3.95 \(\int \frac {\tan ^{12}(c+d x)}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=237 \[ -\frac {3 \tan ^7(c+d x)}{7 a^3 d}-\frac {\tan ^5(c+d x)}{5 a^3 d}+\frac {\tan ^3(c+d x)}{3 a^3 d}-\frac {\tan (c+d x)}{a^3 d}-\frac {125 \tanh ^{-1}(\sin (c+d x))}{128 a^3 d}+\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 a^3 d}-\frac {5 \tan ^3(c+d x) \sec ^3(c+d x)}{48 a^3 d}+\frac {5 \tan (c+d x) \sec ^3(c+d x)}{64 a^3 d}+\frac {\tan ^5(c+d x) \sec (c+d x)}{2 a^3 d}-\frac {5 \tan ^3(c+d x) \sec (c+d x)}{8 a^3 d}+\frac {115 \tan (c+d x) \sec (c+d x)}{128 a^3 d}+\frac {x}{a^3} \]

[Out]

x/a^3-125/128*arctanh(sin(d*x+c))/a^3/d-tan(d*x+c)/a^3/d+115/128*sec(d*x+c)*tan(d*x+c)/a^3/d+5/64*sec(d*x+c)^3
*tan(d*x+c)/a^3/d+1/3*tan(d*x+c)^3/a^3/d-5/8*sec(d*x+c)*tan(d*x+c)^3/a^3/d-5/48*sec(d*x+c)^3*tan(d*x+c)^3/a^3/
d-1/5*tan(d*x+c)^5/a^3/d+1/2*sec(d*x+c)*tan(d*x+c)^5/a^3/d+1/8*sec(d*x+c)^3*tan(d*x+c)^5/a^3/d-3/7*tan(d*x+c)^
7/a^3/d

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Rubi [A]  time = 0.36, antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3888, 3886, 3473, 8, 2611, 3770, 2607, 30, 3768} \[ -\frac {3 \tan ^7(c+d x)}{7 a^3 d}-\frac {\tan ^5(c+d x)}{5 a^3 d}+\frac {\tan ^3(c+d x)}{3 a^3 d}-\frac {\tan (c+d x)}{a^3 d}-\frac {125 \tanh ^{-1}(\sin (c+d x))}{128 a^3 d}+\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 a^3 d}-\frac {5 \tan ^3(c+d x) \sec ^3(c+d x)}{48 a^3 d}+\frac {5 \tan (c+d x) \sec ^3(c+d x)}{64 a^3 d}+\frac {\tan ^5(c+d x) \sec (c+d x)}{2 a^3 d}-\frac {5 \tan ^3(c+d x) \sec (c+d x)}{8 a^3 d}+\frac {115 \tan (c+d x) \sec (c+d x)}{128 a^3 d}+\frac {x}{a^3} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^12/(a + a*Sec[c + d*x])^3,x]

[Out]

x/a^3 - (125*ArcTanh[Sin[c + d*x]])/(128*a^3*d) - Tan[c + d*x]/(a^3*d) + (115*Sec[c + d*x]*Tan[c + d*x])/(128*
a^3*d) + (5*Sec[c + d*x]^3*Tan[c + d*x])/(64*a^3*d) + Tan[c + d*x]^3/(3*a^3*d) - (5*Sec[c + d*x]*Tan[c + d*x]^
3)/(8*a^3*d) - (5*Sec[c + d*x]^3*Tan[c + d*x]^3)/(48*a^3*d) - Tan[c + d*x]^5/(5*a^3*d) + (Sec[c + d*x]*Tan[c +
d*x]^5)/(2*a^3*d) + (Sec[c + d*x]^3*Tan[c + d*x]^5)/(8*a^3*d) - (3*Tan[c + d*x]^7)/(7*a^3*d)

Rule 8

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

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

Rule 3886

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[ExpandI
ntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]

