Integrand size = 29, antiderivative size = 261 \[ \int (a+b x)^m (c+d x)^{-1-m} (e+f x) (g+h x) \, dx=\frac {(a+b x)^{1+m} (c+d x)^{-m} \left (2 b d^2 e g+b c^2 f h (2+m)-c d (2 b (f g+e h)+a f h m)+d (b c-a d) f h m x\right )}{2 b d^2 (b c-a d) m}-\frac {\left (b^2 c^2 f h (1+m) (2+m)-2 b c d (1+m) (b f g+b e h+a f h m)+d^2 \left (2 b^2 e g+2 a b (f g+e h) m-a^2 f h (1-m) m\right )\right ) (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m,1+m,2+m,-\frac {d (a+b x)}{b c-a d}\right )}{2 b^2 d^2 (b c-a d) m (1+m)} \]
1/2*(b*x+a)^(1+m)*(2*b*d^2*e*g+b*c^2*f*h*(2+m)-c*d*(2*b*(e*h+f*g)+a*f*h*m) +d*(-a*d+b*c)*f*h*m*x)/b/d^2/(-a*d+b*c)/m/((d*x+c)^m)-1/2*(b^2*c^2*f*h*(1+ m)*(2+m)-2*b*c*d*(1+m)*(a*f*h*m+b*e*h+b*f*g)+d^2*(2*b^2*e*g+2*a*b*(e*h+f*g )*m-a^2*f*h*(1-m)*m))*(b*x+a)^(1+m)*(b*(d*x+c)/(-a*d+b*c))^m*hypergeom([m, 1+m],[2+m],-d*(b*x+a)/(-a*d+b*c))/b^2/d^2/(-a*d+b*c)/m/(1+m)/((d*x+c)^m)
Time = 0.20 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.85 \[ \int (a+b x)^m (c+d x)^{-1-m} (e+f x) (g+h x) \, dx=\frac {(a+b x)^{1+m} (c+d x)^{-m} \left (b \left (a d f h m (c+d x)-b \left (2 d^2 e g+c^2 f h (2+m)+c d (-2 f g-2 e h+f h m x)\right )\right )+\frac {\left (a^2 d^2 f h (-1+m) m+2 a b d m (d (f g+e h)-c f h (1+m))+b^2 \left (2 d^2 e g-2 c d (f g+e h) (1+m)+c^2 f h \left (2+3 m+m^2\right )\right )\right ) \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m,1+m,2+m,\frac {d (a+b x)}{-b c+a d}\right )}{1+m}\right )}{2 b^2 d^2 (-b c+a d) m} \]
((a + b*x)^(1 + m)*(b*(a*d*f*h*m*(c + d*x) - b*(2*d^2*e*g + c^2*f*h*(2 + m ) + c*d*(-2*f*g - 2*e*h + f*h*m*x))) + ((a^2*d^2*f*h*(-1 + m)*m + 2*a*b*d* m*(d*(f*g + e*h) - c*f*h*(1 + m)) + b^2*(2*d^2*e*g - 2*c*d*(f*g + e*h)*(1 + m) + c^2*f*h*(2 + 3*m + m^2)))*((b*(c + d*x))/(b*c - a*d))^m*Hypergeomet ric2F1[m, 1 + m, 2 + m, (d*(a + b*x))/(-(b*c) + a*d)])/(1 + m)))/(2*b^2*d^ 2*(-(b*c) + a*d)*m*(c + d*x)^m)
Time = 0.36 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {163, 80, 79}
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.
