3.81 \(\int \frac {\sec ^6(c+d x)}{(a+a \sec (c+d x))^5} \, dx\)

Optimal. Leaf size=177 \[ \frac {\tanh ^{-1}(\sin (c+d x))}{a^5 d}-\frac {661 \tan (c+d x)}{315 d \left (a^5 \sec (c+d x)+a^5\right )}+\frac {173 \tan (c+d x)}{315 a^3 d (a \sec (c+d x)+a)^2}-\frac {34 \tan (c+d x) \sec ^2(c+d x)}{105 a^2 d (a \sec (c+d x)+a)^3}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}-\frac {13 \tan (c+d x) \sec ^3(c+d x)}{63 a d (a \sec (c+d x)+a)^4} \]

[Out]

arctanh(sin(d*x+c))/a^5/d-1/9*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+c))^5-13/63*sec(d*x+c)^3*tan(d*x+c)/a/d/(
a+a*sec(d*x+c))^4-34/105*sec(d*x+c)^2*tan(d*x+c)/a^2/d/(a+a*sec(d*x+c))^3+173/315*tan(d*x+c)/a^3/d/(a+a*sec(d*
x+c))^2-661/315*tan(d*x+c)/d/(a^5+a^5*sec(d*x+c))

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Rubi [A]  time = 0.43, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3816, 4019, 4008, 3998, 3770, 3794} \[ \frac {\tanh ^{-1}(\sin (c+d x))}{a^5 d}-\frac {34 \tan (c+d x) \sec ^2(c+d x)}{105 a^2 d (a \sec (c+d x)+a)^3}-\frac {661 \tan (c+d x)}{315 d \left (a^5 \sec (c+d x)+a^5\right )}+\frac {173 \tan (c+d x)}{315 a^3 d (a \sec (c+d x)+a)^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}-\frac {13 \tan (c+d x) \sec ^3(c+d x)}{63 a d (a \sec (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6/(a + a*Sec[c + d*x])^5,x]

[Out]

ArcTanh[Sin[c + d*x]]/(a^5*d) - (Sec[c + d*x]^4*Tan[c + d*x])/(9*d*(a + a*Sec[c + d*x])^5) - (13*Sec[c + d*x]^
3*Tan[c + d*x])/(63*a*d*(a + a*Sec[c + d*x])^4) - (34*Sec[c + d*x]^2*Tan[c + d*x])/(105*a^2*d*(a + a*Sec[c + d
*x])^3) + (173*Tan[c + d*x])/(315*a^3*d*(a + a*Sec[c + d*x])^2) - (661*Tan[c + d*x])/(315*d*(a^5 + a^5*Sec[c +
d*x]))

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{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 3816

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(d^2*
Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 2))/(f*(2*m + 1)), x] + Dist[d^2/(a*b*(2*m + 1)), In
t[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2)*(b*(n - 2) + a*(m - n + 2)*Csc[e + f*x]), x], x] /; Fr
eeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 2] && (IntegersQ[2*m, 2*n] || IntegerQ[m]
)

