3.82 \(\int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^5} \, dx\)

Optimal. Leaf size=159 \[ \frac {4 \tan (c+d x)}{45 d \left (a^5 \sec (c+d x)+a^5\right )}-\frac {32 \tan (c+d x)}{315 a d \left (a^2 \sec (c+d x)+a^2\right )^2}+\frac {4 \tan (c+d x)}{105 a^2 d (a \sec (c+d x)+a)^3}+\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}+\frac {4 \tan (c+d x) \sec ^3(c+d x)}{63 a d (a \sec (c+d x)+a)^4} \]

[Out]

1/9*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+c))^5+4/63*sec(d*x+c)^3*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^4+4/105*tan
(d*x+c)/a^2/d/(a+a*sec(d*x+c))^3-32/315*tan(d*x+c)/a/d/(a^2+a^2*sec(d*x+c))^2+4/45*tan(d*x+c)/d/(a^5+a^5*sec(d
*x+c))

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Rubi [A]  time = 0.22, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3810, 3799, 4000, 3794} \[ \frac {4 \tan (c+d x)}{45 d \left (a^5 \sec (c+d x)+a^5\right )}-\frac {32 \tan (c+d x)}{315 a d \left (a^2 \sec (c+d x)+a^2\right )^2}+\frac {4 \tan (c+d x)}{105 a^2 d (a \sec (c+d x)+a)^3}+\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}+\frac {4 \tan (c+d x) \sec ^3(c+d x)}{63 a d (a \sec (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/(a + a*Sec[c + d*x])^5,x]

[Out]

(Sec[c + d*x]^4*Tan[c + d*x])/(9*d*(a + a*Sec[c + d*x])^5) + (4*Sec[c + d*x]^3*Tan[c + d*x])/(63*a*d*(a + a*Se
c[c + d*x])^4) + (4*Tan[c + d*x])/(105*a^2*d*(a + a*Sec[c + d*x])^3) - (32*Tan[c + d*x])/(315*a*d*(a^2 + a^2*S
ec[c + d*x])^2) + (4*Tan[c + d*x])/(45*d*(a^5 + a^5*Sec[c + d*x]))

Rule 3794

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 3799

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(b*Cot[e + f*x]*(
a + b*Csc[e + f*x])^m)/(a*f*(2*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1)*(a*m - b*(2*m + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)
]

Rule 3810

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(b*d*C
ot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1))/(a*f*(2*m + 1)), x] + Dist[(d*(m + 1))/(b*(2*m +
1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && E
qQ[a^2 - b^2, 0] && EqQ[m + n, 0] && LtQ[m, -2^(-1)] && IntegerQ[2*m]

Rule 4000

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(a*f*(2*m + 1)), x] + Dist[(a*B*m + A*b*
(m + 1))/(a*b*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, A, B, e, f}, x
] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] && LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^5} \, dx &=\frac {\sec ^4(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}+\frac {4 \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^4} \, dx}{9 a}\\ &=\frac {\sec ^4(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}+\frac {4 \sec ^3(c+d x) \tan (c+d x)}{63 a d (a+a \sec (c+d x))^4}+\frac {4 \int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^3} \, dx}{21 a^2}\\ &=\frac {\sec ^4(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}+\frac {4 \sec ^3(c+d x) \tan (c+d x)}{63 a d (a+a \sec (c+d x))^4}+\frac {4 \tan (c+d x)}{105 a^2 d (a+a \sec (c+d x))^3}+\frac {4 \int \frac {\sec (c+d x) (-3 a+5 a \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{105 a^4}\\ &=\frac {\sec ^4(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}+\frac {4 \sec ^3(c+d x) \tan (c+d x)}{63 a d (a+a \sec (c+d x))^4}+\frac {4 \tan (c+d x)}{105 a^2 d (a+a \sec (c+d x))^3}-\frac {32 \tan (c+d x)}{315 a^3 d (a+a \sec (c+d x))^2}+\frac {4 \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{45 a^4}\\ &=\frac {\sec ^4(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}+\frac {4 \sec ^3(c+d x) \tan (c+d x)}{63 a d (a+a \sec (c+d x))^4}+\frac {4 \tan (c+d x)}{105 a^2 d (a+a \sec (c+d x))^3}-\frac {32 \tan (c+d x)}{315 a^3 d (a+a \sec (c+d x))^2}+\frac {4 \tan (c+d x)}{45 d \left (a^5+a^5 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 97, normalized size = 0.61 \[ \frac {\left (126 \sin \left (\frac {1}{2} (c+d x)\right )+84 \sin \left (\frac {3}{2} (c+d x)\right )+36 \sin \left (\frac {5}{2} (c+d x)\right )+9 \sin \left (\frac {7}{2} (c+d x)\right )+\sin \left (\frac {9}{2} (c+d x)\right )\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x)}{315 a^5 d (\sec (c+d x)+1)^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/(a + a*Sec[c + d*x])^5,x]

