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

Optimal. Leaf size=297 \[ -\frac{\left (d^2-a f^2\right )^4 \left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}+d+e x\right )^{n-4}}{16 e f^3 (4-n)}+\frac{\left (d^2-a f^2\right )^3 \left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}+d+e x\right )^{n-2}}{4 e f^3 (2-n)}+\frac{3 \left (d^2-a f^2\right )^2 \left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}+d+e x\right )^n}{8 e f^3 n}-\frac{\left (d^2-a f^2\right ) \left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}+d+e x\right )^{n+2}}{4 e f^3 (n+2)}+\frac{\left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}+d+e x\right )^{n+4}}{16 e f^3 (n+4)} \]

[Out]

-((d^2 - a*f^2)^4*(d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^(-4 + n))/(16*e*f^3*(4 - n)) + ((d^2 -
a*f^2)^3*(d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^(-2 + n))/(4*e*f^3*(2 - n)) + (3*(d^2 - a*f^2)
^2*(d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^n)/(8*e*f^3*n) - ((d^2 - a*f^2)*(d + e*x + f*Sqrt[a +
(2*d*e*x)/f^2 + (e^2*x^2)/f^2])^(2 + n))/(4*e*f^3*(2 + n)) + (d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/
f^2])^(4 + n)/(16*e*f^3*(4 + n))

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Rubi [A]  time = 0.419456, antiderivative size = 297, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 58, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.052, Rules used = {2121, 12, 270} \[ -\frac{\left (d^2-a f^2\right )^4 \left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}+d+e x\right )^{n-4}}{16 e f^3 (4-n)}+\frac{\left (d^2-a f^2\right )^3 \left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}+d+e x\right )^{n-2}}{4 e f^3 (2-n)}+\frac{3 \left (d^2-a f^2\right )^2 \left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}+d+e x\right )^n}{8 e f^3 n}-\frac{\left (d^2-a f^2\right ) \left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}+d+e x\right )^{n+2}}{4 e f^3 (n+2)}+\frac{\left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}+d+e x\right )^{n+4}}{16 e f^3 (n+4)} \]

Antiderivative was successfully verified.

[In]

Int[(a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2)^(3/2)*(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)^4*(d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^(-4 + n))/(16*e*f^3*(4 - n)) + ((d^2 -
a*f^2)^3*(d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^(-2 + n))/(4*e*f^3*(2 - n)) + (3*(d^2 - a*f^2)
^2*(d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^n)/(8*e*f^3*n) - ((d^2 - a*f^2)*(d + e*x + f*Sqrt[a +
(2*d*e*x)/f^2 + (e^2*x^2)/f^2])^(2 + n))/(4*e*f^3*(2 + n)) + (d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/
f^2])^(4 + n)/(16*e*f^3*(4 + 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 )^{3/2} \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^{-5+n} \left (d^2 e-\left (-a e+\frac{2 d^2 e}{f^2}\right ) f^2+e x^2\right )^4}{32 e^5} \, dx,x,d+e x+f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}\right )}{f^3}\\ &=\frac{\operatorname{Subst}\left (\int x^{-5+n} \left (d^2 e-\left (-a e+\frac{2 d^2 e}{f^2}\right ) f^2+e x^2\right )^4 \, dx,x,d+e x+f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}\right )}{16 e^5 f^3}\\ &=\frac{\operatorname{Subst}\left (\int \left (e^4 \left (d^2-a f^2\right )^4 x^{-5+n}-4 e^4 \left (d^2-a f^2\right )^3 x^{-3+n}+6 e^4 \left (d^2-a f^2\right )^2 x^{-1+n}-4 e^4 \left (d^2-a f^2\right ) x^{1+n}+e^4 x^{3+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 )}{16 e^5 f^3}\\ &=-\frac{\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 )^{-4+n}}{16 e f^3 (4-n)}+\frac{\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 )^{-2+n}}{4 e f^3 (2-n)}+\frac{3 \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 )^n}{8 e f^3 n}-\frac{\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 )^{2+n}}{4 e f^3 (2+n)}+\frac{\left (d+e x+f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}\right )^{4+n}}{16 e f^3 (4+n)}\\ \end{align*}

