\(\int \sqrt {a+b x^2} (A+B x+C x^2+D x^3) \, dx\) [55]

Optimal result

Integrand size = 27, antiderivative size = 132 \[ \int \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {(4 A b-a C) x \sqrt {a+b x^2}}{8 b}+\frac {(b B-a D) \left (a+b x^2\right )^{3/2}}{3 b^2}+\frac {C x \left (a+b x^2\right )^{3/2}}{4 b}+\frac {D \left (a+b x^2\right )^{5/2}}{5 b^2}+\frac {a (4 A b-a C) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.82 \[ \int \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {\sqrt {a+b x^2} \left (-16 a^2 D+a b (40 B+x (15 C+8 D x))+2 b^2 x (30 A+x (20 B+3 x (5 C+4 D x)))\right )+15 a \sqrt {b} (-4 A b+a C) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{120 b^2} \] Input:

Integrate[Sqrt[a + b*x^2]*(A + B*x + C*x^2 + D*x^3),x]
 

Output:

(Sqrt[a + b*x^2]*(-16*a^2*D + a*b*(40*B + x*(15*C + 8*D*x)) + 2*b^2*x*(30* 
A + x*(20*B + 3*x*(5*C + 4*D*x)))) + 15*a*Sqrt[b]*(-4*A*b + a*C)*Log[-(Sqr 
t[b]*x) + Sqrt[a + b*x^2]])/(120*b^2)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2346, 2346, 27, 455, 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.

\(\displaystyle \int \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx\)

\(\Big \downarrow \) 2346

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

\(\Big \downarrow \) 2346

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

Int[Sqrt[a + b*x^2]*(A + B*x + C*x^2 + D*x^3),x]
 

Output:

(D*x^2*(a + b*x^2)^(3/2))/(5*b) + ((5*C*x*(a + b*x^2)^(3/2))/4 + ((4*(5*b* 
B - 2*a*D)*(a + b*x^2)^(3/2))/(3*b) + 5*(4*A*b - a*C)*((x*Sqrt[a + b*x^2]) 
/2 + (a*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*Sqrt[b])))/4)/(5*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 455

Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( 
a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c   Int[(a + b*x^2)^p, x], x] 
/; FreeQ[{a, b, c, d, p}, x] &&  !LeQ[p, -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 = 0.60 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.12

method	result	size
default	\(A \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )+\frac {B \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3 b}+C \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )+D \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{5 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{15 b^{2}}\right )\)	\(148\)

Input:

int((b*x^2+a)^(1/2)*(D*x^3+C*x^2+B*x+A),x,method=_RETURNVERBOSE)
 

Output:

A*(1/2*x*(b*x^2+a)^(1/2)+1/2*a/b^(1/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2)))+1/3* 
B*(b*x^2+a)^(3/2)/b+C*(1/4*x*(b*x^2+a)^(3/2)/b-1/4*a/b*(1/2*x*(b*x^2+a)^(1 
/2)+1/2*a/b^(1/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))))+D*(1/5*x^2*(b*x^2+a)^(3/ 
2)/b-2/15*a/b^2*(b*x^2+a)^(3/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.75 \[ \int \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\left [-\frac {15 \, {\left (C a^{2} - 4 \, A a b\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (24 \, D b^{2} x^{4} + 30 \, C b^{2} x^{3} - 16 \, D a^{2} + 40 \, B a b + 8 \, {\left (D a b + 5 \, B b^{2}\right )} x^{2} + 15 \, {\left (C a b + 4 \, A b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{240 \, b^{2}}, \frac {15 \, {\left (C a^{2} - 4 \, A a b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (24 \, D b^{2} x^{4} + 30 \, C b^{2} x^{3} - 16 \, D a^{2} + 40 \, B a b + 8 \, {\left (D a b + 5 \, B b^{2}\right )} x^{2} + 15 \, {\left (C a b + 4 \, A b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{120 \, b^{2}}\right ] \] Input:

integrate((b*x^2+a)^(1/2)*(D*x^3+C*x^2+B*x+A),x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

[-1/240*(15*(C*a^2 - 4*A*a*b)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqr 
t(b)*x - a) - 2*(24*D*b^2*x^4 + 30*C*b^2*x^3 - 16*D*a^2 + 40*B*a*b + 8*(D* 
a*b + 5*B*b^2)*x^2 + 15*(C*a*b + 4*A*b^2)*x)*sqrt(b*x^2 + a))/b^2, 1/120*( 
15*(C*a^2 - 4*A*a*b)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (24*D*b 
^2*x^4 + 30*C*b^2*x^3 - 16*D*a^2 + 40*B*a*b + 8*(D*a*b + 5*B*b^2)*x^2 + 15 
*(C*a*b + 4*A*b^2)*x)*sqrt(b*x^2 + a))/b^2]
 

