\(\int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^2} \, dx\) [554]

Optimal result

Integrand size = 21, antiderivative size = 178 \[ \int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {6 a \left (a^2+b^2\right )^2 \log (a+b \tan (c+d x))}{b^7 d}+\frac {\left (5 a^4+9 a^2 b^2+3 b^4\right ) \tan (c+d x)}{b^6 d}-\frac {a \left (2 a^2+3 b^2\right ) \tan ^2(c+d x)}{b^5 d}+\frac {\left (a^2+b^2\right ) \tan ^3(c+d x)}{b^4 d}-\frac {a \tan ^4(c+d x)}{2 b^3 d}+\frac {\tan ^5(c+d x)}{5 b^2 d}-\frac {\left (a^2+b^2\right )^3}{b^7 d (a+b \tan (c+d x))} \]

[Out]

-6*a*(a^2+b^2)^2*ln(a+b*tan(d*x+c))/b^7/d+(5*a^4+9*a^2*b^2+3*b^4)*tan(d*x+c)/b^6/d-a*(2*a^2+3*b^2)*tan(d*x+c)^
2/b^5/d+(a^2+b^2)*tan(d*x+c)^3/b^4/d-1/2*a*tan(d*x+c)^4/b^3/d+1/5*tan(d*x+c)^5/b^2/d-(a^2+b^2)^3/b^7/d/(a+b*ta
n(d*x+c))

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3587, 711} \[ \int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {\left (a^2+b^2\right )^3}{b^7 d (a+b \tan (c+d x))}-\frac {6 a \left (a^2+b^2\right )^2 \log (a+b \tan (c+d x))}{b^7 d}-\frac {a \left (2 a^2+3 b^2\right ) \tan ^2(c+d x)}{b^5 d}+\frac {\left (a^2+b^2\right ) \tan ^3(c+d x)}{b^4 d}+\frac {\left (5 a^4+9 a^2 b^2+3 b^4\right ) \tan (c+d x)}{b^6 d}-\frac {a \tan ^4(c+d x)}{2 b^3 d}+\frac {\tan ^5(c+d x)}{5 b^2 d} \]

[In]

Int[Sec[c + d*x]^8/(a + b*Tan[c + d*x])^2,x]

[Out]

(-6*a*(a^2 + b^2)^2*Log[a + b*Tan[c + d*x]])/(b^7*d) + ((5*a^4 + 9*a^2*b^2 + 3*b^4)*Tan[c + d*x])/(b^6*d) - (a
*(2*a^2 + 3*b^2)*Tan[c + d*x]^2)/(b^5*d) + ((a^2 + b^2)*Tan[c + d*x]^3)/(b^4*d) - (a*Tan[c + d*x]^4)/(2*b^3*d)
+ Tan[c + d*x]^5/(5*b^2*d) - (a^2 + b^2)^3/(b^7*d*(a + b*Tan[c + d*x]))

