3.61 \(\int (a \sec ^4(x))^{7/2} \, dx\)

Optimal. Leaf size=163 \[ a^3 \sin (x) \cos (x) \sqrt {a \sec ^4(x)}+\frac {1}{13} a^3 \sin ^2(x) \tan ^{11}(x) \sqrt {a \sec ^4(x)}+\frac {6}{11} a^3 \sin ^2(x) \tan ^9(x) \sqrt {a \sec ^4(x)}+\frac {5}{3} a^3 \sin ^2(x) \tan ^7(x) \sqrt {a \sec ^4(x)}+\frac {20}{7} a^3 \sin ^2(x) \tan ^5(x) \sqrt {a \sec ^4(x)}+3 a^3 \sin ^2(x) \tan ^3(x) \sqrt {a \sec ^4(x)}+2 a^3 \sin ^2(x) \tan (x) \sqrt {a \sec ^4(x)} \]

[Out]

a^3*cos(x)*sin(x)*(a*sec(x)^4)^(1/2)+2*a^3*sin(x)^2*(a*sec(x)^4)^(1/2)*tan(x)+3*a^3*sin(x)^2*(a*sec(x)^4)^(1/2
)*tan(x)^3+20/7*a^3*sin(x)^2*(a*sec(x)^4)^(1/2)*tan(x)^5+5/3*a^3*sin(x)^2*(a*sec(x)^4)^(1/2)*tan(x)^7+6/11*a^3
*sin(x)^2*(a*sec(x)^4)^(1/2)*tan(x)^9+1/13*a^3*sin(x)^2*(a*sec(x)^4)^(1/2)*tan(x)^11

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Rubi [A]  time = 0.04, antiderivative size = 163, 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} \[ a^3 \sin (x) \cos (x) \sqrt {a \sec ^4(x)}+\frac {1}{13} a^3 \sin ^2(x) \tan ^{11}(x) \sqrt {a \sec ^4(x)}+\frac {6}{11} a^3 \sin ^2(x) \tan ^9(x) \sqrt {a \sec ^4(x)}+\frac {5}{3} a^3 \sin ^2(x) \tan ^7(x) \sqrt {a \sec ^4(x)}+\frac {20}{7} a^3 \sin ^2(x) \tan ^5(x) \sqrt {a \sec ^4(x)}+3 a^3 \sin ^2(x) \tan ^3(x) \sqrt {a \sec ^4(x)}+2 a^3 \sin ^2(x) \tan (x) \sqrt {a \sec ^4(x)} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sec[x]^4)^(7/2),x]

[Out]

a^3*Cos[x]*Sqrt[a*Sec[x]^4]*Sin[x] + 2*a^3*Sqrt[a*Sec[x]^4]*Sin[x]^2*Tan[x] + 3*a^3*Sqrt[a*Sec[x]^4]*Sin[x]^2*
Tan[x]^3 + (20*a^3*Sqrt[a*Sec[x]^4]*Sin[x]^2*Tan[x]^5)/7 + (5*a^3*Sqrt[a*Sec[x]^4]*Sin[x]^2*Tan[x]^7)/3 + (6*a
^3*Sqrt[a*Sec[x]^4]*Sin[x]^2*Tan[x]^9)/11 + (a^3*Sqrt[a*Sec[x]^4]*Sin[x]^2*Tan[x]^11)/13

