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

Optimal. Leaf size=365 \[ \frac{\left (d^2-a f^2\right )^5 \left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}+d+e x\right )^{n-5}}{32 e f^4 (5-n)}-\frac{5 \left (d^2-a f^2\right )^4 \left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}+d+e x\right )^{n-3}}{32 e f^4 (3-n)}+\frac{5 \left (d^2-a f^2\right )^3 \left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}+d+e x\right )^{n-1}}{16 e f^4 (1-n)}+\frac{5 \left (d^2-a f^2\right )^2 \left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}+d+e x\right )^{n+1}}{16 e f^4 (n+1)}-\frac{5 \left (d^2-a f^2\right ) \left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}+d+e x\right )^{n+3}}{32 e f^4 (n+3)}+\frac{\left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}+d+e x\right )^{n+5}}{32 e f^4 (n+5)} \]

[Out]

((d^2 - a*f^2)^5*(d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^(-5 + n))/(32*e*f^4*(5 - n)) - (5*(d^2
- a*f^2)^4*(d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^(-3 + n))/(32*e*f^4*(3 - n)) + (5*(d^2 - a*f^
2)^3*(d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^(-1 + n))/(16*e*f^4*(1 - n)) + (5*(d^2 - a*f^2)^2*(
d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^(1 + n))/(16*e*f^4*(1 + n)) - (5*(d^2 - a*f^2)*(d + e*x +
f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^(3 + n))/(32*e*f^4*(3 + n)) + (d + e*x + f*Sqrt[a + (2*d*e*x)/f^2
+ (e^2*x^2)/f^2])^(5 + n)/(32*e*f^4*(5 + n))

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Rubi [A]  time = 0.47376, antiderivative size = 365, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 56, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.054, Rules used = {2121, 12, 270} \[ \frac{\left (d^2-a f^2\right )^5 \left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}+d+e x\right )^{n-5}}{32 e f^4 (5-n)}-\frac{5 \left (d^2-a f^2\right )^4 \left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}+d+e x\right )^{n-3}}{32 e f^4 (3-n)}+\frac{5 \left (d^2-a f^2\right )^3 \left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}+d+e x\right )^{n-1}}{16 e f^4 (1-n)}+\frac{5 \left (d^2-a f^2\right )^2 \left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}+d+e x\right )^{n+1}}{16 e f^4 (n+1)}-\frac{5 \left (d^2-a f^2\right ) \left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}+d+e x\right )^{n+3}}{32 e f^4 (n+3)}+\frac{\left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}+d+e x\right )^{n+5}}{32 e f^4 (n+5)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d^2 - a*f^2)^5*(d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^(-5 + n))/(32*e*f^4*(5 - n)) - (5*(d^2
- a*f^2)^4*(d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^(-3 + n))/(32*e*f^4*(3 - n)) + (5*(d^2 - a*f^
2)^3*(d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^(-1 + n))/(16*e*f^4*(1 - n)) + (5*(d^2 - a*f^2)^2*(
d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^(1 + n))/(16*e*f^4*(1 + n)) - (5*(d^2 - a*f^2)*(d + e*x +
f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^(3 + n))/(32*e*f^4*(3 + n)) + (d + e*x + f*Sqrt[a + (2*d*e*x)/f^2
+ (e^2*x^2)/f^2])^(5 + n)/(32*e*f^4*(5 + 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 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \left (a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}\right )^2 \left (d+e x+f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}\right )^n \, dx &=\frac{2 \operatorname{Subst}\left (\int \frac{x^{-6+n} \left (d^2 e-\left (-a e+\frac{2 d^2 e}{f^2}\right ) f^2+e x^2\right )^5}{64 e^6} \, dx,x,d+e x+f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}\right )}{f^4}\\ &=\frac{\operatorname{Subst}\left (\int x^{-6+n} \left (d^2 e-\left (-a e+\frac{2 d^2 e}{f^2}\right ) f^2+e x^2\right )^5 \, dx,x,d+e x+f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}\right )}{32 e^6 f^4}\\ &=\frac{\operatorname{Subst}\left (\int \left (-e^5 \left (d^2-a f^2\right )^5 x^{-6+n}+5 e^5 \left (d^2-a f^2\right )^4 x^{-4+n}-10 e^5 \left (d^2-a f^2\right )^3 x^{-2+n}+10 e^5 \left (d^2-a f^2\right )^2 x^n-5 e^5 \left (d^2-a f^2\right ) x^{2+n}+e^5 x^{4+n}\right ) \, dx,x,d+e x+f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}\right )}{32 e^6 f^4}\\ &=\frac{\left (d^2-a f^2\right )^5 \left (d+e x+f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}\right )^{-5+n}}{32 e f^4 (5-n)}-\frac{5 \left (d^2-a f^2\right )^4 \left (d+e x+f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}\right )^{-3+n}}{32 e f^4 (3-n)}+\frac{5 \left (d^2-a f^2\right )^3 \left (d+e x+f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}\right )^{-1+n}}{16 e f^4 (1-n)}+\frac{5 \left (d^2-a f^2\right )^2 \left (d+e x+f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}\right )^{1+n}}{16 e f^4 (1+n)}-\frac{5 \left (d^2-a f^2\right ) \left (d+e x+f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}\right )^{3+n}}{32 e f^4 (3+n)}+\frac{\left (d+e x+f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}\right )^{5+n}}{32 e f^4 (5+n)}\\ \end{align*}

