\(\int \frac {\tan ^7(c+d x)}{(a+b \sec (c+d x))^2} \, dx\) [300]

Optimal result

Integrand size = 21, antiderivative size = 179 \[ \int \frac {\tan ^7(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {\log (\cos (c+d x))}{a^2 d}+\frac {\left (a^2-b^2\right )^2 \left (5 a^2+b^2\right ) \log (a+b \sec (c+d x))}{a^2 b^6 d}-\frac {2 a \left (2 a^2-3 b^2\right ) \sec (c+d x)}{b^5 d}+\frac {3 \left (a^2-b^2\right ) \sec ^2(c+d x)}{2 b^4 d}-\frac {2 a \sec ^3(c+d x)}{3 b^3 d}+\frac {\sec ^4(c+d x)}{4 b^2 d}+\frac {\left (a^2-b^2\right )^3}{a b^6 d (a+b \sec (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 179, 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} \[ \int \frac {\tan ^7(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {\left (a^2-b^2\right )^3}{a b^6 d (a+b \sec (c+d x))}+\frac {\left (a^2-b^2\right )^2 \left (5 a^2+b^2\right ) \log (a+b \sec (c+d x))}{a^2 b^6 d}-\frac {2 a \left (2 a^2-3 b^2\right ) \sec (c+d x)}{b^5 d}+\frac {3 \left (a^2-b^2\right ) \sec ^2(c+d x)}{2 b^4 d}+\frac {\log (\cos (c+d x))}{a^2 d}-\frac {2 a \sec ^3(c+d x)}{3 b^3 d}+\frac {\sec ^4(c+d x)}{4 b^2 d} \]

[In]

Int[Tan[c + d*x]^7/(a + b*Sec[c + d*x])^2,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 1.51 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.91 \[ \int \frac {\tan ^7(c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {-\frac {b^6 \log (\cos (c+d x))}{a^2}-\frac {\left (a^2-b^2\right )^2 \left (5 a^2+b^2\right ) \log (a+b \sec (c+d x))}{a^2}+2 a b \left (2 a^2-3 b^2\right ) \sec (c+d x)-\frac {3}{2} (a-b) b^2 (a+b) \sec ^2(c+d x)+\frac {2}{3} a b^3 \sec ^3(c+d x)-\frac {1}{4} b^4 \sec ^4(c+d x)-\frac {\left (a^2-b^2\right )^3}{a (a+b \sec (c+d x))}}{b^6 d} \]

[In]

Integrate[Tan[c + d*x]^7/(a + b*Sec[c + d*x])^2,x]

[Out]

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

Maple [A] (verified)

