Integrand size = 22, antiderivative size = 145 \[ \int \frac {(e x)^m (A+B x)}{\left (a+c x^2\right )^{3/2}} \, dx=\frac {A (e x)^{1+m} \sqrt {1+\frac {c x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {c x^2}{a}\right )}{a e (1+m) \sqrt {a+c x^2}}+\frac {B (e x)^{2+m} \sqrt {1+\frac {c x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {2+m}{2},\frac {4+m}{2},-\frac {c x^2}{a}\right )}{a e^2 (2+m) \sqrt {a+c x^2}} \]
A*(e*x)^(1+m)*hypergeom([3/2, 1/2+1/2*m],[3/2+1/2*m],-c*x^2/a)*(1+c*x^2/a) ^(1/2)/a/e/(1+m)/(c*x^2+a)^(1/2)+B*(e*x)^(2+m)*hypergeom([3/2, 1+1/2*m],[2 +1/2*m],-c*x^2/a)*(1+c*x^2/a)^(1/2)/a/e^2/(2+m)/(c*x^2+a)^(1/2)
Time = 0.33 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.77 \[ \int \frac {(e x)^m (A+B x)}{\left (a+c x^2\right )^{3/2}} \, dx=\frac {x (e x)^m \sqrt {1+\frac {c x^2}{a}} \left (B (1+m) x \operatorname {Hypergeometric2F1}\left (\frac {3}{2},1+\frac {m}{2},2+\frac {m}{2},-\frac {c x^2}{a}\right )+A (2+m) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {c x^2}{a}\right )\right )}{a (1+m) (2+m) \sqrt {a+c x^2}} \]
(x*(e*x)^m*Sqrt[1 + (c*x^2)/a]*(B*(1 + m)*x*Hypergeometric2F1[3/2, 1 + m/2 , 2 + m/2, -((c*x^2)/a)] + A*(2 + m)*Hypergeometric2F1[3/2, (1 + m)/2, (3 + m)/2, -((c*x^2)/a)]))/(a*(1 + m)*(2 + m)*Sqrt[a + c*x^2])
Time = 0.24 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {557, 279, 278}
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.
\(\displaystyle \int \frac {(A+B x) (e x)^m}{\left (a+c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 557 |
\(\displaystyle A \int \frac {(e x)^m}{\left (c x^2+a\right )^{3/2}}dx+\frac {B \int \frac {(e x)^{m+1}}{\left (c x^2+a\right )^{3/2}}dx}{e}\) |
\(\Big \downarrow \) 279 |
\(\displaystyle \frac {A \sqrt {\frac {c x^2}{a}+1} \int \frac {(e x)^m}{\left (\frac {c x^2}{a}+1\right )^{3/2}}dx}{a \sqrt {a+c x^2}}+\frac {B \sqrt {\frac {c x^2}{a}+1} \int \frac {(e x)^{m+1}}{\left (\frac {c x^2}{a}+1\right )^{3/2}}dx}{a e \sqrt {a+c x^2}}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {A \sqrt {\frac {c x^2}{a}+1} (e x)^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+1}{2},\frac {m+3}{2},-\frac {c x^2}{a}\right )}{a e (m+1) \sqrt {a+c x^2}}+\frac {B \sqrt {\frac {c x^2}{a}+1} (e x)^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+2}{2},\frac {m+4}{2},-\frac {c x^2}{a}\right )}{a e^2 (m+2) \sqrt {a+c x^2}}\) |
(A*(e*x)^(1 + m)*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[3/2, (1 + m)/2, (3 + m)/2, -((c*x^2)/a)])/(a*e*(1 + m)*Sqrt[a + c*x^2]) + (B*(e*x)^(2 + m)*Sq rt[1 + (c*x^2)/a]*Hypergeometric2F1[3/2, (2 + m)/2, (4 + m)/2, -((c*x^2)/a )])/(a*e^2*(2 + m)*Sqrt[a + c*x^2])
3.5.96.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[(c*x)^m* (1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && !(ILtQ[p, 0] || GtQ[a, 0])
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Sym bol] :> Simp[c Int[(e*x)^m*(a + b*x^2)^p, x], x] + Simp[d/e Int[(e*x)^( m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x]
\[\int \frac {\left (e x \right )^{m} \left (B x +A \right )}{\left (c \,x^{2}+a \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(e x)^m (A+B x)}{\left (a+c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (B x + A\right )} \left (e x\right )^{m}}{{\left (c x^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
Result contains complex when optimal does not.
Time = 5.13 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.76 \[ \int \frac {(e x)^m (A+B x)}{\left (a+c x^2\right )^{3/2}} \, dx=\frac {A e^{m} x^{m + 1} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {m}{2} + \frac {1}{2} \\ \frac {m}{2} + \frac {3}{2} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {B e^{m} x^{m + 2} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {m}{2} + 1 \\ \frac {m}{2} + 2 \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (\frac {m}{2} + 2\right )} \]
A*e**m*x**(m + 1)*gamma(m/2 + 1/2)*hyper((3/2, m/2 + 1/2), (m/2 + 3/2,), c *x**2*exp_polar(I*pi)/a)/(2*a**(3/2)*gamma(m/2 + 3/2)) + B*e**m*x**(m + 2) *gamma(m/2 + 1)*hyper((3/2, m/2 + 1), (m/2 + 2,), c*x**2*exp_polar(I*pi)/a )/(2*a**(3/2)*gamma(m/2 + 2))
\[ \int \frac {(e x)^m (A+B x)}{\left (a+c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (B x + A\right )} \left (e x\right )^{m}}{{\left (c x^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(e x)^m (A+B x)}{\left (a+c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (B x + A\right )} \left (e x\right )^{m}}{{\left (c x^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(e x)^m (A+B x)}{\left (a+c x^2\right )^{3/2}} \, dx=\int \frac {{\left (e\,x\right )}^m\,\left (A+B\,x\right )}{{\left (c\,x^2+a\right )}^{3/2}} \,d x \]