\(\int \frac {(c+d x)^{1+m} (e+f x)}{(a+b x)^2} \, dx\) [243]

Optimal result

Integrand size = 22, antiderivative size = 108 \[ \int \frac {(c+d x)^{1+m} (e+f x)}{(a+b x)^2} \, dx=\frac {f (c+d x)^{2+m}}{b d (1+m) (a+b x)}+\frac {(b c f+b d e (1+m)-a d f (2+m)) (c+d x)^{2+m} \operatorname {Hypergeometric2F1}\left (2,2+m,3+m,\frac {b (c+d x)}{b c-a d}\right )}{b (b c-a d)^2 (1+m) (2+m)} \] Output:

f*(d*x+c)^(2+m)/b/d/(1+m)/(b*x+a)+(b*c*f+b*d*e*(1+m)-a*d*f*(2+m))*(d*x+c)^ 
(2+m)*hypergeom([2, 2+m],[3+m],b*(d*x+c)/(-a*d+b*c))/b/(-a*d+b*c)^2/(1+m)/ 
(2+m)
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.94 \[ \int \frac {(c+d x)^{1+m} (e+f x)}{(a+b x)^2} \, dx=\frac {(c+d x)^{2+m} \left (-\frac {(b c-a d) (b e-a f)}{a+b x}-\frac {(b c f+b d e (1+m)-a d f (2+m)) \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {b (c+d x)}{b c-a d}\right )}{2+m}\right )}{b (b c-a d)^2} \] Input:

Integrate[((c + d*x)^(1 + m)*(e + f*x))/(a + b*x)^2,x]
 

Output:

((c + d*x)^(2 + m)*(-(((b*c - a*d)*(b*e - a*f))/(a + b*x)) - ((b*c*f + b*d 
*e*(1 + m) - a*d*f*(2 + m))*Hypergeometric2F1[1, 2 + m, 3 + m, (b*(c + d*x 
))/(b*c - a*d)])/(2 + m)))/(b*(b*c - a*d)^2)
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.06, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {87, 78}

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.

\(\displaystyle \int \frac {(e+f x) (c+d x)^{m+1}}{(a+b x)^2} \, dx\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {(-a d f (m+2)+b c f+b d e (m+1)) \int \frac {(c+d x)^{m+1}}{a+b x}dx}{b (b c-a d)}-\frac {(b e-a f) (c+d x)^{m+2}}{b (a+b x) (b c-a d)}\)

\(\Big \downarrow \) 78

\(\displaystyle -\frac {(c+d x)^{m+2} (-a d f (m+2)+b c f+b d e (m+1)) \operatorname {Hypergeometric2F1}\left (1,m+2,m+3,\frac {b (c+d x)}{b c-a d}\right )}{b (m+2) (b c-a d)^2}-\frac {(b e-a f) (c+d x)^{m+2}}{b (a+b x) (b c-a d)}\)

Input:

Int[((c + d*x)^(1 + m)*(e + f*x))/(a + b*x)^2,x]
 

Output:

-(((b*e - a*f)*(c + d*x)^(2 + m))/(b*(b*c - a*d)*(a + b*x))) - ((b*c*f + b 
*d*e*(1 + m) - a*d*f*(2 + m))*(c + d*x)^(2 + m)*Hypergeometric2F1[1, 2 + m 
, 3 + m, (b*(c + d*x))/(b*c - a*d)])/(b*(b*c - a*d)^2*(2 + m))
 

Defintions of rubi rules used

rule 78

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b 
*c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m 
 + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] 
 &&  !IntegerQ[m] && IntegerQ[n]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

Maple [F]

\[\int \frac {\left (x d +c \right )^{1+m} \left (f x +e \right )}{\left (b x +a \right )^{2}}d x\]

Input:

int((d*x+c)^(1+m)*(f*x+e)/(b*x+a)^2,x)
 

Output:

int((d*x+c)^(1+m)*(f*x+e)/(b*x+a)^2,x)
 

Fricas [F]

\[ \int \frac {(c+d x)^{1+m} (e+f x)}{(a+b x)^2} \, dx=\int { \frac {{\left (f x + e\right )} {\left (d x + c\right )}^{m + 1}}{{\left (b x + a\right )}^{2}} \,d x } \] Input:

integrate((d*x+c)^(1+m)*(f*x+e)/(b*x+a)^2,x, algorithm="fricas")
 

Output:

integral((f*x + e)*(d*x + c)^(m + 1)/(b^2*x^2 + 2*a*b*x + a^2), x)
 

Sympy [F]

\[ \int \frac {(c+d x)^{1+m} (e+f x)}{(a+b x)^2} \, dx=\int \frac {\left (c + d x\right )^{m + 1} \left (e + f x\right )}{\left (a + b x\right )^{2}}\, dx \] Input:

integrate((d*x+c)**(1+m)*(f*x+e)/(b*x+a)**2,x)
 

Output:

