3.62 \(\int (a \sec ^4(x))^{5/2} \, dx\)

Optimal. Leaf size=117 \[ a^2 \sin (x) \cos (x) \sqrt{a \sec ^4(x)}+\frac{1}{9} a^2 \sin ^2(x) \tan ^7(x) \sqrt{a \sec ^4(x)}+\frac{4}{7} a^2 \sin ^2(x) \tan ^5(x) \sqrt{a \sec ^4(x)}+\frac{6}{5} a^2 \sin ^2(x) \tan ^3(x) \sqrt{a \sec ^4(x)}+\frac{4}{3} a^2 \sin ^2(x) \tan (x) \sqrt{a \sec ^4(x)} \]

[Out]

a^2*Cos[x]*Sqrt[a*Sec[x]^4]*Sin[x] + (4*a^2*Sqrt[a*Sec[x]^4]*Sin[x]^2*Tan[x])/3 + (6*a^2*Sqrt[a*Sec[x]^4]*Sin[
x]^2*Tan[x]^3)/5 + (4*a^2*Sqrt[a*Sec[x]^4]*Sin[x]^2*Tan[x]^5)/7 + (a^2*Sqrt[a*Sec[x]^4]*Sin[x]^2*Tan[x]^7)/9

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Rubi [A]  time = 0.0303353, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4123, 3767} \[ a^2 \sin (x) \cos (x) \sqrt{a \sec ^4(x)}+\frac{1}{9} a^2 \sin ^2(x) \tan ^7(x) \sqrt{a \sec ^4(x)}+\frac{4}{7} a^2 \sin ^2(x) \tan ^5(x) \sqrt{a \sec ^4(x)}+\frac{6}{5} a^2 \sin ^2(x) \tan ^3(x) \sqrt{a \sec ^4(x)}+\frac{4}{3} a^2 \sin ^2(x) \tan (x) \sqrt{a \sec ^4(x)} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sec[x]^4)^(5/2),x]

[Out]

a^2*Cos[x]*Sqrt[a*Sec[x]^4]*Sin[x] + (4*a^2*Sqrt[a*Sec[x]^4]*Sin[x]^2*Tan[x])/3 + (6*a^2*Sqrt[a*Sec[x]^4]*Sin[
x]^2*Tan[x]^3)/5 + (4*a^2*Sqrt[a*Sec[x]^4]*Sin[x]^2*Tan[x]^5)/7 + (a^2*Sqrt[a*Sec[x]^4]*Sin[x]^2*Tan[x]^7)/9

Rule 4123

Int[((b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Dist[(b^IntPart[p]*(b*(c*Sec[e + f*x])^n)^
FracPart[p])/(c*Sec[e + f*x])^(n*FracPart[p]), Int[(c*Sec[e + f*x])^(n*p), x], x] /; FreeQ[{b, c, e, f, n, p},
x] &&  !IntegerQ[p]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin{align*} \int \left (a \sec ^4(x)\right )^{5/2} \, dx &=\left (a^2 \cos ^2(x) \sqrt{a \sec ^4(x)}\right ) \int \sec ^{10}(x) \, dx\\ &=-\left (\left (a^2 \cos ^2(x) \sqrt{a \sec ^4(x)}\right ) \operatorname{Subst}\left (\int \left (1+4 x^2+6 x^4+4 x^6+x^8\right ) \, dx,x,-\tan (x)\right )\right )\\ &=a^2 \cos (x) \sqrt{a \sec ^4(x)} \sin (x)+\frac{4}{3} a^2 \sqrt{a \sec ^4(x)} \sin ^2(x) \tan (x)+\frac{6}{5} a^2 \sqrt{a \sec ^4(x)} \sin ^2(x) \tan ^3(x)+\frac{4}{7} a^2 \sqrt{a \sec ^4(x)} \sin ^2(x) \tan ^5(x)+\frac{1}{9} a^2 \sqrt{a \sec ^4(x)} \sin ^2(x) \tan ^7(x)\\ \end{align*}

