3.3.75 \(\int e^{\frac {x}{2}+x z} x^4 \sin ^4(\pi z) \, dz\) [275]

Optimal. Leaf size=199 \[ \frac {24 e^{\frac {x}{2}+x z} \pi ^4 x^3}{64 \pi ^4+20 \pi ^2 x^2+x^4}-\frac {24 e^{\frac {x}{2}+x z} \pi ^3 x^4 \cos (\pi z) \sin (\pi z)}{64 \pi ^4+20 \pi ^2 x^2+x^4}+\frac {12 e^{\frac {x}{2}+x z} \pi ^2 x^5 \sin ^2(\pi z)}{64 \pi ^4+20 \pi ^2 x^2+x^4}-\frac {4 e^{\frac {x}{2}+x z} \pi x^4 \cos (\pi z) \sin ^3(\pi z)}{16 \pi ^2+x^2}+\frac {e^{\frac {x}{2}+x z} x^5 \sin ^4(\pi z)}{16 \pi ^2+x^2} \]

[Out]

24*exp(1/2*x+x*z)*Pi^4*x^3/(64*Pi^4+20*Pi^2*x^2+x^4)-24*exp(1/2*x+x*z)*Pi^3*x^4*cos(Pi*z)*sin(Pi*z)/(64*Pi^4+2
0*Pi^2*x^2+x^4)+12*exp(1/2*x+x*z)*Pi^2*x^5*sin(Pi*z)^2/(64*Pi^4+20*Pi^2*x^2+x^4)-4*exp(1/2*x+x*z)*Pi*x^4*cos(P
i*z)*sin(Pi*z)^3/(16*Pi^2+x^2)+exp(1/2*x+x*z)*x^5*sin(Pi*z)^4/(16*Pi^2+x^2)

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Rubi [A]
time = 0.07, antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 4519, 2225} \begin {gather*} \frac {x^5 e^{x z+\frac {x}{2}} \sin ^4(\pi z)}{x^2+16 \pi ^2}-\frac {4 \pi x^4 e^{x z+\frac {x}{2}} \sin ^3(\pi z) \cos (\pi z)}{x^2+16 \pi ^2}-\frac {24 \pi ^3 x^4 e^{x z+\frac {x}{2}} \sin (\pi z) \cos (\pi z)}{x^4+20 \pi ^2 x^2+64 \pi ^4}+\frac {12 \pi ^2 x^5 e^{x z+\frac {x}{2}} \sin ^2(\pi z)}{x^4+20 \pi ^2 x^2+64 \pi ^4}+\frac {24 \pi ^4 x^3 e^{x z+\frac {x}{2}}}{x^4+20 \pi ^2 x^2+64 \pi ^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(x/2 + x*z)*x^4*Sin[Pi*z]^4,z]

[Out]

(24*E^(x/2 + x*z)*Pi^4*x^3)/(64*Pi^4 + 20*Pi^2*x^2 + x^4) - (24*E^(x/2 + x*z)*Pi^3*x^4*Cos[Pi*z]*Sin[Pi*z])/(6
4*Pi^4 + 20*Pi^2*x^2 + x^4) + (12*E^(x/2 + x*z)*Pi^2*x^5*Sin[Pi*z]^2)/(64*Pi^4 + 20*Pi^2*x^2 + x^4) - (4*E^(x/
2 + x*z)*Pi*x^4*Cos[Pi*z]*Sin[Pi*z]^3)/(16*Pi^2 + x^2) + (E^(x/2 + x*z)*x^5*Sin[Pi*z]^4)/(16*Pi^2 + x^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 4519

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)]^(n_), x_Symbol] :> Simp[b*c*Log[F]*F^(c*(a + b*x
))*(Sin[d + e*x]^n/(e^2*n^2 + b^2*c^2*Log[F]^2)), x] + (Dist[(n*(n - 1)*e^2)/(e^2*n^2 + b^2*c^2*Log[F]^2), Int
[F^(c*(a + b*x))*Sin[d + e*x]^(n - 2), x], x] - Simp[e*n*F^(c*(a + b*x))*Cos[d + e*x]*(Sin[d + e*x]^(n - 1)/(e
^2*n^2 + b^2*c^2*Log[F]^2)), x]) /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2*n^2 + b^2*c^2*Log[F]^2, 0] && GtQ[
n, 1]

