3.17 \(\int \frac {1}{\sqrt {b \tanh (c+d x)}} \, dx\)

Optimal. Leaf size=57 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{\sqrt {b} d}+\frac {\tan ^{-1}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{\sqrt {b} d} \]

[Out]

arctan((b*tanh(d*x+c))^(1/2)/b^(1/2))/d/b^(1/2)+arctanh((b*tanh(d*x+c))^(1/2)/b^(1/2))/d/b^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3476, 329, 212, 206, 203} \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{\sqrt {b} d}+\frac {\tan ^{-1}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{\sqrt {b} d} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[b*Tanh[c + d*x]],x]

[Out]

ArcTan[Sqrt[b*Tanh[c + d*x]]/Sqrt[b]]/(Sqrt[b]*d) + ArcTanh[Sqrt[b*Tanh[c + d*x]]/Sqrt[b]]/(Sqrt[b]*d)

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])

Rule 206

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

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {b \tanh (c+d x)}} \, dx &=-\frac {b \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (-b^2+x^2\right )} \, dx,x,b \tanh (c+d x)\right )}{d}\\ &=-\frac {(2 b) \operatorname {Subst}\left (\int \frac {1}{-b^2+x^4} \, dx,x,\sqrt {b \tanh (c+d x)}\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{b-x^2} \, dx,x,\sqrt {b \tanh (c+d x)}\right )}{d}+\frac {\operatorname {Subst}\left (\int \frac {1}{b+x^2} \, dx,x,\sqrt {b \tanh (c+d x)}\right )}{d}\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{\sqrt {b} d}+\frac {\tanh ^{-1}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{\sqrt {b} d}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 49, normalized size = 0.86 \[ \frac {\sqrt {\tanh (c+d x)} \left (\tanh ^{-1}\left (\sqrt {\tanh (c+d x)}\right )+\tan ^{-1}\left (\sqrt {\tanh (c+d x)}\right )\right )}{d \sqrt {b \tanh (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[b*Tanh[c + d*x]],x]

[Out]

((ArcTan[Sqrt[Tanh[c + d*x]]] + ArcTanh[Sqrt[Tanh[c + d*x]]])*Sqrt[Tanh[c + d*x]])/(d*Sqrt[b*Tanh[c + d*x]])

