Integrand size = 16, antiderivative size = 157 \[ \int e^{\text {arctanh}(x)} (1-x)^{3/2} \sin (x) \, dx=-2 \sqrt {1+x} \cos (x)+(1+x)^{3/2} \cos (x)+\sqrt {2 \pi } \cos (1) \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1+x}\right )+\frac {3}{2} \sqrt {\frac {\pi }{2}} \cos (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1+x}\right )-\frac {3}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1+x}\right ) \sin (1)+\sqrt {2 \pi } \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {1+x}\right ) \sin (1)-\frac {3}{2} \sqrt {1+x} \sin (x) \]
(1+x)^(3/2)*cos(x)+3/4*cos(1)*FresnelS(2^(1/2)/Pi^(1/2)*(1+x)^(1/2))*2^(1/ 2)*Pi^(1/2)-3/4*FresnelC(2^(1/2)/Pi^(1/2)*(1+x)^(1/2))*sin(1)*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)+Fresne lS(2^(1/2)/Pi^(1/2)*(1+x)^(1/2))*sin(1)*2^(1/2)*Pi^(1/2)-2*cos(x)*(1+x)^(1 /2)-3/2*sin(x)*(1+x)^(1/2)
Result contains complex when optimal does not.
Time = 2.48 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.12 \[ \int e^{\text {arctanh}(x)} (1-x)^{3/2} \sin (x) \, dx=\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) e^{-i x} \sqrt {1-x^2} \left ((2+2 i) \sqrt {-1-x} \left ((-3+2 i)+e^{2 i x} ((3+2 i)-2 i x)-2 i x\right )-(3+4 i) e^{i x} \sqrt {2 \pi } \text {erf}\left (\frac {(1+i) \sqrt {-1-x}}{\sqrt {2}}\right ) (\cos (1)-i \sin (1))+(4+3 i) e^{i x} \sqrt {2 \pi } \text {erfi}\left (\frac {(1+i) \sqrt {-1-x}}{\sqrt {2}}\right ) (-i \cos (1)+\sin (1))\right )}{\sqrt {-1-x} \sqrt {1-x}} \]
((1/16 + I/16)*Sqrt[1 - x^2]*((2 + 2*I)*Sqrt[-1 - x]*((-3 + 2*I) + E^((2*I )*x)*((3 + 2*I) - (2*I)*x) - (2*I)*x) - (3 + 4*I)*E^(I*x)*Sqrt[2*Pi]*Erf[( (1 + I)*Sqrt[-1 - x])/Sqrt[2]]*(Cos[1] - I*Sin[1]) + (4 + 3*I)*E^(I*x)*Sqr t[2*Pi]*Erfi[((1 + I)*Sqrt[-1 - x])/Sqrt[2]]*((-I)*Cos[1] + Sin[1])))/(E^( I*x)*Sqrt[-1 - x]*Sqrt[1 - x])
Time = 0.56 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, 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.
