3.51 \(\int (a \cos ^4(x))^{5/2} \, dx\)

Optimal. Leaf size=132 \[ \frac {21}{128} a^2 \sin (x) \cos (x) \sqrt {a \cos ^4(x)}+\frac {63}{256} a^2 \tan (x) \sqrt {a \cos ^4(x)}+\frac {63}{256} a^2 x \sec ^2(x) \sqrt {a \cos ^4(x)}+\frac {1}{10} a^2 \sin (x) \cos ^7(x) \sqrt {a \cos ^4(x)}+\frac {9}{80} a^2 \sin (x) \cos ^5(x) \sqrt {a \cos ^4(x)}+\frac {21}{160} a^2 \sin (x) \cos ^3(x) \sqrt {a \cos ^4(x)} \]

[Out]

63/256*a^2*x*sec(x)^2*(a*cos(x)^4)^(1/2)+21/128*a^2*cos(x)*sin(x)*(a*cos(x)^4)^(1/2)+21/160*a^2*cos(x)^3*sin(x
)*(a*cos(x)^4)^(1/2)+9/80*a^2*cos(x)^5*sin(x)*(a*cos(x)^4)^(1/2)+1/10*a^2*cos(x)^7*sin(x)*(a*cos(x)^4)^(1/2)+6
3/256*a^2*(a*cos(x)^4)^(1/2)*tan(x)

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Rubi [A]  time = 0.05, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3207, 2635, 8} \[ \frac {1}{10} a^2 \sin (x) \cos ^7(x) \sqrt {a \cos ^4(x)}+\frac {9}{80} a^2 \sin (x) \cos ^5(x) \sqrt {a \cos ^4(x)}+\frac {21}{160} a^2 \sin (x) \cos ^3(x) \sqrt {a \cos ^4(x)}+\frac {21}{128} a^2 \sin (x) \cos (x) \sqrt {a \cos ^4(x)}+\frac {63}{256} a^2 \tan (x) \sqrt {a \cos ^4(x)}+\frac {63}{256} a^2 x \sec ^2(x) \sqrt {a \cos ^4(x)} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cos[x]^4)^(5/2),x]

[Out]

(63*a^2*x*Sqrt[a*Cos[x]^4]*Sec[x]^2)/256 + (21*a^2*Cos[x]*Sqrt[a*Cos[x]^4]*Sin[x])/128 + (21*a^2*Cos[x]^3*Sqrt
[a*Cos[x]^4]*Sin[x])/160 + (9*a^2*Cos[x]^5*Sqrt[a*Cos[x]^4]*Sin[x])/80 + (a^2*Cos[x]^7*Sqrt[a*Cos[x]^4]*Sin[x]
)/10 + (63*a^2*Sqrt[a*Cos[x]^4]*Tan[x])/256

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 3207

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rubi steps

\begin {align*} \int \left (a \cos ^4(x)\right )^{5/2} \, dx &=\left (a^2 \sqrt {a \cos ^4(x)} \sec ^2(x)\right ) \int \cos ^{10}(x) \, dx\\ &=\frac {1}{10} a^2 \cos ^7(x) \sqrt {a \cos ^4(x)} \sin (x)+\frac {1}{10} \left (9 a^2 \sqrt {a \cos ^4(x)} \sec ^2(x)\right ) \int \cos ^8(x) \, dx\\ &=\frac {9}{80} a^2 \cos ^5(x) \sqrt {a \cos ^4(x)} \sin (x)+\frac {1}{10} a^2 \cos ^7(x) \sqrt {a \cos ^4(x)} \sin (x)+\frac {1}{80} \left (63 a^2 \sqrt {a \cos ^4(x)} \sec ^2(x)\right ) \int \cos ^6(x) \, dx\\ &=\frac {21}{160} a^2 \cos ^3(x) \sqrt {a \cos ^4(x)} \sin (x)+\frac {9}{80} a^2 \cos ^5(x) \sqrt {a \cos ^4(x)} \sin (x)+\frac {1}{10} a^2 \cos ^7(x) \sqrt {a \cos ^4(x)} \sin (x)+\frac {1}{32} \left (21 a^2 \sqrt {a \cos ^4(x)} \sec ^2(x)\right ) \int \cos ^4(x) \, dx\\ &=\frac {21}{128} a^2 \cos (x) \sqrt {a \cos ^4(x)} \sin (x)+\frac {21}{160} a^2 \cos ^3(x) \sqrt {a \cos ^4(x)} \sin (x)+\frac {9}{80} a^2 \cos ^5(x) \sqrt {a \cos ^4(x)} \sin (x)+\frac {1}{10} a^2 \cos ^7(x) \sqrt {a \cos ^4(x)} \sin (x)+\frac {1}{128} \left (63 a^2 \sqrt {a \cos ^4(x)} \sec ^2(x)\right ) \int \cos ^2(x) \, dx\\ &=\frac {21}{128} a^2 \cos (x) \sqrt {a \cos ^4(x)} \sin (x)+\frac {21}{160} a^2 \cos ^3(x) \sqrt {a \cos ^4(x)} \sin (x)+\frac {9}{80} a^2 \cos ^5(x) \sqrt {a \cos ^4(x)} \sin (x)+\frac {1}{10} a^2 \cos ^7(x) \sqrt {a \cos ^4(x)} \sin (x)+\frac {63}{256} a^2 \sqrt {a \cos ^4(x)} \tan (x)+\frac {1}{256} \left (63 a^2 \sqrt {a \cos ^4(x)} \sec ^2(x)\right ) \int 1 \, dx\\ &=\frac {63}{256} a^2 x \sqrt {a \cos ^4(x)} \sec ^2(x)+\frac {21}{128} a^2 \cos (x) \sqrt {a \cos ^4(x)} \sin (x)+\frac {21}{160} a^2 \cos ^3(x) \sqrt {a \cos ^4(x)} \sin (x)+\frac {9}{80} a^2 \cos ^5(x) \sqrt {a \cos ^4(x)} \sin (x)+\frac {1}{10} a^2 \cos ^7(x) \sqrt {a \cos ^4(x)} \sin (x)+\frac {63}{256} a^2 \sqrt {a \cos ^4(x)} \tan (x)\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 53, normalized size = 0.40 \[ \frac {a (2520 x+2100 \sin (2 x)+600 \sin (4 x)+150 \sin (6 x)+25 \sin (8 x)+2 \sin (10 x)) \sec ^6(x) \left (a \cos ^4(x)\right )^{3/2}}{10240} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Cos[x]^4)^(5/2),x]

