\(\int e^{\text {arctanh}(x)} \sqrt {1-x} x \sin (x) \, dx\) [832]

Optimal result

Integrand size = 17, antiderivative size = 163 \[ \int e^{\text {arctanh}(x)} \sqrt {1-x} x \sin (x) \, dx=\sqrt {1+x} \cos (x)-(1+x)^{3/2} \cos (x)-\sqrt {\frac {\pi }{2}} \cos (1) \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1+x}\right )-\frac {3}{2} \sqrt {\frac {\pi }{2}} \cos (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1+x}\right )+\frac {3}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1+x}\right ) \sin (1)-\sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1+x}\right ) \sin (1)+\frac {3}{2} \sqrt {1+x} \sin (x) \] Output:

(1+x)^(1/2)*cos(x)-(1+x)^(3/2)*cos(x)-1/2*2^(1/2)*Pi^(1/2)*cos(1)*FresnelC 
(2^(1/2)/Pi^(1/2)*(1+x)^(1/2))-3/4*2^(1/2)*Pi^(1/2)*cos(1)*FresnelS(2^(1/2 
)/Pi^(1/2)*(1+x)^(1/2))+3/4*2^(1/2)*Pi^(1/2)*FresnelC(2^(1/2)/Pi^(1/2)*(1+ 
x)^(1/2))*sin(1)-1/2*2^(1/2)*Pi^(1/2)*FresnelS(2^(1/2)/Pi^(1/2)*(1+x)^(1/2 
))*sin(1)+3/2*(1+x)^(1/2)*sin(x)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.37 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.06 \[ \int e^{\text {arctanh}(x)} \sqrt {1-x} x \sin (x) \, dx=\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) e^{-i x} \sqrt {1-x^2} \left ((2+2 i) \sqrt {-1-x} \left (3+e^{2 i x} (-3+2 i x)+2 i x\right )+(3+2 i) e^{i x} \sqrt {2 \pi } \text {erf}\left (\frac {(1+i) \sqrt {-1-x}}{\sqrt {2}}\right ) (\cos (1)-i \sin (1))-(3-2 i) e^{i x} \sqrt {2 \pi } \text {erfi}\left (\frac {(1+i) \sqrt {-1-x}}{\sqrt {2}}\right ) (\cos (1)+i \sin (1))\right )}{\sqrt {-1-x} \sqrt {1-x}} \] Input:

Integrate[E^ArcTanh[x]*Sqrt[1 - x]*x*Sin[x],x]
 

Output:

((1/16 + I/16)*Sqrt[1 - x^2]*((2 + 2*I)*Sqrt[-1 - x]*(3 + E^((2*I)*x)*(-3 
+ (2*I)*x) + (2*I)*x) + (3 + 2*I)*E^(I*x)*Sqrt[2*Pi]*Erf[((1 + I)*Sqrt[-1 
- x])/Sqrt[2]]*(Cos[1] - I*Sin[1]) - (3 - 2*I)*E^(I*x)*Sqrt[2*Pi]*Erfi[((1 
 + I)*Sqrt[-1 - x])/Sqrt[2]]*(Cos[1] + I*Sin[1])))/(E^(I*x)*Sqrt[-1 - x]*S 
qrt[1 - x])
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {6679, 7267, 25, 7293, 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.

\(\displaystyle \int \sqrt {1-x} x e^{\text {arctanh}(x)} \sin (x) \, dx\)

\(\Big \downarrow \) 6679

\(\displaystyle \int x \sqrt {x+1} \sin (x)dx\)

\(\Big \downarrow \) 7267

\(\displaystyle -2 \int -x (x+1) \sin (x)d\sqrt {x+1}\)

\(\Big \downarrow \) 25

\(\displaystyle 2 \int x (x+1) \sin (x)d\sqrt {x+1}\)

\(\Big \downarrow \) 7293

\(\displaystyle 2 \int \left ((x+1)^2 \sin (x)-(x+1) \sin (x)\right )d\sqrt {x+1}\)

\(\Big \downarrow \) 2009

\(\displaystyle -2 \left (-\frac {3}{4} \sqrt {\frac {\pi }{2}} \sin (1) \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {x+1}\right )+\frac {1}{2} \sqrt {\frac {\pi }{2}} \cos (1) \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {x+1}\right )+\frac {1}{2} \sqrt {\frac {\pi }{2}} \sin (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {x+1}\right )+\frac {3}{4} \sqrt {\frac {\pi }{2}} \cos (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {x+1}\right )-\frac {3}{4} \sqrt {x+1} \sin (x)+\frac {1}{2} (x+1)^{3/2} \cos (x)-\frac {1}{2} \sqrt {x+1} \cos (x)\right )\)

