\(\int \frac {x^2 (c+d x)^2}{\sqrt {a+b x^2}} \, dx\) [1171]

Optimal result

Integrand size = 22, antiderivative size = 138 \[ \int \frac {x^2 (c+d x)^2}{\sqrt {a+b x^2}} \, dx=-\frac {2 a c d \sqrt {a+b x^2}}{b^2}+\frac {\left (4 b c^2-3 a d^2\right ) x \sqrt {a+b x^2}}{8 b^2}+\frac {d^2 x^3 \sqrt {a+b x^2}}{4 b}+\frac {2 c d \left (a+b x^2\right )^{3/2}}{3 b^2}-\frac {a \left (4 b c^2-3 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{5/2}} \] Output:

-2*a*c*d*(b*x^2+a)^(1/2)/b^2+1/8*(-3*a*d^2+4*b*c^2)*x*(b*x^2+a)^(1/2)/b^2+ 
1/4*d^2*x^3*(b*x^2+a)^(1/2)/b+2/3*c*d*(b*x^2+a)^(3/2)/b^2-1/8*a*(-3*a*d^2+ 
4*b*c^2)*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(5/2)
 

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.78 \[ \int \frac {x^2 (c+d x)^2}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {a+b x^2} \left (-32 a c d+12 b c^2 x-9 a d^2 x+16 b c d x^2+6 b d^2 x^3\right )}{24 b^2}+\frac {a \left (-4 b c^2+3 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{4 b^{5/2}} \] Input:

Integrate[(x^2*(c + d*x)^2)/Sqrt[a + b*x^2],x]
 

Output:

(Sqrt[a + b*x^2]*(-32*a*c*d + 12*b*c^2*x - 9*a*d^2*x + 16*b*c*d*x^2 + 6*b* 
d^2*x^3))/(24*b^2) + (a*(-4*b*c^2 + 3*a*d^2)*ArcTanh[(Sqrt[b]*x)/(-Sqrt[a] 
 + Sqrt[a + b*x^2])])/(4*b^(5/2))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {541, 533, 27, 533, 25, 27, 455, 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.

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

\(\Big \downarrow \) 541

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

\(\Big \downarrow \) 533

\(\displaystyle \frac {\frac {8}{3} c d x^2 \sqrt {a+b x^2}-\frac {\int \frac {b x \left (16 a c d-3 \left (4 b c^2-3 a d^2\right ) x\right )}{\sqrt {b x^2+a}}dx}{3 b}}{4 b}+\frac {d^2 x^3 \sqrt {a+b x^2}}{4 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {8}{3} c d x^2 \sqrt {a+b x^2}-\frac {1}{3} \int \frac {x \left (16 a c d-3 \left (4 b c^2-3 a d^2\right ) x\right )}{\sqrt {b x^2+a}}dx}{4 b}+\frac {d^2 x^3 \sqrt {a+b x^2}}{4 b}\)

\(\Big \downarrow \) 533

\(\displaystyle \frac {\frac {1}{3} \left (\frac {\int -\frac {a \left (3 \left (4 b c^2-3 a d^2\right )+32 b c d x\right )}{\sqrt {b x^2+a}}dx}{2 b}+\frac {3 x \sqrt {a+b x^2} \left (4 b c^2-3 a d^2\right )}{2 b}\right )+\frac {8}{3} c d x^2 \sqrt {a+b x^2}}{4 b}+\frac {d^2 x^3 \sqrt {a+b x^2}}{4 b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {1}{3} \left (\frac {3 x \sqrt {a+b x^2} \left (4 b c^2-3 a d^2\right )}{2 b}-\frac {\int \frac {a \left (3 \left (4 b c^2-3 a d^2\right )+32 b c d x\right )}{\sqrt {b x^2+a}}dx}{2 b}\right )+\frac {8}{3} c d x^2 \sqrt {a+b x^2}}{4 b}+\frac {d^2 x^3 \sqrt {a+b x^2}}{4 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{3} \left (\frac {3 x \sqrt {a+b x^2} \left (4 b c^2-3 a d^2\right )}{2 b}-\frac {a \int \frac {3 \left (4 b c^2-3 a d^2\right )+32 b c d x}{\sqrt {b x^2+a}}dx}{2 b}\right )+\frac {8}{3} c d x^2 \sqrt {a+b x^2}}{4 b}+\frac {d^2 x^3 \sqrt {a+b x^2}}{4 b}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {\frac {1}{3} \left (\frac {3 x \sqrt {a+b x^2} \left (4 b c^2-3 a d^2\right )}{2 b}-\frac {a \left (3 \left (4 b c^2-3 a d^2\right ) \int \frac {1}{\sqrt {b x^2+a}}dx+32 c d \sqrt {a+b x^2}\right )}{2 b}\right )+\frac {8}{3} c d x^2 \sqrt {a+b x^2}}{4 b}+\frac {d^2 x^3 \sqrt {a+b x^2}}{4 b}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {1}{3} \left (\frac {3 x \sqrt {a+b x^2} \left (4 b c^2-3 a d^2\right )}{2 b}-\frac {a \left (3 \left (4 b c^2-3 a d^2\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+32 c d \sqrt {a+b x^2}\right )}{2 b}\right )+\frac {8}{3} c d x^2 \sqrt {a+b x^2}}{4 b}+\frac {d^2 x^3 \sqrt {a+b x^2}}{4 b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{3} \left (\frac {3 x \sqrt {a+b x^2} \left (4 b c^2-3 a d^2\right )}{2 b}-\frac {a \left (\frac {3 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (4 b c^2-3 a d^2\right )}{\sqrt {b}}+32 c d \sqrt {a+b x^2}\right )}{2 b}\right )+\frac {8}{3} c d x^2 \sqrt {a+b x^2}}{4 b}+\frac {d^2 x^3 \sqrt {a+b x^2}}{4 b}\)

