3.1.29 \(\int (a+a \sec (c+d x))^2 \tan ^6(c+d x) \, dx\) [29]

Optimal. Leaf size=161 \[ -a^2 x-\frac {5 a^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 \tan (c+d x)}{d}+\frac {5 a^2 \sec (c+d x) \tan (c+d x)}{8 d}-\frac {a^2 \tan ^3(c+d x)}{3 d}-\frac {5 a^2 \sec (c+d x) \tan ^3(c+d x)}{12 d}+\frac {a^2 \tan ^5(c+d x)}{5 d}+\frac {a^2 \sec (c+d x) \tan ^5(c+d x)}{3 d}+\frac {a^2 \tan ^7(c+d x)}{7 d} \]

[Out]

-a^2*x-5/8*a^2*arctanh(sin(d*x+c))/d+a^2*tan(d*x+c)/d+5/8*a^2*sec(d*x+c)*tan(d*x+c)/d-1/3*a^2*tan(d*x+c)^3/d-5
/12*a^2*sec(d*x+c)*tan(d*x+c)^3/d+1/5*a^2*tan(d*x+c)^5/d+1/3*a^2*sec(d*x+c)*tan(d*x+c)^5/d+1/7*a^2*tan(d*x+c)^
7/d

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Rubi [A]
time = 0.14, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3971, 3554, 8, 2691, 3855, 2687, 30} \begin {gather*} \frac {a^2 \tan ^7(c+d x)}{7 d}+\frac {a^2 \tan ^5(c+d x)}{5 d}-\frac {a^2 \tan ^3(c+d x)}{3 d}+\frac {a^2 \tan (c+d x)}{d}-\frac {5 a^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 \tan ^5(c+d x) \sec (c+d x)}{3 d}-\frac {5 a^2 \tan ^3(c+d x) \sec (c+d x)}{12 d}+\frac {5 a^2 \tan (c+d x) \sec (c+d x)}{8 d}-a^2 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^2*Tan[c + d*x]^6,x]

[Out]

-(a^2*x) - (5*a^2*ArcTanh[Sin[c + d*x]])/(8*d) + (a^2*Tan[c + d*x])/d + (5*a^2*Sec[c + d*x]*Tan[c + d*x])/(8*d
) - (a^2*Tan[c + d*x]^3)/(3*d) - (5*a^2*Sec[c + d*x]*Tan[c + d*x]^3)/(12*d) + (a^2*Tan[c + d*x]^5)/(5*d) + (a^
2*Sec[c + d*x]*Tan[c + d*x]^5)/(3*d) + (a^2*Tan[c + d*x]^7)/(7*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 2687

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 2691

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 3554

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 3855

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

Rule 3971

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]

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(337\) vs. \(2(161)=322\).
time = 1.47, size = 337, normalized size = 2.09 \begin {gather*} \frac {a^2 (1+\cos (c+d x))^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^7(c+d x) \left (33600 \cos ^7(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 (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec (c) (-14700 d x \cos (d x)-14700 d x \cos (2 c+d x)-8820 d x \cos (2 c+3 d x)-8820 d x \cos (4 c+3 d x)-2940 d x \cos (4 c+5 d x)-2940 d x \cos (6 c+5 d x)-420 d x \cos (6 c+7 d x)-420 d x \cos (8 c+7 d x)+24640 \sin (d x)-16240 \sin (2 c+d x)+2975 \sin (c+2 d x)+2975 \sin (3 c+2 d x)+14448 \sin (2 c+3 d x)-10080 \sin (4 c+3 d x)+980 \sin (3 c+4 d x)+980 \sin (5 c+4 d x)+6496 \sin (4 c+5 d x)-1680 \sin (6 c+5 d x)+1155 \sin (5 c+6 d x)+1155 \sin (7 c+6 d x)+1168 \sin (6 c+7 d x))\right )}{215040 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^2*Tan[c + d*x]^6,x]

[Out]

