\(\int e^{\text {arctanh}(x)} x (1+x)^{3/2} \sin (x) \, dx\) [849]

Optimal result

Integrand size = 15, antiderivative size = 335 \[ \int e^{\text {arctanh}(x)} x (1+x)^{3/2} \sin (x) \, dx=\frac {17}{4} \sqrt {1-x} \cos (x)-5 (1-x)^{3/2} \cos (x)+(1-x)^{5/2} \cos (x)+\frac {15}{4} \sqrt {\frac {\pi }{2}} \cos (1) \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-4 \sqrt {2 \pi } \cos (1) \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-\frac {15}{2} \sqrt {\frac {\pi }{2}} \cos (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+4 \sqrt {2 \pi } \cos (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+\frac {15}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \sin (1)-4 \sqrt {2 \pi } \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \sin (1)+\frac {15}{4} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \sin (1)-4 \sqrt {2 \pi } \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \sin (1)-\frac {15}{2} \sqrt {1-x} \sin (x)+\frac {5}{2} (1-x)^{3/2} \sin (x) \] Output:

17/4*(1-x)^(1/2)*cos(x)-5*(1-x)^(3/2)*cos(x)+(1-x)^(5/2)*cos(x)-17/8*2^(1/ 
2)*Pi^(1/2)*cos(1)*FresnelC(2^(1/2)/Pi^(1/2)*(1-x)^(1/2))+1/4*2^(1/2)*Pi^( 
1/2)*cos(1)*FresnelS(2^(1/2)/Pi^(1/2)*(1-x)^(1/2))-1/4*2^(1/2)*Pi^(1/2)*Fr 
esnelC(2^(1/2)/Pi^(1/2)*(1-x)^(1/2))*sin(1)-17/8*2^(1/2)*Pi^(1/2)*FresnelS 
(2^(1/2)/Pi^(1/2)*(1-x)^(1/2))*sin(1)-15/2*(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 = 6.57 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.60 \[ \int e^{\text {arctanh}(x)} x (1+x)^{3/2} \sin (x) \, dx=\frac {\left (\frac {1}{32}+\frac {i}{32}\right ) \sqrt {1+x} \left ((-2-17 i) \sqrt {2 \pi } \sqrt {-1+x} \text {erfi}\left (\frac {(1+i) \sqrt {-1+x}}{\sqrt {2}}\right ) (\cos (1)+i \sin (1))-(2-2 i) \left ((-1-20 i)-(11-10 i) x+(8+10 i) x^2+4 x^3\right ) (\cos (x)+i \sin (x))+\left ((2+2 i) \left ((-20-i)+(10-11 i) x+(10+8 i) x^2+4 i x^3\right ) (\cos (1)+i \sin (1))+(17+2 i) \sqrt {2 \pi } \sqrt {-1+x} \text {erf}\left (\frac {(1+i) \sqrt {-1+x}}{\sqrt {2}}\right ) (-i \cos (x)+\sin (x))\right ) (\cos (1+x)-i \sin (1+x))\right )}{\sqrt {1-x^2}} \] Input:

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

Output:

((1/32 + I/32)*Sqrt[1 + x]*((-2 - 17*I)*Sqrt[2*Pi]*Sqrt[-1 + x]*Erfi[((1 + 
 I)*Sqrt[-1 + x])/Sqrt[2]]*(Cos[1] + I*Sin[1]) - (2 - 2*I)*((-1 - 20*I) - 
(11 - 10*I)*x + (8 + 10*I)*x^2 + 4*x^3)*(Cos[x] + I*Sin[x]) + ((2 + 2*I)*( 
(-20 - I) + (10 - 11*I)*x + (10 + 8*I)*x^2 + (4*I)*x^3)*(Cos[1] + I*Sin[1] 
) + (17 + 2*I)*Sqrt[2*Pi]*Sqrt[-1 + x]*Erf[((1 + I)*Sqrt[-1 + x])/Sqrt[2]] 
*((-I)*Cos[x] + Sin[x]))*(Cos[1 + x] - I*Sin[1 + x])))/Sqrt[1 - x^2]
 

Rubi [A] (verified)

Time = 1.42 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6679, 7267, 25, 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.

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

\(\Big \downarrow \) 6679

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

\(\Big \downarrow \) 7267

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 7293

\(\displaystyle -2 \int \left (-\sin (x) (1-x)^3+5 \sin (x) (1-x)^2-8 \sin (x) (1-x)+4 \sin (x)\right )d\sqrt {1-x}\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 \left (-2 \sqrt {2 \pi } \sin (1) \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+\frac {15}{4} \sqrt {\frac {\pi }{2}} \sin (1) \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-2 \sqrt {2 \pi } \cos (1) \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+\frac {15}{8} \sqrt {\frac {\pi }{2}} \cos (1) \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-2 \sqrt {2 \pi } \sin (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+\frac {15}{8} \sqrt {\frac {\pi }{2}} \sin (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+2 \sqrt {2 \pi } \cos (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-\frac {15}{4} \sqrt {\frac {\pi }{2}} \cos (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+\frac {5}{4} (1-x)^{3/2} \sin (x)-\frac {15}{4} \sqrt {1-x} \sin (x)+\frac {1}{2} (1-x)^{5/2} \cos (x)-\frac {5}{2} (1-x)^{3/2} \cos (x)+\frac {17}{8} \sqrt {1-x} \cos (x)\right )\)

Input:

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

Output:

2*((17*Sqrt[1 - x]*Cos[x])/8 - (5*(1 - x)^(3/2)*Cos[x])/2 + ((1 - x)^(5/2) 
*Cos[x])/2 + (15*Sqrt[Pi/2]*Cos[1]*FresnelC[Sqrt[2/Pi]*Sqrt[1 - x]])/8 - 2 
*Sqrt[2*Pi]*Cos[1]*FresnelC[Sqrt[2/Pi]*Sqrt[1 - x]] - (15*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]] + (15*Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Sqrt[1 - x]]*Sin[1 
])/4 - 2*Sqrt[2*Pi]*FresnelC[Sqrt[2/Pi]*Sqrt[1 - x]]*Sin[1] + (15*Sqrt[Pi/ 
2]*FresnelS[Sqrt[2/Pi]*Sqrt[1 - x]]*Sin[1])/8 - 2*Sqrt[2*Pi]*FresnelS[Sqrt 
[2/Pi]*Sqrt[1 - x]]*Sin[1] - (15*Sqrt[1 - x]*Sin[x])/4 + (5*(1 - x)^(3/2)* 
Sin[x])/4)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

integrate((1+x)**(5/2)/(-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 = 0.38 (sec) , antiderivative size = 983, normalized size of antiderivative = 2.93 \[ \int e^{\text {arctanh}(x)} x (1+x)^{3/2} \sin (x) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*((((-I*cos(1) - sin(1))*gamma(7/2, I*x - I) + (I*cos(1) - sin(1))*gam 
ma(7/2, -I*x + I))*cos(7/2*arctan2(x - 1, 0)) - ((cos(1) - I*sin(1))*gamma 
(7/2, I*x - I) + (cos(1) + I*sin(1))*gamma(7/2, -I*x + I))*sin(7/2*arctan2 
(x - 1, 0)))*x^3 - 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) + s 
qrt(pi)*(erf(sqrt(-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))*co 
s(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*abs(x - 1) - 8*((( 
-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 
)))*(x - 1)^2 - (5*(((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))*si 
n(5/2*arctan2(x - 1, 0)))*abs(x - 1) + 3*((-I*cos(1) - sin(1))*gamma(7/2, 
I*x - I) + (I*cos(1) - sin(1))*gamma(7/2, -I*x + I))*cos(7/2*arctan2(x - 1 
, 0)) - 3*((cos(1) - I*sin(1))*gamma(7/2, I*x - I) + (cos(1) + I*sin(1))*g 
amma(7/2, -I*x + I))*sin(7/2*arctan2(x - 1, 0)))*x^2 - (8*(((I*cos(1) + si 
n(1))*gamma(3/2, I*x - I) + (-I*cos(1) + sin(1))*gamma(3/2, -I*x + I))*...
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.60 \[ \int e^{\text {arctanh}(x)} x (1+x)^{3/2} \sin (x) \, dx=-\left (\frac {19}{32} i - \frac {15}{32}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {-x + 1}\right ) e^{i} + \left (\frac {19}{32} i + \frac {15}{32}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {-x + 1}\right ) e^{\left (-i\right )} - \frac {1}{8} i \, {\left (4 i \, {\left (x - 1\right )}^{2} \sqrt {-x + 1} - \left (12 i - 10\right ) \, {\left (-x + 1\right )}^{\frac {3}{2}} - \left (3 i + 18\right ) \, \sqrt {-x + 1}\right )} e^{\left (i \, x\right )} - \frac {1}{2} i \, {\left (-2 i \, {\left (-x + 1\right )}^{\frac {3}{2}} + \left (4 i - 3\right ) \, \sqrt {-x + 1}\right )} e^{\left (i \, x\right )} - \frac {1}{8} i \, {\left (4 i \, {\left (x - 1\right )}^{2} \sqrt {-x + 1} - \left (12 i + 10\right ) \, {\left (-x + 1\right )}^{\frac {3}{2}} - \left (3 i - 18\right ) \, \sqrt {-x + 1}\right )} e^{\left (-i \, x\right )} - \frac {1}{2} i \, {\left (-2 i \, {\left (-x + 1\right )}^{\frac {3}{2}} + \left (4 i + 3\right ) \, \sqrt {-x + 1}\right )} e^{\left (-i \, x\right )} + \frac {1}{2} \, \sqrt {-x + 1} e^{\left (i \, x\right )} + \frac {1}{2} \, \sqrt {-x + 1} e^{\left (-i \, x\right )} \] Input:

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

Output:

-(19/32*I - 15/32)*sqrt(2)*sqrt(pi)*erf(-(1/2*I + 1/2)*sqrt(2)*sqrt(-x + 1 
))*e^I + (19/32*I + 15/32)*sqrt(2)*sqrt(pi)*erf((1/2*I - 1/2)*sqrt(2)*sqrt 
(-x + 1))*e^(-I) - 1/8*I*(4*I*(x - 1)^2*sqrt(-x + 1) - (12*I - 10)*(-x + 1 
)^(3/2) - (3*I + 18)*sqrt(-x + 1))*e^(I*x) - 1/2*I*(-2*I*(-x + 1)^(3/2) + 
(4*I - 3)*sqrt(-x + 1))*e^(I*x) - 1/8*I*(4*I*(x - 1)^2*sqrt(-x + 1) - (12* 
I + 10)*(-x + 1)^(3/2) - (3*I - 18)*sqrt(-x + 1))*e^(-I*x) - 1/2*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. \[ \int e^{\text {arctanh}(x)} x (1+x)^{3/2} \sin (x) \, dx=\int \frac {x\,\sin \left (x\right )\,{\left (x+1\right )}^{5/2}}{\sqrt {1-x^2}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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