3.18.50 \(\int \frac {(a^2+2 a b x+b^2 x^2)^p}{(d+e x)^3} \, dx\) [1750]

Optimal. Leaf size=74 \[ \frac {b^2 (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p \, _2F_1\left (3,1+2 p;2 (1+p);-\frac {e (a+b x)}{b d-a e}\right )}{(b d-a e)^3 (1+2 p)} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 74, normalized size of antiderivative = 1.00, 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 = {660, 70} \begin {gather*} \frac {b^2 (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p \, _2F_1\left (3,2 p+1;2 (p+1);-\frac {e (a+b x)}{b d-a e}\right )}{(2 p+1) (b d-a e)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^p/(d + e*x)^3,x]

[Out]

(b^2*(a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^p*Hypergeometric2F1[3, 1 + 2*p, 2*(1 + p), -((e*(a + b*x))/(b*d - a*e
))])/((b*d - a*e)^3*(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*} \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^p}{(d+e x)^3} \, 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)^3} \, dx\\ &=\frac {b^2 (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p \, _2F_1\left (3,1+2 p;2 (1+p);-\frac {e (a+b x)}{b d-a e}\right )}{(b d-a e)^3 (1+2 p)}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 65, normalized size = 0.88 \begin {gather*} \frac {b^2 (a+b x) \left ((a+b x)^2\right )^p \, _2F_1\left (3,1+2 p;2+2 p;-\frac {e (a+b x)}{b d-a e}\right )}{(b d-a e)^3 (1+2 p)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^p/(d + e*x)^3,x]

[Out]

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

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Maple [F]
time = 0.46, size = 0, normalized size = 0.00 \[\int \frac {\left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p}}{\left (e x +d \right )^{3}}\, dx\]

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)^3,x)

[Out]

int((b^2*x^2+2*a*b*x+a^2)^p/(e*x+d)^3,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)^3,x, algorithm="maxima")

[Out]

integrate((b^2*x^2 + 2*a*b*x + a^2)^p/(x*e + d)^3, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)^3,x, algorithm="fricas")

[Out]

integral((b^2*x^2 + 2*a*b*x + a^2)^p/(x^3*e^3 + 3*d*x^2*e^2 + 3*d^2*x*e + d^3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\left (a + b x\right )^{2}\right )^{p}}{\left (d + e x\right )^{3}}\, dx \end {gather*}

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)**3,x)

[Out]

Integral(((a + b*x)**2)**p/(d + e*x)**3, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)^3,x, algorithm="giac")

[Out]

integrate((b^2*x^2 + 2*a*b*x + a^2)^p/(x*e + d)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^p}{{\left (d+e\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^2 + 2*a*b*x)^p/(d + e*x)^3,x)

[Out]

int((a^2 + b^2*x^2 + 2*a*b*x)^p/(d + e*x)^3, x)

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