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

Optimal. Leaf size=163 \[ \frac{3}{2} \sqrt{\frac{\pi }{2}} \sin (1) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{x+1}\right )-\sqrt{\frac{\pi }{2}} \cos (1) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{x+1}\right )-\sqrt{\frac{\pi }{2}} \sin (1) S\left (\sqrt{\frac{2}{\pi }} \sqrt{x+1}\right )-\frac{3}{2} \sqrt{\frac{\pi }{2}} \cos (1) S\left (\sqrt{\frac{2}{\pi }} \sqrt{x+1}\right )+\frac{3}{2} \sqrt{x+1} \sin (x)+(x+1)^{3/2} (-\cos (x))+\sqrt{x+1} \cos (x) \]

[Out]

Sqrt[1 + x]*Cos[x] - (1 + x)^(3/2)*Cos[x] - Sqrt[Pi/2]*Cos[1]*FresnelC[Sqrt[2/Pi]*Sqrt[1 + x]] - (3*Sqrt[Pi/2]
*Cos[1]*FresnelS[Sqrt[2/Pi]*Sqrt[1 + x]])/2 + (3*Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Sqrt[1 + x]]*Sin[1])/2 - Sqrt[
Pi/2]*FresnelS[Sqrt[2/Pi]*Sqrt[1 + x]]*Sin[1] + (3*Sqrt[1 + x]*Sin[x])/2

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Rubi [A]  time = 0.297927, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.471, Rules used = {6129, 6742, 3385, 3354, 3352, 3351, 3386, 3353} \[ \frac{3}{2} \sqrt{\frac{\pi }{2}} \sin (1) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{x+1}\right )-\sqrt{\frac{\pi }{2}} \cos (1) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{x+1}\right )-\sqrt{\frac{\pi }{2}} \sin (1) S\left (\sqrt{\frac{2}{\pi }} \sqrt{x+1}\right )-\frac{3}{2} \sqrt{\frac{\pi }{2}} \cos (1) S\left (\sqrt{\frac{2}{\pi }} \sqrt{x+1}\right )+\frac{3}{2} \sqrt{x+1} \sin (x)+(x+1)^{3/2} (-\cos (x))+\sqrt{x+1} \cos (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[1 + x]*Cos[x] - (1 + x)^(3/2)*Cos[x] - Sqrt[Pi/2]*Cos[1]*FresnelC[Sqrt[2/Pi]*Sqrt[1 + x]] - (3*Sqrt[Pi/2]
*Cos[1]*FresnelS[Sqrt[2/Pi]*Sqrt[1 + x]])/2 + (3*Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Sqrt[1 + x]]*Sin[1])/2 - Sqrt[
Pi/2]*FresnelS[Sqrt[2/Pi]*Sqrt[1 + x]]*Sin[1] + (3*Sqrt[1 + x]*Sin[x])/2

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 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 3385

Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> -Simp[(e^(n - 1)*(e*x)^(m - n + 1)*Cos[c + d
*x^n])/(d*n), x] + Dist[(e^n*(m - n + 1))/(d*n), Int[(e*x)^(m - n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e},
x] && IGtQ[n, 0] && LtQ[n, m + 1]

Rule 3354

Int[Cos[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Cos[c], Int[Cos[d*(e + f*x)^2], x], x] - Dist[
Sin[c], Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, 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 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 3386

Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[(e^(n - 1)*(e*x)^(m - n + 1)*Sin[c + d*
x^n])/(d*n), x] - Dist[(e^n*(m - n + 1))/(d*n), Int[(e*x)^(m - n)*Sin[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x
] && IGtQ[n, 0] && LtQ[n, m + 1]

Rule 3353

Int[Sin[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Sin[c], Int[Cos[d*(e + f*x)^2], x], x] + Dist[
Cos[c], Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, d, e, f}, x]

Rubi steps

\begin{align*} \int e^{\tanh ^{-1}(x)} \sqrt{1-x} x \sin (x) \, dx &=\int x \sqrt{1+x} \sin (x) \, dx\\ &=-\left (2 \operatorname{Subst}\left (\int x^2 \left (-1+x^2\right ) \sin \left (1-x^2\right ) \, dx,x,\sqrt{1+x}\right )\right )\\ &=-\left (2 \operatorname{Subst}\left (\int \left (-x^2 \sin \left (1-x^2\right )+x^4 \sin \left (1-x^2\right )\right ) \, dx,x,\sqrt{1+x}\right )\right )\\ &=2 \operatorname{Subst}\left (\int x^2 \sin \left (1-x^2\right ) \, dx,x,\sqrt{1+x}\right )-2 \operatorname{Subst}\left (\int x^4 \sin \left (1-x^2\right ) \, dx,x,\sqrt{1+x}\right )\\ &=\sqrt{1+x} \cos (x)-(1+x)^{3/2} \cos (x)+3 \operatorname{Subst}\left (\int x^2 \cos \left (1-x^2\right ) \, dx,x,\sqrt{1+x}\right )-\operatorname{Subst}\left (\int \cos \left (1-x^2\right ) \, dx,x,\sqrt{1+x}\right )\\ &=\sqrt{1+x} \cos (x)-(1+x)^{3/2} \cos (x)+\frac{3}{2} \sqrt{1+x} \sin (x)+\frac{3}{2} \operatorname{Subst}\left (\int \sin \left (1-x^2\right ) \, dx,x,\sqrt{1+x}\right )-\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)-(1+x)^{3/2} \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)+\frac{3}{2} \sqrt{1+x} \sin (x)-\frac{1}{2} (3 \cos (1)) \operatorname{Subst}\left (\int \sin \left (x^2\right ) \, dx,x,\sqrt{1+x}\right )+\frac{1}{2} (3 \sin (1)) \operatorname{Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt{1+x}\right )\\ &=\sqrt{1+x} \cos (x)-(1+x)^{3/2} \cos (x)-\sqrt{\frac{\pi }{2}} \cos (1) C\left (\sqrt{\frac{2}{\pi }} \sqrt{1+x}\right )-\frac{3}{2} \sqrt{\frac{\pi }{2}} \cos (1) S\left (\sqrt{\frac{2}{\pi }} \sqrt{1+x}\right )+\frac{3}{2} \sqrt{\frac{\pi }{2}} C\left (\sqrt{\frac{2}{\pi }} \sqrt{1+x}\right ) \sin (1)-\sqrt{\frac{\pi }{2}} S\left (\sqrt{\frac{2}{\pi }} \sqrt{1+x}\right ) \sin (1)+\frac{3}{2} \sqrt{1+x} \sin (x)\\ \end{align*}

