Integrand size = 14, antiderivative size = 62 \[ \int \frac {e^{\text {arctanh}(x)} \sin (x)}{\sqrt {1+x}} \, dx=\sqrt {2 \pi } \cos (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-\sqrt {2 \pi } \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \sin (1) \] Output:
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)
Result contains complex when optimal does not.
Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.13 \[ \int \frac {e^{\text {arctanh}(x)} \sin (x)}{\sqrt {1+x}} \, dx=-\frac {e^{-i} \left (e^{2 i} \sqrt {-i (-1+x)} \Gamma \left (\frac {1}{2},-i (-1+x)\right )+\sqrt {i (-1+x)} \Gamma \left (\frac {1}{2},i (-1+x)\right )\right )}{2 \sqrt {1-x}} \] Input:
Integrate[(E^ArcTanh[x]*Sin[x])/Sqrt[1 + x],x]
Output:
-1/2*(E^(2*I)*Sqrt[(-I)*(-1 + x)]*Gamma[1/2, (-I)*(-1 + x)] + Sqrt[I*(-1 + x)]*Gamma[1/2, I*(-1 + x)])/(E^I*Sqrt[1 - x])
Time = 0.49 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {6679, 3042, 3787, 25, 3042, 3785, 3786, 3832, 3833}
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.
\(\displaystyle \int \frac {e^{\text {arctanh}(x)} \sin (x)}{\sqrt {x+1}} \, dx\) |
\(\Big \downarrow \) 6679 |
\(\displaystyle \int \frac {\sin (x)}{\sqrt {1-x}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (x)}{\sqrt {1-x}}dx\) |
\(\Big \downarrow \) 3787 |
\(\displaystyle \sin (1) \int \frac {\cos (1-x)}{\sqrt {1-x}}dx+\cos (1) \int -\frac {\sin (1-x)}{\sqrt {1-x}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \sin (1) \int \frac {\cos (1-x)}{\sqrt {1-x}}dx-\cos (1) \int \frac {\sin (1-x)}{\sqrt {1-x}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sin (1) \int \frac {\sin \left (-x+\frac {\pi }{2}+1\right )}{\sqrt {1-x}}dx-\cos (1) \int \frac {\sin (1-x)}{\sqrt {1-x}}dx\) |
\(\Big \downarrow \) 3785 |
\(\displaystyle -2 \sin (1) \int \cos (1-x)d\sqrt {1-x}-\cos (1) \int \frac {\sin (1-x)}{\sqrt {1-x}}dx\) |
\(\Big \downarrow \) 3786 |
\(\displaystyle 2 \cos (1) \int \sin (1-x)d\sqrt {1-x}-2 \sin (1) \int \cos (1-x)d\sqrt {1-x}\) |
\(\Big \downarrow \) 3832 |
\(\displaystyle \sqrt {2 \pi } \cos (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-2 \sin (1) \int \cos (1-x)d\sqrt {1-x}\) |
\(\Big \downarrow \) 3833 |
\(\displaystyle \sqrt {2 \pi } \cos (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-\sqrt {2 \pi } \sin (1) \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )\) |
Input:
Int[(E^ArcTanh[x]*Sin[x])/Sqrt[1 + x],x]
Output:
Sqrt[2*Pi]*Cos[1]*FresnelS[Sqrt[2/Pi]*Sqrt[1 - x]] - Sqrt[2*Pi]*FresnelC[S qrt[2/Pi]*Sqrt[1 - x]]*Sin[1]
Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> S imp[2/d Subst[Int[Cos[f*(x^2/d)], x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]
Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[2/d Subst[Int[Sin[f*(x^2/d)], x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f }, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]
Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Cos [(d*e - c*f)/d] Int[Sin[c*(f/d) + f*x]/Sqrt[c + d*x], x], x] + Simp[Sin[( d*e - c*f)/d] Int[Cos[c*(f/d) + f*x]/Sqrt[c + d*x], x], x] /; FreeQ[{c, d , e, f}, x] && ComplexFreeQ[f] && NeQ[d*e - c*f, 0]
Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
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])
\[\int \frac {\sqrt {1+x}\, \sin \left (x \right )}{\sqrt {-x^{2}+1}}d x\]
Input:
int((1+x)^(1/2)/(-x^2+1)^(1/2)*sin(x),x)
Output:
int((1+x)^(1/2)/(-x^2+1)^(1/2)*sin(x),x)
\[ \int \frac {e^{\text {arctanh}(x)} \sin (x)}{\sqrt {1+x}} \, dx=\int { \frac {\sqrt {x + 1} \sin \left (x\right )}{\sqrt {-x^{2} + 1}} \,d x } \] Input:
integrate((1+x)^(1/2)/(-x^2+1)^(1/2)*sin(x),x, algorithm="fricas")
Output:
integral(-sqrt(-x^2 + 1)*sqrt(x + 1)*sin(x)/(x^2 - 1), x)
\[ \int \frac {e^{\text {arctanh}(x)} \sin (x)}{\sqrt {1+x}} \, dx=\int \frac {\sqrt {x + 1} \sin {\left (x \right )}}{\sqrt {- \left (x - 1\right ) \left (x + 1\right )}}\, dx \] Input:
integrate((1+x)**(1/2)/(-x**2+1)**(1/2)*sin(x),x)
Output:
Integral(sqrt(x + 1)*sin(x)/sqrt(-(x - 1)*(x + 1)), x)
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.71 \[ \int \frac {e^{\text {arctanh}(x)} \sin (x)}{\sqrt {1+x}} \, dx=-\frac {{\left ({\left ({\left (i \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {i \, x - i}\right ) - 1\right )} - i \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {-i \, x + i}\right ) - 1\right )}\right )} \cos \left (1\right ) + {\left (\sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {i \, x - i}\right ) - 1\right )} + \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {-i \, x + i}\right ) - 1\right )}\right )} \sin \left (1\right )\right )} \cos \left (\frac {1}{2} \, \arctan \left (x - 1, 0\right )\right ) + {\left ({\left (\sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {i \, x - i}\right ) - 1\right )} + \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {-i \, x + i}\right ) - 1\right )}\right )} \cos \left (1\right ) + {\left (-i \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {i \, x - i}\right ) - 1\right )} + i \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {-i \, x + i}\right ) - 1\right )}\right )} \sin \left (1\right )\right )} \sin \left (\frac {1}{2} \, \arctan \left (x - 1, 0\right )\right )\right )} \sqrt {-x + 1}}{2 \, \sqrt {{\left | x - 1 \right |}}} \] Input:
integrate((1+x)^(1/2)/(-x^2+1)^(1/2)*sin(x),x, algorithm="maxima")
Output:
-1/2*(((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(sqrt (-I*x + I)) - 1))*sin(1))*cos(1/2*arctan2(x - 1, 0)) + ((sqrt(pi)*(erf(sqr t(I*x - I)) - 1) + sqrt(pi)*(erf(sqrt(-I*x + I)) - 1))*cos(1) + (-I*sqrt(p i)*(erf(sqrt(I*x - I)) - 1) + I*sqrt(pi)*(erf(sqrt(-I*x + I)) - 1))*sin(1) )*sin(1/2*arctan2(x - 1, 0)))*sqrt(-x + 1)/sqrt(abs(x - 1))
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.76 \[ \int \frac {e^{\text {arctanh}(x)} \sin (x)}{\sqrt {1+x}} \, dx=-\left (\frac {1}{4} i + \frac {1}{4}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {-x + 1}\right ) e^{i} + \left (\frac {1}{4} i - \frac {1}{4}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {-x + 1}\right ) e^{\left (-i\right )} \] Input:
integrate((1+x)^(1/2)/(-x^2+1)^(1/2)*sin(x),x, algorithm="giac")
Output:
-(1/4*I + 1/4)*sqrt(2)*sqrt(pi)*erf(-(1/2*I + 1/2)*sqrt(2)*sqrt(-x + 1))*e ^I + (1/4*I - 1/4)*sqrt(2)*sqrt(pi)*erf((1/2*I - 1/2)*sqrt(2)*sqrt(-x + 1) )*e^(-I)
Timed out. \[ \int \frac {e^{\text {arctanh}(x)} \sin (x)}{\sqrt {1+x}} \, dx=\int \frac {\sin \left (x\right )\,\sqrt {x+1}}{\sqrt {1-x^2}} \,d x \] Input:
int((sin(x)*(x + 1)^(1/2))/(1 - x^2)^(1/2),x)
Output:
int((sin(x)*(x + 1)^(1/2))/(1 - x^2)^(1/2), x)
\[ \int \frac {e^{\text {arctanh}(x)} \sin (x)}{\sqrt {1+x}} \, dx=\int \frac {\sin \left (x \right )}{\sqrt {1-x}}d x \] Input:
int((1+x)^(1/2)/(-x^2+1)^(1/2)*sin(x),x)
Output:
int(sin(x)/sqrt( - x + 1),x)