3.80 \(\int \frac {\sec ^7(c+d x)}{(a+a \sec (c+d x))^5} \, dx\)

Optimal. Leaf size=200 \[ \frac {181 \tan (c+d x)}{63 a^5 d}-\frac {5 \tanh ^{-1}(\sin (c+d x))}{a^5 d}+\frac {5 \tan (c+d x)}{d \left (a^5 \sec (c+d x)+a^5\right )}-\frac {67 \tan (c+d x) \sec ^2(c+d x)}{63 a^3 d (a \sec (c+d x)+a)^2}-\frac {29 \tan (c+d x) \sec ^3(c+d x)}{63 a^2 d (a \sec (c+d x)+a)^3}-\frac {\tan (c+d x) \sec ^5(c+d x)}{9 d (a \sec (c+d x)+a)^5}-\frac {5 \tan (c+d x) \sec ^4(c+d x)}{21 a d (a \sec (c+d x)+a)^4} \]

[Out]

-5*arctanh(sin(d*x+c))/a^5/d+181/63*tan(d*x+c)/a^5/d-1/9*sec(d*x+c)^5*tan(d*x+c)/d/(a+a*sec(d*x+c))^5-5/21*sec
(d*x+c)^4*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^4-29/63*sec(d*x+c)^3*tan(d*x+c)/a^2/d/(a+a*sec(d*x+c))^3-67/63*sec(d
*x+c)^2*tan(d*x+c)/a^3/d/(a+a*sec(d*x+c))^2+5*tan(d*x+c)/d/(a^5+a^5*sec(d*x+c))

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Rubi [A]  time = 0.48, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3816, 4019, 4008, 3787, 3770, 3767, 8} \[ \frac {181 \tan (c+d x)}{63 a^5 d}-\frac {5 \tanh ^{-1}(\sin (c+d x))}{a^5 d}-\frac {29 \tan (c+d x) \sec ^3(c+d x)}{63 a^2 d (a \sec (c+d x)+a)^3}-\frac {67 \tan (c+d x) \sec ^2(c+d x)}{63 a^3 d (a \sec (c+d x)+a)^2}+\frac {5 \tan (c+d x)}{d \left (a^5 \sec (c+d x)+a^5\right )}-\frac {\tan (c+d x) \sec ^5(c+d x)}{9 d (a \sec (c+d x)+a)^5}-\frac {5 \tan (c+d x) \sec ^4(c+d x)}{21 a d (a \sec (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^7/(a + a*Sec[c + d*x])^5,x]

[Out]

(-5*ArcTanh[Sin[c + d*x]])/(a^5*d) + (181*Tan[c + d*x])/(63*a^5*d) - (Sec[c + d*x]^5*Tan[c + d*x])/(9*d*(a + a
*Sec[c + d*x])^5) - (5*Sec[c + d*x]^4*Tan[c + d*x])/(21*a*d*(a + a*Sec[c + d*x])^4) - (29*Sec[c + d*x]^3*Tan[c
+ d*x])/(63*a^2*d*(a + a*Sec[c + d*x])^3) - (67*Sec[c + d*x]^2*Tan[c + d*x])/(63*a^3*d*(a + a*Sec[c + d*x])^2
) + (5*Tan[c + d*x])/(d*(a^5 + a^5*Sec[c + d*x]))

