\(\int \sqrt {a+b x^2} (A+B x^2+C x^4+D x^6) \, dx\) [134]

Optimal result

Integrand size = 29, antiderivative size = 191 \[ \int \sqrt {a+b x^2} \left (A+B x^2+C x^4+D x^6\right ) \, dx=\frac {1}{128} \left (64 A-\frac {a \left (16 b^2 B-8 a b C+5 a^2 D\right )}{b^3}\right ) x \sqrt {a+b x^2}+\frac {\left (16 b^2 B-8 a b C+5 a^2 D\right ) x \left (a+b x^2\right )^{3/2}}{64 b^3}+\frac {(8 b C-5 a D) x^3 \left (a+b x^2\right )^{3/2}}{48 b^2}+\frac {D x^5 \left (a+b x^2\right )^{3/2}}{8 b}+\frac {a \left (64 A b^3-a \left (16 b^2 B-8 a b C+5 a^2 D\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{7/2}} \] Output:

1/128*(64*A-a*(16*B*b^2-8*C*a*b+5*D*a^2)/b^3)*x*(b*x^2+a)^(1/2)+1/64*(16*B 
*b^2-8*C*a*b+5*D*a^2)*x*(b*x^2+a)^(3/2)/b^3+1/48*(8*C*b-5*D*a)*x^3*(b*x^2+ 
a)^(3/2)/b^2+1/8*D*x^5*(b*x^2+a)^(3/2)/b+1/128*a*(64*A*b^3-a*(16*B*b^2-8*C 
*a*b+5*D*a^2))*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(7/2)
 

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.80 \[ \int \sqrt {a+b x^2} \left (A+B x^2+C x^4+D x^6\right ) \, dx=\frac {\sqrt {b} x \sqrt {a+b x^2} \left (192 A b^3+15 a^3 D-2 a^2 b \left (12 C+5 D x^2\right )+8 a b^2 \left (6 B+2 C x^2+D x^4\right )+16 b^3 x^2 \left (6 B+4 C x^2+3 D x^4\right )\right )+3 a \left (-64 A b^3+a \left (16 b^2 B-8 a b C+5 a^2 D\right )\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{384 b^{7/2}} \] Input:

Integrate[Sqrt[a + b*x^2]*(A + B*x^2 + C*x^4 + D*x^6),x]
 

Output:

(Sqrt[b]*x*Sqrt[a + b*x^2]*(192*A*b^3 + 15*a^3*D - 2*a^2*b*(12*C + 5*D*x^2 
) + 8*a*b^2*(6*B + 2*C*x^2 + D*x^4) + 16*b^3*x^2*(6*B + 4*C*x^2 + 3*D*x^4) 
) + 3*a*(-64*A*b^3 + a*(16*b^2*B - 8*a*b*C + 5*a^2*D))*Log[-(Sqrt[b]*x) + 
Sqrt[a + b*x^2]])/(384*b^(7/2))
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2346, 1473, 27, 299, 211, 224, 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.

\(\displaystyle \int \sqrt {a+b x^2} \left (A+B x^2+C x^4+D x^6\right ) \, dx\)

\(\Big \downarrow \) 2346

\(\displaystyle \frac {\int \sqrt {b x^2+a} \left ((8 b C-5 a D) x^4+8 b B x^2+8 A b\right )dx}{8 b}+\frac {D x^5 \left (a+b x^2\right )^{3/2}}{8 b}\)

