Integrand size = 15, antiderivative size = 91 \[ \int x^2 \left (a+b x^2\right )^{3/2} \, dx=\frac {a^2 x \sqrt {a+b x^2}}{16 b}+\frac {1}{8} a x^3 \sqrt {a+b x^2}+\frac {1}{6} x^3 \left (a+b x^2\right )^{3/2}-\frac {a^3 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{3/2}} \]
1/6*x^3*(b*x^2+a)^(3/2)-1/16*a^3*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(3/2 )+1/16*a^2*x*(b*x^2+a)^(1/2)/b+1/8*a*x^3*(b*x^2+a)^(1/2)
Time = 0.25 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.90 \[ \int x^2 \left (a+b x^2\right )^{3/2} \, dx=\frac {x \sqrt {a+b x^2} \left (3 a^2+14 a b x^2+8 b^2 x^4\right )}{48 b}-\frac {a^3 \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{8 b^{3/2}} \]
(x*Sqrt[a + b*x^2]*(3*a^2 + 14*a*b*x^2 + 8*b^2*x^4))/(48*b) - (a^3*ArcTanh [(Sqrt[b]*x)/(-Sqrt[a] + Sqrt[a + b*x^2])])/(8*b^(3/2))
Time = 0.19 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {248, 248, 262, 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.
\(\displaystyle \int x^2 \left (a+b x^2\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 248 |
\(\displaystyle \frac {1}{2} a \int x^2 \sqrt {b x^2+a}dx+\frac {1}{6} x^3 \left (a+b x^2\right )^{3/2}\) |
\(\Big \downarrow \) 248 |
\(\displaystyle \frac {1}{2} a \left (\frac {1}{4} a \int \frac {x^2}{\sqrt {b x^2+a}}dx+\frac {1}{4} x^3 \sqrt {a+b x^2}\right )+\frac {1}{6} x^3 \left (a+b x^2\right )^{3/2}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {1}{2} a \left (\frac {1}{4} a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}\right )+\frac {1}{4} x^3 \sqrt {a+b x^2}\right )+\frac {1}{6} x^3 \left (a+b x^2\right )^{3/2}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {1}{2} a \left (\frac {1}{4} a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}\right )+\frac {1}{4} x^3 \sqrt {a+b x^2}\right )+\frac {1}{6} x^3 \left (a+b x^2\right )^{3/2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} a \left (\frac {1}{4} a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{3/2}}\right )+\frac {1}{4} x^3 \sqrt {a+b x^2}\right )+\frac {1}{6} x^3 \left (a+b x^2\right )^{3/2}\) |
(x^3*(a + b*x^2)^(3/2))/6 + (a*((x^3*Sqrt[a + b*x^2])/4 + (a*((x*Sqrt[a + b*x^2])/(2*b) - (a*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*b^(3/2))))/4)) /2
3.4.79.3.1 Defintions of rubi rules used
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_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 2*p + 1))), x] + Simp[2*a*(p/(m + 2*p + 1)) Int[(c*x)^m*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && GtQ[ p, 0] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Time = 1.86 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.68
method | result | size |
risch | \(\frac {x \left (8 b^{2} x^{4}+14 a b \,x^{2}+3 a^{2}\right ) \sqrt {b \,x^{2}+a}}{48 b}-\frac {a^{3} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {3}{2}}}\) | \(62\) |
default | \(\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\) | \(74\) |
pseudoelliptic | \(\frac {8 \sqrt {b \,x^{2}+a}\, b^{\frac {5}{2}} x^{5}+14 a \,b^{\frac {3}{2}} x^{3} \sqrt {b \,x^{2}+a}+3 a^{2} x \sqrt {b \,x^{2}+a}\, \sqrt {b}-3 \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right ) a^{3}}{48 b^{\frac {3}{2}}}\) | \(82\) |
1/48*x*(8*b^2*x^4+14*a*b*x^2+3*a^2)*(b*x^2+a)^(1/2)/b-1/16/b^(3/2)*a^3*ln( x*b^(1/2)+(b*x^2+a)^(1/2))
Time = 0.28 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.59 \[ \int x^2 \left (a+b x^2\right )^{3/2} \, dx=\left [\frac {3 \, a^{3} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (8 \, b^{3} x^{5} + 14 \, a b^{2} x^{3} + 3 \, a^{2} b x\right )} \sqrt {b x^{2} + a}}{96 \, b^{2}}, \frac {3 \, a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (8 \, b^{3} x^{5} + 14 \, a b^{2} x^{3} + 3 \, a^{2} b x\right )} \sqrt {b x^{2} + a}}{48 \, b^{2}}\right ] \]
[1/96*(3*a^3*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*( 8*b^3*x^5 + 14*a*b^2*x^3 + 3*a^2*b*x)*sqrt(b*x^2 + a))/b^2, 1/48*(3*a^3*sq rt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (8*b^3*x^5 + 14*a*b^2*x^3 + 3* a^2*b*x)*sqrt(b*x^2 + a))/b^2]
Time = 3.48 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.31 \[ \int x^2 \left (a+b x^2\right )^{3/2} \, dx=\frac {a^{\frac {5}{2}} x}{16 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {17 a^{\frac {3}{2}} x^{3}}{48 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {11 \sqrt {a} b x^{5}}{24 \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{16 b^{\frac {3}{2}}} + \frac {b^{2} x^{7}}{6 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \]
a**(5/2)*x/(16*b*sqrt(1 + b*x**2/a)) + 17*a**(3/2)*x**3/(48*sqrt(1 + b*x** 2/a)) + 11*sqrt(a)*b*x**5/(24*sqrt(1 + b*x**2/a)) - a**3*asinh(sqrt(b)*x/s qrt(a))/(16*b**(3/2)) + b**2*x**7/(6*sqrt(a)*sqrt(1 + b*x**2/a))
Time = 0.20 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.74 \[ \int x^2 \left (a+b x^2\right )^{3/2} \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} x}{6 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} a x}{24 \, b} - \frac {\sqrt {b x^{2} + a} a^{2} x}{16 \, b} - \frac {a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {3}{2}}} \]
1/6*(b*x^2 + a)^(5/2)*x/b - 1/24*(b*x^2 + a)^(3/2)*a*x/b - 1/16*sqrt(b*x^2 + a)*a^2*x/b - 1/16*a^3*arcsinh(b*x/sqrt(a*b))/b^(3/2)
Time = 0.29 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.69 \[ \int x^2 \left (a+b x^2\right )^{3/2} \, dx=\frac {1}{48} \, {\left (2 \, {\left (4 \, b x^{2} + 7 \, a\right )} x^{2} + \frac {3 \, a^{2}}{b}\right )} \sqrt {b x^{2} + a} x + \frac {a^{3} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {3}{2}}} \]
1/48*(2*(4*b*x^2 + 7*a)*x^2 + 3*a^2/b)*sqrt(b*x^2 + a)*x + 1/16*a^3*log(ab s(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(3/2)
Timed out. \[ \int x^2 \left (a+b x^2\right )^{3/2} \, dx=\int x^2\,{\left (b\,x^2+a\right )}^{3/2} \,d x \]