Rubi steps

\begin {align*} \int e^{\frac {x}{2}+x z} x^4 \sin ^4(\pi z) \, dz &=x^4 \int e^{\frac {x}{2}+x z} \sin ^4(\pi z) \, dz\\ &=-\frac {4 e^{\frac {x}{2}+x z} \pi x^4 \cos (\pi z) \sin ^3(\pi z)}{16 \pi ^2+x^2}+\frac {e^{\frac {x}{2}+x z} x^5 \sin ^4(\pi z)}{16 \pi ^2+x^2}+\frac {\left (12 \pi ^2 x^4\right ) \int e^{\frac {x}{2}+x z} \sin ^2(\pi z) \, dz}{16 \pi ^2+x^2}\\ &=-\frac {24 e^{\frac {x}{2}+x z} \pi ^3 x^4 \cos (\pi z) \sin (\pi z)}{64 \pi ^4+20 \pi ^2 x^2+x^4}+\frac {12 e^{\frac {x}{2}+x z} \pi ^2 x^5 \sin ^2(\pi z)}{64 \pi ^4+20 \pi ^2 x^2+x^4}-\frac {4 e^{\frac {x}{2}+x z} \pi x^4 \cos (\pi z) \sin ^3(\pi z)}{16 \pi ^2+x^2}+\frac {e^{\frac {x}{2}+x z} x^5 \sin ^4(\pi z)}{16 \pi ^2+x^2}+\frac {\left (24 \pi ^4 x^4\right ) \int e^{\frac {x}{2}+x z} \, dz}{64 \pi ^4+20 \pi ^2 x^2+x^4}\\ &=\frac {24 e^{\frac {x}{2}+x z} \pi ^4 x^3}{64 \pi ^4+20 \pi ^2 x^2+x^4}-\frac {24 e^{\frac {x}{2}+x z} \pi ^3 x^4 \cos (\pi z) \sin (\pi z)}{64 \pi ^4+20 \pi ^2 x^2+x^4}+\frac {12 e^{\frac {x}{2}+x z} \pi ^2 x^5 \sin ^2(\pi z)}{64 \pi ^4+20 \pi ^2 x^2+x^4}-\frac {4 e^{\frac {x}{2}+x z} \pi x^4 \cos (\pi z) \sin ^3(\pi z)}{16 \pi ^2+x^2}+\frac {e^{\frac {x}{2}+x z} x^5 \sin ^4(\pi z)}{16 \pi ^2+x^2}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 136, normalized size = 0.68 \begin {gather*} \frac {e^{x \left (\frac {1}{2}+z\right )} x^4 \left (192 \pi ^4+60 \pi ^2 x^2+3 x^4-4 x^2 \left (16 \pi ^2+x^2\right ) \cos (2 \pi z)+x^2 \left (4 \pi ^2+x^2\right ) \cos (4 \pi z)-128 \pi ^3 x \sin (2 \pi z)-8 \pi x^3 \sin (2 \pi z)+16 \pi ^3 x \sin (4 \pi z)+4 \pi x^3 \sin (4 \pi z)\right )}{8 \left (64 \pi ^4 x+20 \pi ^2 x^3+x^5\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(x/2 + x*z)*x^4*Sin[Pi*z]^4,z]

[Out]

(E^(x*(1/2 + z))*x^4*(192*Pi^4 + 60*Pi^2*x^2 + 3*x^4 - 4*x^2*(16*Pi^2 + x^2)*Cos[2*Pi*z] + x^2*(4*Pi^2 + x^2)*
Cos[4*Pi*z] - 128*Pi^3*x*Sin[2*Pi*z] - 8*Pi*x^3*Sin[2*Pi*z] + 16*Pi^3*x*Sin[4*Pi*z] + 4*Pi*x^3*Sin[4*Pi*z]))/(
8*(64*Pi^4*x + 20*Pi^2*x^3 + x^5))

