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

Optimal. Leaf size=107 \[ \frac {\left (d^2-a f^2\right ) \left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^{n-1}}{2 e (1-n)}+\frac {\left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^{n+1}}{2 e (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2116, 12, 14} \[ \frac {\left (d^2-a f^2\right ) \left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^{n-1}}{2 e (1-n)}+\frac {\left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^{n+1}}{2 e (n+1)} \]

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,x]

[Out]

((d^2 - a*f^2)*(d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^(-1 + n))/(2*e*(1 - n)) + (d + e*x + f*Sq
rt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^(1 + n)/(2*e*(1 + 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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2116

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2])^(n_))^(p_.), x_Symbol]
:> Dist[2, Subst[Int[((g + h*x^n)^p*(d^2*e - (b*d - a*e)*f^2 - (2*d*e - b*f^2)*x + e*x^2))/(-2*d*e + b*f^2 +
2*e*x)^2, x], x, d + e*x + f*Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c, d, e, f, g, h, n}, x] && EqQ[e^2 -
c*f^2, 0] && IntegerQ[p]

Rubi steps

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

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Mathematica [A]  time = 0.39, size = 89, normalized size = 0.83 \[ \frac {\left (f \sqrt {a+\frac {e x (2 d+e x)}{f^2}}+d+e x\right )^{n-1} \left (\frac {a f^2-d^2}{n-1}+\frac {\left (f \sqrt {a+\frac {e x (2 d+e x)}{f^2}}+d+e x\right )^2}{n+1}\right )}{2 e} \]

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,x]

[Out]

((d + e*x + f*Sqrt[a + (e*x*(2*d + e*x))/f^2])^(-1 + n)*((-d^2 + a*f^2)/(-1 + n) + (d + e*x + f*Sqrt[a + (e*x*
(2*d + e*x))/f^2])^2/(1 + n)))/(2*e)

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fricas [A]  time = 0.47, size = 80, normalized size = 0.75 \[ \frac {{\left (f n \sqrt {\frac {e^{2} x^{2} + a f^{2} + 2 \, d e x}{f^{2}}} - e x - d\right )} {\left (e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2} + 2 \, d e x}{f^{2}}} + d\right )}^{n}}{e n^{2} - e} \]

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,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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*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 = 0.01, size = 0, normalized size = 0.00 \[ \int \left (e x +d +\sqrt {\frac {e^{2} x^{2}}{f^{2}}+a +\frac {2 d e x}{f^{2}}}\, f \right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((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 \[ \int {\left (e x + \sqrt {\frac {e^{2} x^{2}}{f^{2}} + a + \frac {2 \, d e x}{f^{2}}} f + d\right )}^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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*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 \[ \int {\left (d+f\,\sqrt {a+\frac {e^2\,x^2}{f^2}+\frac {2\,d\,e\,x}{f^2}}+e\,x\right )}^n \,d x \]

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,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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