Integrand size = 33, antiderivative size = 261 \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^3} \, dx=\frac {(b c-a d) (a+b x)^{-1+m} (c+d x)^{2-m}}{8 b d^2 (b c+a d+2 b d x)^2}+\frac {(1-2 m) (a+b x)^{-1+m} (c+d x)^{2-m}}{8 b d^2 (b c+a d+2 b d x)}-\frac {\left (1-4 m+2 m^2\right ) (a+b x)^{-1+m} (c+d x)^{1-m} \operatorname {Hypergeometric2F1}\left (1,-1+m,m,-\frac {d (a+b x)}{b (c+d x)}\right )}{8 b^2 d^2 (1-m)}-\frac {(b c-a d) (a+b x)^{-1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (-1+m,-1+m,m,-\frac {d (a+b x)}{b c-a d}\right )}{8 b^3 d^2 (1-m)} \]
1/8*(-a*d+b*c)*(b*x+a)^(-1+m)*(d*x+c)^(2-m)/b/d^2/(2*b*d*x+a*d+b*c)^2+1/8* (1-2*m)*(b*x+a)^(-1+m)*(d*x+c)^(2-m)/b/d^2/(2*b*d*x+a*d+b*c)-1/8*(2*m^2-4* m+1)*(b*x+a)^(-1+m)*(d*x+c)^(1-m)*hypergeom([1, -1+m],[m],-d*(b*x+a)/b/(d* x+c))/b^2/d^2/(1-m)-1/8*(-a*d+b*c)*(b*x+a)^(-1+m)*(b*(d*x+c)/(-a*d+b*c))^m *hypergeom([-1+m, -1+m],[m],-d*(b*x+a)/(-a*d+b*c))/b^3/d^2/(1-m)/((d*x+c)^ m)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 1.60 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.43 \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^3} \, dx=\frac {(a+b x)^m (c+d x)^{-m} \left (\frac {4 (-b c+a d) \left (\frac {d (a+b x)}{a d+b (c+2 d x)}\right )^{1-m} \left (\frac {b (c+d x)}{a d+b (c+2 d x)}\right )^m \operatorname {AppellF1}\left (1,m,-m,2,\frac {-b c+a d}{a d+b (c+2 d x)},\frac {b c-a d}{b c+a d+2 b d x}\right )}{d^2 (a+b x)}+\frac {-\frac {(b c-a d)^3 \left (\frac {d (a+b x)}{a d+b (c+2 d x)}\right )^{-m} \left (\frac {b (c+d x)}{a d+b (c+2 d x)}\right )^m \operatorname {AppellF1}\left (2,m,-m,3,\frac {-b c+a d}{a d+b (c+2 d x)},\frac {b c-a d}{b c+a d+2 b d x}\right )}{(a d+b (c+2 d x))^2}+\frac {4 b \left (\frac {d (a+b x)}{-b c+a d}\right )^{-m} (c+d x) \operatorname {AppellF1}\left (1-m,-m,1,2-m,\frac {b (c+d x)}{b c-a d},\frac {2 b (c+d x)}{b c-a d}\right )}{-1+m}}{d (b c-a d)}\right )}{16 b^3} \]
((a + b*x)^m*((4*(-(b*c) + a*d)*((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)) + (-(((b*c - a*d)^3*((b*(c + d*x))/(a*d + b*(c + 2*d*x)))^m*Appell F1[2, m, -m, 3, (-(b*c) + a*d)/(a*d + b*(c + 2*d*x)), (b*c - a*d)/(b*c + a *d + 2*b*d*x)])/(((d*(a + b*x))/(a*d + b*(c + 2*d*x)))^m*(a*d + b*(c + 2*d *x))^2)) + (4*b*(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)])/((-1 + m)*((d*(a + b*x))/(-(b*c) + a*d))^m))/(d*(b*c - a*d))))/(16*b^3*(c + d*x)^m)
Time = 0.77 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {140, 80, 79, 2116, 27, 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.
