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

Optimal. Leaf size=335 \[ -4 \sqrt {2 \pi } \sin (1) C\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+\frac {15}{2} \sqrt {\frac {\pi }{2}} \sin (1) C\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-4 \sqrt {2 \pi } \cos (1) C\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+\frac {15}{4} \sqrt {\frac {\pi }{2}} \cos (1) C\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-4 \sqrt {2 \pi } \sin (1) S\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+\frac {15}{4} \sqrt {\frac {\pi }{2}} \sin (1) S\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+4 \sqrt {2 \pi } \cos (1) S\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-\frac {15}{2} \sqrt {\frac {\pi }{2}} \cos (1) S\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+\frac {5}{2} (1-x)^{3/2} \sin (x)-\frac {15}{2} \sqrt {1-x} \sin (x)+(1-x)^{5/2} \cos (x)-5 (1-x)^{3/2} \cos (x)+\frac {17}{4} \sqrt {1-x} \cos (x) \]

[Out]

-5*(1-x)^(3/2)*cos(x)+(1-x)^(5/2)*cos(x)+5/2*(1-x)^(3/2)*sin(x)-17/8*cos(1)*FresnelC(2^(1/2)/Pi^(1/2)*(1-x)^(1
/2))*2^(1/2)*Pi^(1/2)+1/4*cos(1)*FresnelS(2^(1/2)/Pi^(1/2)*(1-x)^(1/2))*2^(1/2)*Pi^(1/2)-1/4*FresnelC(2^(1/2)/
Pi^(1/2)*(1-x)^(1/2))*sin(1)*2^(1/2)*Pi^(1/2)-17/8*FresnelS(2^(1/2)/Pi^(1/2)*(1-x)^(1/2))*sin(1)*2^(1/2)*Pi^(1
/2)+17/4*cos(x)*(1-x)^(1/2)-15/2*sin(x)*(1-x)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.48, antiderivative size = 335, normalized size of antiderivative = 1.00, number of steps used = 22, 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} \[ -4 \sqrt {2 \pi } \sin (1) \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+\frac {15}{2} \sqrt {\frac {\pi }{2}} \sin (1) \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-4 \sqrt {2 \pi } \cos (1) \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+\frac {15}{4} \sqrt {\frac {\pi }{2}} \cos (1) \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-4 \sqrt {2 \pi } \sin (1) S\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+\frac {15}{4} \sqrt {\frac {\pi }{2}} \sin (1) S\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+4 \sqrt {2 \pi } \cos (1) S\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-\frac {15}{2} \sqrt {\frac {\pi }{2}} \cos (1) S\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+\frac {5}{2} (1-x)^{3/2} \sin (x)-\frac {15}{2} \sqrt {1-x} \sin (x)+(1-x)^{5/2} \cos (x)-5 (1-x)^{3/2} \cos (x)+\frac {17}{4} \sqrt {1-x} \cos (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(17*Sqrt[1 - x]*Cos[x])/4 - 5*(1 - x)^(3/2)*Cos[x] + (1 - x)^(5/2)*Cos[x] + (15*Sqrt[Pi/2]*Cos[1]*FresnelC[Sqr
t[2/Pi]*Sqrt[1 - x]])/4 - 4*Sqrt[2*Pi]*Cos[1]*FresnelC[Sqrt[2/Pi]*Sqrt[1 - x]] - (15*Sqrt[Pi/2]*Cos[1]*Fresnel
S[Sqrt[2/Pi]*Sqrt[1 - x]])/2 + 4*Sqrt[2*Pi]*Cos[1]*FresnelS[Sqrt[2/Pi]*Sqrt[1 - x]] + (15*Sqrt[Pi/2]*FresnelC[
Sqrt[2/Pi]*Sqrt[1 - x]]*Sin[1])/2 - 4*Sqrt[2*Pi]*FresnelC[Sqrt[2/Pi]*Sqrt[1 - x]]*Sin[1] + (15*Sqrt[Pi/2]*Fres
