Integrand size = 31, antiderivative size = 394 \[ \int \frac {(e x)^m \left (A+B x^n\right ) \left (c+d x^n\right )^3}{\left (a+b x^n\right )^2} \, dx=-\frac {d^2 (A b (3 b c (1+m+n)-a d (1+m+2 n))-a B (3 b c (1+m+2 n)-a d (1+m+3 n))) x^{1+n} (e x)^m}{a b^3 n (1+m+n)}-\frac {d^3 (A b (1+m+2 n)-a B (1+m+3 n)) x^{1+2 n} (e x)^m}{a b^2 n (1+m+2 n)}-\frac {d \left (A b \left (3 b^2 c^2 (1+m)-3 a b c d (1+m+n)+a^2 d^2 (1+m+2 n)\right )-a B \left (3 b^2 c^2 (1+m+n)-3 a b c d (1+m+2 n)+a^2 d^2 (1+m+3 n)\right )\right ) (e x)^{1+m}}{a b^4 e (1+m) n}+\frac {(A b-a B) (e x)^{1+m} \left (c+d x^n\right )^3}{a b e n \left (a+b x^n\right )}-\frac {(b c-a d)^2 (A b (b c (1+m-n)-a d (1+m+2 n))-a B (b c (1+m)-a d (1+m+3 n))) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )}{a^2 b^4 e (1+m) n} \]
-d^2*(A*b*(3*b*c*(1+m+n)-a*d*(1+m+2*n))-a*B*(3*b*c*(1+m+2*n)-a*d*(1+m+3*n) ))*x^(1+n)*(e*x)^m/a/b^3/n/(1+m+n)-d^3*(A*b*(1+m+2*n)-a*B*(1+m+3*n))*x^(1+ 2*n)*(e*x)^m/a/b^2/n/(1+m+2*n)-d*(A*b*(3*b^2*c^2*(1+m)-3*a*b*c*d*(1+m+n)+a ^2*d^2*(1+m+2*n))-a*B*(3*b^2*c^2*(1+m+n)-3*a*b*c*d*(1+m+2*n)+a^2*d^2*(1+m+ 3*n)))*(e*x)^(1+m)/a/b^4/e/(1+m)/n+(A*b-B*a)*(e*x)^(1+m)*(c+d*x^n)^3/a/b/e /n/(a+b*x^n)-(-a*d+b*c)^2*(A*b*(b*c*(1+m-n)-a*d*(1+m+2*n))-a*B*(b*c*(1+m)- a*d*(1+m+3*n)))*(e*x)^(1+m)*hypergeom([1, (1+m)/n],[(1+m+n)/n],-b*x^n/a)/a ^2/b^4/e/(1+m)/n
Time = 1.06 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.55 \[ \int \frac {(e x)^m \left (A+B x^n\right ) \left (c+d x^n\right )^3}{\left (a+b x^n\right )^2} \, dx=\frac {x (e x)^m \left (\frac {d \left (3 a^2 B d^2+3 b^2 c (B c+A d)-2 a b d (3 B c+A d)\right )}{1+m}+\frac {b d^2 (3 b B c+A b d-2 a B d) x^n}{1+m+n}+\frac {b^2 B d^3 x^{2 n}}{1+m+2 n}+\frac {(b c-a d)^2 (b B c+3 A b d-4 a B d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )}{a (1+m)}+\frac {(-A b+a B) (-b c+a d)^3 \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )}{a^2 (1+m)}\right )}{b^4} \]
(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^n)/(1 + m + n) + (b^2*B*d ^3*x^(2*n))/(1 + m + 2*n) + ((b*c - a*d)^2*(b*B*c + 3*A*b*d - 4*a*B*d)*Hyp ergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((b*x^n)/a)])/(a*(1 + m)) + ( (-(A*b) + a*B)*(-(b*c) + a*d)^3*Hypergeometric2F1[2, (1 + m)/n, (1 + m + n )/n, -((b*x^n)/a)])/(a^2*(1 + m))))/b^4
Time = 0.98 (sec) , antiderivative size = 376, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {1064, 25, 1040, 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.
