3.1.67 \(\int \frac {1}{(a \sec ^4(x))^{5/2}} \, dx\) [67]

Optimal. Leaf size=132 \[ \frac {63 x \sec ^2(x)}{256 a^2 \sqrt {a \sec ^4(x)}}+\frac {21 \cos (x) \sin (x)}{128 a^2 \sqrt {a \sec ^4(x)}}+\frac {21 \cos ^3(x) \sin (x)}{160 a^2 \sqrt {a \sec ^4(x)}}+\frac {9 \cos ^5(x) \sin (x)}{80 a^2 \sqrt {a \sec ^4(x)}}+\frac {\cos ^7(x) \sin (x)}{10 a^2 \sqrt {a \sec ^4(x)}}+\frac {63 \tan (x)}{256 a^2 \sqrt {a \sec ^4(x)}} \]

[Out]

63/256*x*sec(x)^2/a^2/(a*sec(x)^4)^(1/2)+21/128*cos(x)*sin(x)/a^2/(a*sec(x)^4)^(1/2)+21/160*cos(x)^3*sin(x)/a^
2/(a*sec(x)^4)^(1/2)+9/80*cos(x)^5*sin(x)/a^2/(a*sec(x)^4)^(1/2)+1/10*cos(x)^7*sin(x)/a^2/(a*sec(x)^4)^(1/2)+6
3/256*tan(x)/a^2/(a*sec(x)^4)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {4208, 2715, 8} \begin {gather*} \frac {63 x \sec ^2(x)}{256 a^2 \sqrt {a \sec ^4(x)}}+\frac {63 \tan (x)}{256 a^2 \sqrt {a \sec ^4(x)}}+\frac {\sin (x) \cos ^7(x)}{10 a^2 \sqrt {a \sec ^4(x)}}+\frac {9 \sin (x) \cos ^5(x)}{80 a^2 \sqrt {a \sec ^4(x)}}+\frac {21 \sin (x) \cos ^3(x)}{160 a^2 \sqrt {a \sec ^4(x)}}+\frac {21 \sin (x) \cos (x)}{128 a^2 \sqrt {a \sec ^4(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(63*x*Sec[x]^2)/(256*a^2*Sqrt[a*Sec[x]^4]) + (21*Cos[x]*Sin[x])/(128*a^2*Sqrt[a*Sec[x]^4]) + (21*Cos[x]^3*Sin[
x])/(160*a^2*Sqrt[a*Sec[x]^4]) + (9*Cos[x]^5*Sin[x])/(80*a^2*Sqrt[a*Sec[x]^4]) + (Cos[x]^7*Sin[x])/(10*a^2*Sqr
t[a*Sec[x]^4]) + (63*Tan[x])/(256*a^2*Sqrt[a*Sec[x]^4])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 4208

