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/(a*cos(x)^4)^(1/2)+4/3*sin(x)^2*tan(x)/a^2/(a*cos(x)^4)^(1/2)+6/5*sin(x)^2*tan(x)^3/a^2/(a*c
os(x)^4)^(1/2)+4/7*sin(x)^2*tan(x)^5/a^2/(a*cos(x)^4)^(1/2)+1/9*sin(x)^2*tan(x)^7/a^2/(a*cos(x)^4)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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*}

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Mathematica [A]  time = 0.05, size = 47, normalized size = 0.40 \[ \frac {(130 \cos (2 x)+46 \cos (4 x)+10 \cos (6 x)+\cos (8 x)+128) \tan (x) \sec ^6(x)}{315 a^2 \sqrt {a \cos ^4(x)}} \]

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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fricas [A]  time = 0.51, size = 45, normalized size = 0.38 \[ \frac {{\left (128 \, \cos \relax (x)^{8} + 64 \, \cos \relax (x)^{6} + 48 \, \cos \relax (x)^{4} + 40 \, \cos \relax (x)^{2} + 35\right )} \sqrt {a \cos \relax (x)^{4}} \sin \relax (x)}{315 \, a^{3} \cos \relax (x)^{11}} \]

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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giac [A]  time = 0.53, size = 34, normalized size = 0.29 \[ \frac {35 \, \tan \relax (x)^{9} + 180 \, \tan \relax (x)^{7} + 378 \, \tan \relax (x)^{5} + 420 \, \tan \relax (x)^{3} + 315 \, \tan \relax (x)}{315 \, a^{\frac {5}{2}}} \]

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)

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maple [A]  time = 0.15, size = 41, normalized size = 0.35 \[ \frac {\sin \relax (x ) \left (128 \left (\cos ^{8}\relax (x )\right )+64 \left (\cos ^{6}\relax (x )\right )+48 \left (\cos ^{4}\relax (x )\right )+40 \left (\cos ^{2}\relax (x )\right )+35\right ) \cos \relax (x )}{315 \left (a \left (\cos ^{4}\relax (x )\right )\right )^{\frac {5}{2}}} \]

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 = 0.52, size = 34, normalized size = 0.29 \[ \frac {35 \, \tan \relax (x)^{9} + 180 \, \tan \relax (x)^{7} + 378 \, \tan \relax (x)^{5} + 420 \, \tan \relax (x)^{3} + 315 \, \tan \relax (x)}{315 \, a^{\frac {5}{2}}} \]

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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mupad [B]  time = 3.74, size = 306, normalized size = 2.62 \[ \frac {{\mathrm {e}}^{x\,4{}\mathrm {i}}\,\sqrt {a\,{\left (\frac {{\mathrm {e}}^{-x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{x\,1{}\mathrm {i}}}{2}\right )}^4}\,2048{}\mathrm {i}}{5\,a^3\,{\left ({\mathrm {e}}^{x\,2{}\mathrm {i}}+1\right )}^5\,\left ({\mathrm {e}}^{x\,2{}\mathrm {i}}+2\,{\mathrm {e}}^{x\,4{}\mathrm {i}}+{\mathrm {e}}^{x\,6{}\mathrm {i}}\right )}-\frac {{\mathrm {e}}^{x\,4{}\mathrm {i}}\,\sqrt {a\,{\left (\frac {{\mathrm {e}}^{-x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{x\,1{}\mathrm {i}}}{2}\right )}^4}\,4096{}\mathrm {i}}{3\,a^3\,{\left ({\mathrm {e}}^{x\,2{}\mathrm {i}}+1\right )}^6\,\left ({\mathrm {e}}^{x\,2{}\mathrm {i}}+2\,{\mathrm {e}}^{x\,4{}\mathrm {i}}+{\mathrm {e}}^{x\,6{}\mathrm {i}}\right )}+\frac {{\mathrm {e}}^{x\,4{}\mathrm {i}}\,\sqrt {a\,{\left (\frac {{\mathrm {e}}^{-x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{x\,1{}\mathrm {i}}}{2}\right )}^4}\,12288{}\mathrm {i}}{7\,a^3\,{\left ({\mathrm {e}}^{x\,2{}\mathrm {i}}+1\right )}^7\,\left ({\mathrm {e}}^{x\,2{}\mathrm {i}}+2\,{\mathrm {e}}^{x\,4{}\mathrm {i}}+{\mathrm {e}}^{x\,6{}\mathrm {i}}\right )}-\frac {{\mathrm {e}}^{x\,4{}\mathrm {i}}\,\sqrt {a\,{\left (\frac {{\mathrm {e}}^{-x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{x\,1{}\mathrm {i}}}{2}\right )}^4}\,1024{}\mathrm {i}}{a^3\,{\left ({\mathrm {e}}^{x\,2{}\mathrm {i}}+1\right )}^8\,\left ({\mathrm {e}}^{x\,2{}\mathrm {i}}+2\,{\mathrm {e}}^{x\,4{}\mathrm {i}}+{\mathrm {e}}^{x\,6{}\mathrm {i}}\right )}+\frac {{\mathrm {e}}^{x\,4{}\mathrm {i}}\,\sqrt {a\,{\left (\frac {{\mathrm {e}}^{-x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{x\,1{}\mathrm {i}}}{2}\right )}^4}\,2048{}\mathrm {i}}{9\,a^3\,{\left ({\mathrm {e}}^{x\,2{}\mathrm {i}}+1\right )}^9\,\left ({\mathrm {e}}^{x\,2{}\mathrm {i}}+2\,{\mathrm {e}}^{x\,4{}\mathrm {i}}+{\mathrm {e}}^{x\,6{}\mathrm {i}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(x*4i)*(a*(exp(-x*1i)/2 + exp(x*1i)/2)^4)^(1/2)*2048i)/(5*a^3*(exp(x*2i) + 1)^5*(exp(x*2i) + 2*exp(x*4i) +
exp(x*6i))) - (exp(x*4i)*(a*(exp(-x*1i)/2 + exp(x*1i)/2)^4)^(1/2)*4096i)/(3*a^3*(exp(x*2i) + 1)^6*(exp(x*2i)
+ 2*exp(x*4i) + exp(x*6i))) + (exp(x*4i)*(a*(exp(-x*1i)/2 + exp(x*1i)/2)^4)^(1/2)*12288i)/(7*a^3*(exp(x*2i) +
1)^7*(exp(x*2i) + 2*exp(x*4i) + exp(x*6i))) - (exp(x*4i)*(a*(exp(-x*1i)/2 + exp(x*1i)/2)^4)^(1/2)*1024i)/(a^3*
(exp(x*2i) + 1)^8*(exp(x*2i) + 2*exp(x*4i) + exp(x*6i))) + (exp(x*4i)*(a*(exp(-x*1i)/2 + exp(x*1i)/2)^4)^(1/2)
*2048i)/(9*a^3*(exp(x*2i) + 1)^9*(exp(x*2i) + 2*exp(x*4i) + exp(x*6i)))

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

[Out]

Timed out

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