3.1755 \(\int \frac{(a^2+2 a b x+b^2 x^2)^p}{(d+e x)^{5/2}} \, dx\)

Optimal. Leaf size=83 \[ -\frac{2 \left (a^2+2 a b x+b^2 x^2\right )^p \left (-\frac{e (a+b x)}{b d-a e}\right )^{-2 p} \, _2F_1\left (-\frac{3}{2},-2 p;-\frac{1}{2};\frac{b (d+e x)}{b d-a e}\right )}{3 e (d+e x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0400015, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {646, 70, 69} \[ -\frac{2 \left (a^2+2 a b x+b^2 x^2\right )^p \left (-\frac{e (a+b x)}{b d-a e}\right )^{-2 p} \, _2F_1\left (-\frac{3}{2},-2 p;-\frac{1}{2};\frac{b (d+e x)}{b d-a e}\right )}{3 e (d+e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^p/(d + e*x)^(5/2),x]

[Out]

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

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 70

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

Rule 69

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

Rubi steps

\begin{align*} \int \frac{\left (a^2+2 a b x+b^2 x^2\right )^p}{(d+e x)^{5/2}} \, 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)^{5/2}} \, dx\\ &=\left (\left (\frac{e \left (a b+b^2 x\right )}{-b^2 d+a b e}\right )^{-2 p} \left (a^2+2 a b x+b^2 x^2\right )^p\right ) \int \frac{\left (-\frac{a e}{b d-a e}-\frac{b e x}{b d-a e}\right )^{2 p}}{(d+e x)^{5/2}} \, dx\\ &=-\frac{2 \left (-\frac{e (a+b x)}{b d-a e}\right )^{-2 p} \left (a^2+2 a b x+b^2 x^2\right )^p \, _2F_1\left (-\frac{3}{2},-2 p;-\frac{1}{2};\frac{b (d+e x)}{b d-a e}\right )}{3 e (d+e x)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0206526, size = 73, normalized size = 0.88 \[ -\frac{2 \left ((a+b x)^2\right )^p \left (\frac{e (a+b x)}{a e-b d}\right )^{-2 p} \, _2F_1\left (-\frac{3}{2},-2 p;-\frac{1}{2};\frac{b (d+e x)}{b d-a e}\right )}{3 e (d+e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^p/(d + e*x)^(5/2),x]

[Out]

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

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Maple [F]  time = 1.18, size = 0, normalized size = 0. \begin{align*} \int{ \left ({b}^{2}{x}^{2}+2\,abx+{a}^{2} \right ) ^{p} \left ( ex+d \right ) ^{-{\frac{5}{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)^(5/2),x)

[Out]

int((b^2*x^2+2*a*b*x+a^2)^p/(e*x+d)^(5/2),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 )}^{\frac{5}{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)^(5/2),x, algorithm="maxima")

[Out]

integrate((b^2*x^2 + 2*a*b*x + a^2)^p/(e*x + d)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{e x + d}{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p}}{e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}, 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)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(e*x + d)*(b^2*x^2 + 2*a*b*x + a^2)^p/(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3), 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 )^{\frac{5}{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)**(5/2),x)

[Out]

Integral(((a + b*x)**2)**p/(d + e*x)**(5/2), 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 )}^{\frac{5}{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)^(5/2),x, algorithm="giac")

[Out]

integrate((b^2*x^2 + 2*a*b*x + a^2)^p/(e*x + d)^(5/2), x)