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

Optimal. Leaf size=322 \[ \frac{(e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{d x^n}{c}\right ) (a d (b c (m+1)-a d (m-n+1)) (B c (m+1)-A d (m-2 n+1))-b c (a d (m+1)-b c (m+n+1)) (A d (m+1)-B c (m+2 n+1)))}{2 c^3 d^3 e (m+1) n^2}-\frac{(e x)^{m+1} (b c-a d) \left (a (B c (m+1)-A d (m-2 n+1))-b x^n (A d (m+1)-B c (m+2 n+1))\right )}{2 c^2 d^2 e n^2 \left (c+d x^n\right )}+\frac{b (e x)^{m+1} (a d (m+1)-b c (m+n+1)) (A d (m+1)-B c (m+2 n+1))}{2 c^2 d^3 e (m+1) n^2}-\frac{(e x)^{m+1} \left (a+b x^n\right )^2 (B c-A d)}{2 c d e n \left (c+d x^n\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.556817, antiderivative size = 322, 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 = {594, 459, 364} \[ \frac{(e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{d x^n}{c}\right ) (a d (b c (m+1)-a d (m-n+1)) (B c (m+1)-A d (m-2 n+1))-b c (a d (m+1)-b c (m+n+1)) (A d (m+1)-B c (m+2 n+1)))}{2 c^3 d^3 e (m+1) n^2}-\frac{(e x)^{m+1} (b c-a d) \left (a (B c (m+1)-A d (m-2 n+1))-b x^n (A d (m+1)-B c (m+2 n+1))\right )}{2 c^2 d^2 e n^2 \left (c+d x^n\right )}+\frac{b (e x)^{m+1} (a d (m+1)-b c (m+n+1)) (A d (m+1)-B c (m+2 n+1))}{2 c^2 d^3 e (m+1) n^2}-\frac{(e x)^{m+1} \left (a+b x^n\right )^2 (B c-A d)}{2 c d e n \left (c+d x^n\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 594

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, n}, x]
&& 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^n\right )^2 \left (A+B x^n\right )}{\left (c+d x^n\right )^3} \, dx &=-\frac{(B c-A d) (e x)^{1+m} \left (a+b x^n\right )^2}{2 c d e n \left (c+d x^n\right )^2}-\frac{\int \frac{(e x)^m \left (a+b x^n\right ) \left (-a (B c (1+m)-A d (1+m-2 n))+b (A d (1+m)-B c (1+m+2 n)) x^n\right )}{\left (c+d x^n\right )^2} \, dx}{2 c d n}\\ &=-\frac{(B c-A d) (e x)^{1+m} \left (a+b x^n\right )^2}{2 c d e n \left (c+d x^n\right )^2}-\frac{(b c-a d) (e x)^{1+m} \left (a (B c (1+m)-A d (1+m-2 n))-b (A d (1+m)-B c (1+m+2 n)) x^n\right )}{2 c^2 d^2 e n^2 \left (c+d x^n\right )}+\frac{\int \frac{(e x)^m \left (a (B c (1+m)-A d (1+m-2 n)) (b c (1+m)-a d (1+m-n))+b (a d (1+m)-b c (1+m+n)) (A d (1+m)-B c (1+m+2 n)) x^n\right )}{c+d x^n} \, dx}{2 c^2 d^2 n^2}\\ &=\frac{b (a d (1+m)-b c (1+m+n)) (A d (1+m)-B c (1+m+2 n)) (e x)^{1+m}}{2 c^2 d^3 e (1+m) n^2}-\frac{(B c-A d) (e x)^{1+m} \left (a+b x^n\right )^2}{2 c d e n \left (c+d x^n\right )^2}-\frac{(b c-a d) (e x)^{1+m} \left (a (B c (1+m)-A d (1+m-2 n))-b (A d (1+m)-B c (1+m+2 n)) x^n\right )}{2 c^2 d^2 e n^2 \left (c+d x^n\right )}+\frac{\left (a (B c (1+m)-A d (1+m-2 n)) (b c (1+m)-a d (1+m-n))-\frac{b c (a d (1+m)-b c (1+m+n)) (A d (1+m)-B c (1+m+2 n))}{d}\right ) \int \frac{(e x)^m}{c+d x^n} \, dx}{2 c^2 d^2 n^2}\\ &=\frac{b (a d (1+m)-b c (1+m+n)) (A d (1+m)-B c (1+m+2 n)) (e x)^{1+m}}{2 c^2 d^3 e (1+m) n^2}-\frac{(B c-A d) (e x)^{1+m} \left (a+b x^n\right )^2}{2 c d e n \left (c+d x^n\right )^2}-\frac{(b c-a d) (e x)^{1+m} \left (a (B c (1+m)-A d (1+m-2 n))-b (A d (1+m)-B c (1+m+2 n)) x^n\right )}{2 c^2 d^2 e n^2 \left (c+d x^n\right )}+\frac{\left (a (B c (1+m)-A d (1+m-2 n)) (b c (1+m)-a d (1+m-n))-\frac{b c (a d (1+m)-b c (1+m+n)) (A d (1+m)-B c (1+m+2 n))}{d}\right ) (e x)^{1+m} \, _2F_1\left (1,\frac{1+m}{n};\frac{1+m+n}{n};-\frac{d x^n}{c}\right )}{2 c^3 d^2 e (1+m) n^2}\\ \end{align*}

Mathematica [A]  time = 0.279731, size = 172, normalized size = 0.53 \[ \frac{x (e x)^m \left (\frac{(b c-a d) (-a B d-2 A b d+3 b B c) \, _2F_1\left (2,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{d x^n}{c}\right )}{c^2}-\frac{(b c-a d)^2 (B c-A d) \, _2F_1\left (3,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{d x^n}{c}\right )}{c^3}-\frac{b (-2 a B d-A b d+3 b B c) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{d x^n}{c}\right )}{c}+b^2 B\right )}{d^3 (m+1)} \]

Antiderivative was successfully verified.

[In]

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

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Maple [F]  time = 0.542, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex \right ) ^{m} \left ( a+b{x}^{n} \right ) ^{2} \left ( A+B{x}^{n} \right ) }{ \left ( c+d{x}^{n} \right ) ^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B b^{2} x^{3 \, n} + A a^{2} +{\left (2 \, B a b + A b^{2}\right )} x^{2 \, n} +{\left (B a^{2} + 2 \, A a b\right )} x^{n}\right )} \left (e x\right )^{m}}{d^{3} x^{3 \, n} + 3 \, c d^{2} x^{2 \, n} + 3 \, c^{2} d x^{n} + c^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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