\(\int (a+b \sec (c+d x))^2 \tan ^6(c+d x) \, dx\) [279]

Optimal result

Integrand size = 21, antiderivative size = 157 \[ \int (a+b \sec (c+d x))^2 \tan ^6(c+d x) \, dx=-a^2 x-\frac {5 a b \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^2 \tan (c+d x)}{d}+\frac {5 a b \sec (c+d x) \tan (c+d x)}{8 d}-\frac {a^2 \tan ^3(c+d x)}{3 d}-\frac {5 a b \sec (c+d x) \tan ^3(c+d x)}{12 d}+\frac {a^2 \tan ^5(c+d x)}{5 d}+\frac {a b \sec (c+d x) \tan ^5(c+d x)}{3 d}+\frac {b^2 \tan ^7(c+d x)}{7 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3971, 3554, 8, 2691, 3855, 2687, 30} \[ \int (a+b \sec (c+d x))^2 \tan ^6(c+d x) \, dx=\frac {a^2 \tan ^5(c+d x)}{5 d}-\frac {a^2 \tan ^3(c+d x)}{3 d}+\frac {a^2 \tan (c+d x)}{d}-a^2 x-\frac {5 a b \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a b \tan ^5(c+d x) \sec (c+d x)}{3 d}-\frac {5 a b \tan ^3(c+d x) \sec (c+d x)}{12 d}+\frac {5 a b \tan (c+d x) \sec (c+d x)}{8 d}+\frac {b^2 \tan ^7(c+d x)}{7 d} \]

[In]

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

[Out]

-(a^2*x) - (5*a*b*ArcTanh[Sin[c + d*x]])/(8*d) + (a^2*Tan[c + d*x])/d + (5*a*b*Sec[c + d*x]*Tan[c + d*x])/(8*d
) - (a^2*Tan[c + d*x]^3)/(3*d) - (5*a*b*Sec[c + d*x]*Tan[c + d*x]^3)/(12*d) + (a^2*Tan[c + d*x]^5)/(5*d) + (a*
b*Sec[c + d*x]*Tan[c + d*x]^5)/(3*d) + (b^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*} \text {integral}& = \int \left (a^2 \tan ^6(c+d x)+2 a b \sec (c+d x) \tan ^6(c+d x)+b^2 \sec ^2(c+d x) \tan ^6(c+d x)\right ) \, dx \\ & = a^2 \int \tan ^6(c+d x) \, dx+(2 a b) \int \sec (c+d x) \tan ^6(c+d x) \, dx+b^2 \int \sec ^2(c+d x) \tan ^6(c+d x) \, dx \\ & = \frac {a^2 \tan ^5(c+d x)}{5 d}+\frac {a b \sec (c+d x) \tan ^5(c+d x)}{3 d}-a^2 \int \tan ^4(c+d x) \, dx-\frac {1}{3} (5 a b) \int \sec (c+d x) \tan ^4(c+d x) \, dx+\frac {b^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 b \sec (c+d x) \tan ^3(c+d x)}{12 d}+\frac {a^2 \tan ^5(c+d x)}{5 d}+\frac {a b \sec (c+d x) \tan ^5(c+d x)}{3 d}+\frac {b^2 \tan ^7(c+d x)}{7 d}+a^2 \int \tan ^2(c+d x) \, dx+\frac {1}{4} (5 a b) \int \sec (c+d x) \tan ^2(c+d x) \, dx \\ & = \frac {a^2 \tan (c+d x)}{d}+\frac {5 a b \sec (c+d x) \tan (c+d x)}{8 d}-\frac {a^2 \tan ^3(c+d x)}{3 d}-\frac {5 a b \sec (c+d x) \tan ^3(c+d x)}{12 d}+\frac {a^2 \tan ^5(c+d x)}{5 d}+\frac {a b \sec (c+d x) \tan ^5(c+d x)}{3 d}+\frac {b^2 \tan ^7(c+d x)}{7 d}-a^2 \int 1 \, dx-\frac {1}{8} (5 a b) \int \sec (c+d x) \, dx \\ & = -a^2 x-\frac {5 a b \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^2 \tan (c+d x)}{d}+\frac {5 a b \sec (c+d x) \tan (c+d x)}{8 d}-\frac {a^2 \tan ^3(c+d x)}{3 d}-\frac {5 a b \sec (c+d x) \tan ^3(c+d x)}{12 d}+\frac {a^2 \tan ^5(c+d x)}{5 d}+\frac {a b \sec (c+d x) \tan ^5(c+d x)}{3 d}+\frac {b^2 \tan ^7(c+d x)}{7 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.26 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.84 \[ \int (a+b \sec (c+d x))^2 \tan ^6(c+d x) \, dx=\frac {-840 a^2 \arctan (\tan (c+d x))-525 a b \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (175 a b (-1+7 \cos (2 (c+d x))) \sec ^5(c+d x)+105 a b \sec (c+d x) \left (-5+16 \tan ^4(c+d x)\right )+8 \left (105 a^2-35 a^2 \tan ^2(c+d x)+21 a^2 \tan ^4(c+d x)+15 b^2 \tan ^6(c+d x)\right )\right )}{840 d} \]