Rule 3888

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[a^(2*n
)/e^(2*n), Int[(e*Cot[c + d*x])^(m + 2*n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && E
qQ[a^2 - b^2, 0] && ILtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\tan ^{12}(c+d x)}{(a+a \sec (c+d x))^3} \, dx &=\frac {\int (-a+a \sec (c+d x))^3 \tan ^6(c+d x) \, dx}{a^6}\\ &=\frac {\int \left (-a^3 \tan ^6(c+d x)+3 a^3 \sec (c+d x) \tan ^6(c+d x)-3 a^3 \sec ^2(c+d x) \tan ^6(c+d x)+a^3 \sec ^3(c+d x) \tan ^6(c+d x)\right ) \, dx}{a^6}\\ &=-\frac {\int \tan ^6(c+d x) \, dx}{a^3}+\frac {\int \sec ^3(c+d x) \tan ^6(c+d x) \, dx}{a^3}+\frac {3 \int \sec (c+d x) \tan ^6(c+d x) \, dx}{a^3}-\frac {3 \int \sec ^2(c+d x) \tan ^6(c+d x) \, dx}{a^3}\\ &=-\frac {\tan ^5(c+d x)}{5 a^3 d}+\frac {\sec (c+d x) \tan ^5(c+d x)}{2 a^3 d}+\frac {\sec ^3(c+d x) \tan ^5(c+d x)}{8 a^3 d}-\frac {5 \int \sec ^3(c+d x) \tan ^4(c+d x) \, dx}{8 a^3}+\frac {\int \tan ^4(c+d x) \, dx}{a^3}-\frac {5 \int \sec (c+d x) \tan ^4(c+d x) \, dx}{2 a^3}-\frac {3 \operatorname {Subst}\left (\int x^6 \, dx,x,\tan (c+d x)\right )}{a^3 d}\\ &=\frac {\tan ^3(c+d x)}{3 a^3 d}-\frac {5 \sec (c+d x) \tan ^3(c+d x)}{8 a^3 d}-\frac {5 \sec ^3(c+d x) \tan ^3(c+d x)}{48 a^3 d}-\frac {\tan ^5(c+d x)}{5 a^3 d}+\frac {\sec (c+d x) \tan ^5(c+d x)}{2 a^3 d}+\frac {\sec ^3(c+d x) \tan ^5(c+d x)}{8 a^3 d}-\frac {3 \tan ^7(c+d x)}{7 a^3 d}+\frac {5 \int \sec ^3(c+d x) \tan ^2(c+d x) \, dx}{16 a^3}-\frac {\int \tan ^2(c+d x) \, dx}{a^3}+\frac {15 \int \sec (c+d x) \tan ^2(c+d x) \, dx}{8 a^3}\\ &=-\frac {\tan (c+d x)}{a^3 d}+\frac {15 \sec (c+d x) \tan (c+d x)}{16 a^3 d}+\frac {5 \sec ^3(c+d x) \tan (c+d x)}{64 a^3 d}+\frac {\tan ^3(c+d x)}{3 a^3 d}-\frac {5 \sec (c+d x) \tan ^3(c+d x)}{8 a^3 d}-\frac {5 \sec ^3(c+d x) \tan ^3(c+d x)}{48 a^3 d}-\frac {\tan ^5(c+d x)}{5 a^3 d}+\frac {\sec (c+d x) \tan ^5(c+d x)}{2 a^3 d}+\frac {\sec ^3(c+d x) \tan ^5(c+d x)}{8 a^3 d}-\frac {3 \tan ^7(c+d x)}{7 a^3 d}-\frac {5 \int \sec ^3(c+d x) \, dx}{64 a^3}-\frac {15 \int \sec (c+d x) \, dx}{16 a^3}+\frac {\int 1 \, dx}{a^3}\\ &=\frac {x}{a^3}-\frac {15 \tanh ^{-1}(\sin (c+d x))}{16 a^3 d}-\frac {\tan (c+d x)}{a^3 d}+\frac {115 \sec (c+d x) \tan (c+d x)}{128 a^3 d}+\frac {5 \sec ^3(c+d x) \tan (c+d x)}{64 a^3 d}+\frac {\tan ^3(c+d x)}{3 a^3 d}-\frac {5 \sec (c+d x) \tan ^3(c+d x)}{8 a^3 d}-\frac {5 \sec ^3(c+d x) \tan ^3(c+d x)}{48 a^3 d}-\frac {\tan ^5(c+d x)}{5 a^3 d}+\frac {\sec (c+d x) \tan ^5(c+d x)}{2 a^3 d}+\frac {\sec ^3(c+d x) \tan ^5(c+d x)}{8 a^3 d}-\frac {3 \tan ^7(c+d x)}{7 a^3 d}-\frac {5 \int \sec (c+d x) \, dx}{128 a^3}\\ &=\frac {x}{a^3}-\frac {125 \tanh ^{-1}(\sin (c+d x))}{128 a^3 d}-\frac {\tan (c+d x)}{a^3 d}+\frac {115 \sec (c+d x) \tan (c+d x)}{128 a^3 d}+\frac {5 \sec ^3(c+d x) \tan (c+d x)}{64 a^3 d}+\frac {\tan ^3(c+d x)}{3 a^3 d}-\frac {5 \sec (c+d x) \tan ^3(c+d x)}{8 a^3 d}-\frac {5 \sec ^3(c+d x) \tan ^3(c+d x)}{48 a^3 d}-\frac {\tan ^5(c+d x)}{5 a^3 d}+\frac {\sec (c+d x) \tan ^5(c+d x)}{2 a^3 d}+\frac {\sec ^3(c+d x) \tan ^5(c+d x)}{8 a^3 d}-\frac {3 \tan ^7(c+d x)}{7 a^3 d}\\ \end {align*}