Rule 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 4008

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

Rule 4019

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

Rubi steps

\begin {align*} \int \frac {\sec ^6(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 {\int \frac {\sec ^4(c+d x) (4 a-9 a \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx}{9 a^2}\\ &=-\frac {\sec ^4(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}-\frac {13 \sec ^3(c+d x) \tan (c+d x)}{63 a d (a+a \sec (c+d x))^4}-\frac {\int \frac {\sec ^3(c+d x) \left (39 a^2-63 a^2 \sec (c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx}{63 a^4}\\ &=-\frac {\sec ^4(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}-\frac {13 \sec ^3(c+d x) \tan (c+d x)}{63 a d (a+a \sec (c+d x))^4}-\frac {34 \sec ^2(c+d x) \tan (c+d x)}{105 a^2 d (a+a \sec (c+d x))^3}-\frac {\int \frac {\sec ^2(c+d x) \left (204 a^3-315 a^3 \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{315 a^6}\\ &=-\frac {\sec ^4(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}-\frac {13 \sec ^3(c+d x) \tan (c+d x)}{63 a d (a+a \sec (c+d x))^4}-\frac {34 \sec ^2(c+d x) \tan (c+d x)}{105 a^2 d (a+a \sec (c+d x))^3}+\frac {173 \tan (c+d x)}{315 a^3 d (a+a \sec (c+d x))^2}+\frac {\int \frac {\sec (c+d x) \left (-1038 a^4+945 a^4 \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{945 a^8}\\ &=-\frac {\sec ^4(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}-\frac {13 \sec ^3(c+d x) \tan (c+d x)}{63 a d (a+a \sec (c+d x))^4}-\frac {34 \sec ^2(c+d x) \tan (c+d x)}{105 a^2 d (a+a \sec (c+d x))^3}+\frac {173 \tan (c+d x)}{315 a^3 d (a+a \sec (c+d x))^2}+\frac {\int \sec (c+d x) \, dx}{a^5}-\frac {661 \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{315 a^4}\\ &=\frac {\tanh ^{-1}(\sin (c+d x))}{a^5 d}-\frac {\sec ^4(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}-\frac {13 \sec ^3(c+d x) \tan (c+d x)}{63 a d (a+a \sec (c+d x))^4}-\frac {34 \sec ^2(c+d x) \tan (c+d x)}{105 a^2 d (a+a \sec (c+d x))^3}+\frac {173 \tan (c+d x)}{315 a^3 d (a+a \sec (c+d x))^2}-\frac {661 \tan (c+d x)}{315 d \left (a^5+a^5 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 1.96, size = 219, normalized size = 1.24 \[ -\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (\sec \left (\frac {c}{2}\right ) \left (-25515 \sin \left (c+\frac {d x}{2}\right )+29757 \sin \left (c+\frac {3 d x}{2}\right )-11235 \sin \left (2 c+\frac {3 d x}{2}\right )+14733 \sin \left (2 c+\frac {5 d x}{2}\right )-2835 \sin \left (3 c+\frac {5 d x}{2}\right )+4077 \sin \left (3 c+\frac {7 d x}{2}\right )-315 \sin \left (4 c+\frac {7 d x}{2}\right )+488 \sin \left (4 c+\frac {9 d x}{2}\right )+35973 \sin \left (\frac {d x}{2}\right )\right )+80640 \cos ^9\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{2520 a^5 d (\sec (c+d x)+1)^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6/(a + a*Sec[c + d*x])^5,x]

[Out]

-1/2520*(Cos[(c + d*x)/2]*Sec[c + d*x]^5*(80640*Cos[(c + d*x)/2]^9*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] -
Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Sec[c/2]*(35973*Sin[(d*x)/2] - 25515*Sin[c + (d*x)/2] + 29757*Sin
[c + (3*d*x)/2] - 11235*Sin[2*c + (3*d*x)/2] + 14733*Sin[2*c + (5*d*x)/2] - 2835*Sin[3*c + (5*d*x)/2] + 4077*S
in[3*c + (7*d*x)/2] - 315*Sin[4*c + (7*d*x)/2] + 488*Sin[4*c + (9*d*x)/2])))/(a^5*d*(1 + Sec[c + d*x])^5)

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fricas [A]  time = 0.66, size = 246, normalized size = 1.39 \[ \frac {315 \, {\left (\cos \left (d x + c\right )^{5} + 5 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{3} + 10 \, \cos \left (d x + c\right )^{2} + 5 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 315 \, {\left (\cos \left (d x + c\right )^{5} + 5 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{3} + 10 \, \cos \left (d x + c\right )^{2} + 5 \, \cos \left (d x + c\right ) + 1\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (488 \, \cos \left (d x + c\right )^{4} + 2125 \, \cos \left (d x + c\right )^{3} + 3549 \, \cos \left (d x + c\right )^{2} + 2740 \, \cos \left (d x + c\right ) + 863\right )} \sin \left (d x + c\right )}{630 \, {\left (a^{5} d \cos \left (d x + c\right )^{5} + 5 \, a^{5} d \cos \left (d x + c\right )^{4} + 10 \, a^{5} d \cos \left (d x + c\right )^{3} + 10 \, a^{5} d \cos \left (d x + c\right )^{2} + 5 \, a^{5} d \cos \left (d x + c\right ) + a^{5} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(a+a*sec(d*x+c))^5,x, algorithm="fricas")

[Out]