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 \frac {1}{\left (a \sec ^4(x)\right )^{5/2}} \, dx &=\frac {\sec ^2(x) \int \cos ^{10}(x) \, dx}{a^2 \sqrt {a \sec ^4(x)}}\\ &=\frac {\cos ^7(x) \sin (x)}{10 a^2 \sqrt {a \sec ^4(x)}}+\frac {\left (9 \sec ^2(x)\right ) \int \cos ^8(x) \, dx}{10 a^2 \sqrt {a \sec ^4(x)}}\\ &=\frac {9 \cos ^5(x) \sin (x)}{80 a^2 \sqrt {a \sec ^4(x)}}+\frac {\cos ^7(x) \sin (x)}{10 a^2 \sqrt {a \sec ^4(x)}}+\frac {\left (63 \sec ^2(x)\right ) \int \cos ^6(x) \, dx}{80 a^2 \sqrt {a \sec ^4(x)}}\\ &=\frac {21 \cos ^3(x) \sin (x)}{160 a^2 \sqrt {a \sec ^4(x)}}+\frac {9 \cos ^5(x) \sin (x)}{80 a^2 \sqrt {a \sec ^4(x)}}+\frac {\cos ^7(x) \sin (x)}{10 a^2 \sqrt {a \sec ^4(x)}}+\frac {\left (21 \sec ^2(x)\right ) \int \cos ^4(x) \, dx}{32 a^2 \sqrt {a \sec ^4(x)}}\\ &=\frac {21 \cos (x) \sin (x)}{128 a^2 \sqrt {a \sec ^4(x)}}+\frac {21 \cos ^3(x) \sin (x)}{160 a^2 \sqrt {a \sec ^4(x)}}+\frac {9 \cos ^5(x) \sin (x)}{80 a^2 \sqrt {a \sec ^4(x)}}+\frac {\cos ^7(x) \sin (x)}{10 a^2 \sqrt {a \sec ^4(x)}}+\frac {\left (63 \sec ^2(x)\right ) \int \cos ^2(x) \, dx}{128 a^2 \sqrt {a \sec ^4(x)}}\\ &=\frac {21 \cos (x) \sin (x)}{128 a^2 \sqrt {a \sec ^4(x)}}+\frac {21 \cos ^3(x) \sin (x)}{160 a^2 \sqrt {a \sec ^4(x)}}+\frac {9 \cos ^5(x) \sin (x)}{80 a^2 \sqrt {a \sec ^4(x)}}+\frac {\cos ^7(x) \sin (x)}{10 a^2 \sqrt {a \sec ^4(x)}}+\frac {63 \tan (x)}{256 a^2 \sqrt {a \sec ^4(x)}}+\frac {\left (63 \sec ^2(x)\right ) \int 1 \, dx}{256 a^2 \sqrt {a \sec ^4(x)}}\\ &=\frac {63 x \sec ^2(x)}{256 a^2 \sqrt {a \sec ^4(x)}}+\frac {21 \cos (x) \sin (x)}{128 a^2 \sqrt {a \sec ^4(x)}}+\frac {21 \cos ^3(x) \sin (x)}{160 a^2 \sqrt {a \sec ^4(x)}}+\frac {9 \cos ^5(x) \sin (x)}{80 a^2 \sqrt {a \sec ^4(x)}}+\frac {\cos ^7(x) \sin (x)}{10 a^2 \sqrt {a \sec ^4(x)}}+\frac {63 \tan (x)}{256 a^2 \sqrt {a \sec ^4(x)}}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 55, normalized size = 0.42 \begin {gather*} \frac {\cos ^2(x) \sqrt {a \sec ^4(x)} (2520 x+2100 \sin (2 x)+600 \sin (4 x)+150 \sin (6 x)+25 \sin (8 x)+2 \sin (10 x))}{10240 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[x]^2*Sqrt[a*Sec[x]^4]*(2520*x + 2100*Sin[2*x] + 600*Sin[4*x] + 150*Sin[6*x] + 25*Sin[8*x] + 2*Sin[10*x]))
/(10240*a^3)

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Maple [A]
time = 0.45, size = 57, normalized size = 0.43

