\(\int \sqrt {a+b x^2} (A+B x^2+C x^4) \, dx\) [116]

Optimal result

Integrand size = 24, antiderivative size = 134 \[ \int \sqrt {a+b x^2} \left (A+B x^2+C x^4\right ) \, dx=\frac {1}{16} \left (8 A-\frac {a (2 b B-a C)}{b^2}\right ) x \sqrt {a+b x^2}+\frac {(2 b B-a C) x \left (a+b x^2\right )^{3/2}}{8 b^2}+\frac {C x^3 \left (a+b x^2\right )^{3/2}}{6 b}+\frac {a \left (8 A b^2-a (2 b B-a C)\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{5/2}} \] Output:

1/16*(8*A-a*(2*B*b-C*a)/b^2)*x*(b*x^2+a)^(1/2)+1/8*(2*B*b-C*a)*x*(b*x^2+a) 
^(3/2)/b^2+1/6*C*x^3*(b*x^2+a)^(3/2)/b+1/16*a*(8*A*b^2-a*(2*B*b-C*a))*arct 
anh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(5/2)
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.84 \[ \int \sqrt {a+b x^2} \left (A+B x^2+C x^4\right ) \, dx=\frac {x \sqrt {a+b x^2} \left (24 A b^2+6 a b B-3 a^2 C+12 b^2 B x^2+2 a b C x^2+8 b^2 C x^4\right )}{48 b^2}-\frac {a \left (8 A b^2-2 a b B+a^2 C\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{16 b^{5/2}} \] Input:

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

Output:

(x*Sqrt[a + b*x^2]*(24*A*b^2 + 6*a*b*B - 3*a^2*C + 12*b^2*B*x^2 + 2*a*b*C* 
x^2 + 8*b^2*C*x^4))/(48*b^2) - (a*(8*A*b^2 - 2*a*b*B + a^2*C)*Log[-(Sqrt[b 
]*x) + Sqrt[a + b*x^2]])/(16*b^(5/2))
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {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.

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

\(\Big \downarrow \) 1473

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 299

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(C*x^3*(a + b*x^2)^(3/2))/(6*b) + (((2*b*B - a*C)*x*(a + b*x^2)^(3/2))/(4* 
b) + ((8*A*b^2 - a*(2*b*B - a*C))*((x*Sqrt[a + b*x^2])/2 + (a*ArcTanh[(Sqr 
t[b]*x)/Sqrt[a + b*x^2]])/(2*Sqrt[b])))/(4*b))/(2*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]
 

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.72

method	result	size
pseudoelliptic	\(\frac {a \left (b^{2} A -\frac {1}{4} a b B +\frac {1}{8} a^{2} C \right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )+\sqrt {b \,x^{2}+a}\, x \left (\left (\frac {1}{3} C \,x^{4}+\frac {1}{2} x^{2} B +A \right ) b^{\frac {5}{2}}+\frac {a \left (\left (\frac {C \,x^{2}}{3}+B \right ) b^{\frac {3}{2}}-\frac {C a \sqrt {b}}{2}\right )}{4}\right )}{2 b^{\frac {5}{2}}}\)	\(96\)
risch	\(\frac {x \left (8 C \,b^{2} x^{4}+12 b^{2} B \,x^{2}+2 x^{2} a C b +24 b^{2} A +6 a b B -3 a^{2} C \right ) \sqrt {b \,x^{2}+a}}{48 b^{2}}+\frac {a \left (8 b^{2} A -2 a b B +a^{2} C \right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {5}{2}}}\)	\(100\)
default	\(A \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )+C \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )}{2 b}\right )+B \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )\)	\(181\)

Input:

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

Output:

1/2*(a*(b^2*A-1/4*a*b*B+1/8*a^2*C)*arctanh((b*x^2+a)^(1/2)/x/b^(1/2))+(b*x 
^2+a)^(1/2)*x*((1/3*C*x^4+1/2*x^2*B+A)*b^(5/2)+1/4*a*((1/3*C*x^2+B)*b^(3/2 
)-1/2*C*a*b^(1/2))))/b^(5/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.73 \[ \int \sqrt {a+b x^2} \left (A+B x^2+C x^4\right ) \, dx=\left [\frac {3 \, {\left (C a^{3} - 2 \, B a^{2} b + 8 \, A a b^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (8 \, C b^{3} x^{5} + 2 \, {\left (C a b^{2} + 6 \, B b^{3}\right )} x^{3} - 3 \, {\left (C a^{2} b - 2 \, B a b^{2} - 8 \, A b^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{96 \, b^{3}}, -\frac {3 \, {\left (C a^{3} - 2 \, B a^{2} b + 8 \, A a b^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (8 \, C b^{3} x^{5} + 2 \, {\left (C a b^{2} + 6 \, B b^{3}\right )} x^{3} - 3 \, {\left (C a^{2} b - 2 \, B a b^{2} - 8 \, A b^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, b^{3}}\right ] \] Input:

