3.50 \(\int \frac {1}{(a \cos ^3(x))^{5/2}} \, dx\)

Optimal. Leaf size=117 \[ \frac {154 \sin (x) \cos (x)}{195 a^2 \sqrt {a \cos ^3(x)}}+\frac {154 \tan (x)}{585 a^2 \sqrt {a \cos ^3(x)}}-\frac {154 \cos ^{\frac {3}{2}}(x) E\left (\left .\frac {x}{2}\right |2\right )}{195 a^2 \sqrt {a \cos ^3(x)}}+\frac {2 \tan (x) \sec ^4(x)}{13 a^2 \sqrt {a \cos ^3(x)}}+\frac {22 \tan (x) \sec ^2(x)}{117 a^2 \sqrt {a \cos ^3(x)}} \]

[Out]

-154/195*cos(x)^(3/2)*(cos(1/2*x)^2)^(1/2)/cos(1/2*x)*EllipticE(sin(1/2*x),2^(1/2))/a^2/(a*cos(x)^3)^(1/2)+154
/195*cos(x)*sin(x)/a^2/(a*cos(x)^3)^(1/2)+154/585*tan(x)/a^2/(a*cos(x)^3)^(1/2)+22/117*sec(x)^2*tan(x)/a^2/(a*
cos(x)^3)^(1/2)+2/13*sec(x)^4*tan(x)/a^2/(a*cos(x)^3)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3207, 2636, 2639} \[ \frac {154 \sin (x) \cos (x)}{195 a^2 \sqrt {a \cos ^3(x)}}+\frac {154 \tan (x)}{585 a^2 \sqrt {a \cos ^3(x)}}-\frac {154 \cos ^{\frac {3}{2}}(x) E\left (\left .\frac {x}{2}\right |2\right )}{195 a^2 \sqrt {a \cos ^3(x)}}+\frac {2 \tan (x) \sec ^4(x)}{13 a^2 \sqrt {a \cos ^3(x)}}+\frac {22 \tan (x) \sec ^2(x)}{117 a^2 \sqrt {a \cos ^3(x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cos[x]^3)^(-5/2),x]

[Out]

(-154*Cos[x]^(3/2)*EllipticE[x/2, 2])/(195*a^2*Sqrt[a*Cos[x]^3]) + (154*Cos[x]*Sin[x])/(195*a^2*Sqrt[a*Cos[x]^
3]) + (154*Tan[x])/(585*a^2*Sqrt[a*Cos[x]^3]) + (22*Sec[x]^2*Tan[x])/(117*a^2*Sqrt[a*Cos[x]^3]) + (2*Sec[x]^4*
Tan[x])/(13*a^2*Sqrt[a*Cos[x]^3])

Rule 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

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 \frac {1}{\left (a \cos ^3(x)\right )^{5/2}} \, dx &=\frac {\cos ^{\frac {3}{2}}(x) \int \frac {1}{\cos ^{\frac {15}{2}}(x)} \, dx}{a^2 \sqrt {a \cos ^3(x)}}\\ &=\frac {2 \sec ^4(x) \tan (x)}{13 a^2 \sqrt {a \cos ^3(x)}}+\frac {\left (11 \cos ^{\frac {3}{2}}(x)\right ) \int \frac {1}{\cos ^{\frac {11}{2}}(x)} \, dx}{13 a^2 \sqrt {a \cos ^3(x)}}\\ &=\frac {22 \sec ^2(x) \tan (x)}{117 a^2 \sqrt {a \cos ^3(x)}}+\frac {2 \sec ^4(x) \tan (x)}{13 a^2 \sqrt {a \cos ^3(x)}}+\frac {\left (77 \cos ^{\frac {3}{2}}(x)\right ) \int \frac {1}{\cos ^{\frac {7}{2}}(x)} \, dx}{117 a^2 \sqrt {a \cos ^3(x)}}\\ &=\frac {154 \tan (x)}{585 a^2 \sqrt {a \cos ^3(x)}}+\frac {22 \sec ^2(x) \tan (x)}{117 a^2 \sqrt {a \cos ^3(x)}}+\frac {2 \sec ^4(x) \tan (x)}{13 a^2 \sqrt {a \cos ^3(x)}}+\frac {\left (77 \cos ^{\frac {3}{2}}(x)\right ) \int \frac {1}{\cos ^{\frac {3}{2}}(x)} \, dx}{195 a^2 \sqrt {a \cos ^3(x)}}\\ &=\frac {154 \cos (x) \sin (x)}{195 a^2 \sqrt {a \cos ^3(x)}}+\frac {154 \tan (x)}{585 a^2 \sqrt {a \cos ^3(x)}}+\frac {22 \sec ^2(x) \tan (x)}{117 a^2 \sqrt {a \cos ^3(x)}}+\frac {2 \sec ^4(x) \tan (x)}{13 a^2 \sqrt {a \cos ^3(x)}}-\frac {\left (77 \cos ^{\frac {3}{2}}(x)\right ) \int \sqrt {\cos (x)} \, dx}{195 a^2 \sqrt {a \cos ^3(x)}}\\ &=-\frac {154 \cos ^{\frac {3}{2}}(x) E\left (\left .\frac {x}{2}\right |2\right )}{195 a^2 \sqrt {a \cos ^3(x)}}+\frac {154 \cos (x) \sin (x)}{195 a^2 \sqrt {a \cos ^3(x)}}+\frac {154 \tan (x)}{585 a^2 \sqrt {a \cos ^3(x)}}+\frac {22 \sec ^2(x) \tan (x)}{117 a^2 \sqrt {a \cos ^3(x)}}+\frac {2 \sec ^4(x) \tan (x)}{13 a^2 \sqrt {a \cos ^3(x)}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 57, normalized size = 0.49 \[ \frac {-462 \cos ^{\frac {3}{2}}(x) E\left (\left .\frac {x}{2}\right |2\right )+462 \sin (x) \cos (x)+2 \tan (x) \left (45 \sec ^4(x)+55 \sec ^2(x)+77\right )}{585 a^2 \sqrt {a \cos ^3(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Cos[x]^3)^(-5/2),x]