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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 164.25, size = 422, normalized size = 2.12 \begin {gather*} \text {Piecewise}\left [\left \{\left \{\frac {x^4 \left (\frac {I}{16}-\frac {\text {Pi} z}{16}+\frac {I \text {Pi} z \text {Sin}\left [4 \text {Pi} z\right ]}{16}+\frac {\text {Pi} z \text {Cos}\left [2 \text {Pi} z\right ]^2}{8}-\frac {I \text {Cos}\left [2 \text {Pi} z\right ]}{6}-\frac {\text {Sin}\left [4 \text {Pi} z\right ]}{64}+\frac {I \text {Cos}\left [2 \text {Pi} z\right ]^2}{16}+\frac {\text {Sin}\left [2 \text {Pi} z\right ]}{12}\right ) E^{-4 I \text {Pi} z}}{\text {Pi}},x\text {==}0\text {$\vert $$\vert $}x\text {==}-4 I \text {Pi}\right \},\left \{\frac {x^4 \left (-5 I+I 6 \text {Pi} z \text {Sin}\left [2 \text {Pi} z\right ]+6 \text {Pi} z \text {Cos}\left [2 \text {Pi} z\right ]-2 \text {Sin}\left [2 \text {Pi} z\right ]-\text {Sin}\left [4 \text {Pi} z\right ]+I 5 \text {Cos}\left [2 \text {Pi} z\right ]+I \text {Cos}\left [2 \text {Pi} z\right ]^2\right ) E^{-2 I \text {Pi} z}}{24 \text {Pi}},x\text {==}-2 I \text {Pi}\right \},\left \{\frac {x^4 \left (5 I-6 I \text {Pi} z \text {Sin}\left [2 \text {Pi} z\right ]+6 \text {Pi} z \text {Cos}\left [2 \text {Pi} z\right ]-5 I \text {Cos}\left [2 \text {Pi} z\right ]-2 \text {Sin}\left [2 \text {Pi} z\right ]-I \text {Cos}\left [2 \text {Pi} z\right ]^2-\text {Sin}\left [4 \text {Pi} z\right ]\right ) E^{I 2 \text {Pi} z}}{24 \text {Pi}},x\text {==}I 2 \text {Pi}\right \},\left \{\frac {x^4 \left (-\frac {I}{16}-\frac {I \text {Pi} z \text {Sin}\left [4 \text {Pi} z\right ]}{16}-\frac {\text {Pi} z}{16}+\frac {\text {Pi} z \text {Cos}\left [2 \text {Pi} z\right ]^2}{8}-\frac {I \text {Cos}\left [2 \text {Pi} z\right ]^2}{16}-\frac {\text {Sin}\left [4 \text {Pi} z\right ]}{64}+\frac {I \text {Cos}\left [2 \text {Pi} z\right ]}{6}+\frac {\text {Sin}\left [2 \text {Pi} z\right ]}{12}\right ) E^{I 4 \text {Pi} z}}{\text {Pi}},x\text {==}I 4 \text {Pi}\right \},\left \{\frac {x^3 \left (48 \text {Pi}^4+x \left (2 x^3 \text {Sin}\left [\text {Pi} z\right ]^4-2 \text {Pi} x^2 \text {Sin}\left [2 \text {Pi} z\right ]+\text {Pi} x^2 \text {Sin}\left [4 \text {Pi} z\right ]+32 \text {Pi}^2 x \text {Sin}\left [\text {Pi} z\right ]^4-32 \text {Pi}^3 \text {Sin}\left [2 \text {Pi} z\right ]+4 \text {Pi}^3 \text {Sin}\left [4 \text {Pi} z\right ]\right )+3 \text {Pi}^2 x^2 \left (1-\text {Cos}\left [4 \text {Pi} z\right ]\right )\right ) E^{\frac {x \left (1+2 z\right )}{2}}}{2 \left (64 \text {Pi}^4+x^4+20 \text {Pi}^2 x^2\right )},\text {True}\right \}\right \}\right ] \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[Exp[x*z+x/2]*Sin[Pi*z]^4*x^4,z]')

