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

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

[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

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Rubi [A]  time = 0.0510577, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, 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 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
]])

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 8

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

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*}

Mathematica [A]  time = 0.123115, size = 53, normalized size = 0.4 \[ \frac{a (2520 x+2100 \sin (2 x)+600 \sin (4 x)+150 \sin (6 x)+25 \sin (8 x)+2 \sin (10 x)) \sec ^6(x) \left (a \cos ^4(x)\right )^{3/2}}{10240} \]

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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Maple [A]  time = 0.295, size = 57, normalized size = 0.4 \begin{align*}{\frac{128\,\sin \left ( x \right ) \left ( \cos \left ( x \right ) \right ) ^{9}+144\,\sin \left ( x \right ) \left ( \cos \left ( x \right ) \right ) ^{7}+168\,\sin \left ( x \right ) \left ( \cos \left ( x \right ) \right ) ^{5}+210\, \left ( \cos \left ( x \right ) \right ) ^{3}\sin \left ( x \right ) +315\,\cos \left ( x \right ) \sin \left ( x \right ) +315\,x}{1280\, \left ( \cos \left ( x \right ) \right ) ^{10}} \left ( a \left ( \cos \left ( x \right ) \right ) ^{4} \right ) ^{{\frac{5}{2}}}} \end{align*}

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 = 2.47314, size = 115, normalized size = 0.87 \begin{align*} \frac{63}{256} \, a^{\frac{5}{2}} x + \frac{315 \, a^{\frac{5}{2}} \tan \left (x\right )^{9} + 1470 \, a^{\frac{5}{2}} \tan \left (x\right )^{7} + 2688 \, a^{\frac{5}{2}} \tan \left (x\right )^{5} + 2370 \, a^{\frac{5}{2}} \tan \left (x\right )^{3} + 965 \, a^{\frac{5}{2}} \tan \left (x\right )}{1280 \,{\left (\tan \left (x\right )^{10} + 5 \, \tan \left (x\right )^{8} + 10 \, \tan \left (x\right )^{6} + 10 \, \tan \left (x\right )^{4} + 5 \, \tan \left (x\right )^{2} + 1\right )}} \end{align*}

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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Fricas [A]  time = 1.16141, size = 200, normalized size = 1.52 \begin{align*} \frac{\sqrt{a \cos \left (x\right )^{4}}{\left (315 \, a^{2} x +{\left (128 \, a^{2} \cos \left (x\right )^{9} + 144 \, a^{2} \cos \left (x\right )^{7} + 168 \, a^{2} \cos \left (x\right )^{5} + 210 \, a^{2} \cos \left (x\right )^{3} + 315 \, a^{2} \cos \left (x\right )\right )} \sin \left (x\right )\right )}}{1280 \, \cos \left (x\right )^{2}} \end{align*}

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

[Out]

Timed out

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Giac [A]  time = 1.39909, size = 77, normalized size = 0.58 \begin{align*} \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} \end{align*}

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)