Integrand size = 24, antiderivative size = 107 \[ \int (b d+2 c d x)^m \left (a+b x+c x^2\right )^p \, dx=-\frac {2 (b d+2 c d x)^{1+m} \left (\frac {1}{4} \left (4 a-\frac {b^2}{c}\right )+\frac {(b+2 c x)^2}{4 c}\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (3+m+2 p),\frac {3+m}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{\left (b^2-4 a c\right ) d (1+m)} \] Output:
-2*(2*c*d*x+b*d)^(1+m)*(a-1/4*b^2/c+1/4*(2*c*x+b)^2/c)^(p+1)*hypergeom([1, 3/2+1/2*m+p],[3/2+1/2*m],(2*c*x+b)^2/(-4*a*c+b^2))/(-4*a*c+b^2)/d/(1+m)
Time = 0.34 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.06 \[ \int (b d+2 c d x)^m \left (a+b x+c x^2\right )^p \, dx=\frac {2^{-1-2 p} (b+2 c x) (d (b+2 c x))^m (a+x (b+c x))^p \left (\frac {c (a+x (b+c x))}{-b^2+4 a c}\right )^{-p} \, _2F_1\left (\frac {1}{2}+\frac {m}{2},-p;\frac {3}{2}+\frac {m}{2};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{c (1+m)} \] Input:
Integrate[(b*d + 2*c*d*x)^m*(a + b*x + c*x^2)^p,x]
Output:
(2^(-1 - 2*p)*(b + 2*c*x)*(d*(b + 2*c*x))^m*(a + x*(b + c*x))^p*Hypergeome tricPFQ[{1/2 + m/2, -p}, {3/2 + m/2}, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(c*(1 + m)*((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^p)
Time = 0.23 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.34, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1118, 279, 278}
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.
\(\displaystyle \int \left (a+b x+c x^2\right )^p (b d+2 c d x)^m \, dx\) |
\(\Big \downarrow \) 1118 |
\(\displaystyle \frac {\int (b d+2 c x d)^m \left (\frac {(b d+2 c x d)^2}{4 c d^2}+\frac {1}{4} \left (4 a-\frac {b^2}{c}\right )\right )^pd(b d+2 c x d)}{2 c d}\) |
\(\Big \downarrow \) 279 |
\(\displaystyle \frac {\left (\frac {1}{4} \left (4 a-\frac {b^2}{c}\right )+\frac {(b d+2 c d x)^2}{4 c d^2}\right )^p \left (1-\frac {(b d+2 c d x)^2}{d^2 \left (b^2-4 a c\right )}\right )^{-p} \int (b d+2 c x d)^m \left (1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}\right )^pd(b d+2 c x d)}{2 c d}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {(b d+2 c d x)^{m+1} \left (\frac {1}{4} \left (4 a-\frac {b^2}{c}\right )+\frac {(b d+2 c d x)^2}{4 c d^2}\right )^p \left (1-\frac {(b d+2 c d x)^2}{d^2 \left (b^2-4 a c\right )}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {m+1}{2},-p,\frac {m+3}{2},\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}\right )}{2 c d (m+1)}\) |
Input:
Int[(b*d + 2*c*d*x)^m*(a + b*x + c*x^2)^p,x]
Output:
((b*d + 2*c*d*x)^(1 + m)*((4*a - b^2/c)/4 + (b*d + 2*c*d*x)^2/(4*c*d^2))^p *Hypergeometric2F1[(1 + m)/2, -p, (3 + m)/2, (b*d + 2*c*d*x)^2/((b^2 - 4*a *c)*d^2)])/(2*c*d*(1 + m)*(1 - (b*d + 2*c*d*x)^2/((b^2 - 4*a*c)*d^2))^p)
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[(c*x)^m* (1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && !(ILtQ[p, 0] || GtQ[a, 0])
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> Simp[1/e Subst[Int[x^m*(a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[2*c*d - b*e, 0]
\[\int \left (2 c d x +b d \right )^{m} \left (c \,x^{2}+b x +a \right )^{p}d x\]
