\(\int \sin ^3(\tanh (a+b x)) \, dx\) [251]

Optimal result

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

-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+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
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.79 \[ \int \sin ^3(\tanh (a+b x)) \, dx=\frac {-6 \operatorname {CosIntegral}(1-\tanh (a+b x)) \sin (1)-6 \operatorname {CosIntegral}(1+\tanh (a+b x)) \sin (1)+2 \operatorname {CosIntegral}(3-3 \tanh (a+b x)) \sin (3)+2 \operatorname {CosIntegral}(3+3 \tanh (a+b x)) \sin (3)-2 \cos (3) \text {Si}(3-3 \tanh (a+b x))+6 \cos (1) \text {Si}(1-\tanh (a+b x))+6 \cos (1) \text {Si}(1+\tanh (a+b x))-2 \cos (3) \text {Si}(3+3 \tanh (a+b x))}{16 b} \] Input:

Integrate[Sin[Tanh[a + b*x]]^3,x]
 

Output:

(-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*Co 
s[1]*SinIntegral[1 - Tanh[a + b*x]] + 6*Cos[1]*SinIntegral[1 + Tanh[a + b* 
x]] - 2*Cos[3]*SinIntegral[3 + 3*Tanh[a + b*x]])/(16*b)
 

Rubi [A] (verified)

Time = 1.00 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.87, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4853, 7276, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sin ^3(\tanh (a+b x)) \, dx\)

\(\Big \downarrow \) 4853

\(\displaystyle \frac {\int \frac {\sin ^3(\tanh (a+b x))}{1-\tanh ^2(a+b x)}d\tanh (a+b x)}{b}\)

\(\Big \downarrow \) 7276

\(\displaystyle \frac {\int \left (\frac {\sin ^3(\tanh (a+b x))}{2 (\tanh (a+b x)+1)}-\frac {\sin ^3(\tanh (a+b x))}{2 (\tanh (a+b x)-1)}\right )d\tanh (a+b x)}{b}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {1}{8} \sin (3) \operatorname {CosIntegral}(3-3 \tanh (a+b x))+\frac {1}{8} \sin (3) \operatorname {CosIntegral}(3 \tanh (a+b x)+3)-\frac {3}{8} \sin (1) \operatorname {CosIntegral}(1-\tanh (a+b x))-\frac {3}{8} \sin (1) \operatorname {CosIntegral}(\tanh (a+b x)+1)-\frac {1}{8} \cos (3) \text {Si}(3-3 \tanh (a+b x))+\frac {3}{8} \cos (1) \text {Si}(1-\tanh (a+b x))+\frac {3}{8} \cos (1) \text {Si}(\tanh (a+b x)+1)-\frac {1}{8} \cos (3) \text {Si}(3 \tanh (a+b x)+3)}{b}\)

Input:

Int[Sin[Tanh[a + b*x]]^3,x]
 

Output:

((-3*CosIntegral[1 - Tanh[a + b*x]]*Sin[1])/8 - (3*CosIntegral[1 + Tanh[a 
+ b*x]]*Sin[1])/8 + (CosIntegral[3 - 3*Tanh[a + b*x]]*Sin[3])/8 + (CosInte 
gral[3 + 3*Tanh[a + b*x]]*Sin[3])/8 - (Cos[3]*SinIntegral[3 - 3*Tanh[a + b 
*x]])/8 + (3*Cos[1]*SinIntegral[1 - Tanh[a + b*x]])/8 + (3*Cos[1]*SinInteg 
ral[1 + Tanh[a + b*x]])/8 - (Cos[3]*SinIntegral[3 + 3*Tanh[a + b*x]])/8)/b
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4853

Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, Simp[With[{d = FreeFa 
ctors[Tan[v], x]}, d/Coefficient[v, x, 1]   Subst[Int[SubstFor[1/(1 + d^2*x 
^2), Tan[v]/d, u, x], x], x, Tan[v]/d]], x] /;  !FalseQ[v] && FunctionOfQ[N 
onfreeFactors[Tan[v], x], u, x, True] && TryPureTanSubst[ActivateTrig[u], x 
]]
 

rule 7276

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

Maple [A] (verified)

Time = 1.78 (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	\(-\frac {i {\mathrm e}^{-3 i} \operatorname {expIntegral}_{1}\left (-\frac {6 i}{1+{\mathrm e}^{2 b x +2 a}}\right )}{16 b}+\frac {i {\mathrm e}^{3 i} \operatorname {expIntegral}_{1}\left (-\frac {6 i}{1+{\mathrm e}^{2 b x +2 a}}+6 i\right )}{16 b}-\frac {3 i {\mathrm e}^{i} \operatorname {expIntegral}_{1}\left (\frac {2 i}{1+{\mathrm e}^{2 b x +2 a}}\right )}{16 b}+\frac {3 i {\mathrm e}^{-i} \operatorname {expIntegral}_{1}\left (\frac {2 i}{1+{\mathrm e}^{2 b x +2 a}}-2 i\right )}{16 b}+\frac {3 i {\mathrm e}^{-i} \operatorname {expIntegral}_{1}\left (-\frac {2 i}{1+{\mathrm e}^{2 b x +2 a}}\right )}{16 b}-\frac {3 i {\mathrm e}^{i} \operatorname {expIntegral}_{1}\left (-\frac {2 i}{1+{\mathrm e}^{2 b x +2 a}}+2 i\right )}{16 b}+\frac {i {\mathrm e}^{3 i} \operatorname {expIntegral}_{1}\left (\frac {6 i}{1+{\mathrm e}^{2 b x +2 a}}\right )}{16 b}-\frac {i {\mathrm e}^{-3 i} \operatorname {expIntegral}_{1}\left (\frac {6 i}{1+{\mathrm e}^{2 b x +2 a}}-6 i\right )}{16 b}\)	\(230\)

Input:

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

Output:

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*S 
i(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 = 0.13 (sec) , antiderivative size = 697, normalized size of antiderivative = 4.44 \[ \int \sin ^3(\tanh (a+b x)) \, dx=\text {Too large to display} \] Input:

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

Output:

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))*co 
s_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*(co 
s(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*co 
s(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(6/(cosh(b*x + a) 
^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 + 1)) - 3*(-2*I*co...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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