Integrand size = 22, antiderivative size = 281 \[ \int \frac {(a+b x)^m}{(c+d x)^2 (e+f x)^2} \, dx=\frac {f (b d e+b c f-2 a d f) (a+b x)^{1+m}}{(b c-a d) (b e-a f) (d e-c f)^2 (e+f x)}+\frac {d (a+b x)^{1+m}}{(b c-a d) (d e-c f) (c+d x) (e+f x)}+\frac {d^2 (2 a d f-b (c f (2-m)+d e m)) (a+b x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {d (a+b x)}{b c-a d}\right )}{(b c-a d)^2 (d e-c f)^3 (1+m)}-\frac {f^2 (2 a d f-b (d e (2-m)+c f m)) (a+b x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {f (a+b x)}{b e-a f}\right )}{(b e-a f)^2 (d e-c f)^3 (1+m)} \]
f*(-2*a*d*f+b*c*f+b*d*e)*(b*x+a)^(1+m)/(-a*d+b*c)/(-a*f+b*e)/(-c*f+d*e)^2/ (f*x+e)+d*(b*x+a)^(1+m)/(-a*d+b*c)/(-c*f+d*e)/(d*x+c)/(f*x+e)+d^2*(2*a*d*f -b*(c*f*(2-m)+d*e*m))*(b*x+a)^(1+m)*hypergeom([1, 1+m],[2+m],-d*(b*x+a)/(- a*d+b*c))/(-a*d+b*c)^2/(-c*f+d*e)^3/(1+m)-f^2*(2*a*d*f-b*(d*e*(2-m)+c*f*m) )*(b*x+a)^(1+m)*hypergeom([1, 1+m],[2+m],-f*(b*x+a)/(-a*f+b*e))/(-a*f+b*e) ^2/(-c*f+d*e)^3/(1+m)
Time = 0.37 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.88 \[ \int \frac {(a+b x)^m}{(c+d x)^2 (e+f x)^2} \, dx=\frac {(a+b x)^{1+m} \left (-\frac {f (b d e+b c f-2 a d f)}{(b e-a f) (d e-c f) (e+f x)}-\frac {d}{(c+d x) (e+f x)}-\frac {-d^2 (b e-a f)^2 (-2 a d f-b c f (-2+m)+b d e m) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {d (a+b x)}{-b c+a d}\right )+(b c-a d)^2 f^2 (-2 a d f-b d e (-2+m)+b c f m) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {f (a+b x)}{-b e+a f}\right )}{(b c-a d) (b e-a f)^2 (d e-c f)^2 (1+m)}\right )}{(b c-a d) (-d e+c f)} \]
((a + b*x)^(1 + m)*(-((f*(b*d*e + b*c*f - 2*a*d*f))/((b*e - a*f)*(d*e - c* f)*(e + f*x))) - d/((c + d*x)*(e + f*x)) - (-(d^2*(b*e - a*f)^2*(-2*a*d*f - b*c*f*(-2 + m) + b*d*e*m)*Hypergeometric2F1[1, 1 + m, 2 + m, (d*(a + b*x ))/(-(b*c) + a*d)]) + (b*c - a*d)^2*f^2*(-2*a*d*f - b*d*e*(-2 + m) + b*c*f *m)*Hypergeometric2F1[1, 1 + m, 2 + m, (f*(a + b*x))/(-(b*e) + a*f)])/((b* c - a*d)*(b*e - a*f)^2*(d*e - c*f)^2*(1 + m))))/((b*c - a*d)*(-(d*e) + c*f ))
Time = 0.51 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.18, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {114, 168, 174, 78}
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.
