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

Optimal. Leaf size=62 \[ \sqrt {2 \pi } \cos (1) S\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-\sqrt {2 \pi } \sin (1) C\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {6129, 3306, 3305, 3351, 3304, 3352} \[ \sqrt {2 \pi } \cos (1) S\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-\sqrt {2 \pi } \sin (1) \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[x]*Sin[x])/Sqrt[1 + x],x]

[Out]

Sqrt[2*Pi]*Cos[1]*FresnelS[Sqrt[2/Pi]*Sqrt[1 - x]] - Sqrt[2*Pi]*FresnelC[Sqrt[2/Pi]*Sqrt[1 - x]]*Sin[1]

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

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Mathematica [C]  time = 0.03, size = 70, normalized size = 1.13 \[ -\frac {e^{-i} \left (e^{2 i} \sqrt {-i (x-1)} \Gamma \left (\frac {1}{2},-i (x-1)\right )+\sqrt {i (x-1)} \Gamma \left (\frac {1}{2},i (x-1)\right )\right )}{2 \sqrt {1-x}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^ArcTanh[x]*Sin[x])/Sqrt[1 + x],x]

[Out]

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

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fricas [F]  time = 2.00, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-x^{2} + 1} \sqrt {x + 1} \sin \relax (x)}{x^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [C]  time = 0.18, size = 48, normalized size = 0.77 \[ -\left (\frac {1}{4} i + \frac {1}{4}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {-x + 1}\right ) e^{i} + \left (\frac {1}{4} i - \frac {1}{4}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {-x + 1}\right ) e^{\left (-i\right )} + 0.826619654151000 \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(1/4*I + 1/4)*sqrt(2)*sqrt(pi)*erf(-(1/2*I + 1/2)*sqrt(2)*sqrt(-x + 1))*e^I + (1/4*I - 1/4)*sqrt(2)*sqrt(pi)*
erf((1/2*I - 1/2)*sqrt(2)*sqrt(-x + 1))*e^(-I) + 0.826619654151000

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maple [F]  time = 0.36, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {1+x}\, \sin \relax (x )}{\sqrt {-x^{2}+1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.38, size = 168, normalized size = 2.71 \[ -\frac {{\left ({\left ({\left (i \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {i \, x - i}\right ) - 1\right )} - i \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {-i \, x + i}\right ) - 1\right )}\right )} \cos \relax (1) + {\left (\sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {i \, x - i}\right ) - 1\right )} + \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {-i \, x + i}\right ) - 1\right )}\right )} \sin \relax (1)\right )} \cos \left (\frac {1}{2} \, \arctan \left (x - 1, 0\right )\right ) + {\left ({\left (\sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {i \, x - i}\right ) - 1\right )} + \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {-i \, x + i}\right ) - 1\right )}\right )} \cos \relax (1) + {\left (-i \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {i \, x - i}\right ) - 1\right )} + i \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {-i \, x + i}\right ) - 1\right )}\right )} \sin \relax (1)\right )} \sin \left (\frac {1}{2} \, \arctan \left (x - 1, 0\right )\right )\right )} \sqrt {-x + 1}}{2 \, \sqrt {{\left | x - 1 \right |}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\sin \relax (x)\,\sqrt {x+1}}{\sqrt {1-x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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