\(\int \sin ^3(\tanh (a+b x)) \, dx\) [240]

Optimal result

Integrand size = 9, antiderivative size = 157 \[ \int \sin ^3(\tanh (a+b x)) \, dx=-\frac {3 \operatorname {CosIntegral}(1-\tanh (a+b x)) \sin (1)}{8 b}-\frac {3 \operatorname {CosIntegral}(1+\tanh (a+b x)) \sin (1)}{8 b}+\frac {\operatorname {CosIntegral}(3-3 \tanh (a+b x)) \sin (3)}{8 b}+\frac {\operatorname {CosIntegral}(3+3 \tanh (a+b x)) \sin (3)}{8 b}-\frac {\cos (3) \text {Si}(3-3 \tanh (a+b x))}{8 b}+\frac {3 \cos (1) \text {Si}(1-\tanh (a+b x))}{8 b}+\frac {3 \cos (1) \text {Si}(1+\tanh (a+b x))}{8 b}-\frac {\cos (3) \text {Si}(3+3 \tanh (a+b x))}{8 b} \]

[Out]

1/8*cos(3)*Si(-3+3*tanh(b*x+a))/b-3/8*cos(1)*Si(-1+tanh(b*x+a))/b+3/8*cos(1)*Si(1+tanh(b*x+a))/b-1/8*cos(3)*Si
(3+3*tanh(b*x+a))/b-3/8*Ci(1-tanh(b*x+a))*sin(1)/b-3/8*Ci(1+tanh(b*x+a))*sin(1)/b+1/8*Ci(3-3*tanh(b*x+a))*sin(
3)/b+1/8*Ci(3+3*tanh(b*x+a))*sin(3)/b

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {6857, 3393, 3384, 3380, 3383} \[ \int \sin ^3(\tanh (a+b x)) \, dx=\frac {\sin (3) \operatorname {CosIntegral}(3-3 \tanh (a+b x))}{8 b}+\frac {\sin (3) \operatorname {CosIntegral}(3 \tanh (a+b x)+3)}{8 b}-\frac {3 \sin (1) \operatorname {CosIntegral}(1-\tanh (a+b x))}{8 b}-\frac {3 \sin (1) \operatorname {CosIntegral}(\tanh (a+b x)+1)}{8 b}-\frac {\cos (3) \text {Si}(3-3 \tanh (a+b x))}{8 b}+\frac {3 \cos (1) \text {Si}(1-\tanh (a+b x))}{8 b}+\frac {3 \cos (1) \text {Si}(\tanh (a+b x)+1)}{8 b}-\frac {\cos (3) \text {Si}(3 \tanh (a+b x)+3)}{8 b} \]

[In]

Int[Sin[Tanh[a + b*x]]^3,x]

[Out]