Rule 8

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

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 3770

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

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, 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 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 ^7(c+d x)}{(a+a \sec (c+d x))^5} \, dx &=-\frac {\sec ^5(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}-\frac {\int \frac {\sec ^5(c+d x) (5 a-10 a \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx}{9 a^2}\\ &=-\frac {\sec ^5(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}-\frac {5 \sec ^4(c+d x) \tan (c+d x)}{21 a d (a+a \sec (c+d x))^4}-\frac {\int \frac {\sec ^4(c+d x) \left (60 a^2-85 a^2 \sec (c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx}{63 a^4}\\ &=-\frac {\sec ^5(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}-\frac {5 \sec ^4(c+d x) \tan (c+d x)}{21 a d (a+a \sec (c+d x))^4}-\frac {29 \sec ^3(c+d x) \tan (c+d x)}{63 a^2 d (a+a \sec (c+d x))^3}-\frac {\int \frac {\sec ^3(c+d x) \left (435 a^3-570 a^3 \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{315 a^6}\\ &=-\frac {\sec ^5(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}-\frac {5 \sec ^4(c+d x) \tan (c+d x)}{21 a d (a+a \sec (c+d x))^4}-\frac {29 \sec ^3(c+d x) \tan (c+d x)}{63 a^2 d (a+a \sec (c+d x))^3}-\frac {67 \sec ^2(c+d x) \tan (c+d x)}{63 a^3 d (a+a \sec (c+d x))^2}-\frac {\int \frac {\sec ^2(c+d x) \left (2010 a^4-2715 a^4 \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{945 a^8}\\ &=-\frac {\sec ^5(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}-\frac {5 \sec ^4(c+d x) \tan (c+d x)}{21 a d (a+a \sec (c+d x))^4}-\frac {29 \sec ^3(c+d x) \tan (c+d x)}{63 a^2 d (a+a \sec (c+d x))^3}-\frac {67 \sec ^2(c+d x) \tan (c+d x)}{63 a^3 d (a+a \sec (c+d x))^2}+\frac {5 \tan (c+d x)}{d \left (a^5+a^5 \sec (c+d x)\right )}+\frac {\int \sec (c+d x) \left (-4725 a^5+2715 a^5 \sec (c+d x)\right ) \, dx}{945 a^{10}}\\ &=-\frac {\sec ^5(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}-\frac {5 \sec ^4(c+d x) \tan (c+d x)}{21 a d (a+a \sec (c+d x))^4}-\frac {29 \sec ^3(c+d x) \tan (c+d x)}{63 a^2 d (a+a \sec (c+d x))^3}-\frac {67 \sec ^2(c+d x) \tan (c+d x)}{63 a^3 d (a+a \sec (c+d x))^2}+\frac {5 \tan (c+d x)}{d \left (a^5+a^5 \sec (c+d x)\right )}+\frac {181 \int \sec ^2(c+d x) \, dx}{63 a^5}-\frac {5 \int \sec (c+d x) \, dx}{a^5}\\ &=-\frac {5 \tanh ^{-1}(\sin (c+d x))}{a^5 d}-\frac {\sec ^5(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}-\frac {5 \sec ^4(c+d x) \tan (c+d x)}{21 a d (a+a \sec (c+d x))^4}-\frac {29 \sec ^3(c+d x) \tan (c+d x)}{63 a^2 d (a+a \sec (c+d x))^3}-\frac {67 \sec ^2(c+d x) \tan (c+d x)}{63 a^3 d (a+a \sec (c+d x))^2}+\frac {5 \tan (c+d x)}{d \left (a^5+a^5 \sec (c+d x)\right )}-\frac {181 \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{63 a^5 d}\\ &=-\frac {5 \tanh ^{-1}(\sin (c+d x))}{a^5 d}+\frac {181 \tan (c+d x)}{63 a^5 d}-\frac {\sec ^5(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}-\frac {5 \sec ^4(c+d x) \tan (c+d x)}{21 a d (a+a \sec (c+d x))^4}-\frac {29 \sec ^3(c+d x) \tan (c+d x)}{63 a^2 d (a+a \sec (c+d x))^3}-\frac {67 \sec ^2(c+d x) \tan (c+d x)}{63 a^3 d (a+a \sec (c+d x))^2}+\frac {5 \tan (c+d x)}{d \left (a^5+a^5 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [B]  time = 1.97, size = 401, normalized size = 2.00 \[ \frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (\sec \left (\frac {c}{2}\right ) \sec (c) \left (-56952 \sin \left (c-\frac {d x}{2}\right )+43722 \sin \left (c+\frac {d x}{2}\right )-47208 \sin \left (2 c+\frac {d x}{2}\right )-18144 \sin \left (c+\frac {3 d x}{2}\right )+41796 \sin \left (2 c+\frac {3 d x}{2}\right )-28350 \sin \left (3 c+\frac {3 d x}{2}\right )+34578 \sin \left (c+\frac {5 d x}{2}\right )-5691 \sin \left (2 c+\frac {5 d x}{2}\right )+28719 \sin \left (3 c+\frac {5 d x}{2}\right )-11550 \sin \left (4 c+\frac {5 d x}{2}\right )+15517 \sin \left (2 c+\frac {7 d x}{2}\right )-504 \sin \left (3 c+\frac {7 d x}{2}\right )+13186 \sin \left (4 c+\frac {7 d x}{2}\right )-2835 \sin \left (5 c+\frac {7 d x}{2}\right )+4149 \sin \left (3 c+\frac {9 d x}{2}\right )+252 \sin \left (4 c+\frac {9 d x}{2}\right )+3582 \sin \left (5 c+\frac {9 d x}{2}\right )-315 \sin \left (6 c+\frac {9 d x}{2}\right )+496 \sin \left (4 c+\frac {11 d x}{2}\right )+63 \sin \left (5 c+\frac {11 d x}{2}\right )+433 \sin \left (6 c+\frac {11 d x}{2}\right )-33978 \sin \left (\frac {d x}{2}\right )+52002 \sin \left (\frac {3 d x}{2}\right )\right ) \sec (c+d x)+322560 \cos ^9\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{2016 a^5 d (\sec (c+d x)+1)^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^7/(a + a*Sec[c + d*x])^5,x]