\(\Big \downarrow \) 1473

\(\displaystyle \frac {\frac {\int 3 \sqrt {b x^2+a} \left (16 A b^2+\left (5 D a^2-8 b C a+16 b^2 B\right ) x^2\right )dx}{6 b}+\frac {x^3 \left (a+b x^2\right )^{3/2} (8 b C-5 a D)}{6 b}}{8 b}+\frac {D x^5 \left (a+b x^2\right )^{3/2}}{8 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \sqrt {b x^2+a} \left (16 A b^2+\left (5 D a^2-8 b C a+16 b^2 B\right ) x^2\right )dx}{2 b}+\frac {x^3 \left (a+b x^2\right )^{3/2} (8 b C-5 a D)}{6 b}}{8 b}+\frac {D x^5 \left (a+b x^2\right )^{3/2}}{8 b}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {\frac {\left (64 A b^3-a \left (5 a^2 D-8 a b C+16 b^2 B\right )\right ) \int \sqrt {b x^2+a}dx}{4 b}+\frac {x \left (a+b x^2\right )^{3/2} \left (5 a^2 D-8 a b C+16 b^2 B\right )}{4 b}}{2 b}+\frac {x^3 \left (a+b x^2\right )^{3/2} (8 b C-5 a D)}{6 b}}{8 b}+\frac {D x^5 \left (a+b x^2\right )^{3/2}}{8 b}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\frac {\frac {\left (64 A b^3-a \left (5 a^2 D-8 a b C+16 b^2 B\right )\right ) \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )}{4 b}+\frac {x \left (a+b x^2\right )^{3/2} \left (5 a^2 D-8 a b C+16 b^2 B\right )}{4 b}}{2 b}+\frac {x^3 \left (a+b x^2\right )^{3/2} (8 b C-5 a D)}{6 b}}{8 b}+\frac {D x^5 \left (a+b x^2\right )^{3/2}}{8 b}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {\frac {\left (64 A b^3-a \left (5 a^2 D-8 a b C+16 b^2 B\right )\right ) \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )}{4 b}+\frac {x \left (a+b x^2\right )^{3/2} \left (5 a^2 D-8 a b C+16 b^2 B\right )}{4 b}}{2 b}+\frac {x^3 \left (a+b x^2\right )^{3/2} (8 b C-5 a D)}{6 b}}{8 b}+\frac {D x^5 \left (a+b x^2\right )^{3/2}}{8 b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {\left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right ) \left (64 A b^3-a \left (5 a^2 D-8 a b C+16 b^2 B\right )\right )}{4 b}+\frac {x \left (a+b x^2\right )^{3/2} \left (5 a^2 D-8 a b C+16 b^2 B\right )}{4 b}}{2 b}+\frac {x^3 \left (a+b x^2\right )^{3/2} (8 b C-5 a D)}{6 b}}{8 b}+\frac {D x^5 \left (a+b x^2\right )^{3/2}}{8 b}\)

Input:

Int[Sqrt[a + b*x^2]*(A + B*x^2 + C*x^4 + D*x^6),x]
 

Output:

(D*x^5*(a + b*x^2)^(3/2))/(8*b) + (((8*b*C - 5*a*D)*x^3*(a + b*x^2)^(3/2)) 
/(6*b) + (((16*b^2*B - 8*a*b*C + 5*a^2*D)*x*(a + b*x^2)^(3/2))/(4*b) + ((6 
4*A*b^3 - a*(16*b^2*B - 8*a*b*C + 5*a^2*D))*((x*Sqrt[a + b*x^2])/2 + (a*Ar 
cTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*Sqrt[b])))/(4*b))/(2*b))/(8*b)
 

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 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

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 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 299

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

rule 1473

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
x_Symbol] :> Simp[c^p*x^(4*p - 1)*((d + e*x^2)^(q + 1)/(e*(4*p + 2*q + 1))) 
, x] + Simp[1/(e*(4*p + 2*q + 1))   Int[(d + e*x^2)^q*ExpandToSum[e*(4*p + 
2*q + 1)*(a + b*x^2 + c*x^4)^p - d*c^p*(4*p - 1)*x^(4*p - 2) - e*c^p*(4*p + 
 2*q + 1)*x^(4*p), x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b^2 - 
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] &&  !LtQ[q, -1]
 

rule 2346

Int[(Pq_)*((a_) + (b_.)*(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^2)^(p + 1)/(b*( 
q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1))   Int[(a + b*x^2)^p*ExpandToS 
um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], 
x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] &&  !LeQ[p, -1]
 