[In]

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

[Out]

(-840*a^2*ArcTan[Tan[c + d*x]] - 525*a*b*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(175*a*b*(-1 + 7*Cos[2*(c + d*x)
])*Sec[c + d*x]^5 + 105*a*b*Sec[c + d*x]*(-5 + 16*Tan[c + d*x]^4) + 8*(105*a^2 - 35*a^2*Tan[c + d*x]^2 + 21*a^
2*Tan[c + d*x]^4 + 15*b^2*Tan[c + d*x]^6)))/(840*d)

Maple [A] (verified)

Time = 3.66 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.06

method	result	size
parts	\(\frac {a^{2} \left (\frac {\tan \left (d x +c \right )^{5}}{5}-\frac {\tan \left (d x +c \right )^{3}}{3}+\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {b^{2} \tan \left (d x +c \right )^{7}}{7 d}+\frac {2 a b \left (\frac {\sin \left (d x +c \right )^{7}}{6 \cos \left (d x +c \right )^{6}}-\frac {\sin \left (d x +c \right )^{7}}{24 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{7}}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{5}}{16}+\frac {5 \sin \left (d x +c \right )^{3}}{48}+\frac {5 \sin \left (d x +c \right )}{16}-\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}\)	\(167\)
derivativedivides	\(\frac {a^{2} \left (\frac {\tan \left (d x +c \right )^{5}}{5}-\frac {\tan \left (d x +c \right )^{3}}{3}+\tan \left (d x +c \right )-d x -c \right )+2 a b \left (\frac {\sin \left (d x +c \right )^{7}}{6 \cos \left (d x +c \right )^{6}}-\frac {\sin \left (d x +c \right )^{7}}{24 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{7}}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{5}}{16}+\frac {5 \sin \left (d x +c \right )^{3}}{48}+\frac {5 \sin \left (d x +c \right )}{16}-\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+\frac {b^{2} \sin \left (d x +c \right )^{7}}{7 \cos \left (d x +c \right )^{7}}}{d}\)	\(168\)
default	\(\frac {a^{2} \left (\frac {\tan \left (d x +c \right )^{5}}{5}-\frac {\tan \left (d x +c \right )^{3}}{3}+\tan \left (d x +c \right )-d x -c \right )+2 a b \left (\frac {\sin \left (d x +c \right )^{7}}{6 \cos \left (d x +c \right )^{6}}-\frac {\sin \left (d x +c \right )^{7}}{24 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{7}}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{5}}{16}+\frac {5 \sin \left (d x +c \right )^{3}}{48}+\frac {5 \sin \left (d x +c \right )}{16}-\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+\frac {b^{2} \sin \left (d x +c \right )^{7}}{7 \cos \left (d x +c \right )^{7}}}{d}\)	\(168\)
risch	\(-a^{2} x -\frac {i \left (1155 a b \,{\mathrm e}^{13 i \left (d x +c \right )}-2520 a^{2} {\mathrm e}^{12 i \left (d x +c \right )}+840 b^{2} {\mathrm e}^{12 i \left (d x +c \right )}+980 a b \,{\mathrm e}^{11 i \left (d x +c \right )}-10080 a^{2} {\mathrm e}^{10 i \left (d x +c \right )}+2975 a b \,{\mathrm e}^{9 i \left (d x +c \right )}-20440 a^{2} {\mathrm e}^{8 i \left (d x +c \right )}+4200 b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-24640 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-2975 a b \,{\mathrm e}^{5 i \left (d x +c \right )}-16968 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+2520 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-980 a b \,{\mathrm e}^{3 i \left (d x +c \right )}-6496 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-1155 a b \,{\mathrm e}^{i \left (d x +c \right )}-1288 a^{2}+120 b^{2}\right )}{420 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{7}}+\frac {5 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {5 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}\)	\(282\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.17 \[ \int (a+b \sec (c+d x))^2 \tan ^6(c+d x) \, dx=-\frac {1680 \, a^{2} d x \cos \left (d x + c\right )^{7} + 525 \, a b \cos \left (d x + c\right )^{7} \log \left (\sin \left (d x + c\right ) + 1\right ) - 525 \, a b \cos \left (d x + c\right )^{7} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (1155 \, a b \cos \left (d x + c\right )^{5} + 8 \, {\left (161 \, a^{2} - 15 \, b^{2}\right )} \cos \left (d x + c\right )^{6} - 910 \, a b \cos \left (d x + c\right )^{3} - 8 \, {\left (77 \, a^{2} - 45 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + 280 \, a b \cos \left (d x + c\right ) + 24 \, {\left (7 \, a^{2} - 15 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 120 \, b^{2}\right )} \sin \left (d x + c\right )}{1680 \, d \cos \left (d x + c\right )^{7}} \]

[In]

integrate((a+b*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*b*cos(d*x + c)^7*log(sin(d*x + c) + 1) - 525*a*b*cos(d*x + c)^7*l
og(-sin(d*x + c) + 1) - 2*(1155*a*b*cos(d*x + c)^5 + 8*(161*a^2 - 15*b^2)*cos(d*x + c)^6 - 910*a*b*cos(d*x + c
)^3 - 8*(77*a^2 - 45*b^2)*cos(d*x + c)^4 + 280*a*b*cos(d*x + c) + 24*(7*a^2 - 15*b^2)*cos(d*x + c)^2 + 120*b^2
)*sin(d*x + c))/(d*cos(d*x + c)^7)

Sympy [F]

\[ \int (a+b \sec (c+d x))^2 \tan ^6(c+d x) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \tan ^{6}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.96 \[ \int (a+b \sec (c+d x))^2 \tan ^6(c+d x) \, dx=\frac {240 \, b^{2} \tan \left (d x + c\right )^{7} + 112 \, {\left (3 \, \tan \left (d x + c\right )^{5} - 5 \, \tan \left (d x + c\right )^{3} - 15 \, d x - 15 \, c + 15 \, \tan \left (d x + c\right )\right )} a^{2} - 35 \, a b {\left (\frac {2 \, {\left (33 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} + 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{1680 \, d} \]

[In]

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

[Out]

1/1680*(240*b^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*b*(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

Giac [A] (verification not implemented)

none

Time = 2.30 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.80 \[ \int (a+b \sec (c+d x))^2 \tan ^6(c+d x) \, dx=-\frac {840 \, {\left (d x + c\right )} a^{2} + 525 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 525 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (840 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} - 525 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} - 6160 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 3500 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 19768 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 9905 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 28896 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 7680 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 19768 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9905 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6160 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3500 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 840 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 525 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{7}}}{840 \, d} \]

