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

Optimal. Leaf size=53 \[ \frac {8}{15} a^2 \tan (x) \sqrt {a \cos ^2(x)}+\frac {1}{5} \tan (x) \left (a \cos ^2(x)\right )^{5/2}+\frac {4}{15} a \tan (x) \left (a \cos ^2(x)\right )^{3/2} \]

[Out]

4/15*a*(a*cos(x)^2)^(3/2)*tan(x)+1/5*(a*cos(x)^2)^(5/2)*tan(x)+8/15*a^2*(a*cos(x)^2)^(1/2)*tan(x)

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Rubi [A]  time = 0.04, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3203, 3207, 2637} \[ \frac {8}{15} a^2 \tan (x) \sqrt {a \cos ^2(x)}+\frac {1}{5} \tan (x) \left (a \cos ^2(x)\right )^{5/2}+\frac {4}{15} a \tan (x) \left (a \cos ^2(x)\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^2*Sqrt[a*Cos[x]^2]*Tan[x])/15 + (4*a*(a*Cos[x]^2)^(3/2)*Tan[x])/15 + ((a*Cos[x]^2)^(5/2)*Tan[x])/5

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3203

Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> -Simp[(Cot[e + f*x]*(b*Sin[e + f*x]^2)^p)/(2*f*p), x]
+ Dist[(b*(2*p - 1))/(2*p), Int[(b*Sin[e + f*x]^2)^(p - 1), x], x] /; FreeQ[{b, e, f}, x] &&  !IntegerQ[p] &&
GtQ[p, 1]

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 ^2(x)\right )^{5/2} \, dx &=\frac {1}{5} \left (a \cos ^2(x)\right )^{5/2} \tan (x)+\frac {1}{5} (4 a) \int \left (a \cos ^2(x)\right )^{3/2} \, dx\\ &=\frac {4}{15} a \left (a \cos ^2(x)\right )^{3/2} \tan (x)+\frac {1}{5} \left (a \cos ^2(x)\right )^{5/2} \tan (x)+\frac {1}{15} \left (8 a^2\right ) \int \sqrt {a \cos ^2(x)} \, dx\\ &=\frac {4}{15} a \left (a \cos ^2(x)\right )^{3/2} \tan (x)+\frac {1}{5} \left (a \cos ^2(x)\right )^{5/2} \tan (x)+\frac {1}{15} \left (8 a^2 \sqrt {a \cos ^2(x)} \sec (x)\right ) \int \cos (x) \, dx\\ &=\frac {8}{15} a^2 \sqrt {a \cos ^2(x)} \tan (x)+\frac {4}{15} a \left (a \cos ^2(x)\right )^{3/2} \tan (x)+\frac {1}{5} \left (a \cos ^2(x)\right )^{5/2} \tan (x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 36, normalized size = 0.68 \[ \frac {1}{240} a^2 (150 \sin (x)+25 \sin (3 x)+3 \sin (5 x)) \sec (x) \sqrt {a \cos ^2(x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sqrt[a*Cos[x]^2]*Sec[x]*(150*Sin[x] + 25*Sin[3*x] + 3*Sin[5*x]))/240

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fricas [A]  time = 0.42, size = 40, normalized size = 0.75 \[ \frac {{\left (3 \, a^{2} \cos \relax (x)^{4} + 4 \, a^{2} \cos \relax (x)^{2} + 8 \, a^{2}\right )} \sqrt {a \cos \relax (x)^{2}} \sin \relax (x)}{15 \, \cos \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*a^2*cos(x)^4 + 4*a^2*cos(x)^2 + 8*a^2)*sqrt(a*cos(x)^2)*sin(x)/cos(x)

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giac [A]  time = 0.54, size = 34, normalized size = 0.64 \[ \frac {1}{15} \, {\left (3 \, a^{2} \sin \relax (x)^{5} - 10 \, a^{2} \sin \relax (x)^{3} + 15 \, a^{2} \sin \relax (x)\right )} \sqrt {a} \mathrm {sgn}\left (\cos \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*a^2*sin(x)^5 - 10*a^2*sin(x)^3 + 15*a^2*sin(x))*sqrt(a)*sgn(cos(x))

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maple [A]  time = 0.09, size = 32, normalized size = 0.60 \[ \frac {a^{3} \cos \relax (x ) \sin \relax (x ) \left (3 \left (\cos ^{4}\relax (x )\right )+4 \left (\cos ^{2}\relax (x )\right )+8\right )}{15 \sqrt {a \left (\cos ^{2}\relax (x )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*a^3*cos(x)*sin(x)*(3*cos(x)^4+4*cos(x)^2+8)/(a*cos(x)^2)^(1/2)

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maxima [A]  time = 0.68, size = 31, normalized size = 0.58 \[ \frac {1}{240} \, {\left (3 \, a^{2} \sin \left (5 \, x\right ) + 25 \, a^{2} \sin \left (3 \, x\right ) + 150 \, a^{2} \sin \relax (x)\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(3*a^2*sin(5*x) + 25*a^2*sin(3*x) + 150*a^2*sin(x))*sqrt(a)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\left (a\,{\cos \relax (x)}^2\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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