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 )^{7/2} \, dx &=\left (a^3 \cos ^2(x) \sqrt {a \sec ^4(x)}\right ) \int \sec ^{14}(x) \, dx\\ &=-\left (\left (a^3 \cos ^2(x) \sqrt {a \sec ^4(x)}\right ) \operatorname {Subst}\left (\int \left (1+6 x^2+15 x^4+20 x^6+15 x^8+6 x^{10}+x^{12}\right ) \, dx,x,-\tan (x)\right )\right )\\ &=a^3 \cos (x) \sqrt {a \sec ^4(x)} \sin (x)+2 a^3 \sqrt {a \sec ^4(x)} \sin ^2(x) \tan (x)+3 a^3 \sqrt {a \sec ^4(x)} \sin ^2(x) \tan ^3(x)+\frac {20}{7} a^3 \sqrt {a \sec ^4(x)} \sin ^2(x) \tan ^5(x)+\frac {5}{3} a^3 \sqrt {a \sec ^4(x)} \sin ^2(x) \tan ^7(x)+\frac {6}{11} a^3 \sqrt {a \sec ^4(x)} \sin ^2(x) \tan ^9(x)+\frac {1}{13} a^3 \sqrt {a \sec ^4(x)} \sin ^2(x) \tan ^{11}(x)\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 54, normalized size = 0.33 \[ \frac {\sin (x) \cos (x) (2380 \cos (2 x)+1093 \cos (4 x)+378 \cos (6 x)+92 \cos (8 x)+14 \cos (10 x)+\cos (12 x)+2048) \left (a \sec ^4(x)\right )^{7/2}}{6006} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sec[x]^4)^(7/2),x]

[Out]

(Cos[x]*(2048 + 2380*Cos[2*x] + 1093*Cos[4*x] + 378*Cos[6*x] + 92*Cos[8*x] + 14*Cos[10*x] + Cos[12*x])*(a*Sec[
x]^4)^(7/2)*Sin[x])/6006

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fricas [A]  time = 0.68, size = 76, normalized size = 0.47 \[ \frac {{\left (1024 \, a^{3} \cos \relax (x)^{12} + 512 \, a^{3} \cos \relax (x)^{10} + 384 \, a^{3} \cos \relax (x)^{8} + 320 \, a^{3} \cos \relax (x)^{6} + 280 \, a^{3} \cos \relax (x)^{4} + 252 \, a^{3} \cos \relax (x)^{2} + 231 \, a^{3}\right )} \sqrt {\frac {a}{\cos \relax (x)^{4}}} \sin \relax (x)}{3003 \, \cos \relax (x)^{11}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sec(x)^4)^(7/2),x, algorithm="fricas")

[Out]

1/3003*(1024*a^3*cos(x)^12 + 512*a^3*cos(x)^10 + 384*a^3*cos(x)^8 + 320*a^3*cos(x)^6 + 280*a^3*cos(x)^4 + 252*
a^3*cos(x)^2 + 231*a^3)*sqrt(a/cos(x)^4)*sin(x)/cos(x)^11

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giac [A]  time = 0.41, size = 67, normalized size = 0.41 \[ \frac {1}{3003} \, {\left (231 \, a^{3} \tan \relax (x)^{13} + 1638 \, a^{3} \tan \relax (x)^{11} + 5005 \, a^{3} \tan \relax (x)^{9} + 8580 \, a^{3} \tan \relax (x)^{7} + 9009 \, a^{3} \tan \relax (x)^{5} + 6006 \, a^{3} \tan \relax (x)^{3} + 3003 \, a^{3} \tan \relax (x)\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sec(x)^4)^(7/2),x, algorithm="giac")

[Out]

1/3003*(231*a^3*tan(x)^13 + 1638*a^3*tan(x)^11 + 5005*a^3*tan(x)^9 + 8580*a^3*tan(x)^7 + 9009*a^3*tan(x)^5 + 6
006*a^3*tan(x)^3 + 3003*a^3*tan(x))*sqrt(a)

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maple [A]  time = 0.56, size = 53, normalized size = 0.33 \[ \frac {\left (1024 \left (\cos ^{12}\relax (x )\right )+512 \left (\cos ^{10}\relax (x )\right )+384 \left (\cos ^{8}\relax (x )\right )+320 \left (\cos ^{6}\relax (x )\right )+280 \left (\cos ^{4}\relax (x )\right )+252 \left (\cos ^{2}\relax (x )\right )+231\right ) \left (\frac {a}{\cos \relax (x )^{4}}\right )^{\frac {7}{2}} \sin \relax (x ) \cos \relax (x )}{3003} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sec(x)^4)^(7/2),x)

[Out]

