3.3.46 \(\int \cos (\tanh (a+b x)) \, dx\) [246]

Optimal. Leaf size=77 \[ -\frac {\cos (1) \text {CosIntegral}(1-\tanh (a+b x))}{2 b}+\frac {\cos (1) \text {CosIntegral}(1+\tanh (a+b x))}{2 b}-\frac {\sin (1) \text {Si}(1-\tanh (a+b x))}{2 b}+\frac {\sin (1) \text {Si}(1+\tanh (a+b x))}{2 b} \]

[Out]

-1/2*Ci(1-tanh(b*x+a))*cos(1)/b+1/2*Ci(1+tanh(b*x+a))*cos(1)/b+1/2*Si(-1+tanh(b*x+a))*sin(1)/b+1/2*Si(1+tanh(b
*x+a))*sin(1)/b

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Rubi [A]
time = 0.11, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3415, 3384, 3380, 3383} \begin {gather*} -\frac {\cos (1) \text {CosIntegral}(1-\tanh (a+b x))}{2 b}+\frac {\cos (1) \text {CosIntegral}(\tanh (a+b x)+1)}{2 b}-\frac {\sin (1) \text {Si}(1-\tanh (a+b x))}{2 b}+\frac {\sin (1) \text {Si}(\tanh (a+b x)+1)}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[Tanh[a + b*x]],x]

[Out]

-1/2*(Cos[1]*CosIntegral[1 - Tanh[a + b*x]])/b + (Cos[1]*CosIntegral[1 + Tanh[a + b*x]])/(2*b) - (Sin[1]*SinIn
tegral[1 - Tanh[a + b*x]])/(2*b) + (Sin[1]*SinIntegral[1 + Tanh[a + b*x]])/(2*b)

Rule 3380

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3383

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3384

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3415

Int[Cos[(c_.) + (d_.)*(x_)]*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[Cos[c + d*x], (a +
b*x^n)^p, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[p, 0] && IGtQ[n, 0] && (EqQ[n, 2] || EqQ[p, -1])

Rubi steps

\begin {align*} \int \cos (\tanh (a+b x)) \, dx &=\frac {\text {Subst}\left (\int \frac {\cos (x)}{1-x^2} \, dx,x,\tanh (a+b x)\right )}{b}\\ &=\frac {\text {Subst}\left (\int \left (\frac {\cos (x)}{2 (1-x)}+\frac {\cos (x)}{2 (1+x)}\right ) \, dx,x,\tanh (a+b x)\right )}{b}\\ &=\frac {\text {Subst}\left (\int \frac {\cos (x)}{1-x} \, dx,x,\tanh (a+b x)\right )}{2 b}+\frac {\text {Subst}\left (\int \frac {\cos (x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{2 b}\\ &=\frac {\cos (1) \text {Subst}\left (\int \frac {\cos (1-x)}{1-x} \, dx,x,\tanh (a+b x)\right )}{2 b}+\frac {\cos (1) \text {Subst}\left (\int \frac {\cos (1+x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{2 b}+\frac {\sin (1) \text {Subst}\left (\int \frac {\sin (1-x)}{1-x} \, dx,x,\tanh (a+b x)\right )}{2 b}+\frac {\sin (1) \text {Subst}\left (\int \frac {\sin (1+x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{2 b}\\ &=-\frac {\cos (1) \text {Ci}(1-\tanh (a+b x))}{2 b}+\frac {\cos (1) \text {Ci}(1+\tanh (a+b x))}{2 b}-\frac {\sin (1) \text {Si}(1-\tanh (a+b x))}{2 b}+\frac {\sin (1) \text {Si}(1+\tanh (a+b x))}{2 b}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 62, normalized size = 0.81 \begin {gather*} \frac {-\cos (1) \text {CosIntegral}(1-\tanh (a+b x))+\cos (1) \text {CosIntegral}(1+\tanh (a+b x))-\sin (1) \text {Si}(1-\tanh (a+b x))+\sin (1) \text {Si}(1+\tanh (a+b x))}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[Tanh[a + b*x]],x]

[Out]

(-(Cos[1]*CosIntegral[1 - Tanh[a + b*x]]) + Cos[1]*CosIntegral[1 + Tanh[a + b*x]] - Sin[1]*SinIntegral[1 - Tan
h[a + b*x]] + Sin[1]*SinIntegral[1 + Tanh[a + b*x]])/(2*b)