Rule 711

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rule 3587

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (1+\frac {x^2}{b^2}\right )^3}{(a+x)^2} \, dx,x,b \tan (c+d x)\right )}{b d} \\ & = \frac {\text {Subst}\left (\int \left (\frac {5 a^4+9 a^2 b^2+3 b^4}{b^6}-\frac {2 a \left (2 a^2+3 b^2\right ) x}{b^6}+\frac {3 \left (a^2+b^2\right ) x^2}{b^6}-\frac {2 a x^3}{b^6}+\frac {x^4}{b^6}+\frac {\left (a^2+b^2\right )^3}{b^6 (a+x)^2}-\frac {6 a \left (a^2+b^2\right )^2}{b^6 (a+x)}\right ) \, dx,x,b \tan (c+d x)\right )}{b d} \\ & = -\frac {6 a \left (a^2+b^2\right )^2 \log (a+b \tan (c+d x))}{b^7 d}+\frac {\left (5 a^4+9 a^2 b^2+3 b^4\right ) \tan (c+d x)}{b^6 d}-\frac {a \left (2 a^2+3 b^2\right ) \tan ^2(c+d x)}{b^5 d}+\frac {\left (a^2+b^2\right ) \tan ^3(c+d x)}{b^4 d}-\frac {a \tan ^4(c+d x)}{2 b^3 d}+\frac {\tan ^5(c+d x)}{5 b^2 d}-\frac {\left (a^2+b^2\right )^3}{b^7 d (a+b \tan (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.62 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.29 \[ \int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {2 b^6 \sec ^6(c+d x)+b^4 \sec ^4(c+d x) \left (a^2+4 b^2-3 a b \tan (c+d x)\right )-2 \left (8 \left (a^2+b^2\right )^3+30 a^2 \left (a^2+b^2\right )^2 \log (a+b \tan (c+d x))+2 a b \left (-11 a^4-18 a^2 b^2-4 b^4+15 \left (a^2+b^2\right )^2 \log (a+b \tan (c+d x))\right ) \tan (c+d x)-b^2 \left (15 a^4+29 a^2 b^2+8 b^4\right ) \tan ^2(c+d x)+a b^3 \left (5 a^2+7 b^2\right ) \tan ^3(c+d x)-2 a^2 b^4 \tan ^4(c+d x)\right )}{10 b^7 d (a+b \tan (c+d x))} \]

[In]

Integrate[Sec[c + d*x]^8/(a + b*Tan[c + d*x])^2,x]

[Out]

(2*b^6*Sec[c + d*x]^6 + b^4*Sec[c + d*x]^4*(a^2 + 4*b^2 - 3*a*b*Tan[c + d*x]) - 2*(8*(a^2 + b^2)^3 + 30*a^2*(a
^2 + b^2)^2*Log[a + b*Tan[c + d*x]] + 2*a*b*(-11*a^4 - 18*a^2*b^2 - 4*b^4 + 15*(a^2 + b^2)^2*Log[a + b*Tan[c +
d*x]])*Tan[c + d*x] - b^2*(15*a^4 + 29*a^2*b^2 + 8*b^4)*Tan[c + d*x]^2 + a*b^3*(5*a^2 + 7*b^2)*Tan[c + d*x]^3
- 2*a^2*b^4*Tan[c + d*x]^4))/(10*b^7*d*(a + b*Tan[c + d*x]))

Maple [A] (verified)

