\(\int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^3} \, dx\) [566]

Optimal result

Integrand size = 21, antiderivative size = 185 \[ \int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {3 \left (a^2+b^2\right ) \left (5 a^2+b^2\right ) \log (a+b \tan (c+d x))}{b^7 d}-\frac {a \left (10 a^2+9 b^2\right ) \tan (c+d x)}{b^6 d}+\frac {3 \left (2 a^2+b^2\right ) \tan ^2(c+d x)}{2 b^5 d}-\frac {a \tan ^3(c+d x)}{b^4 d}+\frac {\tan ^4(c+d x)}{4 b^3 d}-\frac {\left (a^2+b^2\right )^3}{2 b^7 d (a+b \tan (c+d x))^2}+\frac {6 a \left (a^2+b^2\right )^2}{b^7 d (a+b \tan (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 185, 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} \[ \int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {6 a \left (a^2+b^2\right )^2}{b^7 d (a+b \tan (c+d x))}-\frac {\left (a^2+b^2\right )^3}{2 b^7 d (a+b \tan (c+d x))^2}+\frac {3 \left (a^2+b^2\right ) \left (5 a^2+b^2\right ) \log (a+b \tan (c+d x))}{b^7 d}-\frac {a \left (10 a^2+9 b^2\right ) \tan (c+d x)}{b^6 d}+\frac {3 \left (2 a^2+b^2\right ) \tan ^2(c+d x)}{2 b^5 d}-\frac {a \tan ^3(c+d x)}{b^4 d}+\frac {\tan ^4(c+d x)}{4 b^3 d} \]

[In]

Int[Sec[c + d*x]^8/(a + b*Tan[c + d*x])^3,x]

[Out]

(3*(a^2 + b^2)*(5*a^2 + b^2)*Log[a + b*Tan[c + d*x]])/(b^7*d) - (a*(10*a^2 + 9*b^2)*Tan[c + d*x])/(b^6*d) + (3
*(2*a^2 + b^2)*Tan[c + d*x]^2)/(2*b^5*d) - (a*Tan[c + d*x]^3)/(b^4*d) + Tan[c + d*x]^4/(4*b^3*d) - (a^2 + b^2)
^3/(2*b^7*d*(a + b*Tan[c + d*x])^2) + (6*a*(a^2 + b^2)^2)/(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)^3} \, dx,x,b \tan (c+d x)\right )}{b d} \\ & = \frac {\text {Subst}\left (\int \left (\frac {-10 a^3-9 a b^2}{b^6}+\frac {3 \left (2 a^2+b^2\right ) x}{b^6}-\frac {3 a x^2}{b^6}+\frac {x^3}{b^6}+\frac {\left (a^2+b^2\right )^3}{b^6 (a+x)^3}-\frac {6 a \left (a^2+b^2\right )^2}{b^6 (a+x)^2}+\frac {3 \left (5 a^4+6 a^2 b^2+b^4\right )}{b^6 (a+x)}\right ) \, dx,x,b \tan (c+d x)\right )}{b d} \\ & = \frac {3 \left (a^2+b^2\right ) \left (5 a^2+b^2\right ) \log (a+b \tan (c+d x))}{b^7 d}-\frac {a \left (10 a^2+9 b^2\right ) \tan (c+d x)}{b^6 d}+\frac {3 \left (2 a^2+b^2\right ) \tan ^2(c+d x)}{2 b^5 d}-\frac {a \tan ^3(c+d x)}{b^4 d}+\frac {\tan ^4(c+d x)}{4 b^3 d}-\frac {\left (a^2+b^2\right )^3}{2 b^7 d (a+b \tan (c+d x))^2}+\frac {6 a \left (a^2+b^2\right )^2}{b^7 d (a+b \tan (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.51 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.47 \[ \int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {2 \left (a^2+b^2\right ) \left (19 a^4+16 a^2 b^2-3 b^4+6 a^2 \left (5 a^2+b^2\right ) \log (a+b \tan (c+d x))\right )+b^6 \sec ^6(c+d x)+4 a b \left (4 a^4+17 a^2 b^2+11 b^4+6 \left (5 a^4+6 a^2 b^2+b^4\right ) \log (a+b \tan (c+d x))\right ) \tan (c+d x)+4 b^2 \left (-13 a^4-10 a^2 b^2+3 \left (5 a^4+6 a^2 b^2+b^4\right ) \log (a+b \tan (c+d x))\right ) \tan ^2(c+d x)-20 a b^3 \left (a^2+b^2\right ) \tan ^3(c+d x)+4 a^2 b^4 \tan ^4(c+d x)+b^4 \sec ^4(c+d x) \left (a^2+3 b^2-2 a b \tan (c+d x)\right )}{4 b^7 d (a+b \tan (c+d x))^2} \]

