Integrand size = 24, antiderivative size = 309 \[ \int \frac {(a+b x)^m (c+d x)^{-m}}{(e+f x)^4} \, dx=-\frac {f (a+b x)^{1+m} (c+d x)^{1-m}}{3 (b e-a f) (d e-c f) (e+f x)^3}-\frac {f (b (4 d e-c f (2-m))-a d f (2+m)) (a+b x)^{1+m} (c+d x)^{1-m}}{6 (b e-a f)^2 (d e-c f)^2 (e+f x)^2}-\frac {(b c-a d) \left (2 a b d f (3 d e-c f (1-m)) (1+m)-a^2 d^2 f^2 \left (2+3 m+m^2\right )-b^2 \left (6 d^2 e^2-6 c d e f (1-m)+c^2 f^2 \left (2-3 m+m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-1-m} \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{6 (b e-a f)^4 (d e-c f)^2 (1+m)} \]
-1/3*f*(b*x+a)^(1+m)*(d*x+c)^(1-m)/(-a*f+b*e)/(-c*f+d*e)/(f*x+e)^3-1/6*f*( b*(4*d*e-c*f*(2-m))-a*d*f*(2+m))*(b*x+a)^(1+m)*(d*x+c)^(1-m)/(-a*f+b*e)^2/ (-c*f+d*e)^2/(f*x+e)^2-1/6*(-a*d+b*c)*(2*a*b*d*f*(3*d*e-c*f*(1-m))*(1+m)-a ^2*d^2*f^2*(m^2+3*m+2)-b^2*(6*d^2*e^2-6*c*d*e*f*(1-m)+c^2*f^2*(m^2-3*m+2)) )*(b*x+a)^(1+m)*(d*x+c)^(-1-m)*hypergeom([2, 1+m],[2+m],(-c*f+d*e)*(b*x+a) /(-a*f+b*e)/(d*x+c))/(-a*f+b*e)^4/(-c*f+d*e)^2/(1+m)
Time = 0.25 (sec) , antiderivative size = 255, normalized size of antiderivative = 0.83 \[ \int \frac {(a+b x)^m (c+d x)^{-m}}{(e+f x)^4} \, dx=\frac {(a+b x)^{1+m} (c+d x)^{-1-m} \left (-\frac {2 f (-b e+a f) (-d e+c f) (c+d x)^2}{(e+f x)^3}-\frac {f (4 b d e+b c f (-2+m)-a d f (2+m)) (c+d x)^2}{(e+f x)^2}+\frac {(b c-a d) \left (-2 a b d f (3 d e+c f (-1+m)) (1+m)+a^2 d^2 f^2 \left (2+3 m+m^2\right )+b^2 \left (6 d^2 e^2+6 c d e f (-1+m)+c^2 f^2 \left (2-3 m+m^2\right )\right )\right ) \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(b e-a f)^2 (1+m)}\right )}{6 (b e-a f)^2 (d e-c f)^2} \]
((a + b*x)^(1 + m)*(c + d*x)^(-1 - m)*((-2*f*(-(b*e) + a*f)*(-(d*e) + c*f) *(c + d*x)^2)/(e + f*x)^3 - (f*(4*b*d*e + b*c*f*(-2 + m) - a*d*f*(2 + m))* (c + d*x)^2)/(e + f*x)^2 + ((b*c - a*d)*(-2*a*b*d*f*(3*d*e + c*f*(-1 + m)) *(1 + m) + a^2*d^2*f^2*(2 + 3*m + m^2) + b^2*(6*d^2*e^2 + 6*c*d*e*f*(-1 + m) + c^2*f^2*(2 - 3*m + m^2)))*Hypergeometric2F1[2, 1 + m, 2 + m, ((d*e - c*f)*(a + b*x))/((b*e - a*f)*(c + d*x))])/((b*e - a*f)^2*(1 + m))))/(6*(b* e - a*f)^2*(d*e - c*f)^2)
Time = 0.44 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {114, 25, 168, 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.
