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

Optimal. Leaf size=482 \[ \frac{d (e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{d x^n}{c}\right ) \left (-a^2 d^2 (m-n+1) (B c (m+1)-A d (m-2 n+1))+2 a b c d \left (B c (m+1) (m-3 n+1)-A d \left (m^2+m (2-5 n)+4 n^2-5 n+1\right )\right )+b^2 c^2 (m-3 n+1) (A d (m-4 n+1)-B c (m-2 n+1))\right )}{2 c^3 e (m+1) n^2 (b c-a d)^4}-\frac{d (e x)^{m+1} \left (a^2 d (B c (m+1)-A d (m-2 n+1))-a b c (m-6 n+1) (B c-A d)-2 A b^2 c^2 n\right )}{2 a c^2 e n^2 (b c-a d)^3 \left (c+d x^n\right )}+\frac{b^2 (e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b x^n}{a}\right ) (A b (a d (m-4 n+1)-b c (m-n+1))+a B (b c (m+1)-a d (m-3 n+1)))}{a^2 e (m+1) n (b c-a d)^4}+\frac{d (e x)^{m+1} (a A d-3 a B c+2 A b c)}{2 a c e n (b c-a d)^2 \left (c+d x^n\right )^2}+\frac{(e x)^{m+1} (A b-a B)}{a e n (b c-a d) \left (a+b x^n\right ) \left (c+d x^n\right )^2} \]

[Out]

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

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Rubi [A]  time = 2.07065, antiderivative size = 482, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {595, 597, 364} \[ \frac{d (e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{d x^n}{c}\right ) \left (-a^2 d^2 (m-n+1) (B c (m+1)-A d (m-2 n+1))+2 a b c d \left (B c (m+1) (m-3 n+1)-A d \left (m^2+m (2-5 n)+4 n^2-5 n+1\right )\right )+b^2 c^2 (m-3 n+1) (A d (m-4 n+1)-B c (m-2 n+1))\right )}{2 c^3 e (m+1) n^2 (b c-a d)^4}-\frac{d (e x)^{m+1} \left (a^2 d (B c (m+1)-A d (m-2 n+1))-a b c (m-6 n+1) (B c-A d)-2 A b^2 c^2 n\right )}{2 a c^2 e n^2 (b c-a d)^3 \left (c+d x^n\right )}+\frac{b^2 (e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b x^n}{a}\right ) (A b (a d (m-4 n+1)-b c (m-n+1))+a B (b c (m+1)-a d (m-3 n+1)))}{a^2 e (m+1) n (b c-a d)^4}+\frac{d (e x)^{m+1} (a A d-3 a B c+2 A b c)}{2 a c e n (b c-a d)^2 \left (c+d x^n\right )^2}+\frac{(e x)^{m+1} (A b-a B)}{a e n (b c-a d) \left (a+b x^n\right ) \left (c+d x^n\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 595

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, n, q}, x] && LtQ[p, -1]

Rule 597

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, n, p}, x]

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

Mathematica [A]  time = 0.463409, size = 271, normalized size = 0.56 \[ \frac{x (e x)^m \left (\frac{b^2 (a B-A b) (a d-b c) \, _2F_1\left (2,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b x^n}{a}\right )}{a^2}+\frac{b^2 (2 a B d-3 A b d+b B c) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b x^n}{a}\right )}{a}-\frac{d (b c-a d) (a B d-2 A b d+b B c) \, _2F_1\left (2,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{d x^n}{c}\right )}{c^2}+\frac{d (b c-a d)^2 (A d-B c) \, _2F_1\left (3,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{d x^n}{c}\right )}{c^3}-\frac{b d (2 a B d-3 A b d+b B c) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{d x^n}{c}\right )}{c}\right )}{(m+1) (b c-a d)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.678, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex \right ) ^{m} \left ( A+B{x}^{n} \right ) }{ \left ( a+b{x}^{n} \right ) ^{2} \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)/(a+b*x^n)^2/(c+d*x^n)^3,x)

[Out]

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

[Out]