[Out]

(-462*Cos[x]^(3/2)*EllipticE[x/2, 2] + 462*Cos[x]*Sin[x] + 2*(77 + 55*Sec[x]^2 + 45*Sec[x]^4)*Tan[x])/(585*a^2
*Sqrt[a*Cos[x]^3])

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fricas [F]  time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a \cos \relax (x)^{3}}}{a^{3} \cos \relax (x)^{9}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(x)^3)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(a*cos(x)^3)/(a^3*cos(x)^9), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \cos \relax (x)^{3}\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(x)^3)^(5/2),x, algorithm="giac")

[Out]

integrate((a*cos(x)^3)^(-5/2), x)

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maple [C]  time = 0.38, size = 223, normalized size = 1.91 \[ -\frac {2 \left (\cos \relax (x )+1\right )^{2} \left (-1+\cos \relax (x )\right )^{2} \left (231 i \left (\cos ^{7}\relax (x )\right ) \sin \relax (x ) \sqrt {\frac {1}{\cos \relax (x )+1}}\, \sqrt {\frac {\cos \relax (x )}{\cos \relax (x )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \relax (x )\right )}{\sin \relax (x )}, i\right )-231 i \left (\cos ^{7}\relax (x )\right ) \sin \relax (x ) \sqrt {\frac {1}{\cos \relax (x )+1}}\, \sqrt {\frac {\cos \relax (x )}{\cos \relax (x )+1}}\, \EllipticE \left (\frac {i \left (-1+\cos \relax (x )\right )}{\sin \relax (x )}, i\right )+231 i \left (\cos ^{6}\relax (x )\right ) \sin \relax (x ) \sqrt {\frac {1}{\cos \relax (x )+1}}\, \sqrt {\frac {\cos \relax (x )}{\cos \relax (x )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \relax (x )\right )}{\sin \relax (x )}, i\right )-231 i \left (\cos ^{6}\relax (x )\right ) \sin \relax (x ) \sqrt {\frac {1}{\cos \relax (x )+1}}\, \sqrt {\frac {\cos \relax (x )}{\cos \relax (x )+1}}\, \EllipticE \left (\frac {i \left (-1+\cos \relax (x )\right )}{\sin \relax (x )}, i\right )+231 \left (\cos ^{7}\relax (x )\right )-154 \left (\cos ^{6}\relax (x )\right )-22 \left (\cos ^{4}\relax (x )\right )-10 \left (\cos ^{2}\relax (x )\right )-45\right ) \cos \relax (x )}{585 \sin \relax (x )^{5} \left (a \left (\cos ^{3}\relax (x )\right )\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cos(x)^3)^(5/2),x)

[Out]

-2/585*(cos(x)+1)^2*(-1+cos(x))^2*(231*I*cos(x)^7*sin(x)*(1/(cos(x)+1))^(1/2)*(cos(x)/(cos(x)+1))^(1/2)*Ellipt
icF(I*(-1+cos(x))/sin(x),I)-231*I*cos(x)^7*sin(x)*(1/(cos(x)+1))^(1/2)*(cos(x)/(cos(x)+1))^(1/2)*EllipticE(I*(
-1+cos(x))/sin(x),I)+231*I*cos(x)^6*sin(x)*(1/(cos(x)+1))^(1/2)*(cos(x)/(cos(x)+1))^(1/2)*EllipticF(I*(-1+cos(
x))/sin(x),I)-231*I*cos(x)^6*sin(x)*(1/(cos(x)+1))^(1/2)*(cos(x)/(cos(x)+1))^(1/2)*EllipticE(I*(-1+cos(x))/sin
(x),I)+231*cos(x)^7-154*cos(x)^6-22*cos(x)^4-10*cos(x)^2-45)*cos(x)/sin(x)^5/(a*cos(x)^3)^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \cos \relax (x)^{3}\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(x)^3)^(5/2),x, algorithm="maxima")

[Out]

integrate((a*cos(x)^3)^(-5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a\,{\cos \relax (x)}^3\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cos(x)^3)^(5/2),x)

[Out]

int(1/(a*cos(x)^3)^(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(1/(a*cos(x)**3)**(5/2),x)

[Out]

Timed out

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