[In]

Integrate[Sec[c + d*x]^8/(a + b*Tan[c + d*x])^3,x]

[Out]

(2*(a^2 + b^2)*(19*a^4 + 16*a^2*b^2 - 3*b^4 + 6*a^2*(5*a^2 + b^2)*Log[a + b*Tan[c + d*x]]) + b^6*Sec[c + d*x]^
6 + 4*a*b*(4*a^4 + 17*a^2*b^2 + 11*b^4 + 6*(5*a^4 + 6*a^2*b^2 + b^4)*Log[a + b*Tan[c + d*x]])*Tan[c + d*x] + 4
*b^2*(-13*a^4 - 10*a^2*b^2 + 3*(5*a^4 + 6*a^2*b^2 + b^4)*Log[a + b*Tan[c + d*x]])*Tan[c + d*x]^2 - 20*a*b^3*(a
^2 + b^2)*Tan[c + d*x]^3 + 4*a^2*b^4*Tan[c + d*x]^4 + b^4*Sec[c + d*x]^4*(a^2 + 3*b^2 - 2*a*b*Tan[c + d*x]))/(
4*b^7*d*(a + b*Tan[c + d*x])^2)

Maple [A] (verified)

Time = 1.08 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.05

\[\frac {\frac {\frac {\left (\tan ^{4}\left (d x +c \right )\right ) b^{3}}{4}-a \left (\tan ^{3}\left (d x +c \right )\right ) b^{2}+3 a^{2} b \left (\tan ^{2}\left (d x +c \right )\right )+\frac {3 b^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{2}-10 a^{3} \tan \left (d x +c \right )-9 a \,b^{2} \tan \left (d x +c \right )}{b^{6}}+\frac {6 a \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}{b^{7} \left (a +b \tan \left (d x +c \right )\right )}+\frac {\left (15 a^{4}+18 a^{2} b^{2}+3 b^{4}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{7}}-\frac {a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}}{2 b^{7} \left (a +b \tan \left (d x +c \right )\right )^{2}}}{d}\]

[In]

int(sec(d*x+c)^8/(a+b*tan(d*x+c))^3,x)

[Out]

1/d*(1/b^6*(1/4*tan(d*x+c)^4*b^3-a*tan(d*x+c)^3*b^2+3*a^2*b*tan(d*x+c)^2+3/2*b^3*tan(d*x+c)^2-10*a^3*tan(d*x+c
)-9*a*b^2*tan(d*x+c))+6*a/b^7*(a^4+2*a^2*b^2+b^4)/(a+b*tan(d*x+c))+(15*a^4+18*a^2*b^2+3*b^4)/b^7*ln(a+b*tan(d*
x+c))-1/2/b^7*(a^6+3*a^4*b^2+3*a^2*b^4+b^6)/(a+b*tan(d*x+c))^2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 476 vs. \(2 (179) = 358\).

Time = 0.31 (sec) , antiderivative size = 476, normalized size of antiderivative = 2.57 \[ \int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {8 \, {\left (15 \, a^{4} b^{2} + 13 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{6} + b^{6} - 2 \, {\left (45 \, a^{4} b^{2} + 44 \, a^{2} b^{4} + 3 \, b^{6}\right )} \cos \left (d x + c\right )^{4} + {\left (5 \, a^{2} b^{4} + 3 \, b^{6}\right )} \cos \left (d x + c\right )^{2} + 6 \, {\left ({\left (5 \, a^{6} + a^{4} b^{2} - 5 \, a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )^{6} + 2 \, {\left (5 \, a^{5} b + 6 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )^{5} \sin \left (d x + c\right ) + {\left (5 \, a^{4} b^{2} + 6 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - 6 \, {\left ({\left (5 \, a^{6} + a^{4} b^{2} - 5 \, a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )^{6} + 2 \, {\left (5 \, a^{5} b + 6 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )^{5} \sin \left (d x + c\right ) + {\left (5 \, a^{4} b^{2} + 6 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\cos \left (d x + c\right )^{2}\right ) - 2 \, {\left (a b^{5} \cos \left (d x + c\right ) + 2 \, {\left (15 \, a^{5} b - 2 \, a^{3} b^{3} - 13 \, a b^{5}\right )} \cos \left (d x + c\right )^{5} + 10 \, {\left (a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{4 \, {\left (2 \, a b^{8} d \cos \left (d x + c\right )^{5} \sin \left (d x + c\right ) + b^{9} d \cos \left (d x + c\right )^{4} + {\left (a^{2} b^{7} - b^{9}\right )} d \cos \left (d x + c\right )^{6}\right )}} \]

