3.3.27 \(\int (g x)^m (d+e x)^2 (d^2-e^2 x^2)^{5/2} \, dx\) [227]

Optimal. Leaf size=206 \[ -\frac {(g x)^{1+m} \left (d^2-e^2 x^2\right )^{7/2}}{g (8+m)}+\frac {d^6 (9+2 m) (g x)^{1+m} \sqrt {d^2-e^2 x^2} \, _2F_1\left (-\frac {5}{2},\frac {1+m}{2};\frac {3+m}{2};\frac {e^2 x^2}{d^2}\right )}{g (1+m) (8+m) \sqrt {1-\frac {e^2 x^2}{d^2}}}+\frac {2 d^5 e (g x)^{2+m} \sqrt {d^2-e^2 x^2} \, _2F_1\left (-\frac {5}{2},\frac {2+m}{2};\frac {4+m}{2};\frac {e^2 x^2}{d^2}\right )}{g^2 (2+m) \sqrt {1-\frac {e^2 x^2}{d^2}}} \]

[Out]

-(g*x)^(1+m)*(-e^2*x^2+d^2)^(7/2)/g/(8+m)+d^6*(9+2*m)*(g*x)^(1+m)*hypergeom([-5/2, 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)/(8+m)/(1-e^2*x^2/d^2)^(1/2)+2*d^5*e*(g*x)^(2+m)*hypergeom([-5/2, 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 = 0.14, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1823, 822, 372, 371} \begin {gather*} -\frac {\left (d^2-e^2 x^2\right )^{7/2} (g x)^{m+1}}{g (m+8)}+\frac {d^6 (2 m+9) \sqrt {d^2-e^2 x^2} (g x)^{m+1} \, _2F_1\left (-\frac {5}{2},\frac {m+1}{2};\frac {m+3}{2};\frac {e^2 x^2}{d^2}\right )}{g (m+1) (m+8) \sqrt {1-\frac {e^2 x^2}{d^2}}}+\frac {2 d^5 e \sqrt {d^2-e^2 x^2} (g x)^{m+2} \, _2F_1\left (-\frac {5}{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}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((g*x)^(1 + m)*(d^2 - e^2*x^2)^(7/2))/(g*(8 + m))) + (d^6*(9 + 2*m)*(g*x)^(1 + m)*Sqrt[d^2 - e^2*x^2]*Hyperg
eometric2F1[-5/2, (1 + m)/2, (3 + m)/2, (e^2*x^2)/d^2])/(g*(1 + m)*(8 + m)*Sqrt[1 - (e^2*x^2)/d^2]) + (2*d^5*e
*(g*x)^(2 + m)*Sqrt[d^2 - e^2*x^2]*Hypergeometric2F1[-5/2, (2 + m)/2, (4 + m)/2, (e^2*x^2)/d^2])/(g^2*(2 + m)*
Sqrt[1 - (e^2*x^2)/d^2])