Input:

Int[(x^2*(c + d*x)^2)/Sqrt[a + b*x^2],x]
 

Output:

(d^2*x^3*Sqrt[a + b*x^2])/(4*b) + ((8*c*d*x^2*Sqrt[a + b*x^2])/3 + ((3*(4* 
b*c^2 - 3*a*d^2)*x*Sqrt[a + b*x^2])/(2*b) - (a*(32*c*d*Sqrt[a + b*x^2] + ( 
3*(4*b*c^2 - 3*a*d^2)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/Sqrt[b]))/(2*b 
))/3)/(4*b)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 533

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

rule 541

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Simp[d^n*x^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*(m + n + 2*p + 1))), x 
] + Simp[1/(b*(m + n + 2*p + 1))   Int[x^m*(a + b*x^2)^p*ExpandToSum[b*(m + 
 n + 2*p + 1)*(c + d*x)^n - b*d^n*(m + n + 2*p + 1)*x^n - a*d^n*(m + n - 1) 
*x^(n - 2), x], x], x] /; FreeQ[{a, b, c, d, m, p}, x] && IGtQ[n, 1] && IGt 
Q[m, -2] && GtQ[p, -1] && IntegerQ[2*p]
 

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.64

method	result	size
risch	\(-\frac {\left (-6 b \,d^{2} x^{3}-16 b c d \,x^{2}+9 a \,d^{2} x -12 c^{2} b x +32 a c d \right ) \sqrt {b \,x^{2}+a}}{24 b^{2}}+\frac {a \left (3 a \,d^{2}-4 b \,c^{2}\right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {5}{2}}}\)	\(88\)
default	\(c^{2} \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )+d^{2} \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )+2 c d \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )\)	\(147\)

Input:

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

Output:

-1/24*(-6*b*d^2*x^3-16*b*c*d*x^2+9*a*d^2*x-12*b*c^2*x+32*a*c*d)*(b*x^2+a)^ 
(1/2)/b^2+1/8*a*(3*a*d^2-4*b*c^2)/b^(5/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.56 \[ \int \frac {x^2 (c+d x)^2}{\sqrt {a+b x^2}} \, dx=\left [-\frac {3 \, {\left (4 \, a b c^{2} - 3 \, a^{2} d^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (6 \, b^{2} d^{2} x^{3} + 16 \, b^{2} c d x^{2} - 32 \, a b c d + 3 \, {\left (4 \, b^{2} c^{2} - 3 \, a b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, b^{3}}, \frac {3 \, {\left (4 \, a b c^{2} - 3 \, a^{2} d^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (6 \, b^{2} d^{2} x^{3} + 16 \, b^{2} c d x^{2} - 32 \, a b c d + 3 \, {\left (4 \, b^{2} c^{2} - 3 \, a b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{24 \, b^{3}}\right ] \] Input:

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

Output:

[-1/48*(3*(4*a*b*c^2 - 3*a^2*d^2)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a) 
*sqrt(b)*x - a) - 2*(6*b^2*d^2*x^3 + 16*b^2*c*d*x^2 - 32*a*b*c*d + 3*(4*b^ 
2*c^2 - 3*a*b*d^2)*x)*sqrt(b*x^2 + a))/b^3, 1/24*(3*(4*a*b*c^2 - 3*a^2*d^2 
)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (6*b^2*d^2*x^3 + 16*b^2*c* 
d*x^2 - 32*a*b*c*d + 3*(4*b^2*c^2 - 3*a*b*d^2)*x)*sqrt(b*x^2 + a))/b^3]
 