nelS[Sqrt[2/Pi]*Sqrt[1 - x]]*Sin[1])/4 - 4*Sqrt[2*Pi]*FresnelS[Sqrt[2/Pi]*Sqrt[1 - x]]*Sin[1] - (15*Sqrt[1 - x
]*Sin[x])/2 + (5*(1 - x)^(3/2)*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)} x (1+x)^{3/2} \sin (x) \, dx &=\int \frac {x (1+x)^2 \sin (x)}{\sqrt {1-x}} \, dx\\ &=2 \operatorname {Subst}\left (\int \left (-2+x^2\right )^2 \left (-1+x^2\right ) \sin \left (1-x^2\right ) \, dx,x,\sqrt {1-x}\right )\\ &=2 \operatorname {Subst}\left (\int \left (-4 \sin \left (1-x^2\right )+8 x^2 \sin \left (1-x^2\right )-5 x^4 \sin \left (1-x^2\right )+x^6 \sin \left (1-x^2\right )\right ) \, dx,x,\sqrt {1-x}\right )\\ &=2 \operatorname {Subst}\left (\int x^6 \sin \left (1-x^2\right ) \, dx,x,\sqrt {1-x}\right )-8 \operatorname {Subst}\left (\int \sin \left (1-x^2\right ) \, dx,x,\sqrt {1-x}\right )-10 \operatorname {Subst}\left (\int x^4 \sin \left (1-x^2\right ) \, dx,x,\sqrt {1-x}\right )+16 \operatorname {Subst}\left (\int x^2 \sin \left (1-x^2\right ) \, dx,x,\sqrt {1-x}\right )\\ &=8 \sqrt {1-x} \cos (x)-5 (1-x)^{3/2} \cos (x)+(1-x)^{5/2} \cos (x)-5 \operatorname {Subst}\left (\int x^4 \cos \left (1-x^2\right ) \, dx,x,\sqrt {1-x}\right )-8 \operatorname {Subst}\left (\int \cos \left (1-x^2\right ) \, dx,x,\sqrt {1-x}\right )+15 \operatorname {Subst}\left (\int x^2 \cos \left (1-x^2\right ) \, dx,x,\sqrt {1-x}\right )+(8 \cos (1)) \operatorname {Subst}\left (\int \sin \left (x^2\right ) \, dx,x,\sqrt {1-x}\right )-(8 \sin (1)) \operatorname {Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt {1-x}\right )\\ &=8 \sqrt {1-x} \cos (x)-5 (1-x)^{3/2} \cos (x)+(1-x)^{5/2} \cos (x)+4 \sqrt {2 \pi } \cos (1) S\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-4 \sqrt {2 \pi } C\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \sin (1)-\frac {15}{2} \sqrt {1-x} \sin (x)+\frac {5}{2} (1-x)^{3/2} \sin (x)+\frac {15}{2} \operatorname {Subst}\left (\int \sin \left (1-x^2\right ) \, dx,x,\sqrt {1-x}\right )-\frac {15}{2} \operatorname {Subst}\left (\int x^2 \sin \left (1-x^2\right ) \, dx,x,\sqrt {1-x}\right )-(8 \cos (1)) \operatorname {Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt {1-x}\right )-(8 \sin (1)) \operatorname {Subst}\left (\int \sin \left (x^2\right ) \, dx,x,\sqrt {1-x}\right )\\ &=\frac {17}{4} \sqrt {1-x} \cos (x)-5 (1-x)^{3/2} \cos (x)+(1-x)^{5/2} \cos (x)-4 \sqrt {2 \pi } \cos (1) C\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+4 \sqrt {2 \pi } \cos (1) S\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-4 \sqrt {2 \pi } C\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \sin (1)-4 \sqrt {2 \pi } S\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \sin (1)-\frac {15}{2} \sqrt {1-x} \sin (x)+\frac {5}{2} (1-x)^{3/2} \sin (x)+\frac {15}{4} \operatorname {Subst}\left (\int \cos \left (1-x^2\right ) \, dx,x,\sqrt {1-x}\right )-\frac {1}{2} (15 \cos (1)) \operatorname {Subst}\left (\int \sin \left (x^2\right ) \, dx,x,\sqrt {1-x}\right )+\frac {1}{2} (15 \sin (1)) \operatorname {Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt {1-x}\right )\\ &=\frac {17}{4} \sqrt {1-x} \cos (x)-5 (1-x)^{3/2} \cos (x)+(1-x)^{5/2} \cos (x)-4 \sqrt {2 \pi } \cos (1) C\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-\frac {15}{2} \sqrt {\frac {\pi }{2}} \cos (1) S\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+4 \sqrt {2 \pi } \cos (1) S\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+\frac {15}{2} \sqrt {\frac {\pi }{2}} C\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \sin (1)-4 \sqrt {2 \pi } C\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \sin (1)-4 \sqrt {2 \pi } S\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \sin (1)-\frac {15}{2} \sqrt {1-x} \sin (x)+\frac {5}{2} (1-x)^{3/2} \sin (x)+\frac {1}{4} (15 \cos (1)) \operatorname {Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt {1-x}\right )+\frac {1}{4} (15 \sin (1)) \operatorname {Subst}\left (\int \sin \left (x^2\right ) \, dx,x,\sqrt {1-x}\right )\\ &=\frac {17}{4} \sqrt {1-x} \cos (x)-5 (1-x)^{3/2} \cos (x)+(1-x)^{5/2} \cos (x)+\frac {15}{4} \sqrt {\frac {\pi }{2}} \cos (1) C\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-4 \sqrt {2 \pi } \cos (1) C\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )-\frac {15}{2} \sqrt {\frac {\pi }{2}} \cos (1) S\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+4 \sqrt {2 \pi } \cos (1) S\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right )+\frac {15}{2} \sqrt {\frac {\pi }{2}} C\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \sin (1)-4 \sqrt {2 \pi } C\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \sin (1)+\frac {15}{4} \sqrt {\frac {\pi }{2}} S\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \sin (1)-4 \sqrt {2 \pi } S\left (\sqrt {\frac {2}{\pi }} \sqrt {1-x}\right ) \sin (1)-\frac {15}{2} \sqrt {1-x} \sin (x)+\frac {5}{2} (1-x)^{3/2} \sin (x)\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 9.77, size = 200, normalized size = 0.60 \[ \frac {\left (\frac {1}{32}+\frac {i}{32}\right ) \sqrt {x+1} \left ((\cos (x+1)-i \sin (x+1)) \left ((17+2 i) \sqrt {2 \pi } \sqrt {x-1} \text {erf}\left (\frac {(1+i) \sqrt {x-1}}{\sqrt {2}}\right ) (\sin (x)-i \cos (x))+(2+2 i) \left (4 i x^3+(10+8 i) x^2+(10-11 i) x-(20+i)\right ) (\cos (1)+i \sin (1))\right )+(-2-17 i) \sqrt {2 \pi } \sqrt {x-1} \text {erfi}\left (\frac {(1+i) \sqrt {x-1}}{\sqrt {2}}\right ) (\cos (1)+i \sin (1))-(2-2 i) \left (4 x^3+(8+10 i) x^2-(11-10 i) x-(1+20 i)\right ) (\cos (x)+i \sin (x))\right )}{\sqrt {1-x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1/32 + I/32)*Sqrt[1 + x]*((-2 - 17*I)*Sqrt[2*Pi]*Sqrt[-1 + x]*Erfi[((1 + I)*Sqrt[-1 + x])/Sqrt[2]]*(Cos[1] +
I*Sin[1]) - (2 - 2*I)*((-1 - 20*I) - (11 - 10*I)*x + (8 + 10*I)*x^2 + 4*x^3)*(Cos[x] + I*Sin[x]) + ((2 + 2*I)
*((-20 - I) + (10 - 11*I)*x + (10 + 8*I)*x^2 + (4*I)*x^3)*(Cos[1] + I*Sin[1]) + (17 + 2*I)*Sqrt[2*Pi]*Sqrt[-1
+ x]*Erf[((1 + I)*Sqrt[-1 + x])/Sqrt[2]]*((-I)*Cos[x] + Sin[x]))*(Cos[1 + x] - I*Sin[1 + x])))/Sqrt[1 - x^2]

