Integrand size = 22, antiderivative size = 141 \[ \int (e x)^m (A+B x) \left (a+c x^2\right )^{3/2} \, dx=\frac {a A (e x)^{1+m} \sqrt {a+c x^2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {c x^2}{a}\right )}{e (1+m) \sqrt {1+\frac {c x^2}{a}}}+\frac {a B (e x)^{2+m} \sqrt {a+c x^2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {2+m}{2},\frac {4+m}{2},-\frac {c x^2}{a}\right )}{e^2 (2+m) \sqrt {1+\frac {c x^2}{a}}} \]
a*A*(e*x)^(1+m)*hypergeom([-3/2, 1/2+1/2*m],[3/2+1/2*m],-c*x^2/a)*(c*x^2+a )^(1/2)/e/(1+m)/(1+c*x^2/a)^(1/2)+a*B*(e*x)^(2+m)*hypergeom([-3/2, 1+1/2*m ],[2+1/2*m],-c*x^2/a)*(c*x^2+a)^(1/2)/e^2/(2+m)/(1+c*x^2/a)^(1/2)
Time = 0.29 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.77 \[ \int (e x)^m (A+B x) \left (a+c x^2\right )^{3/2} \, dx=\frac {a x (e x)^m \sqrt {a+c x^2} \left (B (1+m) x \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1+\frac {m}{2},2+\frac {m}{2},-\frac {c x^2}{a}\right )+A (2+m) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {c x^2}{a}\right )\right )}{(1+m) (2+m) \sqrt {1+\frac {c x^2}{a}}} \]
(a*x*(e*x)^m*Sqrt[a + c*x^2]*(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)]))/((1 + m)*(2 + m)*Sqrt[1 + (c*x^2)/a])
Time = 0.25 (sec) , antiderivative size = 141, 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.
\(\displaystyle \int \left (a+c x^2\right )^{3/2} (A+B x) (e x)^m \, dx\) |
\(\Big \downarrow \) 557 |
\(\displaystyle A \int (e x)^m \left (c x^2+a\right )^{3/2}dx+\frac {B \int (e x)^{m+1} \left (c x^2+a\right )^{3/2}dx}{e}\) |
\(\Big \downarrow \) 279 |
\(\displaystyle \frac {a A \sqrt {a+c x^2} \int (e x)^m \left (\frac {c x^2}{a}+1\right )^{3/2}dx}{\sqrt {\frac {c x^2}{a}+1}}+\frac {a B \sqrt {a+c x^2} \int (e x)^{m+1} \left (\frac {c x^2}{a}+1\right )^{3/2}dx}{e \sqrt {\frac {c x^2}{a}+1}}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {a A \sqrt {a+c x^2} (e x)^{m+1} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {m+1}{2},\frac {m+3}{2},-\frac {c x^2}{a}\right )}{e (m+1) \sqrt {\frac {c x^2}{a}+1}}+\frac {a B \sqrt {a+c x^2} (e x)^{m+2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {m+2}{2},\frac {m+4}{2},-\frac {c x^2}{a}\right )}{e^2 (m+2) \sqrt {\frac {c x^2}{a}+1}}\) |
(a*A*(e*x)^(1 + m)*Sqrt[a + c*x^2]*Hypergeometric2F1[-3/2, (1 + m)/2, (3 + m)/2, -((c*x^2)/a)])/(e*(1 + m)*Sqrt[1 + (c*x^2)/a]) + (a*B*(e*x)^(2 + m) *Sqrt[a + c*x^2]*Hypergeometric2F1[-3/2, (2 + m)/2, (4 + m)/2, -((c*x^2)/a )])/(e^2*(2 + m)*Sqrt[1 + (c*x^2)/a])
3.5.93.3.1 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]
\[\int \left (e x \right )^{m} \left (B x +A \right ) \left (c \,x^{2}+a \right )^{\frac {3}{2}}d x\]
\[ \int (e x)^m (A+B x) \left (a+c x^2\right )^{3/2} \, dx=\int { {\left (c x^{2} + a\right )}^{\frac {3}{2}} {\left (B x + A\right )} \left (e x\right )^{m} \,d x } \]
Result contains complex when optimal does not.
Time = 3.76 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.65 \[ \int (e x)^m (A+B x) \left (a+c x^2\right )^{3/2} \, dx=\frac {A a^{\frac {3}{2}} e^{m} x^{m + 1} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + \frac {1}{2} \\ \frac {m}{2} + \frac {3}{2} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {A \sqrt {a} c e^{m} x^{m + 3} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + \frac {3}{2} \\ \frac {m}{2} + \frac {5}{2} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {B a^{\frac {3}{2}} e^{m} x^{m + 2} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + 1 \\ \frac {m}{2} + 2 \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + 2\right )} + \frac {B \sqrt {a} c e^{m} x^{m + 4} \Gamma \left (\frac {m}{2} + 2\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + 2 \\ \frac {m}{2} + 3 \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + 3\right )} \]
A*a**(3/2)*e**m*x**(m + 1)*gamma(m/2 + 1/2)*hyper((-1/2, m/2 + 1/2), (m/2 + 3/2,), c*x**2*exp_polar(I*pi)/a)/(2*gamma(m/2 + 3/2)) + A*sqrt(a)*c*e**m *x**(m + 3)*gamma(m/2 + 3/2)*hyper((-1/2, m/2 + 3/2), (m/2 + 5/2,), c*x**2 *exp_polar(I*pi)/a)/(2*gamma(m/2 + 5/2)) + B*a**(3/2)*e**m*x**(m + 2)*gamm a(m/2 + 1)*hyper((-1/2, m/2 + 1), (m/2 + 2,), c*x**2*exp_polar(I*pi)/a)/(2 *gamma(m/2 + 2)) + B*sqrt(a)*c*e**m*x**(m + 4)*gamma(m/2 + 2)*hyper((-1/2, m/2 + 2), (m/2 + 3,), c*x**2*exp_polar(I*pi)/a)/(2*gamma(m/2 + 3))
\[ \int (e x)^m (A+B x) \left (a+c x^2\right )^{3/2} \, dx=\int { {\left (c x^{2} + a\right )}^{\frac {3}{2}} {\left (B x + A\right )} \left (e x\right )^{m} \,d x } \]
\[ \int (e x)^m (A+B x) \left (a+c x^2\right )^{3/2} \, dx=\int { {\left (c x^{2} + a\right )}^{\frac {3}{2}} {\left (B x + A\right )} \left (e x\right )^{m} \,d x } \]
Timed out. \[ \int (e x)^m (A+B x) \left (a+c x^2\right )^{3/2} \, dx=\int {\left (e\,x\right )}^m\,{\left (c\,x^2+a\right )}^{3/2}\,\left (A+B\,x\right ) \,d x \]