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

Optimal. Leaf size=247 \[ \frac{(e x)^{m+1} (b c-a d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right ) (A b (a d (m+3)+b (c-c m))+a B (b c (m+1)-a d (m+5)))}{2 a^2 b^3 e (m+1)}-\frac{d (e x)^{m+1} (A b (2 b c (m+1)-a d (m+3))-a B (2 b c (m+3)-a d (m+5)))}{2 a b^3 e (m+1)}-\frac{d^2 (e x)^{m+3} (A b (m+3)-a B (m+5))}{2 a b^2 e^3 (m+3)}+\frac{\left (c+d x^2\right )^2 (e x)^{m+1} (A b-a B)}{2 a b e \left (a+b x^2\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.442144, antiderivative size = 247, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {577, 570, 364} \[ \frac{(e x)^{m+1} (b c-a d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right ) (A b (a d (m+3)+b (c-c m))+a B (b c (m+1)-a d (m+5)))}{2 a^2 b^3 e (m+1)}-\frac{d (e x)^{m+1} (A b (2 b c (m+1)-a d (m+3))-a B (2 b c (m+3)-a d (m+5)))}{2 a b^3 e (m+1)}-\frac{d^2 (e x)^{m+3} (A b (m+3)-a B (m+5))}{2 a b^2 e^3 (m+3)}+\frac{\left (c+d x^2\right )^2 (e x)^{m+1} (A b-a B)}{2 a b e \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 577

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)/(a*b*g*n*(p + 1)), x] + Dist[
1/(a*b*n*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(b*e*n*(p + 1) + (b*e - a*f)*(m
+ 1)) + d*(b*e*n*(p + 1) + (b*e - a*f)*(m + n*q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] &&
IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] &&  !(EqQ[q, 1] && SimplerQ[b*c - a*d, b*e - a*f])

Rule 570

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^
(r_.), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*(c + d*x^n)^q*(e + f*x^n)^r, x], x] /; FreeQ[{a,
b, c, d, e, f, g, m, n}, x] && IGtQ[p, -2] && IGtQ[q, 0] && IGtQ[r, 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 (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx &=\frac{(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^2}{2 a b e \left (a+b x^2\right )}-\frac{\int \frac{(e x)^m \left (c+d x^2\right ) \left (-c (A b (1-m)+a B (1+m))+d (A b (3+m)-a B (5+m)) x^2\right )}{a+b x^2} \, dx}{2 a b}\\ &=\frac{(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^2}{2 a b e \left (a+b x^2\right )}-\frac{\int \left (\frac{d (A b (2 b c (1+m)-a d (3+m))-a B (2 b c (3+m)-a d (5+m))) (e x)^m}{b^2}+\frac{d^2 (A b (3+m)-a B (5+m)) (e x)^{2+m}}{b e^2}+\frac{\left (-A b^3 c^2-a b^2 B c^2-2 a A b^2 c d+6 a^2 b B c d+3 a^2 A b d^2-5 a^3 B d^2+A b^3 c^2 m-a b^2 B c^2 m-2 a A b^2 c d m+2 a^2 b B c d m+a^2 A b d^2 m-a^3 B d^2 m\right ) (e x)^m}{b^2 \left (a+b x^2\right )}\right ) \, dx}{2 a b}\\ &=-\frac{d (A b (2 b c (1+m)-a d (3+m))-a B (2 b c (3+m)-a d (5+m))) (e x)^{1+m}}{2 a b^3 e (1+m)}-\frac{d^2 (A b (3+m)-a B (5+m)) (e x)^{3+m}}{2 a b^2 e^3 (3+m)}+\frac{(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^2}{2 a b e \left (a+b x^2\right )}-\frac{\left (-A b^3 c^2-a b^2 B c^2-2 a A b^2 c d+6 a^2 b B c d+3 a^2 A b d^2-5 a^3 B d^2+A b^3 c^2 m-a b^2 B c^2 m-2 a A b^2 c d m+2 a^2 b B c d m+a^2 A b d^2 m-a^3 B d^2 m\right ) \int \frac{(e x)^m}{a+b x^2} \, dx}{2 a b^3}\\ &=-\frac{d (A b (2 b c (1+m)-a d (3+m))-a B (2 b c (3+m)-a d (5+m))) (e x)^{1+m}}{2 a b^3 e (1+m)}-\frac{d^2 (A b (3+m)-a B (5+m)) (e x)^{3+m}}{2 a b^2 e^3 (3+m)}+\frac{(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^2}{2 a b e \left (a+b x^2\right )}+\frac{(b c-a d) (A b (b c (1-m)+a d (3+m))+a B (b c (1+m)-a d (5+m))) (e x)^{1+m} \, _2F_1\left (1,\frac{1+m}{2};\frac{3+m}{2};-\frac{b x^2}{a}\right )}{2 a^2 b^3 e (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.17388, size = 156, normalized size = 0.63 \[ \frac{x (e x)^m \left (\frac{(A b-a B) (b c-a d)^2 \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a^2 (m+1)}+\frac{(b c-a d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right ) (-3 a B d+2 A b d+b B c)}{a (m+1)}+\frac{d (-2 a B d+A b d+2 b B c)}{m+1}+\frac{b B d^2 x^2}{m+3}\right )}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.057, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex \right ) ^{m} \left ( B{x}^{2}+A \right ) \left ( d{x}^{2}+c \right ) ^{2}}{ \left ( b{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^2+A)*(d*x^2+c)^2/(b*x^2+a)^2,x)

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

integral((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)*(e*x)^m/(b^2*x^4 + 2*a*b*x^2 + a^
2), x)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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