(-3*CosIntegral[1 - Tanh[a + b*x]]*Sin[1])/(8*b) - (3*CosIntegral[1 + Tanh[a + b*x]]*Sin[1])/(8*b) + (CosInteg
ral[3 - 3*Tanh[a + b*x]]*Sin[3])/(8*b) + (CosIntegral[3 + 3*Tanh[a + b*x]]*Sin[3])/(8*b) - (Cos[3]*SinIntegral
[3 - 3*Tanh[a + b*x]])/(8*b) + (3*Cos[1]*SinIntegral[1 - Tanh[a + b*x]])/(8*b) + (3*Cos[1]*SinIntegral[1 + Tan
h[a + b*x]])/(8*b) - (Cos[3]*SinIntegral[3 + 3*Tanh[a + b*x]])/(8*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 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\sin ^3(x)}{1-x^2} \, dx,x,\tanh (a+b x)\right )}{b} \\ & = \frac {\text {Subst}\left (\int \left (-\frac {\sin ^3(x)}{2 (-1+x)}+\frac {\sin ^3(x)}{2 (1+x)}\right ) \, dx,x,\tanh (a+b x)\right )}{b} \\ & = -\frac {\text {Subst}\left (\int \frac {\sin ^3(x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{2 b}+\frac {\text {Subst}\left (\int \frac {\sin ^3(x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{2 b} \\ & = -\frac {\text {Subst}\left (\int \left (\frac {3 \sin (x)}{4 (-1+x)}-\frac {\sin (3 x)}{4 (-1+x)}\right ) \, dx,x,\tanh (a+b x)\right )}{2 b}+\frac {\text {Subst}\left (\int \left (\frac {3 \sin (x)}{4 (1+x)}-\frac {\sin (3 x)}{4 (1+x)}\right ) \, dx,x,\tanh (a+b x)\right )}{2 b} \\ & = \frac {\text {Subst}\left (\int \frac {\sin (3 x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}-\frac {\text {Subst}\left (\int \frac {\sin (3 x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}-\frac {3 \text {Subst}\left (\int \frac {\sin (x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}+\frac {3 \text {Subst}\left (\int \frac {\sin (x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{8 b} \\ & = \frac {(3 \cos (1)) \text {Subst}\left (\int \frac {\sin (1-x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}+\frac {(3 \cos (1)) \text {Subst}\left (\int \frac {\sin (1+x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}-\frac {\cos (3) \text {Subst}\left (\int \frac {\sin (3-3 x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}-\frac {\cos (3) \text {Subst}\left (\int \frac {\sin (3+3 x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}-\frac {(3 \sin (1)) \text {Subst}\left (\int \frac {\cos (1-x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}-\frac {(3 \sin (1)) \text {Subst}\left (\int \frac {\cos (1+x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}+\frac {\sin (3) \text {Subst}\left (\int \frac {\cos (3-3 x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}+\frac {\sin (3) \text {Subst}\left (\int \frac {\cos (3+3 x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{8 b} \\ & = -\frac {3 \operatorname {CosIntegral}(1-\tanh (a+b x)) \sin (1)}{8 b}-\frac {3 \operatorname {CosIntegral}(1+\tanh (a+b x)) \sin (1)}{8 b}+\frac {\operatorname {CosIntegral}(3-3 \tanh (a+b x)) \sin (3)}{8 b}+\frac {\operatorname {CosIntegral}(3+3 \tanh (a+b x)) \sin (3)}{8 b}-\frac {\cos (3) \text {Si}(3-3 \tanh (a+b x))}{8 b}+\frac {3 \cos (1) \text {Si}(1-\tanh (a+b x))}{8 b}+\frac {3 \cos (1) \text {Si}(1+\tanh (a+b x))}{8 b}-\frac {\cos (3) \text {Si}(3+3 \tanh (a+b x))}{8 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.79 \[ \int \sin ^3(\tanh (a+b x)) \, dx=\frac {-6 \operatorname {CosIntegral}(1-\tanh (a+b x)) \sin (1)-6 \operatorname {CosIntegral}(1+\tanh (a+b x)) \sin (1)+2 \operatorname {CosIntegral}(3-3 \tanh (a+b x)) \sin (3)+2 \operatorname {CosIntegral}(3+3 \tanh (a+b x)) \sin (3)-2 \cos (3) \text {Si}(3-3 \tanh (a+b x))+6 \cos (1) \text {Si}(1-\tanh (a+b x))+6 \cos (1) \text {Si}(1+\tanh (a+b x))-2 \cos (3) \text {Si}(3+3 \tanh (a+b x))}{16 b} \]

[In]

Integrate[Sin[Tanh[a + b*x]]^3,x]

[Out]

(-6*CosIntegral[1 - Tanh[a + b*x]]*Sin[1] - 6*CosIntegral[1 + Tanh[a + b*x]]*Sin[1] + 2*CosIntegral[3 - 3*Tanh
[a + b*x]]*Sin[3] + 2*CosIntegral[3 + 3*Tanh[a + b*x]]*Sin[3] - 2*Cos[3]*SinIntegral[3 - 3*Tanh[a + b*x]] + 6*
Cos[1]*SinIntegral[1 - Tanh[a + b*x]] + 6*Cos[1]*SinIntegral[1 + Tanh[a + b*x]] - 2*Cos[3]*SinIntegral[3 + 3*T
anh[a + b*x]])/(16*b)

