\(\int \frac {\sec ^6(c+d x)}{(a+a \sec (c+d x))^3} \, dx\) [61]

Optimal result

Integrand size = 21, antiderivative size = 162 \[ \int \frac {\sec ^6(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {13 \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {152 \tan (c+d x)}{15 a^3 d}+\frac {13 \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {\sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {11 \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {76 \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \]

[Out]

13/2*arctanh(sin(d*x+c))/a^3/d-152/15*tan(d*x+c)/a^3/d+13/2*sec(d*x+c)*tan(d*x+c)/a^3/d-1/5*sec(d*x+c)^4*tan(d
*x+c)/d/(a+a*sec(d*x+c))^3-11/15*sec(d*x+c)^3*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^2-76/15*sec(d*x+c)^2*tan(d*x+c)/
d/(a^3+a^3*sec(d*x+c))

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3901, 4104, 3872, 3852, 8, 3853, 3855} \[ \int \frac {\sec ^6(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {13 \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {152 \tan (c+d x)}{15 a^3 d}-\frac {76 \tan (c+d x) \sec ^2(c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {13 \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac {\tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac {11 \tan (c+d x) \sec ^3(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]

[In]

Int[Sec[c + d*x]^6/(a + a*Sec[c + d*x])^3,x]

[Out]

(13*ArcTanh[Sin[c + d*x]])/(2*a^3*d) - (152*Tan[c + d*x])/(15*a^3*d) + (13*Sec[c + d*x]*Tan[c + d*x])/(2*a^3*d
) - (Sec[c + d*x]^4*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) - (11*Sec[c + d*x]^3*Tan[c + d*x])/(15*a*d*(a +
a*Sec[c + d*x])^2) - (76*Sec[c + d*x]^2*Tan[c + d*x])/(15*d*(a^3 + a^3*Sec[c + d*x]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3901

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-d^2)
*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 2)/(f*(2*m + 1))), x] + Dist[d^2/(a*b*(2*m + 1)),
Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2)*(b*(n - 2) + a*(m - n + 2)*Csc[e + f*x]), x], x] /;
FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 2] && (IntegersQ[2*m, 2*n] || IntegerQ[
m])

Rule 4104

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(
a*f*(2*m + 1))), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1)*Simp[A
*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b,
d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && GtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {\sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {\int \frac {\sec ^4(c+d x) (4 a-7 a \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2} \\ & = -\frac {\sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {11 \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {\int \frac {\sec ^3(c+d x) \left (33 a^2-43 a^2 \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4} \\ & = -\frac {\sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {11 \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {76 \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {\int \sec ^2(c+d x) \left (152 a^3-195 a^3 \sec (c+d x)\right ) \, dx}{15 a^6} \\ & = -\frac {\sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {11 \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {76 \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {152 \int \sec ^2(c+d x) \, dx}{15 a^3}+\frac {13 \int \sec ^3(c+d x) \, dx}{a^3} \\ & = \frac {13 \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {\sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {11 \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {76 \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {13 \int \sec (c+d x) \, dx}{2 a^3}+\frac {152 \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d} \\ & = \frac {13 \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {152 \tan (c+d x)}{15 a^3 d}+\frac {13 \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {\sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {11 \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {76 \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.10 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.69 \[ \int \frac {\sec ^6(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {1560 \text {arctanh}(\sin (c+d x)) \cos ^6\left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x)-\frac {1}{8} \sec ^5(c+d x) (1142 \sin (c+d x)+1614 \sin (2 (c+d x))+1414 \sin (3 (c+d x))+717 \sin (4 (c+d x))+152 \sin (5 (c+d x)))}{30 a^3 d (1+\sec (c+d x))^3} \]

[In]

Integrate[Sec[c + d*x]^6/(a + a*Sec[c + d*x])^3,x]

[Out]

(1560*ArcTanh[Sin[c + d*x]]*Cos[(c + d*x)/2]^6*Sec[c + d*x]^3 - (Sec[c + d*x]^5*(1142*Sin[c + d*x] + 1614*Sin[
2*(c + d*x)] + 1414*Sin[3*(c + d*x)] + 717*Sin[4*(c + d*x)] + 152*Sin[5*(c + d*x)]))/8)/(30*a^3*d*(1 + Sec[c +
d*x])^3)