________________________________________________________________________________________

fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (x^{2} + x\right )} \sqrt {-x^{2} + 1} \sqrt {x + 1} \sin \relax (x)}{x - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^(5/2)/(-x^2+1)^(1/2)*x*sin(x),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [C]  time = 0.24, size = 202, normalized size = 0.60 \[ -\left (\frac {19}{32} i - \frac {15}{32}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {-x + 1}\right ) e^{i} + \left (\frac {19}{32} i + \frac {15}{32}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {-x + 1}\right ) e^{\left (-i\right )} - \frac {1}{8} i \, {\left (4 i \, {\left (x - 1\right )}^{2} \sqrt {-x + 1} - \left (12 i - 10\right ) \, {\left (-x + 1\right )}^{\frac {3}{2}} - \left (3 i + 18\right ) \, \sqrt {-x + 1}\right )} e^{\left (i \, x\right )} - \frac {1}{2} i \, {\left (-2 i \, {\left (-x + 1\right )}^{\frac {3}{2}} + \left (4 i - 3\right ) \, \sqrt {-x + 1}\right )} e^{\left (i \, x\right )} - \frac {1}{8} i \, {\left (4 i \, {\left (x - 1\right )}^{2} \sqrt {-x + 1} - \left (12 i + 10\right ) \, {\left (-x + 1\right )}^{\frac {3}{2}} - \left (3 i - 18\right ) \, \sqrt {-x + 1}\right )} e^{\left (-i \, x\right )} - \frac {1}{2} i \, {\left (-2 i \, {\left (-x + 1\right )}^{\frac {3}{2}} + \left (4 i + 3\right ) \, \sqrt {-x + 1}\right )} e^{\left (-i \, x\right )} + \frac {1}{2} \, \sqrt {-x + 1} e^{\left (i \, x\right )} + \frac {1}{2} \, \sqrt {-x + 1} e^{\left (-i \, x\right )} + 3.25954715712000 \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^(5/2)/(-x^2+1)^(1/2)*x*sin(x),x, algorithm="giac")

[Out]

-(19/32*I - 15/32)*sqrt(2)*sqrt(pi)*erf(-(1/2*I + 1/2)*sqrt(2)*sqrt(-x + 1))*e^I + (19/32*I + 15/32)*sqrt(2)*s
qrt(pi)*erf((1/2*I - 1/2)*sqrt(2)*sqrt(-x + 1))*e^(-I) - 1/8*I*(4*I*(x - 1)^2*sqrt(-x + 1) - (12*I - 10)*(-x +
1)^(3/2) - (3*I + 18)*sqrt(-x + 1))*e^(I*x) - 1/2*I*(-2*I*(-x + 1)^(3/2) + (4*I - 3)*sqrt(-x + 1))*e^(I*x) -
1/8*I*(4*I*(x - 1)^2*sqrt(-x + 1) - (12*I + 10)*(-x + 1)^(3/2) - (3*I - 18)*sqrt(-x + 1))*e^(-I*x) - 1/2*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) +
3.25954715712000

________________________________________________________________________________________

maple [F]  time = 0.36, size = 0, normalized size = 0.00 \[ \int \frac {\left (1+x \right )^{\frac {5}{2}} x \sin \relax (x )}{\sqrt {-x^{2}+1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+x)^(5/2)/(-x^2+1)^(1/2)*x*sin(x),x)

[Out]

int((1+x)^(5/2)/(-x^2+1)^(1/2)*x*sin(x),x)

________________________________________________________________________________________

maxima [C]  time = 0.47, size = 1010, normalized size = 3.01 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^(5/2)/(-x^2+1)^(1/2)*x*sin(x),x, algorithm="maxima")

[Out]