(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*Sec[c + d*x]^7*(33600*Cos[c + d*x]^7*(Log[Cos[(c + d*x)/2] - Sin[
(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Sec[c]*(-14700*d*x*Cos[d*x] - 14700*d*x*Cos[2*c +
d*x] - 8820*d*x*Cos[2*c + 3*d*x] - 8820*d*x*Cos[4*c + 3*d*x] - 2940*d*x*Cos[4*c + 5*d*x] - 2940*d*x*Cos[6*c +
5*d*x] - 420*d*x*Cos[6*c + 7*d*x] - 420*d*x*Cos[8*c + 7*d*x] + 24640*Sin[d*x] - 16240*Sin[2*c + d*x] + 2975*Si
n[c + 2*d*x] + 2975*Sin[3*c + 2*d*x] + 14448*Sin[2*c + 3*d*x] - 10080*Sin[4*c + 3*d*x] + 980*Sin[3*c + 4*d*x]
+ 980*Sin[5*c + 4*d*x] + 6496*Sin[4*c + 5*d*x] - 1680*Sin[6*c + 5*d*x] + 1155*Sin[5*c + 6*d*x] + 1155*Sin[7*c
+ 6*d*x] + 1168*Sin[6*c + 7*d*x])))/(215040*d)

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Maple [A]
time = 0.12, size = 169, normalized size = 1.05

method	result	size
derivativedivides	\(\frac {\frac {a^{2} \left (\sin ^{7}\left (d x +c \right )\right )}{7 \cos \left (d x +c \right )^{7}}+2 a^{2} \left (\frac {\sin ^{7}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}-\frac {\sin ^{7}\left (d x +c \right )}{24 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{7}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{16}+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{48}+\frac {5 \sin \left (d x +c \right )}{16}-\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+a^{2} \left (\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}+\tan \left (d x +c \right )-d x -c \right )}{d}\)	\(169\)
default	\(\frac {\frac {a^{2} \left (\sin ^{7}\left (d x +c \right )\right )}{7 \cos \left (d x +c \right )^{7}}+2 a^{2} \left (\frac {\sin ^{7}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}-\frac {\sin ^{7}\left (d x +c \right )}{24 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{7}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{16}+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{48}+\frac {5 \sin \left (d x +c \right )}{16}-\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+a^{2} \left (\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}+\tan \left (d x +c \right )-d x -c \right )}{d}\)	\(169\)
risch	\(-a^{2} x -\frac {i a^{2} \left (1155 \,{\mathrm e}^{13 i \left (d x +c \right )}-1680 \,{\mathrm e}^{12 i \left (d x +c \right )}+980 \,{\mathrm e}^{11 i \left (d x +c \right )}-10080 \,{\mathrm e}^{10 i \left (d x +c \right )}+2975 \,{\mathrm e}^{9 i \left (d x +c \right )}-16240 \,{\mathrm e}^{8 i \left (d x +c \right )}-24640 \,{\mathrm e}^{6 i \left (d x +c \right )}-2975 \,{\mathrm e}^{5 i \left (d x +c \right )}-14448 \,{\mathrm e}^{4 i \left (d x +c \right )}-980 \,{\mathrm e}^{3 i \left (d x +c \right )}-6496 \,{\mathrm e}^{2 i \left (d x +c \right )}-1155 \,{\mathrm e}^{i \left (d x +c \right )}-1168\right )}{420 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{7}}+\frac {5 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {5 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}\)	\(206\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^2*tan(d*x+c)^6,x,method=_RETURNVERBOSE)

[Out]

1/d*(1/7*a^2*sin(d*x+c)^7/cos(d*x+c)^7+2*a^2*(1/6*sin(d*x+c)^7/cos(d*x+c)^6-1/24*sin(d*x+c)^7/cos(d*x+c)^4+1/1
6*sin(d*x+c)^7/cos(d*x+c)^2+1/16*sin(d*x+c)^5+5/48*sin(d*x+c)^3+5/16*sin(d*x+c)-5/16*ln(sec(d*x+c)+tan(d*x+c))
)+a^2*(1/5*tan(d*x+c)^5-1/3*tan(d*x+c)^3+tan(d*x+c)-d*x-c))

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Maxima [A]
time = 0.47, size = 151, normalized size = 0.94 \begin {gather*} \frac {240 \, a^{2} \tan \left (d x + c\right )^{7} + 112 \, {\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^{2} - 35 \, a^{2} {\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 )}}{1680 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*tan(d*x+c)^6,x, algorithm="maxima")

[Out]