Time = 265.50 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.13

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (\tan ^{5}\left (d x +c \right )\right ) b^{4}}{5}-\frac {a \,b^{3} \left (\tan ^{4}\left (d x +c \right )\right )}{2}+a^{2} b^{2} \left (\tan ^{3}\left (d x +c \right )\right )+b^{4} \left (\tan ^{3}\left (d x +c \right )\right )-2 a^{3} b \left (\tan ^{2}\left (d x +c \right )\right )-3 a \,b^{3} \left (\tan ^{2}\left (d x +c \right )\right )+5 a^{4} \tan \left (d x +c \right )+9 a^{2} b^{2} \tan \left (d x +c \right )+3 b^{4} \tan \left (d x +c \right )}{b^{6}}-\frac {6 a \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{7}}-\frac {a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}}{b^{7} \left (a +b \tan \left (d x +c \right )\right )}}{d}\)	\(201\)
default	\(\frac {\frac {\frac {\left (\tan ^{5}\left (d x +c \right )\right ) b^{4}}{5}-\frac {a \,b^{3} \left (\tan ^{4}\left (d x +c \right )\right )}{2}+a^{2} b^{2} \left (\tan ^{3}\left (d x +c \right )\right )+b^{4} \left (\tan ^{3}\left (d x +c \right )\right )-2 a^{3} b \left (\tan ^{2}\left (d x +c \right )\right )-3 a \,b^{3} \left (\tan ^{2}\left (d x +c \right )\right )+5 a^{4} \tan \left (d x +c \right )+9 a^{2} b^{2} \tan \left (d x +c \right )+3 b^{4} \tan \left (d x +c \right )}{b^{6}}-\frac {6 a \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{7}}-\frac {a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}}{b^{7} \left (a +b \tan \left (d x +c \right )\right )}}{d}\)	\(201\)
risch	\(-\frac {4 i \left (25 a^{2} b^{3}+40 b^{5} {\mathrm e}^{4 i \left (d x +c \right )}+32 b^{5} {\mathrm e}^{2 i \left (d x +c \right )}-33 i a \,b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-30 i a^{3} b^{2} {\mathrm e}^{10 i \left (d x +c \right )}-135 i a^{3} b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-15 i a \,b^{4} {\mathrm e}^{10 i \left (d x +c \right )}-60 i a \,b^{4} {\mathrm e}^{8 i \left (d x +c \right )}-250 i a^{3} b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-80 i a \,b^{4} {\mathrm e}^{6 i \left (d x +c \right )}-240 i a^{3} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-60 i a \,b^{4} {\mathrm e}^{4 i \left (d x +c \right )}-120 i a^{3} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-15 i a^{5}+8 b^{5}+15 a^{4} b -25 i a^{3} b^{2}-8 i a \,b^{4}+90 a^{4} b \,{\mathrm e}^{4 i \left (d x +c \right )}+140 a^{2} b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-15 i a^{5} {\mathrm e}^{10 i \left (d x +c \right )}-150 i a^{5} {\mathrm e}^{6 i \left (d x +c \right )}-150 i a^{5} {\mathrm e}^{4 i \left (d x +c \right )}-75 i a^{5} {\mathrm e}^{2 i \left (d x +c \right )}+100 a^{2} b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-75 i a^{5} {\mathrm e}^{8 i \left (d x +c \right )}+60 a^{4} b \,{\mathrm e}^{2 i \left (d x +c \right )}+15 a^{4} b \,{\mathrm e}^{8 i \left (d x +c \right )}+15 a^{2} b^{3} {\mathrm e}^{8 i \left (d x +c \right )}+60 a^{4} b \,{\mathrm e}^{6 i \left (d x +c \right )}+80 a^{2} b^{3} {\mathrm e}^{6 i \left (d x +c \right )}\right )}{5 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right ) b^{6} d}-\frac {6 a^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{b^{7} d}-\frac {12 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{b^{5} d}-\frac {6 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{b^{3} d}+\frac {6 a^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b^{7} d}+\frac {12 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b^{5} d}+\frac {6 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b^{3} d}\)	\(686\)

[In]

int(sec(d*x+c)^8/(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(1/b^6*(1/5*tan(d*x+c)^5*b^4-1/2*a*b^3*tan(d*x+c)^4+a^2*b^2*tan(d*x+c)^3+b^4*tan(d*x+c)^3-2*a^3*b*tan(d*x+
c)^2-3*a*b^3*tan(d*x+c)^2+5*a^4*tan(d*x+c)+9*a^2*b^2*tan(d*x+c)+3*b^4*tan(d*x+c))-6*a/b^7*(a^4+2*a^2*b^2+b^4)*
ln(a+b*tan(d*x+c))-1/b^7*(a^6+3*a^4*b^2+3*a^2*b^4+b^6)/(a+b*tan(d*x+c)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 386 vs. \(2 (174) = 348\).

