\(\int \sec ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx\) [43]

Optimal result

Integrand size = 29, antiderivative size = 134 \[ \int \sec ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3 a (A+B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a (5 A+4 B) \tan (c+d x)}{5 d}+\frac {3 a (A+B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a (A+B) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a B \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {a (5 A+4 B) \tan ^3(c+d x)}{15 d} \]

[Out]

3/8*a*(A+B)*arctanh(sin(d*x+c))/d+1/5*a*(5*A+4*B)*tan(d*x+c)/d+3/8*a*(A+B)*sec(d*x+c)*tan(d*x+c)/d+1/4*a*(A+B)
*sec(d*x+c)^3*tan(d*x+c)/d+1/5*a*B*sec(d*x+c)^4*tan(d*x+c)/d+1/15*a*(5*A+4*B)*tan(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {4082, 3872, 3852, 3853, 3855} \[ \int \sec ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3 a (A+B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a (5 A+4 B) \tan ^3(c+d x)}{15 d}+\frac {a (5 A+4 B) \tan (c+d x)}{5 d}+\frac {a (A+B) \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {3 a (A+B) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a B \tan (c+d x) \sec ^4(c+d x)}{5 d} \]

[In]

Int[Sec[c + d*x]^4*(a + a*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]

[Out]

(3*a*(A + B)*ArcTanh[Sin[c + d*x]])/(8*d) + (a*(5*A + 4*B)*Tan[c + d*x])/(5*d) + (3*a*(A + B)*Sec[c + d*x]*Tan
[c + d*x])/(8*d) + (a*(A + B)*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (a*B*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + (
a*(5*A + 4*B)*Tan[c + d*x]^3)/(15*d)

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 4082

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

Rubi steps \begin{align*} \text {integral}& = \frac {a B \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int \sec ^4(c+d x) (a (5 A+4 B)+5 a (A+B) \sec (c+d x)) \, dx \\ & = \frac {a B \sec ^4(c+d x) \tan (c+d x)}{5 d}+(a (A+B)) \int \sec ^5(c+d x) \, dx+\frac {1}{5} (a (5 A+4 B)) \int \sec ^4(c+d x) \, dx \\ & = \frac {a (A+B) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a B \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{4} (3 a (A+B)) \int \sec ^3(c+d x) \, dx-\frac {(a (5 A+4 B)) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d} \\ & = \frac {a (5 A+4 B) \tan (c+d x)}{5 d}+\frac {3 a (A+B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a (A+B) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a B \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {a (5 A+4 B) \tan ^3(c+d x)}{15 d}+\frac {1}{8} (3 a (A+B)) \int \sec (c+d x) \, dx \\ & = \frac {3 a (A+B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a (5 A+4 B) \tan (c+d x)}{5 d}+\frac {3 a (A+B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a (A+B) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a B \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {a (5 A+4 B) \tan ^3(c+d x)}{15 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.65 \[ \int \sec ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {a \left (45 (A+B) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (45 (A+B) \sec (c+d x)+30 (A+B) \sec ^3(c+d x)+8 \left (15 (A+B)+5 (A+2 B) \tan ^2(c+d x)+3 B \tan ^4(c+d x)\right )\right )\right )}{120 d} \]

[In]

Integrate[Sec[c + d*x]^4*(a + a*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]

[Out]

(a*(45*(A + B)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(45*(A + B)*Sec[c + d*x] + 30*(A + B)*Sec[c + d*x]^3 + 8*(
15*(A + B) + 5*(A + 2*B)*Tan[c + d*x]^2 + 3*B*Tan[c + d*x]^4))))/(120*d)

Maple [A] (verified)

Time = 3.82 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.87

method	result	size
parts	\(\frac {\left (a A +B a \right ) \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {B a \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}-\frac {a A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(117\)
derivativedivides	\(\frac {a A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B a \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )-a A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(154\)
default	\(\frac {a A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B a \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )-a A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(154\)
norman	\(\frac {-\frac {13 a \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {3 a \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}-\frac {4 a \left (25 A +29 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}+\frac {a \left (29 A +13 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}+\frac {a \left (35 A +19 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}-\frac {3 a \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {3 a \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(171\)
parallelrisch	\(\frac {8 \left (-\frac {45 \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{32}+\frac {45 \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{32}+\frac {21 \left (A +B \right ) \sin \left (2 d x +2 c \right )}{16}+\left (\frac {5 A}{4}+B \right ) \sin \left (3 d x +3 c \right )+\frac {9 \left (A +B \right ) \sin \left (4 d x +4 c \right )}{32}+\left (\frac {A}{4}+\frac {B}{5}\right ) \sin \left (5 d x +5 c \right )+\sin \left (d x +c \right ) \left (A +2 B \right )\right ) a}{3 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\)	\(203\)
risch	\(-\frac {i a \left (45 A \,{\mathrm e}^{9 i \left (d x +c \right )}+45 B \,{\mathrm e}^{9 i \left (d x +c \right )}+210 A \,{\mathrm e}^{7 i \left (d x +c \right )}+210 B \,{\mathrm e}^{7 i \left (d x +c \right )}-240 A \,{\mathrm e}^{6 i \left (d x +c \right )}-560 A \,{\mathrm e}^{4 i \left (d x +c \right )}-640 B \,{\mathrm e}^{4 i \left (d x +c \right )}-210 A \,{\mathrm e}^{3 i \left (d x +c \right )}-210 B \,{\mathrm e}^{3 i \left (d x +c \right )}-400 A \,{\mathrm e}^{2 i \left (d x +c \right )}-320 B \,{\mathrm e}^{2 i \left (d x +c \right )}-45 \,{\mathrm e}^{i \left (d x +c \right )} A -45 B \,{\mathrm e}^{i \left (d x +c \right )}-80 A -64 B \right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{8 d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{8 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{8 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{8 d}\)	\(265\)