Input:
int((2*c*d*x+b*d)^m*(c*x^2+b*x+a)^p,x)
Output:
int((2*c*d*x+b*d)^m*(c*x^2+b*x+a)^p,x)
\[ \int (b d+2 c d x)^m \left (a+b x+c x^2\right )^p \, dx=\int { {\left (2 \, c d x + b d\right )}^{m} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \] Input:
integrate((2*c*d*x+b*d)^m*(c*x^2+b*x+a)^p,x, algorithm="fricas")
Output:
integral((2*c*d*x + b*d)^m*(c*x^2 + b*x + a)^p, x)
Timed out. \[ \int (b d+2 c d x)^m \left (a+b x+c x^2\right )^p \, dx=\text {Timed out} \] Input:
integrate((2*c*d*x+b*d)**m*(c*x**2+b*x+a)**p,x)
Output:
Timed out
\[ \int (b d+2 c d x)^m \left (a+b x+c x^2\right )^p \, dx=\int { {\left (2 \, c d x + b d\right )}^{m} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \] Input:
integrate((2*c*d*x+b*d)^m*(c*x^2+b*x+a)^p,x, algorithm="maxima")
Output:
integrate((2*c*d*x + b*d)^m*(c*x^2 + b*x + a)^p, x)
\[ \int (b d+2 c d x)^m \left (a+b x+c x^2\right )^p \, dx=\int { {\left (2 \, c d x + b d\right )}^{m} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \] Input:
integrate((2*c*d*x+b*d)^m*(c*x^2+b*x+a)^p,x, algorithm="giac")
Output:
integrate((2*c*d*x + b*d)^m*(c*x^2 + b*x + a)^p, x)
Timed out. \[ \int (b d+2 c d x)^m \left (a+b x+c x^2\right )^p \, dx=\int {\left (b\,d+2\,c\,d\,x\right )}^m\,{\left (c\,x^2+b\,x+a\right )}^p \,d x \] Input:
int((b*d + 2*c*d*x)^m*(a + b*x + c*x^2)^p,x)
Output:
int((b*d + 2*c*d*x)^m*(a + b*x + c*x^2)^p, x)
\[ \int (b d+2 c d x)^m \left (a+b x+c x^2\right )^p \, dx=\text {too large to display} \] Input:
int((2*c*d*x+b*d)^m*(c*x^2+b*x+a)^p,x)
Output:
(4*(a + b*x + c*x**2)**p*(b*d + 2*c*d*x)**m*a*c*p + (a + b*x + c*x**2)**p* (b*d + 2*c*d*x)**m*b**2*m + (a + b*x + c*x**2)**p*(b*d + 2*c*d*x)**m*b**2* p + 2*(a + b*x + c*x**2)**p*(b*d + 2*c*d*x)**m*b*c*m*x + 4*(a + b*x + c*x* *2)**p*(b*d + 2*c*d*x)**m*b*c*p*x - 8*int(((a + b*x + c*x**2)**p*(b*d + 2* c*d*x)**m*x**2)/(a*b*m**2 + 4*a*b*m*p + a*b*m + 4*a*b*p**2 + 2*a*b*p + 2*a *c*m**2*x + 8*a*c*m*p*x + 2*a*c*m*x + 8*a*c*p**2*x + 4*a*c*p*x + b**2*m**2 *x + 4*b**2*m*p*x + b**2*m*x + 4*b**2*p**2*x + 2*b**2*p*x + 3*b*c*m**2*x** 2 + 12*b*c*m*p*x**2 + 3*b*c*m*x**2 + 12*b*c*p**2*x**2 + 6*b*c*p*x**2 + 2*c **2*m**2*x**3 + 8*c**2*m*p*x**3 + 2*c**2*m*x**3 + 8*c**2*p**2*x**3 + 4*c** 2*p*x**3),x)*a*c**3*m**3*p - 48*int(((a + b*x + c*x**2)**p*(b*d + 2*c*d*x) **m*x**2)/(a*b*m**2 + 4*a*b*m*p + a*b*m + 4*a*b*p**2 + 2*a*b*p + 2*a*c*m** 2*x + 8*a*c*m*p*x + 2*a*c*m*x + 8*a*c*p**2*x + 4*a*c*p*x + b**2*m**2*x + 4 *b**2*m*p*x + b**2*m*x + 4*b**2*p**2*x + 2*b**2*p*x + 3*b*c*m**2*x**2 + 12 *b*c*m*p*x**2 + 3*b*c*m*x**2 + 12*b*c*p**2*x**2 + 6*b*c*p*x**2 + 2*c**2*m* *2*x**3 + 8*c**2*m*p*x**3 + 2*c**2*m*x**3 + 8*c**2*p**2*x**3 + 4*c**2*p*x* *3),x)*a*c**3*m**2*p**2 - 8*int(((a + b*x + c*x**2)**p*(b*d + 2*c*d*x)**m* x**2)/(a*b*m**2 + 4*a*b*m*p + a*b*m + 4*a*b*p**2 + 2*a*b*p + 2*a*c*m**2*x + 8*a*c*m*p*x + 2*a*c*m*x + 8*a*c*p**2*x + 4*a*c*p*x + b**2*m**2*x + 4*b** 2*m*p*x + b**2*m*x + 4*b**2*p**2*x + 2*b**2*p*x + 3*b*c*m**2*x**2 + 12*b*c *m*p*x**2 + 3*b*c*m*x**2 + 12*b*c*p**2*x**2 + 6*b*c*p*x**2 + 2*c**2*m**...