3.9.12 \(\int e^{\tanh ^{-1}(x)} x (1+x)^{3/2} \sin (x) \, dx\) [812]

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

[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)

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Rubi [A]
time = 0.34, 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 = {6264, 6874, 3434, 3433, 3432, 3466, 3435, 3467} \begin {gather*} -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) \end {gather*}

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 3432

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3433

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3434

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 3435

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 3466

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 3467

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 6264

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 6874

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 \text {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 \text {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 \text {Subst}\left (\int x^6 \sin \left (1-x^2\right ) \, dx,x,\sqrt {1-x}\right )-8 \text {Subst}\left (\int \sin \left (1-x^2\right ) \, dx,x,\sqrt {1-x}\right )-10 \text {Subst}\left (\int x^4 \sin \left (1-x^2\right ) \, dx,x,\sqrt {1-x}\right )+16 \text {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 \text {Subst}\left (\int x^4 \cos \left (1-x^2\right ) \, dx,x,\sqrt {1-x}\right )-8 \text {Subst}\left (\int \cos \left (1-x^2\right ) \, dx,x,\sqrt {1-x}\right )+15 \text {Subst}\left (\int x^2 \cos \left (1-x^2\right ) \, dx,x,\sqrt {1-x}\right )+(8 \cos (1)) \text {Subst}\left (\int \sin \left (x^2\right ) \, dx,x,\sqrt {1-x}\right )-(8 \sin (1)) \text {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} \text {Subst}\left (\int \sin \left (1-x^2\right ) \, dx,x,\sqrt {1-x}\right )-\frac {15}{2} \text {Subst}\left (\int x^2 \sin \left (1-x^2\right ) \, dx,x,\sqrt {1-x}\right )-(8 \cos (1)) \text {Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt {1-x}\right )-(8 \sin (1)) \text {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} \text {Subst}\left (\int \cos \left (1-x^2\right ) \, dx,x,\sqrt {1-x}\right )-\frac {1}{2} (15 \cos (1)) \text {Subst}\left (\int \sin \left (x^2\right ) \, dx,x,\sqrt {1-x}\right )+\frac {1}{2} (15 \sin (1)) \text {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)) \text {Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt {1-x}\right )+\frac {1}{4} (15 \sin (1)) \text {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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 3.76, size = 201, normalized size = 0.60 \begin {gather*} \frac {\left (\frac {1}{32}+\frac {i}{32}\right ) \sqrt {1+x} \left ((-2-17 i) \sqrt {2 \pi } \sqrt {-1+x} \text {Erfi}\left (\frac {(1+i) \sqrt {-1+x}}{\sqrt {2}}\right ) (\cos (1)+i \sin (1))-(2-2 i) \left ((-1-20 i)-(11-10 i) x+(8+10 i) x^2+4 x^3\right ) (\cos (x)+i \sin (x))-(1+i) \left (2 \left ((-1+20 i)-(11+10 i) x+(8-10 i) x^2+4 x^3\right ) (-i \cos (1)+\sin (1))+(15+19 i) \sqrt {\frac {\pi }{2}} \sqrt {-1+x} \text {Erf}\left (\frac {(1+i) \sqrt {-1+x}}{\sqrt {2}}\right ) (\cos (x)+i \sin (x))\right ) (\cos (1+x)-i \sin (1+x))\right )}{\sqrt {1-x^2}} \end {gather*}

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]) - (1 + I)*(2
*((-1 + 20*I) - (11 + 10*I)*x + (8 - 10*I)*x^2 + 4*x^3)*((-I)*Cos[1] + Sin[1]) + (15 + 19*I)*Sqrt[Pi/2]*Sqrt[-
1 + x]*Erf[((1 + I)*Sqrt[-1 + x])/Sqrt[2]]*(Cos[x] + I*Sin[x]))*(Cos[1 + x] - I*Sin[1 + x])))/Sqrt[1 - x^2]

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (1+x \right )^{\frac {5}{2}} x \sin \left (x \right )}{\sqrt {-x^{2}+1}}\, dx\]

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)

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Maxima [C] Result contains complex when optimal does not.
time = 0.36, size = 983, normalized size = 2.93 \begin {gather*} \text {Too large to display} \end {gather*}

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) + I*sqrt(pi)*(erf(sqrt(-I*x + I)) - 1))*cos(1)
- (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))
- ((sqrt(pi)*(erf(sqrt(I*x - I)) - 1) + sqrt(pi)*(erf(sqrt(-I*x + I)) - 1))*cos(1) - (I*sqrt(pi)*(erf(sqrt(I*x
- I)) - 1) - I*sqrt(pi)*(erf(sqrt(-I*x + I)) - 1))*sin(1))*sin(1/2*arctan2(x - 1, 0)))*(x - 1)^2*abs(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*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))*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)) + ((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)))*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) - 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)))*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)) + ((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)))*(x - 1)^2 + 10*(((-I*cos(1) - s
in(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)))*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) - I*sin(1))*gamma(7/2, I*x - I) + (cos(1) + I*sin(1))*gamma(7/2, -I*x + I))*sin(7/2*ar
ctan2(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)) + ((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)))*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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

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Giac [C] Result contains complex when optimal does not.
time = 0.42, size = 202, normalized size = 0.60 \begin {gather*} -\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 \end {gather*}

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

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

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)

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