3.41 \(\int \frac{(e x)^m (A+B x^2)}{(a+b x^2) (c+d x^2)^3} \, dx\)

Optimal. Leaf size=333 \[ \frac{(e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right ) \left (-a^2 d^2 (1-m) (A d (3-m)+B c (m+1))+2 a b c d \left (A d \left (m^2-6 m+5\right )+B c \left (-m^2+2 m+3\right )\right )+b^2 c^2 (3-m) (B c (1-m)-A d (5-m))\right )}{8 c^3 e (m+1) (b c-a d)^3}+\frac{b^2 (e x)^{m+1} (A b-a B) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a e (m+1) (b c-a d)^3}+\frac{(e x)^{m+1} (a d (A d (3-m)+B c (m+1))+b c (B c (3-m)-A d (7-m)))}{8 c^2 e \left (c+d x^2\right ) (b c-a d)^2}+\frac{(e x)^{m+1} (B c-A d)}{4 c e \left (c+d x^2\right )^2 (b c-a d)} \]

[Out]

((B*c - A*d)*(e*x)^(1 + m))/(4*c*(b*c - a*d)*e*(c + d*x^2)^2) + ((b*c*(B*c*(3 - m) - A*d*(7 - m)) + a*d*(A*d*(
3 - m) + B*c*(1 + m)))*(e*x)^(1 + m))/(8*c^2*(b*c - a*d)^2*e*(c + d*x^2)) + (b^2*(A*b - a*B)*(e*x)^(1 + m)*Hyp
ergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a*(b*c - a*d)^3*e*(1 + m)) + ((b^2*c^2*(B*c*(1 - m) -
A*d*(5 - m))*(3 - m) - a^2*d^2*(1 - m)*(A*d*(3 - m) + B*c*(1 + m)) + 2*a*b*c*d*(B*c*(3 + 2*m - m^2) + A*d*(5 -
6*m + m^2)))*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(8*c^3*(b*c - a*d)^3*e*(
1 + m))