\(\displaystyle \int \frac {(a+b x)^m (c+d x)^{2-m}}{(a d+b c+2 b d x)^3} \, dx\) |
\(\Big \downarrow \) 140 |
\(\displaystyle \int \frac {(a+b x)^{m-2} (c+d x)^{1-m} \left (-\frac {1}{2} b (b c-a d) x^2-\frac {(b c-a d) (3 b c+a d) x}{4 d}-\frac {(b c-a d) \left (b^2 c^2+4 a b d c-a^2 d^2\right )}{8 b d^2}\right )}{(b c+a d+2 b d x)^3}dx+\frac {\int (a+b x)^{m-2} (c+d x)^{1-m}dx}{8 b d^2}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \int \frac {(a+b x)^{m-2} (c+d x)^{1-m} \left (-\frac {1}{2} b (b c-a d) x^2-\frac {(b c-a d) (3 b c+a d) x}{4 d}-\frac {(b c-a d) \left (b^2 c^2+4 a b d c-a^2 d^2\right )}{8 b d^2}\right )}{(b c+a d+2 b d x)^3}dx+\frac {(b c-a d) (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \int (a+b x)^{m-2} \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{1-m}dx}{8 b^2 d^2}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle \int \frac {(a+b x)^{m-2} (c+d x)^{1-m} \left (-\frac {1}{2} b (b c-a d) x^2-\frac {(b c-a d) (3 b c+a d) x}{4 d}-\frac {(b c-a d) \left (b^2 c^2+4 a b d c-a^2 d^2\right )}{8 b d^2}\right )}{(b c+a d+2 b d x)^3}dx-\frac {(b c-a d) (a+b x)^{m-1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m-1,m-1,m,-\frac {d (a+b x)}{b c-a d}\right )}{8 b^3 d^2 (1-m)}\) |
\(\Big \downarrow \) 2116 |
\(\displaystyle \frac {\int -\frac {(b c-a d)^3 (a+b x)^{m-2} (c+d x)^{1-m} (b c m+a (d-d m)+b d x)}{4 d (b c+a d+2 b d x)^2}dx}{2 b d (b c-a d)^2}-\frac {(b c-a d) (a+b x)^{m-1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m-1,m-1,m,-\frac {d (a+b x)}{b c-a d}\right )}{8 b^3 d^2 (1-m)}+\frac {(b c-a d) (a+b x)^{m-1} (c+d x)^{2-m}}{8 b d^2 (a d+b c+2 b d x)^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {(b c-a d) \int \frac {(a+b x)^{m-2} (c+d x)^{1-m} (b c m+a (d-d m)+b d x)}{(b c+a d+2 b d x)^2}dx}{8 b d^2}-\frac {(b c-a d) (a+b x)^{m-1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m-1,m-1,m,-\frac {d (a+b x)}{b c-a d}\right )}{8 b^3 d^2 (1-m)}+\frac {(b c-a d) (a+b x)^{m-1} (c+d x)^{2-m}}{8 b d^2 (a d+b c+2 b d x)^2}\) |
\(\Big \downarrow \) 168 |
\(\displaystyle -\frac {(b c-a d) \left (\frac {\int -\frac {b d (b c-a d)^2 \left (2 m^2-4 m+1\right ) (a+b x)^{m-2} (c+d x)^{1-m}}{b c+a d+2 b d x}dx}{b d (b c-a d)^2}-\frac {(1-2 m) (a+b x)^{m-1} (c+d x)^{2-m}}{(b c-a d) (a d+b c+2 b d x)}\right )}{8 b d^2}-\frac {(b c-a d) (a+b x)^{m-1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m-1,m-1,m,-\frac {d (a+b x)}{b c-a d}\right )}{8 b^3 d^2 (1-m)}+\frac {(b c-a d) (a+b x)^{m-1} (c+d x)^{2-m}}{8 b d^2 (a d+b c+2 b d x)^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {(b c-a d) \left (-\frac {\int \frac {b d (b c-a d)^2 \left (2 m^2-4 m+1\right ) (a+b x)^{m-2} (c+d x)^{1-m}}{b c+a d+2 b d x}dx}{b d (b c-a d)^2}-\frac {(1-2 m) (a+b x)^{m-1} (c+d x)^{2-m}}{(b c-a d) (a d+b c+2 b d x)}\right )}{8 b d^2}-\frac {(b c-a d) (a+b x)^{m-1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m-1,m-1,m,-\frac {d (a+b x)}{b c-a d}\right )}{8 b^3 d^2 (1-m)}+\frac {(b c-a d) (a+b x)^{m-1} (c+d x)^{2-m}}{8 b d^2 (a d+b c+2 b d x)^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {(b c-a d) \left (-\left (2 m^2-4 m+1\right ) \int \frac {(a+b x)^{m-2} (c+d x)^{1-m}}{b c+a d+2 b d x}dx-\frac {(1-2 m) (a+b x)^{m-1} (c+d x)^{2-m}}{(b c-a d) (a d+b c+2 b d x)}\right )}{8 b d^2}-\frac {(b c-a d) (a+b x)^{m-1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m-1,m-1,m,-\frac {d (a+b x)}{b c-a d}\right )}{8 b^3 d^2 (1-m)}+\frac {(b c-a d) (a+b x)^{m-1} (c+d x)^{2-m}}{8 b d^2 (a d+b c+2 b d x)^2}\) |
