\(\int \frac {\tan ^9(c+d x)}{a+b \sec (c+d x)} \, dx\) [286]

Optimal result

Integrand size = 21, antiderivative size = 250 \[ \int \frac {\tan ^9(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {\log (\cos (c+d x))}{a d}-\frac {\left (a^2-b^2\right )^4 \log (a+b \sec (c+d x))}{a b^8 d}+\frac {\left (a^6-4 a^4 b^2+6 a^2 b^4-4 b^6\right ) \sec (c+d x)}{b^7 d}-\frac {a \left (a^4-4 a^2 b^2+6 b^4\right ) \sec ^2(c+d x)}{2 b^6 d}+\frac {\left (a^4-4 a^2 b^2+6 b^4\right ) \sec ^3(c+d x)}{3 b^5 d}-\frac {a \left (a^2-4 b^2\right ) \sec ^4(c+d x)}{4 b^4 d}+\frac {\left (a^2-4 b^2\right ) \sec ^5(c+d x)}{5 b^3 d}-\frac {a \sec ^6(c+d x)}{6 b^2 d}+\frac {\sec ^7(c+d x)}{7 b d} \]

[Out]

-ln(cos(d*x+c))/a/d-(a^2-b^2)^4*ln(a+b*sec(d*x+c))/a/b^8/d+(a^6-4*a^4*b^2+6*a^2*b^4-4*b^6)*sec(d*x+c)/b^7/d-1/
2*a*(a^4-4*a^2*b^2+6*b^4)*sec(d*x+c)^2/b^6/d+1/3*(a^4-4*a^2*b^2+6*b^4)*sec(d*x+c)^3/b^5/d-1/4*a*(a^2-4*b^2)*se
c(d*x+c)^4/b^4/d+1/5*(a^2-4*b^2)*sec(d*x+c)^5/b^3/d-1/6*a*sec(d*x+c)^6/b^2/d+1/7*sec(d*x+c)^7/b/d

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 250, 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 = {3970, 908} \[ \int \frac {\tan ^9(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {\left (a^2-b^2\right )^4 \log (a+b \sec (c+d x))}{a b^8 d}-\frac {a \left (a^2-4 b^2\right ) \sec ^4(c+d x)}{4 b^4 d}+\frac {\left (a^2-4 b^2\right ) \sec ^5(c+d x)}{5 b^3 d}-\frac {a \left (a^4-4 a^2 b^2+6 b^4\right ) \sec ^2(c+d x)}{2 b^6 d}+\frac {\left (a^4-4 a^2 b^2+6 b^4\right ) \sec ^3(c+d x)}{3 b^5 d}+\frac {\left (a^6-4 a^4 b^2+6 a^2 b^4-4 b^6\right ) \sec (c+d x)}{b^7 d}-\frac {a \sec ^6(c+d x)}{6 b^2 d}-\frac {\log (\cos (c+d x))}{a d}+\frac {\sec ^7(c+d x)}{7 b d} \]

[In]

Int[Tan[c + d*x]^9/(a + b*Sec[c + d*x]),x]

[Out]

-(Log[Cos[c + d*x]]/(a*d)) - ((a^2 - b^2)^4*Log[a + b*Sec[c + d*x]])/(a*b^8*d) + ((a^6 - 4*a^4*b^2 + 6*a^2*b^4
- 4*b^6)*Sec[c + d*x])/(b^7*d) - (a*(a^4 - 4*a^2*b^2 + 6*b^4)*Sec[c + d*x]^2)/(2*b^6*d) + ((a^4 - 4*a^2*b^2 +
6*b^4)*Sec[c + d*x]^3)/(3*b^5*d) - (a*(a^2 - 4*b^2)*Sec[c + d*x]^4)/(4*b^4*d) + ((a^2 - 4*b^2)*Sec[c + d*x]^5
)/(5*b^3*d) - (a*Sec[c + d*x]^6)/(6*b^2*d) + Sec[c + d*x]^7/(7*b*d)