[Out]

Piecewise[{{x ^ 4 (I / 16 - Pi z / 16 + I Pi z Sin[4 Pi z] / 16 + Pi z Cos[2 Pi z] ^ 2 / 8 - I Cos[2 Pi z] / 6
- Sin[4 Pi z] / 64 + I Cos[2 Pi z] ^ 2 / 16 + Sin[2 Pi z] / 12) E ^ (-4 I Pi z) / Pi, x == 0 || x == -4 I Pi}
, {x ^ 4 (-5 I + I 6 Pi z Sin[2 Pi z] + 6 Pi z Cos[2 Pi z] - 2 Sin[2 Pi z] - Sin[4 Pi z] + I 5 Cos[2 Pi z] + I
Cos[2 Pi z] ^ 2) E ^ (-2 I Pi z) / (24 Pi), x == -2 I Pi}, {x ^ 4 (5 I - 6 I Pi z Sin[2 Pi z] + 6 Pi z Cos[2
Pi z] - 5 I Cos[2 Pi z] - 2 Sin[2 Pi z] - I Cos[2 Pi z] ^ 2 - Sin[4 Pi z]) E ^ (I 2 Pi z) / (24 Pi), x == I 2
Pi}, {x ^ 4 (-I / 16 - I Pi z Sin[4 Pi z] / 16 - Pi z / 16 + Pi z Cos[2 Pi z] ^ 2 / 8 - I Cos[2 Pi z] ^ 2 / 16
- Sin[4 Pi z] / 64 + I Cos[2 Pi z] / 6 + Sin[2 Pi z] / 12) E ^ (I 4 Pi z) / Pi, x == I 4 Pi}, {x ^ 3 (48 Pi ^
4 + x (2 x ^ 3 Sin[Pi z] ^ 4 - 2 Pi x ^ 2 Sin[2 Pi z] + Pi x ^ 2 Sin[4 Pi z] + 32 Pi ^ 2 x Sin[Pi z] ^ 4 - 32
Pi ^ 3 Sin[2 Pi z] + 4 Pi ^ 3 Sin[4 Pi z]) + 3 Pi ^ 2 x ^ 2 (1 - Cos[4 Pi z])) E ^ (x (1 + 2 z) / 2) / (2 (64
Pi ^ 4 + x ^ 4 + 20 Pi ^ 2 x ^ 2)), True}}]

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Maple [A]
time = 0.18, size = 128, normalized size = 0.64