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {14}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-26 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {14}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+26 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d \,a^{3}}\)	\(135\)
default	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {14}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-26 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {14}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+26 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d \,a^{3}}\)	\(135\)
parallelrisch	\(\frac {\left (-1560 \cos \left (2 d x +2 c \right )-1560\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (1560 \cos \left (2 d x +2 c \right )+1560\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-2331 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \left (\cos \left (d x +c \right )+\frac {174 \cos \left (2 d x +2 c \right )}{259}+\frac {239 \cos \left (3 d x +3 c \right )}{777}+\frac {152 \cos \left (4 d x +4 c \right )}{2331}+\frac {1354}{2331}\right )}{240 a^{3} d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(138\)
risch	\(-\frac {i \left (195 \,{\mathrm e}^{8 i \left (d x +c \right )}+975 \,{\mathrm e}^{7 i \left (d x +c \right )}+2275 \,{\mathrm e}^{6 i \left (d x +c \right )}+3575 \,{\mathrm e}^{5 i \left (d x +c \right )}+4329 \,{\mathrm e}^{4 i \left (d x +c \right )}+3805 \,{\mathrm e}^{3 i \left (d x +c \right )}+2673 \,{\mathrm e}^{2 i \left (d x +c \right )}+1325 \,{\mathrm e}^{i \left (d x +c \right )}+304\right )}{15 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}-\frac {13 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a^{3} d}+\frac {13 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a^{3} d}\)	\(169\)
norman	\(\frac {\frac {51 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {721 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}+\frac {6613 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}-\frac {1165 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{12 a d}+\frac {475 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 a d}-\frac {59 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{12 a d}-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{12 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{20 a d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} a^{2}}-\frac {13 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3} d}+\frac {13 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3} d}\)	\(212\)

[In]

int(sec(d*x+c)^6/(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/4/d/a^3*(-1/5*tan(1/2*d*x+1/2*c)^5-8/3*tan(1/2*d*x+1/2*c)^3-31*tan(1/2*d*x+1/2*c)+2/(tan(1/2*d*x+1/2*c)-1)^2
+14/(tan(1/2*d*x+1/2*c)-1)-26*ln(tan(1/2*d*x+1/2*c)-1)-2/(tan(1/2*d*x+1/2*c)+1)^2+14/(tan(1/2*d*x+1/2*c)+1)+26
*ln(tan(1/2*d*x+1/2*c)+1))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.27 \[ \int \frac {\sec ^6(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {195 \, {\left (\cos \left (d x + c\right )^{5} + 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 195 \, {\left (\cos \left (d x + c\right )^{5} + 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (304 \, \cos \left (d x + c\right )^{4} + 717 \, \cos \left (d x + c\right )^{3} + 479 \, \cos \left (d x + c\right )^{2} + 45 \, \cos \left (d x + c\right ) - 15\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

integrate(sec(d*x+c)^6/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/60*(195*(cos(d*x + c)^5 + 3*cos(d*x + c)^4 + 3*cos(d*x + c)^3 + cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 195*
(cos(d*x + c)^5 + 3*cos(d*x + c)^4 + 3*cos(d*x + c)^3 + cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(304*cos(d*
x + c)^4 + 717*cos(d*x + c)^3 + 479*cos(d*x + c)^2 + 45*cos(d*x + c) - 15)*sin(d*x + c))/(a^3*d*cos(d*x + c)^5
+ 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)

Sympy [F]

\[ \int \frac {\sec ^6(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {\sec ^{6}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

[In]

integrate(sec(d*x+c)**6/(a+a*sec(d*x+c))**3,x)

[Out]

Integral(sec(c + d*x)**6/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.30 \[ \int \frac {\sec ^6(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} - \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}}{60 \, d} \]

[In]

integrate(sec(d*x+c)^6/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) - 7*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^3 - 2*a^3*sin(d*x + c
)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1) + 4
0*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 390*log(sin(d*x + c)/(cos
(d*x + c) + 1) + 1)/a^3 + 390*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3)/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.86 \[ \int \frac {\sec ^6(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {390 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {390 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {60 \, {\left (7 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}} - \frac {3 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 465 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

[In]

integrate(sec(d*x+c)^6/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(390*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 390*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 + 60*(7*tan(1/2*
d*x + 1/2*c)^3 - 5*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^3) - (3*a^12*tan(1/2*d*x + 1/2*c)^5
+ 40*a^12*tan(1/2*d*x + 1/2*c)^3 + 465*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d

Mupad [B] (verification not implemented)

Time = 12.99 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.87 \[ \int \frac {\sec ^6(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {13\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20\,a^3\,d}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3\,a^3\,d}-\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}-\frac {31\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,a^3\,d} \]

[In]

int(1/(cos(c + d*x)^6*(a + a/cos(c + d*x))^3),x)

[Out]

(13*atanh(tan(c/2 + (d*x)/2)))/(a^3*d) - tan(c/2 + (d*x)/2)^5/(20*a^3*d) - (2*tan(c/2 + (d*x)/2)^3)/(3*a^3*d)
- (5*tan(c/2 + (d*x)/2) - 7*tan(c/2 + (d*x)/2)^3)/(d*(a^3*tan(c/2 + (d*x)/2)^4 - 2*a^3*tan(c/2 + (d*x)/2)^2 +
a^3)) - (31*tan(c/2 + (d*x)/2))/(4*a^3*d)