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

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Rubi [A]  time = 0.03, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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]

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]

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*}

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Mathematica [A]  time = 0.10, 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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fricas [A]  time = 0.63, size = 58, normalized size = 0.50 \[ \frac {{\left (128 \, a^{2} \cos \relax (x)^{8} + 64 \, a^{2} \cos \relax (x)^{6} + 48 \, a^{2} \cos \relax (x)^{4} + 40 \, a^{2} \cos \relax (x)^{2} + 35 \, a^{2}\right )} \sqrt {\frac {a}{\cos \relax (x)^{4}}} \sin \relax (x)}{315 \, \cos \relax (x)^{7}} \]

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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giac [A]  time = 0.90, size = 49, normalized size = 0.42 \[ \frac {1}{315} \, {\left (35 \, a^{2} \tan \relax (x)^{9} + 180 \, a^{2} \tan \relax (x)^{7} + 378 \, a^{2} \tan \relax (x)^{5} + 420 \, a^{2} \tan \relax (x)^{3} + 315 \, a^{2} \tan \relax (x)\right )} \sqrt {a} \]

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)

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maple [A]  time = 0.36, size = 41, normalized size = 0.35 \[ \frac {\left (128 \left (\cos ^{8}\relax (x )\right )+64 \left (\cos ^{6}\relax (x )\right )+48 \left (\cos ^{4}\relax (x )\right )+40 \left (\cos ^{2}\relax (x )\right )+35\right ) \cos \relax (x ) \sin \relax (x ) \left (\frac {a}{\cos \relax (x )^{4}}\right )^{\frac {5}{2}}}{315} \]

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 = 0.75, size = 43, normalized size = 0.37 \[ \frac {1}{9} \, a^{\frac {5}{2}} \tan \relax (x)^{9} + \frac {4}{7} \, a^{\frac {5}{2}} \tan \relax (x)^{7} + \frac {6}{5} \, a^{\frac {5}{2}} \tan \relax (x)^{5} + \frac {4}{3} \, a^{\frac {5}{2}} \tan \relax (x)^{3} + a^{\frac {5}{2}} \tan \relax (x) \]

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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mupad [B]  time = 2.35, size = 119, normalized size = 1.02 \[ \frac {128\,a^{5/2}\,\left ({\mathrm {e}}^{x\,46{}\mathrm {i}}\,1{}\mathrm {i}+{\mathrm {e}}^{x\,48{}\mathrm {i}}\,9{}\mathrm {i}+{\mathrm {e}}^{x\,50{}\mathrm {i}}\,36{}\mathrm {i}+{\mathrm {e}}^{x\,52{}\mathrm {i}}\,84{}\mathrm {i}+{\mathrm {e}}^{x\,54{}\mathrm {i}}\,126{}\mathrm {i}\right )}{315\,\left (\frac {{\mathrm {e}}^{-x\,2{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{x\,2{}\mathrm {i}}}{2}+1\right )\,\left ({\mathrm {e}}^{x\,48{}\mathrm {i}}+7\,{\mathrm {e}}^{x\,50{}\mathrm {i}}+21\,{\mathrm {e}}^{x\,52{}\mathrm {i}}+35\,{\mathrm {e}}^{x\,54{}\mathrm {i}}+35\,{\mathrm {e}}^{x\,56{}\mathrm {i}}+21\,{\mathrm {e}}^{x\,58{}\mathrm {i}}+7\,{\mathrm {e}}^{x\,60{}\mathrm {i}}+{\mathrm {e}}^{x\,62{}\mathrm {i}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a/cos(x)^4)^(5/2),x)

[Out]

(128*a^(5/2)*(exp(x*46i)*1i + exp(x*48i)*9i + exp(x*50i)*36i + exp(x*52i)*84i + exp(x*54i)*126i))/(315*(exp(-x
*2i)/2 + exp(x*2i)/2 + 1)*(exp(x*48i) + 7*exp(x*50i) + 21*exp(x*52i) + 35*exp(x*54i) + 35*exp(x*56i) + 21*exp(
x*58i) + 7*exp(x*60i) + exp(x*62i)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \sec ^{4}{\relax (x )}\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*sec(x)**4)**(5/2), x)

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