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Rubi [A]  time = 0.719435, antiderivative size = 333, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {579, 584, 364} \[ \frac{(e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right ) \left (-a^2 d^2 (1-m) (A d (3-m)+B c (m+1))+2 a b c d \left (A d \left (m^2-6 m+5\right )+B c \left (-m^2+2 m+3\right )\right )+b^2 c^2 (3-m) (B c (1-m)-A d (5-m))\right )}{8 c^3 e (m+1) (b c-a d)^3}+\frac{b^2 (e x)^{m+1} (A b-a B) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a e (m+1) (b c-a d)^3}+\frac{(e x)^{m+1} (a d (A d (3-m)+B c (m+1))+b c (B c (3-m)-A d (7-m)))}{8 c^2 e \left (c+d x^2\right ) (b c-a d)^2}+\frac{(e x)^{m+1} (B c-A d)}{4 c e \left (c+d x^2\right )^2 (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((B*c - A*d)*(e*x)^(1 + m))/(4*c*(b*c - a*d)*e*(c + d*x^2)^2) + ((b*c*(B*c*(3 - m) - A*d*(7 - m)) + a*d*(A*d*(
3 - m) + B*c*(1 + m)))*(e*x)^(1 + m))/(8*c^2*(b*c - a*d)^2*e*(c + d*x^2)) + (b^2*(A*b - a*B)*(e*x)^(1 + m)*Hyp
ergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a*(b*c - a*d)^3*e*(1 + m)) + ((b^2*c^2*(B*c*(1 - m) -
A*d*(5 - m))*(3 - m) - a^2*d^2*(1 - m)*(A*d*(3 - m) + B*c*(1 + m)) + 2*a*b*c*d*(B*c*(3 + 2*m - m^2) + A*d*(5 -
6*m + m^2)))*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(8*c^3*(b*c - a*d)^3*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 )^3} \, dx &=\frac{(B c-A d) (e x)^{1+m}}{4 c (b c-a d) e \left (c+d x^2\right )^2}+\frac{\int \frac{(e x)^m \left (4 A b c-a A d (3-m)-a B c (1+m)+b (B c-A d) (3-m) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx}{4 c (b c-a d)}\\ &=\frac{(B c-A d) (e x)^{1+m}}{4 c (b c-a d) e \left (c+d x^2\right )^2}+\frac{(b c (B c (3-m)-A d (7-m))+a d (A d (3-m)+B c (1+m))) (e x)^{1+m}}{8 c^2 (b c-a d)^2 e \left (c+d x^2\right )}+\frac{\int \frac{(e x)^m \left (a B c (a d (1-m)-b c (5-m)) (1+m)+A \left (8 b^2 c^2-a b c d \left (7-8 m+m^2\right )+a^2 d^2 \left (3-4 m+m^2\right )\right )+b (1-m) (b c (B c (3-m)-A d (7-m))+a d (A d (3-m)+B c (1+m))) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{8 c^2 (b c-a d)^2}\\ &=\frac{(B c-A d) (e x)^{1+m}}{4 c (b c-a d) e \left (c+d x^2\right )^2}+\frac{(b c (B c (3-m)-A d (7-m))+a d (A d (3-m)+B c (1+m))) (e x)^{1+m}}{8 c^2 (b c-a d)^2 e \left (c+d x^2\right )}+\frac{\int \left (\frac{8 b^2 (A b-a B) c^2 (e x)^m}{(b c-a d) \left (a+b x^2\right )}+\frac{\left (b^2 c^2 (B c (1-m)-A d (5-m)) (3-m)-a^2 d^2 (1-m) (A d (3-m)+B c (1+m))+2 a b c d \left (B c \left (3+2 m-m^2\right )+A d \left (5-6 m+m^2\right )\right )\right ) (e x)^m}{(b c-a d) \left (c+d x^2\right )}\right ) \, dx}{8 c^2 (b c-a d)^2}\\ &=\frac{(B c-A d) (e x)^{1+m}}{4 c (b c-a d) e \left (c+d x^2\right )^2}+\frac{(b c (B c (3-m)-A d (7-m))+a d (A d (3-m)+B c (1+m))) (e x)^{1+m}}{8 c^2 (b c-a d)^2 e \left (c+d x^2\right )}+\frac{\left (b^2 (A b-a B)\right ) \int \frac{(e x)^m}{a+b x^2} \, dx}{(b c-a d)^3}+\frac{\left (b^2 c^2 (B c (1-m)-A d (5-m)) (3-m)-a^2 d^2 (1-m) (A d (3-m)+B c (1+m))+2 a b c d \left (B c \left (3+2 m-m^2\right )+A d \left (5-6 m+m^2\right )\right )\right ) \int \frac{(e x)^m}{c+d x^2} \, dx}{8 c^2 (b c-a d)^3}\\ &=\frac{(B c-A d) (e x)^{1+m}}{4 c (b c-a d) e \left (c+d x^2\right )^2}+\frac{(b c (B c (3-m)-A d (7-m))+a d (A d (3-m)+B c (1+m))) (e x)^{1+m}}{8 c^2 (b c-a d)^2 e \left (c+d x^2\right )}+\frac{b^2 (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)^3 e (1+m)}+\frac{\left (b^2 c^2 (B c (1-m)-A d (5-m)) (3-m)-a^2 d^2 (1-m) (A d (3-m)+B c (1+m))+2 a b c d \left (B c \left (3+2 m-m^2\right )+A d \left (5-6 m+m^2\right )\right )\right ) (e x)^{1+m} \, _2F_1\left (1,\frac{1+m}{2};\frac{3+m}{2};-\frac{d x^2}{c}\right )}{8 c^3 (b c-a d)^3 e (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.207456, size = 197, normalized size = 0.59 \[ \frac{x (e x)^m \left (\frac{b^2 (A b-a B) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a}-\frac{d (A b-a B) (b c-a d) \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c^2}+\frac{(b c-a d)^2 (B c-A d) \, _2F_1\left (3,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c^3}-\frac{b d (A b-a B) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c}\right )}{(m+1) (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(e*x)^m*((b^2*(A*b - a*B)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/a - (b*(A*b - a*B)*d*Hy
pergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/c - ((A*b - a*B)*d*(b*c - a*d)*Hypergeometric2F1[2, (1
+ m)/2, (3 + m)/2, -((d*x^2)/c)])/c^2 + ((b*c - a*d)^2*(B*c - A*d)*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2,
-((d*x^2)/c)])/c^3))/((b*c - a*d)^3*(1 + m))

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Maple [F]  time = 0.06, 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 ) ^{3}}}\, 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)^3,x)

[Out]

int((e*x)^m*(B*x^2+A)/(b*x^2+a)/(d*x^2+c)^3,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 )}^{3}}\,{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)^3,x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*(e*x)^m/((b*x^2 + a)*(d*x^2 + c)^3), 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^{3} x^{8} +{\left (3 \, b c d^{2} + a d^{3}\right )} x^{6} + 3 \,{\left (b c^{2} d + a c d^{2}\right )} x^{4} + a c^{3} +{\left (b c^{3} + 3 \, a c^{2} 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)^3,x, algorithm="fricas")

[Out]

integral((B*x^2 + A)*(e*x)^m/(b*d^3*x^8 + (3*b*c*d^2 + a*d^3)*x^6 + 3*(b*c^2*d + a*c*d^2)*x^4 + a*c^3 + (b*c^3
+ 3*a*c^2*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)**3,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 )}^{3}}\,{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)^3,x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*(e*x)^m/((b*x^2 + a)*(d*x^2 + c)^3), x)