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

Optimal. Leaf size=171 \[ -\frac {\left (d^2-a f^2\right )^2 \left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^{n-2}}{4 e f (2-n)}-\frac {\left (d^2-a f^2\right ) \left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^n}{2 e f n}+\frac {\left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^{n+2}}{4 e f (n+2)} \]

[Out]

-1/4*(-a*f^2+d^2)^2*(d+e*x+f*(a+2*d*e*x/f^2+e^2*x^2/f^2)^(1/2))^(-2+n)/e/f/(2-n)-1/2*(-a*f^2+d^2)*(d+e*x+f*(a+
2*d*e*x/f^2+e^2*x^2/f^2)^(1/2))^n/e/f/n+1/4*(d+e*x+f*(a+2*d*e*x/f^2+e^2*x^2/f^2)^(1/2))^(2+n)/e/f/(2+n)

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Rubi [A]  time = 0.32, antiderivative size = 171, normalized size of antiderivative = 1.00, 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} \[ -\frac {\left (d^2-a f^2\right )^2 \left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^{n-2}}{4 e f (2-n)}-\frac {\left (d^2-a f^2\right ) \left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^n}{2 e f n}+\frac {\left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^{n+2}}{4 e f (n+2)} \]

Antiderivative was successfully verified.

[In]

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

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]

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])

Rubi steps

\begin {align*} \int \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^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^{-3+n} \left (d^2 e-\left (-a e+\frac {2 d^2 e}{f^2}\right ) f^2+e x^2\right )^2}{8 e^3} \, dx,x,d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int x^{-3+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 )}{4 e^3 f}\\ &=\frac {\operatorname {Subst}\left (\int \left (e^2 \left (d^2-a f^2\right )^2 x^{-3+n}-2 e^2 \left (d^2-a f^2\right ) x^{-1+n}+e^2 x^{1+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 )}{4 e^3 f}\\ &=-\frac {\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 )^{-2+n}}{4 e f (2-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 )^n}{2 e f n}+\frac {\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 (2+n)}\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 135, normalized size = 0.79 \[ \frac {\left (f \sqrt {a+\frac {e x (2 d+e x)}{f^2}}+d+e x\right )^n \left (\frac {\left (d^2-a f^2\right )^2}{(n-2) \left (f \sqrt {a+\frac {e x (2 d+e x)}{f^2}}+d+e x\right )^2}+\frac {2 \left (a f^2-d^2\right )}{n}+\frac {\left (f \sqrt {a+\frac {e x (2 d+e x)}{f^2}}+d+e x\right )^2}{n+2}\right )}{4 e f} \]

Antiderivative was successfully verified.

[In]

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

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fricas [A]  time = 0.45, size = 122, normalized size = 0.71 \[ \frac {{\left (e^{2} n^{2} x^{2} + a f^{2} n^{2} + 2 \, d e n^{2} x - 2 \, a f^{2} + 2 \, d^{2} - 2 \, {\left (e f n x + d f n\right )} \sqrt {\frac {e^{2} x^{2} + a f^{2} + 2 \, d e x}{f^{2}}}\right )} {\left (e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2} + 2 \, d e x}{f^{2}}} + d\right )}^{n}}{e f n^{3} - 4 \, e f n} \]

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

[Out]

(e^2*n^2*x^2 + a*f^2*n^2 + 2*d*e*n^2*x - 2*a*f^2 + 2*d^2 - 2*(e*f*n*x + d*f*n)*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*n^3 - 4*e*f*n)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\frac {e^{2} x^{2}}{f^{2}} + a + \frac {2 \, d e x}{f^{2}}} {\left (e x + \sqrt {\frac {e^{2} x^{2}}{f^{2}} + a + \frac {2 \, d e x}{f^{2}}} f + d\right )}^{n}\,{d x} \]

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)^(1/2)*(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(sqrt(e^2*x^2/f^2 + a + 2*d*e*x/f^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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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[ \int \sqrt {\frac {e^{2} x^{2}}{f^{2}}+a +\frac {2 d e x}{f^{2}}}\, \left (e x +d +\sqrt {\frac {e^{2} x^{2}}{f^{2}}+a +\frac {2 d e x}{f^{2}}}\, f \right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\frac {e^{2} x^{2}}{f^{2}} + a + \frac {2 \, d e x}{f^{2}}} {\left (e x + \sqrt {\frac {e^{2} x^{2}}{f^{2}} + a + \frac {2 \, d e x}{f^{2}}} f + d\right )}^{n}\,{d x} \]

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)^(1/2)*(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(sqrt(e^2*x^2/f^2 + a + 2*d*e*x/f^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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (d+f\,\sqrt {a+\frac {e^2\,x^2}{f^2}+\frac {2\,d\,e\,x}{f^2}}+e\,x\right )}^n\,\sqrt {a+\frac {e^2\,x^2}{f^2}+\frac {2\,d\,e\,x}{f^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + \frac {2 d e x}{f^{2}} + \frac {e^{2} x^{2}}{f^{2}}} \left (d + e x + f \sqrt {a + \frac {2 d e x}{f^{2}} + \frac {e^{2} x^{2}}{f^{2}}}\right )^{n}\, dx \]

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

[Out]

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

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