Mathematica [A]  time = 0.0911082, size = 42, normalized size = 0.36 \[ \frac{1}{315} \sin (x) \cos (x) (130 \cos (2 x)+46 \cos (4 x)+10 \cos (6 x)+\cos (8 x)+128) \left (a \sec ^4(x)\right )^{5/2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sec[x]^4)^(5/2),x]

[Out]

(Cos[x]*(128 + 130*Cos[2*x] + 46*Cos[4*x] + 10*Cos[6*x] + Cos[8*x])*(a*Sec[x]^4)^(5/2)*Sin[x])/315

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Maple [A]  time = 0.113, size = 41, normalized size = 0.4 \begin{align*}{\frac{ \left ( 128\, \left ( \cos \left ( x \right ) \right ) ^{8}+64\, \left ( \cos \left ( x \right ) \right ) ^{6}+48\, \left ( \cos \left ( x \right ) \right ) ^{4}+40\, \left ( \cos \left ( x \right ) \right ) ^{2}+35 \right ) \cos \left ( x \right ) \sin \left ( x \right ) }{315} \left ({\frac{a}{ \left ( \cos \left ( x \right ) \right ) ^{4}}} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sec(x)^4)^(5/2),x)

[Out]

1/315*(128*cos(x)^8+64*cos(x)^6+48*cos(x)^4+40*cos(x)^2+35)*cos(x)*sin(x)*(a/cos(x)^4)^(5/2)

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Maxima [A]  time = 1.79159, size = 58, normalized size = 0.5 \begin{align*} \frac{1}{9} \, a^{\frac{5}{2}} \tan \left (x\right )^{9} + \frac{4}{7} \, a^{\frac{5}{2}} \tan \left (x\right )^{7} + \frac{6}{5} \, a^{\frac{5}{2}} \tan \left (x\right )^{5} + \frac{4}{3} \, a^{\frac{5}{2}} \tan \left (x\right )^{3} + a^{\frac{5}{2}} \tan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sec(x)^4)^(5/2),x, algorithm="maxima")

[Out]

1/9*a^(5/2)*tan(x)^9 + 4/7*a^(5/2)*tan(x)^7 + 6/5*a^(5/2)*tan(x)^5 + 4/3*a^(5/2)*tan(x)^3 + a^(5/2)*tan(x)

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Fricas [A]  time = 1.4071, size = 165, normalized size = 1.41 \begin{align*} \frac{{\left (128 \, a^{2} \cos \left (x\right )^{8} + 64 \, a^{2} \cos \left (x\right )^{6} + 48 \, a^{2} \cos \left (x\right )^{4} + 40 \, a^{2} \cos \left (x\right )^{2} + 35 \, a^{2}\right )} \sqrt{\frac{a}{\cos \left (x\right )^{4}}} \sin \left (x\right )}{315 \, \cos \left (x\right )^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sec(x)^4)^(5/2),x, algorithm="fricas")

[Out]

1/315*(128*a^2*cos(x)^8 + 64*a^2*cos(x)^6 + 48*a^2*cos(x)^4 + 40*a^2*cos(x)^2 + 35*a^2)*sqrt(a/cos(x)^4)*sin(x
)/cos(x)^7

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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((a*sec(x)**4)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.24393, size = 66, normalized size = 0.56 \begin{align*} \frac{1}{315} \,{\left (35 \, a^{2} \tan \left (x\right )^{9} + 180 \, a^{2} \tan \left (x\right )^{7} + 378 \, a^{2} \tan \left (x\right )^{5} + 420 \, a^{2} \tan \left (x\right )^{3} + 315 \, a^{2} \tan \left (x\right )\right )} \sqrt{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sec(x)^4)^(5/2),x, algorithm="giac")

[Out]

1/315*(35*a^2*tan(x)^9 + 180*a^2*tan(x)^7 + 378*a^2*tan(x)^5 + 420*a^2*tan(x)^3 + 315*a^2*tan(x))*sqrt(a)