3.47 \(\int (a+a \sec (c+d x))^3 \tan ^6(c+d x) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.299389, antiderivative size = 237, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3886, 3473, 8, 2611, 3770, 2607, 30, 3768} \[ \frac{3 a^3 \tan ^7(c+d x)}{7 d}+\frac{a^3 \tan ^5(c+d x)}{5 d}-\frac{a^3 \tan ^3(c+d x)}{3 d}+\frac{a^3 \tan (c+d x)}{d}-\frac{125 a^3 \tanh ^{-1}(\sin (c+d x))}{128 d}+\frac{a^3 \tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac{5 a^3 \tan ^3(c+d x) \sec ^3(c+d x)}{48 d}+\frac{5 a^3 \tan (c+d x) \sec ^3(c+d x)}{64 d}+\frac{a^3 \tan ^5(c+d x) \sec (c+d x)}{2 d}-\frac{5 a^3 \tan ^3(c+d x) \sec (c+d x)}{8 d}+\frac{115 a^3 \tan (c+d x) \sec (c+d x)}{128 d}-a^3 x \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^3*Tan[c + d*x]^6,x]

[Out]

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

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 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 8

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

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 3770

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

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 30

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

Rubi steps

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

Mathematica [A]  time = 2.06622, size = 363, normalized size = 1.53 \[ \frac{a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac{1}{2} (c+d x)\right ) \sec ^8(c+d x) \left (1680000 \cos ^8(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-\sec (c) (-133175 \sin (2 c+d x)-544768 \sin (c+2 d x)+286720 \sin (3 c+2 d x)-63595 \sin (2 c+3 d x)-63595 \sin (4 c+3 d x)-254464 \sin (3 c+4 d x)+161280 \sin (5 c+4 d x)-65135 \sin (4 c+5 d x)-65135 \sin (6 c+5 d x)-118784 \sin (5 c+6 d x)-27195 \sin (6 c+7 d x)-27195 \sin (8 c+7 d x)-14848 \sin (7 c+8 d x)+470400 d x \cos (c)+376320 d x \cos (c+2 d x)+376320 d x \cos (3 c+2 d x)+188160 d x \cos (3 c+4 d x)+188160 d x \cos (5 c+4 d x)+53760 d x \cos (5 c+6 d x)+53760 d x \cos (7 c+6 d x)+6720 d x \cos (7 c+8 d x)+6720 d x \cos (9 c+8 d x)+519680 \sin (c)-133175 \sin (d x))\right )}{13762560 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^3*Tan[c + d*x]^6,x]

[Out]

(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*Sec[c + d*x]^8*(1680000*Cos[c + d*x]^8*(Log[Cos[(c + d*x)/2] - Si
n[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Sec[c]*(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*Co
s[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*S
in[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] - 14
848*Sin[7*c + 8*d*x])))/(13762560*d)

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Maple [A]  time = 0.056, size = 250, normalized size = 1.1 \begin{align*}{\frac{{a}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{5}}{5\,d}}-{\frac{{a}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{{a}^{3}\tan \left ( dx+c \right ) }{d}}-{a}^{3}x-{\frac{{a}^{3}c}{d}}+{\frac{25\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{48\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}-{\frac{25\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{192\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{25\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{128\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{25\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{128\,d}}+{\frac{125\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{384\,d}}+{\frac{125\,{a}^{3}\sin \left ( dx+c \right ) }{128\,d}}-{\frac{125\,{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{128\,d}}+{\frac{3\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{7\,d \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{8}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^3*tan(d*x+c)^6,x)

[Out]

1/5*a^3*tan(d*x+c)^5/d-1/3*a^3*tan(d*x+c)^3/d+a^3*tan(d*x+c)/d-a^3*x-1/d*a^3*c+25/48/d*a^3*sin(d*x+c)^7/cos(d*
x+c)^6-25/192/d*a^3*sin(d*x+c)^7/cos(d*x+c)^4+25/128/d*a^3*sin(d*x+c)^7/cos(d*x+c)^2+25/128/d*a^3*sin(d*x+c)^5
+125/384/d*a^3*sin(d*x+c)^3+125/128/d*a^3*sin(d*x+c)-125/128/d*a^3*ln(sec(d*x+c)+tan(d*x+c))+3/7/d*a^3*sin(d*x
+c)^7/cos(d*x+c)^7+1/8/d*a^3*sin(d*x+c)^7/cos(d*x+c)^8