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Mathematica [A]  time = 1.34, size = 362, normalized size = 1.53 \[ \frac {\cos ^6\left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (1680000 \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec (c) \sec ^8(c+d x) (133175 \sin (2 c+d x)-544768 \sin (c+2 d x)+286720 \sin (3 c+2 d x)+63595 \sin (2 c+3 d x)+63595 \sin (4 c+3 d x)-254464 \sin (3 c+4 d x)+161280 \sin (5 c+4 d x)+65135 \sin (4 c+5 d x)+65135 \sin (6 c+5 d x)-118784 \sin (5 c+6 d x)+27195 \sin (6 c+7 d x)+27195 \sin (8 c+7 d x)-14848 \sin (7 c+8 d x)+470400 d x \cos (c)+376320 d x \cos (c+2 d x)+376320 d x \cos (3 c+2 d x)+188160 d x \cos (3 c+4 d x)+188160 d x \cos (5 c+4 d x)+53760 d x \cos (5 c+6 d x)+53760 d x \cos (7 c+6 d x)+6720 d x \cos (7 c+8 d x)+6720 d x \cos (9 c+8 d x)+519680 \sin (c)+133175 \sin (d x))\right )}{215040 a^3 d (\sec (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^12/(a + a*Sec[c + d*x])^3,x]

[Out]

(Cos[(c + d*x)/2]^6*Sec[c + d*x]^3*(1680000*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] +
Sin[(c + d*x)/2]]) + Sec[c]*Sec[c + d*x]^8*(470400*d*x*Cos[c] + 376320*d*x*Cos[c + 2*d*x] + 376320*d*x*Cos[3*
c + 2*d*x] + 188160*d*x*Cos[3*c + 4*d*x] + 188160*d*x*Cos[5*c + 4*d*x] + 53760*d*x*Cos[5*c + 6*d*x] + 53760*d*
x*Cos[7*c + 6*d*x] + 6720*d*x*Cos[7*c + 8*d*x] + 6720*d*x*Cos[9*c + 8*d*x] + 519680*Sin[c] + 133175*Sin[d*x] +
133175*Sin[2*c + d*x] - 544768*Sin[c + 2*d*x] + 286720*Sin[3*c + 2*d*x] + 63595*Sin[2*c + 3*d*x] + 63595*Sin[
4*c + 3*d*x] - 254464*Sin[3*c + 4*d*x] + 161280*Sin[5*c + 4*d*x] + 65135*Sin[4*c + 5*d*x] + 65135*Sin[6*c + 5*
d*x] - 118784*Sin[5*c + 6*d*x] + 27195*Sin[6*c + 7*d*x] + 27195*Sin[8*c + 7*d*x] - 14848*Sin[7*c + 8*d*x])))/(
215040*a^3*d*(1 + Sec[c + d*x])^3)

