Integrand size = 26, antiderivative size = 190 \[ \int \frac {(a+b x)^m (c+d x)^{1-m}}{(e+f x)^2} \, dx=-\frac {(a+b x)^m (c+d x)^{1-m}}{f (e+f x)}+\frac {(a d f (1-m)-b (d e-c f m)) (a+b x)^m (c+d x)^{-m} \operatorname {Hypergeometric2F1}\left (1,m,1+m,\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{f^2 (b e-a f) m}+\frac {d (a+b x)^m (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m,m,1+m,-\frac {d (a+b x)}{b c-a d}\right )}{f^2 m} \]
-(b*x+a)^m*(d*x+c)^(1-m)/f/(f*x+e)+(a*d*f*(1-m)-b*(-c*f*m+d*e))*(b*x+a)^m* hypergeom([1, m],[1+m],(-c*f+d*e)*(b*x+a)/(-a*f+b*e)/(d*x+c))/f^2/(-a*f+b* e)/m/((d*x+c)^m)+d*(b*x+a)^m*(b*(d*x+c)/(-a*d+b*c))^m*hypergeom([m, m],[1+ m],-d*(b*x+a)/(-a*d+b*c))/f^2/m/((d*x+c)^m)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 0.26 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.07 \[ \int \frac {(a+b x)^m (c+d x)^{1-m}}{(e+f x)^2} \, dx=-\frac {(a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \left (\frac {b (e+f x)}{b e-a f}\right )^{-m} \left (d (e+f x) \left (\frac {b (e+f x)}{b e-a f}\right )^m \operatorname {AppellF1}\left (1+m,m,1,2+m,\frac {d (a+b x)}{-b c+a d},\frac {f (a+b x)}{-b e+a f}\right )+(-d e+c f) \operatorname {Hypergeometric2F1}\left (m,1+m,2+m,\frac {(-d e+c f) (a+b x)}{(b c-a d) (e+f x)}\right )\right )}{f (-b e+a f) (1+m) (e+f x)} \]
-(((a + b*x)^(1 + m)*((b*(c + d*x))/(b*c - a*d))^m*(d*(e + f*x)*((b*(e + f *x))/(b*e - a*f))^m*AppellF1[1 + m, m, 1, 2 + m, (d*(a + b*x))/(-(b*c) + a *d), (f*(a + b*x))/(-(b*e) + a*f)] + (-(d*e) + c*f)*Hypergeometric2F1[m, 1 + m, 2 + m, ((-(d*e) + c*f)*(a + b*x))/((b*c - a*d)*(e + f*x))]))/(f*(-(b *e) + a*f)*(1 + m)*(c + d*x)^m*(e + f*x)*((b*(e + f*x))/(b*e - a*f))^m))
Time = 0.37 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {140, 80, 79, 168, 25, 27, 141}
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.
\(\displaystyle \int \frac {(a+b x)^m (c+d x)^{1-m}}{(e+f x)^2} \, dx\) |
\(\Big \downarrow \) 140 |
\(\displaystyle \int \frac {(a+b x)^{m-1} (c+d x)^{-m} \left (-\frac {b d e^2}{f^2}+a c+\left (a d+b \left (c-\frac {2 d e}{f}\right )\right ) x\right )}{(e+f x)^2}dx+\frac {b d \int (a+b x)^{m-1} (c+d x)^{-m}dx}{f^2}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \int \frac {(a+b x)^{m-1} (c+d x)^{-m} \left (-\frac {b d e^2}{f^2}+a c+\left (a d+b \left (c-\frac {2 d e}{f}\right )\right ) x\right )}{(e+f x)^2}dx+\frac {b d (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \int (a+b x)^{m-1} \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{-m}dx}{f^2}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle \int \frac {(a+b x)^{m-1} (c+d x)^{-m} \left (-\frac {b d e^2}{f^2}+a c+\left (a d+b \left (c-\frac {2 d e}{f}\right )\right ) x\right )}{(e+f x)^2}dx+\frac {d (a+b x)^m (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m,m,m+1,-\frac {d (a+b x)}{b c-a d}\right )}{f^2 m}\) |
\(\Big \downarrow \) 168 |