Time = 2.54 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.12

method	result	size
derivativedivides	\(\frac {-\frac {a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}{a^{2} b^{5} \left (b +a \cos \left (d x +c \right )\right )}+\frac {\left (5 a^{6}-9 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{b^{6} a^{2}}-\frac {-3 a^{2}+3 b^{2}}{2 b^{4} \cos \left (d x +c \right )^{2}}+\frac {\left (-5 a^{4}+9 a^{2} b^{2}-3 b^{4}\right ) \ln \left (\cos \left (d x +c \right )\right )}{b^{6}}+\frac {1}{4 b^{2} \cos \left (d x +c \right )^{4}}-\frac {2 a}{3 b^{3} \cos \left (d x +c \right )^{3}}-\frac {2 a \left (2 a^{2}-3 b^{2}\right )}{b^{5} \cos \left (d x +c \right )}}{d}\)	\(200\)
default	\(\frac {-\frac {a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}{a^{2} b^{5} \left (b +a \cos \left (d x +c \right )\right )}+\frac {\left (5 a^{6}-9 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{b^{6} a^{2}}-\frac {-3 a^{2}+3 b^{2}}{2 b^{4} \cos \left (d x +c \right )^{2}}+\frac {\left (-5 a^{4}+9 a^{2} b^{2}-3 b^{4}\right ) \ln \left (\cos \left (d x +c \right )\right )}{b^{6}}+\frac {1}{4 b^{2} \cos \left (d x +c \right )^{4}}-\frac {2 a}{3 b^{3} \cos \left (d x +c \right )^{3}}-\frac {2 a \left (2 a^{2}-3 b^{2}\right )}{b^{5} \cos \left (d x +c \right )}}{d}\)	\(200\)
risch	\(-\frac {i x}{a^{2}}-\frac {2 i c}{a^{2} d}-\frac {2 \left (-12 b^{6} {\mathrm e}^{3 i \left (d x +c \right )}-3 b^{6} {\mathrm e}^{i \left (d x +c \right )}+15 a^{6} {\mathrm e}^{i \left (d x +c \right )}+60 a^{6} {\mathrm e}^{3 i \left (d x +c \right )}+60 a^{6} {\mathrm e}^{7 i \left (d x +c \right )}+90 a^{6} {\mathrm e}^{5 i \left (d x +c \right )}-3 b^{6} {\mathrm e}^{9 i \left (d x +c \right )}+15 a^{6} {\mathrm e}^{9 i \left (d x +c \right )}-27 a^{4} b^{2} {\mathrm e}^{i \left (d x +c \right )}+9 a^{2} b^{4} {\mathrm e}^{i \left (d x +c \right )}-118 a^{4} b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+54 a^{2} b^{4} {\mathrm e}^{3 i \left (d x +c \right )}+45 a^{5} b \,{\mathrm e}^{4 i \left (d x +c \right )}-71 a^{3} b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-182 a^{4} b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+78 a^{2} b^{4} {\mathrm e}^{5 i \left (d x +c \right )}-118 a^{4} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+15 a^{5} b \,{\mathrm e}^{2 i \left (d x +c \right )}+9 a^{2} b^{4} {\mathrm e}^{9 i \left (d x +c \right )}-71 a^{3} b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+45 a^{5} b \,{\mathrm e}^{6 i \left (d x +c \right )}-27 a^{3} b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-27 a^{4} b^{2} {\mathrm e}^{9 i \left (d x +c \right )}+54 a^{2} b^{4} {\mathrm e}^{7 i \left (d x +c \right )}+15 a^{5} b \,{\mathrm e}^{8 i \left (d x +c \right )}-27 a^{3} b^{3} {\mathrm e}^{8 i \left (d x +c \right )}-12 b^{6} {\mathrm e}^{7 i \left (d x +c \right )}-18 b^{6} {\mathrm e}^{5 i \left (d x +c \right )}\right )}{3 d \,b^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4} a^{2} \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}-\frac {5 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{4}}{b^{6} d}+\frac {9 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{2}}{b^{4} d}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b^{2} d}+\frac {5 a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{b^{6} d}-\frac {9 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{b^{4} d}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{b^{2} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a^{2} d}\)	\(720\)

[In]

int(tan(d*x+c)^7/(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.74 \[ \int \frac {\tan ^7(c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {5 \, a^{3} b^{4} \cos \left (d x + c\right ) - 3 \, a^{2} b^{5} + 12 \, {\left (5 \, a^{6} b - 9 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{4} + 6 \, {\left (5 \, a^{5} b^{2} - 9 \, a^{3} b^{4}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (5 \, a^{4} b^{3} - 9 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2} - 12 \, {\left ({\left (5 \, a^{7} - 9 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{5} + {\left (5 \, a^{6} b - 9 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) + 12 \, {\left ({\left (5 \, a^{7} - 9 \, a^{5} b^{2} + 3 \, a^{3} b^{4}\right )} \cos \left (d x + c\right )^{5} + {\left (5 \, a^{6} b - 9 \, a^{4} b^{3} + 3 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\cos \left (d x + c\right )\right )}{12 \, {\left (a^{3} b^{6} d \cos \left (d x + c\right )^{5} + a^{2} b^{7} d \cos \left (d x + c\right )^{4}\right )}} \]

[In]

integrate(tan(d*x+c)^7/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/12*(5*a^3*b^4*cos(d*x + c) - 3*a^2*b^5 + 12*(5*a^6*b - 9*a^4*b^3 + 3*a^2*b^5 - b^7)*cos(d*x + c)^4 + 6*(5*a
^5*b^2 - 9*a^3*b^4)*cos(d*x + c)^3 - 2*(5*a^4*b^3 - 9*a^2*b^5)*cos(d*x + c)^2 - 12*((5*a^7 - 9*a^5*b^2 + 3*a^3
*b^4 + a*b^6)*cos(d*x + c)^5 + (5*a^6*b - 9*a^4*b^3 + 3*a^2*b^5 + b^7)*cos(d*x + c)^4)*log(a*cos(d*x + c) + b)
+ 12*((5*a^7 - 9*a^5*b^2 + 3*a^3*b^4)*cos(d*x + c)^5 + (5*a^6*b - 9*a^4*b^3 + 3*a^2*b^5)*cos(d*x + c)^4)*log(
-cos(d*x + c)))/(a^3*b^6*d*cos(d*x + c)^5 + a^2*b^7*d*cos(d*x + c)^4)

