Integrand size = 24, antiderivative size = 68 \[ \int \frac {(b d+2 c d x)^m}{a+b x+c x^2} \, dx=-\frac {2 (b d+2 c d x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{\left (b^2-4 a c\right ) d (1+m)} \] Output:
-2*(2*c*d*x+b*d)^(1+m)*hypergeom([1, 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)/d/(1+m)
Time = 0.09 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00 \[ \int \frac {(b d+2 c d x)^m}{a+b x+c x^2} \, dx=-\frac {2 (b+2 c x) (d (b+2 c x))^m \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{\left (b^2-4 a c\right ) (1+m)} \] Input:
Integrate[(b*d + 2*c*d*x)^m/(a + b*x + c*x^2),x]
Output:
(-2*(b + 2*c*x)*(d*(b + 2*c*x))^m*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/ 2, (b + 2*c*x)^2/(b^2 - 4*a*c)])/((b^2 - 4*a*c)*(1 + m))
Time = 0.20 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.09, 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.
\(\displaystyle \int \frac {(b d+2 c d x)^m}{a+b x+c x^2} \, dx\) |
\(\Big \downarrow \) 1118 |
\(\displaystyle \frac {\int \frac {4 c d^2 (b d+2 c x d)^m}{\left (4 a-\frac {b^2}{c}\right ) c d^2+(b d+2 c x d)^2}d(b d+2 c x d)}{2 c d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 d \int -\frac {(b d+2 c x d)^m}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d(b d+2 c x d)\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 d \int \frac {(b d+2 c x d)^m}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d(b d+2 c x d)\) |
\(\Big \downarrow \) 278 |
\(\displaystyle -\frac {2 (b d+2 c d x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}\right )}{d (m+1) \left (b^2-4 a c\right )}\) |
Input:
Int[(b*d + 2*c*d*x)^m/(a + b*x + c*x^2),x]
Output:
(-2*(b*d + 2*c*d*x)^(1 + m)*Hypergeometric2F1[1, (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)*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]
\[\int \frac {\left (2 c d x +b d \right )^{m}}{c \,x^{2}+b x +a}d x\]
Input:
int((2*c*d*x+b*d)^m/(c*x^2+b*x+a),x)
Output:
int((2*c*d*x+b*d)^m/(c*x^2+b*x+a),x)
\[ \int \frac {(b d+2 c d x)^m}{a+b x+c x^2} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{m}}{c x^{2} + b x + a} \,d x } \] Input:
integrate((2*c*d*x+b*d)^m/(c*x^2+b*x+a),x, algorithm="fricas")
Output:
integral((2*c*d*x + b*d)^m/(c*x^2 + b*x + a), x)
\[ \int \frac {(b d+2 c d x)^m}{a+b x+c x^2} \, dx=\int \frac {\left (d \left (b + 2 c x\right )\right )^{m}}{a + b x + c x^{2}}\, dx \] Input:
integrate((2*c*d*x+b*d)**m/(c*x**2+b*x+a),x)
Output:
Integral((d*(b + 2*c*x))**m/(a + b*x + c*x**2), x)
\[ \int \frac {(b d+2 c d x)^m}{a+b x+c x^2} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{m}}{c x^{2} + b x + a} \,d x } \] Input:
integrate((2*c*d*x+b*d)^m/(c*x^2+b*x+a),x, algorithm="maxima")
Output:
integrate((2*c*d*x + b*d)^m/(c*x^2 + b*x + a), x)
\[ \int \frac {(b d+2 c d x)^m}{a+b x+c x^2} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{m}}{c x^{2} + b x + a} \,d x } \] Input:
integrate((2*c*d*x+b*d)^m/(c*x^2+b*x+a),x, algorithm="giac")
Output:
integrate((2*c*d*x + b*d)^m/(c*x^2 + b*x + a), x)
Timed out. \[ \int \frac {(b d+2 c d x)^m}{a+b x+c x^2} \, dx=\int \frac {{\left (b\,d+2\,c\,d\,x\right )}^m}{c\,x^2+b\,x+a} \,d x \] Input:
int((b*d + 2*c*d*x)^m/(a + b*x + c*x^2),x)
Output:
int((b*d + 2*c*d*x)^m/(a + b*x + c*x^2), x)
\[ \int \frac {(b d+2 c d x)^m}{a+b x+c x^2} \, dx=\int \frac {\left (2 c d x +b d \right )^{m}}{c \,x^{2}+b x +a}d x \] Input:
int((2*c*d*x+b*d)^m/(c*x^2+b*x+a),x)
Output:
int((b*d + 2*c*d*x)**m/(a + b*x + c*x**2),x)