[Out]

(Cos[(c + d*x)/2]*Sec[c + d*x]^5*(322560*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]*Sec[c]*Sec[c + d*x]*(-33978*Sin[(d*x)/2] + 52002*Sin[(3*d*x)/2]
- 56952*Sin[c - (d*x)/2] + 43722*Sin[c + (d*x)/2] - 47208*Sin[2*c + (d*x)/2] - 18144*Sin[c + (3*d*x)/2] + 417
96*Sin[2*c + (3*d*x)/2] - 28350*Sin[3*c + (3*d*x)/2] + 34578*Sin[c + (5*d*x)/2] - 5691*Sin[2*c + (5*d*x)/2] +
28719*Sin[3*c + (5*d*x)/2] - 11550*Sin[4*c + (5*d*x)/2] + 15517*Sin[2*c + (7*d*x)/2] - 504*Sin[3*c + (7*d*x)/2
] + 13186*Sin[4*c + (7*d*x)/2] - 2835*Sin[5*c + (7*d*x)/2] + 4149*Sin[3*c + (9*d*x)/2] + 252*Sin[4*c + (9*d*x)
/2] + 3582*Sin[5*c + (9*d*x)/2] - 315*Sin[6*c + (9*d*x)/2] + 496*Sin[4*c + (11*d*x)/2] + 63*Sin[5*c + (11*d*x)
/2] + 433*Sin[6*c + (11*d*x)/2])))/(2016*a^5*d*(1 + Sec[c + d*x])^5)

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fricas [A]  time = 1.31, size = 278, normalized size = 1.39 \[ -\frac {315 \, {\left (\cos \left (d x + c\right )^{6} + 5 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{3} + 5 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 315 \, {\left (\cos \left (d x + c\right )^{6} + 5 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{3} + 5 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (496 \, \cos \left (d x + c\right )^{5} + 2165 \, \cos \left (d x + c\right )^{4} + 3633 \, \cos \left (d x + c\right )^{3} + 2840 \, \cos \left (d x + c\right )^{2} + 946 \, \cos \left (d x + c\right ) + 63\right )} \sin \left (d x + c\right )}{126 \, {\left (a^{5} d \cos \left (d x + c\right )^{6} + 5 \, a^{5} d \cos \left (d x + c\right )^{5} + 10 \, a^{5} d \cos \left (d x + c\right )^{4} + 10 \, a^{5} d \cos \left (d x + c\right )^{3} + 5 \, a^{5} d \cos \left (d x + c\right )^{2} + a^{5} d \cos \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7/(a+a*sec(d*x+c))^5,x, algorithm="fricas")

[Out]

-1/126*(315*(cos(d*x + c)^6 + 5*cos(d*x + c)^5 + 10*cos(d*x + c)^4 + 10*cos(d*x + c)^3 + 5*cos(d*x + c)^2 + co
s(d*x + c))*log(sin(d*x + c) + 1) - 315*(cos(d*x + c)^6 + 5*cos(d*x + c)^5 + 10*cos(d*x + c)^4 + 10*cos(d*x +
c)^3 + 5*cos(d*x + c)^2 + cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(496*cos(d*x + c)^5 + 2165*cos(d*x + c)^4 +
3633*cos(d*x + c)^3 + 2840*cos(d*x + c)^2 + 946*cos(d*x + c) + 63)*sin(d*x + c))/(a^5*d*cos(d*x + c)^6 + 5*a^
5*d*cos(d*x + c)^5 + 10*a^5*d*cos(d*x + c)^4 + 10*a^5*d*cos(d*x + c)^3 + 5*a^5*d*cos(d*x + c)^2 + a^5*d*cos(d*
x + c))