Input:

Int[E^ArcTanh[x]*Sqrt[1 - x]*x*Sin[x],x]
 

Output:

-2*(-1/2*(Sqrt[1 + x]*Cos[x]) + ((1 + x)^(3/2)*Cos[x])/2 + (Sqrt[Pi/2]*Cos 
[1]*FresnelC[Sqrt[2/Pi]*Sqrt[1 + x]])/2 + (3*Sqrt[Pi/2]*Cos[1]*FresnelS[Sq 
rt[2/Pi]*Sqrt[1 + x]])/4 - (3*Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Sqrt[1 + x]]* 
Sin[1])/4 + (Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Sqrt[1 + x]]*Sin[1])/2 - (3*Sq 
rt[1 + x]*Sin[x])/4)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009

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

rule 6679

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol 
] :> Simp[c^p   Int[u*(1 + d*(x/c))^p*((1 + a*x)^(n/2)/(1 - a*x)^(n/2)), x] 
, x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c^2 - d^2, 0] && (IntegerQ[p] 
|| GtQ[c, 0])
 

rule 7267

Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si 
mp[lst[[2]]*lst[[4]]   Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x 
] /;  !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [F]

\[\int \frac {\left (1+x \right ) \sqrt {1-x}\, x \sin \left (x \right )}{\sqrt {-x^{2}+1}}d x\]

Input:

int((1+x)/(-x^2+1)^(1/2)*(1-x)^(1/2)*x*sin(x),x)
 

Output:

int((1+x)/(-x^2+1)^(1/2)*(1-x)^(1/2)*x*sin(x),x)
 

Fricas [F]

\[ \int e^{\text {arctanh}(x)} \sqrt {1-x} x \sin (x) \, dx=\int { \frac {{\left (x + 1\right )} x \sqrt {-x + 1} \sin \left (x\right )}{\sqrt {-x^{2} + 1}} \,d x } \] Input:

integrate((1+x)/(-x^2+1)^(1/2)*(1-x)^(1/2)*x*sin(x),x, algorithm="fricas")
 

Output:

integral(-sqrt(-x^2 + 1)*x*sqrt(-x + 1)*sin(x)/(x - 1), x)
 

Sympy [F]

\[ \int e^{\text {arctanh}(x)} \sqrt {1-x} x \sin (x) \, dx=\int \frac {x \sqrt {1 - x} \left (x + 1\right ) \sin {\left (x \right )}}{\sqrt {- \left (x - 1\right ) \left (x + 1\right )}}\, dx \] Input:

integrate((1+x)/(-x**2+1)**(1/2)*(1-x)**(1/2)*x*sin(x),x)
 

Output:

Integral(x*sqrt(1 - x)*(x + 1)*sin(x)/sqrt(-(x - 1)*(x + 1)), x)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.17 (sec) , antiderivative size = 898, normalized size of antiderivative = 5.51 \[ \int e^{\text {arctanh}(x)} \sqrt {1-x} x \sin (x) \, dx=\text {Too large to display} \] Input:

integrate((1+x)/(-x^2+1)^(1/2)*(1-x)^(1/2)*x*sin(x),x, algorithm="maxima")
 

Output:

-1/2*(((I*sqrt(pi)*(erf(sqrt(I*x + I)) - 1) - I*sqrt(pi)*(erf(sqrt(-I*x - 
I)) - 1))*cos(1) - (sqrt(pi)*(erf(sqrt(I*x + I)) - 1) + sqrt(pi)*(erf(sqrt 
(-I*x - I)) - 1))*sin(1))*abs(x + 1)*cos(1/2*arctan2(x + 1, 0)) + ((sqrt(p 
i)*(erf(sqrt(I*x + I)) - 1) + sqrt(pi)*(erf(sqrt(-I*x - I)) - 1))*cos(1) + 
 (I*sqrt(pi)*(erf(sqrt(I*x + I)) - 1) - I*sqrt(pi)*(erf(sqrt(-I*x - I)) - 
1))*sin(1))*abs(x + 1)*sin(1/2*arctan2(x + 1, 0)) + (((I*cos(1) - sin(1))* 
gamma(3/2, I*x + I) + (-I*cos(1) - sin(1))*gamma(3/2, -I*x - I))*x + (I*co 
s(1) - sin(1))*gamma(3/2, I*x + I) + (-I*cos(1) - sin(1))*gamma(3/2, -I*x 
- I))*cos(3/2*arctan2(x + 1, 0)) + (((cos(1) + I*sin(1))*gamma(3/2, I*x + 
I) + (cos(1) - I*sin(1))*gamma(3/2, -I*x - I))*x + (cos(1) + I*sin(1))*gam 
ma(3/2, I*x + I) + (cos(1) - I*sin(1))*gamma(3/2, -I*x - I))*sin(3/2*arcta 
n2(x + 1, 0)))*sqrt(abs(x + 1))/(x + 1)^(3/2) - 1/2*(((-I*sqrt(pi)*(erf(sq 
rt(I*x + I)) - 1) + I*sqrt(pi)*(erf(sqrt(-I*x - I)) - 1))*cos(1) + (sqrt(p 
i)*(erf(sqrt(I*x + I)) - 1) + sqrt(pi)*(erf(sqrt(-I*x - I)) - 1))*sin(1))* 
(x + 1)^2*cos(1/2*arctan2(x + 1, 0)) - ((sqrt(pi)*(erf(sqrt(I*x + I)) - 1) 
 + sqrt(pi)*(erf(sqrt(-I*x - I)) - 1))*cos(1) - (-I*sqrt(pi)*(erf(sqrt(I*x 
 + I)) - 1) + I*sqrt(pi)*(erf(sqrt(-I*x - I)) - 1))*sin(1))*(x + 1)^2*sin( 
1/2*arctan2(x + 1, 0)) - 2*(((I*cos(1) - sin(1))*gamma(3/2, I*x + I) + (-I 
*cos(1) - sin(1))*gamma(3/2, -I*x - I))*x + (I*cos(1) - sin(1))*gamma(3/2, 
 I*x + I) + (-I*cos(1) - sin(1))*gamma(3/2, -I*x - I))*abs(x + 1)*cos(3...
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.66 \[ \int e^{\text {arctanh}(x)} \sqrt {1-x} x \sin (x) \, dx=\left (\frac {1}{16} i + \frac {5}{16}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {x + 1}\right ) e^{i} - \left (\frac {1}{16} i - \frac {5}{16}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {x + 1}\right ) e^{\left (-i\right )} + \frac {1}{4} i \, {\left (2 i \, {\left (x + 1\right )}^{\frac {3}{2}} - \left (4 i + 3\right ) \, \sqrt {x + 1}\right )} e^{\left (i \, x\right )} + \frac {1}{4} i \, {\left (2 i \, {\left (x + 1\right )}^{\frac {3}{2}} - \left (4 i - 3\right ) \, \sqrt {x + 1}\right )} e^{\left (-i \, x\right )} - \frac {1}{2} \, \sqrt {x + 1} e^{\left (i \, x\right )} - \frac {1}{2} \, \sqrt {x + 1} e^{\left (-i \, x\right )} \] Input:

integrate((1+x)/(-x^2+1)^(1/2)*(1-x)^(1/2)*x*sin(x),x, algorithm="giac")
 

Output:

(1/16*I + 5/16)*sqrt(2)*sqrt(pi)*erf(-(1/2*I + 1/2)*sqrt(2)*sqrt(x + 1))*e 
^I - (1/16*I - 5/16)*sqrt(2)*sqrt(pi)*erf((1/2*I - 1/2)*sqrt(2)*sqrt(x + 1 
))*e^(-I) + 1/4*I*(2*I*(x + 1)^(3/2) - (4*I + 3)*sqrt(x + 1))*e^(I*x) + 1/ 
4*I*(2*I*(x + 1)^(3/2) - (4*I - 3)*sqrt(x + 1))*e^(-I*x) - 1/2*sqrt(x + 1) 
*e^(I*x) - 1/2*sqrt(x + 1)*e^(-I*x)
 

Mupad [F(-1)]

Timed out. \[ \int e^{\text {arctanh}(x)} \sqrt {1-x} x \sin (x) \, dx=\int \frac {x\,\sin \left (x\right )\,\sqrt {1-x}\,\left (x+1\right )}{\sqrt {1-x^2}} \,d x \] Input:

int((x*sin(x)*(1 - x)^(1/2)*(x + 1))/(1 - x^2)^(1/2),x)
 

Output:

int((x*sin(x)*(1 - x)^(1/2)*(x + 1))/(1 - x^2)^(1/2), x)
 

Reduce [F]

\[ \int e^{\text {arctanh}(x)} \sqrt {1-x} x \sin (x) \, dx=\int \frac {\sin \left (x \right ) x^{2}}{\sqrt {x +1}}d x +\int \frac {\sin \left (x \right ) x}{\sqrt {x +1}}d x \] Input:

int((1+x)/(-x^2+1)^(1/2)*(1-x)^(1/2)*x*sin(x),x)
 

Output:

int((sin(x)*x**2)/sqrt(x + 1),x) + int((sin(x)*x)/sqrt(x + 1),x)