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.71

method	result	size
pseudoelliptic	\(\frac {a \left (b^{3} A -\frac {1}{4} a \,b^{2} B +\frac {1}{8} a^{2} b C -\frac {5}{64} a^{3} D\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )+\sqrt {b \,x^{2}+a}\, x \left (\left (\frac {1}{4} D x^{6}+\frac {1}{3} C \,x^{4}+\frac {1}{2} x^{2} B +A \right ) b^{\frac {7}{2}}+\frac {5 \left (\left (\frac {8}{15} D x^{4}+\frac {16}{15} C \,x^{2}+\frac {16}{5} B \right ) b^{\frac {5}{2}}+a \left (\left (-\frac {2 D x^{2}}{3}-\frac {8 C}{5}\right ) b^{\frac {3}{2}}+D a \sqrt {b}\right )\right ) a}{64}\right )}{2 b^{\frac {7}{2}}}\)	\(135\)
default	\(A \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )+C \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )}{2 b}\right )+D \left (\frac {x^{5} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{8 b}-\frac {5 a \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )}{2 b}\right )}{8 b}\right )+B \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )\)	\(288\)

Input:

int((b*x^2+a)^(1/2)*(D*x^6+C*x^4+B*x^2+A),x,method=_RETURNVERBOSE)
 

Output:

1/2*(a*(b^3*A-1/4*a*b^2*B+1/8*a^2*b*C-5/64*a^3*D)*arctanh((b*x^2+a)^(1/2)/ 
x/b^(1/2))+(b*x^2+a)^(1/2)*x*((1/4*D*x^6+1/3*C*x^4+1/2*x^2*B+A)*b^(7/2)+5/ 
64*((8/15*D*x^4+16/15*C*x^2+16/5*B)*b^(5/2)+a*((-2/3*D*x^2-8/5*C)*b^(3/2)+ 
D*a*b^(1/2)))*a))/b^(7/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.71 \[ \int \sqrt {a+b x^2} \left (A+B x^2+C x^4+D x^6\right ) \, dx=\left [-\frac {3 \, {\left (5 \, D a^{4} - 8 \, C a^{3} b + 16 \, B a^{2} b^{2} - 64 \, A a b^{3}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (48 \, D b^{4} x^{7} + 8 \, {\left (D a b^{3} + 8 \, C b^{4}\right )} x^{5} - 2 \, {\left (5 \, D a^{2} b^{2} - 8 \, C a b^{3} - 48 \, B b^{4}\right )} x^{3} + 3 \, {\left (5 \, D a^{3} b - 8 \, C a^{2} b^{2} + 16 \, B a b^{3} + 64 \, A b^{4}\right )} x\right )} \sqrt {b x^{2} + a}}{768 \, b^{4}}, \frac {3 \, {\left (5 \, D a^{4} - 8 \, C a^{3} b + 16 \, B a^{2} b^{2} - 64 \, A a b^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (48 \, D b^{4} x^{7} + 8 \, {\left (D a b^{3} + 8 \, C b^{4}\right )} x^{5} - 2 \, {\left (5 \, D a^{2} b^{2} - 8 \, C a b^{3} - 48 \, B b^{4}\right )} x^{3} + 3 \, {\left (5 \, D a^{3} b - 8 \, C a^{2} b^{2} + 16 \, B a b^{3} + 64 \, A b^{4}\right )} x\right )} \sqrt {b x^{2} + a}}{384 \, b^{4}}\right ] \] Input:

integrate((b*x^2+a)^(1/2)*(D*x^6+C*x^4+B*x^2+A),x, algorithm="fricas")
 

Output:

[-1/768*(3*(5*D*a^4 - 8*C*a^3*b + 16*B*a^2*b^2 - 64*A*a*b^3)*sqrt(b)*log(- 
2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(48*D*b^4*x^7 + 8*(D*a*b^3 
+ 8*C*b^4)*x^5 - 2*(5*D*a^2*b^2 - 8*C*a*b^3 - 48*B*b^4)*x^3 + 3*(5*D*a^3*b 
 - 8*C*a^2*b^2 + 16*B*a*b^3 + 64*A*b^4)*x)*sqrt(b*x^2 + a))/b^4, 1/384*(3* 
(5*D*a^4 - 8*C*a^3*b + 16*B*a^2*b^2 - 64*A*a*b^3)*sqrt(-b)*arctan(sqrt(-b) 
*x/sqrt(b*x^2 + a)) + (48*D*b^4*x^7 + 8*(D*a*b^3 + 8*C*b^4)*x^5 - 2*(5*D*a 
^2*b^2 - 8*C*a*b^3 - 48*B*b^4)*x^3 + 3*(5*D*a^3*b - 8*C*a^2*b^2 + 16*B*a*b 
^3 + 64*A*b^4)*x)*sqrt(b*x^2 + a))/b^4]
 