[Out]

(a*(a*Cos[x]^4)^(3/2)*Sec[x]^6*(2520*x + 2100*Sin[2*x] + 600*Sin[4*x] + 150*Sin[6*x] + 25*Sin[8*x] + 2*Sin[10*
x]))/10240

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fricas [A]  time = 0.53, size = 68, normalized size = 0.52 \[ \frac {\sqrt {a \cos \relax (x)^{4}} {\left (315 \, a^{2} x + {\left (128 \, a^{2} \cos \relax (x)^{9} + 144 \, a^{2} \cos \relax (x)^{7} + 168 \, a^{2} \cos \relax (x)^{5} + 210 \, a^{2} \cos \relax (x)^{3} + 315 \, a^{2} \cos \relax (x)\right )} \sin \relax (x)\right )}}{1280 \, \cos \relax (x)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cos(x)^4)^(5/2),x, algorithm="fricas")

[Out]

1/1280*sqrt(a*cos(x)^4)*(315*a^2*x + (128*a^2*cos(x)^9 + 144*a^2*cos(x)^7 + 168*a^2*cos(x)^5 + 210*a^2*cos(x)^
3 + 315*a^2*cos(x))*sin(x))/cos(x)^2

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giac [A]  time = 0.60, size = 57, normalized size = 0.43 \[ \frac {1}{10240} \, {\left (2520 \, a^{2} x + 2 \, a^{2} \sin \left (10 \, x\right ) + 25 \, a^{2} \sin \left (8 \, x\right ) + 150 \, a^{2} \sin \left (6 \, x\right ) + 600 \, a^{2} \sin \left (4 \, x\right ) + 2100 \, a^{2} \sin \left (2 \, x\right )\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cos(x)^4)^(5/2),x, algorithm="giac")

[Out]

1/10240*(2520*a^2*x + 2*a^2*sin(10*x) + 25*a^2*sin(8*x) + 150*a^2*sin(6*x) + 600*a^2*sin(4*x) + 2100*a^2*sin(2
*x))*sqrt(a)

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maple [A]  time = 0.41, size = 57, normalized size = 0.43 \[ \frac {\left (a \left (\cos ^{4}\relax (x )\right )\right )^{\frac {5}{2}} \left (128 \sin \relax (x ) \left (\cos ^{9}\relax (x )\right )+144 \sin \relax (x ) \left (\cos ^{7}\relax (x )\right )+168 \sin \relax (x ) \left (\cos ^{5}\relax (x )\right )+210 \left (\cos ^{3}\relax (x )\right ) \sin \relax (x )+315 \cos \relax (x ) \sin \relax (x )+315 x \right )}{1280 \cos \relax (x )^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cos(x)^4)^(5/2),x)

[Out]

1/1280*(a*cos(x)^4)^(5/2)*(128*sin(x)*cos(x)^9+144*sin(x)*cos(x)^7+168*sin(x)*cos(x)^5+210*cos(x)^3*sin(x)+315
*cos(x)*sin(x)+315*x)/cos(x)^10

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maxima [A]  time = 0.90, size = 85, normalized size = 0.64 \[ \frac {63}{256} \, a^{\frac {5}{2}} x + \frac {315 \, a^{\frac {5}{2}} \tan \relax (x)^{9} + 1470 \, a^{\frac {5}{2}} \tan \relax (x)^{7} + 2688 \, a^{\frac {5}{2}} \tan \relax (x)^{5} + 2370 \, a^{\frac {5}{2}} \tan \relax (x)^{3} + 965 \, a^{\frac {5}{2}} \tan \relax (x)}{1280 \, {\left (\tan \relax (x)^{10} + 5 \, \tan \relax (x)^{8} + 10 \, \tan \relax (x)^{6} + 10 \, \tan \relax (x)^{4} + 5 \, \tan \relax (x)^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cos(x)^4)^(5/2),x, algorithm="maxima")

[Out]

63/256*a^(5/2)*x + 1/1280*(315*a^(5/2)*tan(x)^9 + 1470*a^(5/2)*tan(x)^7 + 2688*a^(5/2)*tan(x)^5 + 2370*a^(5/2)
*tan(x)^3 + 965*a^(5/2)*tan(x))/(tan(x)^10 + 5*tan(x)^8 + 10*tan(x)^6 + 10*tan(x)^4 + 5*tan(x)^2 + 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a\,{\cos \relax (x)}^4\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cos(x)^4)^(5/2),x)

[Out]

int((a*cos(x)^4)^(5/2), x)

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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((a*cos(x)**4)**(5/2),x)

[Out]

Timed out

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