\(\int e^{\text {arctanh}(x)} (1+x)^{3/2} \sin (x) \, dx\) [835]

Optimal result

Integrand size = 14, antiderivative size = 236 \[ \int e^{\text {arctanh}(x)} (1+x)^{3/2} \sin (x) \, dx=4 \sqrt {1-x} \cos (x)-(1-x)^{3/2} \cos (x)-2 \sqrt {2 \pi } \cos (1) \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-\frac {3}{2} \sqrt {\frac {\pi }{2}} \cos (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+4 \sqrt {2 \pi } \cos (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+\frac {3}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \sin (1)-4 \sqrt {2 \pi } \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \sin (1)-2 \sqrt {2 \pi } \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \sin (1)-\frac {3}{2} \sqrt {1-x} \sin (x) \] Output:

4*(1-x)^(1/2)*cos(x)-(1-x)^(3/2)*cos(x)-2*2^(1/2)*Pi^(1/2)*cos(1)*FresnelC 
(2^(1/2)/Pi^(1/2)*(1-x)^(1/2))+13/4*2^(1/2)*Pi^(1/2)*cos(1)*FresnelS(2^(1/ 
2)/Pi^(1/2)*(1-x)^(1/2))-13/4*2^(1/2)*Pi^(1/2)*FresnelC(2^(1/2)/Pi^(1/2)*( 
1-x)^(1/2))*sin(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 = 0.25 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.75 \[ \int e^{\text {arctanh}(x)} (1+x)^{3/2} \sin (x) \, dx=\frac {i \sqrt {1-x^2} \left ((5+21 i) \sqrt {\frac {\pi }{2}} \text {erfi}\left (\frac {(1+i) \sqrt {-1+x}}{\sqrt {2}}\right ) (-i \cos (1)+\sin (1))+2 \sqrt {-1+x} ((6+3 i)+2 x) (-i \cos (x)+\sin (x))-\left (2 ((3+6 i)+2 i x) \sqrt {-1+x} (\cos (1)+i \sin (1))+(21+5 i) \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {(1+i) \sqrt {-1+x}}{\sqrt {2}}\right ) (-i \cos (x)+\sin (x))\right ) (\cos (1+x)-i \sin (1+x))\right )}{8 \sqrt {-1+x} \sqrt {1+x}} \] Input:

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

Output:

((I/8)*Sqrt[1 - x^2]*((5 + 21*I)*Sqrt[Pi/2]*Erfi[((1 + I)*Sqrt[-1 + x])/Sq 
rt[2]]*((-I)*Cos[1] + Sin[1]) + 2*Sqrt[-1 + x]*((6 + 3*I) + 2*x)*((-I)*Cos 
[x] + Sin[x]) - (2*((3 + 6*I) + (2*I)*x)*Sqrt[-1 + x]*(Cos[1] + I*Sin[1]) 
+ (21 + 5*I)*Sqrt[Pi/2]*Erf[((1 + I)*Sqrt[-1 + x])/Sqrt[2]]*((-I)*Cos[x] + 
 Sin[x]))*(Cos[1 + x] - I*Sin[1 + x])))/(Sqrt[-1 + x]*Sqrt[1 + x])
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, 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.

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

\(\Big \downarrow \) 6679

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

\(\Big \downarrow \) 7267

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

\(\Big \downarrow \) 7293

\(\displaystyle -2 \int \left (\sin (x) (1-x)^2-4 \sin (x) (1-x)+4 \sin (x)\right )d\sqrt {1-x}\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-2*(-2*Sqrt[1 - x]*Cos[x] + ((1 - x)^(3/2)*Cos[x])/2 + Sqrt[2*Pi]*Cos[1]*F 
resnelC[Sqrt[2/Pi]*Sqrt[1 - x]] + (3*Sqrt[Pi/2]*Cos[1]*FresnelS[Sqrt[2/Pi] 
*Sqrt[1 - x]])/4 - 2*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 + 2*Sqrt[2*Pi]*Fr 
esnelC[Sqrt[2/Pi]*Sqrt[1 - x]]*Sin[1] + Sqrt[2*Pi]*FresnelS[Sqrt[2/Pi]*Sqr 
t[1 - x]]*Sin[1] + (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]

\[\int \frac {\left (1+x \right )^{\frac {5}{2}} \sin \left (x \right )}{\sqrt {-x^{2}+1}}d x\]

