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

Optimal. Leaf size=205 \[ \frac{(e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right ) (a d (A d (1-m)+B c (m+1))+b c (B c (1-m)-A d (3-m)))}{2 c^2 e (m+1) (b c-a d)^2}+\frac{b (e x)^{m+1} (A b-a B) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a e (m+1) (b c-a d)^2}+\frac{(e x)^{m+1} (B c-A d)}{2 c e \left (c+d x^2\right ) (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.383297, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {579, 584, 364} \[ \frac{(e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right ) (a d (A d (1-m)+B c (m+1))+b c (B c (1-m)-A d (3-m)))}{2 c^2 e (m+1) (b c-a d)^2}+\frac{b (e x)^{m+1} (A b-a B) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a e (m+1) (b c-a d)^2}+\frac{(e x)^{m+1} (B c-A d)}{2 c e \left (c+d x^2\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 579

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> -Simp[((b*e - a*f)*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*g*n*(b*c - a*d)*(p +
1)), x] + Dist[1/(a*n*(b*c - a*d)*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f)*(
m + 1) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
e, f, g, m, q}, x] && IGtQ[n, 0] && LtQ[p, -1]

Rule 584

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

Mathematica [A]  time = 0.170549, size = 147, normalized size = 0.72 \[ \frac{x (e x)^m \left (b c^2 (A b-a B) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )+a c d (a B-A b) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )+a (b c-a d) (B c-A d) \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )\right )}{a c^2 (m+1) (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.069, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( B{x}^{2}+A \right ) \left ( ex \right ) ^{m}}{ \left ( b{x}^{2}+a \right ) \left ( d{x}^{2}+c \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((e*x)^m*(B*x^2+A)/(b*x^2+a)/(d*x^2+c)^2,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}}{{\left (b x^{2} + a\right )}{\left (d x^{2} + c\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x^2 + A)*(e*x)^m/((b*x^2 + a)*(d*x^2 + c)^2), 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}}{b d^{2} x^{6} +{\left (2 \, b c d + a d^{2}\right )} x^{4} + a c^{2} +{\left (b c^{2} + 2 \, a c d\right )} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*x^2 + A)*(e*x)^m/(b*d^2*x^6 + (2*b*c*d + a*d^2)*x^4 + a*c^2 + (b*c^2 + 2*a*c*d)*x^2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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}}{{\left (b x^{2} + a\right )}{\left (d x^{2} + c\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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