3.1037 \(\int \frac{\tanh (c+d x)}{\sqrt{a \sinh ^2(c+d x)}} \, dx\)

Optimal. Leaf size=30 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a \sinh ^2(c+d x)}}{\sqrt{a}}\right )}{\sqrt{a} d} \]

[Out]

ArcTan[Sqrt[a*Sinh[c + d*x]^2]/Sqrt[a]]/(Sqrt[a]*d)

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Rubi [A]  time = 0.0385723, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3205, 63, 203} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a \sinh ^2(c+d x)}}{\sqrt{a}}\right )}{\sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[c + d*x]/Sqrt[a*Sinh[c + d*x]^2],x]

[Out]

ArcTan[Sqrt[a*Sinh[c + d*x]^2]/Sqrt[a]]/(Sqrt[a]*d)

Rule 3205

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
]

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 203

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

Rubi steps

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

Mathematica [A]  time = 0.0371999, size = 31, normalized size = 1.03 \[ \frac{\sinh (c+d x) \tan ^{-1}(\sinh (c+d x))}{d \sqrt{a \sinh ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[c + d*x]/Sqrt[a*Sinh[c + d*x]^2],x]

[Out]

(ArcTan[Sinh[c + d*x]]*Sinh[c + d*x])/(d*Sqrt[a*Sinh[c + d*x]^2])

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Maple [C]  time = 0.114, size = 39, normalized size = 1.3 \begin{align*}{\frac{1}{d}\mbox{{\tt ` int/indef0`}} \left ({\frac{\sinh \left ( dx+c \right ) }{ \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}{\frac{1}{\sqrt{a \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}}}},\sinh \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(d*x+c)/(a*sinh(d*x+c)^2)^(1/2),x)

[Out]

`int/indef0`(sinh(d*x+c)/cosh(d*x+c)^2/(a*sinh(d*x+c)^2)^(1/2),sinh(d*x+c))/d

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Maxima [A]  time = 1.63042, size = 24, normalized size = 0.8 \begin{align*} \frac{2 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{\sqrt{a} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(e^(-d*x - c))/(sqrt(a)*d)

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Fricas [B]  time = 1.88352, size = 842, normalized size = 28.07 \begin{align*} \left [-\frac{\sqrt{-a} \log \left (-\frac{a \cosh \left (d x + c\right )^{2} + 2 \, \sqrt{a e^{\left (4 \, d x + 4 \, c\right )} - 2 \, a e^{\left (2 \, d x + 2 \, c\right )} + a}{\left (\cosh \left (d x + c\right ) e^{\left (d x + c\right )} + e^{\left (d x + c\right )} \sinh \left (d x + c\right )\right )} \sqrt{-a} e^{\left (-d x - c\right )} -{\left (a e^{\left (2 \, d x + 2 \, c\right )} - a\right )} \sinh \left (d x + c\right )^{2} -{\left (a \cosh \left (d x + c\right )^{2} - a\right )} e^{\left (2 \, d x + 2 \, c\right )} - 2 \,{\left (a \cosh \left (d x + c\right ) e^{\left (2 \, d x + 2 \, c\right )} - a \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - a}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} \sinh \left (d x + c\right )^{2} - \cosh \left (d x + c\right )^{2} +{\left (\cosh \left (d x + c\right )^{2} + 1\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \,{\left (\cosh \left (d x + c\right ) e^{\left (2 \, d x + 2 \, c\right )} - \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - 1}\right )}{a d}, \frac{2 \, \sqrt{a e^{\left (4 \, d x + 4 \, c\right )} - 2 \, a e^{\left (2 \, d x + 2 \, c\right )} + a} \arctan \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right )}{a d e^{\left (2 \, d x + 2 \, c\right )} - a d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh{\left (c + d x \right )}}{\sqrt{a \sinh ^{2}{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)/(a*sinh(d*x+c)**2)**(1/2),x)

[Out]

Integral(tanh(c + d*x)/sqrt(a*sinh(c + d*x)**2), x)

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Giac [A]  time = 1.2259, size = 49, normalized size = 1.63 \begin{align*} \frac{2 \, \arctan \left (e^{\left (d x + c\right )}\right )}{\sqrt{a} d \mathrm{sgn}\left (e^{\left (3 \, d x + 3 \, c\right )} - e^{\left (d x + c\right )}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(e^(d*x + c))/(sqrt(a)*d*sgn(e^(3*d*x + 3*c) - e^(d*x + c)))