\(\int \frac {(b d+2 c d x)^m}{(a+b x+c x^2)^2} \, dx\) [1427]

Optimal result

Integrand size = 24, antiderivative size = 68 \[ \int \frac {(b d+2 c d x)^m}{\left (a+b x+c x^2\right )^2} \, dx=\frac {8 c (d (b+2 c x))^{1+m} \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{\left (b^2-4 a c\right )^2 d (1+m)} \]

[Out]

8*c*(d*(2*c*x+b))^(1+m)*hypergeom([2, 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)^2/d/(1+m)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {708, 371} \[ \int \frac {(b d+2 c d x)^m}{\left (a+b x+c x^2\right )^2} \, dx=\frac {8 c (d (b+2 c x))^{m+1} \operatorname {Hypergeometric2F1}\left (2,\frac {m+1}{2},\frac {m+3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{d (m+1) \left (b^2-4 a c\right )^2} \]

[In]

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

[Out]

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

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 708

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[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] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^m}{\left (a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}\right )^2} \, dx,x,b d+2 c d x\right )}{2 c d} \\ & = \frac {8 c (d (b+2 c x))^{1+m} \, _2F_1\left (2,\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 )^2 d (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.01 \[ \int \frac {(b d+2 c d x)^m}{\left (a+b x+c x^2\right )^2} \, dx=\frac {8 c (b+2 c x) (d (b+2 c x))^m \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{\left (b^2-4 a c\right )^2 (1+m)} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

int((2*c*d*x+b*d)^m/(c*x^2+b*x+a)^2,x)

[Out]

int((2*c*d*x+b*d)^m/(c*x^2+b*x+a)^2,x)

Fricas [F]

\[ \int \frac {(b d+2 c d x)^m}{\left (a+b x+c x^2\right )^2} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{m}}{{\left (c x^{2} + b x + a\right )}^{2}} \,d x } \]

[In]

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

[Out]

integral((2*c*d*x + b*d)^m/(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2), x)

Sympy [F]

\[ \int \frac {(b d+2 c d x)^m}{\left (a+b x+c x^2\right )^2} \, dx=\int \frac {\left (d \left (b + 2 c x\right )\right )^{m}}{\left (a + b x + c x^{2}\right )^{2}}\, dx \]

[In]

integrate((2*c*d*x+b*d)**m/(c*x**2+b*x+a)**2,x)

[Out]

Integral((d*(b + 2*c*x))**m/(a + b*x + c*x**2)**2, x)

Maxima [F]

\[ \int \frac {(b d+2 c d x)^m}{\left (a+b x+c x^2\right )^2} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{m}}{{\left (c x^{2} + b x + a\right )}^{2}} \,d x } \]

[In]

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

[Out]

integrate((2*c*d*x + b*d)^m/(c*x^2 + b*x + a)^2, x)

Giac [F]

\[ \int \frac {(b d+2 c d x)^m}{\left (a+b x+c x^2\right )^2} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{m}}{{\left (c x^{2} + b x + a\right )}^{2}} \,d x } \]

[In]

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

[Out]

integrate((2*c*d*x + b*d)^m/(c*x^2 + b*x + a)^2, x)

Mupad [F(-1)]

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

[In]

int((b*d + 2*c*d*x)^m/(a + b*x + c*x^2)^2,x)

[Out]

int((b*d + 2*c*d*x)^m/(a + b*x + c*x^2)^2, x)