[Out]

(Cos[(c + d*x)/2]*Sec[c + d*x]^5*(126*Sin[(c + d*x)/2] + 84*Sin[(3*(c + d*x))/2] + 36*Sin[(5*(c + d*x))/2] + 9
*Sin[(7*(c + d*x))/2] + Sin[(9*(c + d*x))/2]))/(315*a^5*d*(1 + Sec[c + d*x])^5)

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fricas [A]  time = 1.70, size = 123, normalized size = 0.77 \[ \frac {{\left (8 \, \cos \left (d x + c\right )^{4} + 40 \, \cos \left (d x + c\right )^{3} + 84 \, \cos \left (d x + c\right )^{2} + 100 \, \cos \left (d x + c\right ) + 83\right )} \sin \left (d x + c\right )}{315 \, {\left (a^{5} d \cos \left (d x + c\right )^{5} + 5 \, a^{5} d \cos \left (d x + c\right )^{4} + 10 \, a^{5} d \cos \left (d x + c\right )^{3} + 10 \, a^{5} d \cos \left (d x + c\right )^{2} + 5 \, a^{5} d \cos \left (d x + c\right ) + a^{5} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+a*sec(d*x+c))^5,x, algorithm="fricas")

[Out]

1/315*(8*cos(d*x + c)^4 + 40*cos(d*x + c)^3 + 84*cos(d*x + c)^2 + 100*cos(d*x + c) + 83)*sin(d*x + c)/(a^5*d*c
os(d*x + c)^5 + 5*a^5*d*cos(d*x + c)^4 + 10*a^5*d*cos(d*x + c)^3 + 10*a^5*d*cos(d*x + c)^2 + 5*a^5*d*cos(d*x +
c) + a^5*d)

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giac [A]  time = 1.46, size = 72, normalized size = 0.45 \[ \frac {35 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 180 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 378 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 420 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 315 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{5040 \, a^{5} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+a*sec(d*x+c))^5,x, algorithm="giac")

[Out]

1/5040*(35*tan(1/2*d*x + 1/2*c)^9 + 180*tan(1/2*d*x + 1/2*c)^7 + 378*tan(1/2*d*x + 1/2*c)^5 + 420*tan(1/2*d*x
+ 1/2*c)^3 + 315*tan(1/2*d*x + 1/2*c))/(a^5*d)

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maple [A]  time = 0.41, size = 71, normalized size = 0.45 \[ \frac {\frac {\left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{9}+\frac {4 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {6 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d \,a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5/(a+a*sec(d*x+c))^5,x)

[Out]

1/16/d/a^5*(1/9*tan(1/2*d*x+1/2*c)^9+4/7*tan(1/2*d*x+1/2*c)^7+6/5*tan(1/2*d*x+1/2*c)^5+4/3*tan(1/2*d*x+1/2*c)^
3+tan(1/2*d*x+1/2*c))

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maxima [A]  time = 0.71, size = 107, normalized size = 0.67 \[ \frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {420 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {378 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {180 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {35 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{5040 \, a^{5} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+a*sec(d*x+c))^5,x, algorithm="maxima")

[Out]

1/5040*(315*sin(d*x + c)/(cos(d*x + c) + 1) + 420*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 378*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5 + 180*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 35*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)/(a^5*d)

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mupad [B]  time = 0.76, size = 127, normalized size = 0.80 \[ \frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (315\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+420\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+378\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+180\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+35\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\right )}{5040\,a^5\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^5*(a + a/cos(c + d*x))^5),x)

[Out]

(sin(c/2 + (d*x)/2)*(315*cos(c/2 + (d*x)/2)^8 + 35*sin(c/2 + (d*x)/2)^8 + 180*cos(c/2 + (d*x)/2)^2*sin(c/2 + (
d*x)/2)^6 + 378*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^4 + 420*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^2))/(5
040*a^5*d*cos(c/2 + (d*x)/2)^9)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec ^{5}{\left (c + d x \right )}}{\sec ^{5}{\left (c + d x \right )} + 5 \sec ^{4}{\left (c + d x \right )} + 10 \sec ^{3}{\left (c + d x \right )} + 10 \sec ^{2}{\left (c + d x \right )} + 5 \sec {\left (c + d x \right )} + 1}\, dx}{a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5/(a+a*sec(d*x+c))**5,x)

[Out]

Integral(sec(c + d*x)**5/(sec(c + d*x)**5 + 5*sec(c + d*x)**4 + 10*sec(c + d*x)**3 + 10*sec(c + d*x)**2 + 5*se
c(c + d*x) + 1), x)/a**5

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