\(\displaystyle \int (1-x)^{3/2} e^{\text {arctanh}(x)} \sin (x) \, dx\) |
\(\Big \downarrow \) 6679 |
\(\displaystyle \int (1-x) \sqrt {x+1} \sin (x)dx\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle 2 \int (1-x) (x+1) \sin (x)d\sqrt {x+1}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int -((1-x) (x+1) \sin (x))d\sqrt {x+1}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -2 \int \left ((x+1)^2 \sin (x)-2 (x+1) \sin (x)\right )d\sqrt {x+1}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (-\frac {3}{4} \sqrt {\frac {\pi }{2}} \sin (1) \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {x+1}\right )+\sqrt {\frac {\pi }{2}} \cos (1) \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {x+1}\right )+\sqrt {\frac {\pi }{2}} \sin (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {x+1}\right )+\frac {3}{4} \sqrt {\frac {\pi }{2}} \cos (1) \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {x+1}\right )-\frac {3}{4} \sqrt {x+1} \sin (x)+\frac {1}{2} (x+1)^{3/2} \cos (x)-\sqrt {x+1} \cos (x)\right )\) |
2*(-(Sqrt[1 + x]*Cos[x]) + ((1 + x)^(3/2)*Cos[x])/2 + Sqrt[Pi/2]*Cos[1]*Fr esnelC[Sqrt[2/Pi]*Sqrt[1 + x]] + (3*Sqrt[Pi/2]*Cos[1]*FresnelS[Sqrt[2/Pi]* Sqrt[1 + x]])/4 - (3*Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Sqrt[1 + x]]*Sin[1])/4 + Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Sqrt[1 + x]]*Sin[1] - (3*Sqrt[1 + x]*Sin [x])/4)
3.9.15.3.1 Defintions of rubi rules used
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]]
\[\int \frac {\left (1+x \right ) \left (1-x \right )^{\frac {3}{2}} \sin \left (x \right )}{\sqrt {-x^{2}+1}}d x\]
\[ \int e^{\text {arctanh}(x)} (1-x)^{3/2} \sin (x) \, dx=\int { \frac {{\left (x + 1\right )} {\left (-x + 1\right )}^{\frac {3}{2}} \sin \left (x\right )}{\sqrt {-x^{2} + 1}} \,d x } \]
Timed out. \[ \int e^{\text {arctanh}(x)} (1-x)^{3/2} \sin (x) \, dx=\text {Timed out} \]
Result contains complex when optimal does not.
Time = 0.35 (sec) , antiderivative size = 905, normalized size of antiderivative = 5.76 \[ \int e^{\text {arctanh}(x)} (1-x)^{3/2} \sin (x) \, dx=\text {Too large to display} \]
-1/2*(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(s qrt(-I*x - I)) - 1))*sin(1))*abs(x + 1)*cos(1/2*arctan2(x + 1, 0)) - 2*((s qrt(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))*abs(x + 1)*sin(1/2*arctan2(x + 1, 0)) - (((I*cos(1) - si n(1))*gamma(3/2, I*x + I) + (-I*cos(1) - sin(1))*gamma(3/2, -I*x - I))*x + (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))*x + (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(abs(x + 1))/(x + 1)^(3/2) - 1/2*(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))*s in(1))*(x + 1)^2*cos(1/2*arctan2(x + 1, 0)) + 2*((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))*(x + 1 )^2*sin(1/2*arctan2(x + 1, 0)) + 3*(((I*cos(1) - sin(1))*gamma(3/2, I*x + I) + (-I*cos(1) - sin(1))*gamma(3/2, -I*x - I))*x + (I*cos(1) - sin(1))*ga mma(3/2, I*x + I) + (-I*cos(1) - sin(1))*gamma(3/2, -I*x - I))*abs(x + ...
Result contains complex when optimal does not.
Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.55 \[ \int e^{\text {arctanh}(x)} (1-x)^{3/2} \sin (x) \, dx=\left (\frac {1}{16} i - \frac {7}{16}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {x + 1}\right ) e^{i} - \left (\frac {1}{16} i + \frac {7}{16}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {x + 1}\right ) e^{\left (-i\right )} - \frac {1}{4} i \, {\left (2 i \, {\left (x + 1\right )}^{\frac {3}{2}} - \left (4 i + 3\right ) \, \sqrt {x + 1}\right )} e^{\left (i \, x\right )} - \frac {1}{4} i \, {\left (2 i \, {\left (x + 1\right )}^{\frac {3}{2}} - \left (4 i - 3\right ) \, \sqrt {x + 1}\right )} e^{\left (-i \, x\right )} + 0.197577103470000 \]
(1/16*I - 7/16)*sqrt(2)*sqrt(pi)*erf(-(1/2*I + 1/2)*sqrt(2)*sqrt(x + 1))*e ^I - (1/16*I + 7/16)*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) + 0.1975771034700 00
Timed out. \[ \int e^{\text {arctanh}(x)} (1-x)^{3/2} \sin (x) \, dx=\int \frac {\sin \left (x\right )\,{\left (1-x\right )}^{3/2}\,\left (x+1\right )}{\sqrt {1-x^2}} \,d x \]