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giac [A]  time = 0.98, size = 155, normalized size = 0.78 \[ -\frac {\frac {5040 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{5}} - \frac {5040 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{5}} + \frac {2016 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{5}} - \frac {7 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 72 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 378 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1512 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8127 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{45}}}{1008 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7/(a+a*sec(d*x+c))^5,x, algorithm="giac")

[Out]

-1/1008*(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 + 2016*tan(
1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a^5) - (7*a^40*tan(1/2*d*x + 1/2*c)^9 + 72*a^40*tan(1/2*d*x + 1
/2*c)^7 + 378*a^40*tan(1/2*d*x + 1/2*c)^5 + 1512*a^40*tan(1/2*d*x + 1/2*c)^3 + 8127*a^40*tan(1/2*d*x + 1/2*c))
/a^45)/d

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maple [A]  time = 0.36, size = 177, normalized size = 0.88 \[ \frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{144 d \,a^{5}}+\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{14 d \,a^{5}}+\frac {3 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{5}}+\frac {3 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{5}}+\frac {129 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d \,a^{5}}-\frac {1}{d \,a^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{5}}-\frac {1}{d \,a^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^7/(a+a*sec(d*x+c))^5,x)

[Out]

1/144/d/a^5*tan(1/2*d*x+1/2*c)^9+1/14/d/a^5*tan(1/2*d*x+1/2*c)^7+3/8/d/a^5*tan(1/2*d*x+1/2*c)^5+3/2/d/a^5*tan(
1/2*d*x+1/2*c)^3+129/16/d/a^5*tan(1/2*d*x+1/2*c)-1/d/a^5/(tan(1/2*d*x+1/2*c)-1)+5/d/a^5*ln(tan(1/2*d*x+1/2*c)-
1)-1/d/a^5/(tan(1/2*d*x+1/2*c)+1)-5/d/a^5*ln(tan(1/2*d*x+1/2*c)+1)

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maxima [A]  time = 0.76, size = 206, normalized size = 1.03 \[ \frac {\frac {2016 \, \sin \left (d x + c\right )}{{\left (a^{5} - \frac {a^{5} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {8127 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {1512 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {378 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {72 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {7 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{a^{5}} - \frac {5040 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{5}} + \frac {5040 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{5}}}{1008 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7/(a+a*sec(d*x+c))^5,x, algorithm="maxima")

[Out]

1/1008*(2016*sin(d*x + c)/((a^5 - a^5*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (8127*sin(d*x
+ c)/(cos(d*x + c) + 1) + 1512*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 378*sin(d*x + c)^5/(cos(d*x + c) + 1)^5
+ 72*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 7*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 = 0.72, size = 149, normalized size = 0.74 \[ \frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2\,a^5\,d}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{8\,a^5\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{14\,a^5\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{144\,a^5\,d}-\frac {10\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^5\,d}-\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^5\right )}+\frac {129\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,a^5\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^7*(a + a/cos(c + d*x))^5),x)

[Out]

(3*tan(c/2 + (d*x)/2)^3)/(2*a^5*d) + (3*tan(c/2 + (d*x)/2)^5)/(8*a^5*d) + tan(c/2 + (d*x)/2)^7/(14*a^5*d) + ta
n(c/2 + (d*x)/2)^9/(144*a^5*d) - (10*atanh(tan(c/2 + (d*x)/2)))/(a^5*d) - (2*tan(c/2 + (d*x)/2))/(d*(a^5*tan(c
/2 + (d*x)/2)^2 - a^5)) + (129*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 \[ \frac {\int \frac {\sec ^{7}{\left (c + d x \right )}}{\sec ^{5}{\left (c + d x \right )} + 5 \sec ^{4}{\left (c + d x \right )} + 10 \sec ^{3}{\left (c + d x \right )} + 10 \sec ^{2}{\left (c + d x \right )} + 5 \sec {\left (c + d x \right )} + 1}\, dx}{a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**7/(a+a*sec(d*x+c))**5,x)

[Out]

Integral(sec(c + d*x)**7/(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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