method	result	size
default	\(-\frac {x^{4} \left (-\frac {3 \,{\mathrm e}^{\frac {1}{2} x +x z}}{x}-\frac {x \,{\mathrm e}^{\frac {1}{2} x +x z} \cos \left (4 \pi z \right )}{16 \pi ^{2}+x^{2}}-\frac {4 \pi \,{\mathrm e}^{\frac {1}{2} x +x z} \sin \left (4 \pi z \right )}{16 \pi ^{2}+x^{2}}+\frac {4 x \,{\mathrm e}^{\frac {1}{2} x +x z} \cos \left (2 \pi z \right )}{4 \pi ^{2}+x^{2}}+\frac {8 \pi \,{\mathrm e}^{\frac {1}{2} x +x z} \sin \left (2 \pi z \right )}{4 \pi ^{2}+x^{2}}\right )}{8}\)	\(128\)
risch	\(\frac {3 x^{3} {\mathrm e}^{\frac {x \left (1+2 z \right )}{2}}}{8}+\frac {x^{5} {\mathrm e}^{\frac {1}{2} x +x z} \cos \left (4 \pi z \right )}{128 \pi ^{2}+8 x^{2}}+\frac {x^{4} {\mathrm e}^{\frac {1}{2} x +x z} \pi \sin \left (4 \pi z \right )}{32 \pi ^{2}+2 x^{2}}-\frac {x^{5} {\mathrm e}^{\frac {1}{2} x +x z} \cos \left (2 \pi z \right )}{2 \left (4 \pi ^{2}+x^{2}\right )}-\frac {x^{4} {\mathrm e}^{\frac {1}{2} x +x z} \pi \sin \left (2 \pi z \right )}{4 \pi ^{2}+x^{2}}\)	\(134\)
norman	\(\frac {\frac {24 \,{\mathrm e}^{\frac {1}{2} x +x z} \pi ^{4} x^{3}}{64 \pi ^{4}+20 \pi ^{2} x^{2}+x^{4}}-\frac {48 \pi ^{3} x^{4} {\mathrm e}^{\frac {1}{2} x +x z} \tan \left (\frac {\pi z}{2}\right )}{64 \pi ^{4}+20 \pi ^{2} x^{2}+x^{4}}+\frac {48 \pi ^{3} x^{4} {\mathrm e}^{\frac {1}{2} x +x z} \left (\tan ^{7}\left (\frac {\pi z}{2}\right )\right )}{64 \pi ^{4}+20 \pi ^{2} x^{2}+x^{4}}+\frac {16 \left (9 \pi ^{4}+10 \pi ^{2} x^{2}+x^{4}\right ) x^{3} {\mathrm e}^{\frac {1}{2} x +x z} \left (\tan ^{4}\left (\frac {\pi z}{2}\right )\right )}{64 \pi ^{4}+20 \pi ^{2} x^{2}+x^{4}}+\frac {24 \,{\mathrm e}^{\frac {1}{2} x +x z} \pi ^{4} x^{3} \left (\tan ^{8}\left (\frac {\pi z}{2}\right )\right )}{64 \pi ^{4}+20 \pi ^{2} x^{2}+x^{4}}-\frac {16 \pi \,x^{4} \left (11 \pi ^{2}+2 x^{2}\right ) {\mathrm e}^{\frac {1}{2} x +x z} \left (\tan ^{3}\left (\frac {\pi z}{2}\right )\right )}{64 \pi ^{4}+20 \pi ^{2} x^{2}+x^{4}}+\frac {16 \pi \,x^{4} \left (11 \pi ^{2}+2 x^{2}\right ) {\mathrm e}^{\frac {1}{2} x +x z} \left (\tan ^{5}\left (\frac {\pi z}{2}\right )\right )}{64 \pi ^{4}+20 \pi ^{2} x^{2}+x^{4}}+\frac {48 x^{3} \left (2 \pi ^{2}+x^{2}\right ) \pi ^{2} {\mathrm e}^{\frac {1}{2} x +x z} \left (\tan ^{2}\left (\frac {\pi z}{2}\right )\right )}{64 \pi ^{4}+20 \pi ^{2} x^{2}+x^{4}}+\frac {48 x^{3} \left (2 \pi ^{2}+x^{2}\right ) \pi ^{2} {\mathrm e}^{\frac {1}{2} x +x z} \left (\tan ^{6}\left (\frac {\pi z}{2}\right )\right )}{64 \pi ^{4}+20 \pi ^{2} x^{2}+x^{4}}}{\left (1+\tan ^{2}\left (\frac {\pi z}{2}\right )\right )^{4}}\)	\(433\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*exp(1/2*x+x*z)*sin(Pi*z)^4,z,method=_RETURNVERBOSE)

[Out]

-1/8*x^4*(-3*exp(1/2*x+x*z)/x-x/(16*Pi^2+x^2)*exp(1/2*x+x*z)*cos(4*Pi*z)-4*Pi/(16*Pi^2+x^2)*exp(1/2*x+x*z)*sin
(4*Pi*z)+4*x/(4*Pi^2+x^2)*exp(1/2*x+x*z)*cos(2*Pi*z)+8*Pi/(4*Pi^2+x^2)*exp(1/2*x+x*z)*sin(2*Pi*z))

