3.398 \(\int \frac{(d+e x)^2 (a+b x^2)^p}{x^3} \, dx\)

Optimal. Leaf size=127 \[ -\frac{\left (a+b x^2\right )^{p+1} \left (a e^2+b d^2 p\right ) \, _2F_1\left (1,p+1;p+2;\frac{b x^2}{a}+1\right )}{2 a^2 (p+1)}-\frac{d^2 \left (a+b x^2\right )^{p+1}}{2 a x^2}-\frac{2 d e \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b x^2}{a}\right )}{x} \]

[Out]

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

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Rubi [A]  time = 0.12457, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {1807, 764, 365, 364, 266, 65} \[ -\frac{\left (a+b x^2\right )^{p+1} \left (a e^2+b d^2 p\right ) \, _2F_1\left (1,p+1;p+2;\frac{b x^2}{a}+1\right )}{2 a^2 (p+1)}-\frac{d^2 \left (a+b x^2\right )^{p+1}}{2 a x^2}-\frac{2 d e \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b x^2}{a}\right )}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rule 764

Int[(x_)^(m_.)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[f, Int[x^m*(a + c*x^2)^p, x]
, x] + Dist[g, Int[x^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, f, g, p}, x] && IntegerQ[m] &&  !IntegerQ[2
*p]

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 65

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

Rubi steps

\begin{align*} \int \frac{(d+e x)^2 \left (a+b x^2\right )^p}{x^3} \, dx &=-\frac{d^2 \left (a+b x^2\right )^{1+p}}{2 a x^2}-\frac{\int \frac{\left (-4 a d e-2 \left (a e^2+b d^2 p\right ) x\right ) \left (a+b x^2\right )^p}{x^2} \, dx}{2 a}\\ &=-\frac{d^2 \left (a+b x^2\right )^{1+p}}{2 a x^2}+(2 d e) \int \frac{\left (a+b x^2\right )^p}{x^2} \, dx+\frac{\left (a e^2+b d^2 p\right ) \int \frac{\left (a+b x^2\right )^p}{x} \, dx}{a}\\ &=-\frac{d^2 \left (a+b x^2\right )^{1+p}}{2 a x^2}+\frac{\left (a e^2+b d^2 p\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^p}{x} \, dx,x,x^2\right )}{2 a}+\left (2 d e \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p}\right ) \int \frac{\left (1+\frac{b x^2}{a}\right )^p}{x^2} \, dx\\ &=-\frac{d^2 \left (a+b x^2\right )^{1+p}}{2 a x^2}-\frac{2 d e \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b x^2}{a}\right )}{x}-\frac{\left (a e^2+b d^2 p\right ) \left (a+b x^2\right )^{1+p} \, _2F_1\left (1,1+p;2+p;1+\frac{b x^2}{a}\right )}{2 a^2 (1+p)}\\ \end{align*}

Mathematica [A]  time = 0.0956011, size = 119, normalized size = 0.94 \[ \frac{1}{2} \left (a+b x^2\right )^p \left (-\frac{\left (a+b x^2\right ) \left (a e^2 \, _2F_1\left (1,p+1;p+2;\frac{b x^2}{a}+1\right )-b d^2 \, _2F_1\left (2,p+1;p+2;\frac{b x^2}{a}+1\right )\right )}{a^2 (p+1)}-\frac{4 d e \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b x^2}{a}\right )}{x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.509, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex+d \right ) ^{2} \left ( b{x}^{2}+a \right ) ^{p}}{{x}^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 32.8256, size = 119, normalized size = 0.94 \begin{align*} - \frac{2 a^{p} d e{{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, - p \\ \frac{1}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{x} - \frac{b^{p} d^{2} x^{2 p} \Gamma \left (1 - p\right ){{}_{2}F_{1}\left (\begin{matrix} - p, 1 - p \\ 2 - p \end{matrix}\middle |{\frac{a e^{i \pi }}{b x^{2}}} \right )}}{2 x^{2} \Gamma \left (2 - p\right )} - \frac{b^{p} e^{2} x^{2 p} \Gamma \left (- p\right ){{}_{2}F_{1}\left (\begin{matrix} - p, - p \\ 1 - p \end{matrix}\middle |{\frac{a e^{i \pi }}{b x^{2}}} \right )}}{2 \Gamma \left (1 - p\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a**p*d*e*hyper((-1/2, -p), (1/2,), b*x**2*exp_polar(I*pi)/a)/x - b**p*d**2*x**(2*p)*gamma(1 - p)*hyper((-p,
1 - p), (2 - p,), a*exp_polar(I*pi)/(b*x**2))/(2*x**2*gamma(2 - p)) - b**p*e**2*x**(2*p)*gamma(-p)*hyper((-p,
-p), (1 - p,), a*exp_polar(I*pi)/(b*x**2))/(2*gamma(1 - p))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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