\(\int \frac {(c+d x)^q (a x+b x^2)^p}{x^2} \, dx\) [216]

Optimal result

Integrand size = 22, antiderivative size = 79 \[ \int \frac {(c+d x)^q \left (a x+b x^2\right )^p}{x^2} \, dx=-\frac {\left (1+\frac {b x}{a}\right )^{-p} (c+d x)^q \left (1+\frac {d x}{c}\right )^{-q} \left (a x+b x^2\right )^p \operatorname {AppellF1}\left (-1+p,-p,-q,p,-\frac {b x}{a},-\frac {d x}{c}\right )}{(1-p) x} \] Output:

-(d*x+c)^q*(b*x^2+a*x)^p*AppellF1(-1+p,-p,-q,p,-b*x/a,-d*x/c)/(1-p)/x/((1+ 
b*x/a)^p)/((1+d*x/c)^q)
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.96 \[ \int \frac {(c+d x)^q \left (a x+b x^2\right )^p}{x^2} \, dx=\frac {\left (\frac {a+b x}{a}\right )^{-p} (x (a+b x))^p (c+d x)^q \left (\frac {c+d x}{c}\right )^{-q} \operatorname {AppellF1}\left (-1+p,-p,-q,p,-\frac {b x}{a},-\frac {d x}{c}\right )}{(-1+p) x} \] Input:

Integrate[((c + d*x)^q*(a*x + b*x^2)^p)/x^2,x]
 

Output:

((x*(a + b*x))^p*(c + d*x)^q*AppellF1[-1 + p, -p, -q, p, -((b*x)/a), -((d* 
x)/c)])/((-1 + p)*x*((a + b*x)/a)^p*((c + d*x)/c)^q)
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1261, 152, 152, 150}

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.

\(\displaystyle \int \frac {\left (a x+b x^2\right )^p (c+d x)^q}{x^2} \, dx\)

\(\Big \downarrow \) 1261

\(\displaystyle x^{-p} (a+b x)^{-p} \left (a x+b x^2\right )^p \int x^{p-2} (a+b x)^p (c+d x)^qdx\)

\(\Big \downarrow \) 152

\(\displaystyle x^{-p} \left (\frac {b x}{a}+1\right )^{-p} \left (a x+b x^2\right )^p \int x^{p-2} \left (\frac {b x}{a}+1\right )^p (c+d x)^qdx\)

\(\Big \downarrow \) 152

\(\displaystyle x^{-p} \left (\frac {b x}{a}+1\right )^{-p} \left (a x+b x^2\right )^p (c+d x)^q \left (\frac {d x}{c}+1\right )^{-q} \int x^{p-2} \left (\frac {b x}{a}+1\right )^p \left (\frac {d x}{c}+1\right )^qdx\)

\(\Big \downarrow \) 150

\(\displaystyle -\frac {\left (\frac {b x}{a}+1\right )^{-p} \left (a x+b x^2\right )^p (c+d x)^q \left (\frac {d x}{c}+1\right )^{-q} \operatorname {AppellF1}\left (p-1,-p,-q,p,-\frac {b x}{a},-\frac {d x}{c}\right )}{(1-p) x}\)

Input:

Int[((c + d*x)^q*(a*x + b*x^2)^p)/x^2,x]
 

Output:

-(((c + d*x)^q*(a*x + b*x^2)^p*AppellF1[-1 + p, -p, -q, p, -((b*x)/a), -(( 
d*x)/c)])/((1 - p)*x*(1 + (b*x)/a)^p*(1 + (d*x)/c)^q))
 

Defintions of rubi rules used

rule 150

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 
, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
 

rule 152

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]) 
Int[(b*x)^m*(1 + d*(x/c))^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d, e, f, m, 
 n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]
 

rule 1261

Int[((e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((b_.)*(x_) + (c_.)*(x_)^2) 
^(p_), x_Symbol] :> Simp[(e*x)^m*((b*x + c*x^2)^p/(x^(m + p)*(b + c*x)^p)) 
  Int[x^(m + p)*(f + g*x)^n*(b + c*x)^p, x], x] /; FreeQ[{b, c, e, f, g, m, 
 n}, x] &&  !IGtQ[n, 0]
 

Maple [F]

\[\int \frac {\left (d x +c \right )^{q} \left (b \,x^{2}+a x \right )^{p}}{x^{2}}d x\]

Input:

int((d*x+c)^q*(b*x^2+a*x)^p/x^2,x)
 

Output:

int((d*x+c)^q*(b*x^2+a*x)^p/x^2,x)
 