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fricas [B]  time = 0.58, size = 599, normalized size = 10.51 \[ \left [-\frac {2 \, \sqrt {-b} \arctan \left (\frac {{\left (\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2}\right )} \sqrt {-b} \sqrt {\frac {b \sinh \left (d x + c\right )}{\cosh \left (d x + c\right )}}}{b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} - b}\right ) + \sqrt {-b} \log \left (-\frac {b \cosh \left (d x + c\right )^{4} + 4 \, b \cosh \left (d x + c\right )^{3} \sinh \left (d x + c\right ) + 6 \, b \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{2} + 4 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + b \sinh \left (d x + c\right )^{4} - 2 \, {\left (\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} + 1\right )} \sqrt {-b} \sqrt {\frac {b \sinh \left (d x + c\right )}{\cosh \left (d x + c\right )}} - 2 \, b}{\cosh \left (d x + c\right )^{4} + 4 \, \cosh \left (d x + c\right )^{3} \sinh \left (d x + c\right ) + 6 \, \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{2} + 4 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + \sinh \left (d x + c\right )^{4}}\right )}{4 \, b d}, -\frac {2 \, \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {\frac {b \sinh \left (d x + c\right )}{\cosh \left (d x + c\right )}}}{b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} - b}\right ) - \sqrt {b} \log \left (2 \, b \cosh \left (d x + c\right )^{4} + 8 \, b \cosh \left (d x + c\right )^{3} \sinh \left (d x + c\right ) + 12 \, b \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{2} + 8 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + 2 \, b \sinh \left (d x + c\right )^{4} + 2 \, {\left (\cosh \left (d x + c\right )^{4} + 4 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + \sinh \left (d x + c\right )^{4} + {\left (6 \, \cosh \left (d x + c\right )^{2} + 1\right )} \sinh \left (d x + c\right )^{2} + \cosh \left (d x + c\right )^{2} + 2 \, {\left (2 \, \cosh \left (d x + c\right )^{3} + \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )} \sqrt {b} \sqrt {\frac {b \sinh \left (d x + c\right )}{\cosh \left (d x + c\right )}} - b\right )}{4 \, b d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(2*sqrt(-b)*arctan((cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2)*sqrt(-b)*sqrt(b*s
inh(d*x + c)/cosh(d*x + c))/(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 - b)) + s
qrt(-b)*log(-(b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)^3*sinh(d*x + c) + 6*b*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*
b*cosh(d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 - 2*(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sin
h(d*x + c)^2 + 1)*sqrt(-b)*sqrt(b*sinh(d*x + c)/cosh(d*x + c)) - 2*b)/(cosh(d*x + c)^4 + 4*cosh(d*x + c)^3*sin
h(d*x + c) + 6*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*cosh(d*x + c)*sinh(d*x + c)^3 + sinh(d*x + c)^4)))/(b*d), -
1/4*(2*sqrt(b)*arctan(sqrt(b)*sqrt(b*sinh(d*x + c)/cosh(d*x + c))/(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(
d*x + c) + b*sinh(d*x + c)^2 - b)) - sqrt(b)*log(2*b*cosh(d*x + c)^4 + 8*b*cosh(d*x + c)^3*sinh(d*x + c) + 12*
b*cosh(d*x + c)^2*sinh(d*x + c)^2 + 8*b*cosh(d*x + c)*sinh(d*x + c)^3 + 2*b*sinh(d*x + c)^4 + 2*(cosh(d*x + c)
^4 + 4*cosh(d*x + c)*sinh(d*x + c)^3 + sinh(d*x + c)^4 + (6*cosh(d*x + c)^2 + 1)*sinh(d*x + c)^2 + cosh(d*x +
c)^2 + 2*(2*cosh(d*x + c)^3 + cosh(d*x + c))*sinh(d*x + c))*sqrt(b)*sqrt(b*sinh(d*x + c)/cosh(d*x + c)) - b))/
(b*d)]

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giac [A]  time = 0.28, size = 89, normalized size = 1.56 \[ \frac {\frac {2 \, \arctan \left (-\frac {\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}}{\sqrt {b}}\right )}{\sqrt {b}} - \frac {\log \left ({\left | -\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} + \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b} \right |}\right )}{\sqrt {b}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*arctan(-(sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b))/sqrt(b))/sqrt(b) - log(abs(-sqrt(b)*e^(
2*d*x + 2*c) + sqrt(b*e^(4*d*x + 4*c) - b)))/sqrt(b))/d

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maple [A]  time = 0.09, size = 46, normalized size = 0.81 \[ \frac {\arctan \left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{d \sqrt {b}}+\frac {\arctanh \left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{d \sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan((b*tanh(d*x+c))^(1/2)/b^(1/2))/d/b^(1/2)+arctanh((b*tanh(d*x+c))^(1/2)/b^(1/2))/d/b^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \tanh \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(b*tanh(d*x + c)), x)

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mupad [B]  time = 1.12, size = 38, normalized size = 0.67 \[ \frac {\mathrm {atan}\left (\frac {\sqrt {b\,\mathrm {tanh}\left (c+d\,x\right )}}{\sqrt {b}}\right )+\mathrm {atanh}\left (\frac {\sqrt {b\,\mathrm {tanh}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{\sqrt {b}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tanh(c + d*x))^(1/2),x)

[Out]

(atan((b*tanh(c + d*x))^(1/2)/b^(1/2)) + atanh((b*tanh(c + d*x))^(1/2)/b^(1/2)))/(b^(1/2)*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \tanh {\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(b*tanh(c + d*x)), x)

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