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fricas [A]  time = 0.60, size = 147, normalized size = 0.62 \[ \frac {26880 \, d x \cos \left (d x + c\right )^{8} - 13125 \, \cos \left (d x + c\right )^{8} \log \left (\sin \left (d x + c\right ) + 1\right ) + 13125 \, \cos \left (d x + c\right )^{8} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (14848 \, \cos \left (d x + c\right )^{7} - 27195 \, \cos \left (d x + c\right )^{6} + 7424 \, \cos \left (d x + c\right )^{5} + 17710 \, \cos \left (d x + c\right )^{4} - 14592 \, \cos \left (d x + c\right )^{3} - 1960 \, \cos \left (d x + c\right )^{2} + 5760 \, \cos \left (d x + c\right ) - 1680\right )} \sin \left (d x + c\right )}{26880 \, a^{3} d \cos \left (d x + c\right )^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^12/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/26880*(26880*d*x*cos(d*x + c)^8 - 13125*cos(d*x + c)^8*log(sin(d*x + c) + 1) + 13125*cos(d*x + c)^8*log(-sin
(d*x + c) + 1) - 2*(14848*cos(d*x + c)^7 - 27195*cos(d*x + c)^6 + 7424*cos(d*x + c)^5 + 17710*cos(d*x + c)^4 -
14592*cos(d*x + c)^3 - 1960*cos(d*x + c)^2 + 5760*cos(d*x + c) - 1680)*sin(d*x + c))/(a^3*d*cos(d*x + c)^8)

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giac [A]  time = 116.12, size = 175, normalized size = 0.74 \[ \frac {\frac {13440 \, {\left (d x + c\right )}}{a^{3}} - \frac {13125 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} + \frac {13125 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {2 \, {\left (26565 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{15} - 212625 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 749973 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 550089 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 269879 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 79723 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 11375 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 315 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{8} a^{3}}}{13440 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^12/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/13440*(13440*(d*x + c)/a^3 - 13125*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 + 13125*log(abs(tan(1/2*d*x + 1/2*
c) - 1))/a^3 + 2*(26565*tan(1/2*d*x + 1/2*c)^15 - 212625*tan(1/2*d*x + 1/2*c)^13 + 749973*tan(1/2*d*x + 1/2*c)
^11 - 550089*tan(1/2*d*x + 1/2*c)^9 + 269879*tan(1/2*d*x + 1/2*c)^7 - 79723*tan(1/2*d*x + 1/2*c)^5 + 11375*tan
(1/2*d*x + 1/2*c)^3 - 315*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^8*a^3))/d

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maple [A]  time = 0.85, size = 396, normalized size = 1.67 \[ \frac {1}{8 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{8}}+\frac {13}{14 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{7}}+\frac {65}{24 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{6}}+\frac {143}{40 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}+\frac {79}{64 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {49}{32 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {29}{128 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {253}{128 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {125 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{128 a^{3} d}-\frac {1}{8 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8}}+\frac {13}{14 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {65}{24 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {143}{40 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {79}{64 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {49}{32 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {29}{128 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {253}{128 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {125 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{128 a^{3} d}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^12/(a+a*sec(d*x+c))^3,x)

[Out]

1/8/a^3/d/(tan(1/2*d*x+1/2*c)-1)^8+13/14/a^3/d/(tan(1/2*d*x+1/2*c)-1)^7+65/24/a^3/d/(tan(1/2*d*x+1/2*c)-1)^6+1
43/40/a^3/d/(tan(1/2*d*x+1/2*c)-1)^5+79/64/a^3/d/(tan(1/2*d*x+1/2*c)-1)^4-49/32/a^3/d/(tan(1/2*d*x+1/2*c)-1)^3
-29/128/a^3/d/(tan(1/2*d*x+1/2*c)-1)^2+253/128/a^3/d/(tan(1/2*d*x+1/2*c)-1)+125/128/a^3/d*ln(tan(1/2*d*x+1/2*c
)-1)-1/8/a^3/d/(tan(1/2*d*x+1/2*c)+1)^8+13/14/a^3/d/(tan(1/2*d*x+1/2*c)+1)^7-65/24/a^3/d/(tan(1/2*d*x+1/2*c)+1
)^6+143/40/a^3/d/(tan(1/2*d*x+1/2*c)+1)^5-79/64/a^3/d/(tan(1/2*d*x+1/2*c)+1)^4-49/32/a^3/d/(tan(1/2*d*x+1/2*c)
+1)^3+29/128/a^3/d/(tan(1/2*d*x+1/2*c)+1)^2+253/128/a^3/d/(tan(1/2*d*x+1/2*c)+1)-125/128/a^3/d*ln(tan(1/2*d*x+
1/2*c)+1)+2/d/a^3*arctan(tan(1/2*d*x+1/2*c))

