3.481 \(\int \frac {\tanh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx\)

Optimal. Leaf size=41 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sinh ^2(e+f x)}}{\sqrt {a-b}}\right )}{f \sqrt {a-b}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3194, 63, 208} \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sinh ^2(e+f x)}}{\sqrt {a-b}}\right )}{f \sqrt {a-b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[Sqrt[a + b*Sinh[e + f*x]^2]/Sqrt[a - b]]/(Sqrt[a - b]*f))

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 3194

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

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 44, normalized size = 1.07 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \cosh ^2(e+f x)-b}}{\sqrt {a-b}}\right )}{f \sqrt {a-b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[Sqrt[a - b + b*Cosh[e + f*x]^2]/Sqrt[a - b]]/(Sqrt[a - b]*f))

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fricas [B]  time = 0.83, size = 433, normalized size = 10.56 \[ \left [\frac {\log \left (\frac {b \cosh \left (f x + e\right )^{4} + 4 \, b \cosh \left (f x + e\right ) \sinh \left (f x + e\right )^{3} + b \sinh \left (f x + e\right )^{4} + 2 \, {\left (4 \, a - 3 \, b\right )} \cosh \left (f x + e\right )^{2} + 2 \, {\left (3 \, b \cosh \left (f x + e\right )^{2} + 4 \, a - 3 \, b\right )} \sinh \left (f x + e\right )^{2} - 4 \, \sqrt {2} \sqrt {a - b} \sqrt {\frac {b \cosh \left (f x + e\right )^{2} + b \sinh \left (f x + e\right )^{2} + 2 \, a - b}{\cosh \left (f x + e\right )^{2} - 2 \, \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + \sinh \left (f x + e\right )^{2}}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )} + 4 \, {\left (b \cosh \left (f x + e\right )^{3} + {\left (4 \, a - 3 \, b\right )} \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right ) + b}{\cosh \left (f x + e\right )^{4} + 4 \, \cosh \left (f x + e\right ) \sinh \left (f x + e\right )^{3} + \sinh \left (f x + e\right )^{4} + 2 \, {\left (3 \, \cosh \left (f x + e\right )^{2} + 1\right )} \sinh \left (f x + e\right )^{2} + 2 \, \cosh \left (f x + e\right )^{2} + 4 \, {\left (\cosh \left (f x + e\right )^{3} + \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right ) + 1}\right )}{2 \, \sqrt {a - b} f}, -\frac {\sqrt {-a + b} \arctan \left (-\frac {\sqrt {2} \sqrt {-a + b} \sqrt {\frac {b \cosh \left (f x + e\right )^{2} + b \sinh \left (f x + e\right )^{2} + 2 \, a - b}{\cosh \left (f x + e\right )^{2} - 2 \, \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + \sinh \left (f x + e\right )^{2}}}}{2 \, {\left ({\left (a - b\right )} \cosh \left (f x + e\right ) + {\left (a - b\right )} \sinh \left (f x + e\right )\right )}}\right )}{{\left (a - b\right )} f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*log((b*cosh(f*x + e)^4 + 4*b*cosh(f*x + e)*sinh(f*x + e)^3 + b*sinh(f*x + e)^4 + 2*(4*a - 3*b)*cosh(f*x +
e)^2 + 2*(3*b*cosh(f*x + e)^2 + 4*a - 3*b)*sinh(f*x + e)^2 - 4*sqrt(2)*sqrt(a - b)*sqrt((b*cosh(f*x + e)^2 +
b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2))*(cosh(f*x +
e) + sinh(f*x + e)) + 4*(b*cosh(f*x + e)^3 + (4*a - 3*b)*cosh(f*x + e))*sinh(f*x + e) + b)/(cosh(f*x + e)^4 +
4*cosh(f*x + e)*sinh(f*x + e)^3 + sinh(f*x + e)^4 + 2*(3*cosh(f*x + e)^2 + 1)*sinh(f*x + e)^2 + 2*cosh(f*x + e
)^2 + 4*(cosh(f*x + e)^3 + cosh(f*x + e))*sinh(f*x + e) + 1))/(sqrt(a - b)*f), -sqrt(-a + b)*arctan(-1/2*sqrt(
2)*sqrt(-a + b)*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh
(f*x + e) + sinh(f*x + e)^2))/((a - b)*cosh(f*x + e) + (a - b)*sinh(f*x + e)))/((a - b)*f)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]Error: Bad Argument Type

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maple [C]  time = 0.18, size = 41, normalized size = 1.00 \[ \frac {\mathit {`\,int/indef0`\,}\left (\frac {\sinh \left (f x +e \right )}{\cosh \left (f x +e \right )^{2} \sqrt {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}}, \sinh \left (f x +e \right )\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

`int/indef0`(sinh(f*x+e)/cosh(f*x+e)^2/(a+b*sinh(f*x+e)^2)^(1/2),sinh(f*x+e))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(tanh(f*x + e)/sqrt(b*sinh(f*x + e)^2 + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {tanh}\left (e+f\,x\right )}{\sqrt {b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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