3.8.31 \(\int \frac {e^{3 x}}{\sqrt {25+16 e^{2 x}}} \, dx\) [731]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 33, normalized size of antiderivative = 1.00, 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 = {2280, 327, 221} \begin {gather*} \frac {1}{32} e^x \sqrt {16 e^{2 x}+25}-\frac {25}{128} \sinh ^{-1}\left (\frac {4 e^x}{5}\right ) \end {gather*}

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 221

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]

Rule 327

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 2280

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]

Rubi steps

\begin {align*} \int \frac {e^{3 x}}{\sqrt {25+16 e^{2 x}}} \, dx &=\text {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} \text {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*}

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Mathematica [A]
time = 0.03, size = 33, normalized size = 1.00 \begin {gather*} \frac {1}{32} e^x \sqrt {25+16 e^{2 x}}-\frac {25}{128} \sinh ^{-1}\left (\frac {4 e^x}{5}\right ) \end {gather*}

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 = 0.03, size = 23, normalized size = 0.70

method	result	size
default	\(-\frac {25 \arcsinh \left (\frac {4 \,{\mathrm e}^{x}}{5}\right )}{128}+\frac {{\mathrm e}^{x} \sqrt {25+16 \,{\mathrm e}^{2 x}}}{32}\)	\(23\)
risch	\(-\frac {25 \arcsinh \left (\frac {4 \,{\mathrm e}^{x}}{5}\right )}{128}+\frac {{\mathrm e}^{x} \sqrt {25+16 \,{\mathrm e}^{2 x}}}{32}\)	\(23\)
meijerg	error in int/gbinthm/express: unable to compute coeff\	N/A

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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] Leaf count of result is larger than twice the leaf count of optimal. 74 vs. \(2 (22) = 44\).
time = 0.29, size = 74, normalized size = 2.24 \begin {gather*} \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 {gather*}

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 = 0.38, size = 33, normalized size = 1.00 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{3 x}}{\sqrt {16 e^{2 x} + 25}}\, dx \end {gather*}

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 = 3.54, size = 33, normalized size = 1.00 \begin {gather*} \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 {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{3\,x}}{\sqrt {16\,{\mathrm {e}}^{2\,x}+25}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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