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Maxima [A]  time = 1.75189, size = 354, normalized size = 1.49 \begin{align*} \frac{11520 \, a^{3} \tan \left (d x + c\right )^{7} + 1792 \,{\left (3 \, \tan \left (d x + c\right )^{5} - 5 \, \tan \left (d x + c\right )^{3} - 15 \, d x - 15 \, c + 15 \, \tan \left (d x + c\right )\right )} a^{3} + 35 \, a^{3}{\left (\frac{2 \,{\left (15 \, \sin \left (d x + c\right )^{7} + 73 \, \sin \left (d x + c\right )^{5} - 55 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{8} - 4 \, \sin \left (d x + c\right )^{6} + 6 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{2} + 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 840 \, a^{3}{\left (\frac{2 \,{\left (33 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} + 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{26880 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3*tan(d*x+c)^6,x, algorithm="maxima")

[Out]

1/26880*(11520*a^3*tan(d*x + c)^7 + 1792*(3*tan(d*x + c)^5 - 5*tan(d*x + c)^3 - 15*d*x - 15*c + 15*tan(d*x + c
))*a^3 + 35*a^3*(2*(15*sin(d*x + c)^7 + 73*sin(d*x + c)^5 - 55*sin(d*x + c)^3 + 15*sin(d*x + c))/(sin(d*x + c)
^8 - 4*sin(d*x + c)^6 + 6*sin(d*x + c)^4 - 4*sin(d*x + c)^2 + 1) - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x +
c) - 1)) - 840*a^3*(2*(33*sin(d*x + c)^5 - 40*sin(d*x + c)^3 + 15*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x +
c)^4 + 3*sin(d*x + c)^2 - 1) + 15*log(sin(d*x + c) + 1) - 15*log(sin(d*x + c) - 1)))/d

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Fricas [A]  time = 1.312, size = 501, normalized size = 2.11 \begin{align*} -\frac{26880 \, a^{3} d x \cos \left (d x + c\right )^{8} + 13125 \, a^{3} \cos \left (d x + c\right )^{8} \log \left (\sin \left (d x + c\right ) + 1\right ) - 13125 \, a^{3} \cos \left (d x + c\right )^{8} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (14848 \, a^{3} \cos \left (d x + c\right )^{7} + 27195 \, a^{3} \cos \left (d x + c\right )^{6} + 7424 \, a^{3} \cos \left (d x + c\right )^{5} - 17710 \, a^{3} \cos \left (d x + c\right )^{4} - 14592 \, a^{3} \cos \left (d x + c\right )^{3} + 1960 \, a^{3} \cos \left (d x + c\right )^{2} + 5760 \, a^{3} \cos \left (d x + c\right ) + 1680 \, a^{3}\right )} \sin \left (d x + c\right )}{26880 \, d \cos \left (d x + c\right )^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3*tan(d*x+c)^6,x, algorithm="fricas")

[Out]

-1/26880*(26880*a^3*d*x*cos(d*x + c)^8 + 13125*a^3*cos(d*x + c)^8*log(sin(d*x + c) + 1) - 13125*a^3*cos(d*x +
c)^8*log(-sin(d*x + c) + 1) - 2*(14848*a^3*cos(d*x + c)^7 + 27195*a^3*cos(d*x + c)^6 + 7424*a^3*cos(d*x + c)^5
- 17710*a^3*cos(d*x + c)^4 - 14592*a^3*cos(d*x + c)^3 + 1960*a^3*cos(d*x + c)^2 + 5760*a^3*cos(d*x + c) + 168
0*a^3)*sin(d*x + c))/(d*cos(d*x + c)^8)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int 3 \tan ^{6}{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int 3 \tan ^{6}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \tan ^{6}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int \tan ^{6}{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**3*tan(d*x+c)**6,x)

[Out]

a**3*(Integral(3*tan(c + d*x)**6*sec(c + d*x), x) + Integral(3*tan(c + d*x)**6*sec(c + d*x)**2, x) + Integral(
tan(c + d*x)**6*sec(c + d*x)**3, x) + Integral(tan(c + d*x)**6, x))

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Giac [A]  time = 4.85306, size = 265, normalized size = 1.12 \begin{align*} -\frac{13440 \,{\left (d x + c\right )} a^{3} + 13125 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 13125 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (315 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{15} - 11375 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{13} + 79723 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} - 269879 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 550089 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 749973 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 212625 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 26565 \, a^{3} \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}}}{13440 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3*tan(d*x+c)^6,x, algorithm="giac")

[Out]

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