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

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

[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]

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Rubi [A]  time = 0.122283, antiderivative size = 72, normalized size of antiderivative = 1., 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 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])

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 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 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 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 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 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]

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*}

Mathematica [C]  time = 0.0235258, size = 77, normalized size = 1.07 \[ -\frac{e^{-i} \sqrt{x+1} \text{Gamma}\left (\frac{3}{2},-i (x+1)\right )}{2 \sqrt{-i (x+1)}}-\frac{e^i \sqrt{x+1} \text{Gamma}\left (\frac{3}{2},i (x+1)\right )}{2 \sqrt{i (x+1)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(Sqrt[1 + x]*Gamma[3/2, (-I)*(1 + x)])/(2*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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Maple [F]  time = 0.302, size = 0, normalized size = 0. \begin{align*} \int{ \left ( 1+x \right ) \sin \left ( x \right ) \sqrt{1-x}{\frac{1}{\sqrt{-{x}^{2}+1}}}}\, dx \end{align*}

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 = 1.29654, size = 672, normalized size = 9.33 \begin{align*} \text{result too large to display} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-x^{2} + 1} \sqrt{-x + 1} \sin \left (x\right )}{x - 1}, x\right ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{1 - x} \left (x + 1\right ) \sin{\left (x \right )}}{\sqrt{- \left (x - 1\right ) \left (x + 1\right )}}\, dx \end{align*}

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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Giac [C]  time = 1.1806, size = 89, normalized size = 1.24 \begin{align*} \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 )} - 0.339605729125 \end{align*}

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
125