3.1.40 \(\int \frac {(e x)^m (A+B x^n)}{(a+b x^n)^2 (c+d x^n)^3} \, dx\) [40]

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

[Out]

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

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Rubi [A]
time = 1.36, antiderivative size = 482, normalized size of antiderivative = 1.00, 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 = {609, 611, 371} \begin {gather*} -\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 {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 {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} \end {gather*}

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 609

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 611

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]

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*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(2178\) vs. \(2(482)=964\).
time = 1.85, size = 2178, normalized size = 4.52 \begin {gather*} \text {Result too large to show} \end {gather*}

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*(-(a^2*c^2*d*(b*c - a*d)^2*(B*c - A*d)*(1 + m)*n*(a + b*x^n)) + a^2*c*d*(-(b*c) + a*d)*(1 + m)*(b*c
*(A*d*(1 + m - 6*n) - B*c*(1 + m - 4*n)) + a*d*(B*c*(1 + m) - A*d*(1 + m - 2*n)))*(a + b*x^n)*(c + d*x^n) + 2*
a*b^2*(-(A*b) + a*B)*c^3*(-(b*c) + a*d)*(1 + m)*n*(c + d*x^n)^2 + 2*b^2*c^3*(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*(a + b*x^n)*(c + d*x^n)^2*Hypergeometric2F1[1, (1 + m)/n,
(1 + m + n)/n, -((b*x^n)/a)] - a^2*b^2*B*c^3*d*(a + b*x^n)*(c + d*x^n)^2*Hypergeometric2F1[1, (1 + m)/n, (1 +
m + n)/n, -((d*x^n)/c)] + a^2*A*b^2*c^2*d^2*(a + b*x^n)*(c + d*x^n)^2*Hypergeometric2F1[1, (1 + m)/n, (1 + m +
n)/n, -((d*x^n)/c)] + 2*a^3*b*B*c^2*d^2*(a + b*x^n)*(c + d*x^n)^2*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)
/n, -((d*x^n)/c)] - 2*a^3*A*b*c*d^3*(a + b*x^n)*(c + d*x^n)^2*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -
((d*x^n)/c)] - a^4*B*c*d^3*(a + b*x^n)*(c + d*x^n)^2*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/
c)] + a^4*A*d^4*(a + b*x^n)*(c + d*x^n)^2*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] - 2*a^2
*b^2*B*c^3*d*m*(a + b*x^n)*(c + d*x^n)^2*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] + 2*a^2*
A*b^2*c^2*d^2*m*(a + b*x^n)*(c + d*x^n)^2*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] + 4*a^3
*b*B*c^2*d^2*m*(a + b*x^n)*(c + d*x^n)^2*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] - 4*a^3*
A*b*c*d^3*m*(a + b*x^n)*(c + d*x^n)^2*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] - 2*a^4*B*c
*d^3*m*(a + b*x^n)*(c + d*x^n)^2*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] + 2*a^4*A*d^4*m*
(a + b*x^n)*(c + d*x^n)^2*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] - a^2*b^2*B*c^3*d*m^2*(
a + b*x^n)*(c + d*x^n)^2*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] + a^2*A*b^2*c^2*d^2*m^2*
(a + b*x^n)*(c + d*x^n)^2*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] + 2*a^3*b*B*c^2*d^2*m^2
*(a + b*x^n)*(c + d*x^n)^2*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] - 2*a^3*A*b*c*d^3*m^2*
(a + b*x^n)*(c + d*x^n)^2*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] - a^4*B*c*d^3*m^2*(a +
b*x^n)*(c + d*x^n)^2*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] + a^4*A*d^4*m^2*(a + b*x^n)*
(c + d*x^n)^2*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] + 5*a^2*b^2*B*c^3*d*n*(a + b*x^n)*(
c + d*x^n)^2*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] - 7*a^2*A*b^2*c^2*d^2*n*(a + b*x^n)*
(c + d*x^n)^2*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] - 6*a^3*b*B*c^2*d^2*n*(a + b*x^n)*(
c + d*x^n)^2*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] + 10*a^3*A*b*c*d^3*n*(a + b*x^n)*(c
+ d*x^n)^2*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] + a^4*B*c*d^3*n*(a + b*x^n)*(c + d*x^n
)^2*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] - 3*a^4*A*d^4*n*(a + b*x^n)*(c + d*x^n)^2*Hyp
ergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] + 5*a^2*b^2*B*c^3*d*m*n*(a + b*x^n)*(c + d*x^n)^2*Hy
pergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] - 7*a^2*A*b^2*c^2*d^2*m*n*(a + b*x^n)*(c + d*x^n)^2
*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] - 6*a^3*b*B*c^2*d^2*m*n*(a + b*x^n)*(c + d*x^n)^
2*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] + 10*a^3*A*b*c*d^3*m*n*(a + b*x^n)*(c + d*x^n)^
2*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] + a^4*B*c*d^3*m*n*(a + b*x^n)*(c + d*x^n)^2*Hyp
ergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] - 3*a^4*A*d^4*m*n*(a + b*x^n)*(c + d*x^n)^2*Hypergeo
metric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] - 6*a^2*b^2*B*c^3*d*n^2*(a + b*x^n)*(c + d*x^n)^2*Hyperge
ometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] + 12*a^2*A*b^2*c^2*d^2*n^2*(a + b*x^n)*(c + d*x^n)^2*Hyp
ergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] - 8*a^3*A*b*c*d^3*n^2*(a + b*x^n)*(c + d*x^n)^2*Hype
rgeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] + 2*a^4*A*d^4*n^2*(a + b*x^n)*(c + d*x^n)^2*Hypergeom
etric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)]))/(2*a^2*c^3*(b*c - a*d)^4*(1 + m)*n^2*(a + b*x^n)*(c + d*
x^n)^2)