Fricas [F]

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

integrate((d*x+c)^q*(b*x^2+a*x)^p/x^2,x, algorithm="fricas")
 

Output:

integral((b*x^2 + a*x)^p*(d*x + c)^q/x^2, x)
 

Sympy [F]

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

integrate((d*x+c)**q*(b*x**2+a*x)**p/x**2,x)
 

Output:

Integral((x*(a + b*x))**p*(c + d*x)**q/x**2, x)
 

Maxima [F]

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

integrate((d*x+c)^q*(b*x^2+a*x)^p/x^2,x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a*x)^p*(d*x + c)^q/x^2, x)
 

Giac [F]

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

integrate((d*x+c)^q*(b*x^2+a*x)^p/x^2,x, algorithm="giac")
 

Output:

integrate((b*x^2 + a*x)^p*(d*x + c)^q/x^2, x)
 

Mupad [F(-1)]

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

int(((a*x + b*x^2)^p*(c + d*x)^q)/x^2,x)
 

Output:

int(((a*x + b*x^2)^p*(c + d*x)^q)/x^2, x)
 

Reduce [F]

\[ \int \frac {(c+d x)^q \left (a x+b x^2\right )^p}{x^2} \, dx=\text {too large to display} \] Input:

int((d*x+c)^q*(b*x^2+a*x)^p/x^2,x)
 

Output:

((c + d*x)**q*(a*x + b*x**2)**p - int(((c + d*x)**q*(a*x + b*x**2)**p)/(a* 
*2*c*d*p**2*x + a**2*c*d*p*q*x - 2*a**2*c*d*p*x - a**2*c*d*q*x + a**2*c*d* 
x + a**2*d**2*p**2*x**2 + a**2*d**2*p*q*x**2 - 2*a**2*d**2*p*x**2 - a**2*d 
**2*q*x**2 + a**2*d**2*x**2 + 2*a*b*c**2*p**2*x - 3*a*b*c**2*p*x + a*b*c** 
2*x + 3*a*b*c*d*p**2*x**2 + a*b*c*d*p*q*x**2 - 5*a*b*c*d*p*x**2 - a*b*c*d* 
q*x**2 + 2*a*b*c*d*x**2 + a*b*d**2*p**2*x**3 + a*b*d**2*p*q*x**3 - 2*a*b*d 
**2*p*x**3 - a*b*d**2*q*x**3 + a*b*d**2*x**3 + 2*b**2*c**2*p**2*x**2 - 3*b 
**2*c**2*p*x**2 + b**2*c**2*x**2 + 2*b**2*c*d*p**2*x**3 - 3*b**2*c*d*p*x** 
3 + b**2*c*d*x**3),x)*a**2*d**2*p**2*q*x - int(((c + d*x)**q*(a*x + b*x**2 
)**p)/(a**2*c*d*p**2*x + a**2*c*d*p*q*x - 2*a**2*c*d*p*x - a**2*c*d*q*x + 
a**2*c*d*x + a**2*d**2*p**2*x**2 + a**2*d**2*p*q*x**2 - 2*a**2*d**2*p*x**2 
 - a**2*d**2*q*x**2 + a**2*d**2*x**2 + 2*a*b*c**2*p**2*x - 3*a*b*c**2*p*x 
+ a*b*c**2*x + 3*a*b*c*d*p**2*x**2 + a*b*c*d*p*q*x**2 - 5*a*b*c*d*p*x**2 - 
 a*b*c*d*q*x**2 + 2*a*b*c*d*x**2 + a*b*d**2*p**2*x**3 + a*b*d**2*p*q*x**3 
- 2*a*b*d**2*p*x**3 - a*b*d**2*q*x**3 + a*b*d**2*x**3 + 2*b**2*c**2*p**2*x 
**2 - 3*b**2*c**2*p*x**2 + b**2*c**2*x**2 + 2*b**2*c*d*p**2*x**3 - 3*b**2* 
c*d*p*x**3 + b**2*c*d*x**3),x)*a**2*d**2*p*q**2*x + 2*int(((c + d*x)**q*(a 
*x + b*x**2)**p)/(a**2*c*d*p**2*x + a**2*c*d*p*q*x - 2*a**2*c*d*p*x - a**2 
*c*d*q*x + a**2*c*d*x + a**2*d**2*p**2*x**2 + a**2*d**2*p*q*x**2 - 2*a**2* 
d**2*p*x**2 - a**2*d**2*q*x**2 + a**2*d**2*x**2 + 2*a*b*c**2*p**2*x - 3...