3.766 \(\int \frac{(f+g x)^n (a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}}{(d+e x)^{5/2}} \, dx\)

Optimal. Leaf size=104 \[ -\frac{(f+g x)^{n+1} (a e+c d x) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2} \, _2F_1\left (1,n+\frac{9}{2};n+2;\frac{c d (f+g x)}{c d f-a e g}\right )}{(n+1) (d+e x)^{5/2} (c d f-a e g)} \]

[Out]

-(((a*e + c*d*x)*(f + g*x)^(1 + n)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)*Hypergeometric2F1[1, 9/2 + n,
2 + n, (c*d*(f + g*x))/(c*d*f - a*e*g)])/((c*d*f - a*e*g)*(1 + n)*(d + e*x)^(5/2)))

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Rubi [A]  time = 0.111715, antiderivative size = 122, normalized size of antiderivative = 1.17, number of steps used = 3, number of rules used = 3, integrand size = 46, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {891, 70, 69} \[ \frac{2 (f+g x)^n (a e+c d x)^3 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (\frac{c d (f+g x)}{c d f-a e g}\right )^{-n} \, _2F_1\left (\frac{7}{2},-n;\frac{9}{2};-\frac{g (a e+c d x)}{c d f-a e g}\right )}{7 c d \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

Int[((f + g*x)^n*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2))/(d + e*x)^(5/2),x]

[Out]

(2*(a*e + c*d*x)^3*(f + g*x)^n*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]*Hypergeometric2F1[7/2, -n, 9/2, -((
g*(a*e + c*d*x))/(c*d*f - a*e*g))])/(7*c*d*Sqrt[d + e*x]*((c*d*(f + g*x))/(c*d*f - a*e*g))^n)

Rule 891

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

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

\begin{align*} \int \frac{(f+g x)^n \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{5/2}} \, dx &=\frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \int (a e+c d x)^{5/2} (f+g x)^n \, dx}{\sqrt{a e+c d x} \sqrt{d+e x}}\\ &=\frac{\left ((f+g x)^n \left (\frac{c d (f+g x)}{c d f-a e g}\right )^{-n} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}\right ) \int (a e+c d x)^{5/2} \left (\frac{c d f}{c d f-a e g}+\frac{c d g x}{c d f-a e g}\right )^n \, dx}{\sqrt{a e+c d x} \sqrt{d+e x}}\\ &=\frac{2 (a e+c d x)^3 (f+g x)^n \left (\frac{c d (f+g x)}{c d f-a e g}\right )^{-n} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, _2F_1\left (\frac{7}{2},-n;\frac{9}{2};-\frac{g (a e+c d x)}{c d f-a e g}\right )}{7 c d \sqrt{d+e x}}\\ \end{align*}

Mathematica [A]  time = 0.0723055, size = 110, normalized size = 1.06 \[ \frac{2 (f+g x)^n (a e+c d x)^3 \sqrt{(d+e x) (a e+c d x)} \left (\frac{c d (f+g x)}{c d f-a e g}\right )^{-n} \, _2F_1\left (\frac{7}{2},-n;\frac{9}{2};\frac{g (a e+c d x)}{a e g-c d f}\right )}{7 c d \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

Integrate[((f + g*x)^n*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2))/(d + e*x)^(5/2),x]

[Out]

(2*(a*e + c*d*x)^3*Sqrt[(a*e + c*d*x)*(d + e*x)]*(f + g*x)^n*Hypergeometric2F1[7/2, -n, 9/2, (g*(a*e + c*d*x))
/(-(c*d*f) + a*e*g)])/(7*c*d*Sqrt[d + e*x]*((c*d*(f + g*x))/(c*d*f - a*e*g))^n)

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Maple [F]  time = 1.705, size = 0, normalized size = 0. \begin{align*} \int{ \left ( gx+f \right ) ^{n} \left ( ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+cde{x}^{2} \right ) ^{{\frac{5}{2}}} \left ( ex+d \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)^n*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(5/2),x)

[Out]

int((g*x+f)^n*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(5/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^n*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(5/2),x, algorithm="maxima")

[Out]

integrate((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(5/2)*(g*x + f)^n/(e*x + d)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^n*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(5/2),x, algorithm="fricas")

[Out]

integral((c^2*d^2*x^2 + 2*a*c*d*e*x + a^2*e^2)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(g*x + f)^n/sqrt(e*
x + d), 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+f)**n*(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(5/2)/(e*x+d)**(5/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^n*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(5/2),x, algorithm="giac")

[Out]

integrate((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(5/2)*(g*x + f)^n/(e*x + d)^(5/2), x)