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 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 (g x)^m (d+e x)^2 \left (d^2-e^2 x^2\right )^{5/2} \, dx &=-\frac {(g x)^{1+m} \left (d^2-e^2 x^2\right )^{7/2}}{g (8+m)}-\frac {\int (g x)^m \left (-d^2 e^2 (9+2 m)-2 d e^3 (8+m) x\right ) \left (d^2-e^2 x^2\right )^{5/2} \, dx}{e^2 (8+m)}\\ &=-\frac {(g x)^{1+m} \left (d^2-e^2 x^2\right )^{7/2}}{g (8+m)}+\frac {(2 d e) \int (g x)^{1+m} \left (d^2-e^2 x^2\right )^{5/2} \, dx}{g}+\frac {\left (d^2 (9+2 m)\right ) \int (g x)^m \left (d^2-e^2 x^2\right )^{5/2} \, dx}{8+m}\\ &=-\frac {(g x)^{1+m} \left (d^2-e^2 x^2\right )^{7/2}}{g (8+m)}+\frac {\left (2 d^5 e \sqrt {d^2-e^2 x^2}\right ) \int (g x)^{1+m} \left (1-\frac {e^2 x^2}{d^2}\right )^{5/2} \, dx}{g \sqrt {1-\frac {e^2 x^2}{d^2}}}+\frac {\left (d^6 (9+2 m) \sqrt {d^2-e^2 x^2}\right ) \int (g x)^m \left (1-\frac {e^2 x^2}{d^2}\right )^{5/2} \, dx}{(8+m) \sqrt {1-\frac {e^2 x^2}{d^2}}}\\ &=-\frac {(g x)^{1+m} \left (d^2-e^2 x^2\right )^{7/2}}{g (8+m)}+\frac {d^6 (9+2 m) (g x)^{1+m} \sqrt {d^2-e^2 x^2} \, _2F_1\left (-\frac {5}{2},\frac {1+m}{2};\frac {3+m}{2};\frac {e^2 x^2}{d^2}\right )}{g (1+m) (8+m) \sqrt {1-\frac {e^2 x^2}{d^2}}}+\frac {2 d^5 e (g x)^{2+m} \sqrt {d^2-e^2 x^2} \, _2F_1\left (-\frac {5}{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 = 0.74, size = 174, normalized size = 0.84 \begin {gather*} \frac {d^4 x (g x)^m \sqrt {d^2-e^2 x^2} \left (d^2 \left (6+5 m+m^2\right ) \, _2F_1\left (-\frac {5}{2},\frac {1+m}{2};\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 {5}{2},\frac {2+m}{2};\frac {4+m}{2};\frac {e^2 x^2}{d^2}\right )+e (2+m) x \, _2F_1\left (-\frac {5}{2},\frac {3+m}{2};\frac {5+m}{2};\frac {e^2 x^2}{d^2}\right )\right )\right )}{(1+m) (2+m) (3+m) \sqrt {1-\frac {e^2 x^2}{d^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^4*x*(g*x)^m*Sqrt[d^2 - e^2*x^2]*(d^2*(6 + 5*m + m^2)*Hypergeometric2F1[-5/2, (1 + m)/2, (3 + m)/2, (e^2*x^2
)/d^2] + e*(1 + m)*x*(2*d*(3 + m)*Hypergeometric2F1[-5/2, (2 + m)/2, (4 + m)/2, (e^2*x^2)/d^2] + e*(2 + m)*x*H
ypergeometric2F1[-5/2, (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 = 0.01, size = 0, normalized size = 0.00 \[\int \left (g x \right )^{m} \left (e x +d \right )^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral((d^4*x^2*e^2 + 2*d^5*x*e + d^6 + (x^6*e^2 + 2*d*x^5*e + d^2*x^4)*e^4 - 2*(d^2*x^4*e^2 + 2*d^3*x^3*e +
d^4*x^2)*e^2)*sqrt(-x^2*e^2 + d^2)*(g*x)^m, x)

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Sympy [C] Result contains complex when optimal does not.
time = 15.52, size = 442, normalized size = 2.15 \begin {gather*} \frac {d^{7} 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^{6} 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^{5} 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 )} - \frac {2 d^{4} e^{3} g^{m} x^{4} x^{m} \Gamma \left (\frac {m}{2} + 2\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + 2 \\ \frac {m}{2} + 3 \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{\Gamma \left (\frac {m}{2} + 3\right )} - \frac {d^{3} e^{4} g^{m} x^{5} x^{m} \Gamma \left (\frac {m}{2} + \frac {5}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + \frac {5}{2} \\ \frac {m}{2} + \frac {7}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} + \frac {d^{2} e^{5} g^{m} x^{6} x^{m} \Gamma \left (\frac {m}{2} + 3\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + 3 \\ \frac {m}{2} + 4 \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{\Gamma \left (\frac {m}{2} + 4\right )} + \frac {d e^{6} g^{m} x^{7} x^{m} \Gamma \left (\frac {m}{2} + \frac {7}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + \frac {7}{2} \\ \frac {m}{2} + \frac {9}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {9}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**7*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**6*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**5*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)) - 2*d**4*e**3*g**m*x**4*x**m*gamma(m/2 + 2)*hyper((
-1/2, m/2 + 2), (m/2 + 3,), e**2*x**2*exp_polar(2*I*pi)/d**2)/gamma(m/2 + 3) - d**3*e**4*g**m*x**5*x**m*gamma(
m/2 + 5/2)*hyper((-1/2, m/2 + 5/2), (m/2 + 7/2,), e**2*x**2*exp_polar(2*I*pi)/d**2)/(2*gamma(m/2 + 7/2)) + d**
2*e**5*g**m*x**6*x**m*gamma(m/2 + 3)*hyper((-1/2, m/2 + 3), (m/2 + 4,), e**2*x**2*exp_polar(2*I*pi)/d**2)/gamm
a(m/2 + 4) + d*e**6*g**m*x**7*x**m*gamma(m/2 + 7/2)*hyper((-1/2, m/2 + 7/2), (m/2 + 9/2,), e**2*x**2*exp_polar
(2*I*pi)/d**2)/(2*gamma(m/2 + 9/2))

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (d^2-e^2\,x^2\right )}^{5/2}\,{\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)^(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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