\(\int \frac {(e x)^m (A+B x)}{\sqrt {a+c x^2}} \, dx\) [495]

Optimal result

Integrand size = 22, antiderivative size = 139 \[ \int \frac {(e x)^m (A+B x)}{\sqrt {a+c x^2}} \, dx=\frac {A (e x)^{1+m} \sqrt {1+\frac {c x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {c x^2}{a}\right )}{e (1+m) \sqrt {a+c x^2}}+\frac {B (e x)^{2+m} \sqrt {1+\frac {c x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},-\frac {c x^2}{a}\right )}{e^2 (2+m) \sqrt {a+c x^2}} \]

[Out]

A*(e*x)^(1+m)*hypergeom([1/2, 1/2+1/2*m],[3/2+1/2*m],-c*x^2/a)*(1+c*x^2/a)^(1/2)/e/(1+m)/(c*x^2+a)^(1/2)+B*(e*
x)^(2+m)*hypergeom([1/2, 1+1/2*m],[2+1/2*m],-c*x^2/a)*(1+c*x^2/a)^(1/2)/e^2/(2+m)/(c*x^2+a)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {822, 372, 371} \[ \int \frac {(e x)^m (A+B x)}{\sqrt {a+c x^2}} \, dx=\frac {A \sqrt {\frac {c x^2}{a}+1} (e x)^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},-\frac {c x^2}{a}\right )}{e (m+1) \sqrt {a+c x^2}}+\frac {B \sqrt {\frac {c x^2}{a}+1} (e x)^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},-\frac {c x^2}{a}\right )}{e^2 (m+2) \sqrt {a+c x^2}} \]

[In]

Int[((e*x)^m*(A + B*x))/Sqrt[a + c*x^2],x]

[Out]

(A*(e*x)^(1 + m)*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, -((c*x^2)/a)])/(e*(1 + m)*Sq
rt[a + c*x^2]) + (B*(e*x)^(2 + m)*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, -((c*x^2)/a
)])/(e^2*(2 + m)*Sqrt[a + c*x^2])

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/
(1 + b*(x^n/a))^FracPart[p]), Int[(c*x)^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 822

Int[((e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[f, Int[(e*x)^m*(a + c*
x^2)^p, x], x] + Dist[g/e, Int[(e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, p}, x] &&  !Ration
alQ[m] &&  !IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = A \int \frac {(e x)^m}{\sqrt {a+c x^2}} \, dx+\frac {B \int \frac {(e x)^{1+m}}{\sqrt {a+c x^2}} \, dx}{e} \\ & = \frac {\left (A \sqrt {1+\frac {c x^2}{a}}\right ) \int \frac {(e x)^m}{\sqrt {1+\frac {c x^2}{a}}} \, dx}{\sqrt {a+c x^2}}+\frac {\left (B \sqrt {1+\frac {c x^2}{a}}\right ) \int \frac {(e x)^{1+m}}{\sqrt {1+\frac {c x^2}{a}}} \, dx}{e \sqrt {a+c x^2}} \\ & = \frac {A (e x)^{1+m} \sqrt {1+\frac {c x^2}{a}} \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};-\frac {c x^2}{a}\right )}{e (1+m) \sqrt {a+c x^2}}+\frac {B (e x)^{2+m} \sqrt {1+\frac {c x^2}{a}} \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};-\frac {c x^2}{a}\right )}{e^2 (2+m) \sqrt {a+c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.78 \[ \int \frac {(e x)^m (A+B x)}{\sqrt {a+c x^2}} \, dx=\frac {x (e x)^m \sqrt {1+\frac {c x^2}{a}} \left (B (1+m) x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+\frac {m}{2},2+\frac {m}{2},-\frac {c x^2}{a}\right )+A (2+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {c x^2}{a}\right )\right )}{(1+m) (2+m) \sqrt {a+c x^2}} \]

[In]

Integrate[((e*x)^m*(A + B*x))/Sqrt[a + c*x^2],x]

[Out]

(x*(e*x)^m*Sqrt[1 + (c*x^2)/a]*(B*(1 + m)*x*Hypergeometric2F1[1/2, 1 + m/2, 2 + m/2, -((c*x^2)/a)] + A*(2 + m)
*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, -((c*x^2)/a)]))/((1 + m)*(2 + m)*Sqrt[a + c*x^2])

Maple [F]

\[\int \frac {\left (e x \right )^{m} \left (B x +A \right )}{\sqrt {c \,x^{2}+a}}d x\]

[In]

int((e*x)^m*(B*x+A)/(c*x^2+a)^(1/2),x)

[Out]

int((e*x)^m*(B*x+A)/(c*x^2+a)^(1/2),x)

Fricas [F]

\[ \int \frac {(e x)^m (A+B x)}{\sqrt {a+c x^2}} \, dx=\int { \frac {{\left (B x + A\right )} \left (e x\right )^{m}}{\sqrt {c x^{2} + a}} \,d x } \]

[In]

integrate((e*x)^m*(B*x+A)/(c*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

integral((B*x + A)*(e*x)^m/sqrt(c*x^2 + a), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.45 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.79 \[ \int \frac {(e x)^m (A+B x)}{\sqrt {a+c x^2}} \, dx=\frac {A e^{m} x^{m + 1} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} + \frac {1}{2} \\ \frac {m}{2} + \frac {3}{2} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {B e^{m} x^{m + 2} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} + 1 \\ \frac {m}{2} + 2 \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {m}{2} + 2\right )} \]

[In]

integrate((e*x)**m*(B*x+A)/(c*x**2+a)**(1/2),x)

[Out]

A*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*sqrt(a)*
gamma(m/2 + 3/2)) + B*e**m*x**(m + 2)*gamma(m/2 + 1)*hyper((1/2, m/2 + 1), (m/2 + 2,), c*x**2*exp_polar(I*pi)/
a)/(2*sqrt(a)*gamma(m/2 + 2))

Maxima [F]

\[ \int \frac {(e x)^m (A+B x)}{\sqrt {a+c x^2}} \, dx=\int { \frac {{\left (B x + A\right )} \left (e x\right )^{m}}{\sqrt {c x^{2} + a}} \,d x } \]

[In]

integrate((e*x)^m*(B*x+A)/(c*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)*(e*x)^m/sqrt(c*x^2 + a), x)

Giac [F]

\[ \int \frac {(e x)^m (A+B x)}{\sqrt {a+c x^2}} \, dx=\int { \frac {{\left (B x + A\right )} \left (e x\right )^{m}}{\sqrt {c x^{2} + a}} \,d x } \]

[In]

integrate((e*x)^m*(B*x+A)/(c*x^2+a)^(1/2),x, algorithm="giac")

[Out]

integrate((B*x + A)*(e*x)^m/sqrt(c*x^2 + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^m (A+B x)}{\sqrt {a+c x^2}} \, dx=\int \frac {{\left (e\,x\right )}^m\,\left (A+B\,x\right )}{\sqrt {c\,x^2+a}} \,d x \]

[In]

int(((e*x)^m*(A + B*x))/(a + c*x^2)^(1/2),x)

[Out]

int(((e*x)^m*(A + B*x))/(a + c*x^2)^(1/2), x)