\(\int (a+b x)^m (c+d x)^{2-m} (e+f x)^2 \, dx\) [1768]

Optimal result

Integrand size = 26, antiderivative size = 237 \[ \int (a+b x)^m (c+d x)^{2-m} (e+f x)^2 \, dx=\frac {f (10 b d e-a d f (8-m)-b c f (2+m)) (a+b x)^{1+m} (c+d x)^{3-m}}{20 b^2 d^2}+\frac {f^2 (a+b x)^{2+m} (c+d x)^{3-m}}{5 b^2 d}+\frac {\left (a^2 d^2 f^2 \left (12-7 m+m^2\right )-2 a b d f (3-m) (5 d e-c f (1+m))+b^2 \left (20 d^2 e^2-10 c d e f (1+m)+c^2 f^2 \left (2+3 m+m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{3-m} \operatorname {Hypergeometric2F1}\left (1,4,2+m,-\frac {d (a+b x)}{b c-a d}\right )}{20 b^2 d^2 (b c-a d) (1+m)} \] Output:

1/20*f*(10*b*d*e-a*d*f*(8-m)-b*c*f*(2+m))*(b*x+a)^(1+m)*(d*x+c)^(3-m)/b^2/ 
d^2+1/5*f^2*(b*x+a)^(2+m)*(d*x+c)^(3-m)/b^2/d+1/20*(a^2*d^2*f^2*(m^2-7*m+1 
2)-2*a*b*d*f*(3-m)*(5*d*e-c*f*(1+m))+b^2*(20*d^2*e^2-10*c*d*e*f*(1+m)+c^2* 
f^2*(m^2+3*m+2)))*(b*x+a)^(1+m)*(d*x+c)^(3-m)*hypergeom([1, 4],[2+m],-d*(b 
*x+a)/(-a*d+b*c))/b^2/d^2/(-a*d+b*c)/(1+m)
 

Mathematica [A] (warning: unable to verify)

Time = 0.29 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.95 \[ \int (a+b x)^m (c+d x)^{2-m} (e+f x)^2 \, dx=\frac {(a+b x)^{1+m} (c+d x)^{-m} \left (b^3 f (1+m) (6 b d e+a d f (-4+m)-b c f (2+m)) (c+d x)^3+4 b^4 d f (1+m) (c+d x)^3 (e+f x)+(b c-a d)^2 \left (a^2 d^2 f^2 \left (12-7 m+m^2\right )-2 a b d f (-3+m) (-5 d e+c f (1+m))+b^2 \left (20 d^2 e^2-10 c d e f (1+m)+c^2 f^2 \left (2+3 m+m^2\right )\right )\right ) \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (-2+m,1+m,2+m,\frac {d (a+b x)}{-b c+a d}\right )\right )}{20 b^5 d^2 (1+m)} \] Input:

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

Output:

((a + b*x)^(1 + m)*(b^3*f*(1 + m)*(6*b*d*e + a*d*f*(-4 + m) - b*c*f*(2 + m 
))*(c + d*x)^3 + 4*b^4*d*f*(1 + m)*(c + d*x)^3*(e + f*x) + (b*c - a*d)^2*( 
a^2*d^2*f^2*(12 - 7*m + m^2) - 2*a*b*d*f*(-3 + m)*(-5*d*e + c*f*(1 + m)) + 
 b^2*(20*d^2*e^2 - 10*c*d*e*f*(1 + m) + c^2*f^2*(2 + 3*m + m^2)))*((b*(c + 
 d*x))/(b*c - a*d))^m*Hypergeometric2F1[-2 + m, 1 + m, 2 + m, (d*(a + b*x) 
)/(-(b*c) + a*d)]))/(20*b^5*d^2*(1 + m)*(c + d*x)^m)
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.14, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {101, 25, 90, 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.

