\(\int (a+b x)^m (c+d x)^n (A+B x+C x^2) \, dx\) [191]

Optimal result

Integrand size = 25, antiderivative size = 256 \[ \int (a+b x)^m (c+d x)^n \left (A+B x+C x^2\right ) \, dx=-\frac {(b c C (2+m)-b B d (3+m+n)+a C d (4+m+2 n)) (a+b x)^{1+m} (c+d x)^{1+n}}{b^2 d^2 (2+m+n) (3+m+n)}+\frac {C (a+b x)^{2+m} (c+d x)^{1+n}}{b^2 d (3+m+n)}-\frac {\left (d (2+m+n) \left (a b c C (2+m)+a^2 C d (1+n)-A b^2 d (3+m+n)\right )-(b c (1+m)+a d (1+n)) (b c C (2+m)-b B d (3+m+n)+a C d (4+m+2 n))\right ) (a+b x)^{1+m} (c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,2+m+n,2+m,-\frac {d (a+b x)}{b c-a d}\right )}{b^2 d^2 (b c-a d) (1+m) (2+m+n) (3+m+n)} \] Output:

-(b*c*C*(2+m)-b*B*d*(3+m+n)+a*C*d*(4+m+2*n))*(b*x+a)^(1+m)*(d*x+c)^(1+n)/b 
^2/d^2/(2+m+n)/(3+m+n)+C*(b*x+a)^(2+m)*(d*x+c)^(1+n)/b^2/d/(3+m+n)-(d*(2+m 
+n)*(a*b*c*C*(2+m)+a^2*C*d*(1+n)-A*b^2*d*(3+m+n))-(b*c*(1+m)+a*d*(1+n))*(b 
*c*C*(2+m)-b*B*d*(3+m+n)+a*C*d*(4+m+2*n)))*(b*x+a)^(1+m)*(d*x+c)^(1+n)*hyp 
ergeom([1, 2+m+n],[2+m],-d*(b*x+a)/(-a*d+b*c))/b^2/d^2/(-a*d+b*c)/(1+m)/(2 
+m+n)/(3+m+n)
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.73 \[ \int (a+b x)^m (c+d x)^n \left (A+B x+C x^2\right ) \, dx=\frac {(a+b x)^{1+m} (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \left (C (b c-a d)^2 \operatorname {Hypergeometric2F1}\left (1+m,-2-n,2+m,\frac {d (a+b x)}{-b c+a d}\right )+b \left (-\left ((b c-a d) (2 c C-B d) \operatorname {Hypergeometric2F1}\left (1+m,-1-n,2+m,\frac {d (a+b x)}{-b c+a d}\right )\right )+b \left (c^2 C-B c d+A d^2\right ) \operatorname {Hypergeometric2F1}\left (1+m,-n,2+m,\frac {d (a+b x)}{-b c+a d}\right )\right )\right )}{b^3 d^2 (1+m)} \] Input:

Integrate[(a + b*x)^m*(c + d*x)^n*(A + B*x + C*x^2),x]
 

Output:

((a + b*x)^(1 + m)*(c + d*x)^n*(C*(b*c - a*d)^2*Hypergeometric2F1[1 + m, - 
2 - n, 2 + m, (d*(a + b*x))/(-(b*c) + a*d)] + b*(-((b*c - a*d)*(2*c*C - B* 
d)*Hypergeometric2F1[1 + m, -1 - n, 2 + m, (d*(a + b*x))/(-(b*c) + a*d)]) 
+ b*(c^2*C - B*c*d + A*d^2)*Hypergeometric2F1[1 + m, -n, 2 + m, (d*(a + b* 
x))/(-(b*c) + a*d)])))/(b^3*d^2*(1 + m)*((b*(c + d*x))/(b*c - a*d))^n)
 

Rubi [A] (warning: unable to verify)

Time = 0.44 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1194, 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.