Mathematica [A]  time = 1.24453, size = 228, normalized size = 0.77 \[ \frac{\left (f \sqrt{a+\frac{e x (2 d+e x)}{f^2}}+d+e x\right )^n \left (\frac{\left (d^2-a f^2\right )^4}{(n-4) \left (f \sqrt{a+\frac{e x (2 d+e x)}{f^2}}+d+e x\right )^4}-\frac{4 \left (d^2-a f^2\right )^3}{(n-2) \left (f \sqrt{a+\frac{e x (2 d+e x)}{f^2}}+d+e x\right )^2}-\frac{4 \left (d^2-a f^2\right ) \left (f \sqrt{a+\frac{e x (2 d+e x)}{f^2}}+d+e x\right )^2}{n+2}+\frac{6 \left (d^2-a f^2\right )^2}{n}+\frac{\left (f \sqrt{a+\frac{e x (2 d+e x)}{f^2}}+d+e x\right )^4}{n+4}\right )}{16 e f^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2)^(3/2)*(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])^n*((6*(d^2 - a*f^2)^2)/n + (d^2 - a*f^2)^4/((-4 + n)*(d + e*x +
f*Sqrt[a + (e*x*(2*d + e*x))/f^2])^4) - (4*(d^2 - a*f^2)^3)/((-2 + n)*(d + e*x + f*Sqrt[a + (e*x*(2*d + e*x))
/f^2])^2) - (4*(d^2 - a*f^2)*(d + e*x + f*Sqrt[a + (e*x*(2*d + e*x))/f^2])^2)/(2 + n) + (d + e*x + f*Sqrt[a +
(e*x*(2*d + e*x))/f^2])^4/(4 + n)))/(16*e*f^3)

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Maple [F]  time = 0.069, size = 0, normalized size = 0. \begin{align*} \int \left ( a+2\,{\frac{dex}{{f}^{2}}}+{\frac{{e}^{2}{x}^{2}}{{f}^{2}}} \right ) ^{{\frac{3}{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)^(3/2)*(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)^(3/2)*(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 )}^{\frac{3}{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)^(3/2)*(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)^(3/2)*(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 = 1.16936, size = 778, normalized size = 2.62 \begin{align*} \frac{{\left (a^{2} f^{4} n^{4} + 24 \, a^{2} f^{4} - 48 \, a d^{2} f^{2} +{\left (e^{4} n^{4} - 4 \, e^{4} n^{2}\right )} x^{4} + 24 \, d^{4} + 4 \,{\left (d e^{3} n^{4} - 4 \, d e^{3} n^{2}\right )} x^{3} - 4 \,{\left (4 \, a^{2} f^{4} - 3 \, a d^{2} f^{2}\right )} n^{2} + 2 \,{\left ({\left (a e^{2} f^{2} + 2 \, d^{2} e^{2}\right )} n^{4} - 2 \,{\left (5 \, a e^{2} f^{2} + d^{2} e^{2}\right )} n^{2}\right )} x^{2} + 4 \,{\left (a d e f^{2} n^{4} - 2 \,{\left (5 \, a d e f^{2} - 3 \, d^{3} e\right )} n^{2}\right )} x - 4 \,{\left (a d f^{3} n^{3} +{\left (e^{3} f n^{3} - 4 \, e^{3} f n\right )} x^{3} + 3 \,{\left (d e^{2} f n^{3} - 4 \, d e^{2} f n\right )} x^{2} - 2 \,{\left (5 \, a d f^{3} - 3 \, d^{3} f\right )} n +{\left ({\left (a e f^{3} + 2 \, d^{2} e f\right )} n^{3} - 2 \,{\left (5 \, a e f^{3} + d^{2} 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^{3} n^{5} - 20 \, e f^{3} n^{3} + 64 \, e f^{3} n} \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)^(3/2)*(d+e*x+f*(a+2*d*e*x/f^2+e^2*x^2/f^2)^(1/2))^n,x, algorithm="fricas
")

[Out]

(a^2*f^4*n^4 + 24*a^2*f^4 - 48*a*d^2*f^2 + (e^4*n^4 - 4*e^4*n^2)*x^4 + 24*d^4 + 4*(d*e^3*n^4 - 4*d*e^3*n^2)*x^
3 - 4*(4*a^2*f^4 - 3*a*d^2*f^2)*n^2 + 2*((a*e^2*f^2 + 2*d^2*e^2)*n^4 - 2*(5*a*e^2*f^2 + d^2*e^2)*n^2)*x^2 + 4*
(a*d*e*f^2*n^4 - 2*(5*a*d*e*f^2 - 3*d^3*e)*n^2)*x - 4*(a*d*f^3*n^3 + (e^3*f*n^3 - 4*e^3*f*n)*x^3 + 3*(d*e^2*f*
n^3 - 4*d*e^2*f*n)*x^2 - 2*(5*a*d*f^3 - 3*d^3*f)*n + ((a*e*f^3 + 2*d^2*e*f)*n^3 - 2*(5*a*e*f^3 + d^2*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^3*n^5 - 20
*e*f^3*n^3 + 64*e*f^3*n)

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