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(\frac {-\frac {\operatorname {Si}\left (3+3 \tanh \left (b x +a \right )\right ) \cos \left (3\right )}{8}+\frac {\operatorname {Ci}\left (3+3 \tanh \left (b x +a \right )\right ) \sin \left (3\right )}{8}+\frac {\operatorname {Si}\left (-3+3 \tanh \left (b x +a \right )\right ) \cos \left (3\right )}{8}+\frac {\operatorname {Ci}\left (-3+3 \tanh \left (b x +a \right )\right ) \sin \left (3\right )}{8}+\frac {3 \,\operatorname {Si}\left (1+\tanh \left (b x +a \right )\right ) \cos \left (1\right )}{8}-\frac {3 \,\operatorname {Ci}\left (1+\tanh \left (b x +a \right )\right ) \sin \left (1\right )}{8}-\frac {3 \,\operatorname {Si}\left (-1+\tanh \left (b x +a \right )\right ) \cos \left (1\right )}{8}-\frac {3 \,\operatorname {Ci}\left (-1+\tanh \left (b x +a \right )\right ) \sin \left (1\right )}{8}}{b}\)	\(118\)
default	\(\frac {-\frac {\operatorname {Si}\left (3+3 \tanh \left (b x +a \right )\right ) \cos \left (3\right )}{8}+\frac {\operatorname {Ci}\left (3+3 \tanh \left (b x +a \right )\right ) \sin \left (3\right )}{8}+\frac {\operatorname {Si}\left (-3+3 \tanh \left (b x +a \right )\right ) \cos \left (3\right )}{8}+\frac {\operatorname {Ci}\left (-3+3 \tanh \left (b x +a \right )\right ) \sin \left (3\right )}{8}+\frac {3 \,\operatorname {Si}\left (1+\tanh \left (b x +a \right )\right ) \cos \left (1\right )}{8}-\frac {3 \,\operatorname {Ci}\left (1+\tanh \left (b x +a \right )\right ) \sin \left (1\right )}{8}-\frac {3 \,\operatorname {Si}\left (-1+\tanh \left (b x +a \right )\right ) \cos \left (1\right )}{8}-\frac {3 \,\operatorname {Ci}\left (-1+\tanh \left (b x +a \right )\right ) \sin \left (1\right )}{8}}{b}\)	\(118\)
risch	\(-\frac {i {\mathrm e}^{-3 i} \operatorname {Ei}_{1}\left (\frac {6 i}{1+{\mathrm e}^{2 b x +2 a}}-6 i\right )}{16 b}-\frac {i {\mathrm e}^{-3 i} \operatorname {Ei}_{1}\left (-\frac {6 i}{1+{\mathrm e}^{2 b x +2 a}}\right )}{16 b}+\frac {i {\mathrm e}^{3 i} \operatorname {Ei}_{1}\left (\frac {6 i}{1+{\mathrm e}^{2 b x +2 a}}\right )}{16 b}+\frac {i {\mathrm e}^{3 i} \operatorname {Ei}_{1}\left (-\frac {6 i}{1+{\mathrm e}^{2 b x +2 a}}+6 i\right )}{16 b}-\frac {3 i {\mathrm e}^{i} \operatorname {Ei}_{1}\left (-\frac {2 i}{1+{\mathrm e}^{2 b x +2 a}}+2 i\right )}{16 b}-\frac {3 i {\mathrm e}^{i} \operatorname {Ei}_{1}\left (\frac {2 i}{1+{\mathrm e}^{2 b x +2 a}}\right )}{16 b}+\frac {3 i {\mathrm e}^{-i} \operatorname {Ei}_{1}\left (-\frac {2 i}{1+{\mathrm e}^{2 b x +2 a}}\right )}{16 b}+\frac {3 i {\mathrm e}^{-i} \operatorname {Ei}_{1}\left (\frac {2 i}{1+{\mathrm e}^{2 b x +2 a}}-2 i\right )}{16 b}\)	\(230\)

[In]

int(sin(tanh(b*x+a))^3,x,method=_RETURNVERBOSE)

[Out]

1/b*(-1/8*Si(3+3*tanh(b*x+a))*cos(3)+1/8*Ci(3+3*tanh(b*x+a))*sin(3)+1/8*Si(-3+3*tanh(b*x+a))*cos(3)+1/8*Ci(-3+
3*tanh(b*x+a))*sin(3)+3/8*Si(1+tanh(b*x+a))*cos(1)-3/8*Ci(1+tanh(b*x+a))*sin(1)-3/8*Si(-1+tanh(b*x+a))*cos(1)-
3/8*Ci(-1+tanh(b*x+a))*sin(1))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 697, normalized size of antiderivative = 4.44 \[ \int \sin ^3(\tanh (a+b x)) \, dx=\text {Too large to display} \]

[In]

integrate(sin(tanh(b*x+a))^3,x, algorithm="fricas")

[Out]