Sympy [F]

\[ \int \frac {\tan ^7(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\tan ^{7}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]

[In]

integrate(tan(d*x+c)**7/(a+b*sec(d*x+c))**2,x)

[Out]

Integral(tan(c + d*x)**7/(a + b*sec(c + d*x))**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.27 \[ \int \frac {\tan ^7(c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {\frac {5 \, a^{3} b^{3} \cos \left (d x + c\right ) - 3 \, a^{2} b^{4} + 12 \, {\left (5 \, a^{6} - 9 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )^{4} + 6 \, {\left (5 \, a^{5} b - 9 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (5 \, a^{4} b^{2} - 9 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2}}{a^{3} b^{5} \cos \left (d x + c\right )^{5} + a^{2} b^{6} \cos \left (d x + c\right )^{4}} + \frac {12 \, {\left (5 \, a^{4} - 9 \, a^{2} b^{2} + 3 \, b^{4}\right )} \log \left (\cos \left (d x + c\right )\right )}{b^{6}} - \frac {12 \, {\left (5 \, a^{6} - 9 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{2} b^{6}}}{12 \, d} \]

[In]

integrate(tan(d*x+c)^7/(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/12*((5*a^3*b^3*cos(d*x + c) - 3*a^2*b^4 + 12*(5*a^6 - 9*a^4*b^2 + 3*a^2*b^4 - b^6)*cos(d*x + c)^4 + 6*(5*a^
5*b - 9*a^3*b^3)*cos(d*x + c)^3 - 2*(5*a^4*b^2 - 9*a^2*b^4)*cos(d*x + c)^2)/(a^3*b^5*cos(d*x + c)^5 + a^2*b^6*
cos(d*x + c)^4) + 12*(5*a^4 - 9*a^2*b^2 + 3*b^4)*log(cos(d*x + c))/b^6 - 12*(5*a^6 - 9*a^4*b^2 + 3*a^2*b^4 + b
^6)*log(a*cos(d*x + c) + b)/(a^2*b^6))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1023 vs. \(2 (173) = 346\).

Time = 3.43 (sec) , antiderivative size = 1023, normalized size of antiderivative = 5.72 \[ \int \frac {\tan ^7(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\text {Too large to display} \]

[In]

integrate(tan(d*x+c)^7/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/12*(12*(5*a^7 - 5*a^6*b - 9*a^5*b^2 + 9*a^4*b^3 + 3*a^3*b^4 - 3*a^2*b^5 + a*b^6 - b^7)*log(abs(a + b + a*(co
s(d*x + c) - 1)/(cos(d*x + c) + 1) - b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1)))/(a^3*b^6 - a^2*b^7) - 12*log(ab
s(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/a^2 - 12*(5*a^4 - 9*a^2*b^2 + 3*b^4)*log(abs(-(cos(d*x + c) - 1
)/(cos(d*x + c) + 1) - 1))/b^6 - 12*(5*a^7 + 7*a^6*b - 7*a^5*b^2 - 13*a^4*b^3 - a^3*b^4 + 5*a^2*b^5 + 3*a*b^6
+ b^7 + 5*a^7*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 5*a^6*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 9*a^5*b^
2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 9*a^4*b^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 3*a^3*b^4*(cos(d*x
+ c) - 1)/(cos(d*x + c) + 1) - 3*a^2*b^5*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + a*b^6*(cos(d*x + c) - 1)/(co
s(d*x + c) + 1) - b^7*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))/((a + b + a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1)
- b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))*a^2*b^6) + (125*a^4 - 96*a^3*b - 225*a^2*b^2 + 128*a*b^3 + 75*b^4
+ 500*a^4*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 288*a^3*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 972*a^2*b^
2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 416*a*b^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 348*b^4*(cos(d*x +
c) - 1)/(cos(d*x + c) + 1) + 750*a^4*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 288*a^3*b*(cos(d*x + c) - 1)
^2/(cos(d*x + c) + 1)^2 - 1494*a^2*b^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 384*a*b^3*(cos(d*x + c) - 1
)^2/(cos(d*x + c) + 1)^2 + 594*b^4*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 500*a^4*(cos(d*x + c) - 1)^3/(c
os(d*x + c) + 1)^3 - 96*a^3*b*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 972*a^2*b^2*(cos(d*x + c) - 1)^3/(co
s(d*x + c) + 1)^3 + 96*a*b^3*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 348*b^4*(cos(d*x + c) - 1)^3/(cos(d*x
+ c) + 1)^3 + 125*a^4*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 - 225*a^2*b^2*(cos(d*x + c) - 1)^4/(cos(d*x +
c) + 1)^4 + 75*b^4*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4)/(b^6*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1
)^4))/d