\(\displaystyle \int (a+b x)^m \left (A+B x+C x^2\right ) (c+d x)^n \, dx\)

\(\Big \downarrow \) 1194

\(\displaystyle \frac {\int -(a+b x)^m (c+d x)^n \left (C d (n+1) a^2+b c C (m+2) a-A b^2 d (m+n+3)+b (b c C (m+2)-b B d (m+n+3)+a C d (m+2 n+4)) x\right )dx}{b^2 d (m+n+3)}+\frac {C (a+b x)^{m+2} (c+d x)^{n+1}}{b^2 d (m+n+3)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {C (a+b x)^{m+2} (c+d x)^{n+1}}{b^2 d (m+n+3)}-\frac {\int (a+b x)^m (c+d x)^n \left (C d (n+1) a^2+b c C (m+2) a-A b^2 d (m+n+3)+b (b c C (m+2)-b B d (m+n+3)+a C d (m+2 n+4)) x\right )dx}{b^2 d (m+n+3)}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {C (a+b x)^{m+2} (c+d x)^{n+1}}{b^2 d (m+n+3)}-\frac {\left (a^2 C d (n+1)-\frac {(a d (n+1)+b c (m+1)) (a C d (m+2 n+4)-b B d (m+n+3)+b c C (m+2))}{d (m+n+2)}+a b c C (m+2)-A b^2 d (m+n+3)\right ) \int (a+b x)^m (c+d x)^ndx+\frac {(a+b x)^{m+1} (c+d x)^{n+1} (a C d (m+2 n+4)-b B d (m+n+3)+b c C (m+2))}{d (m+n+2)}}{b^2 d (m+n+3)}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {C (a+b x)^{m+2} (c+d x)^{n+1}}{b^2 d (m+n+3)}-\frac {(c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \left (a^2 C d (n+1)-\frac {(a d (n+1)+b c (m+1)) (a C d (m+2 n+4)-b B d (m+n+3)+b c C (m+2))}{d (m+n+2)}+a b c C (m+2)-A b^2 d (m+n+3)\right ) \int (a+b x)^m \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^ndx+\frac {(a+b x)^{m+1} (c+d x)^{n+1} (a C d (m+2 n+4)-b B d (m+n+3)+b c C (m+2))}{d (m+n+2)}}{b^2 d (m+n+3)}\)

\(\Big \downarrow \) 79

\(\displaystyle \frac {C (a+b x)^{m+2} (c+d x)^{n+1}}{b^2 d (m+n+3)}-\frac {\frac {(a+b x)^{m+1} (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \operatorname {Hypergeometric2F1}\left (m+1,-n,m+2,-\frac {d (a+b x)}{b c-a d}\right ) \left (a^2 C d (n+1)-\frac {(a d (n+1)+b c (m+1)) (a C d (m+2 n+4)-b B d (m+n+3)+b c C (m+2))}{d (m+n+2)}+a b c C (m+2)-A b^2 d (m+n+3)\right )}{b (m+1)}+\frac {(a+b x)^{m+1} (c+d x)^{n+1} (a C d (m+2 n+4)-b B d (m+n+3)+b c C (m+2))}{d (m+n+2)}}{b^2 d (m+n+3)}\)

Input:

Int[(a + b*x)^m*(c + d*x)^n*(A + B*x + C*x^2),x]
 

Output:

(C*(a + b*x)^(2 + m)*(c + d*x)^(1 + n))/(b^2*d*(3 + m + n)) - (((b*c*C*(2 
+ m) - b*B*d*(3 + m + n) + a*C*d*(4 + m + 2*n))*(a + b*x)^(1 + m)*(c + d*x 
)^(1 + n))/(d*(2 + m + n)) + ((a*b*c*C*(2 + m) + a^2*C*d*(1 + n) - A*b^2*d 
*(3 + m + n) - ((b*c*(1 + m) + a*d*(1 + n))*(b*c*C*(2 + m) - b*B*d*(3 + m 
+ n) + a*C*d*(4 + m + 2*n)))/(d*(2 + m + n)))*(a + b*x)^(1 + m)*(c + d*x)^ 
n*Hypergeometric2F1[1 + m, -n, 2 + m, -((d*(a + b*x))/(b*c - a*d))])/(b*(1 
 + m)*((b*(c + d*x))/(b*c - a*d))^n))/(b^2*d*(3 + m + n))
 