method	result	size
default	\(\frac {128 \left (\cos ^{9}\left (x \right )\right ) \sin \left (x \right )+144 \left (\cos ^{7}\left (x \right )\right ) \sin \left (x \right )+168 \left (\cos ^{5}\left (x \right )\right ) \sin \left (x \right )+210 \left (\cos ^{3}\left (x \right )\right ) \sin \left (x \right )+315 \cos \left (x \right ) \sin \left (x \right )+315 x}{1280 \cos \left (x \right )^{10} \left (\frac {a}{\cos \left (x \right )^{4}}\right )^{\frac {5}{2}}}\)	\(57\)
risch	\(\frac {63 \,{\mathrm e}^{2 i x} x}{256 a^{2} \left ({\mathrm e}^{2 i x}+1\right )^{2} \sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{4}}}}-\frac {i {\mathrm e}^{12 i x}}{10240 a^{2} \left ({\mathrm e}^{2 i x}+1\right )^{2} \sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{4}}}}-\frac {5 i {\mathrm e}^{10 i x}}{4096 a^{2} \left ({\mathrm e}^{2 i x}+1\right )^{2} \sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{4}}}}-\frac {105 i {\mathrm e}^{4 i x}}{1024 a^{2} \left ({\mathrm e}^{2 i x}+1\right )^{2} \sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{4}}}}+\frac {105 i}{1024 a^{2} \left ({\mathrm e}^{2 i x}+1\right )^{2} \sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{4}}}}+\frac {15 i {\mathrm e}^{-2 i x}}{512 a^{2} \left ({\mathrm e}^{2 i x}+1\right )^{2} \sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{4}}}}+\frac {15 i {\mathrm e}^{-4 i x}}{2048 a^{2} \left ({\mathrm e}^{2 i x}+1\right )^{2} \sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{4}}}}-\frac {37 i \cos \left (8 x \right )}{5120 a^{2} \left ({\mathrm e}^{2 i x}+1\right )^{2} \sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{4}}}}+\frac {19 \sin \left (8 x \right )}{2560 a^{2} \left ({\mathrm e}^{2 i x}+1\right )^{2} \sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{4}}}}-\frac {115 i \cos \left (6 x \right )}{4096 a^{2} \left ({\mathrm e}^{2 i x}+1\right )^{2} \sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{4}}}}+\frac {125 \sin \left (6 x \right )}{4096 a^{2} \left ({\mathrm e}^{2 i x}+1\right )^{2} \sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{4}}}}\)	\(409\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*sec(x)^4)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/1280*(128*cos(x)^9*sin(x)+144*cos(x)^7*sin(x)+168*cos(x)^5*sin(x)+210*cos(x)^3*sin(x)+315*cos(x)*sin(x)+315*
x)/cos(x)^10/(a/cos(x)^4)^(5/2)

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Maxima [A]
time = 0.51, size = 88, normalized size = 0.67 \begin {gather*} \frac {315 \, \tan \left (x\right )^{9} + 1470 \, \tan \left (x\right )^{7} + 2688 \, \tan \left (x\right )^{5} + 2370 \, \tan \left (x\right )^{3} + 965 \, \tan \left (x\right )}{1280 \, {\left (a^{\frac {5}{2}} \tan \left (x\right )^{10} + 5 \, a^{\frac {5}{2}} \tan \left (x\right )^{8} + 10 \, a^{\frac {5}{2}} \tan \left (x\right )^{6} + 10 \, a^{\frac {5}{2}} \tan \left (x\right )^{4} + 5 \, a^{\frac {5}{2}} \tan \left (x\right )^{2} + a^{\frac {5}{2}}\right )}} + \frac {63 \, x}{256 \, a^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sec(x)^4)^(5/2),x, algorithm="maxima")

[Out]

1/1280*(315*tan(x)^9 + 1470*tan(x)^7 + 2688*tan(x)^5 + 2370*tan(x)^3 + 965*tan(x))/(a^(5/2)*tan(x)^10 + 5*a^(5
/2)*tan(x)^8 + 10*a^(5/2)*tan(x)^6 + 10*a^(5/2)*tan(x)^4 + 5*a^(5/2)*tan(x)^2 + a^(5/2)) + 63/256*x/a^(5/2)

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Fricas [A]
time = 3.23, size = 55, normalized size = 0.42 \begin {gather*} \frac {{\left (315 \, x \cos \left (x\right )^{2} + {\left (128 \, \cos \left (x\right )^{11} + 144 \, \cos \left (x\right )^{9} + 168 \, \cos \left (x\right )^{7} + 210 \, \cos \left (x\right )^{5} + 315 \, \cos \left (x\right )^{3}\right )} \sin \left (x\right )\right )} \sqrt {\frac {a}{\cos \left (x\right )^{4}}}}{1280 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sec(x)^4)^(5/2),x, algorithm="fricas")

[Out]

1/1280*(315*x*cos(x)^2 + (128*cos(x)^11 + 144*cos(x)^9 + 168*cos(x)^7 + 210*cos(x)^5 + 315*cos(x)^3)*sin(x))*s
qrt(a/cos(x)^4)/a^3

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a \sec ^{4}{\left (x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sec(x)**4)**(5/2),x)

[Out]

Integral((a*sec(x)**4)**(-5/2), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sec(x)^4)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (\frac {a}{{\cos \left (x\right )}^4}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a/cos(x)^4)^(5/2),x)

[Out]

int(1/(a/cos(x)^4)^(5/2), x)

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