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

Optimal. Leaf size=68 \[ \frac{1}{2} f x \sqrt{a+\frac{e^2 x^2}{f^2}}+\frac{a f^2 \tanh ^{-1}\left (\frac{e x}{f \sqrt{a+\frac{e^2 x^2}{f^2}}}\right )}{2 e}+d x+\frac{e x^2}{2} \]

[Out]

d*x + (e*x^2)/2 + (f*x*Sqrt[a + (e^2*x^2)/f^2])/2 + (a*f^2*ArcTanh[(e*x)/(f*Sqrt[a + (e^2*x^2)/f^2])])/(2*e)

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Rubi [A]  time = 0.0339407, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {195, 217, 206} \[ \frac{1}{2} f x \sqrt{a+\frac{e^2 x^2}{f^2}}+\frac{a f^2 \tanh ^{-1}\left (\frac{e x}{f \sqrt{a+\frac{e^2 x^2}{f^2}}}\right )}{2 e}+d x+\frac{e x^2}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

d*x + (e*x^2)/2 + (f*x*Sqrt[a + (e^2*x^2)/f^2])/2 + (a*f^2*ArcTanh[(e*x)/(f*Sqrt[a + (e^2*x^2)/f^2])])/(2*e)

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0449172, size = 81, normalized size = 1.19 \[ \frac{1}{2} f x \sqrt{\frac{a f^2+e^2 x^2}{f^2}}+\frac{a f^2 \log \left (e f \sqrt{\frac{a f^2+e^2 x^2}{f^2}}+e^2 x\right )}{2 e}+d x+\frac{e x^2}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

d*x + (e*x^2)/2 + (f*x*Sqrt[(a*f^2 + e^2*x^2)/f^2])/2 + (a*f^2*Log[e^2*x + e*f*Sqrt[(a*f^2 + e^2*x^2)/f^2]])/(
2*e)

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Maple [A]  time = 0.003, size = 75, normalized size = 1.1 \begin{align*} dx+{\frac{e{x}^{2}}{2}}+{\frac{fx}{2}\sqrt{a+{\frac{{e}^{2}{x}^{2}}{{f}^{2}}}}}+{\frac{af}{2}\ln \left ({\frac{{e}^{2}x}{{f}^{2}}{\frac{1}{\sqrt{{\frac{{e}^{2}}{{f}^{2}}}}}}}+\sqrt{a+{\frac{{e}^{2}{x}^{2}}{{f}^{2}}}} \right ){\frac{1}{\sqrt{{\frac{{e}^{2}}{{f}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.984576, size = 153, normalized size = 2.25 \begin{align*} \frac{e^{2} x^{2} - a f^{2} \log \left (-e x + f \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}}\right ) + e f x \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}} + 2 \, d e x}{2 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.15634, size = 54, normalized size = 0.79 \begin{align*} d x + \frac{e x^{2}}{2} + f \left (\frac{\sqrt{a} x \sqrt{1 + \frac{e^{2} x^{2}}{a f^{2}}}}{2} + \frac{a f \operatorname{asinh}{\left (\frac{e x}{\sqrt{a} f} \right )}}{2 e}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*x + e*x**2/2 + f*(sqrt(a)*x*sqrt(1 + e**2*x**2/(a*f**2))/2 + a*f*asinh(e*x/(sqrt(a)*f))/(2*e))

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Giac [A]  time = 1.15571, size = 88, normalized size = 1.29 \begin{align*} \frac{1}{2} \, x^{2} e + d x - \frac{{\left (a f^{2} e^{\left (-1\right )} \log \left ({\left | -x e + \sqrt{a f^{2} + x^{2} e^{2}} \right |}\right ) - \sqrt{a f^{2} + x^{2} e^{2}} x\right )}{\left | f \right |}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*e + d*x - 1/2*(a*f^2*e^(-1)*log(abs(-x*e + sqrt(a*f^2 + x^2*e^2))) - sqrt(a*f^2 + x^2*e^2)*x)*abs(f)/f