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 1194

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) 
 + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[c^p*(d + e*x)^(m + 2*p)*((f + g*x 
)^(n + 1)/(g*e^(2*p)*(m + n + 2*p + 1))), x] + Simp[1/(g*e^(2*p)*(m + n + 2 
*p + 1))   Int[(d + e*x)^m*(f + g*x)^n*ExpandToSum[g*(m + n + 2*p + 1)*(e^( 
2*p)*(a + b*x + c*x^2)^p - c^p*(d + e*x)^(2*p)) - c^p*(e*f - d*g)*(m + 2*p) 
*(d + e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ 
[p, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && NeQ[m + n + 2*p + 1, 0]
 

Maple [F]

\[\int \left (b x +a \right )^{m} \left (x d +c \right )^{n} \left (C \,x^{2}+B x +A \right )d x\]

Input:

int((b*x+a)^m*(d*x+c)^n*(C*x^2+B*x+A),x)
 

Output:

int((b*x+a)^m*(d*x+c)^n*(C*x^2+B*x+A),x)
 

Fricas [F]

\[ \int (a+b x)^m (c+d x)^n \left (A+B x+C x^2\right ) \, dx=\int { {\left (C x^{2} + B x + A\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{n} \,d x } \] Input:

integrate((b*x+a)^m*(d*x+c)^n*(C*x^2+B*x+A),x, algorithm="fricas")
 

Output:

integral((C*x^2 + B*x + A)*(b*x + a)^m*(d*x + c)^n, x)
 

Sympy [F(-2)]

Exception generated. \[ \int (a+b x)^m (c+d x)^n \left (A+B x+C x^2\right ) \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:

integrate((b*x+a)**m*(d*x+c)**n*(C*x**2+B*x+A),x)
 

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

\[ \int (a+b x)^m (c+d x)^n \left (A+B x+C x^2\right ) \, dx=\int { {\left (C x^{2} + B x + A\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{n} \,d x } \] Input:

integrate((b*x+a)^m*(d*x+c)^n*(C*x^2+B*x+A),x, algorithm="maxima")
 

Output:

integrate((C*x^2 + B*x + A)*(b*x + a)^m*(d*x + c)^n, x)
 

Giac [F]

\[ \int (a+b x)^m (c+d x)^n \left (A+B x+C x^2\right ) \, dx=\int { {\left (C x^{2} + B x + A\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{n} \,d x } \] Input:

integrate((b*x+a)^m*(d*x+c)^n*(C*x^2+B*x+A),x, algorithm="giac")
 

Output:

integrate((C*x^2 + B*x + A)*(b*x + a)^m*(d*x + c)^n, x)
 

Mupad [F(-1)]

Timed out. \[ \int (a+b x)^m (c+d x)^n \left (A+B x+C x^2\right ) \, dx=\int {\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^n\,\left (C\,x^2+B\,x+A\right ) \,d x \] Input:

int((a + b*x)^m*(c + d*x)^n*(A + B*x + C*x^2),x)
 

Output:

int((a + b*x)^m*(c + d*x)^n*(A + B*x + C*x^2), x)
 

Reduce [F]

\[ \int (a+b x)^m (c+d x)^n \left (A+B x+C x^2\right ) \, dx=\int \left (b x +a \right )^{m} \left (d x +c \right )^{n} \left (C \,x^{2}+B x +A \right )d x \] Input:

int((b*x+a)^m*(d*x+c)^n*(C*x^2+B*x+A),x)
 

Output:

int((b*x+a)^m*(d*x+c)^n*(C*x^2+B*x+A),x)