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

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

[Out]

(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)+Fres
nelS(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)

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Rubi [A]  time = 0.23, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 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 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]

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 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 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 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]]

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*}

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Mathematica [C]  time = 7.20, 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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fricas [F]  time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {-x^{2} + 1} \sqrt {-x + 1} \sin \relax (x), x\right ) \]

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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giac [C]  time = 0.19, size = 86, normalized size = 0.55 \[ \left (\frac {1}{16} i - \frac {7}{16}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {x + 1}\right ) e^{i} - \left (\frac {1}{16} i + \frac {7}{16}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {x + 1}\right ) e^{\left (-i\right )} - \frac {1}{4} i \, {\left (2 i \, {\left (x + 1\right )}^{\frac {3}{2}} - \left (4 i + 3\right ) \, \sqrt {x + 1}\right )} e^{\left (i \, x\right )} - \frac {1}{4} i \, {\left (2 i \, {\left (x + 1\right )}^{\frac {3}{2}} - \left (4 i - 3\right ) \, \sqrt {x + 1}\right )} e^{\left (-i \, x\right )} + 0.197577103470000 \]

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) + 0.197577103470000

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maple [F]  time = 0.38, size = 0, normalized size = 0.00 \[ \int \frac {\left (1+x \right ) \left (1-x \right )^{\frac {3}{2}} \sin \relax (x )}{\sqrt {-x^{2}+1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+x)/(-x^2+1)^(1/2)*(1-x)^(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 = 0.50, size = 912, normalized size = 5.81 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)/(-x^2+1)^(1/2)*(1-x)^(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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sin \relax (x)\,{\left (1-x\right )}^{3/2}\,\left (x+1\right )}{\sqrt {1-x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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