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

Output:

[1/96*(3*(C*a^3 - 2*B*a^2*b + 8*A*a*b^2)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x 
^2 + a)*sqrt(b)*x - a) + 2*(8*C*b^3*x^5 + 2*(C*a*b^2 + 6*B*b^3)*x^3 - 3*(C 
*a^2*b - 2*B*a*b^2 - 8*A*b^3)*x)*sqrt(b*x^2 + a))/b^3, -1/48*(3*(C*a^3 - 2 
*B*a^2*b + 8*A*a*b^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (8*C*b 
^3*x^5 + 2*(C*a*b^2 + 6*B*b^3)*x^3 - 3*(C*a^2*b - 2*B*a*b^2 - 8*A*b^3)*x)* 
sqrt(b*x^2 + a))/b^3]
 

Sympy [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.14 \[ \int \sqrt {a+b x^2} \left (A+B x^2+C x^4\right ) \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {C x^{5}}{6} + \frac {x^{3} \left (B b + \frac {C a}{6}\right )}{4 b} + \frac {x \left (A b + B a - \frac {3 a \left (B b + \frac {C a}{6}\right )}{4 b}\right )}{2 b}\right ) + \left (A a - \frac {a \left (A b + B a - \frac {3 a \left (B b + \frac {C a}{6}\right )}{4 b}\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\\sqrt {a} \left (A x + \frac {B x^{3}}{3} + \frac {C x^{5}}{5}\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((sqrt(a + b*x**2)*(C*x**5/6 + x**3*(B*b + C*a/6)/(4*b) + x*(A*b 
+ B*a - 3*a*(B*b + C*a/6)/(4*b))/(2*b)) + (A*a - a*(A*b + B*a - 3*a*(B*b + 
 C*a/6)/(4*b))/(2*b))*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/s 
qrt(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), True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.14 \[ \int \sqrt {a+b x^2} \left (A+B x^2+C x^4\right ) \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} C x^{3}}{6 \, b} + \frac {1}{2} \, \sqrt {b x^{2} + a} A x - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} C a x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} C a^{2} x}{16 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B x}{4 \, b} - \frac {\sqrt {b x^{2} + a} B a x}{8 \, b} + \frac {C a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {5}{2}}} - \frac {B a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {3}{2}}} + \frac {A a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {b}} \] Input:

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

Output:

1/6*(b*x^2 + a)^(3/2)*C*x^3/b + 1/2*sqrt(b*x^2 + a)*A*x - 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 + 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/sq 
rt(a*b))/sqrt(b)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.84 \[ \int \sqrt {a+b x^2} \left (A+B x^2+C x^4\right ) \, dx=\frac {1}{48} \, {\left (2 \, {\left (4 \, C x^{2} + \frac {C a b^{3} + 6 \, B b^{4}}{b^{4}}\right )} x^{2} - \frac {3 \, {\left (C a^{2} b^{2} - 2 \, B a b^{3} - 8 \, A b^{4}\right )}}{b^{4}}\right )} \sqrt {b x^{2} + a} x - \frac {{\left (C a^{3} - 2 \, B a^{2} b + 8 \, A a b^{2}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {5}{2}}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.07 \[ \int \sqrt {a+b x^2} \left (A+B x^2+C x^4\right ) \, dx=\frac {-3 \sqrt {b \,x^{2}+a}\, a^{2} b c x +30 \sqrt {b \,x^{2}+a}\, a \,b^{3} x +2 \sqrt {b \,x^{2}+a}\, a \,b^{2} c \,x^{3}+12 \sqrt {b \,x^{2}+a}\, b^{4} x^{3}+8 \sqrt {b \,x^{2}+a}\, b^{3} c \,x^{5}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} c +18 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b^{2}}{48 b^{3}} \] Input:

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

Output:

( - 3*sqrt(a + b*x**2)*a**2*b*c*x + 30*sqrt(a + b*x**2)*a*b**3*x + 2*sqrt( 
a + b*x**2)*a*b**2*c*x**3 + 12*sqrt(a + b*x**2)*b**4*x**3 + 8*sqrt(a + b*x 
**2)*b**3*c*x**5 + 3*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a 
**3*c + 18*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*b**2)/ 
(48*b**3)