| \(\displaystyle \int \frac {(e x)^m \left (A+B x^n\right ) \left (c+d x^n\right )^3}{\left (a+b x^n\right )^2} \, dx\) |
| \(\Big \downarrow \) 1064 |
| \(\displaystyle \frac {(e x)^{m+1} (A b-a B) \left (c+d x^n\right )^3}{a b e n \left (a+b x^n\right )}-\frac {\int -\frac {(e x)^m \left (d x^n+c\right )^2 \left (c (a B (m+1)-A b (m-n+1))-d (A b (m+2 n+1)-a B (m+3 n+1)) x^n\right )}{b x^n+a}dx}{a b n}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {(e x)^m \left (d x^n+c\right )^2 \left (c (a B (m+1)-A b (m-n+1))-d (A b (m+2 n+1)-a B (m+3 n+1)) x^n\right )}{b x^n+a}dx}{a b n}+\frac {(e x)^{m+1} (A b-a B) \left (c+d x^n\right )^3}{a b e n \left (a+b x^n\right )}\) |
| \(\Big \downarrow \) 1040 |
| \(\displaystyle \frac {\int \left (\frac {d^2 (a B (3 b c (m+2 n+1)-a d (m+3 n+1))-A b (3 b c (m+n+1)-a d (m+2 n+1))) x^n (e x)^m}{b^2}+\frac {d^3 (a B (m+3 n+1)-A b (m+2 n+1)) x^{2 n} (e x)^m}{b}+\frac {d \left (a B \left (3 b^2 (m+n+1) c^2-3 a b d (m+2 n+1) c+a^2 d^2 (m+3 n+1)\right )-A b \left (3 b^2 (m+1) c^2-3 a b d (m+n+1) c+a^2 d^2 (m+2 n+1)\right )\right ) (e x)^m}{b^3}+\frac {(b c-a d)^2 (a B (b c (m+1)-a d (m+3 n+1))-A b (b c (m-n+1)-a d (m+2 n+1))) (e x)^m}{b^3 \left (b x^n+a\right )}\right )dx}{a b n}+\frac {(e x)^{m+1} (A b-a B) \left (c+d x^n\right )^3}{a b e n \left (a+b x^n\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {d (e x)^{m+1} \left (A b \left (a^2 d^2 (m+2 n+1)-3 a b c d (m+n+1)+3 b^2 c^2 (m+1)\right )-a B \left (a^2 d^2 (m+3 n+1)-3 a b c d (m+2 n+1)+3 b^2 c^2 (m+n+1)\right )\right )}{b^3 e (m+1)}-\frac {(e x)^{m+1} (b c-a d)^2 \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {b x^n}{a}\right ) (A b (b c (m-n+1)-a d (m+2 n+1))-a B (b c (m+1)-a d (m+3 n+1)))}{a b^3 e (m+1)}-\frac {d^2 x^{n+1} (e x)^m (A b (3 b c (m+n+1)-a d (m+2 n+1))-a B (3 b c (m+2 n+1)-a d (m+3 n+1)))}{b^2 (m+n+1)}-d^3 x^{2 n+1} (e x)^m \left (A-\frac {a B (m+3 n+1)}{b (m+2 n+1)}\right )}{a b n}+\frac {(e x)^{m+1} (A b-a B) \left (c+d x^n\right )^3}{a b e n \left (a+b x^n\right )}\) |
((A*b - a*B)*(e*x)^(1 + m)*(c + d*x^n)^3)/(a*b*e*n*(a + b*x^n)) + (-((d^2* (A*b*(3*b*c*(1 + m + n) - a*d*(1 + m + 2*n)) - a*B*(3*b*c*(1 + m + 2*n) - a*d*(1 + m + 3*n)))*x^(1 + n)*(e*x)^m)/(b^2*(1 + m + n))) - d^3*(A - (a*B* (1 + m + 3*n))/(b*(1 + m + 2*n)))*x^(1 + 2*n)*(e*x)^m - (d*(A*b*(3*b^2*c^2 *(1 + m) - 3*a*b*c*d*(1 + m + n) + a^2*d^2*(1 + m + 2*n)) - a*B*(3*b^2*c^2 *(1 + m + n) - 3*a*b*c*d*(1 + m + 2*n) + a^2*d^2*(1 + m + 3*n)))*(e*x)^(1 + m))/(b^3*e*(1 + m)) - ((b*c - a*d)^2*(A*b*(b*c*(1 + m - n) - a*d*(1 + m + 2*n)) - a*B*(b*c*(1 + m) - a*d*(1 + m + 3*n)))*(e*x)^(1 + m)*Hypergeomet ric2F1[1, (1 + m)/n, (1 + m + n)/n, -((b*x^n)/a)])/(a*b^3*e*(1 + m)))/(a*b *n)
3.1.20.3.1 Defintions of rubi rules used
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _))^(q_.)*((e_) + (f_.)*(x_)^(n_))^(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]