| \(\displaystyle \int (e+f x) (g+h x) (a+b x)^m (c+d x)^{-m-1} \, dx\) |
| \(\Big \downarrow \) 163 |
| \(\displaystyle \frac {(a+b x)^{m+1} (c+d x)^{-m} \left (-c d (a f h m+2 b (e h+f g))+d f h m x (b c-a d)+b c^2 f h (m+2)+2 b d^2 e g\right )}{2 b d^2 m (b c-a d)}-\frac {\left (d^2 \left (a^2 (-f) h (1-m) m+2 a b m (e h+f g)+2 b^2 e g\right )-2 b c d (m+1) (a f h m+b e h+b f g)+b^2 c^2 f h (m+1) (m+2)\right ) \int (a+b x)^m (c+d x)^{-m}dx}{2 b d^2 m (b c-a d)}\) |
| \(\Big \downarrow \) 80 |
| \(\displaystyle \frac {(a+b x)^{m+1} (c+d x)^{-m} \left (-c d (a f h m+2 b (e h+f g))+d f h m x (b c-a d)+b c^2 f h (m+2)+2 b d^2 e g\right )}{2 b d^2 m (b c-a d)}-\frac {(c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \left (d^2 \left (a^2 (-f) h (1-m) m+2 a b m (e h+f g)+2 b^2 e g\right )-2 b c d (m+1) (a f h m+b e h+b f g)+b^2 c^2 f h (m+1) (m+2)\right ) \int (a+b x)^m \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{-m}dx}{2 b d^2 m (b c-a d)}\) |
| \(\Big \downarrow \) 79 |
| \(\displaystyle \frac {(a+b x)^{m+1} (c+d x)^{-m} \left (-c d (a f h m+2 b (e h+f g))+d f h m x (b c-a d)+b c^2 f h (m+2)+2 b d^2 e g\right )}{2 b d^2 m (b c-a d)}-\frac {(a+b x)^{m+1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m,m+1,m+2,-\frac {d (a+b x)}{b c-a d}\right ) \left (d^2 \left (a^2 (-f) h (1-m) m+2 a b m (e h+f g)+2 b^2 e g\right )-2 b c d (m+1) (a f h m+b e h+b f g)+b^2 c^2 f h (m+1) (m+2)\right )}{2 b^2 d^2 m (m+1) (b c-a d)}\) |
((a + b*x)^(1 + m)*(2*b*d^2*e*g + b*c^2*f*h*(2 + m) - c*d*(2*b*(f*g + e*h) + a*f*h*m) + d*(b*c - a*d)*f*h*m*x))/(2*b*d^2*(b*c - a*d)*m*(c + d*x)^m) - ((b^2*c^2*f*h*(1 + m)*(2 + m) - 2*b*c*d*(1 + m)*(b*f*g + b*e*h + a*f*h*m ) + d^2*(2*b^2*e*g + 2*a*b*(f*g + e*h)*m - a^2*f*h*(1 - m)*m))*(a + b*x)^( 1 + m)*((b*(c + d*x))/(b*c - a*d))^m*Hypergeometric2F1[m, 1 + m, 2 + m, -( (d*(a + b*x))/(b*c - a*d))])/(2*b^2*d^2*(b*c - a*d)*m*(1 + m)*(c + d*x)^m)
3.2.27.3.1 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_) )*((g_.) + (h_.)*(x_)), x_] :> Simp[((a^2*d*f*h*(n + 2) + b^2*d*e*g*(m + n + 3) + a*b*(c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b*f*h*(b*c - a*d)* (m + 1)*x)/(b^2*d*(b*c - a*d)*(m + 1)*(m + n + 3)))*(a + b*x)^(m + 1)*(c + d*x)^(n + 1), x] - Simp[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f *h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c* d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2* d*(b*c - a*d)*(m + 1)*(m + n + 3)) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && ((GeQ[m, -2] && LtQ[m, - 1]) || SumSimplerQ[m, 1]) && NeQ[m, -1] && NeQ[m + n + 3, 0]
\[\int \left (b x +a \right )^{m} \left (d x +c \right )^{-1-m} \left (f x +e \right ) \left (h x +g \right )d x\]
\[ \int (a+b x)^m (c+d x)^{-1-m} (e+f x) (g+h x) \, dx=\int { {\left (f x + e\right )} {\left (h x + g\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 1} \,d x } \]
Exception generated. \[ \int (a+b x)^m (c+d x)^{-1-m} (e+f x) (g+h x) \, dx=\text {Exception raised: HeuristicGCDFailed} \]
\[ \int (a+b x)^m (c+d x)^{-1-m} (e+f x) (g+h x) \, dx=\int { {\left (f x + e\right )} {\left (h x + g\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 1} \,d x } \]
\[ \int (a+b x)^m (c+d x)^{-1-m} (e+f x) (g+h x) \, dx=\int { {\left (f x + e\right )} {\left (h x + g\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 1} \,d x } \]
Timed out. \[ \int (a+b x)^m (c+d x)^{-1-m} (e+f x) (g+h x) \, dx=\int \frac {\left (e+f\,x\right )\,\left (g+h\,x\right )\,{\left (a+b\,x\right )}^m}{{\left (c+d\,x\right )}^{m+1}} \,d x \]