3.4.11 \(\int \frac {(g x)^m (d^2-e^2 x^2)^p}{(d+e x)^2} \, dx\) [311]

Optimal. Leaf size=214 \[ \frac {(g x)^{1+m} \left (d^2-e^2 x^2\right )^{-1+p}}{g (1-m-2 p)}-\frac {2 (m+p) (g x)^{1+m} \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {1+m}{2},2-p;\frac {3+m}{2};\frac {e^2 x^2}{d^2}\right )}{d^2 g (1+m) (1-m-2 p)}-\frac {2 e (g x)^{2+m} \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {2+m}{2},2-p;\frac {4+m}{2};\frac {e^2 x^2}{d^2}\right )}{d^3 g^2 (2+m)} \]

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {866, 1823, 822, 372, 371} \begin {gather*} -\frac {2 (m+p) (g x)^{m+1} \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {m+1}{2},2-p;\frac {m+3}{2};\frac {e^2 x^2}{d^2}\right )}{d^2 g (m+1) (-m-2 p+1)}+\frac {(g x)^{m+1} \left (d^2-e^2 x^2\right )^{p-1}}{g (-m-2 p+1)}-\frac {2 e (g x)^{m+2} \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {m+2}{2},2-p;\frac {m+4}{2};\frac {e^2 x^2}{d^2}\right )}{d^3 g^2 (m+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((g*x)^(1 + m)*(d^2 - e^2*x^2)^(-1 + p))/(g*(1 - m - 2*p)) - (2*(m + p)*(g*x)^(1 + m)*(d^2 - e^2*x^2)^p*Hyperg
eometric2F1[(1 + m)/2, 2 - p, (3 + m)/2, (e^2*x^2)/d^2])/(d^2*g*(1 + m)*(1 - m - 2*p)*(1 - (e^2*x^2)/d^2)^p) -
(2*e*(g*x)^(2 + m)*(d^2 - e^2*x^2)^p*Hypergeometric2F1[(2 + m)/2, 2 - p, (4 + m)/2, (e^2*x^2)/d^2])/(d^3*g^2*
(2 + m)*(1 - (e^2*x^2)/d^2)^p)

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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 822

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

Rule 866

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

Rule 1823

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[f*(c*x)^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*c^(q - 1)*(m + q + 2*p + 1))), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rubi steps

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

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Mathematica [A]
time = 0.16, size = 180, normalized size = 0.84 \begin {gather*} \frac {x (g x)^m \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2 \left (6+5 m+m^2\right ) \, _2F_1\left (\frac {1+m}{2},2-p;\frac {3+m}{2};\frac {e^2 x^2}{d^2}\right )-e (1+m) x \left (2 d (3+m) \, _2F_1\left (\frac {2+m}{2},2-p;\frac {4+m}{2};\frac {e^2 x^2}{d^2}\right )-e (2+m) x \, _2F_1\left (\frac {3+m}{2},2-p;\frac {5+m}{2};\frac {e^2 x^2}{d^2}\right )\right )\right )}{d^4 (1+m) (2+m) (3+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(g*x)^m*(d^2 - e^2*x^2)^p*(d^2*(6 + 5*m + m^2)*Hypergeometric2F1[(1 + m)/2, 2 - p, (3 + m)/2, (e^2*x^2)/d^2
] - e*(1 + m)*x*(2*d*(3 + m)*Hypergeometric2F1[(2 + m)/2, 2 - p, (4 + m)/2, (e^2*x^2)/d^2] - e*(2 + m)*x*Hyper
geometric2F1[(3 + m)/2, 2 - p, (5 + m)/2, (e^2*x^2)/d^2])))/(d^4*(1 + m)*(2 + m)*(3 + m)*(1 - (e^2*x^2)/d^2)^p
)

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (g x \right )^{m} \left (-e^{2} x^{2}+d^{2}\right )^{p}}{\left (e x +d \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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