Integrand size = 15, antiderivative size = 140 \[ \int \frac {e^{\text {arctanh}(x)} x \sin (x)}{\sqrt {1+x}} \, dx=\sqrt {1-x} \cos (x)-\sqrt {\frac {\pi }{2}} \cos (1) \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+\sqrt {2 \pi } \cos (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-\sqrt {2 \pi } \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \sin (1)-\sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \sin (1) \] Output:
(1-x)^(1/2)*cos(x)-1/2*2^(1/2)*Pi^(1/2)*cos(1)*FresnelC(2^(1/2)/Pi^(1/2)*( 1-x)^(1/2))+2^(1/2)*Pi^(1/2)*cos(1)*FresnelS(2^(1/2)/Pi^(1/2)*(1-x)^(1/2)) -2^(1/2)*Pi^(1/2)*FresnelC(2^(1/2)/Pi^(1/2)*(1-x)^(1/2))*sin(1)-1/2*2^(1/2 )*Pi^(1/2)*FresnelS(2^(1/2)/Pi^(1/2)*(1-x)^(1/2))*sin(1)
Result contains complex when optimal does not.
Time = 4.94 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.16 \[ \int \frac {e^{\text {arctanh}(x)} x \sin (x)}{\sqrt {1+x}} \, dx=\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \sqrt {1+x} \left ((-2-i) \sqrt {2 \pi } \sqrt {-1+x} \text {erfi}\left (\frac {(1+i) \sqrt {-1+x}}{\sqrt {2}}\right ) (\cos (1)+i \sin (1))-(2-2 i) (-1+x) (\cos (x)+i \sin (x))+\left ((-2+2 i) (-1+x) (\cos (1)+i \sin (1))+(1+2 i) \sqrt {2 \pi } \sqrt {-1+x} \text {erf}\left (\frac {(1+i) \sqrt {-1+x}}{\sqrt {2}}\right ) (-i \cos (x)+\sin (x))\right ) (\cos (1+x)-i \sin (1+x))\right )}{\sqrt {1-x^2}} \] Input:
Integrate[(E^ArcTanh[x]*x*Sin[x])/Sqrt[1 + x],x]
Output:
((1/8 + I/8)*Sqrt[1 + x]*((-2 - I)*Sqrt[2*Pi]*Sqrt[-1 + x]*Erfi[((1 + I)*S qrt[-1 + x])/Sqrt[2]]*(Cos[1] + I*Sin[1]) - (2 - 2*I)*(-1 + x)*(Cos[x] + I *Sin[x]) + ((-2 + 2*I)*(-1 + x)*(Cos[1] + I*Sin[1]) + (1 + 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]
Time = 0.80 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.09, 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.
\(\displaystyle \int \frac {x e^{\text {arctanh}(x)} \sin (x)}{\sqrt {x+1}} \, dx\) |
\(\Big \downarrow \) 6679 |
\(\displaystyle \int \frac {x \sin (x)}{\sqrt {1-x}}dx\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle 2 \int -x \sin (x)d\sqrt {1-x}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int x \sin (x)d\sqrt {1-x}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -2 \int (\sin (x)-(1-x) \sin (x))d\sqrt {1-x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (-\sqrt {\frac {\pi }{2}} \sin (1) \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-\frac {1}{2} \sqrt {\frac {\pi }{2}} \cos (1) \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-\frac {1}{2} \sqrt {\frac {\pi }{2}} \sin (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+\sqrt {\frac {\pi }{2}} \cos (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+\frac {1}{2} \sqrt {1-x} \cos (x)\right )\) |
Input:
Int[(E^ArcTanh[x]*x*Sin[x])/Sqrt[1 + x],x]
Output:
2*((Sqrt[1 - x]*Cos[x])/2 - (Sqrt[Pi/2]*Cos[1]*FresnelC[Sqrt[2/Pi]*Sqrt[1 - x]])/2 + Sqrt[Pi/2]*Cos[1]*FresnelS[Sqrt[2/Pi]*Sqrt[1 - x]] - Sqrt[Pi/2] *FresnelC[Sqrt[2/Pi]*Sqrt[1 - x]]*Sin[1] - (Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi] *Sqrt[1 - x]]*Sin[1])/2)
