3.264 \(\int \frac{(d+e x)^3 (d^2-e^2 x^2)^p}{x} \, dx\)

Optimal. Leaf size=171 \[ \frac{2 d^2 e (3 p+5) x \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )}{2 p+3}-\frac{d \left (d^2-e^2 x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;1-\frac{e^2 x^2}{d^2}\right )}{2 (p+1)}-\frac{e x \left (d^2-e^2 x^2\right )^{p+1}}{2 p+3}-\frac{3 d \left (d^2-e^2 x^2\right )^{p+1}}{2 (p+1)} \]

[Out]

(-3*d*(d^2 - e^2*x^2)^(1 + p))/(2*(1 + p)) - (e*x*(d^2 - e^2*x^2)^(1 + p))/(3 + 2*p) + (2*d^2*e*(5 + 3*p)*x*(d
^2 - e^2*x^2)^p*Hypergeometric2F1[1/2, -p, 3/2, (e^2*x^2)/d^2])/((3 + 2*p)*(1 - (e^2*x^2)/d^2)^p) - (d*(d^2 -
e^2*x^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 - (e^2*x^2)/d^2])/(2*(1 + p))

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Rubi [A]  time = 0.128589, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {1652, 446, 80, 65, 388, 246, 245} \[ \frac{2 d^2 e (3 p+5) x \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )}{2 p+3}-\frac{d \left (d^2-e^2 x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;1-\frac{e^2 x^2}{d^2}\right )}{2 (p+1)}-\frac{e x \left (d^2-e^2 x^2\right )^{p+1}}{2 p+3}-\frac{3 d \left (d^2-e^2 x^2\right )^{p+1}}{2 (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*d*(d^2 - e^2*x^2)^(1 + p))/(2*(1 + p)) - (e*x*(d^2 - e^2*x^2)^(1 + p))/(3 + 2*p) + (2*d^2*e*(5 + 3*p)*x*(d
^2 - e^2*x^2)^p*Hypergeometric2F1[1/2, -p, 3/2, (e^2*x^2)/d^2])/((3 + 2*p)*(1 - (e^2*x^2)/d^2)^p) - (d*(d^2 -
e^2*x^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 - (e^2*x^2)/d^2])/(2*(1 + p))

Rule 1652

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], k}, Int[x^m*Sum[Coeff[
Pq, x, 2*k]*x^(2*k), {k, 0, q/2}]*(a + b*x^2)^p, x] + Int[x^(m + 1)*Sum[Coeff[Pq, x, 2*k + 1]*x^(2*k), {k, 0,
(q - 1)/2}]*(a + b*x^2)^p, x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] &&  !PolyQ[Pq, x^2] && IGtQ[m, -2] &&  !
IntegerQ[2*p]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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])

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 245

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

Rubi steps

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

Mathematica [A]  time = 0.152175, size = 169, normalized size = 0.99 \[ \frac{1}{6} \left (d^2-e^2 x^2\right )^p \left (18 d^2 e x \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )+2 e^3 x^3 \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};\frac{e^2 x^2}{d^2}\right )-\frac{3 d \left (d^2-e^2 x^2\right ) \, _2F_1\left (1,p+1;p+2;1-\frac{e^2 x^2}{d^2}\right )}{p+1}-\frac{9 d \left (d^2-e^2 x^2\right )}{p+1}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d^2 - e^2*x^2)^p*((-9*d*(d^2 - e^2*x^2))/(1 + p) + (18*d^2*e*x*Hypergeometric2F1[1/2, -p, 3/2, (e^2*x^2)/d^2
])/(1 - (e^2*x^2)/d^2)^p - (3*d*(d^2 - e^2*x^2)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 - (e^2*x^2)/d^2])/(1 + p)
+ (2*e^3*x^3*Hypergeometric2F1[3/2, -p, 5/2, (e^2*x^2)/d^2])/(1 - (e^2*x^2)/d^2)^p))/6

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Maple [F]  time = 0.59, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex+d \right ) ^{3} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{p}}{x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 11.8515, size = 178, normalized size = 1.04 \begin{align*} - \frac{d^{3} e^{2 p} x^{2 p} e^{i \pi p} \Gamma \left (- p\right ){{}_{2}F_{1}\left (\begin{matrix} - p, - p \\ 1 - p \end{matrix}\middle |{\frac{d^{2}}{e^{2} x^{2}}} \right )}}{2 \Gamma \left (1 - p\right )} + 3 d^{2} d^{2 p} e x{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, - p \\ \frac{3}{2} \end{matrix}\middle |{\frac{e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )} + 3 d e^{2} \left (\begin{cases} \frac{x^{2} \left (d^{2}\right )^{p}}{2} & \text{for}\: e^{2} = 0 \\- \frac{\begin{cases} \frac{\left (d^{2} - e^{2} x^{2}\right )^{p + 1}}{p + 1} & \text{for}\: p \neq -1 \\\log{\left (d^{2} - e^{2} x^{2} \right )} & \text{otherwise} \end{cases}}{2 e^{2}} & \text{otherwise} \end{cases}\right ) + \frac{d^{2 p} e^{3} x^{3}{{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, - p \\ \frac{5}{2} \end{matrix}\middle |{\frac{e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-d**3*e**(2*p)*x**(2*p)*exp(I*pi*p)*gamma(-p)*hyper((-p, -p), (1 - p,), d**2/(e**2*x**2))/(2*gamma(1 - p)) + 3
*d**2*d**(2*p)*e*x*hyper((1/2, -p), (3/2,), e**2*x**2*exp_polar(2*I*pi)/d**2) + 3*d*e**2*Piecewise((x**2*(d**2
)**p/2, Eq(e**2, 0)), (-Piecewise(((d**2 - e**2*x**2)**(p + 1)/(p + 1), Ne(p, -1)), (log(d**2 - e**2*x**2), Tr
ue))/(2*e**2), True)) + d**(2*p)*e**3*x**3*hyper((3/2, -p), (5/2,), e**2*x**2*exp_polar(2*I*pi)/d**2)/3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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