3.517 \(\int \frac{(d+e x+f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}})^n}{(a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2})^{3/2}} \, dx\)

Optimal. Leaf size=122 \[ \frac{4 f^3 \left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}+d+e x\right )^{n+2} \, _2F_1\left (2,\frac{n+2}{2};\frac{n+4}{2};\frac{\left (d+e x+f \sqrt{\frac{e^2 x^2}{f^2}+\frac{2 d e x}{f^2}+a}\right )^2}{d^2-a f^2}\right )}{e (n+2) \left (d^2-a f^2\right )^2} \]

[Out]

(4*f^3*(d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^(2 + n)*Hypergeometric2F1[2, (2 + n)/2, (4 + n)/2
, (d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^2/(d^2 - a*f^2)])/(e*(d^2 - a*f^2)^2*(2 + n))

________________________________________________________________________________________

Rubi [A]  time = 0.302176, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 58, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.052, Rules used = {2121, 12, 364} \[ \frac{4 f^3 \left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}+d+e x\right )^{n+2} \, _2F_1\left (2,\frac{n+2}{2};\frac{n+4}{2};\frac{\left (d+e x+f \sqrt{\frac{e^2 x^2}{f^2}+\frac{2 d e x}{f^2}+a}\right )^2}{d^2-a f^2}\right )}{e (n+2) \left (d^2-a f^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^n/(a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2)^(3/2),x]

[Out]

(4*f^3*(d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^(2 + n)*Hypergeometric2F1[2, (2 + n)/2, (4 + n)/2
, (d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^2/(d^2 - a*f^2)])/(e*(d^2 - a*f^2)^2*(2 + n))

Rule 2121

Int[((g_.) + (h_.)*(x_) + (i_.)*(x_)^2)^(m_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)
^2])^(n_.), x_Symbol] :> Dist[(2*(i/c)^m)/f^(2*m), Subst[Int[(x^n*(d^2*e - (b*d - a*e)*f^2 - (2*d*e - b*f^2)*x
+ e*x^2)^(2*m + 1))/(-2*d*e + b*f^2 + 2*e*x)^(2*(m + 1)), x], x, d + e*x + f*Sqrt[a + b*x + c*x^2]], x] /; Fr
eeQ[{a, b, c, d, e, f, g, h, i, n}, x] && EqQ[e^2 - c*f^2, 0] && EqQ[c*g - a*i, 0] && EqQ[c*h - b*i, 0] && Int
egerQ[2*m] && (IntegerQ[m] || GtQ[i/c, 0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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{\left (d+e x+f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}\right )^n}{\left (a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}\right )^{3/2}} \, dx &=\left (2 f^3\right ) \operatorname{Subst}\left (\int \frac{2 e x^{1+n}}{\left (d^2 e-\left (-a e+\frac{2 d^2 e}{f^2}\right ) f^2+e x^2\right )^2} \, dx,x,d+e x+f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}\right )\\ &=\left (4 e f^3\right ) \operatorname{Subst}\left (\int \frac{x^{1+n}}{\left (d^2 e-\left (-a e+\frac{2 d^2 e}{f^2}\right ) f^2+e x^2\right )^2} \, dx,x,d+e x+f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}\right )\\ &=\frac{4 f^3 \left (d+e x+f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}\right )^{2+n} \, _2F_1\left (2,\frac{2+n}{2};\frac{4+n}{2};\frac{\left (d+e x+f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}\right )^2}{d^2-a f^2}\right )}{e \left (d^2-a f^2\right )^2 (2+n)}\\ \end{align*}

Mathematica [A]  time = 0.149754, size = 112, normalized size = 0.92 \[ \frac{4 f^3 \left (f \sqrt{a+\frac{e x (2 d+e x)}{f^2}}+d+e x\right )^{n+2} \, _2F_1\left (2,\frac{n+2}{2};\frac{n+4}{2};\frac{\left (d+e x+f \sqrt{a+\frac{e x (2 d+e x)}{f^2}}\right )^2}{d^2-a f^2}\right )}{e (n+2) \left (d^2-a f^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^n/(a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2)^(3/2),x]

[Out]

