\(\int e^{\text {arctanh}(x)} \sqrt {1-x} \sin (x) \, dx\) [848]

Optimal result

Integrand size = 16, antiderivative size = 72 \[ \int e^{\text {arctanh}(x)} \sqrt {1-x} \sin (x) \, dx=-\sqrt {1+x} \cos (x)+\sqrt {\frac {\pi }{2}} \cos (1) \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1+x}\right )+\sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1+x}\right ) \sin (1) \] Output:

-(1+x)^(1/2)*cos(x)+1/2*2^(1/2)*Pi^(1/2)*cos(1)*FresnelC(2^(1/2)/Pi^(1/2)* 
(1+x)^(1/2))+1/2*2^(1/2)*Pi^(1/2)*FresnelS(2^(1/2)/Pi^(1/2)*(1+x)^(1/2))*s 
in(1)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.99 \[ \int e^{\text {arctanh}(x)} \sqrt {1-x} \sin (x) \, dx=\frac {i e^{-i} \left (-\sqrt {-i (1+x)} \Gamma \left (\frac {3}{2},-i (1+x)\right )+e^{2 i} \sqrt {i (1+x)} \Gamma \left (\frac {3}{2},i (1+x)\right )\right )}{2 \sqrt {1+x}} \] Input:

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

Output:

((I/2)*(-(Sqrt[(-I)*(1 + x)]*Gamma[3/2, (-I)*(1 + x)]) + E^(2*I)*Sqrt[I*(1 
 + x)]*Gamma[3/2, I*(1 + x)]))/(E^I*Sqrt[1 + x])
 

Rubi [A] (verified)

Time = 0.91 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {6679, 3042, 3777, 3042, 3787, 3042, 3785, 3786, 3832, 3833}

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.

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

\(\Big \downarrow \) 6679

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3777

\(\displaystyle \frac {1}{2} \int \frac {\cos (x)}{\sqrt {x+1}}dx-\sqrt {x+1} \cos (x)\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \int \frac {\sin \left (x+\frac {\pi }{2}\right )}{\sqrt {x+1}}dx-\sqrt {x+1} \cos (x)\)

\(\Big \downarrow \) 3787

\(\displaystyle \frac {1}{2} \left (\sin (1) \int \frac {\sin (x+1)}{\sqrt {x+1}}dx+\cos (1) \int \frac {\cos (x+1)}{\sqrt {x+1}}dx\right )-\sqrt {x+1} \cos (x)\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\sin (1) \int \frac {\sin (x+1)}{\sqrt {x+1}}dx+\cos (1) \int \frac {\sin \left (x+\frac {\pi }{2}+1\right )}{\sqrt {x+1}}dx\right )-\sqrt {x+1} \cos (x)\)

\(\Big \downarrow \) 3785

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

\(\Big \downarrow \) 3786

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

\(\Big \downarrow \) 3832

\(\displaystyle \frac {1}{2} \left (2 \cos (1) \int \cos (x+1)d\sqrt {x+1}+\sqrt {2 \pi } \sin (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {x+1}\right )\right )-\sqrt {x+1} \cos (x)\)

\(\Big \downarrow \) 3833

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

Input:

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

Output:

-(Sqrt[1 + x]*Cos[x]) + (Sqrt[2*Pi]*Cos[1]*FresnelC[Sqrt[2/Pi]*Sqrt[1 + x] 
] + Sqrt[2*Pi]*FresnelS[Sqrt[2/Pi]*Sqrt[1 + x]]*Sin[1])/2
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3777

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( 
-(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1)*C 
os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
 

rule 3785

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> S 
imp[2/d   Subst[Int[Cos[f*(x^2/d)], x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, 
d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]
 

rule 3786

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

rule 3787

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

rule 3832

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ 
d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
 

rule 3833

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ 
d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
 

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])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral(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 = 0.20 (sec) , antiderivative size = 498, normalized size of antiderivative = 6.92 \[ \int e^{\text {arctanh}(x)} \sqrt {1-x} \sin (x) \, dx=\text {Too large to display} \] Input:

integrate((1+x)/(-x^2+1)^(1/2)*(1-x)^(1/2)*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(sqr 
t(-I*x - I)) - 1))*sin(1))*cos(1/2*arctan2(x + 1, 0)) - ((sqrt(pi)*(erf(sq 
rt(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 
))*sin(1/2*arctan2(x + 1, 0)))*sqrt(x + 1)/sqrt(abs(x + 1)) - 1/2*(((I*sqr 
t(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(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))*a 
bs(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*cos(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))*gamma(3/2, I*x + 
 I) + (cos(1) - I*sin(1))*gamma(3/2, -I*x - I))*sin(3/2*arctan2(x + 1, 0)) 
)*sqrt(abs(x + 1))/(x + 1)^(3/2)
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.90 \[ \int e^{\text {arctanh}(x)} \sqrt {1-x} \sin (x) \, dx=\left (\frac {1}{8} i - \frac {1}{8}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {x + 1}\right ) e^{i} - \left (\frac {1}{8} i + \frac {1}{8}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {x + 1}\right ) e^{\left (-i\right )} - \frac {1}{2} \, \sqrt {x + 1} e^{\left (i \, x\right )} - \frac {1}{2} \, \sqrt {x + 1} e^{\left (-i \, x\right )} \] Input:

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

Output:

(1/8*I - 1/8)*sqrt(2)*sqrt(pi)*erf(-(1/2*I + 1/2)*sqrt(2)*sqrt(x + 1))*e^I 
 - (1/8*I + 1/8)*sqrt(2)*sqrt(pi)*erf((1/2*I - 1/2)*sqrt(2)*sqrt(x + 1))*e 
^(-I) - 1/2*sqrt(x + 1)*e^(I*x) - 1/2*sqrt(x + 1)*e^(-I*x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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