Rule 908

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 3970

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[-(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[(b^2 - x^2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (b^2-x^2\right )^4}{x (a+x)} \, dx,x,b \sec (c+d x)\right )}{b^8 d} \\ & = \frac {\text {Subst}\left (\int \left (a^6 \left (1+\frac {-4 a^4 b^2+6 a^2 b^4-4 b^6}{a^6}\right )+\frac {b^8}{a x}-a \left (a^4-4 a^2 b^2+6 b^4\right ) x+\left (a^4-4 a^2 b^2+6 b^4\right ) x^2-a \left (a^2-4 b^2\right ) x^3+\left (a^2-4 b^2\right ) x^4-a x^5+x^6-\frac {\left (a^2-b^2\right )^4}{a (a+x)}\right ) \, dx,x,b \sec (c+d x)\right )}{b^8 d} \\ & = -\frac {\log (\cos (c+d x))}{a d}-\frac {\left (a^2-b^2\right )^4 \log (a+b \sec (c+d x))}{a b^8 d}+\frac {\left (a^6-4 a^4 b^2+6 a^2 b^4-4 b^6\right ) \sec (c+d x)}{b^7 d}-\frac {a \left (a^4-4 a^2 b^2+6 b^4\right ) \sec ^2(c+d x)}{2 b^6 d}+\frac {\left (a^4-4 a^2 b^2+6 b^4\right ) \sec ^3(c+d x)}{3 b^5 d}-\frac {a \left (a^2-4 b^2\right ) \sec ^4(c+d x)}{4 b^4 d}+\frac {\left (a^2-4 b^2\right ) \sec ^5(c+d x)}{5 b^3 d}-\frac {a \sec ^6(c+d x)}{6 b^2 d}+\frac {\sec ^7(c+d x)}{7 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.91 (sec) , antiderivative size = 228, normalized size of antiderivative = 0.91 \[ \int \frac {\tan ^9(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {-\frac {b^8 \log (\cos (c+d x))}{a}-\frac {\left (a^2-b^2\right )^4 \log (a+b \sec (c+d x))}{a}+b \left (a^6-4 a^4 b^2+6 a^2 b^4-4 b^6\right ) \sec (c+d x)-\frac {1}{2} a b^2 \left (a^4-4 a^2 b^2+6 b^4\right ) \sec ^2(c+d x)+\frac {1}{3} b^3 \left (a^4-4 a^2 b^2+6 b^4\right ) \sec ^3(c+d x)-\frac {1}{4} a b^4 \left (a^2-4 b^2\right ) \sec ^4(c+d x)+\frac {1}{5} b^5 \left (a^2-4 b^2\right ) \sec ^5(c+d x)-\frac {1}{6} a b^6 \sec ^6(c+d x)+\frac {1}{7} b^7 \sec ^7(c+d x)}{b^8 d} \]

[In]

Integrate[Tan[c + d*x]^9/(a + b*Sec[c + d*x]),x]

[Out]

(-((b^8*Log[Cos[c + d*x]])/a) - ((a^2 - b^2)^4*Log[a + b*Sec[c + d*x]])/a + b*(a^6 - 4*a^4*b^2 + 6*a^2*b^4 - 4
*b^6)*Sec[c + d*x] - (a*b^2*(a^4 - 4*a^2*b^2 + 6*b^4)*Sec[c + d*x]^2)/2 + (b^3*(a^4 - 4*a^2*b^2 + 6*b^4)*Sec[c
+ d*x]^3)/3 - (a*b^4*(a^2 - 4*b^2)*Sec[c + d*x]^4)/4 + (b^5*(a^2 - 4*b^2)*Sec[c + d*x]^5)/5 - (a*b^6*Sec[c +
d*x]^6)/6 + (b^7*Sec[c + d*x]^7)/7)/(b^8*d)

Maple [A] (verified)

