\(\int \frac {(b d+2 c d x)^m}{a+b x+c x^2} \, dx\) [1426]

Optimal result

Integrand size = 24, antiderivative size = 67 \[ \int \frac {(b d+2 c d x)^m}{a+b x+c x^2} \, dx=-\frac {2 (d (b+2 c x))^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{\left (b^2-4 a c\right ) d (1+m)} \]

[Out]

-2*(d*(2*c*x+b))^(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)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 67, 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} \[ \int \frac {(b d+2 c d x)^m}{a+b x+c x^2} \, dx=-\frac {2 (d (b+2 c x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{d (m+1) \left (b^2-4 a c\right )} \]

[In]

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

[Out]

(-2*(d*(b + 2*c*x))^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, (b + 2*c*x)^2/(b^2 - 4*a*c)])/((b^2 - 4
*a*c)*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}{a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}} \, dx,x,b d+2 c d x\right )}{2 c d} \\ & = -\frac {2 (d (b+2 c x))^{1+m} \, _2F_1\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)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.01 \[ \int \frac {(b d+2 c d x)^m}{a+b x+c x^2} \, dx=-\frac {2 (b+2 c x) (d (b+2 c x))^m \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{\left (b^2-4 a c\right ) (1+m)} \]

[In]

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

[Out]

(-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))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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