\(\int \frac {1}{(a \cos ^2(x))^{5/2}} \, dx\) [44]

Optimal result

Integrand size = 10, antiderivative size = 61 \[ \int \frac {1}{\left (a \cos ^2(x)\right )^{5/2}} \, dx=\frac {3 \text {arctanh}(\sin (x)) \cos (x)}{8 a^2 \sqrt {a \cos ^2(x)}}+\frac {\tan (x)}{4 a \left (a \cos ^2(x)\right )^{3/2}}+\frac {3 \tan (x)}{8 a^2 \sqrt {a \cos ^2(x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3283, 3286, 3855} \[ \int \frac {1}{\left (a \cos ^2(x)\right )^{5/2}} \, dx=\frac {3 \cos (x) \text {arctanh}(\sin (x))}{8 a^2 \sqrt {a \cos ^2(x)}}+\frac {3 \tan (x)}{8 a^2 \sqrt {a \cos ^2(x)}}+\frac {\tan (x)}{4 a \left (a \cos ^2(x)\right )^{3/2}} \]

[In]

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

[Out]

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

Rule 3283

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

Rule 3286

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 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\tan (x)}{4 a \left (a \cos ^2(x)\right )^{3/2}}+\frac {3 \int \frac {1}{\left (a \cos ^2(x)\right )^{3/2}} \, dx}{4 a} \\ & = \frac {\tan (x)}{4 a \left (a \cos ^2(x)\right )^{3/2}}+\frac {3 \tan (x)}{8 a^2 \sqrt {a \cos ^2(x)}}+\frac {3 \int \frac {1}{\sqrt {a \cos ^2(x)}} \, dx}{8 a^2} \\ & = \frac {\tan (x)}{4 a \left (a \cos ^2(x)\right )^{3/2}}+\frac {3 \tan (x)}{8 a^2 \sqrt {a \cos ^2(x)}}+\frac {(3 \cos (x)) \int \sec (x) \, dx}{8 a^2 \sqrt {a \cos ^2(x)}} \\ & = \frac {3 \text {arctanh}(\sin (x)) \cos (x)}{8 a^2 \sqrt {a \cos ^2(x)}}+\frac {\tan (x)}{4 a \left (a \cos ^2(x)\right )^{3/2}}+\frac {3 \tan (x)}{8 a^2 \sqrt {a \cos ^2(x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.59 \[ \int \frac {1}{\left (a \cos ^2(x)\right )^{5/2}} \, dx=\frac {3 \text {arctanh}(\sin (x)) \cos (x)+\left (3+2 \sec ^2(x)\right ) \tan (x)}{8 a^2 \sqrt {a \cos ^2(x)}} \]

[In]

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

[Out]

(3*ArcTanh[Sin[x]]*Cos[x] + (3 + 2*Sec[x]^2)*Tan[x])/(8*a^2*Sqrt[a*Cos[x]^2])

Maple [A] (verified)

Time = 1.08 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.46

method	result	size
default	\(\frac {\sqrt {a \left (\sin ^{2}\left (x \right )\right )}\, \left (3 \ln \left (\frac {2 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (x \right )\right )}+2 a}{\cos \left (x \right )}\right ) a \left (\cos ^{4}\left (x \right )\right )+3 \sqrt {a \left (\sin ^{2}\left (x \right )\right )}\, \left (\cos ^{2}\left (x \right )\right ) \sqrt {a}+2 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (x \right )\right )}\right )}{8 a^{\frac {7}{2}} \cos \left (x \right )^{3} \sin \left (x \right ) \sqrt {a \left (\cos ^{2}\left (x \right )\right )}}\)	\(89\)
risch	\(-\frac {i \left (3 \,{\mathrm e}^{6 i x}+11 \,{\mathrm e}^{4 i x}-11 \,{\mathrm e}^{2 i x}-3\right )}{4 a^{2} \left ({\mathrm e}^{2 i x}+1\right )^{3} \sqrt {a \left ({\mathrm e}^{2 i x}+1\right )^{2} {\mathrm e}^{-2 i x}}}+\frac {3 \ln \left ({\mathrm e}^{i x}+i\right ) \cos \left (x \right )}{4 a^{2} \sqrt {a \left ({\mathrm e}^{2 i x}+1\right )^{2} {\mathrm e}^{-2 i x}}}-\frac {3 \ln \left ({\mathrm e}^{i x}-i\right ) \cos \left (x \right )}{4 a^{2} \sqrt {a \left ({\mathrm e}^{2 i x}+1\right )^{2} {\mathrm e}^{-2 i x}}}\)	\(126\)

[In]

int(1/(a*cos(x)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/8/a^(7/2)/cos(x)^3*(a*sin(x)^2)^(1/2)*(3*ln(2*(a^(1/2)*(a*sin(x)^2)^(1/2)+a)/cos(x))*a*cos(x)^4+3*(a*sin(x)^
2)^(1/2)*cos(x)^2*a^(1/2)+2*a^(1/2)*(a*sin(x)^2)^(1/2))/sin(x)/(a*cos(x)^2)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.80 \[ \int \frac {1}{\left (a \cos ^2(x)\right )^{5/2}} \, dx=-\frac {{\left (3 \, \cos \left (x\right )^{4} \log \left (-\frac {\sin \left (x\right ) - 1}{\sin \left (x\right ) + 1}\right ) - 2 \, {\left (3 \, \cos \left (x\right )^{2} + 2\right )} \sin \left (x\right )\right )} \sqrt {a \cos \left (x\right )^{2}}}{16 \, a^{3} \cos \left (x\right )^{5}} \]

[In]

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

[Out]

-1/16*(3*cos(x)^4*log(-(sin(x) - 1)/(sin(x) + 1)) - 2*(3*cos(x)^2 + 2)*sin(x))*sqrt(a*cos(x)^2)/(a^3*cos(x)^5)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\left (a \cos ^2(x)\right )^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 933 vs. \(2 (49) = 98\).