\(\displaystyle \int \frac {(a+b x)^m (c+d x)^{-m}}{(e+f x)^4} \, dx\) |
\(\Big \downarrow \) 114 |
\(\displaystyle -\frac {\int -\frac {(a+b x)^m (c+d x)^{-m} (b (3 d e-c f (2-m))-a d f (m+2)-b d f x)}{(e+f x)^3}dx}{3 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{1-m}}{3 (e+f x)^3 (b e-a f) (d e-c f)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {(a+b x)^m (c+d x)^{-m} (3 b d e-b c f (2-m)-a d f (m+2)-b d f x)}{(e+f x)^3}dx}{3 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{1-m}}{3 (e+f x)^3 (b e-a f) (d e-c f)}\) |
\(\Big \downarrow \) 168 |
\(\displaystyle \frac {-\frac {\int \frac {\left (-\left (\left (6 d^2 e^2-6 c d f (1-m) e+c^2 f^2 \left (m^2-3 m+2\right )\right ) b^2\right )+2 a d f (3 d e-c f (1-m)) (m+1) b-a^2 d^2 f^2 \left (m^2+3 m+2\right )\right ) (a+b x)^m (c+d x)^{-m}}{(e+f x)^2}dx}{2 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{1-m} (-a d f (m+2)-b c f (2-m)+4 b d e)}{2 (e+f x)^2 (b e-a f) (d e-c f)}}{3 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{1-m}}{3 (e+f x)^3 (b e-a f) (d e-c f)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {-\frac {\left (-a^2 d^2 f^2 \left (m^2+3 m+2\right )+2 a b d f (m+1) (3 d e-c f (1-m))-\left (b^2 \left (c^2 f^2 \left (m^2-3 m+2\right )-6 c d e f (1-m)+6 d^2 e^2\right )\right )\right ) \int \frac {(a+b x)^m (c+d x)^{-m}}{(e+f x)^2}dx}{2 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{1-m} (-a d f (m+2)-b c f (2-m)+4 b d e)}{2 (e+f x)^2 (b e-a f) (d e-c f)}}{3 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{1-m}}{3 (e+f x)^3 (b e-a f) (d e-c f)}\) |
\(\Big \downarrow \) 141 |
\(\displaystyle \frac {-\frac {(b c-a d) (a+b x)^{m+1} (c+d x)^{-m-1} \left (-a^2 d^2 f^2 \left (m^2+3 m+2\right )+2 a b d f (m+1) (3 d e-c f (1-m))-\left (b^2 \left (c^2 f^2 \left (m^2-3 m+2\right )-6 c d e f (1-m)+6 d^2 e^2\right )\right )\right ) \operatorname {Hypergeometric2F1}\left (2,m+1,m+2,\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{2 (m+1) (b e-a f)^3 (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{1-m} (-a d f (m+2)-b c f (2-m)+4 b d e)}{2 (e+f x)^2 (b e-a f) (d e-c f)}}{3 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{1-m}}{3 (e+f x)^3 (b e-a f) (d e-c f)}\) |
-1/3*(f*(a + b*x)^(1 + m)*(c + d*x)^(1 - m))/((b*e - a*f)*(d*e - c*f)*(e + f*x)^3) + (-1/2*(f*(4*b*d*e - b*c*f*(2 - m) - a*d*f*(2 + m))*(a + b*x)^(1 + m)*(c + d*x)^(1 - m))/((b*e - a*f)*(d*e - c*f)*(e + f*x)^2) - ((b*c - a *d)*(2*a*b*d*f*(3*d*e - c*f*(1 - m))*(1 + m) - a^2*d^2*f^2*(2 + 3*m + m^2) - b^2*(6*d^2*e^2 - 6*c*d*e*f*(1 - m) + c^2*f^2*(2 - 3*m + m^2)))*(a + b*x )^(1 + m)*(c + d*x)^(-1 - m)*Hypergeometric2F1[2, 1 + m, 2 + m, ((d*e - c* f)*(a + b*x))/((b*e - a*f)*(c + d*x))])/(2*(b*e - a*f)^3*(d*e - c*f)*(1 + m)))/(3*(b*e - a*f)*(d*e - c*f))
3.31.68.3.1 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_)*((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_), 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]
\[\int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{-m}}{\left (f x +e \right )^{4}}d x\]
\[ \int \frac {(a+b x)^m (c+d x)^{-m}}{(e+f x)^4} \, dx=\int { \frac {{\left (b x + a\right )}^{m}}{{\left (f x + e\right )}^{4} {\left (d x + c\right )}^{m}} \,d x } \]
integral((b*x + a)^m/((f^4*x^4 + 4*e*f^3*x^3 + 6*e^2*f^2*x^2 + 4*e^3*f*x + e^4)*(d*x + c)^m), x)
Timed out. \[ \int \frac {(a+b x)^m (c+d x)^{-m}}{(e+f x)^4} \, dx=\text {Timed out} \]
\[ \int \frac {(a+b x)^m (c+d x)^{-m}}{(e+f x)^4} \, dx=\int { \frac {{\left (b x + a\right )}^{m}}{{\left (f x + e\right )}^{4} {\left (d x + c\right )}^{m}} \,d x } \]
\[ \int \frac {(a+b x)^m (c+d x)^{-m}}{(e+f x)^4} \, dx=\int { \frac {{\left (b x + a\right )}^{m}}{{\left (f x + e\right )}^{4} {\left (d x + c\right )}^{m}} \,d x } \]
Timed out. \[ \int \frac {(a+b x)^m (c+d x)^{-m}}{(e+f x)^4} \, dx=\int \frac {{\left (a+b\,x\right )}^m}{{\left (e+f\,x\right )}^4\,{\left (c+d\,x\right )}^m} \,d x \]