-1/2*((((-I*cos(1) - sin(1))*gamma(7/2, I*x - I) + (I*cos(1) - sin(1))*gamma(7/2, -I*x + I))*cos(7/2*arctan2(x
- 1, 0)) - ((cos(1) - I*sin(1))*gamma(7/2, I*x - I) + (cos(1) + I*sin(1))*gamma(7/2, -I*x + I))*sin(7/2*arcta
n2(x - 1, 0)))*x^3 + (((4*I*sqrt(pi)*(erf(sqrt(I*x - I)) - 1) - 4*I*sqrt(pi)*(erf(sqrt(-I*x + I)) - 1))*cos(1)
+ 4*(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
)) + (4*(sqrt(pi)*(erf(sqrt(I*x - I)) - 1) + sqrt(pi)*(erf(sqrt(-I*x + I)) - 1))*cos(1) + (-4*I*sqrt(pi)*(erf(
sqrt(I*x - I)) - 1) + 4*I*sqrt(pi)*(erf(sqrt(-I*x + I)) - 1))*sin(1))*sin(1/2*arctan2(x - 1, 0)))*(x - 1)^2*ab
s(x - 1) - (8*((-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*ar
ctan2(x - 1, 0)) - ((8*cos(1) - 8*I*sin(1))*gamma(3/2, I*x - I) + (8*cos(1) + 8*I*sin(1))*gamma(3/2, -I*x + I)
)*sin(3/2*arctan2(x - 1, 0)))*(x - 1)^2 - ((5*((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)) + ((5*cos(1) - 5*I*sin(1))*gamma(5/2, I*x - I) + (5*cos(1) +
5*I*sin(1))*gamma(5/2, -I*x + I))*sin(5/2*arctan2(x - 1, 0)))*abs(x - 1) + 3*((-I*cos(1) - sin(1))*gamma(7/2,
I*x - I) + (I*cos(1) - sin(1))*gamma(7/2, -I*x + I))*cos(7/2*arctan2(x - 1, 0)) - ((3*cos(1) - 3*I*sin(1))*gam
ma(7/2, I*x - I) + (3*cos(1) + 3*I*sin(1))*gamma(7/2, -I*x + I))*sin(7/2*arctan2(x - 1, 0)))*x^2 - ((8*((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)) + ((
8*cos(1) - 8*I*sin(1))*gamma(3/2, I*x - I) + (8*cos(1) + 8*I*sin(1))*gamma(3/2, -I*x + I))*sin(3/2*arctan2(x -
1, 0)))*(x - 1)^2 + (10*((-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)) - ((10*cos(1) - 10*I*sin(1))*gamma(5/2, I*x - I) + (10*cos(1) + 10*I*sin(1))*gamma
(5/2, -I*x + I))*sin(5/2*arctan2(x - 1, 0)))*abs(x - 1) + 3*((I*cos(1) + sin(1))*gamma(7/2, I*x - I) + (-I*cos
(1) + sin(1))*gamma(7/2, -I*x + I))*cos(7/2*arctan2(x - 1, 0)) + ((3*cos(1) - 3*I*sin(1))*gamma(7/2, I*x - I)
+ (3*cos(1) + 3*I*sin(1))*gamma(7/2, -I*x + I))*sin(7/2*arctan2(x - 1, 0)))*x - (5*((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)) + ((5*cos(1) - 5*I*sin(1
))*gamma(5/2, I*x - I) + (5*cos(1) + 5*I*sin(1))*gamma(5/2, -I*x + I))*sin(5/2*arctan2(x - 1, 0)))*abs(x - 1)
+ ((I*cos(1) + sin(1))*gamma(7/2, I*x - I) + (-I*cos(1) + sin(1))*gamma(7/2, -I*x + I))*cos(7/2*arctan2(x - 1,
0)) + ((cos(1) - I*sin(1))*gamma(7/2, I*x - I) + (cos(1) + I*sin(1))*gamma(7/2, -I*x + I))*sin(7/2*arctan2(x
- 1, 0)))*sqrt(-x + 1)*sqrt(abs(x - 1))/(x - 1)^4

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x\,\sin \relax (x)\,{\left (x+1\right )}^{5/2}}{\sqrt {1-x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*sin(x)*(x + 1)^(5/2))/(1 - x^2)^(1/2),x)

[Out]

int((x*sin(x)*(x + 1)^(5/2))/(1 - x^2)^(1/2), x)

________________________________________________________________________________________

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)**(5/2)/(-x**2+1)**(1/2)*x*sin(x),x)

[Out]

Timed out

________________________________________________________________________________________