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

Optimal. Leaf size=240 \[ -2 \sqrt{2 \pi } \sin (1) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{1-x}\right )+\frac{3}{2} \sqrt{\frac{\pi }{2}} \sin (1) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{1-x}\right )-3 \sqrt{\frac{\pi }{2}} \cos (1) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{1-x}\right )-3 \sqrt{\frac{\pi }{2}} \sin (1) S\left (\sqrt{\frac{2}{\pi }} \sqrt{1-x}\right )+2 \sqrt{2 \pi } \cos (1) S\left (\sqrt{\frac{2}{\pi }} \sqrt{1-x}\right )-\frac{3}{2} \sqrt{\frac{\pi }{2}} \cos (1) S\left (\sqrt{\frac{2}{\pi }} \sqrt{1-x}\right )-\frac{3}{2} \sqrt{1-x} \sin (x)+(1-x)^{3/2} (-\cos (x))+3 \sqrt{1-x} \cos (x) \]

[Out]

3*Sqrt[1 - x]*Cos[x] - (1 - x)^(3/2)*Cos[x] - 3*Sqrt[Pi/2]*Cos[1]*FresnelC[Sqrt[2/Pi]*Sqrt[1 - x]] - (3*Sqrt[P
i/2]*Cos[1]*FresnelS[Sqrt[2/Pi]*Sqrt[1 - x]])/2 + 2*Sqrt[2*Pi]*Cos[1]*FresnelS[Sqrt[2/Pi]*Sqrt[1 - x]] + (3*Sq
rt[Pi/2]*FresnelC[Sqrt[2/Pi]*Sqrt[1 - x]]*Sin[1])/2 - 2*Sqrt[2*Pi]*FresnelC[Sqrt[2/Pi]*Sqrt[1 - x]]*Sin[1] - 3
*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 = 0.396753, antiderivative size = 240, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 8, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.533, Rules used = {6129, 6742, 3353, 3352, 3351, 3385, 3354, 3386} \[ -2 \sqrt{2 \pi } \sin (1) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{1-x}\right )+\frac{3}{2} \sqrt{\frac{\pi }{2}} \sin (1) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{1-x}\right )-3 \sqrt{\frac{\pi }{2}} \cos (1) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{1-x}\right )-3 \sqrt{\frac{\pi }{2}} \sin (1) S\left (\sqrt{\frac{2}{\pi }} \sqrt{1-x}\right )+2 \sqrt{2 \pi } \cos (1) S\left (\sqrt{\frac{2}{\pi }} \sqrt{1-x}\right )-\frac{3}{2} \sqrt{\frac{\pi }{2}} \cos (1) S\left (\sqrt{\frac{2}{\pi }} \sqrt{1-x}\right )-\frac{3}{2} \sqrt{1-x} \sin (x)+(1-x)^{3/2} (-\cos (x))+3 \sqrt{1-x} \cos (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

