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

Optimal. Leaf size=292 \[ \frac{(e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right ) (a d (a d (1-m)+b c (m+1)) (A d (3-m)+B c (m+1))-b c (a d (m+1)-b c (m+3)) (A d (m+1)-B c (m+5)))}{8 c^3 d^3 e (m+1)}-\frac{(e x)^{m+1} (b c-a d) \left (a (A d (3-m)+B c (m+1))-b x^2 (A d (m+1)-B c (m+5))\right )}{8 c^2 d^2 e \left (c+d x^2\right )}+\frac{b (e x)^{m+1} (a d (m+1)-b c (m+3)) (A d (m+1)-B c (m+5))}{8 c^2 d^3 e (m+1)}-\frac{\left (a+b x^2\right )^2 (e x)^{m+1} (B c-A d)}{4 c d e \left (c+d x^2\right )^2} \]

[Out]

(b*(a*d*(1 + m) - b*c*(3 + m))*(A*d*(1 + m) - B*c*(5 + m))*(e*x)^(1 + m))/(8*c^2*d^3*e*(1 + m)) - ((B*c - A*d)
*(e*x)^(1 + m)*(a + b*x^2)^2)/(4*c*d*e*(c + d*x^2)^2) - ((b*c - a*d)*(e*x)^(1 + m)*(a*(A*d*(3 - m) + B*c*(1 +
m)) - b*(A*d*(1 + m) - B*c*(5 + m))*x^2))/(8*c^2*d^2*e*(c + d*x^2)) + ((a*d*(a*d*(1 - m) + b*c*(1 + m))*(A*d*(
3 - m) + B*c*(1 + m)) - b*c*(a*d*(1 + m) - b*c*(3 + m))*(A*d*(1 + m) - B*c*(5 + m)))*(e*x)^(1 + m)*Hypergeomet
ric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(8*c^3*d^3*e*(1 + m))

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Rubi [A]  time = 0.406722, antiderivative size = 292, 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, 459, 364} \[ \frac{(e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right ) (a d (a d (1-m)+b c (m+1)) (A d (3-m)+B c (m+1))-b c (a d (m+1)-b c (m+3)) (A d (m+1)-B c (m+5)))}{8 c^3 d^3 e (m+1)}-\frac{(e x)^{m+1} (b c-a d) \left (a (A d (3-m)+B c (m+1))-b x^2 (A d (m+1)-B c (m+5))\right )}{8 c^2 d^2 e \left (c+d x^2\right )}+\frac{b (e x)^{m+1} (a d (m+1)-b c (m+3)) (A d (m+1)-B c (m+5))}{8 c^2 d^3 e (m+1)}-\frac{\left (a+b x^2\right )^2 (e x)^{m+1} (B c-A d)}{4 c d e \left (c+d x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(a*d*(1 + m) - b*c*(3 + m))*(A*d*(1 + m) - B*c*(5 + m))*(e*x)^(1 + m))/(8*c^2*d^3*e*(1 + m)) - ((B*c - A*d)
*(e*x)^(1 + m)*(a + b*x^2)^2)/(4*c*d*e*(c + d*x^2)^2) - ((b*c - a*d)*(e*x)^(1 + m)*(a*(A*d*(3 - m) + B*c*(1 +
m)) - b*(A*d*(1 + m) - B*c*(5 + m))*x^2))/(8*c^2*d^2*e*(c + d*x^2)) + ((a*d*(a*d*(1 - m) + b*c*(1 + m))*(A*d*(
3 - m) + B*c*(1 + m)) - b*c*(a*d*(1 + m) - b*c*(3 + m))*(A*d*(1 + m) - B*c*(5 + m)))*(e*x)^(1 + m)*Hypergeomet
ric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(8*c^3*d^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 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 )^2 \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx &=-\frac{(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^2}{4 c d e \left (c+d x^2\right )^2}-\frac{\int \frac{(e x)^m \left (a+b x^2\right ) \left (-a (A d (3-m)+B c (1+m))+b (A d (1+m)-B c (5+m)) x^2\right )}{\left (c+d x^2\right )^2} \, dx}{4 c d}\\ &=-\frac{(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^2}{4 c d e \left (c+d x^2\right )^2}-\frac{(b c-a d) (e x)^{1+m} \left (a (A d (3-m)+B c (1+m))-b (A d (1+m)-B c (5+m)) x^2\right )}{8 c^2 d^2 e \left (c+d x^2\right )}+\frac{\int \frac{(e x)^m \left (a (a d (1-m)+b c (1+m)) (A d (3-m)+B c (1+m))+b (a d (1+m)-b c (3+m)) (A d (1+m)-B c (5+m)) x^2\right )}{c+d x^2} \, dx}{8 c^2 d^2}\\ &=\frac{b (a d (1+m)-b c (3+m)) (A d (1+m)-B c (5+m)) (e x)^{1+m}}{8 c^2 d^3 e (1+m)}-\frac{(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^2}{4 c d e \left (c+d x^2\right )^2}-\frac{(b c-a d) (e x)^{1+m} \left (a (A d (3-m)+B c (1+m))-b (A d (1+m)-B c (5+m)) x^2\right )}{8 c^2 d^2 e \left (c+d x^2\right )}+\frac{\left (a (a d (1-m)+b c (1+m)) (A d (3-m)+B c (1+m))-\frac{b c (a d (1+m)-b c (3+m)) (A d (1+m)-B c (5+m))}{d}\right ) \int \frac{(e x)^m}{c+d x^2} \, dx}{8 c^2 d^2}\\ &=\frac{b (a d (1+m)-b c (3+m)) (A d (1+m)-B c (5+m)) (e x)^{1+m}}{8 c^2 d^3 e (1+m)}-\frac{(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^2}{4 c d e \left (c+d x^2\right )^2}-\frac{(b c-a d) (e x)^{1+m} \left (a (A d (3-m)+B c (1+m))-b (A d (1+m)-B c (5+m)) x^2\right )}{8 c^2 d^2 e \left (c+d x^2\right )}+\frac{\left (a (a d (1-m)+b c (1+m)) (A d (3-m)+B c (1+m))-\frac{b c (a d (1+m)-b c (3+m)) (A d (1+m)-B c (5+m))}{d}\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 d^2 e (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.185872, size = 169, normalized size = 0.58 \[ \frac{x (e x)^m \left (\frac{(b c-a d) \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right ) (-a B d-2 A b d+3 b B c)}{c^2}-\frac{(b c-a d)^2 (B c-A d) \, _2F_1\left (3,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c^3}-\frac{b \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right ) (-2 a B d-A b d+3 b B c)}{c}+b^2 B\right )}{d^3 (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.067, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( B{x}^{2}+A \right ) \left ( b{x}^{2}+a \right ) ^{2} \left ( ex \right ) ^{m}}{ \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)^2*(B*x^2+A)/(d*x^2+c)^3,x)

[Out]

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

[Out]

integrate((B*x^2 + A)*(b*x^2 + a)^2*(e*x)^m/(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 b^{2} x^{6} +{\left (2 \, B a b + A b^{2}\right )} x^{4} + A a^{2} +{\left (B a^{2} + 2 \, A a b\right )} x^{2}\right )} \left (e x\right )^{m}}{d^{3} x^{6} + 3 \, c d^{2} x^{4} + 3 \, c^{2} d x^{2} + c^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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