Integrand size = 20, antiderivative size = 88 \[ \int \frac {x^2 (A+B x)}{\sqrt {a+b x^2}} \, dx=-\frac {a B \sqrt {a+b x^2}}{b^2}+\frac {A x \sqrt {a+b x^2}}{2 b}+\frac {B \left (a+b x^2\right )^{3/2}}{3 b^2}-\frac {a A \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{3/2}} \] Output:
-a*B*(b*x^2+a)^(1/2)/b^2+1/2*A*x*(b*x^2+a)^(1/2)/b+1/3*B*(b*x^2+a)^(3/2)/b ^2-1/2*a*A*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(3/2)
Time = 0.33 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.84 \[ \int \frac {x^2 (A+B x)}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {a+b x^2} \left (-4 a B+3 A b x+2 b B x^2\right )}{6 b^2}-\frac {a A \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{b^{3/2}} \] Input:
Integrate[(x^2*(A + B*x))/Sqrt[a + b*x^2],x]
Output:
(Sqrt[a + b*x^2]*(-4*a*B + 3*A*b*x + 2*b*B*x^2))/(6*b^2) - (a*A*ArcTanh[(S qrt[b]*x)/(-Sqrt[a] + Sqrt[a + b*x^2])])/b^(3/2)
Time = 0.37 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {533, 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.
\(\displaystyle \int \frac {x^2 (A+B x)}{\sqrt {a+b x^2}} \, dx\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {B x^2 \sqrt {a+b x^2}}{3 b}-\frac {\int \frac {x (2 a B-3 A b x)}{\sqrt {b x^2+a}}dx}{3 b}\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {B x^2 \sqrt {a+b x^2}}{3 b}-\frac {-\frac {\int -\frac {a b (3 A+4 B x)}{\sqrt {b x^2+a}}dx}{2 b}-\frac {3}{2} A x \sqrt {a+b x^2}}{3 b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {B x^2 \sqrt {a+b x^2}}{3 b}-\frac {\frac {\int \frac {a b (3 A+4 B x)}{\sqrt {b x^2+a}}dx}{2 b}-\frac {3}{2} A x \sqrt {a+b x^2}}{3 b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {B x^2 \sqrt {a+b x^2}}{3 b}-\frac {\frac {1}{2} a \int \frac {3 A+4 B x}{\sqrt {b x^2+a}}dx-\frac {3}{2} A x \sqrt {a+b x^2}}{3 b}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {B x^2 \sqrt {a+b x^2}}{3 b}-\frac {\frac {1}{2} a \left (3 A \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {4 B \sqrt {a+b x^2}}{b}\right )-\frac {3}{2} A x \sqrt {a+b x^2}}{3 b}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {B x^2 \sqrt {a+b x^2}}{3 b}-\frac {\frac {1}{2} a \left (3 A \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {4 B \sqrt {a+b x^2}}{b}\right )-\frac {3}{2} A x \sqrt {a+b x^2}}{3 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {B x^2 \sqrt {a+b x^2}}{3 b}-\frac {\frac {1}{2} a \left (\frac {3 A \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}+\frac {4 B \sqrt {a+b x^2}}{b}\right )-\frac {3}{2} A x \sqrt {a+b x^2}}{3 b}\) |
Input:
Int[(x^2*(A + B*x))/Sqrt[a + b*x^2],x]
Output:
(B*x^2*Sqrt[a + b*x^2])/(3*b) - ((-3*A*x*Sqrt[a + b*x^2])/2 + (a*((4*B*Sqr t[a + b*x^2])/b + (3*A*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/Sqrt[b]))/2)/ (3*b)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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]
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]
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]
Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.64
method | result | size |
risch | \(\frac {\left (2 B b \,x^{2}+3 A b x -4 B a \right ) \sqrt {b \,x^{2}+a}}{6 b^{2}}-\frac {a A \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\) | \(56\) |
default | \(A \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )+B \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )\) | \(77\) |
Input:
int(x^2*(B*x+A)/(b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/6*(2*B*b*x^2+3*A*b*x-4*B*a)*(b*x^2+a)^(1/2)/b^2-1/2*a/b^(3/2)*A*ln(b^(1/ 2)*x+(b*x^2+a)^(1/2))
