3.1.16 \(\int \sqrt {b \tanh (c+d x)} \, dx\) [16]

Optimal. Leaf size=58 \[ -\frac {\sqrt {b} \text {ArcTan}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{d}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{d} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 58, 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 = {3557, 335, 304, 209, 212} \begin {gather*} \frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{d}-\frac {\sqrt {b} \text {ArcTan}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*Tanh[c + d*x]],x]

[Out]

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

Rule 209

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

Rule 212

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

Rule 304

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

Rule 335

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 3557

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

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Mathematica [A]
time = 0.03, size = 51, normalized size = 0.88 \begin {gather*} \frac {\left (-\text {ArcTan}\left (\sqrt {\tanh (c+d x)}\right )+\tanh ^{-1}\left (\sqrt {\tanh (c+d x)}\right )\right ) \sqrt {b \tanh (c+d x)}}{d \sqrt {\tanh (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*Tanh[c + d*x]],x]

[Out]

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

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Maple [A]
time = 2.09, size = 48, normalized size = 0.83

method	result	size
derivativedivides	\(-\frac {2 b \left (\frac {\arctan \left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{2 \sqrt {b}}-\frac {\arctanh \left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{2 \sqrt {b}}\right )}{d}\)	\(48\)
default	\(-\frac {2 b \left (\frac {\arctan \left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{2 \sqrt {b}}-\frac {\arctanh \left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{2 \sqrt {b}}\right )}{d}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*tanh(d*x + c)), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (46) = 92\).
time = 0.41, size = 593, normalized size = 10.22 \begin {gather*} \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 \, 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 \, d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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)))/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*cos
h(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))/d]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {b \tanh {\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(b*tanh(c + d*x)), x)

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Giac [A]
time = 0.44, size = 88, normalized size = 1.52 \begin {gather*} -\frac {2 \, \sqrt {b} \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} \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 )}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.08, size = 41, normalized size = 0.71 \begin {gather*} -\frac {\sqrt {b}\,\left (\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 )\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tanh(c + d*x))^(1/2),x)

[Out]

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

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