1/630*(315*(cos(d*x + c)^5 + 5*cos(d*x + c)^4 + 10*cos(d*x + c)^3 + 10*cos(d*x + c)^2 + 5*cos(d*x + c) + 1)*lo
g(sin(d*x + c) + 1) - 315*(cos(d*x + c)^5 + 5*cos(d*x + c)^4 + 10*cos(d*x + c)^3 + 10*cos(d*x + c)^2 + 5*cos(d
*x + c) + 1)*log(-sin(d*x + c) + 1) - 2*(488*cos(d*x + c)^4 + 2125*cos(d*x + c)^3 + 3549*cos(d*x + c)^2 + 2740
*cos(d*x + c) + 863)*sin(d*x + c))/(a^5*d*cos(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 = 0.88, size = 126, normalized size = 0.71 \[ \frac {\frac {5040 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{5}} - \frac {5040 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{5}} - \frac {35 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 270 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1008 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2730 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9765 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{45}}}{5040 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(a+a*sec(d*x+c))^5,x, algorithm="giac")

[Out]

1/5040*(5040*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^5 - 5040*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^5 - (35*a^40*t
an(1/2*d*x + 1/2*c)^9 + 270*a^40*tan(1/2*d*x + 1/2*c)^7 + 1008*a^40*tan(1/2*d*x + 1/2*c)^5 + 2730*a^40*tan(1/2
*d*x + 1/2*c)^3 + 9765*a^40*tan(1/2*d*x + 1/2*c))/a^45)/d

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maple [A]  time = 0.42, size = 134, normalized size = 0.76 \[ -\frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{144 d \,a^{5}}-\frac {3 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{5}}-\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{5 d \,a^{5}}-\frac {13 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d \,a^{5}}-\frac {31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d \,a^{5}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{5}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6/(a+a*sec(d*x+c))^5,x)

[Out]

-1/144/d/a^5*tan(1/2*d*x+1/2*c)^9-3/56/d/a^5*tan(1/2*d*x+1/2*c)^7-1/5/d/a^5*tan(1/2*d*x+1/2*c)^5-13/24/d/a^5*t
an(1/2*d*x+1/2*c)^3-31/16/d/a^5*tan(1/2*d*x+1/2*c)-1/d/a^5*ln(tan(1/2*d*x+1/2*c)-1)+1/d/a^5*ln(tan(1/2*d*x+1/2
*c)+1)

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maxima [A]  time = 0.69, size = 159, normalized size = 0.90 \[ -\frac {\frac {\frac {9765 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {2730 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {1008 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {270 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {35 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{a^{5}} - \frac {5040 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{5}} + \frac {5040 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{5}}}{5040 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(a+a*sec(d*x+c))^5,x, algorithm="maxima")

[Out]

-1/5040*((9765*sin(d*x + c)/(cos(d*x + c) + 1) + 2730*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 1008*sin(d*x + c)^
5/(cos(d*x + c) + 1)^5 + 270*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
- 5040*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^5 + 5040*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^5)/d

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mupad [B]  time = 0.69, size = 99, normalized size = 0.56 \[ -\frac {\frac {13\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^5}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5\,a^5}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{56\,a^5}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{144\,a^5}-\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^5}+\frac {31\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,a^5}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^6*(a + a/cos(c + d*x))^5),x)

[Out]

-((13*tan(c/2 + (d*x)/2)^3)/(24*a^5) + tan(c/2 + (d*x)/2)^5/(5*a^5) + (3*tan(c/2 + (d*x)/2)^7)/(56*a^5) + tan(
c/2 + (d*x)/2)^9/(144*a^5) - (2*atanh(tan(c/2 + (d*x)/2)))/a^5 + (31*tan(c/2 + (d*x)/2))/(16*a^5))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec ^{6}{\left (c + d x \right )}}{\sec ^{5}{\left (c + d x \right )} + 5 \sec ^{4}{\left (c + d x \right )} + 10 \sec ^{3}{\left (c + d x \right )} + 10 \sec ^{2}{\left (c + d x \right )} + 5 \sec {\left (c + d x \right )} + 1}\, dx}{a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**6/(a+a*sec(d*x+c))**5,x)

[Out]

Integral(sec(c + d*x)**6/(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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