Mathematica [A]  time = 2.98739, size = 280, normalized size = 0.77 \[ \frac{\left (f \sqrt{a+\frac{e x (2 d+e x)}{f^2}}+d+e x\right )^{n-5} \left (-\frac{5 \left (d^2-a f^2\right ) \left (f \sqrt{a+\frac{e x (2 d+e x)}{f^2}}+d+e x\right )^8}{n+3}+\frac{10 \left (d^2-a f^2\right )^2 \left (f \sqrt{a+\frac{e x (2 d+e x)}{f^2}}+d+e x\right )^6}{n+1}-\frac{10 \left (d^2-a f^2\right )^3 \left (f \sqrt{a+\frac{e x (2 d+e x)}{f^2}}+d+e x\right )^4}{n-1}+\frac{5 \left (d^2-a f^2\right )^4 \left (f \sqrt{a+\frac{e x (2 d+e x)}{f^2}}+d+e x\right )^2}{n-3}-\frac{\left (d^2-a f^2\right )^5}{n-5}+\frac{\left (f \sqrt{a+\frac{e x (2 d+e x)}{f^2}}+d+e x\right )^{10}}{n+5}\right )}{32 e f^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x + f*Sqrt[a + (e*x*(2*d + e*x))/f^2])^(-5 + n)*(-((d^2 - a*f^2)^5/(-5 + n)) + (5*(d^2 - a*f^2)^4*(d +
e*x + f*Sqrt[a + (e*x*(2*d + e*x))/f^2])^2)/(-3 + n) - (10*(d^2 - a*f^2)^3*(d + e*x + f*Sqrt[a + (e*x*(2*d +
e*x))/f^2])^4)/(-1 + n) + (10*(d^2 - a*f^2)^2*(d + e*x + f*Sqrt[a + (e*x*(2*d + e*x))/f^2])^6)/(1 + n) - (5*(d
^2 - a*f^2)*(d + e*x + f*Sqrt[a + (e*x*(2*d + e*x))/f^2])^8)/(3 + n) + (d + e*x + f*Sqrt[a + (e*x*(2*d + e*x))
/f^2])^10/(5 + n)))/(32*e*f^4)

