(q + 2*p + 1))   Int[(a 
+ b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b 
*e*(q + p)*x^(q - 1) - c*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c 
, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]
 

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.67

method	result	size
risch	\(\frac {\left (48 b \,d^{3} x^{3}+8 x^{2} e b \,d^{2}+96 a \,d^{3} x +24 b c \,d^{2} x -10 b d \,e^{2} x +48 d^{2} e a -52 b c d e +15 b \,e^{3}\right ) \sqrt {d \,x^{2}+e x +c}\, \sqrt {\left (b \,x^{2}+a \right )^{2}}}{192 d^{3} \left (b \,x^{2}+a \right )}+\frac {\left (64 a c \,d^{3}-16 d^{2} e^{2} a -16 b \,c^{2} d^{2}+24 b c d \,e^{2}-5 b \,e^{4}\right ) \ln \left (\frac {\frac {e}{2}+d x}{\sqrt {d}}+\sqrt {d \,x^{2}+e x +c}\right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{128 d^{\frac {7}{2}} \left (b \,x^{2}+a \right )}\)	\(191\)
default	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (96 d^{\frac {7}{2}} \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} b x -80 d^{\frac {5}{2}} \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} b e +192 d^{\frac {9}{2}} \sqrt {d \,x^{2}+e x +c}\, a x -48 d^{\frac {7}{2}} \sqrt {d \,x^{2}+e x +c}\, b c x +60 d^{\frac {5}{2}} \sqrt {d \,x^{2}+e x +c}\, b \,e^{2} x +96 d^{\frac {7}{2}} \sqrt {d \,x^{2}+e x +c}\, a e -24 d^{\frac {5}{2}} \sqrt {d \,x^{2}+e x +c}\, b c e +30 d^{\frac {3}{2}} \sqrt {d \,x^{2}+e x +c}\, b \,e^{3}+192 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) a c \,d^{4}-48 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) a \,d^{3} e^{2}-48 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) b \,c^{2} d^{3}+72 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) b c \,d^{2} e^{2}-15 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) b d \,e^{4}\right )}{384 \left (b \,x^{2}+a \right ) d^{\frac {9}{2}}}\)	\(373\)

Input:

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

Output:

1/192*(48*b*d^3*x^3+8*b*d^2*e*x^2+96*a*d^3*x+24*b*c*d^2*x-10*b*d*e^2*x+48* 
a*d^2*e-52*b*c*d*e+15*b*e^3)*(d*x^2+e*x+c)^(1/2)/d^3*((b*x^2+a)^2)^(1/2)/( 
b*x^2+a)+1/128*(64*a*c*d^3-16*a*d^2*e^2-16*b*c^2*d^2+24*b*c*d*e^2-5*b*e^4) 
/d^(7/2)*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 = 0.10 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.28 \[ \int \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\left [\frac {3 \, {\left (16 \, b c^{2} d^{2} - 64 \, a c d^{3} + 5 \, b e^{4} - 8 \, {\left (3 \, b c d - 2 \, a d^{2}\right )} e^{2}\right )} \sqrt {d} \log \left (8 \, d^{2} x^{2} + 8 \, d e x - 4 \, \sqrt {d x^{2} + e x + c} {\left (2 \, d x + e\right )} \sqrt {d} + 4 \, c d + e^{2}\right ) + 4 \, {\left (48 \, b d^{4} x^{3} + 8 \, b d^{3} e x^{2} + 15 \, b d e^{3} - 4 \, {\left (13 \, b c d^{2} - 12 \, a d^{3}\right )} e + 2 \, {\left (12 \, b c d^{3} + 48 \, a d^{4} - 5 \, b d^{2} e^{2}\right )} x\right )} \sqrt {d x^{2} + e x + c}}{768 \, d^{4}}, \frac {3 \, {\left (16 \, b c^{2} d^{2} - 64 \, a c d^{3} + 5 \, b e^{4} - 8 \, {\left (3 \, b c d - 2 \, a d^{2}\right )} e^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {d x^{2} + e x + c} {\left (2 \, d x + e\right )} \sqrt {-d}}{2 \, {\left (d^{2} x^{2} + d e x + c d\right )}}\right ) + 2 \, {\left (48 \, b d^{4} x^{3} + 8 \, b d^{3} e x^{2} + 15 \, b d e^{3} - 4 \, {\left (13 \, b c d^{2} - 12 \, a d^{3}\right )} e + 2 \, {\left (12 \, b c d^{3} + 48 \, a d^{4} - 5 \, b d^{2} e^{2}\right )} x\right )} \sqrt {d x^{2} + e x + c}}{384 \, d^{4}}\right ] \] Input:

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

Output:

[1/768*(3*(16*b*c^2*d^2 - 64*a*c*d^3 + 5*b*e^4 - 8*(3*b*c*d - 2*a*d^2)*e^2 
)*sqrt(d)*log(8*d^2*x^2 + 8*d*e*x - 4*sqrt(d*x^2 + e*x + c)*(2*d*x + e)*sq 
rt(d) + 4*c*d + e^2) + 4*(48*b*d^4*x^3 + 8*b*d^3*e*x^2 + 15*b*d*e^3 - 4*(1 
3*b*c*d^2 - 12*a*d^3)*e + 2*(12*b*c*d^3 + 48*a*d^4 - 5*b*d^2*e^2)*x)*sqrt( 
d*x^2 + e*x + c))/d^4, 1/384*(3*(16*b*c^2*d^2 - 64*a*c*d^3 + 5*b*e^4 - 8*( 
3*b*c*d - 2*a*d^2)*e^2)*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*(48*b*d^4*x^3 + 8*b*d^3*e*x^2 + 
15*b*d*e^3 - 4*(13*b*c*d^2 - 12*a*d^3)*e + 2*(12*b*c*d^3 + 48*a*d^4 - 5*b* 
d^2*e^2)*x)*sqrt(d*x^2 + e*x + c))/d^4]
 

Sympy [F]

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

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

Output:

Integral(sqrt(c + d*x**2 + e*x)*sqrt((a + b*x**2)**2), x)
 

Maxima [F(-2)]

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

integrate((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 = 0.14 (sec) , antiderivative size = 262, normalized size of antiderivative = 0.93 \[ \int \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {1}{192} \, \sqrt {d x^{2} + e x + c} {\left (2 \, {\left (4 \, {\left (6 \, b x \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {b e \mathrm {sgn}\left (b x^{2} + a\right )}{d}\right )} x + \frac {12 \, b c d^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 48 \, a d^{3} \mathrm {sgn}\left (b x^{2} + a\right ) - 5 \, b d e^{2} \mathrm {sgn}\left (b x^{2} + a\right )}{d^{3}}\right )} x - \frac {52 \, b c d e \mathrm {sgn}\left (b x^{2} + a\right ) - 48 \, a d^{2} e \mathrm {sgn}\left (b x^{2} + a\right ) - 15 \, b e^{3} \mathrm {sgn}\left (b x^{2} + a\right )}{d^{3}}\right )} + \frac {{\left (16 \, b c^{2} d^{2} \mathrm {sgn}\left (b x^{2} + a\right ) - 64 \, a c d^{3} \mathrm {sgn}\left (b x^{2} + a\right ) - 24 \, b c d e^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 16 \, a d^{2} e^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 5 \, b e^{4} \mathrm {sgn}\left (b x^{2} + a\right )\right )} \log \left ({\left | 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + e x + c}\right )} \sqrt {d} + e \right |}\right )}{128 \, d^{\frac {7}{2}}} \] Input:

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

Output:

1/192*sqrt(d*x^2 + e*x + c)*(2*(4*(6*b*x*sgn(b*x^2 + a) + b*e*sgn(b*x^2 + 
a)/d)*x + (12*b*c*d^2*sgn(b*x^2 + a) + 48*a*d^3*sgn(b*x^2 + a) - 5*b*d*e^2 
*sgn(b*x^2 + a))/d^3)*x - (52*b*c*d*e*sgn(b*x^2 + a) - 48*a*d^2*e*sgn(b*x^ 
2 + a) - 15*b*e^3*sgn(b*x^2 + a))/d^3) + 1/128*(16*b*c^2*d^2*sgn(b*x^2 + a 
) - 64*a*c*d^3*sgn(b*x^2 + a) - 24*b*c*d*e^2*sgn(b*x^2 + a) + 16*a*d^2*e^2 
*sgn(b*x^2 + a) + 5*b*e^4*sgn(b*x^2 + a))*log(abs(2*(sqrt(d)*x - sqrt(d*x^ 
2 + e*x + c))*sqrt(d) + e))/d^(7/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 389, normalized size of antiderivative = 1.37 \[ \int \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {192 \sqrt {d \,x^{2}+e x +c}\, a \,d^{4} x +96 \sqrt {d \,x^{2}+e x +c}\, a \,d^{3} e +48 \sqrt {d \,x^{2}+e x +c}\, b c \,d^{3} x -104 \sqrt {d \,x^{2}+e x +c}\, b c \,d^{2} e +96 \sqrt {d \,x^{2}+e x +c}\, b \,d^{4} x^{3}+16 \sqrt {d \,x^{2}+e x +c}\, b \,d^{3} e \,x^{2}-20 \sqrt {d \,x^{2}+e x +c}\, b \,d^{2} e^{2} x +30 \sqrt {d \,x^{2}+e x +c}\, b d \,e^{3}+192 \sqrt {d}\, \mathrm {log}\left (\frac {2 \sqrt {d}\, \sqrt {d \,x^{2}+e x +c}+2 d x +e}{\sqrt {4 c d -e^{2}}}\right ) a c \,d^{3}-48 \sqrt {d}\, \mathrm {log}\left (\frac {2 \sqrt {d}\, \sqrt {d \,x^{2}+e x +c}+2 d x +e}{\sqrt {4 c d -e^{2}}}\right ) a \,d^{2} e^{2}-48 \sqrt {d}\, \mathrm {log}\left (\frac {2 \sqrt {d}\, \sqrt {d \,x^{2}+e x +c}+2 d x +e}{\sqrt {4 c d -e^{2}}}\right ) b \,c^{2} d^{2}+72 \sqrt {d}\, \mathrm {log}\left (\frac {2 \sqrt {d}\, \sqrt {d \,x^{2}+e x +c}+2 d x +e}{\sqrt {4 c d -e^{2}}}\right ) b c d \,e^{2}-15 \sqrt {d}\, \mathrm {log}\left (\frac {2 \sqrt {d}\, \sqrt {d \,x^{2}+e x +c}+2 d x +e}{\sqrt {4 c d -e^{2}}}\right ) b \,e^{4}}{384 d^{4}} \] Input:

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

Output:

(192*sqrt(c + d*x**2 + e*x)*a*d**4*x + 96*sqrt(c + d*x**2 + e*x)*a*d**3*e 
+ 48*sqrt(c + d*x**2 + e*x)*b*c*d**3*x - 104*sqrt(c + d*x**2 + e*x)*b*c*d* 
*2*e + 96*sqrt(c + d*x**2 + e*x)*b*d**4*x**3 + 16*sqrt(c + d*x**2 + e*x)*b 
*d**3*e*x**2 - 20*sqrt(c + d*x**2 + e*x)*b*d**2*e**2*x + 30*sqrt(c + d*x** 
2 + e*x)*b*d*e**3 + 192*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 - 48*sqrt(d)*log((2*sqrt(d)*sqrt(c + 
 d*x**2 + e*x) + 2*d*x + e)/sqrt(4*c*d - e**2))*a*d**2*e**2 - 48*sqrt(d)*l 
og((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 + 72*sqrt(d)*log((2*sqrt(d)*sqrt(c + d*x**2 + e*x) + 2*d*x + e)/sq 
rt(4*c*d - e**2))*b*c*d*e**2 - 15*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**4)/(384*d**4)