3.5.28 \(\int \sqrt {a+a \sinh ^2(e+f x)} \tanh (e+f x) \, dx\) [428]

Optimal. Leaf size=18 \[ \frac {\sqrt {a \cosh ^2(e+f x)}}{f} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3255, 3284, 16, 32} \begin {gather*} \frac {\sqrt {a \cosh ^2(e+f x)}}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Sinh[e + f*x]^2]*Tanh[e + f*x],x]

[Out]

Sqrt[a*Cosh[e + f*x]^2]/f

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 3255

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*cos[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]

Rule 3284

Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFact
ors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[x^((m - 1)/2)*((b*ff^(n/2)*x^(n/2))^p/(1 - ff*x)
^((m + 1)/2)), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{b, e, f, p}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2
]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 18, normalized size = 1.00 \begin {gather*} \frac {\sqrt {a \cosh ^2(e+f x)}}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + a*Sinh[e + f*x]^2]*Tanh[e + f*x],x]

[Out]

Sqrt[a*Cosh[e + f*x]^2]/f

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Maple [A]
time = 0.55, size = 19, normalized size = 1.06

method	result	size
derivativedivides	\(\frac {\sqrt {a +a \left (\sinh ^{2}\left (f x +e \right )\right )}}{f}\)	\(19\)
default	\(\frac {\sqrt {a +a \left (\sinh ^{2}\left (f x +e \right )\right )}}{f}\)	\(19\)
risch	\(\frac {\sqrt {\left ({\mathrm e}^{2 f x +2 e}+1\right )^{2} a \,{\mathrm e}^{-2 f x -2 e}}\, {\mathrm e}^{2 f x +2 e}}{2 f \left ({\mathrm e}^{2 f x +2 e}+1\right )}+\frac {\sqrt {\left ({\mathrm e}^{2 f x +2 e}+1\right )^{2} a \,{\mathrm e}^{-2 f x -2 e}}}{2 f \left ({\mathrm e}^{2 f x +2 e}+1\right )}\)	\(99\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sinh(f*x+e)^2)^(1/2)*tanh(f*x+e),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.50, size = 34, normalized size = 1.89 \begin {gather*} \frac {\sqrt {a} e^{\left (f x + e\right )}}{2 \, f} + \frac {\sqrt {a} e^{\left (-f x - e\right )}}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 139 vs. \(2 (16) = 32\).
time = 0.54, size = 139, normalized size = 7.72 \begin {gather*} \frac {{\left (2 \, \cosh \left (f x + e\right ) e^{\left (f x + e\right )} \sinh \left (f x + e\right ) + e^{\left (f x + e\right )} \sinh \left (f x + e\right )^{2} + {\left (\cosh \left (f x + e\right )^{2} + 1\right )} e^{\left (f x + e\right )}\right )} \sqrt {a e^{\left (4 \, f x + 4 \, e\right )} + 2 \, a e^{\left (2 \, f x + 2 \, e\right )} + a} e^{\left (-f x - e\right )}}{2 \, {\left (f \cosh \left (f x + e\right ) e^{\left (2 \, f x + 2 \, e\right )} + f \cosh \left (f x + e\right ) + {\left (f e^{\left (2 \, f x + 2 \, e\right )} + f\right )} \sinh \left (f x + e\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*cosh(f*x + e)*e^(f*x + e)*sinh(f*x + e) + e^(f*x + e)*sinh(f*x + e)^2 + (cosh(f*x + e)^2 + 1)*e^(f*x +
e))*sqrt(a*e^(4*f*x + 4*e) + 2*a*e^(2*f*x + 2*e) + a)*e^(-f*x - e)/(f*cosh(f*x + e)*e^(2*f*x + 2*e) + f*cosh(f
*x + e) + (f*e^(2*f*x + 2*e) + f)*sinh(f*x + e))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a \left (\sinh ^{2}{\left (e + f x \right )} + 1\right )} \tanh {\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a*(sinh(e + f*x)**2 + 1))*tanh(e + f*x), x)

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Giac [A]
time = 0.41, size = 24, normalized size = 1.33 \begin {gather*} \frac {\sqrt {a} {\left (e^{\left (f x + e\right )} + e^{\left (-f x - e\right )}\right )}}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.92, size = 18, normalized size = 1.00 \begin {gather*} \frac {\sqrt {a\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a + a*sinh(e + f*x)^2)^(1/2)/f

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