3.56 \(\int \frac{1}{(a \cos ^4(x))^{5/2}} \, dx\)

Optimal. Leaf size=117 \[ \frac{\sin (x) \cos (x)}{a^2 \sqrt{a \cos ^4(x)}}+\frac{\sin ^2(x) \tan ^7(x)}{9 a^2 \sqrt{a \cos ^4(x)}}+\frac{4 \sin ^2(x) \tan ^5(x)}{7 a^2 \sqrt{a \cos ^4(x)}}+\frac{6 \sin ^2(x) \tan ^3(x)}{5 a^2 \sqrt{a \cos ^4(x)}}+\frac{4 \sin ^2(x) \tan (x)}{3 a^2 \sqrt{a \cos ^4(x)}} \]

[Out]

(Cos[x]*Sin[x])/(a^2*Sqrt[a*Cos[x]^4]) + (4*Sin[x]^2*Tan[x])/(3*a^2*Sqrt[a*Cos[x]^4]) + (6*Sin[x]^2*Tan[x]^3)/
(5*a^2*Sqrt[a*Cos[x]^4]) + (4*Sin[x]^2*Tan[x]^5)/(7*a^2*Sqrt[a*Cos[x]^4]) + (Sin[x]^2*Tan[x]^7)/(9*a^2*Sqrt[a*
Cos[x]^4])

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Rubi [A]  time = 0.0321276, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3207, 3767} \[ \frac{\sin (x) \cos (x)}{a^2 \sqrt{a \cos ^4(x)}}+\frac{\sin ^2(x) \tan ^7(x)}{9 a^2 \sqrt{a \cos ^4(x)}}+\frac{4 \sin ^2(x) \tan ^5(x)}{7 a^2 \sqrt{a \cos ^4(x)}}+\frac{6 \sin ^2(x) \tan ^3(x)}{5 a^2 \sqrt{a \cos ^4(x)}}+\frac{4 \sin ^2(x) \tan (x)}{3 a^2 \sqrt{a \cos ^4(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[x]*Sin[x])/(a^2*Sqrt[a*Cos[x]^4]) + (4*Sin[x]^2*Tan[x])/(3*a^2*Sqrt[a*Cos[x]^4]) + (6*Sin[x]^2*Tan[x]^3)/
(5*a^2*Sqrt[a*Cos[x]^4]) + (4*Sin[x]^2*Tan[x]^5)/(7*a^2*Sqrt[a*Cos[x]^4]) + (Sin[x]^2*Tan[x]^7)/(9*a^2*Sqrt[a*
Cos[x]^4])

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 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin{align*} \int \frac{1}{\left (a \cos ^4(x)\right )^{5/2}} \, dx &=\frac{\cos ^2(x) \int \sec ^{10}(x) \, dx}{a^2 \sqrt{a \cos ^4(x)}}\\ &=-\frac{\cos ^2(x) \operatorname{Subst}\left (\int \left (1+4 x^2+6 x^4+4 x^6+x^8\right ) \, dx,x,-\tan (x)\right )}{a^2 \sqrt{a \cos ^4(x)}}\\ &=\frac{\cos (x) \sin (x)}{a^2 \sqrt{a \cos ^4(x)}}+\frac{4 \sin ^2(x) \tan (x)}{3 a^2 \sqrt{a \cos ^4(x)}}+\frac{6 \sin ^2(x) \tan ^3(x)}{5 a^2 \sqrt{a \cos ^4(x)}}+\frac{4 \sin ^2(x) \tan ^5(x)}{7 a^2 \sqrt{a \cos ^4(x)}}+\frac{\sin ^2(x) \tan ^7(x)}{9 a^2 \sqrt{a \cos ^4(x)}}\\ \end{align*}

Mathematica [A]  time = 0.051605, size = 47, normalized size = 0.4 \[ \frac{(130 \cos (2 x)+46 \cos (4 x)+10 \cos (6 x)+\cos (8 x)+128) \tan (x) \sec ^6(x)}{315 a^2 \sqrt{a \cos ^4(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((128 + 130*Cos[2*x] + 46*Cos[4*x] + 10*Cos[6*x] + Cos[8*x])*Sec[x]^6*Tan[x])/(315*a^2*Sqrt[a*Cos[x]^4])

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Maple [A]  time = 0.165, size = 41, normalized size = 0.4 \begin{align*}{\frac{\sin \left ( x \right ) \left ( 128\, \left ( \cos \left ( x \right ) \right ) ^{8}+64\, \left ( \cos \left ( x \right ) \right ) ^{6}+48\, \left ( \cos \left ( x \right ) \right ) ^{4}+40\, \left ( \cos \left ( x \right ) \right ) ^{2}+35 \right ) \cos \left ( x \right ) }{315} \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(1/(a*cos(x)^4)^(5/2),x)

[Out]

1/315*sin(x)*(128*cos(x)^8+64*cos(x)^6+48*cos(x)^4+40*cos(x)^2+35)*cos(x)/(a*cos(x)^4)^(5/2)

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Maxima [A]  time = 1.88253, size = 46, normalized size = 0.39 \begin{align*} \frac{35 \, \tan \left (x\right )^{9} + 180 \, \tan \left (x\right )^{7} + 378 \, \tan \left (x\right )^{5} + 420 \, \tan \left (x\right )^{3} + 315 \, \tan \left (x\right )}{315 \, a^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(x)^4)^(5/2),x, algorithm="maxima")

[Out]

1/315*(35*tan(x)^9 + 180*tan(x)^7 + 378*tan(x)^5 + 420*tan(x)^3 + 315*tan(x))/a^(5/2)

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Fricas [A]  time = 1.08485, size = 147, normalized size = 1.26 \begin{align*} \frac{{\left (128 \, \cos \left (x\right )^{8} + 64 \, \cos \left (x\right )^{6} + 48 \, \cos \left (x\right )^{4} + 40 \, \cos \left (x\right )^{2} + 35\right )} \sqrt{a \cos \left (x\right )^{4}} \sin \left (x\right )}{315 \, a^{3} \cos \left (x\right )^{11}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(x)^4)^(5/2),x, algorithm="fricas")

[Out]

1/315*(128*cos(x)^8 + 64*cos(x)^6 + 48*cos(x)^4 + 40*cos(x)^2 + 35)*sqrt(a*cos(x)^4)*sin(x)/(a^3*cos(x)^11)

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

[Out]

Timed out

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Giac [A]  time = 1.30331, size = 46, normalized size = 0.39 \begin{align*} \frac{35 \, \tan \left (x\right )^{9} + 180 \, \tan \left (x\right )^{7} + 378 \, \tan \left (x\right )^{5} + 420 \, \tan \left (x\right )^{3} + 315 \, \tan \left (x\right )}{315 \, a^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(x)^4)^(5/2),x, algorithm="giac")

[Out]

1/315*(35*tan(x)^9 + 180*tan(x)^7 + 378*tan(x)^5 + 420*tan(x)^3 + 315*tan(x))/a^(5/2)