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Maxima [A]
time = 0.28, size = 160, normalized size = 0.80 \begin {gather*} \frac {{\left ({\left (4 \, \pi ^{2} x^{2} + x^{4}\right )} \cos \left (4 \, \pi z\right ) e^{\left (x z + \frac {1}{2} \, x\right )} - 4 \, {\left (16 \, \pi ^{2} x^{2} + x^{4}\right )} \cos \left (2 \, \pi z\right ) e^{\left (x z + \frac {1}{2} \, x\right )} + 4 \, {\left (4 \, \pi ^{3} x + \pi x^{3}\right )} e^{\left (x z + \frac {1}{2} \, x\right )} \sin \left (4 \, \pi z\right ) - 8 \, {\left (16 \, \pi ^{3} x + \pi x^{3}\right )} e^{\left (x z + \frac {1}{2} \, x\right )} \sin \left (2 \, \pi z\right ) + 3 \, {\left (64 \, \pi ^{4} + 20 \, \pi ^{2} x^{2} + x^{4}\right )} e^{\left (x z + \frac {1}{2} \, x\right )}\right )} x^{4}}{8 \, {\left (64 \, \pi ^{4} x + 20 \, \pi ^{2} x^{3} + x^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*exp(1/2*x+x*z)*sin(pi*z)^4,z, algorithm="maxima")

[Out]

1/8*((4*pi^2*x^2 + x^4)*cos(4*pi*z)*e^(x*z + 1/2*x) - 4*(16*pi^2*x^2 + x^4)*cos(2*pi*z)*e^(x*z + 1/2*x) + 4*(4
*pi^3*x + pi*x^3)*e^(x*z + 1/2*x)*sin(4*pi*z) - 8*(16*pi^3*x + pi*x^3)*e^(x*z + 1/2*x)*sin(2*pi*z) + 3*(64*pi^
4 + 20*pi^2*x^2 + x^4)*e^(x*z + 1/2*x))*x^4/(64*pi^4*x + 20*pi^2*x^3 + x^5)

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Fricas [A]
time = 0.32, size = 145, normalized size = 0.73 \begin {gather*} \frac {4 \, {\left ({\left (4 \, \pi ^{3} x^{4} + \pi x^{6}\right )} \cos \left (\pi z\right )^{3} - {\left (10 \, \pi ^{3} x^{4} + \pi x^{6}\right )} \cos \left (\pi z\right )\right )} e^{\left (x z + \frac {1}{2} \, x\right )} \sin \left (\pi z\right ) + {\left (24 \, \pi ^{4} x^{3} + 16 \, \pi ^{2} x^{5} + x^{7} + {\left (4 \, \pi ^{2} x^{5} + x^{7}\right )} \cos \left (\pi z\right )^{4} - 2 \, {\left (10 \, \pi ^{2} x^{5} + x^{7}\right )} \cos \left (\pi z\right )^{2}\right )} e^{\left (x z + \frac {1}{2} \, x\right )}}{64 \, \pi ^{4} + 20 \, \pi ^{2} x^{2} + x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*exp(1/2*x+x*z)*sin(pi*z)^4,z, algorithm="fricas")

[Out]

(4*((4*pi^3*x^4 + pi*x^6)*cos(pi*z)^3 - (10*pi^3*x^4 + pi*x^6)*cos(pi*z))*e^(x*z + 1/2*x)*sin(pi*z) + (24*pi^4
*x^3 + 16*pi^2*x^5 + x^7 + (4*pi^2*x^5 + x^7)*cos(pi*z)^4 - 2*(10*pi^2*x^5 + x^7)*cos(pi*z)^2)*e^(x*z + 1/2*x)
)/(64*pi^4 + 20*pi^2*x^2 + x^4)

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Sympy [A]
time = 135.78, size = 1277, normalized size = 6.42

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*exp(1/2*x+x*z)*sin(pi*z)**4,z)

[Out]