Sympy [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.11 \[ \int \frac {x^2 (c+d x)^2}{\sqrt {a+b x^2}} \, dx=\begin {cases} - \frac {a \left (- \frac {3 a d^{2}}{4 b} + c^{2}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2 b} + \sqrt {a + b x^{2}} \left (- \frac {4 a c d}{3 b^{2}} + \frac {2 c d x^{2}}{3 b} + \frac {d^{2} x^{3}}{4 b} + \frac {x \left (- \frac {3 a d^{2}}{4 b} + c^{2}\right )}{2 b}\right ) & \text {for}\: b \neq 0 \\\frac {\frac {c^{2} x^{3}}{3} + \frac {c d x^{4}}{2} + \frac {d^{2} x^{5}}{5}}{\sqrt {a}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((-a*(-3*a*d**2/(4*b) + c**2)*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))/(2*b) + 
 sqrt(a + b*x**2)*(-4*a*c*d/(3*b**2) + 2*c*d*x**2/(3*b) + d**2*x**3/(4*b) 
+ x*(-3*a*d**2/(4*b) + c**2)/(2*b)), Ne(b, 0)), ((c**2*x**3/3 + c*d*x**4/2 
 + d**2*x**5/5)/sqrt(a), True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.96 \[ \int \frac {x^2 (c+d x)^2}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} d^{2} x^{3}}{4 \, b} + \frac {2 \, \sqrt {b x^{2} + a} c d x^{2}}{3 \, b} + \frac {\sqrt {b x^{2} + a} c^{2} x}{2 \, b} - \frac {3 \, \sqrt {b x^{2} + a} a d^{2} x}{8 \, b^{2}} - \frac {a c^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}}} + \frac {3 \, a^{2} d^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} - \frac {4 \, \sqrt {b x^{2} + a} a c d}{3 \, b^{2}} \] Input:

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

Output:

1/4*sqrt(b*x^2 + a)*d^2*x^3/b + 2/3*sqrt(b*x^2 + a)*c*d*x^2/b + 1/2*sqrt(b 
*x^2 + a)*c^2*x/b - 3/8*sqrt(b*x^2 + a)*a*d^2*x/b^2 - 1/2*a*c^2*arcsinh(b* 
x/sqrt(a*b))/b^(3/2) + 3/8*a^2*d^2*arcsinh(b*x/sqrt(a*b))/b^(5/2) - 4/3*sq 
rt(b*x^2 + a)*a*c*d/b^2
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.77 \[ \int \frac {x^2 (c+d x)^2}{\sqrt {a+b x^2}} \, dx=\frac {1}{24} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left (\frac {3 \, d^{2} x}{b} + \frac {8 \, c d}{b}\right )} x + \frac {3 \, {\left (4 \, b^{3} c^{2} - 3 \, a b^{2} d^{2}\right )}}{b^{4}}\right )} x - \frac {32 \, a c d}{b^{2}}\right )} + \frac {{\left (4 \, a b c^{2} - 3 \, a^{2} d^{2}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {5}{2}}} \] Input:

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

Output:

1/24*sqrt(b*x^2 + a)*((2*(3*d^2*x/b + 8*c*d/b)*x + 3*(4*b^3*c^2 - 3*a*b^2* 
d^2)/b^4)*x - 32*a*c*d/b^2) + 1/8*(4*a*b*c^2 - 3*a^2*d^2)*log(abs(-sqrt(b) 
*x + sqrt(b*x^2 + a)))/b^(5/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.07 \[ \int \frac {x^2 (c+d x)^2}{\sqrt {a+b x^2}} \, dx=\frac {-32 \sqrt {b \,x^{2}+a}\, a b c d -9 \sqrt {b \,x^{2}+a}\, a b \,d^{2} x +12 \sqrt {b \,x^{2}+a}\, b^{2} c^{2} x +16 \sqrt {b \,x^{2}+a}\, b^{2} c d \,x^{2}+6 \sqrt {b \,x^{2}+a}\, b^{2} d^{2} x^{3}+9 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} d^{2}-12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b \,c^{2}}{24 b^{3}} \] Input:

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

Output:

( - 32*sqrt(a + b*x**2)*a*b*c*d - 9*sqrt(a + b*x**2)*a*b*d**2*x + 12*sqrt( 
a + b*x**2)*b**2*c**2*x + 16*sqrt(a + b*x**2)*b**2*c*d*x**2 + 6*sqrt(a + b 
*x**2)*b**2*d**2*x**3 + 9*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt( 
a))*a**2*d**2 - 12*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b 
*c**2)/(24*b**3)