| \(\displaystyle \int \frac {(a+b x)^m}{(c+d x)^2 (e+f x)^2} \, dx\) |
| \(\Big \downarrow \) 114 |
| \(\displaystyle \frac {\int \frac {(a+b x)^m (2 a d f+b d (1-m) x f-b (c f+d e m))}{(c+d x) (e+f x)^2}dx}{(b c-a d) (d e-c f)}+\frac {d (a+b x)^{m+1}}{(c+d x) (e+f x) (b c-a d) (d e-c f)}\) |
| \(\Big \downarrow \) 168 |
| \(\displaystyle \frac {\frac {f (a+b x)^{m+1} (-2 a d f+b c f+b d e)}{(e+f x) (b e-a f) (d e-c f)}-\frac {\int \frac {(a+b x)^m \left (\left (d^2 m e^2+2 c d f e+c^2 f^2 m\right ) b^2-a d f (d e+c f) (m+2) b+d f (b d e+b c f-2 a d f) m x b+2 a^2 d^2 f^2\right )}{(c+d x) (e+f x)}dx}{(b e-a f) (d e-c f)}}{(b c-a d) (d e-c f)}+\frac {d (a+b x)^{m+1}}{(c+d x) (e+f x) (b c-a d) (d e-c f)}\) |
| \(\Big \downarrow \) 174 |
| \(\displaystyle \frac {\frac {f (a+b x)^{m+1} (-2 a d f+b c f+b d e)}{(e+f x) (b e-a f) (d e-c f)}-\frac {\frac {f^2 (b c-a d) (2 a d f-b c f m-b d e (2-m)) \int \frac {(a+b x)^m}{e+f x}dx}{d e-c f}-\frac {d^2 (b e-a f) (2 a d f-b c f (2-m)-b d e m) \int \frac {(a+b x)^m}{c+d x}dx}{d e-c f}}{(b e-a f) (d e-c f)}}{(b c-a d) (d e-c f)}+\frac {d (a+b x)^{m+1}}{(c+d x) (e+f x) (b c-a d) (d e-c f)}\) |
| \(\Big \downarrow \) 78 |
| \(\displaystyle \frac {\frac {f (a+b x)^{m+1} (-2 a d f+b c f+b d e)}{(e+f x) (b e-a f) (d e-c f)}-\frac {\frac {f^2 (b c-a d) (a+b x)^{m+1} (2 a d f-b c f m-b d e (2-m)) \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {f (a+b x)}{b e-a f}\right )}{(m+1) (b e-a f) (d e-c f)}-\frac {d^2 (b e-a f) (a+b x)^{m+1} (2 a d f-b c f (2-m)-b d e m) \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {d (a+b x)}{b c-a d}\right )}{(m+1) (b c-a d) (d e-c f)}}{(b e-a f) (d e-c f)}}{(b c-a d) (d e-c f)}+\frac {d (a+b x)^{m+1}}{(c+d x) (e+f x) (b c-a d) (d e-c f)}\) |
(d*(a + b*x)^(1 + m))/((b*c - a*d)*(d*e - c*f)*(c + d*x)*(e + f*x)) + ((f* (b*d*e + b*c*f - 2*a*d*f)*(a + b*x)^(1 + m))/((b*e - a*f)*(d*e - c*f)*(e + f*x)) - (-((d^2*(b*e - a*f)*(2*a*d*f - b*c*f*(2 - m) - b*d*e*m)*(a + b*x) ^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -((d*(a + b*x))/(b*c - a*d))]) /((b*c - a*d)*(d*e - c*f)*(1 + m))) + ((b*c - a*d)*f^2*(2*a*d*f - b*d*e*(2 - m) - b*c*f*m)*(a + b*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -((f *(a + b*x))/(b*e - a*f))])/((b*e - a*f)*(d*e - c*f)*(1 + m)))/((b*e - a*f) *(d*e - c*f)))/((b*c - a*d)*(d*e - c*f))
3.33.1.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b *c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] && !IntegerQ[m] && IntegerQ[n]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] && (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n *(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* ((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d) Int[(e + f*x)^ p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d) Int[(e + f*x)^p/(c + d *x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
\[\int \frac {\left (b x +a \right )^{m}}{\left (d x +c \right )^{2} \left (f x +e \right )^{2}}d x\]
\[ \int \frac {(a+b x)^m}{(c+d x)^2 (e+f x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{m}}{{\left (d x + c\right )}^{2} {\left (f x + e\right )}^{2}} \,d x } \]
integral((b*x + a)^m/(d^2*f^2*x^4 + c^2*e^2 + 2*(d^2*e*f + c*d*f^2)*x^3 + (d^2*e^2 + 4*c*d*e*f + c^2*f^2)*x^2 + 2*(c*d*e^2 + c^2*e*f)*x), x)
Timed out. \[ \int \frac {(a+b x)^m}{(c+d x)^2 (e+f x)^2} \, dx=\text {Timed out} \]
\[ \int \frac {(a+b x)^m}{(c+d x)^2 (e+f x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{m}}{{\left (d x + c\right )}^{2} {\left (f x + e\right )}^{2}} \,d x } \]
\[ \int \frac {(a+b x)^m}{(c+d x)^2 (e+f x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{m}}{{\left (d x + c\right )}^{2} {\left (f x + e\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(a+b x)^m}{(c+d x)^2 (e+f x)^2} \, dx=\int \frac {{\left (a+b\,x\right )}^m}{{\left (e+f\,x\right )}^2\,{\left (c+d\,x\right )}^2} \,d x \]