x**4*Piecewise((3*z*sin(pi*z)**4/8 + 3*z*sin(pi*z)**2*cos(pi*z)**2/4 + 3*z*cos(pi*z)**4/8 - 5*sin(pi*z)**3*cos
(pi*z)/(8*pi) - 3*sin(pi*z)*cos(pi*z)**3/(8*pi), Eq(x, 0)), (z*exp(-4*I*pi*z)*sin(pi*z)**4/16 - I*z*exp(-4*I*p
i*z)*sin(pi*z)**3*cos(pi*z)/4 - 3*z*exp(-4*I*pi*z)*sin(pi*z)**2*cos(pi*z)**2/8 + I*z*exp(-4*I*pi*z)*sin(pi*z)*
cos(pi*z)**3/4 + z*exp(-4*I*pi*z)*cos(pi*z)**4/16 + 7*I*exp(-4*I*pi*z)*sin(pi*z)**4/(24*pi) + 11*exp(-4*I*pi*z
)*sin(pi*z)**3*cos(pi*z)/(48*pi) + 5*exp(-4*I*pi*z)*sin(pi*z)*cos(pi*z)**3/(48*pi) - I*exp(-4*I*pi*z)*cos(pi*z
)**4/(24*pi), Eq(x, -4*I*pi)), (-z*exp(-2*I*pi*z)*sin(pi*z)**4/4 + I*z*exp(-2*I*pi*z)*sin(pi*z)**3*cos(pi*z)/2
+ I*z*exp(-2*I*pi*z)*sin(pi*z)*cos(pi*z)**3/2 + z*exp(-2*I*pi*z)*cos(pi*z)**4/4 - 3*I*exp(-2*I*pi*z)*sin(pi*z
)**4/(8*pi) - I*exp(-2*I*pi*z)*sin(pi*z)**2*cos(pi*z)**2/(2*pi) - exp(-2*I*pi*z)*sin(pi*z)*cos(pi*z)**3/(3*pi)
+ I*exp(-2*I*pi*z)*cos(pi*z)**4/(24*pi), Eq(x, -2*I*pi)), (-z*exp(2*I*pi*z)*sin(pi*z)**4/4 - I*z*exp(2*I*pi*z
)*sin(pi*z)**3*cos(pi*z)/2 - I*z*exp(2*I*pi*z)*sin(pi*z)*cos(pi*z)**3/2 + z*exp(2*I*pi*z)*cos(pi*z)**4/4 + 3*I
*exp(2*I*pi*z)*sin(pi*z)**4/(8*pi) + I*exp(2*I*pi*z)*sin(pi*z)**2*cos(pi*z)**2/(2*pi) - exp(2*I*pi*z)*sin(pi*z
)*cos(pi*z)**3/(3*pi) - I*exp(2*I*pi*z)*cos(pi*z)**4/(24*pi), Eq(x, 2*I*pi)), (z*exp(4*I*pi*z)*sin(pi*z)**4/16
+ I*z*exp(4*I*pi*z)*sin(pi*z)**3*cos(pi*z)/4 - 3*z*exp(4*I*pi*z)*sin(pi*z)**2*cos(pi*z)**2/8 - I*z*exp(4*I*pi
*z)*sin(pi*z)*cos(pi*z)**3/4 + z*exp(4*I*pi*z)*cos(pi*z)**4/16 - 7*I*exp(4*I*pi*z)*sin(pi*z)**4/(24*pi) + 11*e
xp(4*I*pi*z)*sin(pi*z)**3*cos(pi*z)/(48*pi) + 5*exp(4*I*pi*z)*sin(pi*z)*cos(pi*z)**3/(48*pi) + I*exp(4*I*pi*z)
*cos(pi*z)**4/(24*pi), Eq(x, 4*I*pi)), (x**4*exp(x/2)*exp(x*z)*sin(pi*z)**4/(x**5 + 20*pi**2*x**3 + 64*pi**4*x
) - 4*pi*x**3*exp(x/2)*exp(x*z)*sin(pi*z)**3*cos(pi*z)/(x**5 + 20*pi**2*x**3 + 64*pi**4*x) + 16*pi**2*x**2*exp
(x/2)*exp(x*z)*sin(pi*z)**4/(x**5 + 20*pi**2*x**3 + 64*pi**4*x) + 12*pi**2*x**2*exp(x/2)*exp(x*z)*sin(pi*z)**2
*cos(pi*z)**2/(x**5 + 20*pi**2*x**3 + 64*pi**4*x) - 40*pi**3*x*exp(x/2)*exp(x*z)*sin(pi*z)**3*cos(pi*z)/(x**5
+ 20*pi**2*x**3 + 64*pi**4*x) - 24*pi**3*x*exp(x/2)*exp(x*z)*sin(pi*z)*cos(pi*z)**3/(x**5 + 20*pi**2*x**3 + 64
*pi**4*x) + 24*pi**4*exp(x/2)*exp(x*z)*sin(pi*z)**4/(x**5 + 20*pi**2*x**3 + 64*pi**4*x) + 48*pi**4*exp(x/2)*ex
p(x*z)*sin(pi*z)**2*cos(pi*z)**2/(x**5 + 20*pi**2*x**3 + 64*pi**4*x) + 24*pi**4*exp(x/2)*exp(x*z)*cos(pi*z)**4
/(x**5 + 20*pi**2*x**3 + 64*pi**4*x), True))