Time = 0.62 (sec) , antiderivative size = 933, normalized size of antiderivative = 15.30 \[ \int \frac {1}{\left (a \cos ^2(x)\right )^{5/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/16*(4*(3*sin(7*x) + 11*sin(5*x) - 11*sin(3*x) - 3*sin(x))*cos(8*x) - 24*(2*sin(6*x) + 3*sin(4*x) + 2*sin(2*x
))*cos(7*x) + 16*(11*sin(5*x) - 11*sin(3*x) - 3*sin(x))*cos(6*x) - 88*(3*sin(4*x) + 2*sin(2*x))*cos(5*x) - 24*
(11*sin(3*x) + 3*sin(x))*cos(4*x) + 3*(2*(4*cos(6*x) + 6*cos(4*x) + 4*cos(2*x) + 1)*cos(8*x) + cos(8*x)^2 + 8*
(6*cos(4*x) + 4*cos(2*x) + 1)*cos(6*x) + 16*cos(6*x)^2 + 12*(4*cos(2*x) + 1)*cos(4*x) + 36*cos(4*x)^2 + 16*cos
(2*x)^2 + 4*(2*sin(6*x) + 3*sin(4*x) + 2*sin(2*x))*sin(8*x) + sin(8*x)^2 + 16*(3*sin(4*x) + 2*sin(2*x))*sin(6*
x) + 16*sin(6*x)^2 + 36*sin(4*x)^2 + 48*sin(4*x)*sin(2*x) + 16*sin(2*x)^2 + 8*cos(2*x) + 1)*log(cos(x)^2 + sin
(x)^2 + 2*sin(x) + 1) - 3*(2*(4*cos(6*x) + 6*cos(4*x) + 4*cos(2*x) + 1)*cos(8*x) + cos(8*x)^2 + 8*(6*cos(4*x)
+ 4*cos(2*x) + 1)*cos(6*x) + 16*cos(6*x)^2 + 12*(4*cos(2*x) + 1)*cos(4*x) + 36*cos(4*x)^2 + 16*cos(2*x)^2 + 4*
(2*sin(6*x) + 3*sin(4*x) + 2*sin(2*x))*sin(8*x) + sin(8*x)^2 + 16*(3*sin(4*x) + 2*sin(2*x))*sin(6*x) + 16*sin(
6*x)^2 + 36*sin(4*x)^2 + 48*sin(4*x)*sin(2*x) + 16*sin(2*x)^2 + 8*cos(2*x) + 1)*log(cos(x)^2 + sin(x)^2 - 2*si
n(x) + 1) - 4*(3*cos(7*x) + 11*cos(5*x) - 11*cos(3*x) - 3*cos(x))*sin(8*x) + 12*(4*cos(6*x) + 6*cos(4*x) + 4*c
os(2*x) + 1)*sin(7*x) - 16*(11*cos(5*x) - 11*cos(3*x) - 3*cos(x))*sin(6*x) + 44*(6*cos(4*x) + 4*cos(2*x) + 1)*
sin(5*x) + 24*(11*cos(3*x) + 3*cos(x))*sin(4*x) - 44*(4*cos(2*x) + 1)*sin(3*x) + 176*cos(3*x)*sin(2*x) + 48*co
s(x)*sin(2*x) - 48*cos(2*x)*sin(x) - 12*sin(x))/((a^2*cos(8*x)^2 + 16*a^2*cos(6*x)^2 + 36*a^2*cos(4*x)^2 + 16*
a^2*cos(2*x)^2 + a^2*sin(8*x)^2 + 16*a^2*sin(6*x)^2 + 36*a^2*sin(4*x)^2 + 48*a^2*sin(4*x)*sin(2*x) + 16*a^2*si
n(2*x)^2 + 8*a^2*cos(2*x) + a^2 + 2*(4*a^2*cos(6*x) + 6*a^2*cos(4*x) + 4*a^2*cos(2*x) + a^2)*cos(8*x) + 8*(6*a
^2*cos(4*x) + 4*a^2*cos(2*x) + a^2)*cos(6*x) + 12*(4*a^2*cos(2*x) + a^2)*cos(4*x) + 4*(2*a^2*sin(6*x) + 3*a^2*
sin(4*x) + 2*a^2*sin(2*x))*sin(8*x) + 16*(3*a^2*sin(4*x) + 2*a^2*sin(2*x))*sin(6*x))*sqrt(a))

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.48 \[ \int \frac {1}{\left (a \cos ^2(x)\right )^{5/2}} \, dx=-\frac {3 \, \sin \left (x\right )^{3} - 5 \, \sin \left (x\right )}{8 \, {\left (\sin \left (x\right )^{2} - 1\right )}^{2} a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (x\right )\right )} \]

[In]

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

[Out]

-1/8*(3*sin(x)^3 - 5*sin(x))/((sin(x)^2 - 1)^2*a^(5/2)*sgn(cos(x)))

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a \cos ^2(x)\right )^{5/2}} \, dx=\int \frac {1}{{\left (a\,{\cos \left (x\right )}^2\right )}^{5/2}} \,d x \]

[In]

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

[Out]

int(1/(a*cos(x)^2)^(5/2), x)