3.731 \(\int \frac{e^{3 x}}{\sqrt{25+16 e^{2 x}}} \, dx\)

Optimal. Leaf size=33 \[ \frac{1}{32} e^x \sqrt{16 e^{2 x}+25}-\frac{25}{128} \sinh ^{-1}\left (\frac{4 e^x}{5}\right ) \]

[Out]

(E^x*Sqrt[25 + 16*E^(2*x)])/32 - (25*ArcSinh[(4*E^x)/5])/128

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Rubi [A]  time = 0.032377, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2248, 321, 215} \[ \frac{1}{32} e^x \sqrt{16 e^{2 x}+25}-\frac{25}{128} \sinh ^{-1}\left (\frac{4 e^x}{5}\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^(3*x)/Sqrt[25 + 16*E^(2*x)],x]

[Out]

(E^x*Sqrt[25 + 16*E^(2*x)])/32 - (25*ArcSinh[(4*E^x)/5])/128

Rule 2248

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Wit
h[{m = FullSimplify[(g*h*Log[G])/(d*e*Log[F])]}, Dist[(Denominator[m]*G^(f*h - (c*g*h)/d))/(d*e*Log[F]), Subst
[Int[x^(Numerator[m] - 1)*(a + b*x^Denominator[m])^p, x], x, F^((e*(c + d*x))/Denominator[m])], x] /; LeQ[m, -
1] || GeQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 215

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

Rubi steps

\begin{align*} \int \frac{e^{3 x}}{\sqrt{25+16 e^{2 x}}} \, dx &=\operatorname{Subst}\left (\int \frac{x^2}{\sqrt{25+16 x^2}} \, dx,x,e^x\right )\\ &=\frac{1}{32} e^x \sqrt{25+16 e^{2 x}}-\frac{25}{32} \operatorname{Subst}\left (\int \frac{1}{\sqrt{25+16 x^2}} \, dx,x,e^x\right )\\ &=\frac{1}{32} e^x \sqrt{25+16 e^{2 x}}-\frac{25}{128} \sinh ^{-1}\left (\frac{4 e^x}{5}\right )\\ \end{align*}

Mathematica [A]  time = 0.0156171, size = 33, normalized size = 1. \[ \frac{1}{32} e^x \sqrt{16 e^{2 x}+25}-\frac{25}{128} \sinh ^{-1}\left (\frac{4 e^x}{5}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^(3*x)/Sqrt[25 + 16*E^(2*x)],x]

[Out]

(E^x*Sqrt[25 + 16*E^(2*x)])/32 - (25*ArcSinh[(4*E^x)/5])/128

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Maple [A]  time = 0.065, size = 23, normalized size = 0.7 \begin{align*}{\frac{{{\rm e}^{x}}}{32}\sqrt{25+16\, \left ({{\rm e}^{x}} \right ) ^{2}}}-{\frac{25}{128}{\it Arcsinh} \left ({\frac{4\,{{\rm e}^{x}}}{5}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(3*x)/(25+16*exp(2*x))^(1/2),x)

[Out]

1/32*exp(x)*(25+16*exp(x)^2)^(1/2)-25/128*arcsinh(4/5*exp(x))

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Maxima [B]  time = 1.01677, size = 100, normalized size = 3.03 \begin{align*} \frac{25 \, \sqrt{16 \, e^{\left (2 \, x\right )} + 25} e^{\left (-x\right )}}{32 \,{\left ({\left (16 \, e^{\left (2 \, x\right )} + 25\right )} e^{\left (-2 \, x\right )} - 16\right )}} - \frac{25}{256} \, \log \left (\sqrt{16 \, e^{\left (2 \, x\right )} + 25} e^{\left (-x\right )} + 4\right ) + \frac{25}{256} \, \log \left (\sqrt{16 \, e^{\left (2 \, x\right )} + 25} e^{\left (-x\right )} - 4\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(3*x)/(25+16*exp(2*x))^(1/2),x, algorithm="maxima")

[Out]

25/32*sqrt(16*e^(2*x) + 25)*e^(-x)/((16*e^(2*x) + 25)*e^(-2*x) - 16) - 25/256*log(sqrt(16*e^(2*x) + 25)*e^(-x)
+ 4) + 25/256*log(sqrt(16*e^(2*x) + 25)*e^(-x) - 4)

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Fricas [A]  time = 0.886718, size = 103, normalized size = 3.12 \begin{align*} \frac{1}{32} \, \sqrt{16 \, e^{\left (2 \, x\right )} + 25} e^{x} + \frac{25}{128} \, \log \left (\sqrt{16 \, e^{\left (2 \, x\right )} + 25} - 4 \, e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(3*x)/(25+16*exp(2*x))^(1/2),x, algorithm="fricas")

[Out]

1/32*sqrt(16*e^(2*x) + 25)*e^x + 25/128*log(sqrt(16*e^(2*x) + 25) - 4*e^x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{3 x}}{\sqrt{16 e^{2 x} + 25}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(3*x)/(25+16*exp(2*x))**(1/2),x)

[Out]

Integral(exp(3*x)/sqrt(16*exp(2*x) + 25), x)

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Giac [A]  time = 1.24746, size = 45, normalized size = 1.36 \begin{align*} \frac{1}{32} \, \sqrt{16 \, e^{\left (2 \, x\right )} + 25} e^{x} + \frac{25}{128} \, \log \left (\sqrt{16 \, e^{\left (2 \, x\right )} + 25} - 4 \, e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(3*x)/(25+16*exp(2*x))^(1/2),x, algorithm="giac")

[Out]

1/32*sqrt(16*e^(2*x) + 25)*e^x + 25/128*log(sqrt(16*e^(2*x) + 25) - 4*e^x)