3.9.11 \(\int e^{\text {arctanh}(x)} \sqrt {1-x} \sin (x) \, dx\) [811]

3.9.11.1 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

1/2*cos(1)*FresnelC(2^(1/2)/Pi^(1/2)*(1+x)^(1/2))*2^(1/2)*Pi^(1/2)+1/2*Fre 
snelS(2^(1/2)/Pi^(1/2)*(1+x)^(1/2))*sin(1)*2^(1/2)*Pi^(1/2)-cos(x)*(1+x)^( 
1/2)
 

3.9.11.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.02 (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])
 

3.9.11.3 Rubi [A] (verified)

Time = 0.54 (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
 

3.9.11.3.1 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])
 

3.9.11.4 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)
 

3.9.11.5 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)
 

3.9.11.6 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)
 

3.9.11.7 Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.31 (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)
 

3.9.11.8 Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.92 \[ \int e^{\text {arctanh}(x)} \sqrt {1-x} \sin (x) \, dx=\left (\frac {1}{8} i - \frac {1}{8}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {x + 1}\right ) e^{i} - \left (\frac {1}{8} i + \frac {1}{8}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {x + 1}\right ) e^{\left (-i\right )} - \frac {1}{2} \, \sqrt {x + 1} e^{\left (i \, x\right )} - \frac {1}{2} \, \sqrt {x + 1} e^{\left (-i \, x\right )} - 0.339605729125000 \]

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) - 0.33960572912 
5000
 

3.9.11.9 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)