Integrand size = 21, antiderivative size = 152 \[ \int x^2 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\frac {2 a x \left (c \sqrt {a+b x^2}\right )^{3/2}}{15 b}+\frac {2}{9} x^3 \left (c \sqrt {a+b x^2}\right )^{3/2}-\frac {4 a^2 x \left (c \sqrt {a+b x^2}\right )^{3/2}}{15 b \left (a+b x^2\right )}+\frac {4 a^{3/2} \left (c \sqrt {a+b x^2}\right )^{3/2} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{15 b^{3/2} \left (1+\frac {b x^2}{a}\right )^{3/4}} \]
2/15*a*x*(c*(b*x^2+a)^(1/2))^(3/2)/b+2/9*x^3*(c*(b*x^2+a)^(1/2))^(3/2)-4/1 5*a^2*x*(c*(b*x^2+a)^(1/2))^(3/2)/b/(b*x^2+a)+4/15*a^(3/2)*(cos(1/2*arctan (x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arctan(x*b^(1/2)/a^(1/2)))*EllipticE (sin(1/2*arctan(x*b^(1/2)/a^(1/2))),2^(1/2))*(c*(b*x^2+a)^(1/2))^(3/2)/b^( 3/2)/(1+b*x^2/a)^(3/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.05 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.45 \[ \int x^2 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\frac {2 x \left (c \sqrt {a+b x^2}\right )^{3/2} \left (a+b x^2-\frac {a \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )}{\left (1+\frac {b x^2}{a}\right )^{3/4}}\right )}{9 b} \]
(2*x*(c*Sqrt[a + b*x^2])^(3/2)*(a + b*x^2 - (a*Hypergeometric2F1[-3/4, 1/2 , 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^(3/4)))/(9*b)
Time = 0.24 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2045, 248, 262, 225, 212}
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.
\(\displaystyle \int x^2 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 2045 |
\(\displaystyle \frac {\left (c \sqrt {a+b x^2}\right )^{3/2} \int x^2 \left (\frac {b x^2}{a}+1\right )^{3/4}dx}{\left (\frac {b x^2}{a}+1\right )^{3/4}}\) |
\(\Big \downarrow \) 248 |
\(\displaystyle \frac {\left (c \sqrt {a+b x^2}\right )^{3/2} \left (\frac {1}{3} \int \frac {x^2}{\sqrt [4]{\frac {b x^2}{a}+1}}dx+\frac {2}{9} x^3 \left (\frac {b x^2}{a}+1\right )^{3/4}\right )}{\left (\frac {b x^2}{a}+1\right )^{3/4}}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {\left (c \sqrt {a+b x^2}\right )^{3/2} \left (\frac {1}{3} \left (\frac {2 a x \left (\frac {b x^2}{a}+1\right )^{3/4}}{5 b}-\frac {2 a \int \frac {1}{\sqrt [4]{\frac {b x^2}{a}+1}}dx}{5 b}\right )+\frac {2}{9} x^3 \left (\frac {b x^2}{a}+1\right )^{3/4}\right )}{\left (\frac {b x^2}{a}+1\right )^{3/4}}\) |
\(\Big \downarrow \) 225 |
\(\displaystyle \frac {\left (c \sqrt {a+b x^2}\right )^{3/2} \left (\frac {1}{3} \left (\frac {2 a x \left (\frac {b x^2}{a}+1\right )^{3/4}}{5 b}-\frac {2 a \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{5/4}}dx\right )}{5 b}\right )+\frac {2}{9} x^3 \left (\frac {b x^2}{a}+1\right )^{3/4}\right )}{\left (\frac {b x^2}{a}+1\right )^{3/4}}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle \frac {\left (c \sqrt {a+b x^2}\right )^{3/2} \left (\frac {1}{3} \left (\frac {2 a x \left (\frac {b x^2}{a}+1\right )^{3/4}}{5 b}-\frac {2 a \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\frac {2 \sqrt {a} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b}}\right )}{5 b}\right )+\frac {2}{9} x^3 \left (\frac {b x^2}{a}+1\right )^{3/4}\right )}{\left (\frac {b x^2}{a}+1\right )^{3/4}}\) |
((c*Sqrt[a + b*x^2])^(3/2)*((2*x^3*(1 + (b*x^2)/a)^(3/4))/9 + ((2*a*x*(1 + (b*x^2)/a)^(3/4))/(5*b) - (2*a*((2*x)/(1 + (b*x^2)/a)^(1/4) - (2*Sqrt[a]* EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/Sqrt[b]))/(5*b))/3))/(1 + (b* x^2)/a)^(3/4)
3.3.55.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) )*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a , 0] && PosQ[b/a]
Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[2*(x/(a + b*x^2)^(1/4)) , x] - Simp[a Int[1/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b}, x] && GtQ[ a, 0] && PosQ[b/a]
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]
Int[(u_.)*((c_.)*((a_) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Simp[Si mp[(c*(a + b*x^n)^q)^p/(1 + b*(x^n/a))^(p*q)] Int[u*(1 + b*(x^n/a))^(p*q) , x], x] /; FreeQ[{a, b, c, n, p, q}, x] && !GeQ[a, 0]
\[\int x^{2} \left (c \sqrt {b \,x^{2}+a}\right )^{\frac {3}{2}}d x\]
\[ \int x^2 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\int { \left (\sqrt {b x^{2} + a} c\right )^{\frac {3}{2}} x^{2} \,d x } \]
\[ \int x^2 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\int x^{2} \left (c \sqrt {a + b x^{2}}\right )^{\frac {3}{2}}\, dx \]
\[ \int x^2 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\int { \left (\sqrt {b x^{2} + a} c\right )^{\frac {3}{2}} x^{2} \,d x } \]
\[ \int x^2 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\int { \left (\sqrt {b x^{2} + a} c\right )^{\frac {3}{2}} x^{2} \,d x } \]
Timed out. \[ \int x^2 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\int x^2\,{\left (c\,\sqrt {b\,x^2+a}\right )}^{3/2} \,d x \]