3.231 \(\int \frac{(g x)^m (d^2-e^2 x^2)^{5/2}}{(d+e x)^2} \, dx\)

Optimal. Leaf size=204 \[ -\frac{2 d e \sqrt{d^2-e^2 x^2} (g x)^{m+2} \, _2F_1\left (-\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\frac{e^2 x^2}{d^2}\right )}{g^2 (m+2) \sqrt{1-\frac{e^2 x^2}{d^2}}}+\frac{d^2 (2 m+5) \sqrt{d^2-e^2 x^2} (g x)^{m+1} \, _2F_1\left (-\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\frac{e^2 x^2}{d^2}\right )}{g (m+1) (m+4) \sqrt{1-\frac{e^2 x^2}{d^2}}}-\frac{\left (d^2-e^2 x^2\right )^{3/2} (g x)^{m+1}}{g (m+4)} \]

[Out]

-(((g*x)^(1 + m)*(d^2 - e^2*x^2)^(3/2))/(g*(4 + m))) + (d^2*(5 + 2*m)*(g*x)^(1 + m)*Sqrt[d^2 - e^2*x^2]*Hyperg
eometric2F1[-1/2, (1 + m)/2, (3 + m)/2, (e^2*x^2)/d^2])/(g*(1 + m)*(4 + m)*Sqrt[1 - (e^2*x^2)/d^2]) - (2*d*e*(
g*x)^(2 + m)*Sqrt[d^2 - e^2*x^2]*Hypergeometric2F1[-1/2, (2 + m)/2, (4 + m)/2, (e^2*x^2)/d^2])/(g^2*(2 + m)*Sq
rt[1 - (e^2*x^2)/d^2])

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Rubi [A]  time = 0.215488, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {852, 1809, 808, 365, 364} \[ -\frac{2 d e \sqrt{d^2-e^2 x^2} (g x)^{m+2} \, _2F_1\left (-\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\frac{e^2 x^2}{d^2}\right )}{g^2 (m+2) \sqrt{1-\frac{e^2 x^2}{d^2}}}+\frac{d^2 (2 m+5) \sqrt{d^2-e^2 x^2} (g x)^{m+1} \, _2F_1\left (-\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\frac{e^2 x^2}{d^2}\right )}{g (m+1) (m+4) \sqrt{1-\frac{e^2 x^2}{d^2}}}-\frac{\left (d^2-e^2 x^2\right )^{3/2} (g x)^{m+1}}{g (m+4)} \]

Antiderivative was successfully verified.

[In]

Int[((g*x)^m*(d^2 - e^2*x^2)^(5/2))/(d + e*x)^2,x]

[Out]

-(((g*x)^(1 + m)*(d^2 - e^2*x^2)^(3/2))/(g*(4 + m))) + (d^2*(5 + 2*m)*(g*x)^(1 + m)*Sqrt[d^2 - e^2*x^2]*Hyperg
eometric2F1[-1/2, (1 + m)/2, (3 + m)/2, (e^2*x^2)/d^2])/(g*(1 + m)*(4 + m)*Sqrt[1 - (e^2*x^2)/d^2]) - (2*d*e*(
g*x)^(2 + m)*Sqrt[d^2 - e^2*x^2]*Hypergeometric2F1[-1/2, (2 + m)/2, (4 + m)/2, (e^2*x^2)/d^2])/(g^2*(2 + m)*Sq
rt[1 - (e^2*x^2)/d^2])

Rule 852

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 1809

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])

Rule 808

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 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])

Rubi steps

\begin{align*} \int \frac{(g x)^m \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx &=\int (g x)^m (d-e x)^2 \sqrt{d^2-e^2 x^2} \, dx\\ &=-\frac{(g x)^{1+m} \left (d^2-e^2 x^2\right )^{3/2}}{g (4+m)}-\frac{\int (g x)^m \left (-d^2 e^2 (5+2 m)+2 d e^3 (4+m) x\right ) \sqrt{d^2-e^2 x^2} \, dx}{e^2 (4+m)}\\ &=-\frac{(g x)^{1+m} \left (d^2-e^2 x^2\right )^{3/2}}{g (4+m)}-\frac{(2 d e) \int (g x)^{1+m} \sqrt{d^2-e^2 x^2} \, dx}{g}+\frac{\left (d^2 (5+2 m)\right ) \int (g x)^m \sqrt{d^2-e^2 x^2} \, dx}{4+m}\\ &=-\frac{(g x)^{1+m} \left (d^2-e^2 x^2\right )^{3/2}}{g (4+m)}-\frac{\left (2 d e \sqrt{d^2-e^2 x^2}\right ) \int (g x)^{1+m} \sqrt{1-\frac{e^2 x^2}{d^2}} \, dx}{g \sqrt{1-\frac{e^2 x^2}{d^2}}}+\frac{\left (d^2 (5+2 m) \sqrt{d^2-e^2 x^2}\right ) \int (g x)^m \sqrt{1-\frac{e^2 x^2}{d^2}} \, dx}{(4+m) \sqrt{1-\frac{e^2 x^2}{d^2}}}\\ &=-\frac{(g x)^{1+m} \left (d^2-e^2 x^2\right )^{3/2}}{g (4+m)}+\frac{d^2 (5+2 m) (g x)^{1+m} \sqrt{d^2-e^2 x^2} \, _2F_1\left (-\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};\frac{e^2 x^2}{d^2}\right )}{g (1+m) (4+m) \sqrt{1-\frac{e^2 x^2}{d^2}}}-\frac{2 d e (g x)^{2+m} \sqrt{d^2-e^2 x^2} \, _2F_1\left (-\frac{1}{2},\frac{2+m}{2};\frac{4+m}{2};\frac{e^2 x^2}{d^2}\right )}{g^2 (2+m) \sqrt{1-\frac{e^2 x^2}{d^2}}}\\ \end{align*}

Mathematica [A]  time = 0.121718, size = 173, normalized size = 0.85 \[ \frac{x \sqrt{d^2-e^2 x^2} (g x)^m \left (d^2 \left (m^2+5 m+6\right ) \, _2F_1\left (-\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\frac{e^2 x^2}{d^2}\right )-e (m+1) x \left (2 d (m+3) \, _2F_1\left (-\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\frac{e^2 x^2}{d^2}\right )-e (m+2) x \, _2F_1\left (-\frac{1}{2},\frac{m+3}{2};\frac{m+5}{2};\frac{e^2 x^2}{d^2}\right )\right )\right )}{(m+1) (m+2) (m+3) \sqrt{1-\frac{e^2 x^2}{d^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[((g*x)^m*(d^2 - e^2*x^2)^(5/2))/(d + e*x)^2,x]

[Out]

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

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Maple [F]  time = 0.606, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( gx \right ) ^{m}}{ \left ( ex+d \right ) ^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x)^m*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^2,x)

[Out]

int((g*x)^m*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} \left (g x\right )^{m}}{{\left (e x + d\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x)^m*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^2,x, algorithm="maxima")

[Out]

integrate((-e^2*x^2 + d^2)^(5/2)*(g*x)^m/(e*x + d)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x)^m*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^2,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x)**m*(-e**2*x**2+d**2)**(5/2)/(e*x+d)**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} \left (g x\right )^{m}}{{\left (e x + d\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x)^m*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^2,x, algorithm="giac")

[Out]

integrate((-e^2*x^2 + d^2)^(5/2)*(g*x)^m/(e*x + d)^2, x)