Input:

int((1+x)^(5/2)/(-x^2+1)^(1/2)*sin(x),x)
 

Output:

int((1+x)^(5/2)/(-x^2+1)^(1/2)*sin(x),x)
 

Fricas [F]

\[ \int e^{\text {arctanh}(x)} (1+x)^{3/2} \sin (x) \, dx=\int { \frac {{\left (x + 1\right )}^{\frac {5}{2}} \sin \left (x\right )}{\sqrt {-x^{2} + 1}} \,d x } \] Input:

integrate((1+x)^(5/2)/(-x^2+1)^(1/2)*sin(x),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

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

integrate((1+x)**(5/2)/(-x**2+1)**(1/2)*sin(x),x)
 

Output:

Timed out
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 628, normalized size of antiderivative = 2.66 \[ \int e^{\text {arctanh}(x)} (1+x)^{3/2} \sin (x) \, dx=\text {Too large to display} \] Input:

integrate((1+x)^(5/2)/(-x^2+1)^(1/2)*sin(x),x, algorithm="maxima")
 

Output:

1/2*(4*(((-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 + 2*(2*(((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))*ga 
mma(5/2, -I*x + I))*cos(5/2*arctan2(x - 1, 0)) - ((cos(1) - I*sin(1))*gamm 
a(5/2, I*x - I) + (cos(1) + I*sin(1))*gamma(5/2, -I*x + I))*sin(5/2*arctan 
2(x - 1, 0)))*x + 4*(((-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))*s 
in(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 = 0.15 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.51 \[ \int e^{\text {arctanh}(x)} (1+x)^{3/2} \sin (x) \, dx=-\left (\frac {21}{16} i + \frac {5}{16}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {-x + 1}\right ) e^{i} + \left (\frac {21}{16} i - \frac {5}{16}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {-x + 1}\right ) e^{\left (-i\right )} - \frac {1}{4} i \, {\left (-2 i \, {\left (-x + 1\right )}^{\frac {3}{2}} + \left (4 i - 3\right ) \, \sqrt {-x + 1}\right )} e^{\left (i \, x\right )} - \frac {1}{4} i \, {\left (-2 i \, {\left (-x + 1\right )}^{\frac {3}{2}} + \left (4 i + 3\right ) \, \sqrt {-x + 1}\right )} e^{\left (-i \, x\right )} + \sqrt {-x + 1} e^{\left (i \, x\right )} + \sqrt {-x + 1} e^{\left (-i \, x\right )} \] Input:

integrate((1+x)^(5/2)/(-x^2+1)^(1/2)*sin(x),x, algorithm="giac")
 

Output:

-(21/16*I + 5/16)*sqrt(2)*sqrt(pi)*erf(-(1/2*I + 1/2)*sqrt(2)*sqrt(-x + 1) 
)*e^I + (21/16*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) + sqrt 
(-x + 1)*e^(I*x) + sqrt(-x + 1)*e^(-I*x)
 

Mupad [F(-1)]

Timed out. \[ \int e^{\text {arctanh}(x)} (1+x)^{3/2} \sin (x) \, dx=\int \frac {\sin \left (x\right )\,{\left (x+1\right )}^{5/2}}{\sqrt {1-x^2}} \,d x \] Input:

int((sin(x)*(x + 1)^(5/2))/(1 - x^2)^(1/2),x)
 

Output:

int((sin(x)*(x + 1)^(5/2))/(1 - x^2)^(1/2), x)
 

Reduce [F]

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

int((1+x)^(5/2)/(-x^2+1)^(1/2)*sin(x),x)
 

Output:

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