Integrand size = 31, antiderivative size = 366 \[ \int \frac {(e x)^m \left (A+B x^n\right )}{\left (a+b x^n\right ) \left (c+d x^n\right )^3} \, dx=\frac {(B c-A d) (e x)^{1+m}}{2 c (b c-a d) e n \left (c+d x^n\right )^2}+\frac {(b c (A d (1+m-4 n)-B c (1+m-2 n))+a d (B c (1+m)-A d (1+m-2 n))) (e x)^{1+m}}{2 c^2 (b c-a d)^2 e n^2 \left (c+d x^n\right )}+\frac {b^2 (A b-a B) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )}{a (b c-a d)^3 e (1+m)}-\frac {\left (b^2 c^2 (A d (1+m-3 n)-B c (1+m-n)) (1+m-2 n)-a^2 d^2 (B c (1+m)-A d (1+m-2 n)) (1+m-n)+2 a b c d \left (B c (1+m) (1+m-2 n)-A d \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )\right )\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{2 c^3 (b c-a d)^3 e (1+m) n^2} \]
1/2*(-A*d+B*c)*(e*x)^(1+m)/c/(-a*d+b*c)/e/n/(c+d*x^n)^2+1/2*(b*c*(A*d*(1+m -4*n)-B*c*(1+m-2*n))+a*d*(B*c*(1+m)-A*d*(1+m-2*n)))*(e*x)^(1+m)/c^2/(-a*d+ b*c)^2/e/n^2/(c+d*x^n)+b^2*(A*b-B*a)*(e*x)^(1+m)*hypergeom([1, (1+m)/n],[( 1+m+n)/n],-b*x^n/a)/a/(-a*d+b*c)^3/e/(1+m)-1/2*(b^2*c^2*(A*d*(1+m-3*n)-B*c *(1+m-n))*(1+m-2*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-2*n)-A*d*(1+m^2+m*(2-4*n)-4*n+3*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)^3/e/(1+m)/n^2
Time = 0.50 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.55 \[ \int \frac {(e x)^m \left (A+B x^n\right )}{\left (a+b x^n\right ) \left (c+d x^n\right )^3} \, dx=\frac {x (e x)^m \left (\frac {b^2 (A b-a B) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )}{a}-\frac {b (A b-a B) d \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{c}-\frac {(A b-a B) d (b c-a d) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{c^2}+\frac {(b c-a d)^2 (B c-A d) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{c^3}\right )}{(b c-a d)^3 (1+m)} \]
(x*(e*x)^m*((b^2*(A*b - a*B)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n , -((b*x^n)/a)])/a - (b*(A*b - a*B)*d*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)])/c - ((A*b - a*B)*d*(b*c - a*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))/(( b*c - a*d)^3*(1 + m))
Time = 1.14 (sec) , antiderivative size = 404, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {1065, 25, 1065, 1067, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(e x)^m \left (A+B x^n\right )}{\left (a+b x^n\right ) \left (c+d x^n\right )^3} \, dx\) |
\(\Big \downarrow \) 1065 |
\(\displaystyle \frac {\int -\frac {(e x)^m \left (b (B c-A d) (m-2 n+1) x^n+a B c (m+1)-a A d (m-2 n+1)-2 A b c n\right )}{\left (b x^n+a\right ) \left (d x^n+c\right )^2}dx}{2 c n (b c-a d)}+\frac {(e x)^{m+1} (B c-A d)}{2 c e n (b c-a d) \left (c+d x^n\right )^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {(e x)^{m+1} (B c-A d)}{2 c e n (b c-a d) \left (c+d x^n\right )^2}-\frac {\int \frac {(e x)^m \left (b (B c-A d) (m-2 n+1) x^n+a B c (m+1)-a A d (m-2 n+1)-2 A b c n\right )}{\left (b x^n+a\right ) \left (d x^n+c\right )^2}dx}{2 c n (b c-a d)}\) |
