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)*(-e^2*x^2+d^2)^(3/2)/g/(4+m)+d^2*(5+2*m)*(g*x)^(1+m)*hypergeom([-1/2, 1/2+1/2*m],[3/2+1/2*m],e^2*
x^2/d^2)*(-e^2*x^2+d^2)^(1/2)/g/(1+m)/(4+m)/(1-e^2*x^2/d^2)^(1/2)-2*d*e*(g*x)^(2+m)*hypergeom([-1/2, 1+1/2*m],
[2+1/2*m],e^2*x^2/d^2)*(-e^2*x^2+d^2)^(1/2)/g^2/(2+m)/(1-e^2*x^2/d^2)^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 204, normalized size of antiderivative = 1.00, 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 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 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 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 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])

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*}

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Mathematica [A]  time = 0.11, 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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fricas [F]  time = 0.96, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (e^{2} x^{2} - 2 \, d e x + d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}} \left (g x\right )^{m}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} \left (g x\right )^{m}}{{\left (e x + d\right )}^{2}}\,{d x} \]

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)

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[ \int \frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} \left (g x \right )^{m}}{\left (e x +d \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x)^m*(-e^2*x^2+d^2)^(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} \left (g x\right )^{m}}{{\left (e x + d\right )}^{2}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (d^2-e^2\,x^2\right )}^{5/2}\,{\left (g\,x\right )}^m}{{\left (d+e\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 91.47, size = 185, normalized size = 0.91 \[ \frac {d^{3} g^{m} x x^{m} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + \frac {1}{2} \\ \frac {m}{2} + \frac {3}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} - \frac {d^{2} e g^{m} x^{2} x^{m} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + 1 \\ \frac {m}{2} + 2 \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{\Gamma \left (\frac {m}{2} + 2\right )} + \frac {d e^{2} g^{m} x^{3} x^{m} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + \frac {3}{2} \\ \frac {m}{2} + \frac {5}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} \]

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]

d**3*g**m*x*x**m*gamma(m/2 + 1/2)*hyper((-1/2, m/2 + 1/2), (m/2 + 3/2,), e**2*x**2*exp_polar(2*I*pi)/d**2)/(2*
gamma(m/2 + 3/2)) - d**2*e*g**m*x**2*x**m*gamma(m/2 + 1)*hyper((-1/2, m/2 + 1), (m/2 + 2,), e**2*x**2*exp_pola
r(2*I*pi)/d**2)/gamma(m/2 + 2) + d*e**2*g**m*x**3*x**m*gamma(m/2 + 3/2)*hyper((-1/2, m/2 + 3/2), (m/2 + 5/2,),
e**2*x**2*exp_polar(2*I*pi)/d**2)/(2*gamma(m/2 + 5/2))

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