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])
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]]
\[\int \frac {\sqrt {1+x}\, x \sin \left (x \right )}{\sqrt {-x^{2}+1}}d x\]
Input:
int((1+x)^(1/2)/(-x^2+1)^(1/2)*x*sin(x),x)
Output:
int((1+x)^(1/2)/(-x^2+1)^(1/2)*x*sin(x),x)
\[ \int \frac {e^{\text {arctanh}(x)} x \sin (x)}{\sqrt {1+x}} \, dx=\int { \frac {\sqrt {x + 1} x \sin \left (x\right )}{\sqrt {-x^{2} + 1}} \,d x } \] Input:
integrate((1+x)^(1/2)/(-x^2+1)^(1/2)*x*sin(x),x, algorithm="fricas")
Output:
integral(-sqrt(-x^2 + 1)*sqrt(x + 1)*x*sin(x)/(x^2 - 1), x)
\[ \int \frac {e^{\text {arctanh}(x)} x \sin (x)}{\sqrt {1+x}} \, dx=\int \frac {x \sqrt {x + 1} \sin {\left (x \right )}}{\sqrt {- \left (x - 1\right ) \left (x + 1\right )}}\, dx \] Input:
integrate((1+x)**(1/2)/(-x**2+1)**(1/2)*x*sin(x),x)
Output:
Integral(x*sqrt(x + 1)*sin(x)/sqrt(-(x - 1)*(x + 1)), x)
Result contains complex when optimal does not.
Time = 0.28 (sec) , antiderivative size = 347, normalized size of antiderivative = 2.48 \[ \int \frac {e^{\text {arctanh}(x)} x \sin (x)}{\sqrt {1+x}} \, dx =\text {Too large to display} \] Input:
integrate((1+x)^(1/2)/(-x^2+1)^(1/2)*x*sin(x),x, algorithm="maxima")
Output:
-1/2*((((-I*cos(1) - sin(1))*gamma(3/2, I*x - I) + (I*cos(1) - sin(1))*gam ma(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 + (((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(p i)*(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))*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)))*abs(x - 1) + ((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)))*sqrt(-x + 1)* sqrt(abs(x - 1))/(x - 1)^2
Result contains complex when optimal does not.
Time = 0.12 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.52 \[ \int \frac {e^{\text {arctanh}(x)} x \sin (x)}{\sqrt {1+x}} \, dx=-\left (\frac {3}{8} i + \frac {1}{8}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {-x + 1}\right ) e^{i} + \left (\frac {3}{8} i - \frac {1}{8}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {-x + 1}\right ) e^{\left (-i\right )} + \frac {1}{2} \, \sqrt {-x + 1} e^{\left (i \, x\right )} + \frac {1}{2} \, \sqrt {-x + 1} e^{\left (-i \, x\right )} \] Input:
integrate((1+x)^(1/2)/(-x^2+1)^(1/2)*x*sin(x),x, algorithm="giac")
Output:
-(3/8*I + 1/8)*sqrt(2)*sqrt(pi)*erf(-(1/2*I + 1/2)*sqrt(2)*sqrt(-x + 1))*e ^I + (3/8*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)
Timed out. \[ \int \frac {e^{\text {arctanh}(x)} x \sin (x)}{\sqrt {1+x}} \, dx=\int \frac {x\,\sin \left (x\right )\,\sqrt {x+1}}{\sqrt {1-x^2}} \,d x \] Input:
int((x*sin(x)*(x + 1)^(1/2))/(1 - x^2)^(1/2),x)
Output:
int((x*sin(x)*(x + 1)^(1/2))/(1 - x^2)^(1/2), x)
\[ \int \frac {e^{\text {arctanh}(x)} x \sin (x)}{\sqrt {1+x}} \, dx=\int \frac {\sin \left (x \right ) x}{\sqrt {1-x}}d x \] Input:
int((1+x)^(1/2)/(-x^2+1)^(1/2)*x*sin(x),x)
Output:
int((sin(x)*x)/sqrt( - x + 1),x)