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Giac [A]
time = 0.00, size = 127, normalized size = 0.64 \begin {gather*} x^{4} \left (\frac {3 \mathrm {e}^{\frac {2 x z+x}{2}}}{8 x}+\mathrm {e}^{\frac {2 x z+x}{2}} \left (-\frac {2 x \cos \left (2 \pi z\right )}{\left (2 x\right )^{2}+\left (4 \pi \right )^{2}}-\frac {4 \pi \sin \left (2 \pi z\right )}{\left (2 x\right )^{2}+\left (4 \pi \right )^{2}}\right )+\mathrm {e}^{\frac {2 x z+x}{2}} \left (\frac {8 x \cos \left (4 \pi z\right )}{\left (8 x\right )^{2}+\left (32 \pi \right )^{2}}+\frac {32 \pi \sin \left (4 \pi z\right )}{\left (8 x\right )^{2}+\left (32 \pi \right )^{2}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*exp(1/2*x+x*z)*sin(pi*z)^4,z)

[Out]

1/8*(e^(x*z + 1/2*x)*(x*cos(4*pi*z)/(16*pi^2 + x^2) + 4*pi*sin(4*pi*z)/(16*pi^2 + x^2)) - 4*e^(x*z + 1/2*x)*(x
*cos(2*pi*z)/(4*pi^2 + x^2) + 2*pi*sin(2*pi*z)/(4*pi^2 + x^2)) + 3*e^(x*z + 1/2*x)/x)*x^4

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Mupad [B]
time = 1.11, size = 140, normalized size = 0.70 \begin {gather*} \frac {x^3\,{\mathrm {e}}^{\frac {x}{2}+x\,z}\,\left (24\,\Pi ^4-\frac {x^4\,\cos \left (2\,\Pi \,z\right )}{2}+\frac {x^4\,\cos \left (4\,\Pi \,z\right )}{8}+\frac {3\,x^4}{8}+\frac {15\,\Pi ^2\,x^2}{2}-\Pi \,x^3\,\sin \left (2\,\Pi \,z\right )-16\,\Pi ^3\,x\,\sin \left (2\,\Pi \,z\right )+\frac {\Pi \,x^3\,\sin \left (4\,\Pi \,z\right )}{2}+2\,\Pi ^3\,x\,\sin \left (4\,\Pi \,z\right )-8\,\Pi ^2\,x^2\,\cos \left (2\,\Pi \,z\right )+\frac {\Pi ^2\,x^2\,\cos \left (4\,\Pi \,z\right )}{2}\right )}{64\,\Pi ^4+20\,\Pi ^2\,x^2+x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*exp(x/2 + x*z)*sin(Pi*z)^4,z)

[Out]

(x^3*exp(x/2 + x*z)*(24*Pi^4 - (x^4*cos(2*Pi*z))/2 + (x^4*cos(4*Pi*z))/8 + (3*x^4)/8 + (15*Pi^2*x^2)/2 - Pi*x^
3*sin(2*Pi*z) - 16*Pi^3*x*sin(2*Pi*z) + (Pi*x^3*sin(4*Pi*z))/2 + 2*Pi^3*x*sin(4*Pi*z) - 8*Pi^2*x^2*cos(2*Pi*z)
+ (Pi^2*x^2*cos(4*Pi*z))/2))/(64*Pi^4 + x^4 + 20*Pi^2*x^2)

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