Integral((c + d*x)**(m + 1)*(e + f*x)/(a + b*x)**2, x)
 

Maxima [F]

\[ \int \frac {(c+d x)^{1+m} (e+f x)}{(a+b x)^2} \, dx=\int { \frac {{\left (f x + e\right )} {\left (d x + c\right )}^{m + 1}}{{\left (b x + a\right )}^{2}} \,d x } \] Input:

integrate((d*x+c)^(1+m)*(f*x+e)/(b*x+a)^2,x, algorithm="maxima")
 

Output:

integrate((f*x + e)*(d*x + c)^(m + 1)/(b*x + a)^2, x)
 

Giac [F]

\[ \int \frac {(c+d x)^{1+m} (e+f x)}{(a+b x)^2} \, dx=\int { \frac {{\left (f x + e\right )} {\left (d x + c\right )}^{m + 1}}{{\left (b x + a\right )}^{2}} \,d x } \] Input:

integrate((d*x+c)^(1+m)*(f*x+e)/(b*x+a)^2,x, algorithm="giac")
 

Output:

integrate((f*x + e)*(d*x + c)^(m + 1)/(b*x + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{1+m} (e+f x)}{(a+b x)^2} \, dx=\int \frac {\left (e+f\,x\right )\,{\left (c+d\,x\right )}^{m+1}}{{\left (a+b\,x\right )}^2} \,d x \] Input:

int(((e + f*x)*(c + d*x)^(m + 1))/(a + b*x)^2,x)
 

Output:

int(((e + f*x)*(c + d*x)^(m + 1))/(a + b*x)^2, x)
 

Reduce [F]

\[ \int \frac {(c+d x)^{1+m} (e+f x)}{(a+b x)^2} \, dx=\text {too large to display} \] Input:

int((d*x+c)^(1+m)*(f*x+e)/(b*x+a)^2,x)
 

Output:

((c + d*x)**m*a**2*c*d*f*m + 2*(c + d*x)**m*a**2*c*d*f - (c + d*x)**m*a**2 
*d**2*f*m**2*x - 2*(c + d*x)**m*a**2*d**2*f*m*x - 2*(c + d*x)**m*a*b*c**2* 
f*m - (c + d*x)**m*a*b*c**2*f - (c + d*x)**m*a*b*c*d*e*m - (c + d*x)**m*a* 
b*c*d*e + 2*(c + d*x)**m*a*b*c*d*f*m**2*x + 2*(c + d*x)**m*a*b*c*d*f*m*x + 
 2*(c + d*x)**m*a*b*c*d*f*x + (c + d*x)**m*a*b*d**2*e*m**2*x + (c + d*x)** 
m*a*b*d**2*e*m*x + (c + d*x)**m*a*b*d**2*f*m**2*x**2 + (c + d*x)**m*b**2*c 
**2*e*m**2 + (c + d*x)**m*b**2*c**2*e*m - 2*(c + d*x)**m*b**2*c**2*f*m*x - 
 (c + d*x)**m*b**2*c**2*f*x - (c + d*x)**m*b**2*c*d*e*m*x - (c + d*x)**m*b 
**2*c*d*e*x - (c + d*x)**m*b**2*c*d*f*m*x**2 + int(((c + d*x)**m*x)/(a**3* 
c*d*m + a**3*d**2*m*x - a**2*b*c**2 + 2*a**2*b*c*d*m*x - a**2*b*c*d*x + 2* 
a**2*b*d**2*m*x**2 - 2*a*b**2*c**2*x + a*b**2*c*d*m*x**2 - 2*a*b**2*c*d*x* 
*2 + a*b**2*d**2*m*x**3 - b**3*c**2*x**2 - b**3*c*d*x**3),x)*a**5*d**4*f*m 
**4 + 3*int(((c + d*x)**m*x)/(a**3*c*d*m + a**3*d**2*m*x - a**2*b*c**2 + 2 
*a**2*b*c*d*m*x - a**2*b*c*d*x + 2*a**2*b*d**2*m*x**2 - 2*a*b**2*c**2*x + 
a*b**2*c*d*m*x**2 - 2*a*b**2*c*d*x**2 + a*b**2*d**2*m*x**3 - b**3*c**2*x** 
2 - b**3*c*d*x**3),x)*a**5*d**4*f*m**3 + 2*int(((c + d*x)**m*x)/(a**3*c*d* 
m + a**3*d**2*m*x - a**2*b*c**2 + 2*a**2*b*c*d*m*x - a**2*b*c*d*x + 2*a**2 
*b*d**2*m*x**2 - 2*a*b**2*c**2*x + a*b**2*c*d*m*x**2 - 2*a*b**2*c*d*x**2 + 
 a*b**2*d**2*m*x**3 - b**3*c**2*x**2 - b**3*c*d*x**3),x)*a**5*d**4*f*m**2 
- 2*int(((c + d*x)**m*x)/(a**3*c*d*m + a**3*d**2*m*x - a**2*b*c**2 + 2*...