3.1749 \(\int \frac{(a^2+2 a b x+b^2 x^2)^p}{(d+e x)^2} \, dx\)

Optimal. Leaf size=72 \[ \frac{b (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p \, _2F_1\left (2,2 p+1;2 (p+1);-\frac{e (a+b x)}{b d-a e}\right )}{(2 p+1) (b d-a e)^2} \]

[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))

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Rubi [A]  time = 0.0304151, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {646, 68} \[ \frac{b (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p \, _2F_1\left (2,2 p+1;2 (p+1);-\frac{e (a+b x)}{b d-a e}\right )}{(2 p+1) (b d-a e)^2} \]

Antiderivative was successfully verified.

[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 646

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]

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{\left (a^2+2 a b x+b^2 x^2\right )^p}{(d+e x)^2} \, dx &=\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]  time = 0.0160252, size = 63, normalized size = 0.88 \[ \frac{b (a+b x) \left ((a+b x)^2\right )^p \, _2F_1\left (2,2 p+1;2 p+2;-\frac{e (a+b x)}{b d-a e}\right )}{(2 p+1) (b d-a e)^2} \]

Antiderivative was successfully verified.

[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))

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Maple [F]  time = 1.188, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ({b}^{2}{x}^{2}+2\,abx+{a}^{2} \right ) ^{p}}{ \left ( ex+d \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p}}{{\left (e x + d\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p}}{e^{2} x^{2} + 2 \, d e x + d^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\left (a + b x\right )^{2}\right )^{p}}{\left (d + e x\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p}}{{\left (e x + d\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)