1/16*((-I*cos(3)^2*cos(1) - (-I*cos(1) + sin(1))*sin(3)^2 - 2*I*(I*cos(3)*cos(1) - cos(3)*sin(1))*sin(3) + I*(
-I*cos(3)^2 + I)*sin(1) + I*cos(1))*cos_integral(3*(cosh(b*x + a) + sinh(b*x + a))/cosh(b*x + a)) - 3*(2*cos(3
)*cos(1)*sin(1) + I*cos(3)*sin(1)^2 + (-I*cos(1)^2 + I)*cos(3) + I*(-I*cos(1)^2 + 2*cos(1)*sin(1) + I*sin(1)^2
+ I)*sin(3))*cos_integral((cosh(b*x + a) + sinh(b*x + a))/cosh(b*x + a)) + (-I*cos(3)^2*cos(1) - (-I*cos(1) +
sin(1))*sin(3)^2 - 2*I*(I*cos(3)*cos(1) - cos(3)*sin(1))*sin(3) + I*(-I*cos(3)^2 + I)*sin(1) + I*cos(1))*cos_
integral(6/(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 + 1)) - 3*(2*cos(3)*cos(1)*sin(1
) + I*cos(3)*sin(1)^2 + (-I*cos(1)^2 + I)*cos(3) + I*(-I*cos(1)^2 + 2*cos(1)*sin(1) + I*sin(1)^2 + I)*sin(3))*
cos_integral(2/(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 + 1)) - (cos(3)^2*cos(1) - (
cos(1) + I*sin(1))*sin(3)^2 + 2*I*(cos(3)*cos(1) + I*cos(3)*sin(1))*sin(3) + I*(cos(3)^2 + 1)*sin(1) + cos(1))
*sin_integral(3*(cosh(b*x + a) + sinh(b*x + a))/cosh(b*x + a)) - 3*(-2*I*cos(3)*cos(1)*sin(1) + cos(3)*sin(1)^
2 - (cos(1)^2 + 1)*cos(3) - I*(cos(1)^2 + 2*I*cos(1)*sin(1) - sin(1)^2 + 1)*sin(3))*sin_integral((cosh(b*x + a
) + sinh(b*x + a))/cosh(b*x + a)) - (cos(3)^2*cos(1) - (cos(1) + I*sin(1))*sin(3)^2 + 2*I*(cos(3)*cos(1) + I*c
os(3)*sin(1))*sin(3) + I*(cos(3)^2 + 1)*sin(1) + cos(1))*sin_integral(6/(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sin
h(b*x + a) + sinh(b*x + a)^2 + 1)) - 3*(-2*I*cos(3)*cos(1)*sin(1) + cos(3)*sin(1)^2 - (cos(1)^2 + 1)*cos(3) -
I*(cos(1)^2 + 2*I*cos(1)*sin(1) - sin(1)^2 + 1)*sin(3))*sin_integral(2/(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh
(b*x + a) + sinh(b*x + a)^2 + 1)))/(b*cos(3)*cos(1) + I*b*cos(3)*sin(1) + I*(b*cos(1) + I*b*sin(1))*sin(3))

Sympy [F]

\[ \int \sin ^3(\tanh (a+b x)) \, dx=\int \sin ^{3}{\left (\tanh {\left (a + b x \right )} \right )}\, dx \]

[In]

integrate(sin(tanh(b*x+a))**3,x)

[Out]

Integral(sin(tanh(a + b*x))**3, x)

Maxima [F]

\[ \int \sin ^3(\tanh (a+b x)) \, dx=\int { \sin \left (\tanh \left (b x + a\right )\right )^{3} \,d x } \]

[In]

integrate(sin(tanh(b*x+a))^3,x, algorithm="maxima")

[Out]

integrate(sin(tanh(b*x + a))^3, x)

Giac [F]

\[ \int \sin ^3(\tanh (a+b x)) \, dx=\int { \sin \left (\tanh \left (b x + a\right )\right )^{3} \,d x } \]

[In]

integrate(sin(tanh(b*x+a))^3,x, algorithm="giac")

[Out]

integrate(sin(tanh(b*x + a))^3, x)

Mupad [F(-1)]

Timed out. \[ \int \sin ^3(\tanh (a+b x)) \, dx=\int {\sin \left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3 \,d x \]

[In]

int(sin(tanh(a + b*x))^3,x)

[Out]

int(sin(tanh(a + b*x))^3, x)