Time = 2.18 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(\frac {\frac {\left (-a^{8}+4 a^{6} b^{2}-6 a^{4} b^{4}+4 a^{2} b^{6}-b^{8}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{b^{8} a}-\frac {a}{6 b^{2} \cos \left (d x +c \right )^{6}}-\frac {-a^{2}+4 b^{2}}{5 b^{3} \cos \left (d x +c \right )^{5}}-\frac {-a^{4}+4 a^{2} b^{2}-6 b^{4}}{3 b^{5} \cos \left (d x +c \right )^{3}}-\frac {-a^{6}+4 a^{4} b^{2}-6 a^{2} b^{4}+4 b^{6}}{b^{7} \cos \left (d x +c \right )}-\frac {\left (a^{2}-4 b^{2}\right ) a}{4 b^{4} \cos \left (d x +c \right )^{4}}-\frac {\left (a^{4}-4 a^{2} b^{2}+6 b^{4}\right ) a}{2 b^{6} \cos \left (d x +c \right )^{2}}+\frac {\left (a^{6}-4 a^{4} b^{2}+6 a^{2} b^{4}-4 b^{6}\right ) a \ln \left (\cos \left (d x +c \right )\right )}{b^{8}}+\frac {1}{7 b \cos \left (d x +c \right )^{7}}}{d}\)	\(273\)
default	\(\frac {\frac {\left (-a^{8}+4 a^{6} b^{2}-6 a^{4} b^{4}+4 a^{2} b^{6}-b^{8}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{b^{8} a}-\frac {a}{6 b^{2} \cos \left (d x +c \right )^{6}}-\frac {-a^{2}+4 b^{2}}{5 b^{3} \cos \left (d x +c \right )^{5}}-\frac {-a^{4}+4 a^{2} b^{2}-6 b^{4}}{3 b^{5} \cos \left (d x +c \right )^{3}}-\frac {-a^{6}+4 a^{4} b^{2}-6 a^{2} b^{4}+4 b^{6}}{b^{7} \cos \left (d x +c \right )}-\frac {\left (a^{2}-4 b^{2}\right ) a}{4 b^{4} \cos \left (d x +c \right )^{4}}-\frac {\left (a^{4}-4 a^{2} b^{2}+6 b^{4}\right ) a}{2 b^{6} \cos \left (d x +c \right )^{2}}+\frac {\left (a^{6}-4 a^{4} b^{2}+6 a^{2} b^{4}-4 b^{6}\right ) a \ln \left (\cos \left (d x +c \right )\right )}{b^{8}}+\frac {1}{7 b \cos \left (d x +c \right )^{7}}}{d}\)	\(273\)
risch	\(\text {Expression too large to display}\)	\(1031\)

[In]

int(tan(d*x+c)^9/(a+b*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*((-a^8+4*a^6*b^2-6*a^4*b^4+4*a^2*b^6-b^8)/b^8/a*ln(b+a*cos(d*x+c))-1/6/b^2*a/cos(d*x+c)^6-1/5*(-a^2+4*b^2)
/b^3/cos(d*x+c)^5-1/3*(-a^4+4*a^2*b^2-6*b^4)/b^5/cos(d*x+c)^3-(-a^6+4*a^4*b^2-6*a^2*b^4+4*b^6)/b^7/cos(d*x+c)-
1/4*(a^2-4*b^2)/b^4*a/cos(d*x+c)^4-1/2*(a^4-4*a^2*b^2+6*b^4)/b^6*a/cos(d*x+c)^2+(a^6-4*a^4*b^2+6*a^2*b^4-4*b^6
)/b^8*a*ln(cos(d*x+c))+1/7/b/cos(d*x+c)^7)

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.17 \[ \int \frac {\tan ^9(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {70 \, a^{2} b^{6} \cos \left (d x + c\right ) + 420 \, {\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} \cos \left (d x + c\right )^{7} \log \left (a \cos \left (d x + c\right ) + b\right ) - 420 \, {\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6}\right )} \cos \left (d x + c\right )^{7} \log \left (-\cos \left (d x + c\right )\right ) - 60 \, a b^{7} - 420 \, {\left (a^{7} b - 4 \, a^{5} b^{3} + 6 \, a^{3} b^{5} - 4 \, a b^{7}\right )} \cos \left (d x + c\right )^{6} + 210 \, {\left (a^{6} b^{2} - 4 \, a^{4} b^{4} + 6 \, a^{2} b^{6}\right )} \cos \left (d x + c\right )^{5} - 140 \, {\left (a^{5} b^{3} - 4 \, a^{3} b^{5} + 6 \, a b^{7}\right )} \cos \left (d x + c\right )^{4} + 105 \, {\left (a^{4} b^{4} - 4 \, a^{2} b^{6}\right )} \cos \left (d x + c\right )^{3} - 84 \, {\left (a^{3} b^{5} - 4 \, a b^{7}\right )} \cos \left (d x + c\right )^{2}}{420 \, a b^{8} d \cos \left (d x + c\right )^{7}} \]

