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

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

[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]))

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Rubi [A]  time = 0.427265, antiderivative size = 177, normalized size of antiderivative = 1., 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 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 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]

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 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 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]

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*}

Mathematica [A]  time = 1.96751, 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]

-(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[Co
s[(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*Sin[3*c
+ (7*d*x)/2] - 315*Sin[4*c + (7*d*x)/2] + 488*Sin[4*c + (9*d*x)/2])))/(2520*a^5*d*(1 + Sec[c + d*x])^5)

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Maple [A]  time = 0.043, size = 134, normalized size = 0.8 \begin{align*} -{\frac{1}{144\,d{a}^{5}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{9}}-{\frac{3}{56\,d{a}^{5}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}-{\frac{1}{5\,d{a}^{5}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{13}{24\,d{a}^{5}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{31}{16\,d{a}^{5}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{d{a}^{5}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{1}{d{a}^{5}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) } \end{align*}

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 = 1.179, size = 215, normalized size = 1.21 \begin{align*} -\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} \end{align*}

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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Fricas [A]  time = 1.78017, size = 671, normalized size = 3.79 \begin{align*} \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 )}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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}} \end{align*}

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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Giac [A]  time = 1.39204, size = 170, normalized size = 0.96 \begin{align*} \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} \end{align*}

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