3.40 \(\int \sec ^2(x)^{5/2} \, dx\)

Optimal. Leaf size=36 \[ \frac{1}{4} \tan (x) \sec ^2(x)^{3/2}+\frac{3}{8} \tan (x) \sqrt{\sec ^2(x)}+\frac{3}{8} \sinh ^{-1}(\tan (x)) \]

[Out]

(3*ArcSinh[Tan[x]])/8 + (3*Sqrt[Sec[x]^2]*Tan[x])/8 + ((Sec[x]^2)^(3/2)*Tan[x])/4

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Rubi [A]  time = 0.0124014, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {4122, 195, 215} \[ \frac{1}{4} \tan (x) \sec ^2(x)^{3/2}+\frac{3}{8} \tan (x) \sqrt{\sec ^2(x)}+\frac{3}{8} \sinh ^{-1}(\tan (x)) \]

Antiderivative was successfully verified.

[In]

Int[(Sec[x]^2)^(5/2),x]

[Out]

(3*ArcSinh[Tan[x]])/8 + (3*Sqrt[Sec[x]^2]*Tan[x])/8 + ((Sec[x]^2)^(3/2)*Tan[x])/4

Rule 4122

Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(b*ff)
/f, Subst[Int[(b + b*ff^2*x^2)^(p - 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{b, e, f, p}, x] &&  !IntegerQ[p
]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

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

Mathematica [A]  time = 0.1402, size = 68, normalized size = 1.89 \[ \frac{1}{16} \cos (x) \sqrt{\sec ^2(x)} \left (\frac{1}{2} (11 \sin (x)+3 \sin (3 x)) \sec ^4(x)-6 \log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )+6 \log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[x]^2)^(5/2),x]

[Out]

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

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Maple [B]  time = 0.091, size = 64, normalized size = 1.8 \begin{align*}{\frac{\cos \left ( x \right ) }{8} \left ( 3\, \left ( \cos \left ( x \right ) \right ) ^{4}\ln \left ( -{\frac{-1+\cos \left ( x \right ) -\sin \left ( x \right ) }{\sin \left ( x \right ) }} \right ) -3\, \left ( \cos \left ( x \right ) \right ) ^{4}\ln \left ( -{\frac{-1+\cos \left ( x \right ) +\sin \left ( x \right ) }{\sin \left ( x \right ) }} \right ) +3\, \left ( \cos \left ( x \right ) \right ) ^{2}\sin \left ( x \right ) +2\,\sin \left ( x \right ) \right ) \left ( \left ( \cos \left ( x \right ) \right ) ^{-2} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sec(x)^2)^(5/2),x)

[Out]

1/8*(3*cos(x)^4*ln(-(-1+cos(x)-sin(x))/sin(x))-3*cos(x)^4*ln(-(-1+cos(x)+sin(x))/sin(x))+3*cos(x)^2*sin(x)+2*s
in(x))*cos(x)*(1/cos(x)^2)^(5/2)

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Maxima [A]  time = 1.64798, size = 41, normalized size = 1.14 \begin{align*} \frac{1}{4} \,{\left (\tan \left (x\right )^{2} + 1\right )}^{\frac{3}{2}} \tan \left (x\right ) + \frac{3}{8} \, \sqrt{\tan \left (x\right )^{2} + 1} \tan \left (x\right ) + \frac{3}{8} \, \operatorname{arsinh}\left (\tan \left (x\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sec(x)^2)^(5/2),x, algorithm="maxima")

[Out]

1/4*(tan(x)^2 + 1)^(3/2)*tan(x) + 3/8*sqrt(tan(x)^2 + 1)*tan(x) + 3/8*arcsinh(tan(x))

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Fricas [A]  time = 1.40946, size = 139, normalized size = 3.86 \begin{align*} -\frac{3 \, \cos \left (x\right )^{4} \log \left (\sin \left (x\right ) + 1\right ) - 3 \, \cos \left (x\right )^{4} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \,{\left (3 \, \cos \left (x\right )^{2} + 2\right )} \sin \left (x\right )}{16 \, \cos \left (x\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sec(x)^2)^(5/2),x, algorithm="fricas")

[Out]

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

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

[Out]

Timed out

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Giac [B]  time = 1.27814, size = 72, normalized size = 2. \begin{align*} \frac{3 \, \log \left (\sin \left (x\right ) + 1\right )}{16 \, \mathrm{sgn}\left (\cos \left (x\right )\right )} - \frac{3 \, \log \left (-\sin \left (x\right ) + 1\right )}{16 \, \mathrm{sgn}\left (\cos \left (x\right )\right )} - \frac{3 \, \sin \left (x\right )^{3} - 5 \, \sin \left (x\right )}{8 \,{\left (\sin \left (x\right )^{2} - 1\right )}^{2} \mathrm{sgn}\left (\cos \left (x\right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sec(x)^2)^(5/2),x, algorithm="giac")

[Out]

3/16*log(sin(x) + 1)/sgn(cos(x)) - 3/16*log(-sin(x) + 1)/sgn(cos(x)) - 1/8*(3*sin(x)^3 - 5*sin(x))/((sin(x)^2
- 1)^2*sgn(cos(x)))