\(\int e^{\text {arctanh}(x)} (1-x)^{3/2} x \sin (x) \, dx\) [836]

Optimal result

Integrand size = 17, antiderivative size = 193 \[ \int e^{\text {arctanh}(x)} (1-x)^{3/2} x \sin (x) \, dx=-\frac {7}{4} \sqrt {1+x} \cos (x)-3 (1+x)^{3/2} \cos (x)+(1+x)^{5/2} \cos (x)+\frac {7}{4} \sqrt {\frac {\pi }{2}} \cos (1) \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1+x}\right )-\frac {9}{2} \sqrt {\frac {\pi }{2}} \cos (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1+x}\right )+\frac {9}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1+x}\right ) \sin (1)+\frac {7}{4} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1+x}\right ) \sin (1)+\frac {9}{2} \sqrt {1+x} \sin (x)-\frac {5}{2} (1+x)^{3/2} \sin (x) \] Output:

-7/4*(1+x)^(1/2)*cos(x)-3*(1+x)^(3/2)*cos(x)+(1+x)^(5/2)*cos(x)+7/8*2^(1/2 
)*Pi^(1/2)*cos(1)*FresnelC(2^(1/2)/Pi^(1/2)*(1+x)^(1/2))-9/4*2^(1/2)*Pi^(1 
/2)*cos(1)*FresnelS(2^(1/2)/Pi^(1/2)*(1+x)^(1/2))+9/4*2^(1/2)*Pi^(1/2)*Fre 
snelC(2^(1/2)/Pi^(1/2)*(1+x)^(1/2))*sin(1)+7/8*2^(1/2)*Pi^(1/2)*FresnelS(2 
^(1/2)/Pi^(1/2)*(1+x)^(1/2))*sin(1)+9/2*(1+x)^(1/2)*sin(x)-5/2*(1+x)^(3/2) 
*sin(x)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.21 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.00 \[ \int e^{\text {arctanh}(x)} (1-x)^{3/2} x \sin (x) \, dx=\frac {\left (\frac {1}{32}+\frac {i}{32}\right ) \sqrt {1-x} \left ((2+2 i) e^{-i x} \left ((8+15 i)-(2-19 i) x-10 x^2-4 i x^3\right )+(18+7 i) e^i \sqrt {2 \pi } \sqrt {-1-x} \text {erfi}\left (\frac {(1+i) \sqrt {-1-x}}{\sqrt {2}}\right )+(\cos (1)-i \sin (1)) \left ((-18+7 i) \sqrt {2 \pi } \sqrt {-1-x} \text {erf}\left (\frac {(1+i) \sqrt {-1-x}}{\sqrt {2}}\right )+(2+2 i) \left ((-8+15 i)+(2+19 i) x+10 x^2-4 i x^3\right ) (\cos (1+x)+i \sin (1+x))\right )\right )}{\sqrt {1-x^2}} \] Input:

Integrate[E^ArcTanh[x]*(1 - x)^(3/2)*x*Sin[x],x]
 

Output:

((1/32 + I/32)*Sqrt[1 - x]*(((2 + 2*I)*((8 + 15*I) - (2 - 19*I)*x - 10*x^2 
 - (4*I)*x^3))/E^(I*x) + (18 + 7*I)*E^I*Sqrt[2*Pi]*Sqrt[-1 - x]*Erfi[((1 + 
 I)*Sqrt[-1 - x])/Sqrt[2]] + (Cos[1] - I*Sin[1])*((-18 + 7*I)*Sqrt[2*Pi]*S 
qrt[-1 - x]*Erf[((1 + I)*Sqrt[-1 - x])/Sqrt[2]] + (2 + 2*I)*((-8 + 15*I) + 
 (2 + 19*I)*x + 10*x^2 - (4*I)*x^3)*(Cos[1 + x] + I*Sin[1 + x]))))/Sqrt[1 
- x^2]
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, 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.

\(\displaystyle \int (1-x)^{3/2} x e^{\text {arctanh}(x)} \sin (x) \, dx\)

\(\Big \downarrow \) 6679

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

\(\Big \downarrow \) 7267

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle 2 \left (\frac {9}{4} \sqrt {\frac {\pi }{2}} \sin (1) \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {x+1}\right )+\frac {7}{8} \sqrt {\frac {\pi }{2}} \cos (1) \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {x+1}\right )+\frac {7}{8} \sqrt {\frac {\pi }{2}} \sin (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {x+1}\right )-\frac {9}{4} \sqrt {\frac {\pi }{2}} \cos (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {x+1}\right )-\frac {5}{4} (x+1)^{3/2} \sin (x)+\frac {9}{4} \sqrt {x+1} \sin (x)+\frac {1}{2} (x+1)^{5/2} \cos (x)-\frac {3}{2} (x+1)^{3/2} \cos (x)-\frac {7}{8} \sqrt {x+1} \cos (x)\right )\)

Input:

Int[E^ArcTanh[x]*(1 - x)^(3/2)*x*Sin[x],x]
 

Output:

2*((-7*Sqrt[1 + x]*Cos[x])/8 - (3*(1 + x)^(3/2)*Cos[x])/2 + ((1 + x)^(5/2) 
*Cos[x])/2 + (7*Sqrt[Pi/2]*Cos[1]*FresnelC[Sqrt[2/Pi]*Sqrt[1 + x]])/8 - (9 
*Sqrt[Pi/2]*Cos[1]*FresnelS[Sqrt[2/Pi]*Sqrt[1 + x]])/4 + (9*Sqrt[Pi/2]*Fre 
snelC[Sqrt[2/Pi]*Sqrt[1 + x]]*Sin[1])/4 + (7*Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi 
]*Sqrt[1 + x]]*Sin[1])/8 + (9*Sqrt[1 + x]*Sin[x])/4 - (5*(1 + x)^(3/2)*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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

Timed out. \[ \int e^{\text {arctanh}(x)} (1-x)^{3/2} x \sin (x) \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.21 (sec) , antiderivative size = 1457, normalized size of antiderivative = 7.55 \[ \int e^{\text {arctanh}(x)} (1-x)^{3/2} x \sin (x) \, dx=\text {Too large to display} \] Input:

integrate((1+x)/(-x^2+1)^(1/2)*(1-x)^(3/2)*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(sq 
rt(-I*x - I)) - 1))*sin(1))*(x + 1)^2*cos(1/2*arctan2(x + 1, 0)) + 2*((sqr 
t(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)) + 3*(((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(3/2*arctan2(x + 1, 0)) + 3*(((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))*abs(x + 1)*sin(3/2*arctan2(x + 1, 0)) - (((I*cos(1) - sin(1))*gamma(5/ 
2, I*x + I) + (-I*cos(1) - sin(1))*gamma(5/2, -I*x - I))*x^2 - 2*((-I*cos( 
1) + sin(1))*gamma(5/2, I*x + I) + (I*cos(1) + sin(1))*gamma(5/2, -I*x - I 
))*x + (I*cos(1) - sin(1))*gamma(5/2, I*x + I) + (-I*cos(1) - sin(1))*gamm 
a(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))*x^2 + 2*((cos(1 
) + I*sin(1))*gamma(5/2, I*x + I) + (cos(1) - I*sin(1))*gamma(5/2, -I*x - 
I))*x + (cos(1) + I*sin(1))*gamma(5/2, I*x + I) + (cos(1) - I*sin(1))*gamm 
a(5/2, -I*x - I))*sin(5/2*arctan2(x + 1, 0)))/((x + 1)^(3/2)*sqrt(abs(x...
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.63 \[ \int e^{\text {arctanh}(x)} (1-x)^{3/2} x \sin (x) \, dx=\left (\frac {25}{32} i + \frac {11}{32}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {x + 1}\right ) e^{i} - \left (\frac {25}{32} i - \frac {11}{32}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {x + 1}\right ) e^{\left (-i\right )} - \frac {1}{8} i \, {\left (4 i \, {\left (x + 1\right )}^{\frac {5}{2}} - \left (12 i + 10\right ) \, {\left (x + 1\right )}^{\frac {3}{2}} - \left (3 i - 18\right ) \, \sqrt {x + 1}\right )} e^{\left (i \, x\right )} - \frac {1}{8} i \, {\left (4 i \, {\left (x + 1\right )}^{\frac {5}{2}} - \left (12 i - 10\right ) \, {\left (x + 1\right )}^{\frac {3}{2}} - \left (3 i + 18\right ) \, \sqrt {x + 1}\right )} e^{\left (-i \, x\right )} - \frac {1}{2} \, \sqrt {x + 1} e^{\left (i \, x\right )} - \frac {1}{2} \, \sqrt {x + 1} e^{\left (-i \, x\right )} \] Input:

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

Output:

(25/32*I + 11/32)*sqrt(2)*sqrt(pi)*erf(-(1/2*I + 1/2)*sqrt(2)*sqrt(x + 1)) 
*e^I - (25/32*I - 11/32)*sqrt(2)*sqrt(pi)*erf((1/2*I - 1/2)*sqrt(2)*sqrt(x 
 + 1))*e^(-I) - 1/8*I*(4*I*(x + 1)^(5/2) - (12*I + 10)*(x + 1)^(3/2) - (3* 
I - 18)*sqrt(x + 1))*e^(I*x) - 1/8*I*(4*I*(x + 1)^(5/2) - (12*I - 10)*(x + 
 1)^(3/2) - (3*I + 18)*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. \[ \int e^{\text {arctanh}(x)} (1-x)^{3/2} x \sin (x) \, dx=\int \frac {x\,\sin \left (x\right )\,{\left (1-x\right )}^{3/2}\,\left (x+1\right )}{\sqrt {1-x^2}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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