Time = 0.29 (sec) , antiderivative size = 386, normalized size of antiderivative = 2.17 \[ \int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {4 \, {\left (15 \, a^{4} b^{2} + 25 \, a^{2} b^{4} + 8 \, b^{6}\right )} \cos \left (d x + c\right )^{6} - 2 \, b^{6} - 2 \, {\left (15 \, a^{4} b^{2} + 25 \, a^{2} b^{4} + 8 \, b^{6}\right )} \cos \left (d x + c\right )^{4} - {\left (5 \, a^{2} b^{4} + 4 \, b^{6}\right )} \cos \left (d x + c\right )^{2} + 30 \, {\left ({\left (a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )^{6} + {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )^{5} \sin \left (d x + c\right )\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - 30 \, {\left ({\left (a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )^{6} + {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )^{5} \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right )^{2}\right ) + {\left (3 \, a b^{5} \cos \left (d x + c\right ) - 4 \, {\left (15 \, a^{5} b + 25 \, a^{3} b^{3} + 8 \, a b^{5}\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (5 \, a^{3} b^{3} + 7 \, a b^{5}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{10 \, {\left (a b^{7} d \cos \left (d x + c\right )^{6} + b^{8} d \cos \left (d x + c\right )^{5} \sin \left (d x + c\right )\right )}} \]

[In]

integrate(sec(d*x+c)^8/(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/10*(4*(15*a^4*b^2 + 25*a^2*b^4 + 8*b^6)*cos(d*x + c)^6 - 2*b^6 - 2*(15*a^4*b^2 + 25*a^2*b^4 + 8*b^6)*cos(d*
x + c)^4 - (5*a^2*b^4 + 4*b^6)*cos(d*x + c)^2 + 30*((a^6 + 2*a^4*b^2 + a^2*b^4)*cos(d*x + c)^6 + (a^5*b + 2*a^
3*b^3 + a*b^5)*cos(d*x + c)^5*sin(d*x + c))*log(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 +
b^2) - 30*((a^6 + 2*a^4*b^2 + a^2*b^4)*cos(d*x + c)^6 + (a^5*b + 2*a^3*b^3 + a*b^5)*cos(d*x + c)^5*sin(d*x +
c))*log(cos(d*x + c)^2) + (3*a*b^5*cos(d*x + c) - 4*(15*a^5*b + 25*a^3*b^3 + 8*a*b^5)*cos(d*x + c)^5 + 2*(5*a^
3*b^3 + 7*a*b^5)*cos(d*x + c)^3)*sin(d*x + c))/(a*b^7*d*cos(d*x + c)^6 + b^8*d*cos(d*x + c)^5*sin(d*x + c))

Sympy [F]

\[ \int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\int \frac {\sec ^{8}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{2}}\, dx \]

[In]

integrate(sec(d*x+c)**8/(a+b*tan(d*x+c))**2,x)

[Out]

Integral(sec(c + d*x)**8/(a + b*tan(c + d*x))**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.04 \[ \int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {\frac {10 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}}{b^{8} \tan \left (d x + c\right ) + a b^{7}} - \frac {2 \, b^{4} \tan \left (d x + c\right )^{5} - 5 \, a b^{3} \tan \left (d x + c\right )^{4} + 10 \, {\left (a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )^{3} - 10 \, {\left (2 \, a^{3} b + 3 \, a b^{3}\right )} \tan \left (d x + c\right )^{2} + 10 \, {\left (5 \, a^{4} + 9 \, a^{2} b^{2} + 3 \, b^{4}\right )} \tan \left (d x + c\right )}{b^{6}} + \frac {60 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{b^{7}}}{10 \, d} \]

[In]

integrate(sec(d*x+c)^8/(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/10*(10*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/(b^8*tan(d*x + c) + a*b^7) - (2*b^4*tan(d*x + c)^5 - 5*a*b^3*tan
(d*x + c)^4 + 10*(a^2*b^2 + b^4)*tan(d*x + c)^3 - 10*(2*a^3*b + 3*a*b^3)*tan(d*x + c)^2 + 10*(5*a^4 + 9*a^2*b^
2 + 3*b^4)*tan(d*x + c))/b^6 + 60*(a^5 + 2*a^3*b^2 + a*b^4)*log(b*tan(d*x + c) + a)/b^7)/d