(((m^2 - m*(7*n - 2) + 12*n^2 - 7*n + 1)*b^2*c^2*d^2*e^m - 2*(m^2 - m*(5*n - 2) + 4*n^2 - 5*n + 1)*a*b*c*d^3*e
^m + (m^2 - m*(3*n - 2) + 2*n^2 - 3*n + 1)*a^2*d^4*e^m)*A - ((m^2 - m*(5*n - 2) + 6*n^2 - 5*n + 1)*b^2*c^3*d*e
^m - 2*(m^2 - m*(3*n - 2) - 3*n + 1)*a*b*c^2*d^2*e^m + (m^2 - m*(n - 2) - n + 1)*a^2*c*d^3*e^m)*B)*integrate(1
/2*x^m/(b^4*c^7*n^2 - 4*a*b^3*c^6*d*n^2 + 6*a^2*b^2*c^5*d^2*n^2 - 4*a^3*b*c^4*d^3*n^2 + a^4*c^3*d^4*n^2 + (b^4
*c^6*d*n^2 - 4*a*b^3*c^5*d^2*n^2 + 6*a^2*b^2*c^4*d^3*n^2 - 4*a^3*b*c^3*d^4*n^2 + a^4*c^2*d^5*n^2)*x^n), x) - (
(b^4*c*e^m*(m - n + 1) - a*b^3*d*e^m*(m - 4*n + 1))*A + (a^2*b^2*d*e^m*(m - 3*n + 1) - a*b^3*c*e^m*(m + 1))*B)
*integrate(x^m/(a^2*b^4*c^4*n - 4*a^3*b^3*c^3*d*n + 6*a^4*b^2*c^2*d^2*n - 4*a^5*b*c*d^3*n + a^6*d^4*n + (a*b^5
*c^4*n - 4*a^2*b^4*c^3*d*n + 6*a^3*b^3*c^2*d^2*n - 4*a^4*b^2*c*d^3*n + a^5*b*d^4*n)*x^n), x) + 1/2*(((a^3*c*d^
3*e^m*(m - 3*n + 1) - a^2*b*c^2*d^2*e^m*(m - 7*n + 1) + 2*b^3*c^4*e^m*n)*A - (a^3*c^2*d^2*e^m*(m - n + 1) - a^
2*b*c^3*d*e^m*(m - 5*n + 1) + 2*a*b^2*c^4*e^m*n)*B)*x*x^m + ((a^2*b*d^4*e^m*(m - 2*n + 1) - a*b^2*c*d^3*e^m*(m
- 6*n + 1) + 2*b^3*c^2*d^2*e^m*n)*A + (a*b^2*c^2*d^2*e^m*(m - 6*n + 1) - a^2*b*c*d^3*e^m*(m + 1))*B)*x*e^(m*l
og(x) + 2*n*log(x)) + ((a^3*d^4*e^m*(m - 2*n + 1) - a*b^2*c^2*d^2*e^m*(m - 7*n + 1) + 4*b^3*c^3*d*e^m*n + 3*a^
2*b*c*d^3*e^m*n)*A + (a*b^2*c^3*d*e^m*(m - 9*n + 1) - a^3*c*d^3*e^m*(m + 1) - 3*a^2*b*c^2*d^2*e^m*n)*B)*x*e^(m
*log(x) + n*log(x)))/(a^2*b^3*c^7*n^2 - 3*a^3*b^2*c^6*d*n^2 + 3*a^4*b*c^5*d^2*n^2 - a^5*c^4*d^3*n^2 + (a*b^4*c
^5*d^2*n^2 - 3*a^2*b^3*c^4*d^3*n^2 + 3*a^3*b^2*c^3*d^4*n^2 - a^4*b*c^2*d^5*n^2)*x^(3*n) + (2*a*b^4*c^6*d*n^2 -
5*a^2*b^3*c^5*d^2*n^2 + 3*a^3*b^2*c^4*d^3*n^2 + a^4*b*c^3*d^4*n^2 - a^5*c^2*d^5*n^2)*x^(2*n) + (a*b^4*c^7*n^2
- a^2*b^3*c^6*d*n^2 - 3*a^3*b^2*c^5*d^2*n^2 + 5*a^4*b*c^4*d^3*n^2 - 2*a^5*c^3*d^4*n^2)*x^n)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B x^{n} + A\right )} \left (e x\right )^{m}}{b^{2} d^{3} x^{5 \, n} + a^{2} c^{3} +{\left (3 \, b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{4 \, n} +{\left (3 \, b^{2} c^{2} d + 6 \, a b c d^{2} + a^{2} d^{3}\right )} x^{3 \, n} +{\left (b^{2} c^{3} + 6 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{2 \, n} +{\left (2 \, a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{n}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*x^n + A)*(e*x)^m/(b^2*d^3*x^(5*n) + a^2*c^3 + (3*b^2*c*d^2 + 2*a*b*d^3)*x^(4*n) + (3*b^2*c^2*d + 6
*a*b*c*d^2 + a^2*d^3)*x^(3*n) + (b^2*c^3 + 6*a*b*c^2*d + 3*a^2*c*d^2)*x^(2*n) + (2*a*b*c^3 + 3*a^2*c^2*d)*x^n)
, 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)/(a+b*x**n)**2/(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 (e x\right )^{m}}{{\left (b x^{n} + a\right )}^{2}{\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)/(a+b*x^n)^2/(c+d*x^n)^3,x, algorithm="giac")

[Out]

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