\(\Big \downarrow \) 1065 |
\(\displaystyle \frac {(e x)^{m+1} (B c-A d)}{2 c e n (b c-a d) \left (c+d x^n\right )^2}-\frac {\frac {\int \frac {(e x)^m \left (b (b c (A d (m-4 n+1)-B c (m-2 n+1))+a d (B c (m+1)-A d (m-2 n+1))) (m-n+1) x^n+a (m+1) (b c (A d (m-4 n+1)-B c (m-2 n+1))+a d (B c (m+1)-A d (m-2 n+1)))+(b c-a d) n (a B c (m+1)-a A d (m-2 n+1)-2 A b c n)\right )}{\left (b x^n+a\right ) \left (d x^n+c\right )}dx}{c n (b c-a d)}-\frac {(e x)^{m+1} (a d (B c (m+1)-A d (m-2 n+1))+b c (A d (m-4 n+1)-B c (m-2 n+1)))}{c e n (b c-a d) \left (c+d x^n\right )}}{2 c n (b c-a d)}\) |
\(\Big \downarrow \) 1067 |
\(\displaystyle \frac {(e x)^{m+1} (B c-A d)}{2 c e n (b c-a d) \left (c+d x^n\right )^2}-\frac {\frac {\int \left (\frac {\left (b^2 (A d (m-3 n+1)-B c (m-n+1)) (m-2 n+1) c^2+2 a b d \left (B c (m+1) (m-2 n+1)-A d \left (m^2+(2-4 n) m+3 n^2-4 n+1\right )\right ) c-a^2 d^2 (B c (m+1)-A d (m-2 n+1)) (m-n+1)\right ) (e x)^m}{(b c-a d) \left (d x^n+c\right )}-\frac {2 b^2 (A b-a B) c^2 n^2 (e x)^m}{(b c-a d) \left (b x^n+a\right )}\right )dx}{c n (b c-a d)}-\frac {(e x)^{m+1} (a d (B c (m+1)-A d (m-2 n+1))+b c (A d (m-4 n+1)-B c (m-2 n+1)))}{c e n (b c-a d) \left (c+d x^n\right )}}{2 c n (b c-a d)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(e x)^{m+1} (B c-A d)}{2 c e n (b c-a d) \left (c+d x^n\right )^2}-\frac {\frac {\frac {(e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {d x^n}{c}\right ) \left (-a^2 d^2 (m-n+1) (B c (m+1)-A d (m-2 n+1))+2 a b c d \left (B c (m+1) (m-2 n+1)-A d \left (m^2+m (2-4 n)+3 n^2-4 n+1\right )\right )+b^2 c^2 (m-2 n+1) (A d (m-3 n+1)-B c (m-n+1))\right )}{c e (m+1) (b c-a d)}-\frac {2 b^2 c^2 n^2 (e x)^{m+1} (A b-a B) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {b x^n}{a}\right )}{a e (m+1) (b c-a d)}}{c n (b c-a d)}-\frac {(e x)^{m+1} (a d (B c (m+1)-A d (m-2 n+1))+b c (A d (m-4 n+1)-B c (m-2 n+1)))}{c e n (b c-a d) \left (c+d x^n\right )}}{2 c n (b c-a d)}\) |
((B*c - A*d)*(e*x)^(1 + m))/(2*c*(b*c - a*d)*e*n*(c + d*x^n)^2) - (-(((b*c *(A*d*(1 + m - 4*n) - B*c*(1 + m - 2*n)) + a*d*(B*c*(1 + m) - A*d*(1 + m - 2*n)))*(e*x)^(1 + m))/(c*(b*c - a*d)*e*n*(c + d*x^n))) + ((-2*b^2*(A*b - a*B)*c^2*n^2*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((b*x^n)/a)])/(a*(b*c - a*d)*e*(1 + m)) + ((b^2*c^2*(A*d*(1 + m - 3*n) - B*c*(1 + m - n))*(1 + m - 2*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 - 2*n) - A*d*(1 + m^2 + m*(2 - 4*n) - 4*n + 3*n^2)))*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)])/(c*(b*c - a*d)*e*(1 + m)))/(c*(b*c - a*d)*n))/(2 *c*(b*c - a*d)*n)