\(\displaystyle -\frac {\int -\frac {(b e-a f) (d e-c f) (a d f (1-m)-b (d e-c f m)) (a+b x)^{m-1} (c+d x)^{-m}}{f^2 (e+f x)}dx}{(b e-a f) (d e-c f)}-\frac {(a+b x)^m (c+d x)^{1-m}}{f (e+f x)}+\frac {d (a+b x)^m (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m,m,m+1,-\frac {d (a+b x)}{b c-a d}\right )}{f^2 m}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {(b e-a f) (d e-c f) (a d f (1-m)-b (d e-c f m)) (a+b x)^{m-1} (c+d x)^{-m}}{f^2 (e+f x)}dx}{(b e-a f) (d e-c f)}-\frac {(a+b x)^m (c+d x)^{1-m}}{f (e+f x)}+\frac {d (a+b x)^m (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m,m,m+1,-\frac {d (a+b x)}{b c-a d}\right )}{f^2 m}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a d f (1-m)-b (d e-c f m)) \int \frac {(a+b x)^{m-1} (c+d x)^{-m}}{e+f x}dx}{f^2}-\frac {(a+b x)^m (c+d x)^{1-m}}{f (e+f x)}+\frac {d (a+b x)^m (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m,m,m+1,-\frac {d (a+b x)}{b c-a d}\right )}{f^2 m}\) |
\(\Big \downarrow \) 141 |
\(\displaystyle \frac {(a+b x)^m (c+d x)^{-m} (a d f (1-m)-b (d e-c f m)) \operatorname {Hypergeometric2F1}\left (1,m,m+1,\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{f^2 m (b e-a f)}-\frac {(a+b x)^m (c+d x)^{1-m}}{f (e+f x)}+\frac {d (a+b x)^m (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m,m,m+1,-\frac {d (a+b x)}{b c-a d}\right )}{f^2 m}\) |
-(((a + b*x)^m*(c + d*x)^(1 - m))/(f*(e + f*x))) + ((a*d*f*(1 - m) - b*(d* e - c*f*m))*(a + b*x)^m*Hypergeometric2F1[1, m, 1 + m, ((d*e - c*f)*(a + b *x))/((b*e - a*f)*(c + d*x))])/(f^2*(b*e - a*f)*m*(c + d*x)^m) + (d*(a + b *x)^m*((b*(c + d*x))/(b*c - a*d))^m*Hypergeometric2F1[m, m, 1 + m, -((d*(a + b*x))/(b*c - a*d))])/(f^2*m*(c + d*x)^m)
3.32.17.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) ^FracPart[n]) Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) ), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !Integ erQ[n] && (RationalQ[m] || !SimplerQ[n + 1, m + 1])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*d^(m + n)*f^p Int[(a + b*x)^(m - 1)/(c + d*x)^m, x] , x] + Int[(a + b*x)^(m - 1)*((e + f*x)^p/(c + d*x)^m)*ExpandToSum[(a + b*x )*(c + d*x)^(-p - 1) - (b*d^(-p - 1)*f^p)/(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[m + n + p + 1, 0] && ILtQ[p, 0] && (GtQ[m, 0] || SumSimplerQ[m, -1] || !(GtQ[n, 0] || SumSimplerQ[n, -1]))
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/((m + 1)*(b*e - a*f)^( n + 1)*(e + f*x)^(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2, (-(d*e - c*f ))*((a + b*x)/((b*c - a*d)*(e + f*x)))], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && ILtQ[n, 0] && (SumSimplerQ[m, 1] || !Su mSimplerQ[p, 1]) && !ILtQ[m, 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]
\[\int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{1-m}}{\left (f x +e \right )^{2}}d x\]
\[ \int \frac {(a+b x)^m (c+d x)^{1-m}}{(e+f x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 1}}{{\left (f x + e\right )}^{2}} \,d x } \]
Exception generated. \[ \int \frac {(a+b x)^m (c+d x)^{1-m}}{(e+f x)^2} \, dx=\text {Exception raised: HeuristicGCDFailed} \]
\[ \int \frac {(a+b x)^m (c+d x)^{1-m}}{(e+f x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 1}}{{\left (f x + e\right )}^{2}} \,d x } \]
\[ \int \frac {(a+b x)^m (c+d x)^{1-m}}{(e+f x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 1}}{{\left (f x + e\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(a+b x)^m (c+d x)^{1-m}}{(e+f x)^2} \, dx=\int \frac {{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^{1-m}}{{\left (e+f\,x\right )}^2} \,d x \]