[In]

integrate(sec(d*x+c)^8/(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/4*(8*(15*a^4*b^2 + 13*a^2*b^4)*cos(d*x + c)^6 + b^6 - 2*(45*a^4*b^2 + 44*a^2*b^4 + 3*b^6)*cos(d*x + c)^4 + (
5*a^2*b^4 + 3*b^6)*cos(d*x + c)^2 + 6*((5*a^6 + a^4*b^2 - 5*a^2*b^4 - b^6)*cos(d*x + c)^6 + 2*(5*a^5*b + 6*a^3
*b^3 + a*b^5)*cos(d*x + c)^5*sin(d*x + c) + (5*a^4*b^2 + 6*a^2*b^4 + b^6)*cos(d*x + c)^4)*log(2*a*b*cos(d*x +
c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2) - 6*((5*a^6 + a^4*b^2 - 5*a^2*b^4 - b^6)*cos(d*x + c)^6 +
2*(5*a^5*b + 6*a^3*b^3 + a*b^5)*cos(d*x + c)^5*sin(d*x + c) + (5*a^4*b^2 + 6*a^2*b^4 + b^6)*cos(d*x + c)^4)*lo
g(cos(d*x + c)^2) - 2*(a*b^5*cos(d*x + c) + 2*(15*a^5*b - 2*a^3*b^3 - 13*a*b^5)*cos(d*x + c)^5 + 10*(a^3*b^3 +
a*b^5)*cos(d*x + c)^3)*sin(d*x + c))/(2*a*b^8*d*cos(d*x + c)^5*sin(d*x + c) + b^9*d*cos(d*x + c)^4 + (a^2*b^7
- b^9)*d*cos(d*x + c)^6)

Sympy [F]

\[ \int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\int \frac {\sec ^{8}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{3}}\, dx \]

[In]

integrate(sec(d*x+c)**8/(a+b*tan(d*x+c))**3,x)

[Out]

Integral(sec(c + d*x)**8/(a + b*tan(c + d*x))**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.08 \[ \int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\frac {2 \, {\left (11 \, a^{6} + 21 \, a^{4} b^{2} + 9 \, a^{2} b^{4} - b^{6} + 12 \, {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \tan \left (d x + c\right )\right )}}{b^{9} \tan \left (d x + c\right )^{2} + 2 \, a b^{8} \tan \left (d x + c\right ) + a^{2} b^{7}} + \frac {b^{3} \tan \left (d x + c\right )^{4} - 4 \, a b^{2} \tan \left (d x + c\right )^{3} + 6 \, {\left (2 \, a^{2} b + b^{3}\right )} \tan \left (d x + c\right )^{2} - 4 \, {\left (10 \, a^{3} + 9 \, a b^{2}\right )} \tan \left (d x + c\right )}{b^{6}} + \frac {12 \, {\left (5 \, a^{4} + 6 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{b^{7}}}{4 \, d} \]

[In]

integrate(sec(d*x+c)^8/(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/4*(2*(11*a^6 + 21*a^4*b^2 + 9*a^2*b^4 - b^6 + 12*(a^5*b + 2*a^3*b^3 + a*b^5)*tan(d*x + c))/(b^9*tan(d*x + c)
^2 + 2*a*b^8*tan(d*x + c) + a^2*b^7) + (b^3*tan(d*x + c)^4 - 4*a*b^2*tan(d*x + c)^3 + 6*(2*a^2*b + b^3)*tan(d*
x + c)^2 - 4*(10*a^3 + 9*a*b^2)*tan(d*x + c))/b^6 + 12*(5*a^4 + 6*a^2*b^2 + b^4)*log(b*tan(d*x + c) + a)/b^7)/
d