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Maple [F]
time = 0.12, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{m} \left (A +B \,x^{n}\right )}{\left (a +b \,x^{n}\right )^{2} \left (c +d \,x^{n}\right )^{3}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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*e^m - m*(7*n - 2)*e^m + (12*n^2 - 7*n + 1)*e^m)*b^2*c^2*d^2 - 2*(m^2*e^m - m*(5*n - 2)*e^m + (4*n^2 - 5
*n + 1)*e^m)*a*b*c*d^3 + (m^2*e^m - m*(3*n - 2)*e^m + (2*n^2 - 3*n + 1)*e^m)*a^2*d^4)*A - ((m^2*e^m - m*(5*n -
2)*e^m + (6*n^2 - 5*n + 1)*e^m)*b^2*c^3*d - 2*(m^2*e^m - m*(3*n - 2)*e^m - (3*n - 1)*e^m)*a*b*c^2*d^2 + (m^2*
e^m - m*(n - 2)*e^m - (n - 1)*e^m)*a^2*c*d^3)*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) - (((m*e^m - (n - 1)*e^m)*b^4*c - (m*e^m - (4*n - 1)*
e^m)*a*b^3*d)*A - ((m*e^m + e^m)*a*b^3*c - (m*e^m - (3*n - 1)*e^m)*a^2*b^2*d)*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*(((2*b^3*c^4*n*e^m - (m*e^m - (7*n - 1)*e
^m)*a^2*b*c^2*d^2 + (m*e^m - (3*n - 1)*e^m)*a^3*c*d^3)*A - (2*a*b^2*c^4*n*e^m - (m*e^m - (5*n - 1)*e^m)*a^2*b*
c^3*d + (m*e^m - (n - 1)*e^m)*a^3*c^2*d^2)*B)*x*x^m + ((2*b^3*c^2*d^2*n*e^m - (m*e^m - (6*n - 1)*e^m)*a*b^2*c*
d^3 + (m*e^m - (2*n - 1)*e^m)*a^2*b*d^4)*A + ((m*e^m - (6*n - 1)*e^m)*a*b^2*c^2*d^2 - (m*e^m + e^m)*a^2*b*c*d^
3)*B)*x*e^(m*log(x) + 2*n*log(x)) + ((4*b^3*c^3*d*n*e^m + 3*a^2*b*c*d^3*n*e^m - (m*e^m - (7*n - 1)*e^m)*a*b^2*
c^2*d^2 + (m*e^m - (2*n - 1)*e^m)*a^3*d^4)*A - (3*a^2*b*c^2*d^2*n*e^m - (m*e^m - (9*n - 1)*e^m)*a*b^2*c^3*d +
(m*e^m + e^m)*a^3*c*d^3)*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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)*(x*e)^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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (e\,x\right )}^m\,\left (A+B\,x^n\right )}{{\left (a+b\,x^n\right )}^2\,{\left (c+d\,x^n\right )}^3} \,d x \end {gather*}

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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