3.811 \(\int e^{\tanh ^{-1}(x)} \sqrt {1-x} \sin (x) \, dx\)

Optimal. Leaf size=72 \[ \sqrt {\frac {\pi }{2}} \cos (1) C\left (\sqrt {\frac {2}{\pi }} \sqrt {x+1}\right )+\sqrt {\frac {\pi }{2}} \sin (1) S\left (\sqrt {\frac {2}{\pi }} \sqrt {x+1}\right )-\sqrt {x+1} \cos (x) \]

[Out]

1/2*cos(1)*FresnelC(2^(1/2)/Pi^(1/2)*(1+x)^(1/2))*2^(1/2)*Pi^(1/2)+1/2*FresnelS(2^(1/2)/Pi^(1/2)*(1+x)^(1/2))*
sin(1)*2^(1/2)*Pi^(1/2)-cos(x)*(1+x)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {6129, 3296, 3306, 3305, 3351, 3304, 3352} \[ \sqrt {\frac {\pi }{2}} \cos (1) \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {x+1}\right )+\sqrt {\frac {\pi }{2}} \sin (1) S\left (\sqrt {\frac {2}{\pi }} \sqrt {x+1}\right )-\sqrt {x+1} \cos (x) \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[x]*Sqrt[1 - x]*Sin[x],x]

[Out]

-(Sqrt[1 + x]*Cos[x]) + Sqrt[Pi/2]*Cos[1]*FresnelC[Sqrt[2/Pi]*Sqrt[1 + x]] + Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Sq
rt[1 + x]]*Sin[1]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3304

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[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]

Rule 3305

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[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]

Rule 3306

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d +
f*x]/Sqrt[c + d*x], x], x] + Dist[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]

Rule 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 6129

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> Dist[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])

Rubi steps

\begin {align*} \int e^{\tanh ^{-1}(x)} \sqrt {1-x} \sin (x) \, dx &=\int \sqrt {1+x} \sin (x) \, dx\\ &=-\sqrt {1+x} \cos (x)+\frac {1}{2} \int \frac {\cos (x)}{\sqrt {1+x}} \, dx\\ &=-\sqrt {1+x} \cos (x)+\frac {1}{2} \cos (1) \int \frac {\cos (1+x)}{\sqrt {1+x}} \, dx+\frac {1}{2} \sin (1) \int \frac {\sin (1+x)}{\sqrt {1+x}} \, dx\\ &=-\sqrt {1+x} \cos (x)+\cos (1) \operatorname {Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt {1+x}\right )+\sin (1) \operatorname {Subst}\left (\int \sin \left (x^2\right ) \, dx,x,\sqrt {1+x}\right )\\ &=-\sqrt {1+x} \cos (x)+\sqrt {\frac {\pi }{2}} \cos (1) C\left (\sqrt {\frac {2}{\pi }} \sqrt {1+x}\right )+\sqrt {\frac {\pi }{2}} S\left (\sqrt {\frac {2}{\pi }} \sqrt {1+x}\right ) \sin (1)\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 77, normalized size = 1.07 \[ -\frac {e^{-i} \sqrt {x+1} \Gamma \left (\frac {3}{2},-i (x+1)\right )}{2 \sqrt {-i (x+1)}}-\frac {e^i \sqrt {x+1} \Gamma \left (\frac {3}{2},i (x+1)\right )}{2 \sqrt {i (x+1)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcTanh[x]*Sqrt[1 - x]*Sin[x],x]

[Out]

-1/2*(Sqrt[1 + x]*Gamma[3/2, (-I)*(1 + x)])/(E^I*Sqrt[(-I)*(1 + x)]) - (E^I*Sqrt[1 + x]*Gamma[3/2, I*(1 + x)])
/(2*Sqrt[I*(1 + x)])

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-x^{2} + 1} \sqrt {-x + 1} \sin \relax (x)}{x - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [C]  time = 0.18, size = 66, normalized size = 0.92 \[ \left (\frac {1}{8} i - \frac {1}{8}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {x + 1}\right ) e^{i} - \left (\frac {1}{8} i + \frac {1}{8}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {x + 1}\right ) e^{\left (-i\right )} - \frac {1}{2} \, \sqrt {x + 1} e^{\left (i \, x\right )} - \frac {1}{2} \, \sqrt {x + 1} e^{\left (-i \, x\right )} - 0.339605729125000 \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/8*I - 1/8)*sqrt(2)*sqrt(pi)*erf(-(1/2*I + 1/2)*sqrt(2)*sqrt(x + 1))*e^I - (1/8*I + 1/8)*sqrt(2)*sqrt(pi)*er
f((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) - 0.339605729
125000

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maple [F]  time = 0.38, size = 0, normalized size = 0.00 \[ \int \frac {\left (1+x \right ) \sqrt {1-x}\, \sin \relax (x )}{\sqrt {-x^{2}+1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.44, size = 498, normalized size = 6.92 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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(s
qrt(I*x + I)) - 1) + sqrt(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))*cos(1) - (-I*sqrt(pi)*(erf(sqrt(I*x + I)) - 1) + I*s
qrt(pi)*(erf(sqrt(-I*x - I)) - 1))*sin(1))*sin(1/2*arctan2(x + 1, 0)))*sqrt(x + 1)/sqrt(abs(x + 1)) - 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))*abs(x + 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*s
qrt(pi)*(erf(sqrt(-I*x - I)) - 1))*sin(1))*abs(x + 1)*sin(1/2*arctan2(x + 1, 0)) + (((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))*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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sin \relax (x)\,\sqrt {1-x}\,\left (x+1\right )}{\sqrt {1-x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {1 - x} \left (x + 1\right ) \sin {\relax (x )}}{\sqrt {- \left (x - 1\right ) \left (x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)/(-x**2+1)**(1/2)*(1-x)**(1/2)*sin(x),x)

[Out]

Integral(sqrt(1 - x)*(x + 1)*sin(x)/sqrt(-(x - 1)*(x + 1)), x)

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