Integrand size = 14, antiderivative size = 141 \[ \int e^{\text {arctanh}(x)} \sqrt {1+x} \sin (x) \, dx=\sqrt {1-x} \cos (x)-\sqrt {\frac {\pi }{2}} \cos (1) \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+2 \sqrt {2 \pi } \cos (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-2 \sqrt {2 \pi } \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \sin (1)-\sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \sin (1) \] 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*2^(1/2)*Pi^(1/2)*cos(1)*FresnelS(2^(1/2)/Pi^(1/2)*(1-x)^(1/2 ))-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 = 0.44 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.95 \[ \int e^{\text {arctanh}(x)} \sqrt {1+x} \sin (x) \, dx=\left (\frac {1}{8}+\frac {i}{8}\right ) \left ((-4+i) e^{-i} \sqrt {2 \pi } \text {erfi}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {2-2 x}\right )+\frac {e^{-i x} \sqrt {1-x^2} \left ((2-2 i) \left (1+e^{2 i x}\right ) \sqrt {-1+x}+(4+i) e^{i (1+x)} \sqrt {2 \pi } \text {erfi}\left (\frac {(1+i) \sqrt {-1+x}}{\sqrt {2}}\right )\right )}{\sqrt {-1+x} \sqrt {1+x}}\right ) \] Input:
Integrate[E^ArcTanh[x]*Sqrt[1 + x]*Sin[x],x]
Output:
(1/8 + I/8)*(((-4 + I)*Sqrt[2*Pi]*Erfi[(1/2 + I/2)*Sqrt[2 - 2*x]])/E^I + ( Sqrt[1 - x^2]*((2 - 2*I)*(1 + E^((2*I)*x))*Sqrt[-1 + x] + (4 + I)*E^(I*(1 + x))*Sqrt[2*Pi]*Erfi[((1 + I)*Sqrt[-1 + x])/Sqrt[2]]))/(E^(I*x)*Sqrt[-1 + x]*Sqrt[1 + x]))
Time = 0.49 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, 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.
\(\displaystyle \int \sqrt {x+1} e^{\text {arctanh}(x)} \sin (x) \, dx\) |
\(\Big \downarrow \) 6679 |
\(\displaystyle \int \frac {(x+1) \sin (x)}{\sqrt {1-x}}dx\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle 2 \int -((x+1) \sin (x))d\sqrt {1-x}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int (x+1) \sin (x)d\sqrt {1-x}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -2 \int (2 \sin (x)-(1-x) \sin (x))d\sqrt {1-x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (-\sqrt {2 \pi } \sin (1) \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-\frac {1}{2} \sqrt {\frac {\pi }{2}} \cos (1) \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-\frac {1}{2} \sqrt {\frac {\pi }{2}} \sin (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+\sqrt {2 \pi } \cos (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+\frac {1}{2} \sqrt {1-x} \cos (x)\right )\) |
Input:
Int[E^ArcTanh[x]*Sqrt[1 + x]*Sin[x],x]
Output:
2*((Sqrt[1 - x]*Cos[x])/2 - (Sqrt[Pi/2]*Cos[1]*FresnelC[Sqrt[2/Pi]*Sqrt[1 - x]])/2 + Sqrt[2*Pi]*Cos[1]*FresnelS[Sqrt[2/Pi]*Sqrt[1 - x]] - Sqrt[2*Pi] *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]]
\[\int \frac {\left (1+x \right )^{\frac {3}{2}} \sin \left (x \right )}{\sqrt {-x^{2}+1}}d x\]
Input:
int((1+x)^(3/2)/(-x^2+1)^(1/2)*sin(x),x)
Output:
int((1+x)^(3/2)/(-x^2+1)^(1/2)*sin(x),x)
\[ \int e^{\text {arctanh}(x)} \sqrt {1+x} \sin (x) \, dx=\int { \frac {{\left (x + 1\right )}^{\frac {3}{2}} \sin \left (x\right )}{\sqrt {-x^{2} + 1}} \,d x } \] Input:
integrate((1+x)^(3/2)/(-x^2+1)^(1/2)*sin(x),x, algorithm="fricas")
Output:
integral(-sqrt(-x^2 + 1)*sqrt(x + 1)*sin(x)/(x - 1), x)
\[ \int e^{\text {arctanh}(x)} \sqrt {1+x} \sin (x) \, dx=\int \frac {\left (x + 1\right )^{\frac {3}{2}} \sin {\left (x \right )}}{\sqrt {- \left (x - 1\right ) \left (x + 1\right )}}\, dx \] Input:
integrate((1+x)**(3/2)/(-x**2+1)**(1/2)*sin(x),x)
Output:
Integral((x + 1)**(3/2)*sin(x)/sqrt(-(x - 1)*(x + 1)), x)
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 351, normalized size of antiderivative = 2.49 \[ \int e^{\text {arctanh}(x)} \sqrt {1+x} \sin (x) \, dx =\text {Too large to display} \] Input:
integrate((1+x)^(3/2)/(-x^2+1)^(1/2)*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 - 2*(((-I*sqrt(pi)*(erf(sqrt(I*x - I)) - 1) + I*sqrt(pi)*(e rf(sqrt(-I*x + I)) - 1))*cos(1) - (sqrt(pi)*(erf(sqrt(I*x - I)) - 1) + sqr t(pi)*(erf(sqrt(-I*x + I)) - 1))*sin(1))*cos(1/2*arctan2(x - 1, 0)) - ((sq rt(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 \[ \int e^{\text {arctanh}(x)} \sqrt {1+x} \sin (x) \, dx=-\left (\frac {5}{8} i + \frac {3}{8}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {-x + 1}\right ) e^{i} + \left (\frac {5}{8} i - \frac {3}{8}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {-x + 1}\right ) e^{\left (-i\right )} + \frac {1}{2} \, \sqrt {-x + 1} e^{\left (i \, x\right )} + \frac {1}{2} \, \sqrt {-x + 1} e^{\left (-i \, x\right )} \] Input:
integrate((1+x)^(3/2)/(-x^2+1)^(1/2)*sin(x),x, algorithm="giac")
Output:
-(5/8*I + 3/8)*sqrt(2)*sqrt(pi)*erf(-(1/2*I + 1/2)*sqrt(2)*sqrt(-x + 1))*e ^I + (5/8*I - 3/8)*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. \[ \int e^{\text {arctanh}(x)} \sqrt {1+x} \sin (x) \, dx=\int \frac {\sin \left (x\right )\,{\left (x+1\right )}^{3/2}}{\sqrt {1-x^2}} \,d x \] Input:
int((sin(x)*(x + 1)^(3/2))/(1 - x^2)^(1/2),x)
Output:
int((sin(x)*(x + 1)^(3/2))/(1 - x^2)^(1/2), x)
\[ \int e^{\text {arctanh}(x)} \sqrt {1+x} \sin (x) \, dx=\int \frac {\sin \left (x \right )}{\sqrt {1-x}}d x +\int \frac {\sin \left (x \right ) x}{\sqrt {1-x}}d x \] Input:
int((1+x)^(3/2)/(-x^2+1)^(1/2)*sin(x),x)
Output:
int(sin(x)/sqrt( - x + 1),x) + int((sin(x)*x)/sqrt( - x + 1),x)