[In]

integrate(tan(d*x+c)^9/(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

-1/420*(70*a^2*b^6*cos(d*x + c) + 420*(a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*cos(d*x + c)^7*log(a*cos
(d*x + c) + b) - 420*(a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6)*cos(d*x + c)^7*log(-cos(d*x + c)) - 60*a*b^7 -
420*(a^7*b - 4*a^5*b^3 + 6*a^3*b^5 - 4*a*b^7)*cos(d*x + c)^6 + 210*(a^6*b^2 - 4*a^4*b^4 + 6*a^2*b^6)*cos(d*x +
c)^5 - 140*(a^5*b^3 - 4*a^3*b^5 + 6*a*b^7)*cos(d*x + c)^4 + 105*(a^4*b^4 - 4*a^2*b^6)*cos(d*x + c)^3 - 84*(a^
3*b^5 - 4*a*b^7)*cos(d*x + c)^2)/(a*b^8*d*cos(d*x + c)^7)

Sympy [F]

\[ \int \frac {\tan ^9(c+d x)}{a+b \sec (c+d x)} \, dx=\int \frac {\tan ^{9}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]

[In]

integrate(tan(d*x+c)**9/(a+b*sec(d*x+c)),x)

[Out]

Integral(tan(c + d*x)**9/(a + b*sec(c + d*x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.07 \[ \int \frac {\tan ^9(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\frac {420 \, {\left (a^{7} - 4 \, a^{5} b^{2} + 6 \, a^{3} b^{4} - 4 \, a b^{6}\right )} \log \left (\cos \left (d x + c\right )\right )}{b^{8}} - \frac {420 \, {\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a b^{8}} - \frac {70 \, a b^{5} \cos \left (d x + c\right ) - 420 \, {\left (a^{6} - 4 \, a^{4} b^{2} + 6 \, a^{2} b^{4} - 4 \, b^{6}\right )} \cos \left (d x + c\right )^{6} - 60 \, b^{6} + 210 \, {\left (a^{5} b - 4 \, a^{3} b^{3} + 6 \, a b^{5}\right )} \cos \left (d x + c\right )^{5} - 140 \, {\left (a^{4} b^{2} - 4 \, a^{2} b^{4} + 6 \, b^{6}\right )} \cos \left (d x + c\right )^{4} + 105 \, {\left (a^{3} b^{3} - 4 \, a b^{5}\right )} \cos \left (d x + c\right )^{3} - 84 \, {\left (a^{2} b^{4} - 4 \, b^{6}\right )} \cos \left (d x + c\right )^{2}}{b^{7} \cos \left (d x + c\right )^{7}}}{420 \, d} \]

[In]

integrate(tan(d*x+c)^9/(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/420*(420*(a^7 - 4*a^5*b^2 + 6*a^3*b^4 - 4*a*b^6)*log(cos(d*x + c))/b^8 - 420*(a^8 - 4*a^6*b^2 + 6*a^4*b^4 -
4*a^2*b^6 + b^8)*log(a*cos(d*x + c) + b)/(a*b^8) - (70*a*b^5*cos(d*x + c) - 420*(a^6 - 4*a^4*b^2 + 6*a^2*b^4 -
4*b^6)*cos(d*x + c)^6 - 60*b^6 + 210*(a^5*b - 4*a^3*b^3 + 6*a*b^5)*cos(d*x + c)^5 - 140*(a^4*b^2 - 4*a^2*b^4
+ 6*b^6)*cos(d*x + c)^4 + 105*(a^3*b^3 - 4*a*b^5)*cos(d*x + c)^3 - 84*(a^2*b^4 - 4*b^6)*cos(d*x + c)^2)/(b^7*c
os(d*x + c)^7))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1768 vs. \(2 (238) = 476\).

