Integrand size = 33, antiderivative size = 144 \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^2} \, dx=-\frac {(b c-a d) (a+b x)^m (c+d x)^{-m} \operatorname {Hypergeometric2F1}\left (2,m,1+m,-\frac {d (a+b x)}{b (c+d x)}\right )}{4 b^3 d m}+\frac {(b c-a d) (a+b x)^m (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (-1+m,m,1+m,-\frac {d (a+b x)}{b c-a d}\right )}{4 b^3 d m} \]
-1/4*(-a*d+b*c)*(b*x+a)^m*hypergeom([2, m],[1+m],-d*(b*x+a)/b/(d*x+c))/b^3 /d/m/((d*x+c)^m)+1/4*(-a*d+b*c)*(b*x+a)^m*(b*(d*x+c)/(-a*d+b*c))^m*hyperge om([m, -1+m],[1+m],-d*(b*x+a)/(-a*d+b*c))/b^3/d/m/((d*x+c)^m)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 1.83 (sec) , antiderivative size = 308, normalized size of antiderivative = 2.14 \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^2} \, dx=\frac {(a+b x)^m (c+d x)^{-m} \left (-\frac {(b c-a d)^2 \left (\frac {d (a+b x)}{a d+b (c+2 d x)}\right )^{1-m} \left (\frac {b (c+d x)}{a d+b (c+2 d x)}\right )^m \operatorname {AppellF1}\left (1,m,-m,2,\frac {-b c+a d}{a d+b (c+2 d x)},\frac {b c-a d}{b c+a d+2 b d x}\right )}{d^2 (a+b x)}+\frac {\left (\frac {d (a+b x)}{-b c+a d}\right )^{-m} \left (4 b (1+m) (c+d x) \operatorname {AppellF1}\left (1-m,-m,1,2-m,\frac {b (c+d x)}{b c-a d},\frac {2 b (c+d x)}{b c-a d}\right )+2 d (-1+m) (a+b x) \left (-\frac {b d (a+b x) (c+d x)}{(b c-a d)^2}\right )^m \operatorname {Hypergeometric2F1}\left (m,1+m,2+m,\frac {d (a+b x)}{-b c+a d}\right )\right )}{d (-1+m) (1+m)}\right )}{8 b^3} \]
((a + b*x)^m*(-(((b*c - a*d)^2*((d*(a + b*x))/(a*d + b*(c + 2*d*x)))^(1 - m)*((b*(c + d*x))/(a*d + b*(c + 2*d*x)))^m*AppellF1[1, m, -m, 2, (-(b*c) + a*d)/(a*d + b*(c + 2*d*x)), (b*c - a*d)/(b*c + a*d + 2*b*d*x)])/(d^2*(a + b*x))) + (4*b*(1 + m)*(c + d*x)*AppellF1[1 - m, -m, 1, 2 - m, (b*(c + d*x ))/(b*c - a*d), (2*b*(c + d*x))/(b*c - a*d)] + 2*d*(-1 + m)*(a + b*x)*(-(( b*d*(a + b*x)*(c + d*x))/(b*c - a*d)^2))^m*Hypergeometric2F1[m, 1 + m, 2 + m, (d*(a + b*x))/(-(b*c) + a*d)])/(d*(-1 + m)*(1 + m)*((d*(a + b*x))/(-(b *c) + a*d))^m)))/(8*b^3*(c + d*x)^m)
Time = 0.27 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {138, 80, 79, 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.
\(\displaystyle \int \frac {(a+b x)^m (c+d x)^{2-m}}{(a d+b c+2 b d x)^2} \, dx\) |
\(\Big \downarrow \) 138 |
\(\displaystyle \frac {\int (a+b x)^{m-1} (c+d x)^{1-m}dx}{4 b d}-\frac {(b c-a d)^2 \int \frac {(a+b x)^{m-1} (c+d x)^{1-m}}{(b c+a d+2 b d x)^2}dx}{4 b d}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {(b c-a d) (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \int (a+b x)^{m-1} \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{1-m}dx}{4 b^2 d}-\frac {(b c-a d)^2 \int \frac {(a+b x)^{m-1} (c+d x)^{1-m}}{(b c+a d+2 b d x)^2}dx}{4 b d}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle \frac {(b c-a d) (a+b x)^m (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m-1,m,m+1,-\frac {d (a+b x)}{b c-a d}\right )}{4 b^3 d m}-\frac {(b c-a d)^2 \int \frac {(a+b x)^{m-1} (c+d x)^{1-m}}{(b c+a d+2 b d x)^2}dx}{4 b d}\) |
\(\Big \downarrow \) 141 |
\(\displaystyle \frac {(b c-a d) (a+b x)^m (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m-1,m,m+1,-\frac {d (a+b x)}{b c-a d}\right )}{4 b^3 d m}-\frac {(b c-a d) (a+b x)^m (c+d x)^{-m} \operatorname {Hypergeometric2F1}\left (2,m,m+1,-\frac {d (a+b x)}{b (c+d x)}\right )}{4 b^3 d m}\) |
-1/4*((b*c - a*d)*(a + b*x)^m*Hypergeometric2F1[2, m, 1 + m, -((d*(a + b*x ))/(b*(c + d*x)))])/(b^3*d*m*(c + d*x)^m) + ((b*c - a*d)*(a + b*x)^m*((b*( c + d*x))/(b*c - a*d))^m*Hypergeometric2F1[-1 + m, m, 1 + m, -((d*(a + b*x ))/(b*c - a*d))])/(4*b^3*d*m*(c + d*x)^m)
3.32.42.3.1 Defintions of rubi rules used
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 _))^2, x_] :> Simp[b*(d/f^2) Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1), x], x] + Simp[(b*e - a*f)*((d*e - c*f)/f^2) Int[(a + b*x)^(m - 1)*((c + d*x) ^(n - 1)/(e + f*x)^2), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ [m + n, 0] && EqQ[2*b*d*e - f*(b*c + a*d), 0]
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]
\[\int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{2-m}}{\left (2 b d x +a d +b c \right )^{2}}d x\]
\[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2}}{{\left (2 \, b d x + b c + a d\right )}^{2}} \,d x } \]
integral((b*x + a)^m*(d*x + c)^(-m + 2)/(4*b^2*d^2*x^2 + b^2*c^2 + 2*a*b*c *d + a^2*d^2 + 4*(b^2*c*d + a*b*d^2)*x), x)
Exception generated. \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^2} \, dx=\text {Exception raised: HeuristicGCDFailed} \]
\[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2}}{{\left (2 \, b d x + b c + a d\right )}^{2}} \,d x } \]
\[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2}}{{\left (2 \, b d x + b c + a d\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^2} \, dx=\int \frac {{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^{2-m}}{{\left (a\,d+b\,c+2\,b\,d\,x\right )}^2} \,d x \]