3.3.99 \(\int \frac {x (d^2-e^2 x^2)^p}{(d+e x)^4} \, dx\) [299]

Optimal. Leaf size=118 \[ \frac {\left (d^2-e^2 x^2\right )^{1+p}}{2 e^2 (3-p) (d+e x)^4}-\frac {2^{-2+p} \left (1+\frac {e x}{d}\right )^{-1-p} \left (d^2-e^2 x^2\right )^{1+p} \, _2F_1\left (3-p,1+p;2+p;\frac {d-e x}{2 d}\right )}{d^4 e^2 (3-p) (1+p)} \]

[Out]

1/2*(-e^2*x^2+d^2)^(1+p)/e^2/(3-p)/(e*x+d)^4-2^(-2+p)*(1+e*x/d)^(-1-p)*(-e^2*x^2+d^2)^(1+p)*hypergeom([1+p, 3-
p],[2+p],1/2*(-e*x+d)/d)/d^4/e^2/(-p^2+2*p+3)

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Rubi [A]
time = 0.03, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {807, 692, 71} \begin {gather*} \frac {\left (d^2-e^2 x^2\right )^{p+1}}{2 e^2 (3-p) (d+e x)^4}-\frac {2^{p-2} \left (\frac {e x}{d}+1\right )^{-p-1} \left (d^2-e^2 x^2\right )^{p+1} \, _2F_1\left (3-p,p+1;p+2;\frac {d-e x}{2 d}\right )}{d^4 e^2 (3-p) (p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(d^2 - e^2*x^2)^p)/(d + e*x)^4,x]

[Out]

(d^2 - e^2*x^2)^(1 + p)/(2*e^2*(3 - p)*(d + e*x)^4) - (2^(-2 + p)*(1 + (e*x)/d)^(-1 - p)*(d^2 - e^2*x^2)^(1 +
p)*Hypergeometric2F1[3 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/(d^4*e^2*(3 - p)*(1 + p))

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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]))

Rule 692

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^(m - 1)*((a + c*x^2)^(p + 1)/((1
+ e*(x/d))^(p + 1)*(a/d + (c*x)/e)^(p + 1))), Int[(1 + e*(x/d))^(m + p)*(a/d + (c/e)*x)^p, x], x] /; FreeQ[{a,
c, d, e, m}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && (IntegerQ[m] || GtQ[d, 0]) &&  !(IGtQ[m, 0] && (
IntegerQ[3*p] || IntegerQ[4*p]))

Rule 807

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d
+ e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d*(m + p + 1))), x] + Dist[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d
)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2
+ a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) &&
NeQ[m + p + 1, 0]

Rubi steps

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

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Mathematica [A]
time = 0.29, size = 102, normalized size = 0.86 \begin {gather*} \frac {2^{-4+p} (d-e x) \left (1+\frac {e x}{d}\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (-2 \, _2F_1\left (3-p,1+p;2+p;\frac {d-e x}{2 d}\right )+\, _2F_1\left (4-p,1+p;2+p;\frac {d-e x}{2 d}\right )\right )}{d^3 e^2 (1+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x*(d^2 - e^2*x^2)^p)/(d + e*x)^4,x]

[Out]

(2^(-4 + p)*(d - e*x)*(d^2 - e^2*x^2)^p*(-2*Hypergeometric2F1[3 - p, 1 + p, 2 + p, (d - e*x)/(2*d)] + Hypergeo
metric2F1[4 - p, 1 + p, 2 + p, (d - e*x)/(2*d)]))/(d^3*e^2*(1 + p)*(1 + (e*x)/d)^p)

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {x \left (-e^{2} x^{2}+d^{2}\right )^{p}}{\left (e x +d \right )^{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(-e^2*x^2+d^2)^p/(e*x+d)^4,x)

[Out]

int(x*(-e^2*x^2+d^2)^p/(e*x+d)^4,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(x*(-e^2*x^2+d^2)^p/(e*x+d)^4,x, algorithm="maxima")

[Out]

integrate((-x^2*e^2 + d^2)^p*x/(x*e + d)^4, 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(x*(-e^2*x^2+d^2)^p/(e*x+d)^4,x, algorithm="fricas")

[Out]

integral((-x^2*e^2 + d^2)^p*x/(x^4*e^4 + 4*d*x^3*e^3 + 6*d^2*x^2*e^2 + 4*d^3*x*e + d^4), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{\left (d + e x\right )^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-e**2*x**2+d**2)**p/(e*x+d)**4,x)

[Out]

Integral(x*(-(-d + e*x)*(d + e*x))**p/(d + e*x)**4, 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(x*(-e^2*x^2+d^2)^p/(e*x+d)^4,x, algorithm="giac")

[Out]

integrate((-x^2*e^2 + d^2)^p*x/(x*e + d)^4, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x\,{\left (d^2-e^2\,x^2\right )}^p}{{\left (d+e\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(d^2 - e^2*x^2)^p)/(d + e*x)^4,x)

[Out]

int((x*(d^2 - e^2*x^2)^p)/(d + e*x)^4, x)

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