\(\Big \downarrow \) 141 |
\(\displaystyle -\frac {(b c-a d) (a+b x)^{m-1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m-1,m-1,m,-\frac {d (a+b x)}{b c-a d}\right )}{8 b^3 d^2 (1-m)}-\frac {(b c-a d) \left (\frac {\left (2 m^2-4 m+1\right ) (a+b x)^{m-1} (c+d x)^{1-m} \operatorname {Hypergeometric2F1}\left (1,m-1,m,-\frac {d (a+b x)}{b (c+d x)}\right )}{b (1-m) (b c-a d)}-\frac {(1-2 m) (a+b x)^{m-1} (c+d x)^{2-m}}{(b c-a d) (a d+b c+2 b d x)}\right )}{8 b d^2}+\frac {(b c-a d) (a+b x)^{m-1} (c+d x)^{2-m}}{8 b d^2 (a d+b c+2 b d x)^2}\) |
((b*c - a*d)*(a + b*x)^(-1 + m)*(c + d*x)^(2 - m))/(8*b*d^2*(b*c + a*d + 2 *b*d*x)^2) - ((b*c - a*d)*(-(((1 - 2*m)*(a + b*x)^(-1 + m)*(c + d*x)^(2 - m))/((b*c - a*d)*(b*c + a*d + 2*b*d*x))) + ((1 - 4*m + 2*m^2)*(a + b*x)^(- 1 + m)*(c + d*x)^(1 - m)*Hypergeometric2F1[1, -1 + m, m, -((d*(a + b*x))/( b*(c + d*x)))])/(b*(b*c - a*d)*(1 - m))))/(8*b*d^2) - ((b*c - a*d)*(a + b* x)^(-1 + m)*((b*(c + d*x))/(b*c - a*d))^m*Hypergeometric2F1[-1 + m, -1 + m , m, -((d*(a + b*x))/(b*c - a*d))])/(8*b^3*d^2*(1 - m)*(c + d*x)^m)
3.32.43.3.1 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[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_ .)*(x_))^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[Px, a + b*x, x], R = PolynomialRemainder[Px, a + b*x, x]}, Simp[b*R*(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] + Si mp[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*ExpandToSum[(m + 1)*(b*c - a*d)*(b*e - a*f)*Qx + a*d*f*R*(m + 1 ) - b*R*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*R*(m + n + p + 3)*x, x] , x], x]] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && PolyQ[Px, x] && ILtQ[m, -1]
\[\int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{2-m}}{\left (2 b d x +a d +b c \right )^{3}}d x\]
\[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^3} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2}}{{\left (2 \, b d x + b c + a d\right )}^{3}} \,d x } \]
integral((b*x + a)^m*(d*x + c)^(-m + 2)/(8*b^3*d^3*x^3 + b^3*c^3 + 3*a*b^2 *c^2*d + 3*a^2*b*c*d^2 + a^3*d^3 + 12*(b^3*c*d^2 + a*b^2*d^3)*x^2 + 6*(b^3 *c^2*d + 2*a*b^2*c*d^2 + a^2*b*d^3)*x), x)
Exception generated. \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^3} \, dx=\text {Exception raised: HeuristicGCDFailed} \]
\[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^3} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2}}{{\left (2 \, b d x + b c + a d\right )}^{3}} \,d x } \]
\[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^3} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2}}{{\left (2 \, b d x + b c + a d\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^3} \, dx=\int \frac {{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^{2-m}}{{\left (a\,d+b\,c+2\,b\,d\,x\right )}^3} \,d x \]