1/1680*(240*a^2*tan(d*x + c)^7 + 112*(3*tan(d*x + c)^5 - 5*tan(d*x + c)^3 - 15*d*x - 15*c + 15*tan(d*x + c))*a
^2 - 35*a^2*(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 = 2.91, size = 165, normalized size = 1.02 \begin {gather*} -\frac {1680 \, a^{2} d x \cos \left (d x + c\right )^{7} + 525 \, a^{2} \cos \left (d x + c\right )^{7} \log \left (\sin \left (d x + c\right ) + 1\right ) - 525 \, a^{2} \cos \left (d x + c\right )^{7} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (1168 \, a^{2} \cos \left (d x + c\right )^{6} + 1155 \, a^{2} \cos \left (d x + c\right )^{5} - 256 \, a^{2} \cos \left (d x + c\right )^{4} - 910 \, a^{2} \cos \left (d x + c\right )^{3} - 192 \, a^{2} \cos \left (d x + c\right )^{2} + 280 \, a^{2} \cos \left (d x + c\right ) + 120 \, a^{2}\right )} \sin \left (d x + c\right )}{1680 \, d \cos \left (d x + c\right )^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*tan(d*x+c)^6,x, algorithm="fricas")

[Out]

-1/1680*(1680*a^2*d*x*cos(d*x + c)^7 + 525*a^2*cos(d*x + c)^7*log(sin(d*x + c) + 1) - 525*a^2*cos(d*x + c)^7*l
og(-sin(d*x + c) + 1) - 2*(1168*a^2*cos(d*x + c)^6 + 1155*a^2*cos(d*x + c)^5 - 256*a^2*cos(d*x + c)^4 - 910*a^
2*cos(d*x + c)^3 - 192*a^2*cos(d*x + c)^2 + 280*a^2*cos(d*x + c) + 120*a^2)*sin(d*x + c))/(d*cos(d*x + c)^7)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{2} \left (\int 2 \tan ^{6}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \tan ^{6}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \tan ^{6}{\left (c + d x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 2.66, size = 180, normalized size = 1.12 \begin {gather*} -\frac {840 \, {\left (d x + c\right )} a^{2} + 525 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 525 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (315 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} - 2660 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 9863 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 21216 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 29673 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9660 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1365 \, a^{2} \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 )}^{7}}}{840 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*tan(d*x+c)^6,x, algorithm="giac")

[Out]

-1/840*(840*(d*x + c)*a^2 + 525*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 525*a^2*log(abs(tan(1/2*d*x + 1/2*c)
- 1)) + 2*(315*a^2*tan(1/2*d*x + 1/2*c)^13 - 2660*a^2*tan(1/2*d*x + 1/2*c)^11 + 9863*a^2*tan(1/2*d*x + 1/2*c)^
9 - 21216*a^2*tan(1/2*d*x + 1/2*c)^7 + 29673*a^2*tan(1/2*d*x + 1/2*c)^5 - 9660*a^2*tan(1/2*d*x + 1/2*c)^3 + 13
65*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^7)/d

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Mupad [B]
time = 2.65, size = 234, normalized size = 1.45 \begin {gather*} -a^2\,x-\frac {5\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}-\frac {\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{4}-\frac {19\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{3}+\frac {1409\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{60}-\frac {1768\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{35}+\frac {1413\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}-23\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {13\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}-7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^6*(a + a/cos(c + d*x))^2,x)

[Out]

- a^2*x - (5*a^2*atanh(tan(c/2 + (d*x)/2)))/(4*d) - ((1413*a^2*tan(c/2 + (d*x)/2)^5)/20 - 23*a^2*tan(c/2 + (d*
x)/2)^3 - (1768*a^2*tan(c/2 + (d*x)/2)^7)/35 + (1409*a^2*tan(c/2 + (d*x)/2)^9)/60 - (19*a^2*tan(c/2 + (d*x)/2)
^11)/3 + (3*a^2*tan(c/2 + (d*x)/2)^13)/4 + (13*a^2*tan(c/2 + (d*x)/2))/4)/(d*(7*tan(c/2 + (d*x)/2)^2 - 21*tan(
c/2 + (d*x)/2)^4 + 35*tan(c/2 + (d*x)/2)^6 - 35*tan(c/2 + (d*x)/2)^8 + 21*tan(c/2 + (d*x)/2)^10 - 7*tan(c/2 +
(d*x)/2)^12 + tan(c/2 + (d*x)/2)^14 - 1))

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