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Maple [F]  time = 0.089, size = 0, normalized size = 0. \begin{align*} \int \left ( a+2\,{\frac{dex}{{f}^{2}}}+{\frac{{e}^{2}{x}^{2}}{{f}^{2}}} \right ) ^{2} \left ( d+ex+f\sqrt{a+2\,{\frac{dex}{{f}^{2}}}+{\frac{{e}^{2}{x}^{2}}{{f}^{2}}}} \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.22951, size = 1384, normalized size = 3.79 \begin{align*} -\frac{{\left (5 \, a^{2} d f^{4} n^{4} + 225 \, a^{2} d f^{4} - 300 \, a d^{3} f^{2} + 5 \,{\left (e^{5} n^{4} - 10 \, e^{5} n^{2} + 9 \, e^{5}\right )} x^{5} + 120 \, d^{5} + 25 \,{\left (d e^{4} n^{4} - 10 \, d e^{4} n^{2} + 9 \, d e^{4}\right )} x^{4} + 10 \,{\left (15 \, a e^{3} f^{2} + 30 \, d^{2} e^{3} +{\left (a e^{3} f^{2} + 4 \, d^{2} e^{3}\right )} n^{4} - 2 \,{\left (8 \, a e^{3} f^{2} + 17 \, d^{2} e^{3}\right )} n^{2}\right )} x^{3} - 10 \,{\left (11 \, a^{2} d f^{4} - 6 \, a d^{3} f^{2}\right )} n^{2} + 10 \,{\left (45 \, a d e^{2} f^{2} +{\left (3 \, a d e^{2} f^{2} + 2 \, d^{3} e^{2}\right )} n^{4} - 2 \,{\left (24 \, a d e^{2} f^{2} + d^{3} e^{2}\right )} n^{2}\right )} x^{2} + 5 \,{\left (45 \, a^{2} e f^{4} +{\left (a^{2} e f^{4} + 4 \, a d^{2} e f^{2}\right )} n^{4} - 2 \,{\left (11 \, a^{2} e f^{4} + 26 \, a d^{2} e f^{2} - 12 \, d^{4} e\right )} n^{2}\right )} x -{\left (a^{2} f^{5} n^{5} +{\left (e^{4} f n^{5} - 10 \, e^{4} f n^{3} + 9 \, e^{4} f n\right )} x^{4} - 10 \,{\left (3 \, a^{2} f^{5} - 2 \, a d^{2} f^{3}\right )} n^{3} + 4 \,{\left (d e^{3} f n^{5} - 10 \, d e^{3} f n^{3} + 9 \, d e^{3} f n\right )} x^{3} + 2 \,{\left ({\left (a e^{2} f^{3} + 2 \, d^{2} e^{2} f\right )} n^{5} - 10 \,{\left (2 \, a e^{2} f^{3} + d^{2} e^{2} f\right )} n^{3} +{\left (19 \, a e^{2} f^{3} + 8 \, d^{2} e^{2} f\right )} n\right )} x^{2} +{\left (149 \, a^{2} f^{5} - 260 \, a d^{2} f^{3} + 120 \, d^{4} f\right )} n + 4 \,{\left (a d e f^{3} n^{5} - 10 \,{\left (2 \, a d e f^{3} - d^{3} e f\right )} n^{3} +{\left (19 \, a d e f^{3} - 10 \, d^{3} e f\right )} n\right )} x\right )} \sqrt{\frac{e^{2} x^{2} + a f^{2} + 2 \, d e x}{f^{2}}}\right )}{\left (e x + f \sqrt{\frac{e^{2} x^{2} + a f^{2} + 2 \, d e x}{f^{2}}} + d\right )}^{n}}{e f^{4} n^{6} - 35 \, e f^{4} n^{4} + 259 \, e f^{4} n^{2} - 225 \, e f^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(5*a^2*d*f^4*n^4 + 225*a^2*d*f^4 - 300*a*d^3*f^2 + 5*(e^5*n^4 - 10*e^5*n^2 + 9*e^5)*x^5 + 120*d^5 + 25*(d*e^4
*n^4 - 10*d*e^4*n^2 + 9*d*e^4)*x^4 + 10*(15*a*e^3*f^2 + 30*d^2*e^3 + (a*e^3*f^2 + 4*d^2*e^3)*n^4 - 2*(8*a*e^3*
f^2 + 17*d^2*e^3)*n^2)*x^3 - 10*(11*a^2*d*f^4 - 6*a*d^3*f^2)*n^2 + 10*(45*a*d*e^2*f^2 + (3*a*d*e^2*f^2 + 2*d^3
*e^2)*n^4 - 2*(24*a*d*e^2*f^2 + d^3*e^2)*n^2)*x^2 + 5*(45*a^2*e*f^4 + (a^2*e*f^4 + 4*a*d^2*e*f^2)*n^4 - 2*(11*
a^2*e*f^4 + 26*a*d^2*e*f^2 - 12*d^4*e)*n^2)*x - (a^2*f^5*n^5 + (e^4*f*n^5 - 10*e^4*f*n^3 + 9*e^4*f*n)*x^4 - 10
*(3*a^2*f^5 - 2*a*d^2*f^3)*n^3 + 4*(d*e^3*f*n^5 - 10*d*e^3*f*n^3 + 9*d*e^3*f*n)*x^3 + 2*((a*e^2*f^3 + 2*d^2*e^
2*f)*n^5 - 10*(2*a*e^2*f^3 + d^2*e^2*f)*n^3 + (19*a*e^2*f^3 + 8*d^2*e^2*f)*n)*x^2 + (149*a^2*f^5 - 260*a*d^2*f
^3 + 120*d^4*f)*n + 4*(a*d*e*f^3*n^5 - 10*(2*a*d*e*f^3 - d^3*e*f)*n^3 + (19*a*d*e*f^3 - 10*d^3*e*f)*n)*x)*sqrt
((e^2*x^2 + a*f^2 + 2*d*e*x)/f^2))*(e*x + f*sqrt((e^2*x^2 + a*f^2 + 2*d*e*x)/f^2) + d)^n/(e*f^4*n^6 - 35*e*f^4
*n^4 + 259*e*f^4*n^2 - 225*e*f^4)

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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