Mupad [B] (verification not implemented)

Time = 15.73 (sec) , antiderivative size = 505, normalized size of antiderivative = 2.82 \[ \int \frac {\tan ^7(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (20\,a^5+5\,a^4\,b-31\,a^3\,b^2-4\,a^2\,b^3+8\,a\,b^4+4\,b^5\right )}{a\,b^5}-\frac {2\,\left (15\,a^5+15\,a^4\,b-22\,a^3\,b^2-22\,a^2\,b^3+3\,a\,b^4+3\,b^5\right )}{3\,a\,b^5}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (60\,a^5+45\,a^4\,b-83\,a^3\,b^2-66\,a^2\,b^3+6\,a\,b^4+12\,b^5\right )}{3\,a\,b^5}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (90\,a^5+45\,a^4\,b-127\,a^3\,b^2-56\,a^2\,b^3+6\,a\,b^4+18\,b^5\right )}{3\,a\,b^5}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (a-b\right )\,\left (-5\,a^4-5\,a^3\,b+4\,a^2\,b^2+4\,a\,b^3+b^4\right )}{a\,b^5}}{d\,\left (\left (b-a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\left (5\,a-3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\left (2\,b-10\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (10\,a+2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (-5\,a-3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a+b\right )}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a^2\,d}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )\,\left (5\,a^4-9\,a^2\,b^2+3\,b^4\right )}{b^6\,d}+\frac {\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )\,{\left (a^2-b^2\right )}^2\,\left (5\,a^2+b^2\right )}{a^2\,b^6\,d} \]

[In]

int(tan(c + d*x)^7/(a + b/cos(c + d*x))^2,x)

[Out]

((2*tan(c/2 + (d*x)/2)^6*(8*a*b^4 + 5*a^4*b + 20*a^5 + 4*b^5 - 4*a^2*b^3 - 31*a^3*b^2))/(a*b^5) - (2*(3*a*b^4
+ 15*a^4*b + 15*a^5 + 3*b^5 - 22*a^2*b^3 - 22*a^3*b^2))/(3*a*b^5) + (2*tan(c/2 + (d*x)/2)^2*(6*a*b^4 + 45*a^4*
b + 60*a^5 + 12*b^5 - 66*a^2*b^3 - 83*a^3*b^2))/(3*a*b^5) - (2*tan(c/2 + (d*x)/2)^4*(6*a*b^4 + 45*a^4*b + 90*a
^5 + 18*b^5 - 56*a^2*b^3 - 127*a^3*b^2))/(3*a*b^5) + (2*tan(c/2 + (d*x)/2)^8*(a - b)*(4*a*b^3 - 5*a^3*b - 5*a^
4 + b^4 + 4*a^2*b^2))/(a*b^5))/(d*(a + b - tan(c/2 + (d*x)/2)^10*(a - b) - tan(c/2 + (d*x)/2)^2*(5*a + 3*b) +
tan(c/2 + (d*x)/2)^4*(10*a + 2*b) + tan(c/2 + (d*x)/2)^8*(5*a - 3*b) - tan(c/2 + (d*x)/2)^6*(10*a - 2*b))) - l
og(tan(c/2 + (d*x)/2)^2 + 1)/(a^2*d) - (log(tan(c/2 + (d*x)/2)^2 - 1)*(5*a^4 + 3*b^4 - 9*a^2*b^2))/(b^6*d) + (
log(a + b - a*tan(c/2 + (d*x)/2)^2 + b*tan(c/2 + (d*x)/2)^2)*(a^2 - b^2)^2*(5*a^2 + b^2))/(a^2*b^6*d)