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maxima [A]  time = 0.62, size = 429, normalized size = 1.81 \[ -\frac {\frac {2 \, {\left (\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {11375 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {79723 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {269879 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {550089 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {749973 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} + \frac {212625 \, \sin \left (d x + c\right )^{13}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{13}} - \frac {26565 \, \sin \left (d x + c\right )^{15}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{15}}\right )}}{a^{3} - \frac {8 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {28 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {56 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {70 \, a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {56 \, a^{3} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {28 \, a^{3} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}} - \frac {8 \, a^{3} \sin \left (d x + c\right )^{14}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{14}} + \frac {a^{3} \sin \left (d x + c\right )^{16}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{16}}} - \frac {26880 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} + \frac {13125 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} - \frac {13125 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}}{13440 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^12/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/13440*(2*(315*sin(d*x + c)/(cos(d*x + c) + 1) - 11375*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 79723*sin(d*x +
c)^5/(cos(d*x + c) + 1)^5 - 269879*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 550089*sin(d*x + c)^9/(cos(d*x + c)
+ 1)^9 - 749973*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 212625*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 - 26565*s
in(d*x + c)^15/(cos(d*x + c) + 1)^15)/(a^3 - 8*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 28*a^3*sin(d*x + c)^4
/(cos(d*x + c) + 1)^4 - 56*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 70*a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^
8 - 56*a^3*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 28*a^3*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 - 8*a^3*sin(d*
x + c)^14/(cos(d*x + c) + 1)^14 + a^3*sin(d*x + c)^16/(cos(d*x + c) + 1)^16) - 26880*arctan(sin(d*x + c)/(cos(
d*x + c) + 1))/a^3 + 13125*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 - 13125*log(sin(d*x + c)/(cos(d*x + c)
+ 1) - 1)/a^3)/d

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mupad [B]  time = 2.56, size = 265, normalized size = 1.12 \[ \frac {x}{a^3}-\frac {125\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{64\,a^3\,d}-\frac {-\frac {253\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{64}+\frac {2025\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{64}-\frac {35713\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{320}+\frac {183363\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{2240}-\frac {269879\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{6720}+\frac {11389\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{960}-\frac {325\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{192}+\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}-8\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+28\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-56\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+70\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-56\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+28\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-8\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^12/(a + a/cos(c + d*x))^3,x)

[Out]

x/a^3 - (125*atanh(tan(c/2 + (d*x)/2)))/(64*a^3*d) - ((3*tan(c/2 + (d*x)/2))/64 - (325*tan(c/2 + (d*x)/2)^3)/1
92 + (11389*tan(c/2 + (d*x)/2)^5)/960 - (269879*tan(c/2 + (d*x)/2)^7)/6720 + (183363*tan(c/2 + (d*x)/2)^9)/224
0 - (35713*tan(c/2 + (d*x)/2)^11)/320 + (2025*tan(c/2 + (d*x)/2)^13)/64 - (253*tan(c/2 + (d*x)/2)^15)/64)/(d*(
28*a^3*tan(c/2 + (d*x)/2)^4 - 8*a^3*tan(c/2 + (d*x)/2)^2 - 56*a^3*tan(c/2 + (d*x)/2)^6 + 70*a^3*tan(c/2 + (d*x
)/2)^8 - 56*a^3*tan(c/2 + (d*x)/2)^10 + 28*a^3*tan(c/2 + (d*x)/2)^12 - 8*a^3*tan(c/2 + (d*x)/2)^14 + a^3*tan(c
/2 + (d*x)/2)^16 + a^3))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\tan ^{12}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**12/(a+a*sec(d*x+c))**3,x)

[Out]

Integral(tan(c + d*x)**12/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3

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