3.1.39.3.1 Defintions of rubi rules used
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] + Simp[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]
Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n _)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> 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]
\[\int \frac {\left (e x \right )^{m} \left (A +B \,x^{n}\right )}{\left (a +b \,x^{n}\right ) \left (c +d \,x^{n}\right )^{3}}d x\]
\[ \int \frac {(e x)^m \left (A+B x^n\right )}{\left (a+b x^n\right ) \left (c+d x^n\right )^3} \, dx=\int { \frac {{\left (B x^{n} + A\right )} \left (e x\right )^{m}}{{\left (b x^{n} + a\right )} {\left (d x^{n} + c\right )}^{3}} \,d x } \]
integral((B*x^n + A)*(e*x)^m/(b*d^3*x^(4*n) + a*c^3 + (3*b*c*d^2 + a*d^3)* x^(3*n) + 3*(b*c^2*d + a*c*d^2)*x^(2*n) + (b*c^3 + 3*a*c^2*d)*x^n), x)
Exception generated. \[ \int \frac {(e x)^m \left (A+B x^n\right )}{\left (a+b x^n\right ) \left (c+d x^n\right )^3} \, dx=\text {Exception raised: HeuristicGCDFailed} \]
\[ \int \frac {(e x)^m \left (A+B x^n\right )}{\left (a+b x^n\right ) \left (c+d x^n\right )^3} \, dx=\int { \frac {{\left (B x^{n} + A\right )} \left (e x\right )^{m}}{{\left (b x^{n} + a\right )} {\left (d x^{n} + c\right )}^{3}} \,d x } \]
(((m^2 - m*(5*n - 2) + 6*n^2 - 5*n + 1)*b^2*c^2*d*e^m - 2*(m^2 - 2*m*(2*n - 1) + 3*n^2 - 4*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 - 2*m*(n - 1) - 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/(b^3*c^6*n^2 - 3*a*b^2*c^5*d*n^2 + 3* a^2*b*c^4*d^2*n^2 - a^3*c^3*d^3*n^2 + (b^3*c^5*d*n^2 - 3*a*b^2*c^4*d^2*n^2 + 3*a^2*b*c^3*d^3*n^2 - a^3*c^2*d^4*n^2)*x^n), x) + (B*a*b^2*e^m - A*b^3* e^m)*integrate(-x^m/(a*b^3*c^3 - 3*a^2*b^2*c^2*d + 3*a^3*b*c*d^2 - a^4*d^3 + (b^4*c^3 - 3*a*b^3*c^2*d + 3*a^2*b^2*c*d^2 - a^3*b*d^3)*x^n), x) - 1/2* (((a*c*d^2*e^m*(m - 3*n + 1) - b*c^2*d*e^m*(m - 5*n + 1))*A - (a*c^2*d*e^m *(m - n + 1) - b*c^3*e^m*(m - 3*n + 1))*B)*x*x^m + ((a*d^3*e^m*(m - 2*n + 1) - b*c*d^2*e^m*(m - 4*n + 1))*A + (b*c^2*d*e^m*(m - 2*n + 1) - a*c*d^2*e ^m*(m + 1))*B)*x*e^(m*log(x) + n*log(x)))/(b^2*c^6*n^2 - 2*a*b*c^5*d*n^2 + a^2*c^4*d^2*n^2 + (b^2*c^4*d^2*n^2 - 2*a*b*c^3*d^3*n^2 + a^2*c^2*d^4*n^2) *x^(2*n) + 2*(b^2*c^5*d*n^2 - 2*a*b*c^4*d^2*n^2 + a^2*c^3*d^3*n^2)*x^n)
\[ \int \frac {(e x)^m \left (A+B x^n\right )}{\left (a+b x^n\right ) \left (c+d x^n\right )^3} \, dx=\int { \frac {{\left (B x^{n} + A\right )} \left (e x\right )^{m}}{{\left (b x^{n} + a\right )} {\left (d x^{n} + c\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {(e x)^m \left (A+B x^n\right )}{\left (a+b x^n\right ) \left (c+d x^n\right )^3} \, dx=\int \frac {{\left (e\,x\right )}^m\,\left (A+B\,x^n\right )}{\left (a+b\,x^n\right )\,{\left (c+d\,x^n\right )}^3} \,d x \]