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

Optimal. Leaf size=163 \[ \frac{d^3 \sqrt{d^2-e^2 x^2} (g x)^{m+1} \, _2F_1\left (-\frac{3}{2},\frac{m+1}{2};\frac{m+3}{2};\frac{e^2 x^2}{d^2}\right )}{g (m+1) \sqrt{1-\frac{e^2 x^2}{d^2}}}-\frac{d^2 e \sqrt{d^2-e^2 x^2} (g x)^{m+2} \, _2F_1\left (-\frac{3}{2},\frac{m+2}{2};\frac{m+4}{2};\frac{e^2 x^2}{d^2}\right )}{g^2 (m+2) \sqrt{1-\frac{e^2 x^2}{d^2}}} \]

[Out]

(d^3*(g*x)^(1 + m)*Sqrt[d^2 - e^2*x^2]*Hypergeometric2F1[-3/2, (1 + m)/2, (3 + m)/2, (e^2*x^2)/d^2])/(g*(1 + m
)*Sqrt[1 - (e^2*x^2)/d^2]) - (d^2*e*(g*x)^(2 + m)*Sqrt[d^2 - e^2*x^2]*Hypergeometric2F1[-3/2, (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 = 0.1405, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {892, 82, 126, 365, 364} \[ \frac{d^3 \sqrt{d^2-e^2 x^2} (g x)^{m+1} \, _2F_1\left (-\frac{3}{2},\frac{m+1}{2};\frac{m+3}{2};\frac{e^2 x^2}{d^2}\right )}{g (m+1) \sqrt{1-\frac{e^2 x^2}{d^2}}}-\frac{d^2 e \sqrt{d^2-e^2 x^2} (g x)^{m+2} \, _2F_1\left (-\frac{3}{2},\frac{m+2}{2};\frac{m+4}{2};\frac{e^2 x^2}{d^2}\right )}{g^2 (m+2) \sqrt{1-\frac{e^2 x^2}{d^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^3*(g*x)^(1 + m)*Sqrt[d^2 - e^2*x^2]*Hypergeometric2F1[-3/2, (1 + m)/2, (3 + m)/2, (e^2*x^2)/d^2])/(g*(1 + m
)*Sqrt[1 - (e^2*x^2)/d^2]) - (d^2*e*(g*x)^(2 + m)*Sqrt[d^2 - e^2*x^2]*Hypergeometric2F1[-3/2, (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 892

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

Rule 82

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

Rule 126

Int[((f_.)*(x_))^(p_.)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Dist[((a + b*x)^Fra
cPart[m]*(c + d*x)^FracPart[m])/(a*c + b*d*x^2)^FracPart[m], Int[(a*c + b*d*x^2)^m*(f*x)^p, x], x] /; FreeQ[{a
, b, c, d, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[m - n, 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} \, dx &=\frac{\sqrt{d^2-e^2 x^2} \int (g x)^m (d-e x)^{5/2} (d+e x)^{3/2} \, dx}{\sqrt{d-e x} \sqrt{d+e x}}\\ &=\frac{\left (d \sqrt{d^2-e^2 x^2}\right ) \int (g x)^m (d-e x)^{3/2} (d+e x)^{3/2} \, dx}{\sqrt{d-e x} \sqrt{d+e x}}-\frac{\left (e \sqrt{d^2-e^2 x^2}\right ) \int (g x)^{1+m} (d-e x)^{3/2} (d+e x)^{3/2} \, dx}{g \sqrt{d-e x} \sqrt{d+e x}}\\ &=d \int (g x)^m \left (d^2-e^2 x^2\right )^{3/2} \, dx-\frac{e \int (g x)^{1+m} \left (d^2-e^2 x^2\right )^{3/2} \, dx}{g}\\ &=\frac{\left (d^3 \sqrt{d^2-e^2 x^2}\right ) \int (g x)^m \left (1-\frac{e^2 x^2}{d^2}\right )^{3/2} \, dx}{\sqrt{1-\frac{e^2 x^2}{d^2}}}-\frac{\left (d^2 e \sqrt{d^2-e^2 x^2}\right ) \int (g x)^{1+m} \left (1-\frac{e^2 x^2}{d^2}\right )^{3/2} \, dx}{g \sqrt{1-\frac{e^2 x^2}{d^2}}}\\ &=\frac{d^3 (g x)^{1+m} \sqrt{d^2-e^2 x^2} \, _2F_1\left (-\frac{3}{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^2 e (g x)^{2+m} \sqrt{d^2-e^2 x^2} \, _2F_1\left (-\frac{3}{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 = 0.0591541, size = 122, normalized size = 0.75 \[ \frac{d^2 x \sqrt{d^2-e^2 x^2} (g x)^m \left (d (m+2) \, _2F_1\left (-\frac{3}{2},\frac{m+1}{2};\frac{m+3}{2};\frac{e^2 x^2}{d^2}\right )-e (m+1) x \, _2F_1\left (-\frac{3}{2},\frac{m}{2}+1;\frac{m}{2}+2;\frac{e^2 x^2}{d^2}\right )\right )}{(m+1) (m+2) \sqrt{1-\frac{e^2 x^2}{d^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*x*(g*x)^m*Sqrt[d^2 - e^2*x^2]*(-(e*(1 + m)*x*Hypergeometric2F1[-3/2, 1 + m/2, 2 + m/2, (e^2*x^2)/d^2]) +
d*(2 + m)*Hypergeometric2F1[-3/2, (1 + m)/2, (3 + 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 = 0.574, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( gx \right ) ^{m}}{ex+d} \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),x)

[Out]

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

[Out]

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

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

[Out]

integral((e^3*x^3 - d*e^2*x^2 - d^2*e*x + d^3)*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),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}}{e x + d}\,{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),x, algorithm="giac")

[Out]

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