Integrand size = 24, antiderivative size = 71 \[ \int \frac {(b d+2 c d x)^m}{\left (a+b x+c x^2\right )^3} \, dx=-\frac {32 c^2 (b d+2 c d x)^{1+m} \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{\left (b^2-4 a c\right )^3 d (1+m)} \] Output:
-32*c^2*(2*c*d*x+b*d)^(1+m)*hypergeom([3, 1/2+1/2*m],[3/2+1/2*m],(2*c*x+b) ^2/(-4*a*c+b^2))/(-4*a*c+b^2)^3/d/(1+m)
Time = 0.11 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00 \[ \int \frac {(b d+2 c d x)^m}{\left (a+b x+c x^2\right )^3} \, dx=-\frac {32 c^2 (b+2 c x) (d (b+2 c x))^m \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{\left (b^2-4 a c\right )^3 (1+m)} \] Input:
Integrate[(b*d + 2*c*d*x)^m/(a + b*x + c*x^2)^3,x]
Output:
(-32*c^2*(b + 2*c*x)*(d*(b + 2*c*x))^m*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, (b + 2*c*x)^2/(b^2 - 4*a*c)])/((b^2 - 4*a*c)^3*(1 + m))
Time = 0.20 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1118, 27, 25, 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.
\(\displaystyle \int \frac {(b d+2 c d x)^m}{\left (a+b x+c x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 1118 |
\(\displaystyle \frac {\int \frac {64 c^3 d^6 (b d+2 c x d)^m}{\left (\left (4 a-\frac {b^2}{c}\right ) c d^2+(b d+2 c x d)^2\right )^3}d(b d+2 c x d)}{2 c d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 32 c^2 d^5 \int -\frac {(b d+2 c x d)^m}{\left (\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2\right )^3}d(b d+2 c x d)\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -32 c^2 d^5 \int \frac {(b d+2 c x d)^m}{\left (\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2\right )^3}d(b d+2 c x d)\) |
\(\Big \downarrow \) 278 |
\(\displaystyle -\frac {32 c^2 (b d+2 c d x)^{m+1} \operatorname {Hypergeometric2F1}\left (3,\frac {m+1}{2},\frac {m+3}{2},\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}\right )}{d (m+1) \left (b^2-4 a c\right )^3}\) |
Input:
Int[(b*d + 2*c*d*x)^m/(a + b*x + c*x^2)^3,x]
Output:
(-32*c^2*(b*d + 2*c*d*x)^(1 + m)*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2 , (b*d + 2*c*d*x)^2/((b^2 - 4*a*c)*d^2)])/((b^2 - 4*a*c)^3*d*(1 + m))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[((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]
\[\int \frac {\left (2 c d x +b d \right )^{m}}{\left (c \,x^{2}+b x +a \right )^{3}}d x\]
Input:
int((2*c*d*x+b*d)^m/(c*x^2+b*x+a)^3,x)
Output:
int((2*c*d*x+b*d)^m/(c*x^2+b*x+a)^3,x)
\[ \int \frac {(b d+2 c d x)^m}{\left (a+b x+c x^2\right )^3} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{m}}{{\left (c x^{2} + b x + a\right )}^{3}} \,d x } \] Input:
integrate((2*c*d*x+b*d)^m/(c*x^2+b*x+a)^3,x, algorithm="fricas")
Output:
integral((2*c*d*x + b*d)^m/(c^3*x^6 + 3*b*c^2*x^5 + 3*(b^2*c + a*c^2)*x^4 + 3*a^2*b*x + (b^3 + 6*a*b*c)*x^3 + a^3 + 3*(a*b^2 + a^2*c)*x^2), x)
Timed out. \[ \int \frac {(b d+2 c d x)^m}{\left (a+b x+c x^2\right )^3} \, dx=\text {Timed out} \] Input:
integrate((2*c*d*x+b*d)**m/(c*x**2+b*x+a)**3,x)
Output:
Timed out
\[ \int \frac {(b d+2 c d x)^m}{\left (a+b x+c x^2\right )^3} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{m}}{{\left (c x^{2} + b x + a\right )}^{3}} \,d x } \] Input:
integrate((2*c*d*x+b*d)^m/(c*x^2+b*x+a)^3,x, algorithm="maxima")
Output:
integrate((2*c*d*x + b*d)^m/(c*x^2 + b*x + a)^3, x)
\[ \int \frac {(b d+2 c d x)^m}{\left (a+b x+c x^2\right )^3} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{m}}{{\left (c x^{2} + b x + a\right )}^{3}} \,d x } \] Input:
integrate((2*c*d*x+b*d)^m/(c*x^2+b*x+a)^3,x, algorithm="giac")
Output:
integrate((2*c*d*x + b*d)^m/(c*x^2 + b*x + a)^3, x)
Timed out. \[ \int \frac {(b d+2 c d x)^m}{\left (a+b x+c x^2\right )^3} \, dx=\int \frac {{\left (b\,d+2\,c\,d\,x\right )}^m}{{\left (c\,x^2+b\,x+a\right )}^3} \,d x \] Input:
int((b*d + 2*c*d*x)^m/(a + b*x + c*x^2)^3,x)
Output:
int((b*d + 2*c*d*x)^m/(a + b*x + c*x^2)^3, x)
\[ \int \frac {(b d+2 c d x)^m}{\left (a+b x+c x^2\right )^3} \, dx=\int \frac {\left (2 c d x +b d \right )^{m}}{c^{3} x^{6}+3 b \,c^{2} x^{5}+3 a \,c^{2} x^{4}+3 b^{2} c \,x^{4}+6 a b c \,x^{3}+b^{3} x^{3}+3 a^{2} c \,x^{2}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}}d x \] Input:
int((2*c*d*x+b*d)^m/(c*x^2+b*x+a)^3,x)
Output:
int((b*d + 2*c*d*x)**m/(a**3 + 3*a**2*b*x + 3*a**2*c*x**2 + 3*a*b**2*x**2 + 6*a*b*c*x**3 + 3*a*c**2*x**4 + b**3*x**3 + 3*b**2*c*x**4 + 3*b*c**2*x**5 + c**3*x**6),x)