3.490 \(\int \frac{(e x)^m (A+B x)}{(a+c x^2)^2} \, dx\)

Optimal. Leaf size=91 \[ \frac{A (e x)^{m+1} \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{c x^2}{a}\right )}{a^2 e (m+1)}+\frac{B (e x)^{m+2} \, _2F_1\left (2,\frac{m+2}{2};\frac{m+4}{2};-\frac{c x^2}{a}\right )}{a^2 e^2 (m+2)} \]

[Out]

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

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Rubi [A]  time = 0.039647, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {808, 364} \[ \frac{A (e x)^{m+1} \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{c x^2}{a}\right )}{a^2 e (m+1)}+\frac{B (e x)^{m+2} \, _2F_1\left (2,\frac{m+2}{2};\frac{m+4}{2};-\frac{c x^2}{a}\right )}{a^2 e^2 (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.026722, size = 82, normalized size = 0.9 \[ \frac{x (e x)^m \left (A (m+2) \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{c x^2}{a}\right )+B (m+1) x \, _2F_1\left (2,\frac{m}{2}+1;\frac{m}{2}+2;-\frac{c x^2}{a}\right )\right )}{a^2 (m+1) (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.036, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex \right ) ^{m} \left ( Bx+A \right ) }{ \left ( c{x}^{2}+a \right ) ^{2}}}\, 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)^2,x)

[Out]

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

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B x + A\right )} \left (e x\right )^{m}}{c^{2} x^{4} + 2 \, a c x^{2} + a^{2}}, 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)^2,x, algorithm="fricas")

[Out]

integral((B*x + A)*(e*x)^m/(c^2*x^4 + 2*a*c*x^2 + a^2), x)

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Sympy [C]  time = 39.3932, size = 770, normalized size = 8.46 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(-a*e**m*m**2*x*x**m*lerchphi(c*x**2*exp_polar(I*pi)/a, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8*a**3*gamma(m/2 + 3
/2) + 8*a**2*c*x**2*gamma(m/2 + 3/2)) + 2*a*e**m*m*x*x**m*gamma(m/2 + 1/2)/(8*a**3*gamma(m/2 + 3/2) + 8*a**2*c
*x**2*gamma(m/2 + 3/2)) + a*e**m*x*x**m*lerchphi(c*x**2*exp_polar(I*pi)/a, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8*a
**3*gamma(m/2 + 3/2) + 8*a**2*c*x**2*gamma(m/2 + 3/2)) + 2*a*e**m*x*x**m*gamma(m/2 + 1/2)/(8*a**3*gamma(m/2 +
3/2) + 8*a**2*c*x**2*gamma(m/2 + 3/2)) - c*e**m*m**2*x**3*x**m*lerchphi(c*x**2*exp_polar(I*pi)/a, 1, m/2 + 1/2
)*gamma(m/2 + 1/2)/(8*a**3*gamma(m/2 + 3/2) + 8*a**2*c*x**2*gamma(m/2 + 3/2)) + c*e**m*x**3*x**m*lerchphi(c*x*
*2*exp_polar(I*pi)/a, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8*a**3*gamma(m/2 + 3/2) + 8*a**2*c*x**2*gamma(m/2 + 3/2)
)) + B*(-a*e**m*m**2*x**2*x**m*lerchphi(c*x**2*exp_polar(I*pi)/a, 1, m/2 + 1)*gamma(m/2 + 1)/(8*a**3*gamma(m/2
+ 2) + 8*a**2*c*x**2*gamma(m/2 + 2)) - 2*a*e**m*m*x**2*x**m*lerchphi(c*x**2*exp_polar(I*pi)/a, 1, m/2 + 1)*ga
mma(m/2 + 1)/(8*a**3*gamma(m/2 + 2) + 8*a**2*c*x**2*gamma(m/2 + 2)) + 2*a*e**m*m*x**2*x**m*gamma(m/2 + 1)/(8*a
**3*gamma(m/2 + 2) + 8*a**2*c*x**2*gamma(m/2 + 2)) + 4*a*e**m*x**2*x**m*gamma(m/2 + 1)/(8*a**3*gamma(m/2 + 2)
+ 8*a**2*c*x**2*gamma(m/2 + 2)) - c*e**m*m**2*x**4*x**m*lerchphi(c*x**2*exp_polar(I*pi)/a, 1, m/2 + 1)*gamma(m
/2 + 1)/(8*a**3*gamma(m/2 + 2) + 8*a**2*c*x**2*gamma(m/2 + 2)) - 2*c*e**m*m*x**4*x**m*lerchphi(c*x**2*exp_pola
r(I*pi)/a, 1, m/2 + 1)*gamma(m/2 + 1)/(8*a**3*gamma(m/2 + 2) + 8*a**2*c*x**2*gamma(m/2 + 2)))

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

[Out]

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