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

Optimal. Leaf size=347 \[ -\frac{d (e x)^{m+1} \left (A b \left (a^2 d^2 (m+5)-3 a b c d (m+3)+3 b^2 c^2 (m+1)\right )-a B \left (a^2 d^2 (m+7)-3 a b c d (m+5)+3 b^2 c^2 (m+3)\right )\right )}{2 a b^4 e (m+1)}+\frac{(e x)^{m+1} (b c-a d)^2 \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right ) (A b (a d (m+5)+b (c-c m))+a B (b c (m+1)-a d (m+7)))}{2 a^2 b^4 e (m+1)}-\frac{d^2 (e x)^{m+3} (A b (3 b c (m+3)-a d (m+5))-a B (3 b c (m+5)-a d (m+7)))}{2 a b^3 e^3 (m+3)}-\frac{d^3 (e x)^{m+5} (A b (m+5)-a B (m+7))}{2 a b^2 e^5 (m+5)}+\frac{\left (c+d x^2\right )^3 (e x)^{m+1} (A b-a B)}{2 a b e \left (a+b x^2\right )} \]

[Out]

-(d*(A*b*(3*b^2*c^2*(1 + m) - 3*a*b*c*d*(3 + m) + a^2*d^2*(5 + m)) - a*B*(3*b^2*c^2*(3 + m) - 3*a*b*c*d*(5 + m
) + a^2*d^2*(7 + m)))*(e*x)^(1 + m))/(2*a*b^4*e*(1 + m)) - (d^2*(A*b*(3*b*c*(3 + m) - a*d*(5 + m)) - a*B*(3*b*
c*(5 + m) - a*d*(7 + m)))*(e*x)^(3 + m))/(2*a*b^3*e^3*(3 + m)) - (d^3*(A*b*(5 + m) - a*B*(7 + m))*(e*x)^(5 + m
))/(2*a*b^2*e^5*(5 + m)) + ((A*b - a*B)*(e*x)^(1 + m)*(c + d*x^2)^3)/(2*a*b*e*(a + b*x^2)) + ((b*c - a*d)^2*(a
*B*(b*c*(1 + m) - a*d*(7 + m)) + A*b*(a*d*(5 + 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^4*e*(1 + m))

