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

Optimal. Leaf size=162 \[ \frac{d^4 e \sqrt{d^2-e^2 x^2} (g x)^{m+2} \, _2F_1\left (-\frac{5}{2},\frac{m+2}{2};\frac{m+4}{2};\frac{e^2 x^2}{d^2}\right )}{g^2 (m+2) \sqrt{1-\frac{e^2 x^2}{d^2}}}+\frac{d^5 \sqrt{d^2-e^2 x^2} (g x)^{m+1} \, _2F_1\left (-\frac{5}{2},\frac{m+1}{2};\frac{m+3}{2};\frac{e^2 x^2}{d^2}\right )}{g (m+1) \sqrt{1-\frac{e^2 x^2}{d^2}}} \]

[Out]

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

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Rubi [A]  time = 0.0835558, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {808, 365, 364} \[ \frac{d^4 e \sqrt{d^2-e^2 x^2} (g x)^{m+2} \, _2F_1\left (-\frac{5}{2},\frac{m+2}{2};\frac{m+4}{2};\frac{e^2 x^2}{d^2}\right )}{g^2 (m+2) \sqrt{1-\frac{e^2 x^2}{d^2}}}+\frac{d^5 \sqrt{d^2-e^2 x^2} (g x)^{m+1} \, _2F_1\left (-\frac{5}{2},\frac{m+1}{2};\frac{m+3}{2};\frac{e^2 x^2}{d^2}\right )}{g (m+1) \sqrt{1-\frac{e^2 x^2}{d^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^5*(g*x)^(1 + m)*Sqrt[d^2 - e^2*x^2]*Hypergeometric2F1[-5/2, (1 + m)/2, (3 + m)/2, (e^2*x^2)/d^2])/(g*(1 + m
)*Sqrt[1 - (e^2*x^2)/d^2]) + (d^4*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 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 (g x)^m (d+e x) \left (d^2-e^2 x^2\right )^{5/2} \, dx &=d \int (g x)^m \left (d^2-e^2 x^2\right )^{5/2} \, dx+\frac{e \int (g x)^{1+m} \left (d^2-e^2 x^2\right )^{5/2} \, dx}{g}\\ &=\frac{\left (d^5 \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}{\sqrt{1-\frac{e^2 x^2}{d^2}}}+\frac{\left (d^4 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{d^5 (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) \sqrt{1-\frac{e^2 x^2}{d^2}}}+\frac{d^4 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*}

Mathematica [A]  time = 0.0575928, size = 121, normalized size = 0.75 \[ \frac{d^4 x \sqrt{d^2-e^2 x^2} (g x)^m \left (d (m+2) \, _2F_1\left (-\frac{5}{2},\frac{m+1}{2};\frac{m+3}{2};\frac{e^2 x^2}{d^2}\right )+e (m+1) x \, _2F_1\left (-\frac{5}{2},\frac{m+2}{2};\frac{m+4}{2};\frac{e^2 x^2}{d^2}\right )\right )}{(m+1) (m+2) \sqrt{1-\frac{e^2 x^2}{d^2}}} \]

Antiderivative was successfully verified.

[In]

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

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Maple [F]  time = 0.364, size = 0, normalized size = 0. \begin{align*} \int \left ( gx \right ) ^{m} \left ( ex+d \right ) \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*x+d)*(-e^2*x^2+d^2)^(5/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (e^{5} x^{5} + d e^{4} x^{4} - 2 \, d^{2} e^{3} x^{3} - 2 \, d^{3} e^{2} x^{2} + d^{4} e x + d^{5}\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*x+d)*(-e^2*x^2+d^2)^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 62.4386, size = 374, normalized size = 2.31 \begin{align*} \frac{d^{6} 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^{5} 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 )}}{2 \Gamma \left (\frac{m}{2} + 2\right )} - \frac{d^{4} 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 )}}{\Gamma \left (\frac{m}{2} + \frac{5}{2}\right )} - \frac{d^{3} 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^{2} 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 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 )}}{2 \Gamma \left (\frac{m}{2} + 4\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**6*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**5*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)/(2*gamma(m/2 + 2)) - d**4*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)/gamma(m/2 + 5/2) - d**3*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**2*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*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)/(2*gamma(
m/2 + 4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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