3.39 \(\int \sec ^2(x)^{7/2} \, dx\)

Optimal. Leaf size=50 \[ \frac{1}{6} \tan (x) \sec ^2(x)^{5/2}+\frac{5}{24} \tan (x) \sec ^2(x)^{3/2}+\frac{5}{16} \tan (x) \sqrt{\sec ^2(x)}+\frac{5}{16} \sinh ^{-1}(\tan (x)) \]

[Out]

(5*ArcSinh[Tan[x]])/16 + (5*Sqrt[Sec[x]^2]*Tan[x])/16 + (5*(Sec[x]^2)^(3/2)*Tan[x])/24 + ((Sec[x]^2)^(5/2)*Tan
[x])/6

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Rubi [A]  time = 0.016905, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {4122, 195, 215} \[ \frac{1}{6} \tan (x) \sec ^2(x)^{5/2}+\frac{5}{24} \tan (x) \sec ^2(x)^{3/2}+\frac{5}{16} \tan (x) \sqrt{\sec ^2(x)}+\frac{5}{16} \sinh ^{-1}(\tan (x)) \]

Antiderivative was successfully verified.

[In]

Int[(Sec[x]^2)^(7/2),x]

[Out]

(5*ArcSinh[Tan[x]])/16 + (5*Sqrt[Sec[x]^2]*Tan[x])/16 + (5*(Sec[x]^2)^(3/2)*Tan[x])/24 + ((Sec[x]^2)^(5/2)*Tan
[x])/6

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)^{7/2} \, dx &=\operatorname{Subst}\left (\int \left (1+x^2\right )^{5/2} \, dx,x,\tan (x)\right )\\ &=\frac{1}{6} \sec ^2(x)^{5/2} \tan (x)+\frac{5}{6} \operatorname{Subst}\left (\int \left (1+x^2\right )^{3/2} \, dx,x,\tan (x)\right )\\ &=\frac{5}{24} \sec ^2(x)^{3/2} \tan (x)+\frac{1}{6} \sec ^2(x)^{5/2} \tan (x)+\frac{5}{8} \operatorname{Subst}\left (\int \sqrt{1+x^2} \, dx,x,\tan (x)\right )\\ &=\frac{5}{16} \sqrt{\sec ^2(x)} \tan (x)+\frac{5}{24} \sec ^2(x)^{3/2} \tan (x)+\frac{1}{6} \sec ^2(x)^{5/2} \tan (x)+\frac{5}{16} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x^2}} \, dx,x,\tan (x)\right )\\ &=\frac{5}{16} \sinh ^{-1}(\tan (x))+\frac{5}{16} \sqrt{\sec ^2(x)} \tan (x)+\frac{5}{24} \sec ^2(x)^{3/2} \tan (x)+\frac{1}{6} \sec ^2(x)^{5/2} \tan (x)\\ \end{align*}

Mathematica [A]  time = 0.290101, size = 74, normalized size = 1.48 \[ \frac{1}{96} \cos (x) \sqrt{\sec ^2(x)} \left (\frac{1}{8} (198 \sin (x)+85 \sin (3 x)+15 \sin (5 x)) \sec ^6(x)-30 \log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )+30 \log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[x]^2)^(7/2),x]

[Out]

(Cos[x]*Sqrt[Sec[x]^2]*(-30*Log[Cos[x/2] - Sin[x/2]] + 30*Log[Cos[x/2] + Sin[x/2]] + (Sec[x]^6*(198*Sin[x] + 8
5*Sin[3*x] + 15*Sin[5*x]))/8))/96

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Maple [A]  time = 0.159, size = 72, normalized size = 1.4 \begin{align*} -{\frac{\cos \left ( x \right ) }{48} \left ( 15\,\ln \left ( -{\frac{-1+\cos \left ( x \right ) +\sin \left ( x \right ) }{\sin \left ( x \right ) }} \right ) \left ( \cos \left ( x \right ) \right ) ^{6}-15\,\ln \left ( -{\frac{-1+\cos \left ( x \right ) -\sin \left ( x \right ) }{\sin \left ( x \right ) }} \right ) \left ( \cos \left ( x \right ) \right ) ^{6}-15\, \left ( \cos \left ( x \right ) \right ) ^{4}\sin \left ( x \right ) -10\, \left ( \cos \left ( x \right ) \right ) ^{2}\sin \left ( x \right ) -8\,\sin \left ( x \right ) \right ) \left ( \left ( \cos \left ( x \right ) \right ) ^{-2} \right ) ^{{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sec(x)^2)^(7/2),x)

[Out]

-1/48*(15*ln(-(-1+cos(x)+sin(x))/sin(x))*cos(x)^6-15*ln(-(-1+cos(x)-sin(x))/sin(x))*cos(x)^6-15*cos(x)^4*sin(x
)-10*cos(x)^2*sin(x)-8*sin(x))*cos(x)*(1/cos(x)^2)^(7/2)

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Maxima [A]  time = 1.74061, size = 57, normalized size = 1.14 \begin{align*} \frac{1}{6} \,{\left (\tan \left (x\right )^{2} + 1\right )}^{\frac{5}{2}} \tan \left (x\right ) + \frac{5}{24} \,{\left (\tan \left (x\right )^{2} + 1\right )}^{\frac{3}{2}} \tan \left (x\right ) + \frac{5}{16} \, \sqrt{\tan \left (x\right )^{2} + 1} \tan \left (x\right ) + \frac{5}{16} \, \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)^(7/2),x, algorithm="maxima")

[Out]

1/6*(tan(x)^2 + 1)^(5/2)*tan(x) + 5/24*(tan(x)^2 + 1)^(3/2)*tan(x) + 5/16*sqrt(tan(x)^2 + 1)*tan(x) + 5/16*arc
sinh(tan(x))

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Fricas [A]  time = 1.36736, size = 162, normalized size = 3.24 \begin{align*} -\frac{15 \, \cos \left (x\right )^{6} \log \left (\sin \left (x\right ) + 1\right ) - 15 \, \cos \left (x\right )^{6} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \,{\left (15 \, \cos \left (x\right )^{4} + 10 \, \cos \left (x\right )^{2} + 8\right )} \sin \left (x\right )}{96 \, \cos \left (x\right )^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sec(x)^2)^(7/2),x, algorithm="fricas")

[Out]

-1/96*(15*cos(x)^6*log(sin(x) + 1) - 15*cos(x)^6*log(-sin(x) + 1) + 2*(15*cos(x)^4 + 10*cos(x)^2 + 8)*sin(x))/
cos(x)^6

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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)**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 1.28619, size = 80, normalized size = 1.6 \begin{align*} \frac{5 \, \log \left (\sin \left (x\right ) + 1\right )}{32 \, \mathrm{sgn}\left (\cos \left (x\right )\right )} - \frac{5 \, \log \left (-\sin \left (x\right ) + 1\right )}{32 \, \mathrm{sgn}\left (\cos \left (x\right )\right )} - \frac{15 \, \sin \left (x\right )^{5} - 40 \, \sin \left (x\right )^{3} + 33 \, \sin \left (x\right )}{48 \,{\left (\sin \left (x\right )^{2} - 1\right )}^{3} \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)^(7/2),x, algorithm="giac")

[Out]

5/32*log(sin(x) + 1)/sgn(cos(x)) - 5/32*log(-sin(x) + 1)/sgn(cos(x)) - 1/48*(15*sin(x)^5 - 40*sin(x)^3 + 33*si
n(x))/((sin(x)^2 - 1)^3*sgn(cos(x)))