1/3003*(1024*cos(x)^12+512*cos(x)^10+384*cos(x)^8+320*cos(x)^6+280*cos(x)^4+252*cos(x)^2+231)*(a/cos(x)^4)^(7/
2)*sin(x)*cos(x)

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maxima [A]  time = 1.14, size = 61, normalized size = 0.37 \[ \frac {1}{13} \, a^{\frac {7}{2}} \tan \relax (x)^{13} + \frac {6}{11} \, a^{\frac {7}{2}} \tan \relax (x)^{11} + \frac {5}{3} \, a^{\frac {7}{2}} \tan \relax (x)^{9} + \frac {20}{7} \, a^{\frac {7}{2}} \tan \relax (x)^{7} + 3 \, a^{\frac {7}{2}} \tan \relax (x)^{5} + 2 \, a^{\frac {7}{2}} \tan \relax (x)^{3} + a^{\frac {7}{2}} \tan \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sec(x)^4)^(7/2),x, algorithm="maxima")

[Out]

1/13*a^(7/2)*tan(x)^13 + 6/11*a^(7/2)*tan(x)^11 + 5/3*a^(7/2)*tan(x)^9 + 20/7*a^(7/2)*tan(x)^7 + 3*a^(7/2)*tan
(x)^5 + 2*a^(7/2)*tan(x)^3 + a^(7/2)*tan(x)

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mupad [B]  time = 4.68, size = 589, normalized size = 3.61 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a/cos(x)^4)^(7/2),x)

[Out]

(a^3*(a/(exp(-x*1i)/2 + exp(x*1i)/2)^4)^(1/2)*(4*exp(x*2i) + 6*exp(x*4i) + 4*exp(x*6i) + exp(x*8i) + 1)*2048i)
/(7*(exp(x*2i) + 1)^7*(exp(x*2i) + 2*exp(x*4i) + exp(x*6i))) - (a^3*(a/(exp(-x*1i)/2 + exp(x*1i)/2)^4)^(1/2)*(
4*exp(x*2i) + 6*exp(x*4i) + 4*exp(x*6i) + exp(x*8i) + 1)*1536i)/((exp(x*2i) + 1)^8*(exp(x*2i) + 2*exp(x*4i) +
exp(x*6i))) + (a^3*(a/(exp(-x*1i)/2 + exp(x*1i)/2)^4)^(1/2)*(4*exp(x*2i) + 6*exp(x*4i) + 4*exp(x*6i) + exp(x*8
i) + 1)*10240i)/(3*(exp(x*2i) + 1)^9*(exp(x*2i) + 2*exp(x*4i) + exp(x*6i))) - (a^3*(a/(exp(-x*1i)/2 + exp(x*1i
)/2)^4)^(1/2)*(4*exp(x*2i) + 6*exp(x*4i) + 4*exp(x*6i) + exp(x*8i) + 1)*4096i)/((exp(x*2i) + 1)^10*(exp(x*2i)
+ 2*exp(x*4i) + exp(x*6i))) + (a^3*(a/(exp(-x*1i)/2 + exp(x*1i)/2)^4)^(1/2)*(4*exp(x*2i) + 6*exp(x*4i) + 4*exp
(x*6i) + exp(x*8i) + 1)*30720i)/(11*(exp(x*2i) + 1)^11*(exp(x*2i) + 2*exp(x*4i) + exp(x*6i))) - (a^3*(a/(exp(-
x*1i)/2 + exp(x*1i)/2)^4)^(1/2)*(4*exp(x*2i) + 6*exp(x*4i) + 4*exp(x*6i) + exp(x*8i) + 1)*1024i)/((exp(x*2i) +
1)^12*(exp(x*2i) + 2*exp(x*4i) + exp(x*6i))) + (a^3*(a/(exp(-x*1i)/2 + exp(x*1i)/2)^4)^(1/2)*(4*exp(x*2i) + 6
*exp(x*4i) + 4*exp(x*6i) + exp(x*8i) + 1)*2048i)/(13*(exp(x*2i) + 1)^13*(exp(x*2i) + 2*exp(x*4i) + exp(x*6i)))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sec(x)**4)**(7/2),x)

[Out]

Timed out

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