Mathematica [C]  time = 8.52325, size = 168, normalized size = 1.03 \[ \frac{\left (\frac{1}{16}+\frac{i}{16}\right ) e^{-i (x+1)} \sqrt{1-x} \left (e^i \left ((3-2 i) \sqrt{2 \pi } e^{i (x+1)} \sqrt{-x-1} \text{Erfi}\left (\frac{(1+i) \sqrt{-x-1}}{\sqrt{2}}\right )+(2+2 i) \left (e^{2 i x} (-3+2 i x)+2 i x+3\right ) (x+1)\right )-(3+2 i) \sqrt{2 \pi } e^{i x} \sqrt{-x-1} \text{Erf}\left (\frac{(1+i) \sqrt{-x-1}}{\sqrt{2}}\right )\right )}{\sqrt{1-x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1/16 + I/16)*Sqrt[1 - x]*((-3 - 2*I)*E^(I*x)*Sqrt[2*Pi]*Sqrt[-1 - x]*Erf[((1 + I)*Sqrt[-1 - x])/Sqrt[2]] + E
^I*((2 + 2*I)*(3 + E^((2*I)*x)*(-3 + (2*I)*x) + (2*I)*x)*(1 + x) + (3 - 2*I)*E^(I*(1 + x))*Sqrt[2*Pi]*Sqrt[-1
- x]*Erfi[((1 + I)*Sqrt[-1 - x])/Sqrt[2]])))/(E^(I*(1 + x))*Sqrt[1 - x^2])

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Maple [F]  time = 0.305, size = 0, normalized size = 0. \begin{align*} \int{x \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)*x*sin(x),x)

[Out]

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

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Maxima [C]  time = 1.46475, size = 1227, normalized size = 7.53 \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)*x*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(sq
rt(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*sqrt(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) - 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))*(x + 1)^2*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*sqrt(pi)*(
erf(sqrt(-I*x - I)) - 1))*sin(1))*(x + 1)^2*sin(1/2*arctan2(x + 1, 0)) - 2*(((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))*abs(x + 1)*cos(3/2*arctan2(x + 1, 0)) - (((2*cos(1) + 2*I*sin(1))*gamma(3/2, I*
x + I) + (2*cos(1) - 2*I*sin(1))*gamma(3/2, -I*x - I))*x + (2*cos(1) + 2*I*sin(1))*gamma(3/2, I*x + I) + (2*co
s(1) - 2*I*sin(1))*gamma(3/2, -I*x - I))*abs(x + 1)*sin(3/2*arctan2(x + 1, 0)) + (((I*cos(1) - sin(1))*gamma(5
/2, I*x + I) + (-I*cos(1) - sin(1))*gamma(5/2, -I*x - I))*x^2 - 2*((-I*cos(1) + sin(1))*gamma(5/2, I*x + I) +
(I*cos(1) + sin(1))*gamma(5/2, -I*x - I))*x + (I*cos(1) - sin(1))*gamma(5/2, I*x + I) + (-I*cos(1) - sin(1))*g
amma(5/2, -I*x - I))*cos(5/2*arctan2(x + 1, 0)) + (((cos(1) + I*sin(1))*gamma(5/2, I*x + I) + (cos(1) - I*sin(
1))*gamma(5/2, -I*x - I))*x^2 + ((2*cos(1) + 2*I*sin(1))*gamma(5/2, I*x + I) + (2*cos(1) - 2*I*sin(1))*gamma(5
/2, -I*x - I))*x + (cos(1) + I*sin(1))*gamma(5/2, I*x + I) + (cos(1) - I*sin(1))*gamma(5/2, -I*x - I))*sin(5/2
*arctan2(x + 1, 0)))/((x + 1)^(3/2)*sqrt(abs(x + 1)))

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)*x*sin(x),x)

[Out]

Timed out

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Giac [C]  time = 1.2461, size = 146, normalized size = 0.9 \begin{align*} \left (\frac{1}{16} i + \frac{5}{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{5}{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 )} - \frac{1}{2} \, \sqrt{x + 1} e^{\left (i \, x\right )} - \frac{1}{2} \, \sqrt{x + 1} e^{\left (-i \, x\right )} - 0.537182832596 \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)*x*sin(x),x, algorithm="giac")

[Out]

(1/16*I + 5/16)*sqrt(2)*sqrt(pi)*erf(-(1/2*I + 1/2)*sqrt(2)*sqrt(x + 1))*e^I - (1/16*I - 5/16)*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) - 1/2*sqrt(x + 1)*e^(I*x) - 1/2*sqrt(x + 1)*e^(-I*x
) - 0.537182832596