3.52 \(\int (a \cos ^4(x))^{3/2} \, dx\)

Optimal. Leaf size=78 \[ \frac {5}{24} a \sin (x) \cos (x) \sqrt {a \cos ^4(x)}+\frac {5}{16} a \tan (x) \sqrt {a \cos ^4(x)}+\frac {5}{16} a x \sec ^2(x) \sqrt {a \cos ^4(x)}+\frac {1}{6} a \sin (x) \cos ^3(x) \sqrt {a \cos ^4(x)} \]

[Out]

5/16*a*x*sec(x)^2*(a*cos(x)^4)^(1/2)+5/24*a*cos(x)*sin(x)*(a*cos(x)^4)^(1/2)+1/6*a*cos(x)^3*sin(x)*(a*cos(x)^4
)^(1/2)+5/16*a*(a*cos(x)^4)^(1/2)*tan(x)

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Rubi [A]  time = 0.03, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3207, 2635, 8} \[ \frac {1}{6} a \sin (x) \cos ^3(x) \sqrt {a \cos ^4(x)}+\frac {5}{24} a \sin (x) \cos (x) \sqrt {a \cos ^4(x)}+\frac {5}{16} a \tan (x) \sqrt {a \cos ^4(x)}+\frac {5}{16} a x \sec ^2(x) \sqrt {a \cos ^4(x)} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cos[x]^4)^(3/2),x]

[Out]

(5*a*x*Sqrt[a*Cos[x]^4]*Sec[x]^2)/16 + (5*a*Cos[x]*Sqrt[a*Cos[x]^4]*Sin[x])/24 + (a*Cos[x]^3*Sqrt[a*Cos[x]^4]*
Sin[x])/6 + (5*a*Sqrt[a*Cos[x]^4]*Tan[x])/16

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 )^{3/2} \, dx &=\left (a \sqrt {a \cos ^4(x)} \sec ^2(x)\right ) \int \cos ^6(x) \, dx\\ &=\frac {1}{6} a \cos ^3(x) \sqrt {a \cos ^4(x)} \sin (x)+\frac {1}{6} \left (5 a \sqrt {a \cos ^4(x)} \sec ^2(x)\right ) \int \cos ^4(x) \, dx\\ &=\frac {5}{24} a \cos (x) \sqrt {a \cos ^4(x)} \sin (x)+\frac {1}{6} a \cos ^3(x) \sqrt {a \cos ^4(x)} \sin (x)+\frac {1}{8} \left (5 a \sqrt {a \cos ^4(x)} \sec ^2(x)\right ) \int \cos ^2(x) \, dx\\ &=\frac {5}{24} a \cos (x) \sqrt {a \cos ^4(x)} \sin (x)+\frac {1}{6} a \cos ^3(x) \sqrt {a \cos ^4(x)} \sin (x)+\frac {5}{16} a \sqrt {a \cos ^4(x)} \tan (x)+\frac {1}{16} \left (5 a \sqrt {a \cos ^4(x)} \sec ^2(x)\right ) \int 1 \, dx\\ &=\frac {5}{16} a x \sqrt {a \cos ^4(x)} \sec ^2(x)+\frac {5}{24} a \cos (x) \sqrt {a \cos ^4(x)} \sin (x)+\frac {1}{6} a \cos ^3(x) \sqrt {a \cos ^4(x)} \sin (x)+\frac {5}{16} a \sqrt {a \cos ^4(x)} \tan (x)\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 38, normalized size = 0.49 \[ \frac {1}{192} (60 x+45 \sin (2 x)+9 \sin (4 x)+\sin (6 x)) \sec ^6(x) \left (a \cos ^4(x)\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Cos[x]^4)^(3/2),x]

[Out]

((a*Cos[x]^4)^(3/2)*Sec[x]^6*(60*x + 45*Sin[2*x] + 9*Sin[4*x] + Sin[6*x]))/192

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fricas [A]  time = 0.52, size = 42, normalized size = 0.54 \[ \frac {\sqrt {a \cos \relax (x)^{4}} {\left (15 \, a x + {\left (8 \, a \cos \relax (x)^{5} + 10 \, a \cos \relax (x)^{3} + 15 \, a \cos \relax (x)\right )} \sin \relax (x)\right )}}{48 \, \cos \relax (x)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cos(x)^4)^(3/2),x, algorithm="fricas")

[Out]

1/48*sqrt(a*cos(x)^4)*(15*a*x + (8*a*cos(x)^5 + 10*a*cos(x)^3 + 15*a*cos(x))*sin(x))/cos(x)^2

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giac [A]  time = 0.55, size = 25, normalized size = 0.32 \[ \frac {1}{192} \, a^{\frac {3}{2}} {\left (60 \, x + \sin \left (6 \, x\right ) + 9 \, \sin \left (4 \, x\right ) + 45 \, \sin \left (2 \, x\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cos(x)^4)^(3/2),x, algorithm="giac")

[Out]

1/192*a^(3/2)*(60*x + sin(6*x) + 9*sin(4*x) + 45*sin(2*x))

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maple [A]  time = 0.17, size = 41, normalized size = 0.53 \[ \frac {\left (a \left (\cos ^{4}\relax (x )\right )\right )^{\frac {3}{2}} \left (8 \sin \relax (x ) \left (\cos ^{5}\relax (x )\right )+10 \left (\cos ^{3}\relax (x )\right ) \sin \relax (x )+15 \cos \relax (x ) \sin \relax (x )+15 x \right )}{48 \cos \relax (x )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cos(x)^4)^(3/2),x)

[Out]

1/48*(a*cos(x)^4)^(3/2)*(8*sin(x)*cos(x)^5+10*cos(x)^3*sin(x)+15*cos(x)*sin(x)+15*x)/cos(x)^6

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maxima [A]  time = 0.67, size = 55, normalized size = 0.71 \[ \frac {5}{16} \, a^{\frac {3}{2}} x + \frac {15 \, a^{\frac {3}{2}} \tan \relax (x)^{5} + 40 \, a^{\frac {3}{2}} \tan \relax (x)^{3} + 33 \, a^{\frac {3}{2}} \tan \relax (x)}{48 \, {\left (\tan \relax (x)^{6} + 3 \, \tan \relax (x)^{4} + 3 \, \tan \relax (x)^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cos(x)^4)^(3/2),x, algorithm="maxima")

[Out]

5/16*a^(3/2)*x + 1/48*(15*a^(3/2)*tan(x)^5 + 40*a^(3/2)*tan(x)^3 + 33*a^(3/2)*tan(x))/(tan(x)^6 + 3*tan(x)^4 +
3*tan(x)^2 + 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a\,{\cos \relax (x)}^4\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cos(x)^4)^(3/2),x)

[Out]

int((a*cos(x)^4)^(3/2), 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)**(3/2),x)

[Out]

Timed out

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