Giac [A] (verification not implemented)

none

Time = 0.62 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.31 \[ \int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\frac {12 \, {\left (5 \, a^{4} + 6 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{b^{7}} - \frac {2 \, {\left (45 \, a^{4} b^{2} \tan \left (d x + c\right )^{2} + 54 \, a^{2} b^{4} \tan \left (d x + c\right )^{2} + 9 \, b^{6} \tan \left (d x + c\right )^{2} + 78 \, a^{5} b \tan \left (d x + c\right ) + 84 \, a^{3} b^{3} \tan \left (d x + c\right ) + 6 \, a b^{5} \tan \left (d x + c\right ) + 34 \, a^{6} + 33 \, a^{4} b^{2} + b^{6}\right )}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{2} b^{7}} + \frac {b^{9} \tan \left (d x + c\right )^{4} - 4 \, a b^{8} \tan \left (d x + c\right )^{3} + 12 \, a^{2} b^{7} \tan \left (d x + c\right )^{2} + 6 \, b^{9} \tan \left (d x + c\right )^{2} - 40 \, a^{3} b^{6} \tan \left (d x + c\right ) - 36 \, a b^{8} \tan \left (d x + c\right )}{b^{12}}}{4 \, d} \]

[In]

integrate(sec(d*x+c)^8/(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/4*(12*(5*a^4 + 6*a^2*b^2 + b^4)*log(abs(b*tan(d*x + c) + a))/b^7 - 2*(45*a^4*b^2*tan(d*x + c)^2 + 54*a^2*b^4
*tan(d*x + c)^2 + 9*b^6*tan(d*x + c)^2 + 78*a^5*b*tan(d*x + c) + 84*a^3*b^3*tan(d*x + c) + 6*a*b^5*tan(d*x + c
) + 34*a^6 + 33*a^4*b^2 + b^6)/((b*tan(d*x + c) + a)^2*b^7) + (b^9*tan(d*x + c)^4 - 4*a*b^8*tan(d*x + c)^3 + 1
2*a^2*b^7*tan(d*x + c)^2 + 6*b^9*tan(d*x + c)^2 - 40*a^3*b^6*tan(d*x + c) - 36*a*b^8*tan(d*x + c))/b^12)/d

Mupad [B] (verification not implemented)

Time = 4.49 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.26 \[ \int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\frac {11\,a^6+21\,a^4\,b^2+9\,a^2\,b^4-b^6}{2\,b}+\mathrm {tan}\left (c+d\,x\right )\,\left (6\,a^5+12\,a^3\,b^2+6\,a\,b^4\right )}{d\,\left (a^2\,b^6+2\,a\,b^7\,\mathrm {tan}\left (c+d\,x\right )+b^8\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {3}{2\,b^3}+\frac {3\,a^2}{b^5}\right )}{d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^4}{4\,b^3\,d}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {8\,a^3}{b^6}-\frac {3\,a\,\left (\frac {3}{b^3}+\frac {6\,a^2}{b^5}\right )}{b}\right )}{d}-\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^3}{b^4\,d}+\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (15\,a^4+18\,a^2\,b^2+3\,b^4\right )}{b^7\,d} \]

[In]

int(1/(cos(c + d*x)^8*(a + b*tan(c + d*x))^3),x)

[Out]

((11*a^6 - b^6 + 9*a^2*b^4 + 21*a^4*b^2)/(2*b) + tan(c + d*x)*(6*a*b^4 + 6*a^5 + 12*a^3*b^2))/(d*(a^2*b^6 + b^
8*tan(c + d*x)^2 + 2*a*b^7*tan(c + d*x))) + (tan(c + d*x)^2*(3/(2*b^3) + (3*a^2)/b^5))/d + tan(c + d*x)^4/(4*b
^3*d) + (tan(c + d*x)*((8*a^3)/b^6 - (3*a*(3/b^3 + (6*a^2)/b^5))/b))/d - (a*tan(c + d*x)^3)/(b^4*d) + (log(a +
b*tan(c + d*x))*(15*a^4 + 3*b^4 + 18*a^2*b^2))/(b^7*d)