Time = 5.76 (sec) , antiderivative size = 1768, normalized size of antiderivative = 7.07 \[ \int \frac {\tan ^9(c+d x)}{a+b \sec (c+d x)} \, dx=\text {Too large to display} \]

[In]

integrate(tan(d*x+c)^9/(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

-1/420*(210*(a^7 - 4*a^5*b^2 + 6*a^3*b^4 - 4*a*b^6)*log(abs(a + b - 2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1)
- a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2))/b^8 - 420*(a^7 -
4*a^5*b^2 + 6*a^3*b^4 - 4*a*b^6)*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1))/b^8 - 210*(a^8 - 4*a^6*
b^2 + 6*a^4*b^4 - 4*a^2*b^6 + 2*b^8)*log(abs(2*b + 2*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 2*b*(cos(d*x +
c) - 1)/(cos(d*x + c) + 1) - 2*abs(a))/abs(2*b + 2*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 2*b*(cos(d*x + c)
- 1)/(cos(d*x + c) + 1) + 2*abs(a)))/(b^8*abs(a)) + (1089*a^7 - 840*a^6*b - 4356*a^5*b^2 + 3080*a^4*b^3 + 653
4*a^3*b^4 - 4088*a^2*b^5 - 4356*a*b^6 + 2232*b^7 + 7623*a^7*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 5040*a^6*b
*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 31332*a^5*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 19040*a^4*b^3*(
cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 48258*a^3*b^4*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 26096*a^2*b^5*(co
s(d*x + c) - 1)/(cos(d*x + c) + 1) - 33012*a*b^6*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 14784*b^7*(cos(d*x +
c) - 1)/(cos(d*x + c) + 1) + 22869*a^7*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 12600*a^6*b*(cos(d*x + c) -
1)^2/(cos(d*x + c) + 1)^2 - 95676*a^5*b^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 47880*a^4*b^3*(cos(d*x
+ c) - 1)^2/(cos(d*x + c) + 1)^2 + 151494*a^3*b^4*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 67368*a^2*b^5*(c
os(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 107436*a*b^6*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 40152*b^7*(
cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 38115*a^7*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 16800*a^6*b*(
cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 160860*a^5*b^2*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 62720*a^
4*b^3*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 258930*a^3*b^4*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 8
6240*a^2*b^5*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 192220*a*b^6*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^
3 + 53760*b^7*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 38115*a^7*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4
- 12600*a^6*b*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 - 160860*a^5*b^2*(cos(d*x + c) - 1)^4/(cos(d*x + c) +
1)^4 + 45080*a^4*b^3*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 + 258930*a^3*b^4*(cos(d*x + c) - 1)^4/(cos(d*x
+ c) + 1)^4 - 56840*a^2*b^5*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 - 192220*a*b^6*(cos(d*x + c) - 1)^4/(cos
(d*x + c) + 1)^4 + 24360*b^7*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 + 22869*a^7*(cos(d*x + c) - 1)^5/(cos(d
*x + c) + 1)^5 - 5040*a^6*b*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 - 95676*a^5*b^2*(cos(d*x + c) - 1)^5/(co
s(d*x + c) + 1)^5 + 16800*a^4*b^3*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 + 151494*a^3*b^4*(cos(d*x + c) - 1
)^5/(cos(d*x + c) + 1)^5 - 18480*a^2*b^5*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 - 107436*a*b^6*(cos(d*x + c
) - 1)^5/(cos(d*x + c) + 1)^5 + 6720*b^7*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 + 7623*a^7*(cos(d*x + c) -
1)^6/(cos(d*x + c) + 1)^6 - 840*a^6*b*(cos(d*x + c) - 1)^6/(cos(d*x + c) + 1)^6 - 31332*a^5*b^2*(cos(d*x + c)
- 1)^6/(cos(d*x + c) + 1)^6 + 2520*a^4*b^3*(cos(d*x + c) - 1)^6/(cos(d*x + c) + 1)^6 + 48258*a^3*b^4*(cos(d*x
+ c) - 1)^6/(cos(d*x + c) + 1)^6 - 2520*a^2*b^5*(cos(d*x + c) - 1)^6/(cos(d*x + c) + 1)^6 - 33012*a*b^6*(cos(d
*x + c) - 1)^6/(cos(d*x + c) + 1)^6 + 840*b^7*(cos(d*x + c) - 1)^6/(cos(d*x + c) + 1)^6 + 1089*a^7*(cos(d*x +
c) - 1)^7/(cos(d*x + c) + 1)^7 - 4356*a^5*b^2*(cos(d*x + c) - 1)^7/(cos(d*x + c) + 1)^7 + 6534*a^3*b^4*(cos(d*
x + c) - 1)^7/(cos(d*x + c) + 1)^7 - 4356*a*b^6*(cos(d*x + c) - 1)^7/(cos(d*x + c) + 1)^7)/(b^8*((cos(d*x + c)
- 1)/(cos(d*x + c) + 1) + 1)^7))/d

