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

Optimal. Leaf size=77 \[ \frac{B (e x)^{m+1}}{d e (m+1)}-\frac{(e x)^{m+1} (B c-A d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c d e (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.0383385, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {459, 364} \[ \frac{B (e x)^{m+1}}{d e (m+1)}-\frac{(e x)^{m+1} (B c-A d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c d e (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 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 \left (A+B x^2\right )}{c+d x^2} \, dx &=\frac{B (e x)^{1+m}}{d e (1+m)}-\frac{(B c (1+m)-A d (1+m)) \int \frac{(e x)^m}{c+d x^2} \, dx}{d (1+m)}\\ &=\frac{B (e x)^{1+m}}{d e (1+m)}-\frac{(B c-A d) (e x)^{1+m} \, _2F_1\left (1,\frac{1+m}{2};\frac{3+m}{2};-\frac{d x^2}{c}\right )}{c d e (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.0481277, size = 56, normalized size = 0.73 \[ \frac{x (e x)^m \left ((A d-B c) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )+B c\right )}{c d (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.033, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( B{x}^{2}+A \right ) \left ( ex \right ) ^{m}}{d{x}^{2}+c}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 5.30542, size = 204, normalized size = 2.65 \begin{align*} \frac{A e^{m} m x x^{m} \Phi \left (\frac{d x^{2} e^{i \pi }}{c}, 1, \frac{m}{2} + \frac{1}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{1}{2}\right )}{4 c \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )} + \frac{A e^{m} x x^{m} \Phi \left (\frac{d x^{2} e^{i \pi }}{c}, 1, \frac{m}{2} + \frac{1}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{1}{2}\right )}{4 c \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )} + \frac{B e^{m} m x^{3} x^{m} \Phi \left (\frac{d x^{2} e^{i \pi }}{c}, 1, \frac{m}{2} + \frac{3}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )}{4 c \Gamma \left (\frac{m}{2} + \frac{5}{2}\right )} + \frac{3 B e^{m} x^{3} x^{m} \Phi \left (\frac{d x^{2} e^{i \pi }}{c}, 1, \frac{m}{2} + \frac{3}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )}{4 c \Gamma \left (\frac{m}{2} + \frac{5}{2}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*e**m*m*x*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(4*c*gamma(m/2 + 3/2)) + A*e
**m*x*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(4*c*gamma(m/2 + 3/2)) + B*e**m*m
*x**3*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(4*c*gamma(m/2 + 5/2)) + 3*B*e**m
*x**3*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(4*c*gamma(m/2 + 5/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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