(4*f^3*(d + e*x + f*Sqrt[a + (e*x*(2*d + e*x))/f^2])^(2 + n)*Hypergeometric2F1[2, (2 + n)/2, (4 + n)/2, (d + e
*x + f*Sqrt[a + (e*x*(2*d + e*x))/f^2])^2/(d^2 - a*f^2)])/(e*(d^2 - a*f^2)^2*(2 + n))

________________________________________________________________________________________

Maple [F]  time = 0.07, size = 0, normalized size = 0. \begin{align*} \int{ \left ( d+ex+f\sqrt{a+2\,{\frac{dex}{{f}^{2}}}+{\frac{{e}^{2}{x}^{2}}{{f}^{2}}}} \right ) ^{n} \left ( a+2\,{\frac{dex}{{f}^{2}}}+{\frac{{e}^{2}{x}^{2}}{{f}^{2}}} \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d+e*x+f*(a+2*d*e*x/f^2+e^2*x^2/f^2)^(1/2))^n/(a+2*d*e*x/f^2+e^2*x^2/f^2)^(3/2),x)

[Out]

int((d+e*x+f*(a+2*d*e*x/f^2+e^2*x^2/f^2)^(1/2))^n/(a+2*d*e*x/f^2+e^2*x^2/f^2)^(3/2),x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + \sqrt{\frac{e^{2} x^{2}}{f^{2}} + a + \frac{2 \, d e x}{f^{2}}} f + d\right )}^{n}}{{\left (\frac{e^{2} x^{2}}{f^{2}} + a + \frac{2 \, d e x}{f^{2}}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+e*x+f*(a+2*d*e*x/f^2+e^2*x^2/f^2)^(1/2))^n/(a+2*d*e*x/f^2+e^2*x^2/f^2)^(3/2),x, algorithm="maxima
")

[Out]

integrate((e*x + sqrt(e^2*x^2/f^2 + a + 2*d*e*x/f^2)*f + d)^n/(e^2*x^2/f^2 + a + 2*d*e*x/f^2)^(3/2), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (e x + f \sqrt{\frac{e^{2} x^{2} + a f^{2} + 2 \, d e x}{f^{2}}} + d\right )}^{n} f^{4} \sqrt{\frac{e^{2} x^{2} + a f^{2} + 2 \, d e x}{f^{2}}}}{e^{4} x^{4} + 4 \, d e^{3} x^{3} + a^{2} f^{4} + 4 \, a d e f^{2} x + 2 \,{\left (a e^{2} f^{2} + 2 \, d^{2} e^{2}\right )} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+e*x+f*(a+2*d*e*x/f^2+e^2*x^2/f^2)^(1/2))^n/(a+2*d*e*x/f^2+e^2*x^2/f^2)^(3/2),x, algorithm="fricas
")

[Out]

integral((e*x + f*sqrt((e^2*x^2 + a*f^2 + 2*d*e*x)/f^2) + d)^n*f^4*sqrt((e^2*x^2 + a*f^2 + 2*d*e*x)/f^2)/(e^4*
x^4 + 4*d*e^3*x^3 + a^2*f^4 + 4*a*d*e*f^2*x + 2*(a*e^2*f^2 + 2*d^2*e^2)*x^2), x)

________________________________________________________________________________________

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((d+e*x+f*(a+2*d*e*x/f**2+e**2*x**2/f**2)**(1/2))**n/(a+2*d*e*x/f**2+e**2*x**2/f**2)**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + \sqrt{\frac{e^{2} x^{2}}{f^{2}} + a + \frac{2 \, d e x}{f^{2}}} f + d\right )}^{n}}{{\left (\frac{e^{2} x^{2}}{f^{2}} + a + \frac{2 \, d e x}{f^{2}}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+e*x+f*(a+2*d*e*x/f^2+e^2*x^2/f^2)^(1/2))^n/(a+2*d*e*x/f^2+e^2*x^2/f^2)^(3/2),x, algorithm="giac")

[Out]

integrate((e*x + sqrt(e^2*x^2/f^2 + a + 2*d*e*x/f^2)*f + d)^n/(e^2*x^2/f^2 + a + 2*d*e*x/f^2)^(3/2), x)