Mupad [B] (verification not implemented)

Time = 15.15 (sec) , antiderivative size = 631, normalized size of antiderivative = 2.52 \[ \int \frac {\tan ^9(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d}-\frac {\frac {2\,\left (105\,a^6-385\,a^4\,b^2+511\,a^2\,b^4-279\,b^6\right )}{105\,b^7}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,\left (a^6+a^5\,b-3\,a^4\,b^2-3\,a^3\,b^3+3\,a^2\,b^4+3\,a\,b^5-b^6\right )}{b^7}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (6\,a^6+5\,a^5\,b-20\,a^4\,b^2-17\,a^3\,b^3+22\,a^2\,b^4+19\,a\,b^5-8\,b^6\right )}{b^7}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (30\,a^6+15\,a^5\,b-112\,a^4\,b^2-54\,a^3\,b^3+154\,a^2\,b^4+71\,a\,b^5-96\,b^6\right )}{3\,b^7}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (45\,a^6+30\,a^5\,b-161\,a^4\,b^2-108\,a^3\,b^3+203\,a^2\,b^4+142\,a\,b^5-87\,b^6\right )}{3\,b^7}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (75\,a^6+25\,a^5\,b-285\,a^4\,b^2-85\,a^3\,b^3+401\,a^2\,b^4+95\,a\,b^5-239\,b^6\right )}{5\,b^7}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (90\,a^6+15\,a^5\,b-340\,a^4\,b^2-45\,a^3\,b^3+466\,a^2\,b^4+45\,a\,b^5-264\,b^6\right )}{15\,b^7}}{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 {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )\,\left (-a^7+4\,a^5\,b^2-6\,a^3\,b^4+4\,a\,b^6\right )}{b^8\,d}-\frac {\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )\,{\left (a^2-b^2\right )}^4}{a\,b^8\,d} \]

[In]

int(tan(c + d*x)^9/(a + b/cos(c + d*x)),x)

[Out]

log(tan(c/2 + (d*x)/2)^2 + 1)/(a*d) - ((2*(105*a^6 - 279*b^6 + 511*a^2*b^4 - 385*a^4*b^2))/(105*b^7) + (2*tan(
c/2 + (d*x)/2)^12*(3*a*b^5 + a^5*b + a^6 - b^6 + 3*a^2*b^4 - 3*a^3*b^3 - 3*a^4*b^2))/b^7 - (2*tan(c/2 + (d*x)/
2)^10*(19*a*b^5 + 5*a^5*b + 6*a^6 - 8*b^6 + 22*a^2*b^4 - 17*a^3*b^3 - 20*a^4*b^2))/b^7 - (4*tan(c/2 + (d*x)/2)
^6*(71*a*b^5 + 15*a^5*b + 30*a^6 - 96*b^6 + 154*a^2*b^4 - 54*a^3*b^3 - 112*a^4*b^2))/(3*b^7) + (2*tan(c/2 + (d
*x)/2)^8*(142*a*b^5 + 30*a^5*b + 45*a^6 - 87*b^6 + 203*a^2*b^4 - 108*a^3*b^3 - 161*a^4*b^2))/(3*b^7) + (2*tan(
c/2 + (d*x)/2)^4*(95*a*b^5 + 25*a^5*b + 75*a^6 - 239*b^6 + 401*a^2*b^4 - 85*a^3*b^3 - 285*a^4*b^2))/(5*b^7) -
(2*tan(c/2 + (d*x)/2)^2*(45*a*b^5 + 15*a^5*b + 90*a^6 - 264*b^6 + 466*a^2*b^4 - 45*a^3*b^3 - 340*a^4*b^2))/(15
*b^7))/(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)) - (log(tan(c/2 + (d*x)/2)
^2 - 1)*(4*a*b^6 - a^7 - 6*a^3*b^4 + 4*a^5*b^2))/(b^8*d) - (log(a + b - a*tan(c/2 + (d*x)/2)^2 + b*tan(c/2 + (
d*x)/2)^2)*(a^2 - b^2)^4)/(a*b^8*d)