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

\(\Big \downarrow \) 101

\(\displaystyle \frac {\int -(a+b x)^m (c+d x)^{2-m} (a f (c f+d e (3-m))-b e (5 d e-c f (m+1))-f (6 b d e-a d f (4-m)-b c f (m+2)) x)dx}{5 b d}+\frac {f (e+f x) (a+b x)^{m+1} (c+d x)^{3-m}}{5 b d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {f (e+f x) (a+b x)^{m+1} (c+d x)^{3-m}}{5 b d}-\frac {\int (a+b x)^m (c+d x)^{2-m} (a f (c f+d e (3-m))-b e (5 d e-c f (m+1))-f (6 b d e-a d f (4-m)-b c f (m+2)) x)dx}{5 b d}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {f (e+f x) (a+b x)^{m+1} (c+d x)^{3-m}}{5 b d}-\frac {\frac {(f (a d (3-m)+b c (m+1)) (-a d f (4-m)-b c f (m+2)+6 b d e)+4 b d (a f (c f+d e (3-m))-b e (5 d e-c f (m+1)))) \int (a+b x)^m (c+d x)^{2-m}dx}{4 b d}-\frac {f (a+b x)^{m+1} (c+d x)^{3-m} (-a d f (4-m)-b c f (m+2)+6 b d e)}{4 b d}}{5 b d}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {f (e+f x) (a+b x)^{m+1} (c+d x)^{3-m}}{5 b d}-\frac {\frac {(b c-a d)^2 (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m (f (a d (3-m)+b c (m+1)) (-a d f (4-m)-b c f (m+2)+6 b d e)+4 b d (a f (c f+d e (3-m))-b e (5 d e-c f (m+1)))) \int (a+b x)^m \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{2-m}dx}{4 b^3 d}-\frac {f (a+b x)^{m+1} (c+d x)^{3-m} (-a d f (4-m)-b c f (m+2)+6 b d e)}{4 b d}}{5 b d}\)

\(\Big \downarrow \) 79

\(\displaystyle \frac {f (e+f x) (a+b x)^{m+1} (c+d x)^{3-m}}{5 b d}-\frac {\frac {(b c-a d)^2 (a+b x)^{m+1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m (f (a d (3-m)+b c (m+1)) (-a d f (4-m)-b c f (m+2)+6 b d e)+4 b d (a f (c f+d e (3-m))-b e (5 d e-c f (m+1)))) \operatorname {Hypergeometric2F1}\left (m-2,m+1,m+2,-\frac {d (a+b x)}{b c-a d}\right )}{4 b^4 d (m+1)}-\frac {f (a+b x)^{m+1} (c+d x)^{3-m} (-a d f (4-m)-b c f (m+2)+6 b d e)}{4 b d}}{5 b d}\)

Input:

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

Output:

(f*(a + b*x)^(1 + m)*(c + d*x)^(3 - m)*(e + f*x))/(5*b*d) - (-1/4*(f*(6*b* 
d*e - a*d*f*(4 - m) - b*c*f*(2 + m))*(a + b*x)^(1 + m)*(c + d*x)^(3 - m))/ 
(b*d) + ((b*c - a*d)^2*(f*(a*d*(3 - m) + b*c*(1 + m))*(6*b*d*e - a*d*f*(4 
- m) - b*c*f*(2 + m)) + 4*b*d*(a*f*(c*f + d*e*(3 - m)) - b*e*(5*d*e - c*f* 
(1 + m))))*(a + b*x)^(1 + m)*((b*(c + d*x))/(b*c - a*d))^m*Hypergeometric2 
F1[-2 + m, 1 + m, 2 + m, -((d*(a + b*x))/(b*c - a*d))])/(4*b^4*d*(1 + m)*( 
c + d*x)^m))/(5*b*d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 79

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]))
 

rule 80

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])
 

rule 90

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

rule 101

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + 
 p + 3))), x] + Simp[1/(d*f*(n + p + 3))   Int[(c + d*x)^n*(e + f*x)^p*Simp 
[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f 
*(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, 
 c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-2)]

Exception generated. \[ \int (a+b x)^m (c+d x)^{2-m} (e+f x)^2 \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:

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

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (a+b x)^m (c+d x)^{2-m} (e+f x)^2 \, dx=\left (\int \frac {\left (b x +a \right )^{m}}{\left (d x +c \right )^{m}}d x \right ) c^{2} e^{2}+\left (\int \frac {\left (b x +a \right )^{m} x^{4}}{\left (d x +c \right )^{m}}d x \right ) d^{2} f^{2}+2 \left (\int \frac {\left (b x +a \right )^{m} x^{3}}{\left (d x +c \right )^{m}}d x \right ) c d \,f^{2}+2 \left (\int \frac {\left (b x +a \right )^{m} x^{3}}{\left (d x +c \right )^{m}}d x \right ) d^{2} e f +\left (\int \frac {\left (b x +a \right )^{m} x^{2}}{\left (d x +c \right )^{m}}d x \right ) c^{2} f^{2}+4 \left (\int \frac {\left (b x +a \right )^{m} x^{2}}{\left (d x +c \right )^{m}}d x \right ) c d e f +\left (\int \frac {\left (b x +a \right )^{m} x^{2}}{\left (d x +c \right )^{m}}d x \right ) d^{2} e^{2}+2 \left (\int \frac {\left (b x +a \right )^{m} x}{\left (d x +c \right )^{m}}d x \right ) c^{2} e f +2 \left (\int \frac {\left (b x +a \right )^{m} x}{\left (d x +c \right )^{m}}d x \right ) c d \,e^{2} \] Input:

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

Output:

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