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] + Simp[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])
\[\int \frac {\left (e x \right )^{m} \left (A +B \,x^{n}\right ) \left (c +d \,x^{n}\right )^{3}}{\left (a +b \,x^{n}\right )^{2}}d x\]
\[ \int \frac {(e x)^m \left (A+B x^n\right ) \left (c+d x^n\right )^3}{\left (a+b x^n\right )^2} \, dx=\int { \frac {{\left (B x^{n} + A\right )} {\left (d x^{n} + c\right )}^{3} \left (e x\right )^{m}}{{\left (b x^{n} + a\right )}^{2}} \,d x } \]
integral((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)*(e*x)^m/(b^2*x^(2*n) + 2*a* b*x^n + a^2), x)
Timed out. \[ \int \frac {(e x)^m \left (A+B x^n\right ) \left (c+d x^n\right )^3}{\left (a+b x^n\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {(e x)^m \left (A+B x^n\right ) \left (c+d x^n\right )^3}{\left (a+b x^n\right )^2} \, dx=\int { \frac {{\left (B x^{n} + A\right )} {\left (d x^{n} + c\right )}^{3} \left (e x\right )^{m}}{{\left (b x^{n} + a\right )}^{2}} \,d x } \]
((a^3*b*d^3*e^m*(m + 2*n + 1) - 3*a^2*b^2*c*d^2*e^m*(m + n + 1) - b^4*c^3* e^m*(m - n + 1) + 3*a*b^3*c^2*d*e^m*(m + 1))*A - (a^4*d^3*e^m*(m + 3*n + 1 ) - 3*a^3*b*c*d^2*e^m*(m + 2*n + 1) + 3*a^2*b^2*c^2*d*e^m*(m + n + 1) - a* b^3*c^3*e^m*(m + 1))*B)*integrate(x^m/(a*b^5*n*x^n + a^2*b^4*n), x) + ((m^ 2*n + (n^2 + 2*n)*m + n^2 + n)*B*a*b^3*d^3*e^m*x*e^(m*log(x) + 3*n*log(x)) + (((m^3 + 3*m^2*(n + 1) + (2*n^2 + 6*n + 3)*m + 2*n^2 + 3*n + 1)*b^4*c^3 *e^m - 3*(m^3 + 3*m^2*(n + 1) + (2*n^2 + 6*n + 3)*m + 2*n^2 + 3*n + 1)*a*b ^3*c^2*d*e^m + 3*(m^3 + m^2*(4*n + 3) + 2*n^3 + (5*n^2 + 8*n + 3)*m + 5*n^ 2 + 4*n + 1)*a^2*b^2*c*d^2*e^m - (m^3 + m^2*(5*n + 3) + 4*n^3 + (8*n^2 + 1 0*n + 3)*m + 8*n^2 + 5*n + 1)*a^3*b*d^3*e^m)*A - ((m^3 + 3*m^2*(n + 1) + ( 2*n^2 + 6*n + 3)*m + 2*n^2 + 3*n + 1)*a*b^3*c^3*e^m - 3*(m^3 + m^2*(4*n + 3) + 2*n^3 + (5*n^2 + 8*n + 3)*m + 5*n^2 + 4*n + 1)*a^2*b^2*c^2*d*e^m + 3* (m^3 + m^2*(5*n + 3) + 4*n^3 + (8*n^2 + 10*n + 3)*m + 8*n^2 + 5*n + 1)*a^3 *b*c*d^2*e^m - (m^3 + 3*m^2*(2*n + 1) + 6*n^3 + (11*n^2 + 12*n + 3)*m + 11 *n^2 + 6*n + 1)*a^4*d^3*e^m)*B)*x*x^m + ((m^2*n + 2*(n^2 + n)*m + 2*n^2 + n)*A*a*b^3*d^3*e^m + (3*(m^2*n + 2*(n^2 + n)*m + 2*n^2 + n)*a*b^3*c*d^2*e^ m - (m^2*n + (3*n^2 + 2*n)*m + 3*n^2 + n)*a^2*b^2*d^3*e^m)*B)*x*e^(m*log(x ) + 2*n*log(x)) + ((3*(m^2*n + 2*n^3 + (3*n^2 + 2*n)*m + 3*n^2 + n)*a*b^3* c*d^2*e^m - (m^2*n + 4*n^3 + 2*(2*n^2 + n)*m + 4*n^2 + n)*a^2*b^2*d^3*e^m) *A + (3*(m^2*n + 2*n^3 + (3*n^2 + 2*n)*m + 3*n^2 + n)*a*b^3*c^2*d*e^m -...
\[ \int \frac {(e x)^m \left (A+B x^n\right ) \left (c+d x^n\right )^3}{\left (a+b x^n\right )^2} \, dx=\int { \frac {{\left (B x^{n} + A\right )} {\left (d x^{n} + c\right )}^{3} \left (e x\right )^{m}}{{\left (b x^{n} + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(e x)^m \left (A+B x^n\right ) \left (c+d x^n\right )^3}{\left (a+b x^n\right )^2} \, dx=\int \frac {{\left (e\,x\right )}^m\,\left (A+B\,x^n\right )\,{\left (c+d\,x^n\right )}^3}{{\left (a+b\,x^n\right )}^2} \,d x \]