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

Optimal. Leaf size=61 \[ \frac{3 \tan (x)}{8 a^2 \sqrt{a \cos ^2(x)}}+\frac{3 \cos (x) \tanh ^{-1}(\sin (x))}{8 a^2 \sqrt{a \cos ^2(x)}}+\frac{\tan (x)}{4 a \left (a \cos ^2(x)\right )^{3/2}} \]

[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])

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Rubi [A]  time = 0.0437965, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {3204, 3207, 3770} \[ \frac{3 \tan (x)}{8 a^2 \sqrt{a \cos ^2(x)}}+\frac{3 \cos (x) \tanh ^{-1}(\sin (x))}{8 a^2 \sqrt{a \cos ^2(x)}}+\frac{\tan (x)}{4 a \left (a \cos ^2(x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[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 3204

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 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 3770

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

Rubi steps

\begin{align*} \int \frac{1}{\left (a \cos ^2(x)\right )^{5/2}} \, dx &=\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 \tanh ^{-1}(\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]  time = 0.156865, size = 72, normalized size = 1.18 \[ \frac{\cos ^5(x) \left (\frac{1}{2} (11 \sin (x)+3 \sin (3 x)) \sec ^4(x)-6 \log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )+6 \log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )\right )}{16 \left (a \cos ^2(x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[x]^5*(-6*Log[Cos[x/2] - Sin[x/2]] + 6*Log[Cos[x/2] + Sin[x/2]] + (Sec[x]^4*(11*Sin[x] + 3*Sin[3*x]))/2))/
(16*(a*Cos[x]^2)^(5/2))

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Maple [A]  time = 1.112, size = 89, normalized size = 1.5 \begin{align*}{\frac{1}{8\, \left ( \cos \left ( x \right ) \right ) ^{3}\sin \left ( x \right ) }\sqrt{a \left ( \sin \left ( x \right ) \right ) ^{2}} \left ( 3\,\ln \left ( 2\,{\frac{\sqrt{a}\sqrt{a \left ( \sin \left ( x \right ) \right ) ^{2}}+a}{\cos \left ( x \right ) }} \right ) \left ( \cos \left ( x \right ) \right ) ^{4}a+3\,\sqrt{a \left ( \sin \left ( x \right ) \right ) ^{2}} \left ( \cos \left ( x \right ) \right ) ^{2}\sqrt{a}+2\,\sqrt{a}\sqrt{a \left ( \sin \left ( x \right ) \right ) ^{2}} \right ){a}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{a \left ( \cos \left ( x \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 3.66056, size = 1260, normalized size = 20.66 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Fricas [A]  time = 1.14584, size = 151, normalized size = 2.48 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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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)**2)**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.56893, size = 170, normalized size = 2.79 \begin{align*} \frac{\frac{3 \, \log \left ({\left | \frac{1}{\tan \left (\frac{1}{2} \, x\right )} + \tan \left (\frac{1}{2} \, x\right ) + 2 \right |}\right )}{\mathrm{sgn}\left (-\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right )} - \frac{3 \, \log \left ({\left | \frac{1}{\tan \left (\frac{1}{2} \, x\right )} + \tan \left (\frac{1}{2} \, x\right ) - 2 \right |}\right )}{\mathrm{sgn}\left (-\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right )} + \frac{4 \,{\left (5 \,{\left (\frac{1}{\tan \left (\frac{1}{2} \, x\right )} + \tan \left (\frac{1}{2} \, x\right )\right )}^{3} - \frac{12}{\tan \left (\frac{1}{2} \, x\right )} - 12 \, \tan \left (\frac{1}{2} \, x\right )\right )}}{{\left ({\left (\frac{1}{\tan \left (\frac{1}{2} \, x\right )} + \tan \left (\frac{1}{2} \, x\right )\right )}^{2} - 4\right )}^{2} \mathrm{sgn}\left (-\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right )}}{16 \, a^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(3*log(abs(1/tan(1/2*x) + tan(1/2*x) + 2))/sgn(-tan(1/2*x)^2 + 1) - 3*log(abs(1/tan(1/2*x) + tan(1/2*x) -
2))/sgn(-tan(1/2*x)^2 + 1) + 4*(5*(1/tan(1/2*x) + tan(1/2*x))^3 - 12/tan(1/2*x) - 12*tan(1/2*x))/(((1/tan(1/2
*x) + tan(1/2*x))^2 - 4)^2*sgn(-tan(1/2*x)^2 + 1)))/a^(5/2)