\(\int e^{\text {arctanh}(x)} x \sqrt {1+x} \sin (x) \, dx\) [845]

Optimal result

Integrand size = 15, antiderivative size = 240 \[ \int e^{\text {arctanh}(x)} x \sqrt {1+x} \sin (x) \, dx=3 \sqrt {1-x} \cos (x)-(1-x)^{3/2} \cos (x)-3 \sqrt {\frac {\pi }{2}} \cos (1) \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-\frac {3}{2} \sqrt {\frac {\pi }{2}} \cos (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+2 \sqrt {2 \pi } \cos (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+\frac {3}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \sin (1)-2 \sqrt {2 \pi } \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \sin (1)-3 \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \sin (1)-\frac {3}{2} \sqrt {1-x} \sin (x) \] Output:

3*(1-x)^(1/2)*cos(x)-(1-x)^(3/2)*cos(x)-3/2*2^(1/2)*Pi^(1/2)*cos(1)*Fresne 
lC(2^(1/2)/Pi^(1/2)*(1-x)^(1/2))+5/4*2^(1/2)*Pi^(1/2)*cos(1)*FresnelS(2^(1 
/2)/Pi^(1/2)*(1-x)^(1/2))-5/4*2^(1/2)*Pi^(1/2)*FresnelC(2^(1/2)/Pi^(1/2)*( 
1-x)^(1/2))*sin(1)-3/2*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 = 6.12 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.77 \[ \int e^{\text {arctanh}(x)} x \sqrt {1+x} \sin (x) \, dx=-\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) \sqrt {1+x} \left ((-6+5 i) \sqrt {2 \pi } \sqrt {-1+x} \text {erfi}\left (\frac {(1+i) \sqrt {-1+x}}{\sqrt {2}}\right ) (\cos (1)+i \sin (1))+(2+2 i) \left ((-4-3 i)+(2+3 i) x+2 x^2\right ) (\cos (x)+i \sin (x))+\left ((2+2 i) \left ((-4+3 i)+(2-3 i) x+2 x^2\right ) (\cos (1)+i \sin (1))-(6+5 i) \sqrt {2 \pi } \sqrt {-1+x} \text {erf}\left (\frac {(1+i) \sqrt {-1+x}}{\sqrt {2}}\right ) (\cos (x)+i \sin (x))\right ) (\cos (1+x)-i \sin (1+x))\right )}{\sqrt {1-x^2}} \] Input:

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

Output:

((-1/16 + I/16)*Sqrt[1 + x]*((-6 + 5*I)*Sqrt[2*Pi]*Sqrt[-1 + x]*Erfi[((1 + 
 I)*Sqrt[-1 + x])/Sqrt[2]]*(Cos[1] + I*Sin[1]) + (2 + 2*I)*((-4 - 3*I) + ( 
2 + 3*I)*x + 2*x^2)*(Cos[x] + I*Sin[x]) + ((2 + 2*I)*((-4 + 3*I) + (2 - 3* 
I)*x + 2*x^2)*(Cos[1] + I*Sin[1]) - (6 + 5*I)*Sqrt[2*Pi]*Sqrt[-1 + x]*Erf[ 
((1 + I)*Sqrt[-1 + x])/Sqrt[2]]*(Cos[x] + I*Sin[x]))*(Cos[1 + x] - I*Sin[1 
 + x])))/Sqrt[1 - x^2]
 

Rubi [A] (verified)

Time = 1.20 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {6679, 7267, 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.

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

\(\Big \downarrow \) 6679

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

\(\Big \downarrow \) 7267

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

\(\Big \downarrow \) 7293

\(\displaystyle -2 \int \left (\sin (x) (1-x)^2-3 \sin (x) (1-x)+2 \sin (x)\right )d\sqrt {1-x}\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

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]

\[\int \frac {\left (1+x \right )^{\frac {3}{2}} x \sin \left (x \right )}{\sqrt {-x^{2}+1}}d x\]

Input:

int((1+x)^(3/2)/(-x^2+1)^(1/2)*x*sin(x),x)
 

Output:

int((1+x)^(3/2)/(-x^2+1)^(1/2)*x*sin(x),x)
 

Fricas [F]

\[ \int e^{\text {arctanh}(x)} x \sqrt {1+x} \sin (x) \, dx=\int { \frac {{\left (x + 1\right )}^{\frac {3}{2}} x \sin \left (x\right )}{\sqrt {-x^{2} + 1}} \,d x } \] Input:

integrate((1+x)^(3/2)/(-x^2+1)^(1/2)*x*sin(x),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

