\(\int \frac {(d+e x)^m}{(a^2+2 a b x+b^2 x^2)^2} \, dx\) [1735]

Optimal result

Integrand size = 26, antiderivative size = 53 \[ \int \frac {(d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {e^3 (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (4,1+m,2+m,\frac {b (d+e x)}{b d-a e}\right )}{(b d-a e)^4 (1+m)} \]

[Out]

e^3*(e*x+d)^(1+m)*hypergeom([4, 1+m],[2+m],b*(e*x+d)/(-a*e+b*d))/(-a*e+b*d)^4/(1+m)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {27, 70} \[ \int \frac {(d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {e^3 (d+e x)^{m+1} \operatorname {Hypergeometric2F1}\left (4,m+1,m+2,\frac {b (d+e x)}{b d-a e}\right )}{(m+1) (b d-a e)^4} \]

[In]

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

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \int \frac {(d+e x)^m}{(a+b x)^4} \, dx \\ & = \frac {e^3 (d+e x)^{1+m} \, _2F_1\left (4,1+m;2+m;\frac {b (d+e x)}{b d-a e}\right )}{(b d-a e)^4 (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.02 \[ \int \frac {(d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {e^3 (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (4,1+m,2+m,-\frac {b (d+e x)}{-b d+a e}\right )}{(-b d+a e)^4 (1+m)} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((e*x + d)^m/(b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3*b*x + a^4), x)

Sympy [F]

\[ \int \frac {(d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\int \frac {\left (d + e x\right )^{m}}{\left (a + b x\right )^{4}}\, dx \]

[In]

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

[Out]

Integral((d + e*x)**m/(a + b*x)**4, x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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