Sympy [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.13 \[ \int \sqrt {a+b x^2} \left (A+B x^2+C x^4+D x^6\right ) \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {D x^{7}}{8} + \frac {x^{5} \left (C b + \frac {D a}{8}\right )}{6 b} + \frac {x^{3} \left (B b + C a - \frac {5 a \left (C b + \frac {D a}{8}\right )}{6 b}\right )}{4 b} + \frac {x \left (A b + B a - \frac {3 a \left (B b + C a - \frac {5 a \left (C b + \frac {D a}{8}\right )}{6 b}\right )}{4 b}\right )}{2 b}\right ) + \left (A a - \frac {a \left (A b + B a - \frac {3 a \left (B b + C a - \frac {5 a \left (C b + \frac {D a}{8}\right )}{6 b}\right )}{4 b}\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\\sqrt {a} \left (A x + \frac {B x^{3}}{3} + \frac {C x^{5}}{5} + \frac {D x^{7}}{7}\right ) & \text {otherwise} \end {cases} \] Input:

integrate((b*x**2+a)**(1/2)*(D*x**6+C*x**4+B*x**2+A),x)
 

Output:

Piecewise((sqrt(a + b*x**2)*(D*x**7/8 + x**5*(C*b + D*a/8)/(6*b) + x**3*(B 
*b + C*a - 5*a*(C*b + D*a/8)/(6*b))/(4*b) + x*(A*b + B*a - 3*a*(B*b + C*a 
- 5*a*(C*b + D*a/8)/(6*b))/(4*b))/(2*b)) + (A*a - a*(A*b + B*a - 3*a*(B*b 
+ C*a - 5*a*(C*b + D*a/8)/(6*b))/(4*b))/(2*b))*Piecewise((log(2*sqrt(b)*sq 
rt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/sqrt(b*x**2), True)) 
, Ne(b, 0)), (sqrt(a)*(A*x + B*x**3/3 + C*x**5/5 + D*x**7/7), True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.29 \[ \int \sqrt {a+b x^2} \left (A+B x^2+C x^4+D x^6\right ) \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} D x^{5}}{8 \, b} - \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} D a x^{3}}{48 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} C x^{3}}{6 \, b} + \frac {1}{2} \, \sqrt {b x^{2} + a} A x + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} D a^{2} x}{64 \, b^{3}} - \frac {5 \, \sqrt {b x^{2} + a} D a^{3} x}{128 \, b^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} C a x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} C a^{2} x}{16 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B x}{4 \, b} - \frac {\sqrt {b x^{2} + a} B a x}{8 \, b} - \frac {5 \, D a^{4} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {7}{2}}} + \frac {C a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {5}{2}}} - \frac {B a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {3}{2}}} + \frac {A a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {b}} \] Input:

integrate((b*x^2+a)^(1/2)*(D*x^6+C*x^4+B*x^2+A),x, algorithm="maxima")
 

Output:

1/8*(b*x^2 + a)^(3/2)*D*x^5/b - 5/48*(b*x^2 + a)^(3/2)*D*a*x^3/b^2 + 1/6*( 
b*x^2 + a)^(3/2)*C*x^3/b + 1/2*sqrt(b*x^2 + a)*A*x + 5/64*(b*x^2 + a)^(3/2 
)*D*a^2*x/b^3 - 5/128*sqrt(b*x^2 + a)*D*a^3*x/b^3 - 1/8*(b*x^2 + a)^(3/2)* 
C*a*x/b^2 + 1/16*sqrt(b*x^2 + a)*C*a^2*x/b^2 + 1/4*(b*x^2 + a)^(3/2)*B*x/b 
 - 1/8*sqrt(b*x^2 + a)*B*a*x/b - 5/128*D*a^4*arcsinh(b*x/sqrt(a*b))/b^(7/2 
) + 1/16*C*a^3*arcsinh(b*x/sqrt(a*b))/b^(5/2) - 1/8*B*a^2*arcsinh(b*x/sqrt 
(a*b))/b^(3/2) + 1/2*A*a*arcsinh(b*x/sqrt(a*b))/sqrt(b)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.87 \[ \int \sqrt {a+b x^2} \left (A+B x^2+C x^4+D x^6\right ) \, dx=\frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (6 \, D x^{2} + \frac {D a b^{5} + 8 \, C b^{6}}{b^{6}}\right )} x^{2} - \frac {5 \, D a^{2} b^{4} - 8 \, C a b^{5} - 48 \, B b^{6}}{b^{6}}\right )} x^{2} + \frac {3 \, {\left (5 \, D a^{3} b^{3} - 8 \, C a^{2} b^{4} + 16 \, B a b^{5} + 64 \, A b^{6}\right )}}{b^{6}}\right )} \sqrt {b x^{2} + a} x + \frac {{\left (5 \, D a^{4} - 8 \, C a^{3} b + 16 \, B a^{2} b^{2} - 64 \, A a b^{3}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{128 \, b^{\frac {7}{2}}} \] Input:

integrate((b*x^2+a)^(1/2)*(D*x^6+C*x^4+B*x^2+A),x, algorithm="giac")
 

Output:

1/384*(2*(4*(6*D*x^2 + (D*a*b^5 + 8*C*b^6)/b^6)*x^2 - (5*D*a^2*b^4 - 8*C*a 
*b^5 - 48*B*b^6)/b^6)*x^2 + 3*(5*D*a^3*b^3 - 8*C*a^2*b^4 + 16*B*a*b^5 + 64 
*A*b^6)/b^6)*sqrt(b*x^2 + a)*x + 1/128*(5*D*a^4 - 8*C*a^3*b + 16*B*a^2*b^2 
 - 64*A*a*b^3)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt {a+b x^2} \left (A+B x^2+C x^4+D x^6\right ) \, dx=\int \sqrt {b\,x^2+a}\,\left (A+B\,x^2+C\,x^4+x^6\,D\right ) \,d x \] Input:

int((a + b*x^2)^(1/2)*(A + B*x^2 + C*x^4 + x^6*D),x)
 

Output:

int((a + b*x^2)^(1/2)*(A + B*x^2 + C*x^4 + x^6*D), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.28 \[ \int \sqrt {a+b x^2} \left (A+B x^2+C x^4+D x^6\right ) \, dx=\frac {15 \sqrt {b \,x^{2}+a}\, a^{3} b d x -24 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} c x -10 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} d \,x^{3}+240 \sqrt {b \,x^{2}+a}\, a \,b^{4} x +16 \sqrt {b \,x^{2}+a}\, a \,b^{3} c \,x^{3}+8 \sqrt {b \,x^{2}+a}\, a \,b^{3} d \,x^{5}+96 \sqrt {b \,x^{2}+a}\, b^{5} x^{3}+64 \sqrt {b \,x^{2}+a}\, b^{4} c \,x^{5}+48 \sqrt {b \,x^{2}+a}\, b^{4} d \,x^{7}-15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{4} d +24 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} b c +144 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b^{3}}{384 b^{4}} \] Input:

int((b*x^2+a)^(1/2)*(D*x^6+C*x^4+B*x^2+A),x)
 

Output:

(15*sqrt(a + b*x**2)*a**3*b*d*x - 24*sqrt(a + b*x**2)*a**2*b**2*c*x - 10*s 
qrt(a + b*x**2)*a**2*b**2*d*x**3 + 240*sqrt(a + b*x**2)*a*b**4*x + 16*sqrt 
(a + b*x**2)*a*b**3*c*x**3 + 8*sqrt(a + b*x**2)*a*b**3*d*x**5 + 96*sqrt(a 
+ b*x**2)*b**5*x**3 + 64*sqrt(a + b*x**2)*b**4*c*x**5 + 48*sqrt(a + b*x**2 
)*b**4*d*x**7 - 15*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a** 
4*d + 24*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*b*c + 14 
4*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*b**3)/(384*b**4 
)