Time = 0.10 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.44 \[ \int \frac {x^2 (A+B x)}{\sqrt {a+b x^2}} \, dx=\left [\frac {3 \, A a \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (2 \, B b x^{2} + 3 \, A b x - 4 \, B a\right )} \sqrt {b x^{2} + a}}{12 \, b^{2}}, \frac {3 \, A a \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (2 \, B b x^{2} + 3 \, A b x - 4 \, B a\right )} \sqrt {b x^{2} + a}}{6 \, b^{2}}\right ] \] Input:
integrate(x^2*(B*x+A)/(b*x^2+a)^(1/2),x, algorithm="fricas")
Output:
[1/12*(3*A*a*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*( 2*B*b*x^2 + 3*A*b*x - 4*B*a)*sqrt(b*x^2 + a))/b^2, 1/6*(3*A*a*sqrt(-b)*arc tan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (2*B*b*x^2 + 3*A*b*x - 4*B*a)*sqrt(b*x^2 + a))/b^2]
Time = 0.44 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.16 \[ \int \frac {x^2 (A+B x)}{\sqrt {a+b x^2}} \, dx=\begin {cases} - \frac {A a \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2 b} + \sqrt {a + b x^{2}} \left (\frac {A x}{2 b} - \frac {2 B a}{3 b^{2}} + \frac {B x^{2}}{3 b}\right ) & \text {for}\: b \neq 0 \\\frac {\frac {A x^{3}}{3} + \frac {B x^{4}}{4}}{\sqrt {a}} & \text {otherwise} \end {cases} \] Input:
integrate(x**2*(B*x+A)/(b*x**2+a)**(1/2),x)
Output:
Piecewise((-A*a*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)*(A*x/ (2*b) - 2*B*a/(3*b**2) + B*x**2/(3*b)), Ne(b, 0)), ((A*x**3/3 + B*x**4/4)/ sqrt(a), True))
Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.76 \[ \int \frac {x^2 (A+B x)}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} B x^{2}}{3 \, b} + \frac {\sqrt {b x^{2} + a} A x}{2 \, b} - \frac {A a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}}} - \frac {2 \, \sqrt {b x^{2} + a} B a}{3 \, b^{2}} \] Input:
integrate(x^2*(B*x+A)/(b*x^2+a)^(1/2),x, algorithm="maxima")
Output:
1/3*sqrt(b*x^2 + a)*B*x^2/b + 1/2*sqrt(b*x^2 + a)*A*x/b - 1/2*A*a*arcsinh( b*x/sqrt(a*b))/b^(3/2) - 2/3*sqrt(b*x^2 + a)*B*a/b^2
Time = 0.13 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.69 \[ \int \frac {x^2 (A+B x)}{\sqrt {a+b x^2}} \, dx=\frac {1}{6} \, \sqrt {b x^{2} + a} {\left ({\left (\frac {2 \, B x}{b} + \frac {3 \, A}{b}\right )} x - \frac {4 \, B a}{b^{2}}\right )} + \frac {A a \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {3}{2}}} \] Input:
integrate(x^2*(B*x+A)/(b*x^2+a)^(1/2),x, algorithm="giac")
Output:
1/6*sqrt(b*x^2 + a)*((2*B*x/b + 3*A/b)*x - 4*B*a/b^2) + 1/2*A*a*log(abs(-s qrt(b)*x + sqrt(b*x^2 + a)))/b^(3/2)
Time = 9.73 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.06 \[ \int \frac {x^2 (A+B x)}{\sqrt {a+b x^2}} \, dx=\left \{\begin {array}{cl} \frac {3\,B\,x^4+4\,A\,x^3}{12\,\sqrt {a}} & \text {\ if\ \ }b=0\\ \frac {A\,x\,\sqrt {b\,x^2+a}}{2\,b}-\frac {A\,a\,\ln \left (2\,\sqrt {b}\,x+2\,\sqrt {b\,x^2+a}\right )}{2\,b^{3/2}}-\frac {B\,\sqrt {b\,x^2+a}\,\left (2\,a-b\,x^2\right )}{3\,b^2} & \text {\ if\ \ }b\neq 0 \end {array}\right . \] Input:
int((x^2*(A + B*x))/(a + b*x^2)^(1/2),x)
Output:
piecewise(b == 0, (4*A*x^3 + 3*B*x^4)/(12*a^(1/2)), b ~= 0, - (A*a*log(2*b ^(1/2)*x + 2*(a + b*x^2)^(1/2)))/(2*b^(3/2)) + (A*x*(a + b*x^2)^(1/2))/(2* b) - (B*(a + b*x^2)^(1/2)*(2*a - b*x^2))/(3*b^2))
Time = 0.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.83 \[ \int \frac {x^2 (A+B x)}{\sqrt {a+b x^2}} \, dx=\frac {3 \sqrt {b \,x^{2}+a}\, a b x -4 \sqrt {b \,x^{2}+a}\, a b +2 \sqrt {b \,x^{2}+a}\, b^{2} x^{2}-3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2}}{6 b^{2}} \] Input:
int(x^2*(B*x+A)/(b*x^2+a)^(1/2),x)
Output:
(3*sqrt(a + b*x**2)*a*b*x - 4*sqrt(a + b*x**2)*a*b + 2*sqrt(a + b*x**2)*b* *2*x**2 - 3*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2)/(6*b **2)