\(\int \frac {(a^2+2 a b x+b^2 x^2)^p}{(d+e x)^2} \, dx\) [1749]

Optimal result

Integrand size = 26, antiderivative size = 72 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^p}{(d+e x)^2} \, dx=\frac {b (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (2,1+2 p,2 (1+p),-\frac {e (a+b x)}{b d-a e}\right )}{(b d-a e)^2 (1+2 p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 72, 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 = {660, 70} \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^p}{(d+e x)^2} \, dx=\frac {b (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (2,2 p+1,2 (p+1),-\frac {e (a+b x)}{b d-a e}\right )}{(2 p+1) (b d-a e)^2} \]

[In]

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

[Out]

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

Rule 660

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^p}{(d+e x)^2} \, dx=\frac {b (a+b x) \left ((a+b x)^2\right )^p \operatorname {Hypergeometric2F1}\left (2,1+2 p,2+2 p,-\frac {e (a+b x)}{b d-a e}\right )}{(b d-a e)^2 (1+2 p)} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((b^2*x^2 + 2*a*b*x + a^2)^p/(e^2*x^2 + 2*d*e*x + d^2), x)

Sympy [F]

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

[In]

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

[Out]

Integral(((a + b*x)**2)**p/(d + e*x)**2, x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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