[In]

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

[Out]

-1/840*(840*(d*x + c)*a^2 + 525*a*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 525*a*b*log(abs(tan(1/2*d*x + 1/2*c)
- 1)) + 2*(840*a^2*tan(1/2*d*x + 1/2*c)^13 - 525*a*b*tan(1/2*d*x + 1/2*c)^13 - 6160*a^2*tan(1/2*d*x + 1/2*c)^1
1 + 3500*a*b*tan(1/2*d*x + 1/2*c)^11 + 19768*a^2*tan(1/2*d*x + 1/2*c)^9 - 9905*a*b*tan(1/2*d*x + 1/2*c)^9 - 28
896*a^2*tan(1/2*d*x + 1/2*c)^7 + 7680*b^2*tan(1/2*d*x + 1/2*c)^7 + 19768*a^2*tan(1/2*d*x + 1/2*c)^5 + 9905*a*b
*tan(1/2*d*x + 1/2*c)^5 - 6160*a^2*tan(1/2*d*x + 1/2*c)^3 - 3500*a*b*tan(1/2*d*x + 1/2*c)^3 + 840*a^2*tan(1/2*
d*x + 1/2*c) + 525*a*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^7)/d

Mupad [B] (verification not implemented)

Time = 15.42 (sec) , antiderivative size = 403, normalized size of antiderivative = 2.57 \[ \int (a+b \sec (c+d x))^2 \tan ^6(c+d x) \, dx=\frac {\left (\frac {5\,a\,b}{4}-2\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}+\left (\frac {44\,a^2}{3}-\frac {25\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {283\,a\,b}{12}-\frac {706\,a^2}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {344\,a^2}{5}-\frac {128\,b^2}{7}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (-\frac {706\,a^2}{15}-\frac {283\,b\,a}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {44\,a^2}{3}+\frac {25\,b\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-2\,a^2-\frac {5\,b\,a}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}-7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {2\,a^2\,\mathrm {atan}\left (\frac {64\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,a^6+25\,a^4\,b^2}+\frac {25\,a^4\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,a^6+25\,a^4\,b^2}\right )}{d}-\frac {5\,a\,b\,\mathrm {atanh}\left (\frac {40\,a^5\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{40\,a^5\,b+\frac {125\,a^3\,b^3}{8}}+\frac {125\,a^3\,b^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,\left (40\,a^5\,b+\frac {125\,a^3\,b^3}{8}\right )}\right )}{4\,d} \]

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^7*((344*a^2)/5 - (128*b^2)/7) + tan(c/2 + (d*x)/2)^13*((5*a*b)/4 - 2*a^2) + tan(c/2 + (d*x
)/2)^3*((25*a*b)/3 + (44*a^2)/3) - tan(c/2 + (d*x)/2)^11*((25*a*b)/3 - (44*a^2)/3) - tan(c/2 + (d*x)/2)^5*((28
3*a*b)/12 + (706*a^2)/15) + tan(c/2 + (d*x)/2)^9*((283*a*b)/12 - (706*a^2)/15) - tan(c/2 + (d*x)/2)*((5*a*b)/4
+ 2*a^2))/(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)) - (2*a^2*atan((64*a^6
*tan(c/2 + (d*x)/2))/(64*a^6 + 25*a^4*b^2) + (25*a^4*b^2*tan(c/2 + (d*x)/2))/(64*a^6 + 25*a^4*b^2)))/d - (5*a*
b*atanh((40*a^5*b*tan(c/2 + (d*x)/2))/(40*a^5*b + (125*a^3*b^3)/8) + (125*a^3*b^3*tan(c/2 + (d*x)/2))/(8*(40*a
^5*b + (125*a^3*b^3)/8))))/(4*d)