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

integrate((1+x)**(3/2)/(-x**2+1)**(1/2)*x*sin(x),x)
 

Output:

Timed out
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 628, normalized size of antiderivative = 2.62 \[ \int e^{\text {arctanh}(x)} x \sqrt {1+x} \sin (x) \, dx=\text {Too large to display} \] Input:

integrate((1+x)^(3/2)/(-x^2+1)^(1/2)*x*sin(x),x, algorithm="maxima")
 

Output:

1/2*(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(s 
qrt(-I*x + I)) - 1))*sin(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))*sin(1/2*arctan2(x - 1, 0)))*(x - 1)^2 - (((-I*cos(1) - sin(1))*gamma(5 
/2, I*x - I) + (I*cos(1) - sin(1))*gamma(5/2, -I*x + I))*cos(5/2*arctan2(x 
 - 1, 0)) - ((cos(1) - I*sin(1))*gamma(5/2, I*x - I) + (cos(1) + I*sin(1)) 
*gamma(5/2, -I*x + I))*sin(5/2*arctan2(x - 1, 0)))*x^2 + (3*(((I*cos(1) + 
sin(1))*gamma(3/2, I*x - I) + (-I*cos(1) + sin(1))*gamma(3/2, -I*x + I))*c 
os(3/2*arctan2(x - 1, 0)) + ((cos(1) - I*sin(1))*gamma(3/2, I*x - I) + (co 
s(1) + I*sin(1))*gamma(3/2, -I*x + I))*sin(3/2*arctan2(x - 1, 0)))*abs(x - 
 1) + 2*((-I*cos(1) - sin(1))*gamma(5/2, I*x - I) + (I*cos(1) - sin(1))*ga 
mma(5/2, -I*x + I))*cos(5/2*arctan2(x - 1, 0)) - 2*((cos(1) - I*sin(1))*ga 
mma(5/2, I*x - I) + (cos(1) + I*sin(1))*gamma(5/2, -I*x + I))*sin(5/2*arct 
an2(x - 1, 0)))*x + 3*(((-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)) 
*sin(3/2*arctan2(x - 1, 0)))*abs(x - 1) - ((-I*cos(1) - sin(1))*gamma(5/2, 
 I*x - I) + (I*cos(1) - sin(1))*gamma(5/2, -I*x + I))*cos(5/2*arctan2(x...
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.51 \[ \int e^{\text {arctanh}(x)} x \sqrt {1+x} \sin (x) \, dx=-\left (\frac {11}{16} i - \frac {1}{16}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {-x + 1}\right ) e^{i} + \left (\frac {11}{16} i + \frac {1}{16}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {-x + 1}\right ) e^{\left (-i\right )} - \frac {1}{4} i \, {\left (-2 i \, {\left (-x + 1\right )}^{\frac {3}{2}} + \left (4 i - 3\right ) \, \sqrt {-x + 1}\right )} e^{\left (i \, x\right )} - \frac {1}{4} i \, {\left (-2 i \, {\left (-x + 1\right )}^{\frac {3}{2}} + \left (4 i + 3\right ) \, \sqrt {-x + 1}\right )} e^{\left (-i \, x\right )} + \frac {1}{2} \, \sqrt {-x + 1} e^{\left (i \, x\right )} + \frac {1}{2} \, \sqrt {-x + 1} e^{\left (-i \, x\right )} \] Input:

integrate((1+x)^(3/2)/(-x^2+1)^(1/2)*x*sin(x),x, algorithm="giac")
 

Output:

-(11/16*I - 1/16)*sqrt(2)*sqrt(pi)*erf(-(1/2*I + 1/2)*sqrt(2)*sqrt(-x + 1) 
)*e^I + (11/16*I + 1/16)*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. \[ \int e^{\text {arctanh}(x)} x \sqrt {1+x} \sin (x) \, dx=\int \frac {x\,\sin \left (x\right )\,{\left (x+1\right )}^{3/2}}{\sqrt {1-x^2}} \,d x \] Input:

int((x*sin(x)*(x + 1)^(3/2))/(1 - x^2)^(1/2),x)
 

Output:

int((x*sin(x)*(x + 1)^(3/2))/(1 - x^2)^(1/2), x)
 

Reduce [F]

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

int((1+x)^(3/2)/(-x^2+1)^(1/2)*x*sin(x),x)
 

Output:

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