3.502 \(\int (e x)^m (A+B x) (a+c x^2)^p \, dx\)

Optimal. Leaf size=135 \[ \frac{A (e x)^{m+1} \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{m+1}{2},-p;\frac{m+3}{2};-\frac{c x^2}{a}\right )}{e (m+1)}+\frac{B (e x)^{m+2} \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{m+2}{2},-p;\frac{m+4}{2};-\frac{c x^2}{a}\right )}{e^2 (m+2)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0552938, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {808, 365, 364} \[ \frac{A (e x)^{m+1} \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{m+1}{2},-p;\frac{m+3}{2};-\frac{c x^2}{a}\right )}{e (m+1)}+\frac{B (e x)^{m+2} \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{m+2}{2},-p;\frac{m+4}{2};-\frac{c x^2}{a}\right )}{e^2 (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 808

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]

Rule 365

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 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

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

Mathematica [A]  time = 0.0383302, size = 106, normalized size = 0.79 \[ \frac{x (e x)^m \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} \left (A (m+2) \, _2F_1\left (\frac{m+1}{2},-p;\frac{m+3}{2};-\frac{c x^2}{a}\right )+B (m+1) x \, _2F_1\left (\frac{m}{2}+1,-p;\frac{m}{2}+2;-\frac{c x^2}{a}\right )\right )}{(m+1) (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(e*x)^m*(a + c*x^2)^p*(B*(1 + m)*x*Hypergeometric2F1[1 + m/2, -p, 2 + m/2, -((c*x^2)/a)] + A*(2 + m)*Hyperg
eometric2F1[(1 + m)/2, -p, (3 + m)/2, -((c*x^2)/a)]))/((1 + m)*(2 + m)*(1 + (c*x^2)/a)^p)

________________________________________________________________________________________

Maple [F]  time = 0.068, size = 0, normalized size = 0. \begin{align*} \int \left ( ex \right ) ^{m} \left ( Bx+A \right ) \left ( c{x}^{2}+a \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B x + A\right )}{\left (c x^{2} + a\right )}^{p} \left (e x\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B x + A\right )}{\left (c x^{2} + a\right )}^{p} \left (e x\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [C]  time = 142.652, size = 109, normalized size = 0.81 \begin{align*} \frac{A a^{p} e^{m} x x^{m} \Gamma \left (\frac{m}{2} + \frac{1}{2}\right ){{}_{2}F_{1}\left (\begin{matrix} - p, \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{B a^{p} e^{m} x^{2} x^{m} \Gamma \left (\frac{m}{2} + 1\right ){{}_{2}F_{1}\left (\begin{matrix} - p, \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B x + A\right )}{\left (c x^{2} + a\right )}^{p} \left (e x\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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