[In]

int(sec(d*x+c)^4*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

(A*a+B*a)/d*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))-B*a/d*(-8/15-1/5*se
c(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)-a*A/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.02 \[ \int \sec ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {45 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (5 \, A + 4 \, B\right )} a \cos \left (d x + c\right )^{4} + 45 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, A + 4 \, B\right )} a \cos \left (d x + c\right )^{2} + 30 \, {\left (A + B\right )} a \cos \left (d x + c\right ) + 24 \, B a\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

[In]

integrate(sec(d*x+c)^4*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/240*(45*(A + B)*a*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 45*(A + B)*a*cos(d*x + c)^5*log(-sin(d*x + c) + 1)
+ 2*(16*(5*A + 4*B)*a*cos(d*x + c)^4 + 45*(A + B)*a*cos(d*x + c)^3 + 8*(5*A + 4*B)*a*cos(d*x + c)^2 + 30*(A +
B)*a*cos(d*x + c) + 24*B*a)*sin(d*x + c))/(d*cos(d*x + c)^5)

Sympy [F]

\[ \int \sec ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=a \left (\int A \sec ^{4}{\left (c + d x \right )}\, dx + \int A \sec ^{5}{\left (c + d x \right )}\, dx + \int B \sec ^{5}{\left (c + d x \right )}\, dx + \int B \sec ^{6}{\left (c + d x \right )}\, dx\right ) \]

[In]

integrate(sec(d*x+c)**4*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x)

[Out]

a*(Integral(A*sec(c + d*x)**4, x) + Integral(A*sec(c + d*x)**5, x) + Integral(B*sec(c + d*x)**5, x) + Integral
(B*sec(c + d*x)**6, x))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.49 \[ \int \sec ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} B a - 15 \, A a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 15 \, B a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]

[In]

integrate(sec(d*x+c)^4*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/240*(80*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a + 16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*
B*a - 15*A*a*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x +
c) + 1) + 3*log(sin(d*x + c) - 1)) - 15*B*a*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x
+ c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)))/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.60 \[ \int \sec ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {45 \, {\left (A a + B a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 45 \, {\left (A a + B a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (45 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 45 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 290 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 130 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 400 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 464 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 350 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 190 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 195 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 195 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]

[In]

integrate(sec(d*x+c)^4*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/120*(45*(A*a + B*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 45*(A*a + B*a)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) -
2*(45*A*a*tan(1/2*d*x + 1/2*c)^9 + 45*B*a*tan(1/2*d*x + 1/2*c)^9 - 290*A*a*tan(1/2*d*x + 1/2*c)^7 - 130*B*a*t
an(1/2*d*x + 1/2*c)^7 + 400*A*a*tan(1/2*d*x + 1/2*c)^5 + 464*B*a*tan(1/2*d*x + 1/2*c)^5 - 350*A*a*tan(1/2*d*x
+ 1/2*c)^3 - 190*B*a*tan(1/2*d*x + 1/2*c)^3 + 195*A*a*tan(1/2*d*x + 1/2*c) + 195*B*a*tan(1/2*d*x + 1/2*c))/(ta
n(1/2*d*x + 1/2*c)^2 - 1)^5)/d

Mupad [B] (verification not implemented)

Time = 16.30 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.48 \[ \int \sec ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3\,a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A+B\right )}{4\,d}-\frac {\left (\frac {3\,A\,a}{4}+\frac {3\,B\,a}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {29\,A\,a}{6}-\frac {13\,B\,a}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,A\,a}{3}+\frac {116\,B\,a}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {35\,A\,a}{6}-\frac {19\,B\,a}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {13\,A\,a}{4}+\frac {13\,B\,a}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

[In]

int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x)))/cos(c + d*x)^4,x)

[Out]

(3*a*atanh(tan(c/2 + (d*x)/2))*(A + B))/(4*d) - (tan(c/2 + (d*x)/2)*((13*A*a)/4 + (13*B*a)/4) + tan(c/2 + (d*x
)/2)^9*((3*A*a)/4 + (3*B*a)/4) - tan(c/2 + (d*x)/2)^7*((29*A*a)/6 + (13*B*a)/6) - tan(c/2 + (d*x)/2)^3*((35*A*
a)/6 + (19*B*a)/6) + tan(c/2 + (d*x)/2)^5*((20*A*a)/3 + (116*B*a)/15))/(d*(5*tan(c/2 + (d*x)/2)^2 - 10*tan(c/2
+ (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1))