3.815 \(\int e^{\tanh ^{-1}(x)} (1-x)^{3/2} \sin (x) \, dx\)

Optimal. Leaf size=157 \[ -\frac{3}{2} \sqrt{\frac{\pi }{2}} \sin (1) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{x+1}\right )+\sqrt{2 \pi } \cos (1) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{x+1}\right )+\sqrt{2 \pi } \sin (1) S\left (\sqrt{\frac{2}{\pi }} \sqrt{x+1}\right )+\frac{3}{2} \sqrt{\frac{\pi }{2}} \cos (1) S\left (\sqrt{\frac{2}{\pi }} \sqrt{x+1}\right )-\frac{3}{2} \sqrt{x+1} \sin (x)+(x+1)^{3/2} \cos (x)-2 \sqrt{x+1} \cos (x) \]

[Out]

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

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Rubi [A]  time = 0.232529, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6129, 6742, 3385, 3354, 3352, 3351, 3386, 3353} \[ -\frac{3}{2} \sqrt{\frac{\pi }{2}} \sin (1) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{x+1}\right )+\sqrt{2 \pi } \cos (1) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{x+1}\right )+\sqrt{2 \pi } \sin (1) S\left (\sqrt{\frac{2}{\pi }} \sqrt{x+1}\right )+\frac{3}{2} \sqrt{\frac{\pi }{2}} \cos (1) S\left (\sqrt{\frac{2}{\pi }} \sqrt{x+1}\right )-\frac{3}{2} \sqrt{x+1} \sin (x)+(x+1)^{3/2} \cos (x)-2 \sqrt{x+1} \cos (x) \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[x]*(1 - x)^(3/2)*Sin[x],x]

[Out]

-2*Sqrt[1 + x]*Cos[x] + (1 + x)^(3/2)*Cos[x] + Sqrt[2*Pi]*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 + Sq
rt[2*Pi]*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)} (1-x)^{3/2} \sin (x) \, dx &=\int (1-x) \sqrt{1+x} \sin (x) \, dx\\ &=2 \operatorname{Subst}\left (\int x^2 \left (-2+x^2\right ) \sin \left (1-x^2\right ) \, dx,x,\sqrt{1+x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (-2 x^2 \sin \left (1-x^2\right )+x^4 \sin \left (1-x^2\right )\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 )-4 \operatorname{Subst}\left (\int x^2 \sin \left (1-x^2\right ) \, dx,x,\sqrt{1+x}\right )\\ &=-2 \sqrt{1+x} \cos (x)+(1+x)^{3/2} \cos (x)+2 \operatorname{Subst}\left (\int \cos \left (1-x^2\right ) \, dx,x,\sqrt{1+x}\right )-3 \operatorname{Subst}\left (\int x^2 \cos \left (1-x^2\right ) \, dx,x,\sqrt{1+x}\right )\\ &=-2 \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 )+(2 \cos (1)) \operatorname{Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt{1+x}\right )+(2 \sin (1)) \operatorname{Subst}\left (\int \sin \left (x^2\right ) \, dx,x,\sqrt{1+x}\right )\\ &=-2 \sqrt{1+x} \cos (x)+(1+x)^{3/2} \cos (x)+\sqrt{2 \pi } \cos (1) C\left (\sqrt{\frac{2}{\pi }} \sqrt{1+x}\right )+\sqrt{2 \pi } 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 )\\ &=-2 \sqrt{1+x} \cos (x)+(1+x)^{3/2} \cos (x)+\sqrt{2 \pi } \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{2 \pi } 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 = 7.01448, size = 176, normalized size = 1.12 \[ \frac{\left (\frac{1}{16}+\frac{i}{16}\right ) e^{-i x} \sqrt{1-x^2} \left (-(3+4 i) \sqrt{2 \pi } e^{i x} \text{Erf}\left (\frac{(1+i) \sqrt{-x-1}}{\sqrt{2}}\right ) (\cos (1)-i \sin (1))+(4+3 i) \sqrt{2 \pi } e^{i x} \text{Erfi}\left (\frac{(1+i) \sqrt{-x-1}}{\sqrt{2}}\right ) (\sin (1)-i \cos (1))+(2+2 i) \sqrt{-x-1} \left (e^{2 i x} ((3+2 i)-2 i x)-2 i x-(3-2 i)\right )\right )}{\sqrt{-x-1} \sqrt{1-x}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcTanh[x]*(1 - x)^(3/2)*Sin[x],x]

[Out]

((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)*S
qrt[2*Pi]*Erfi[((1 + I)*Sqrt[-1 - x])/Sqrt[2]]*((-I)*Cos[1] + Sin[1])))/(E^(I*x)*Sqrt[-1 - x]*Sqrt[1 - x])

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Maple [F]  time = 0.284, size = 0, normalized size = 0. \begin{align*} \int{ \left ( 1+x \right ) \sin \left ( x \right ) \left ( 1-x \right ) ^{{\frac{3}{2}}}{\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)^(3/2)*sin(x),x)

[Out]

int((1+x)/(-x^2+1)^(1/2)*(1-x)^(3/2)*sin(x),x)

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Maxima [C]  time = 1.48217, size = 1231, normalized size = 7.84 \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)^(3/2)*sin(x),x, algorithm="maxima")

[Out]

1/2*(((2*I*sqrt(pi)*(erf(sqrt(I*x + I)) - 1) - 2*I*sqrt(pi)*(erf(sqrt(-I*x - I)) - 1))*cos(1) - 2*(sqrt(pi)*(e
rf(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)) + (
2*(sqrt(pi)*(erf(sqrt(I*x + I)) - 1) + sqrt(pi)*(erf(sqrt(-I*x - I)) - 1))*cos(1) + (2*I*sqrt(pi)*(erf(sqrt(I*
x + I)) - 1) - 2*I*sqrt(pi)*(erf(sqrt(-I*x - I)) - 1))*sin(1))*abs(x + 1)*sin(1/2*arctan2(x + 1, 0)) + (((I*co
s(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*(((
-2*I*sqrt(pi)*(erf(sqrt(I*x + I)) - 1) + 2*I*sqrt(pi)*(erf(sqrt(-I*x - I)) - 1))*cos(1) + 2*(sqrt(pi)*(erf(sqr
t(I*x + I)) - 1) + sqrt(pi)*(erf(sqrt(-I*x - I)) - 1))*sin(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) - (-2*I*sqrt(pi)*(erf(sqrt(I*x + I)
) - 1) + 2*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))*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)) - (((3*cos(1) + 3*
I*sin(1))*gamma(3/2, I*x + I) + (3*cos(1) - 3*I*sin(1))*gamma(3/2, -I*x - I))*x + (3*cos(1) + 3*I*sin(1))*gamm
a(3/2, I*x + I) + (3*cos(1) - 3*I*sin(1))*gamma(3/2, -I*x - I))*abs(x + 1)*sin(3/2*arctan2(x + 1, 0)) + (((I*c
os(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))*gamma(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 (\sqrt{-x^{2} + 1} \sqrt{-x + 1} \sin \left (x\right ), 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)^(3/2)*sin(x),x, algorithm="fricas")

[Out]

integral(sqrt(-x^2 + 1)*sqrt(-x + 1)*sin(x), 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)**(3/2)*sin(x),x)

[Out]

Timed out

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Giac [C]  time = 1.20836, size = 116, normalized size = 0.74 \begin{align*} \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 )} + 0.19757710347 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)/(-x^2+1)^(1/2)*(1-x)^(3/2)*sin(x),x, algorithm="giac")

[Out]

(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) + 0.19757710347