Giac [A] (verification not implemented)

none

Time = 0.50 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.42 \[ \int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {\frac {60 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{b^{7}} - \frac {10 \, {\left (6 \, a^{5} b \tan \left (d x + c\right ) + 12 \, a^{3} b^{3} \tan \left (d x + c\right ) + 6 \, a b^{5} \tan \left (d x + c\right ) + 5 \, a^{6} + 9 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )}}{{\left (b \tan \left (d x + c\right ) + a\right )} b^{7}} - \frac {2 \, b^{8} \tan \left (d x + c\right )^{5} - 5 \, a b^{7} \tan \left (d x + c\right )^{4} + 10 \, a^{2} b^{6} \tan \left (d x + c\right )^{3} + 10 \, b^{8} \tan \left (d x + c\right )^{3} - 20 \, a^{3} b^{5} \tan \left (d x + c\right )^{2} - 30 \, a b^{7} \tan \left (d x + c\right )^{2} + 50 \, a^{4} b^{4} \tan \left (d x + c\right ) + 90 \, a^{2} b^{6} \tan \left (d x + c\right ) + 30 \, b^{8} \tan \left (d x + c\right )}{b^{10}}}{10 \, d} \]

[In]

integrate(sec(d*x+c)^8/(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/10*(60*(a^5 + 2*a^3*b^2 + a*b^4)*log(abs(b*tan(d*x + c) + a))/b^7 - 10*(6*a^5*b*tan(d*x + c) + 12*a^3*b^3*t
an(d*x + c) + 6*a*b^5*tan(d*x + c) + 5*a^6 + 9*a^4*b^2 + 3*a^2*b^4 - b^6)/((b*tan(d*x + c) + a)*b^7) - (2*b^8*
tan(d*x + c)^5 - 5*a*b^7*tan(d*x + c)^4 + 10*a^2*b^6*tan(d*x + c)^3 + 10*b^8*tan(d*x + c)^3 - 20*a^3*b^5*tan(d
*x + c)^2 - 30*a*b^7*tan(d*x + c)^2 + 50*a^4*b^4*tan(d*x + c) + 90*a^2*b^6*tan(d*x + c) + 30*b^8*tan(d*x + c))
/b^10)/d

Mupad [B] (verification not implemented)

Time = 4.14 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.45 \[ \int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {a^3}{b^5}-\frac {a\,\left (\frac {3}{b^2}+\frac {3\,a^2}{b^4}\right )}{b}\right )}{d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^5}{5\,b^2\,d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {1}{b^2}+\frac {a^2}{b^4}\right )}{d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {a^2\,\left (\frac {3}{b^2}+\frac {3\,a^2}{b^4}\right )}{b^2}-\frac {3}{b^2}+\frac {2\,a\,\left (\frac {2\,a^3}{b^5}-\frac {2\,a\,\left (\frac {3}{b^2}+\frac {3\,a^2}{b^4}\right )}{b}\right )}{b}\right )}{d}-\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^4}{2\,b^3\,d}-\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (6\,a^5+12\,a^3\,b^2+6\,a\,b^4\right )}{b^7\,d}-\frac {a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6}{b\,d\,\left (\mathrm {tan}\left (c+d\,x\right )\,b^7+a\,b^6\right )} \]

[In]

int(1/(cos(c + d*x)^8*(a + b*tan(c + d*x))^2),x)

[Out]

(tan(c + d*x)^2*(a^3/b^5 - (a*(3/b^2 + (3*a^2)/b^4))/b))/d + tan(c + d*x)^5/(5*b^2*d) + (tan(c + d*x)^3*(1/b^2
+ a^2/b^4))/d - (tan(c + d*x)*((a^2*(3/b^2 + (3*a^2)/b^4))/b^2 - 3/b^2 + (2*a*((2*a^3)/b^5 - (2*a*(3/b^2 + (3
*a^2)/b^4))/b))/b))/d - (a*tan(c + d*x)^4)/(2*b^3*d) - (log(a + b*tan(c + d*x))*(6*a*b^4 + 6*a^5 + 12*a^3*b^2)
)/(b^7*d) - (a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)/(b*d*(a*b^6 + b^7*tan(c + d*x)))