Sympy [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.21 \[ \int \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {C x^{3}}{4} + \frac {D x^{4}}{5} + \frac {x^{2} \left (B b + \frac {D a}{5}\right )}{3 b} + \frac {x \left (A b + \frac {C a}{4}\right )}{2 b} + \frac {B a - \frac {2 a \left (B b + \frac {D a}{5}\right )}{3 b}}{b}\right ) + \left (A a - \frac {a \left (A b + \frac {C a}{4}\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\\sqrt {a} \left (A x + \frac {B x^{2}}{2} + \frac {C x^{3}}{3} + \frac {D x^{4}}{4}\right ) & \text {otherwise} \end {cases} \] Input:

integrate((b*x**2+a)**(1/2)*(D*x**3+C*x**2+B*x+A),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.98 \[ \int \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} D x^{2}}{5 \, b} + \frac {1}{2} \, \sqrt {b x^{2} + a} A x + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} C x}{4 \, b} - \frac {\sqrt {b x^{2} + a} C a x}{8 \, b} - \frac {C a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {3}{2}}} + \frac {A a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {b}} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} D a}{15 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B}{3 \, b} \] Input:

integrate((b*x^2+a)^(1/2)*(D*x^3+C*x^2+B*x+A),x, algorithm="maxima")
 

Output:

1/5*(b*x^2 + a)^(3/2)*D*x^2/b + 1/2*sqrt(b*x^2 + a)*A*x + 1/4*(b*x^2 + a)^ 
(3/2)*C*x/b - 1/8*sqrt(b*x^2 + a)*C*a*x/b - 1/8*C*a^2*arcsinh(b*x/sqrt(a*b 
))/b^(3/2) + 1/2*A*a*arcsinh(b*x/sqrt(a*b))/sqrt(b) - 2/15*(b*x^2 + a)^(3/ 
2)*D*a/b^2 + 1/3*(b*x^2 + a)^(3/2)*B/b
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.92 \[ \int \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{120} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left (3 \, {\left (4 \, D x + 5 \, C\right )} x + \frac {4 \, {\left (D a b^{2} + 5 \, B b^{3}\right )}}{b^{3}}\right )} x + \frac {15 \, {\left (C a b^{2} + 4 \, A b^{3}\right )}}{b^{3}}\right )} x - \frac {8 \, {\left (2 \, D a^{2} b - 5 \, B a b^{2}\right )}}{b^{3}}\right )} + \frac {{\left (C a^{2} - 4 \, A a b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {3}{2}}} \] Input:

integrate((b*x^2+a)^(1/2)*(D*x^3+C*x^2+B*x+A),x, algorithm="giac")
 

Output:

1/120*sqrt(b*x^2 + a)*((2*(3*(4*D*x + 5*C)*x + 4*(D*a*b^2 + 5*B*b^3)/b^3)* 
x + 15*(C*a*b^2 + 4*A*b^3)/b^3)*x - 8*(2*D*a^2*b - 5*B*a*b^2)/b^3) + 1/8*( 
C*a^2 - 4*A*a*b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(3/2)
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\int \sqrt {b\,x^2+a}\,\left (A+B\,x+C\,x^2+x^3\,D\right ) \,d x \] Input:

int((a + b*x^2)^(1/2)*(A + B*x + C*x^2 + x^3*D),x)
 

Output:

int((a + b*x^2)^(1/2)*(A + B*x + C*x^2 + x^3*D), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.39 \[ \int \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {-16 \sqrt {b \,x^{2}+a}\, a^{2} d +60 \sqrt {b \,x^{2}+a}\, a \,b^{2} x +40 \sqrt {b \,x^{2}+a}\, a \,b^{2}+15 \sqrt {b \,x^{2}+a}\, a b c x +8 \sqrt {b \,x^{2}+a}\, a b d \,x^{2}+40 \sqrt {b \,x^{2}+a}\, b^{3} x^{2}+30 \sqrt {b \,x^{2}+a}\, b^{2} c \,x^{3}+24 \sqrt {b \,x^{2}+a}\, b^{2} d \,x^{4}+60 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b -15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} c}{120 b^{2}} \] Input:

int((b*x^2+a)^(1/2)*(D*x^3+C*x^2+B*x+A),x)
 

Output:

( - 16*sqrt(a + b*x**2)*a**2*d + 60*sqrt(a + b*x**2)*a*b**2*x + 40*sqrt(a 
+ b*x**2)*a*b**2 + 15*sqrt(a + b*x**2)*a*b*c*x + 8*sqrt(a + b*x**2)*a*b*d* 
x**2 + 40*sqrt(a + b*x**2)*b**3*x**2 + 30*sqrt(a + b*x**2)*b**2*c*x**3 + 2 
4*sqrt(a + b*x**2)*b**2*d*x**4 + 60*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b 
)*x)/sqrt(a))*a**2*b - 15*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt( 
a))*a**2*c)/(120*b**2)