\(\int (a+b x^2)^{3/2} (A+B x+C x^2) \, dx\) [31]

Optimal result

Integrand size = 22, antiderivative size = 137 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right ) \, dx=\frac {a (6 A b-a C) x \sqrt {a+b x^2}}{16 b}+\frac {(6 A b-a C) x \left (a+b x^2\right )^{3/2}}{24 b}+\frac {B \left (a+b x^2\right )^{5/2}}{5 b}+\frac {C x \left (a+b x^2\right )^{5/2}}{6 b}+\frac {a^2 (6 A b-a C) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{3/2}} \] Output:

1/16*a*(6*A*b-C*a)*x*(b*x^2+a)^(1/2)/b+1/24*(6*A*b-C*a)*x*(b*x^2+a)^(3/2)/ 
b+1/5*B*(b*x^2+a)^(5/2)/b+1/6*C*x*(b*x^2+a)^(5/2)/b+1/16*a^2*(6*A*b-C*a)*a 
rctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(3/2)
 

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.85 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right ) \, dx=\frac {\sqrt {b} \sqrt {a+b x^2} \left (3 a^2 (16 B+5 C x)+4 b^2 x^3 (15 A+2 x (6 B+5 C x))+2 a b x (75 A+x (48 B+35 C x))\right )+15 a^2 (-6 A b+a C) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{240 b^{3/2}} \] Input:

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

Output:

(Sqrt[b]*Sqrt[a + b*x^2]*(3*a^2*(16*B + 5*C*x) + 4*b^2*x^3*(15*A + 2*x*(6* 
B + 5*C*x)) + 2*a*b*x*(75*A + x*(48*B + 35*C*x))) + 15*a^2*(-6*A*b + a*C)* 
Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(240*b^(3/2))
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.90, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2346, 455, 211, 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.

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

\(\Big \downarrow \) 2346

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(C*x*(a + b*x^2)^(5/2))/(6*b) + ((6*B*(a + b*x^2)^(5/2))/5 + (6*A*b - a*C) 
*((x*(a + b*x^2)^(3/2))/4 + (3*a*((x*Sqrt[a + b*x^2])/2 + (a*ArcTanh[(Sqrt 
[b]*x)/Sqrt[a + b*x^2]])/(2*Sqrt[b])))/4))/(6*b)
 

Defintions of rubi rules used

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 = 0.53 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.82

method	result	size
risch	\(\frac {\left (40 b^{2} C \,x^{5}+48 b^{2} B \,x^{4}+60 A \,b^{2} x^{3}+70 C a b \,x^{3}+96 B a b \,x^{2}+150 a A b x +15 C \,a^{2} x +48 a^{2} B \right ) \sqrt {b \,x^{2}+a}}{240 b}+\frac {a^{2} \left (6 A b -C a \right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {3}{2}}}\)	\(112\)
default	\(A \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )+\frac {B \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5 b}+C \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )\)	\(145\)

Input:

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

Output:

1/240/b*(40*C*b^2*x^5+48*B*b^2*x^4+60*A*b^2*x^3+70*C*a*b*x^3+96*B*a*b*x^2+ 
150*A*a*b*x+15*C*a^2*x+48*B*a^2)*(b*x^2+a)^(1/2)+1/16*a^2*(6*A*b-C*a)/b^(3 
/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.89 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right ) \, dx=\left [-\frac {15 \, {\left (C a^{3} - 6 \, A a^{2} b\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (40 \, C b^{3} x^{5} + 48 \, B b^{3} x^{4} + 96 \, B a b^{2} x^{2} + 48 \, B a^{2} b + 10 \, {\left (7 \, C a b^{2} + 6 \, A b^{3}\right )} x^{3} + 15 \, {\left (C a^{2} b + 10 \, A a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{480 \, b^{2}}, \frac {15 \, {\left (C a^{3} - 6 \, A a^{2} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (40 \, C b^{3} x^{5} + 48 \, B b^{3} x^{4} + 96 \, B a b^{2} x^{2} + 48 \, B a^{2} b + 10 \, {\left (7 \, C a b^{2} + 6 \, A b^{3}\right )} x^{3} + 15 \, {\left (C a^{2} b + 10 \, A a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{240 \, b^{2}}\right ] \] Input:

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

Output:

[-1/480*(15*(C*a^3 - 6*A*a^2*b)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*s 
qrt(b)*x - a) - 2*(40*C*b^3*x^5 + 48*B*b^3*x^4 + 96*B*a*b^2*x^2 + 48*B*a^2 
*b + 10*(7*C*a*b^2 + 6*A*b^3)*x^3 + 15*(C*a^2*b + 10*A*a*b^2)*x)*sqrt(b*x^ 
2 + a))/b^2, 1/240*(15*(C*a^3 - 6*A*a^2*b)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt 
(b*x^2 + a)) + (40*C*b^3*x^5 + 48*B*b^3*x^4 + 96*B*a*b^2*x^2 + 48*B*a^2*b 
+ 10*(7*C*a*b^2 + 6*A*b^3)*x^3 + 15*(C*a^2*b + 10*A*a*b^2)*x)*sqrt(b*x^2 + 
 a))/b^2]
 

Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.53 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right ) \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {B a^{2}}{5 b} + \frac {2 B a x^{2}}{5} + \frac {B b x^{4}}{5} + \frac {C b x^{5}}{6} + \frac {x^{3} \left (A b^{2} + \frac {7 C a b}{6}\right )}{4 b} + \frac {x \left (2 A a b + C a^{2} - \frac {3 a \left (A b^{2} + \frac {7 C a b}{6}\right )}{4 b}\right )}{2 b}\right ) + \left (A a^{2} - \frac {a \left (2 A a b + C a^{2} - \frac {3 a \left (A b^{2} + \frac {7 C a b}{6}\right )}{4 b}\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (A x + \frac {B x^{2}}{2} + \frac {C x^{3}}{3}\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.96 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right ) \, dx=\frac {1}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A x + \frac {3}{8} \, \sqrt {b x^{2} + a} A a x + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} C x}{6 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} C a x}{24 \, b} - \frac {\sqrt {b x^{2} + a} C a^{2} x}{16 \, b} - \frac {C a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {3}{2}}} + \frac {3 \, A a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {b}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B}{5 \, b} \] Input:

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

Output:

1/4*(b*x^2 + a)^(3/2)*A*x + 3/8*sqrt(b*x^2 + a)*A*a*x + 1/6*(b*x^2 + a)^(5 
/2)*C*x/b - 1/24*(b*x^2 + a)^(3/2)*C*a*x/b - 1/16*sqrt(b*x^2 + a)*C*a^2*x/ 
b - 1/16*C*a^3*arcsinh(b*x/sqrt(a*b))/b^(3/2) + 3/8*A*a^2*arcsinh(b*x/sqrt 
(a*b))/sqrt(b) + 1/5*(b*x^2 + a)^(5/2)*B/b
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.91 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right ) \, dx=\frac {1}{240} \, \sqrt {b x^{2} + a} {\left (\frac {48 \, B a^{2}}{b} + {\left (2 \, {\left (48 \, B a + {\left (4 \, {\left (5 \, C b x + 6 \, B b\right )} x + \frac {5 \, {\left (7 \, C a b^{4} + 6 \, A b^{5}\right )}}{b^{4}}\right )} x\right )} x + \frac {15 \, {\left (C a^{2} b^{3} + 10 \, A a b^{4}\right )}}{b^{4}}\right )} x\right )} + \frac {{\left (C a^{3} - 6 \, A a^{2} b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {3}{2}}} \] Input:

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

Output:

1/240*sqrt(b*x^2 + a)*(48*B*a^2/b + (2*(48*B*a + (4*(5*C*b*x + 6*B*b)*x + 
5*(7*C*a*b^4 + 6*A*b^5)/b^4)*x)*x + 15*(C*a^2*b^3 + 10*A*a*b^4)/b^4)*x) + 
1/16*(C*a^3 - 6*A*a^2*b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(3/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.42 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right ) \, dx=\frac {150 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} x +48 \sqrt {b \,x^{2}+a}\, a^{2} b^{2}+15 \sqrt {b \,x^{2}+a}\, a^{2} b c x +60 \sqrt {b \,x^{2}+a}\, a \,b^{3} x^{3}+96 \sqrt {b \,x^{2}+a}\, a \,b^{3} x^{2}+70 \sqrt {b \,x^{2}+a}\, a \,b^{2} c \,x^{3}+48 \sqrt {b \,x^{2}+a}\, b^{4} x^{4}+40 \sqrt {b \,x^{2}+a}\, b^{3} c \,x^{5}+90 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} b -15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} c}{240 b^{2}} \] Input:

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

Output:

(150*sqrt(a + b*x**2)*a**2*b**2*x + 48*sqrt(a + b*x**2)*a**2*b**2 + 15*sqr 
t(a + b*x**2)*a**2*b*c*x + 60*sqrt(a + b*x**2)*a*b**3*x**3 + 96*sqrt(a + b 
*x**2)*a*b**3*x**2 + 70*sqrt(a + b*x**2)*a*b**2*c*x**3 + 48*sqrt(a + b*x** 
2)*b**4*x**4 + 40*sqrt(a + b*x**2)*b**3*c*x**5 + 90*sqrt(b)*log((sqrt(a + 
b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*b - 15*sqrt(b)*log((sqrt(a + b*x**2) + 
sqrt(b)*x)/sqrt(a))*a**3*c)/(240*b**2)