\begin{align*} \int e^{\tanh ^{-1}(x)} x \sqrt{1+x} \sin (x) \, dx &=\int \frac{x (1+x) \sin (x)}{\sqrt{1-x}} \, dx\\ &=-\left (2 \operatorname{Subst}\left (\int \left (-2+x^2\right ) \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 (2 \sin \left (1-x^2\right )-3 x^2 \sin \left (1-x^2\right )+x^4 \sin \left (1-x^2\right )\right ) \, dx,x,\sqrt{1-x}\right )\right )\\ &=-\left (2 \operatorname{Subst}\left (\int x^4 \sin \left (1-x^2\right ) \, dx,x,\sqrt{1-x}\right )\right )-4 \operatorname{Subst}\left (\int \sin \left (1-x^2\right ) \, dx,x,\sqrt{1-x}\right )+6 \operatorname{Subst}\left (\int x^2 \sin \left (1-x^2\right ) \, dx,x,\sqrt{1-x}\right )\\ &=3 \sqrt{1-x} \cos (x)-(1-x)^{3/2} \cos (x)-3 \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 )+(4 \cos (1)) \operatorname{Subst}\left (\int \sin \left (x^2\right ) \, dx,x,\sqrt{1-x}\right )-(4 \sin (1)) \operatorname{Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt{1-x}\right )\\ &=3 \sqrt{1-x} \cos (x)-(1-x)^{3/2} \cos (x)+2 \sqrt{2 \pi } \cos (1) S\left (\sqrt{\frac{2}{\pi }} \sqrt{1-x}\right )-2 \sqrt{2 \pi } C\left (\sqrt{\frac{2}{\pi }} \sqrt{1-x}\right ) \sin (1)-\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 )-(3 \cos (1)) \operatorname{Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt{1-x}\right )-(3 \sin (1)) \operatorname{Subst}\left (\int \sin \left (x^2\right ) \, dx,x,\sqrt{1-x}\right )\\ &=3 \sqrt{1-x} \cos (x)-(1-x)^{3/2} \cos (x)-3 \sqrt{\frac{\pi }{2}} \cos (1) C\left (\sqrt{\frac{2}{\pi }} \sqrt{1-x}\right )+2 \sqrt{2 \pi } \cos (1) S\left (\sqrt{\frac{2}{\pi }} \sqrt{1-x}\right )-2 \sqrt{2 \pi } C\left (\sqrt{\frac{2}{\pi }} \sqrt{1-x}\right ) \sin (1)-3 \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 )\\ &=3 \sqrt{1-x} \cos (x)-(1-x)^{3/2} \cos (x)-3 \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 )+2 \sqrt{2 \pi } \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)-2 \sqrt{2 \pi } C\left (\sqrt{\frac{2}{\pi }} \sqrt{1-x}\right ) \sin (1)-3 \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 = 8.95077, size = 184, normalized size = 0.77 \[ -\frac{\left (\frac{1}{16}-\frac{i}{16}\right ) \sqrt{x+1} \left ((\cos (x+1)-i \sin (x+1)) \left ((2+2 i) \left (2 x^2+(2-3 i) x-(4-3 i)\right ) (\cos (1)+i \sin (1))-(6+5 i) \sqrt{2 \pi } \sqrt{x-1} \text{Erf}\left (\frac{(1+i) \sqrt{x-1}}{\sqrt{2}}\right ) (\cos (x)+i \sin (x))\right )+(-6+5 i) \sqrt{2 \pi } \sqrt{x-1} \text{Erfi}\left (\frac{(1+i) \sqrt{x-1}}{\sqrt{2}}\right ) (\cos (1)+i \sin (1))+(2+2 i) \left (2 x^2+(2+3 i) x-(4+3 i)\right ) (\cos (x)+i \sin (x))\right )}{\sqrt{1-x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-1/16 + I/16)*Sqrt[1 + x]*((-6 + 5*I)*Sqrt[2*Pi]*Sqrt[-1 + x]*Erfi[((1 + I)*Sqrt[-1 + x])/Sqrt[2]]*(Cos[1] +
I*Sin[1]) + (2 + 2*I)*((-4 - 3*I) + (2 + 3*I)*x + 2*x^2)*(Cos[x] + I*Sin[x]) + ((2 + 2*I)*((-4 + 3*I) + (2 -
3*I)*x + 2*x^2)*(Cos[1] + I*Sin[1]) - (6 + 5*I)*Sqrt[2*Pi]*Sqrt[-1 + x]*Erf[((1 + I)*Sqrt[-1 + x])/Sqrt[2]]*(C
os[x] + I*Sin[x]))*(Cos[1 + x] - I*Sin[1 + x])))/Sqrt[1 - x^2]

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Maple [F]  time = 0.276, size = 0, normalized size = 0. \begin{align*} \int{x\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)^(3/2)/(-x^2+1)^(1/2)*x*sin(x),x)

[Out]

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

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Maxima [C]  time = 1.31122, size = 863, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^(3/2)/(-x^2+1)^(1/2)*x*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)*
(erf(sqrt(I*x - I)) - 1) + sqrt(pi)*(erf(sqrt(-I*x + I)) - 1))*sin(1))*cos(1/2*arctan2(x - 1, 0)) + (2*(sqrt(p
i)*(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))*sin(1/2*arctan2(x - 1, 0)))*(x - 1)^2 + (((-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))*sin(5/2*arctan2(x - 1, 0)))*x^2 -
((3*((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)) + ((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))*sin(3/2*
arctan2(x - 1, 0)))*abs(x - 1) + 2*((-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)) - ((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))*sin(5/2*arctan2(x - 1, 0)))*x - (3*((-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)) - ((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))*sin(3/2*arctan2(x - 1, 0)))*abs(x - 1) + ((-I*cos(1) - sin(1))*g
amma(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))*sin(5/2*arctan2(x - 1, 0)))*sqrt(-x + 1)/((
x - 1)^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} \sqrt{x + 1} x \sin \left (x\right )}{x - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [C]  time = 1.23661, size = 167, normalized size = 0.7 \begin{align*} -\left (\frac{11}{16} i - \frac{1}{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{11}{16} i + \frac{1}{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 )} + 1.79526793396 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(11/16*I - 1/16)*sqrt(2)*sqrt(pi)*erf(-(1/2*I + 1/2)*sqrt(2)*sqrt(-x + 1))*e^I + (11/16*I + 1/16)*sqrt(2)*sqr
t(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) + 1.79526793396