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Rubi [A]  time = 0.660891, antiderivative size = 347, 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} \[ -\frac{d (e x)^{m+1} \left (A b \left (a^2 d^2 (m+5)-3 a b c d (m+3)+3 b^2 c^2 (m+1)\right )-a B \left (a^2 d^2 (m+7)-3 a b c d (m+5)+3 b^2 c^2 (m+3)\right )\right )}{2 a b^4 e (m+1)}+\frac{(e x)^{m+1} (b c-a d)^2 \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right ) (A b (a d (m+5)+b (c-c m))+a B (b c (m+1)-a d (m+7)))}{2 a^2 b^4 e (m+1)}-\frac{d^2 (e x)^{m+3} (A b (3 b c (m+3)-a d (m+5))-a B (3 b c (m+5)-a d (m+7)))}{2 a b^3 e^3 (m+3)}-\frac{d^3 (e x)^{m+5} (A b (m+5)-a B (m+7))}{2 a b^2 e^5 (m+5)}+\frac{\left (c+d x^2\right )^3 (e x)^{m+1} (A b-a B)}{2 a b e \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*(A*b*(3*b^2*c^2*(1 + m) - 3*a*b*c*d*(3 + m) + a^2*d^2*(5 + m)) - a*B*(3*b^2*c^2*(3 + m) - 3*a*b*c*d*(5 + m
) + a^2*d^2*(7 + m)))*(e*x)^(1 + m))/(2*a*b^4*e*(1 + m)) - (d^2*(A*b*(3*b*c*(3 + m) - a*d*(5 + m)) - a*B*(3*b*
c*(5 + m) - a*d*(7 + m)))*(e*x)^(3 + m))/(2*a*b^3*e^3*(3 + m)) - (d^3*(A*b*(5 + m) - a*B*(7 + m))*(e*x)^(5 + m
))/(2*a*b^2*e^5*(5 + m)) + ((A*b - a*B)*(e*x)^(1 + m)*(c + d*x^2)^3)/(2*a*b*e*(a + b*x^2)) + ((b*c - a*d)^2*(a
*B*(b*c*(1 + m) - a*d*(7 + m)) + A*b*(a*d*(5 + 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^4*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 )^3}{\left (a+b x^2\right )^2} \, dx &=\frac{(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^3}{2 a b e \left (a+b x^2\right )}-\frac{\int \frac{(e x)^m \left (c+d x^2\right )^2 \left (-c (A b (1-m)+a B (1+m))+d (A b (5+m)-a B (7+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 )^3}{2 a b e \left (a+b x^2\right )}-\frac{\int \left (\frac{d \left (A b \left (3 b^2 c^2 (1+m)-3 a b c d (3+m)+a^2 d^2 (5+m)\right )-a B \left (3 b^2 c^2 (3+m)-3 a b c d (5+m)+a^2 d^2 (7+m)\right )\right ) (e x)^m}{b^3}+\frac{d^2 (A b (3 b c (3+m)-a d (5+m))-a B (3 b c (5+m)-a d (7+m))) (e x)^{2+m}}{b^2 e^2}+\frac{d^3 (A b (5+m)-a B (7+m)) (e x)^{4+m}}{b e^4}+\frac{\left (-A b^4 c^3-a b^3 B c^3-3 a A b^3 c^2 d+9 a^2 b^2 B c^2 d+9 a^2 A b^2 c d^2-15 a^3 b B c d^2-5 a^3 A b d^3+7 a^4 B d^3+A b^4 c^3 m-a b^3 B c^3 m-3 a A b^3 c^2 d m+3 a^2 b^2 B c^2 d m+3 a^2 A b^2 c d^2 m-3 a^3 b B c d^2 m-a^3 A b d^3 m+a^4 B d^3 m\right ) (e x)^m}{b^3 \left (a+b x^2\right )}\right ) \, dx}{2 a b}\\ &=-\frac{d \left (A b \left (3 b^2 c^2 (1+m)-3 a b c d (3+m)+a^2 d^2 (5+m)\right )-a B \left (3 b^2 c^2 (3+m)-3 a b c d (5+m)+a^2 d^2 (7+m)\right )\right ) (e x)^{1+m}}{2 a b^4 e (1+m)}-\frac{d^2 (A b (3 b c (3+m)-a d (5+m))-a B (3 b c (5+m)-a d (7+m))) (e x)^{3+m}}{2 a b^3 e^3 (3+m)}-\frac{d^3 (A b (5+m)-a B (7+m)) (e x)^{5+m}}{2 a b^2 e^5 (5+m)}+\frac{(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^3}{2 a b e \left (a+b x^2\right )}-\frac{\left (-A b^4 c^3-a b^3 B c^3-3 a A b^3 c^2 d+9 a^2 b^2 B c^2 d+9 a^2 A b^2 c d^2-15 a^3 b B c d^2-5 a^3 A b d^3+7 a^4 B d^3+A b^4 c^3 m-a b^3 B c^3 m-3 a A b^3 c^2 d m+3 a^2 b^2 B c^2 d m+3 a^2 A b^2 c d^2 m-3 a^3 b B c d^2 m-a^3 A b d^3 m+a^4 B d^3 m\right ) \int \frac{(e x)^m}{a+b x^2} \, dx}{2 a b^4}\\ &=-\frac{d \left (A b \left (3 b^2 c^2 (1+m)-3 a b c d (3+m)+a^2 d^2 (5+m)\right )-a B \left (3 b^2 c^2 (3+m)-3 a b c d (5+m)+a^2 d^2 (7+m)\right )\right ) (e x)^{1+m}}{2 a b^4 e (1+m)}-\frac{d^2 (A b (3 b c (3+m)-a d (5+m))-a B (3 b c (5+m)-a d (7+m))) (e x)^{3+m}}{2 a b^3 e^3 (3+m)}-\frac{d^3 (A b (5+m)-a B (7+m)) (e x)^{5+m}}{2 a b^2 e^5 (5+m)}+\frac{(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^3}{2 a b e \left (a+b x^2\right )}+\frac{(b c-a d)^2 (A b (b c (1-m)+a d (5+m))+a B (b c (1+m)-a d (7+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^4 e (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.312722, size = 209, normalized size = 0.6 \[ \frac{x (e x)^m \left (\frac{d \left (3 a^2 B d^2-2 a b d (A d+3 B c)+3 b^2 c (A d+B c)\right )}{m+1}+\frac{(a B-A b) (a d-b c)^3 \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a^2 (m+1)}+\frac{b d^2 x^2 (-2 a B d+A b d+3 b B c)}{m+3}+\frac{(b c-a d)^2 \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right ) (-4 a B d+3 A b d+b B c)}{a (m+1)}+\frac{b^2 B d^3 x^4}{m+5}\right )}{b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(e*x)^m*((d*(3*a^2*B*d^2 + 3*b^2*c*(B*c + A*d) - 2*a*b*d*(3*B*c + A*d)))/(1 + m) + (b*d^2*(3*b*B*c + A*b*d
- 2*a*B*d)*x^2)/(3 + m) + (b^2*B*d^3*x^4)/(5 + m) + ((b*c - a*d)^2*(b*B*c + 3*A*b*d - 4*a*B*d)*Hypergeometric2
F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a*(1 + m)) + ((-(A*b) + a*B)*(-(b*c) + a*d)^3*Hypergeometric2F1[2,
(1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a^2*(1 + m))))/b^4

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Maple [F]  time = 0.047, 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 ) ^{3}}{ \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)^3/(b*x^2+a)^2,x)

[Out]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (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}\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)^3/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

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

[Out]

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