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Maple [A]
time = 0.99, size = 58, normalized size = 0.75

method	result	size
derivativedivides	\(\frac {\frac {\sinIntegral \left (-1+\tanh \left (b x +a \right )\right ) \sin \left (1\right )}{2}-\frac {\cosineIntegral \left (-1+\tanh \left (b x +a \right )\right ) \cos \left (1\right )}{2}+\frac {\sinIntegral \left (\tanh \left (b x +a \right )+1\right ) \sin \left (1\right )}{2}+\frac {\cosineIntegral \left (\tanh \left (b x +a \right )+1\right ) \cos \left (1\right )}{2}}{b}\)	\(58\)
default	\(\frac {\frac {\sinIntegral \left (-1+\tanh \left (b x +a \right )\right ) \sin \left (1\right )}{2}-\frac {\cosineIntegral \left (-1+\tanh \left (b x +a \right )\right ) \cos \left (1\right )}{2}+\frac {\sinIntegral \left (\tanh \left (b x +a \right )+1\right ) \sin \left (1\right )}{2}+\frac {\cosineIntegral \left (\tanh \left (b x +a \right )+1\right ) \cos \left (1\right )}{2}}{b}\)	\(58\)
risch	\(\frac {{\mathrm e}^{-i} \expIntegral \left (1, -\frac {2 i}{{\mathrm e}^{2 b x +2 a}+1}\right )}{4 b}-\frac {{\mathrm e}^{i} \expIntegral \left (1, -\frac {2 i}{{\mathrm e}^{2 b x +2 a}+1}+2 i\right )}{4 b}+\frac {{\mathrm e}^{i} \expIntegral \left (1, \frac {2 i}{{\mathrm e}^{2 b x +2 a}+1}\right )}{4 b}-\frac {{\mathrm e}^{-i} \expIntegral \left (1, \frac {2 i}{{\mathrm e}^{2 b x +2 a}+1}-2 i\right )}{4 b}\)	\(112\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(tanh(b*x+a)),x,method=_RETURNVERBOSE)

[Out]

1/b*(1/2*Si(-1+tanh(b*x+a))*sin(1)-1/2*Ci(-1+tanh(b*x+a))*cos(1)+1/2*Si(tanh(b*x+a)+1)*sin(1)+1/2*Ci(tanh(b*x+
a)+1)*cos(1))

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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(cos(tanh(b*x+a)),x, algorithm="maxima")

[Out]

integrate(cos(tanh(b*x + a)), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (67) = 134\).
time = 0.37, size = 151, normalized size = 1.96 \begin {gather*} \frac {\cos \left (1\right ) \operatorname {Ci}\left (\frac {2 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) + \cos \left (1\right ) \operatorname {Ci}\left (-\frac {2 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) - \cos \left (1\right ) \operatorname {Ci}\left (\frac {2}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) - \cos \left (1\right ) \operatorname {Ci}\left (-\frac {2}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) + 2 \, \sin \left (1\right ) \operatorname {Si}\left (\frac {2 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) - 2 \, \sin \left (1\right ) \operatorname {Si}\left (\frac {2}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right )}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(tanh(b*x+a)),x, algorithm="fricas")

[Out]

1/4*(cos(1)*cos_integral(2*e^(2*b*x + 2*a)/(e^(2*b*x + 2*a) + 1)) + cos(1)*cos_integral(-2*e^(2*b*x + 2*a)/(e^
(2*b*x + 2*a) + 1)) - cos(1)*cos_integral(2/(e^(2*b*x + 2*a) + 1)) - cos(1)*cos_integral(-2/(e^(2*b*x + 2*a) +
1)) + 2*sin(1)*sin_integral(2*e^(2*b*x + 2*a)/(e^(2*b*x + 2*a) + 1)) - 2*sin(1)*sin_integral(2/(e^(2*b*x + 2*
a) + 1)))/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(tanh(b*x+a)),x)

[Out]

Integral(cos(tanh(a + b*x)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(tanh(b*x+a)),x, algorithm="giac")

[Out]

integrate(cos(tanh(b*x + a)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \cos \left (\mathrm {tanh}\left (a+b\,x\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(tanh(a + b*x)),x)

[Out]

int(cos(tanh(a + b*x)), x)

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