\(\int \frac {\sec ^5(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx\) [99]

Optimal result

Integrand size = 31, antiderivative size = 202 \[ \int \frac {\sec ^5(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {(6 A-13 B) \text {arctanh}(\sin (c+d x))}{2 a^3 d}+\frac {8 (9 A-19 B) \tan (c+d x)}{15 a^3 d}-\frac {(6 A-13 B) \sec (c+d x) \tan (c+d x)}{2 a^3 d}+\frac {(A-B) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(6 A-11 B) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {4 (9 A-19 B) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \]

[Out]

-1/2*(6*A-13*B)*arctanh(sin(d*x+c))/a^3/d+8/15*(9*A-19*B)*tan(d*x+c)/a^3/d-1/2*(6*A-13*B)*sec(d*x+c)*tan(d*x+c
)/a^3/d+1/5*(A-B)*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+c))^3+1/15*(6*A-11*B)*sec(d*x+c)^3*tan(d*x+c)/a/d/(a+
a*sec(d*x+c))^2+4/15*(9*A-19*B)*sec(d*x+c)^2*tan(d*x+c)/d/(a^3+a^3*sec(d*x+c))

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4104, 3872, 3852, 8, 3853, 3855} \[ \int \frac {\sec ^5(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {(6 A-13 B) \text {arctanh}(\sin (c+d x))}{2 a^3 d}+\frac {8 (9 A-19 B) \tan (c+d x)}{15 a^3 d}+\frac {4 (9 A-19 B) \tan (c+d x) \sec ^2(c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {(6 A-13 B) \tan (c+d x) \sec (c+d x)}{2 a^3 d}+\frac {(A-B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {(6 A-11 B) \tan (c+d x) \sec ^3(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]

[In]

Int[(Sec[c + d*x]^5*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^3,x]

[Out]

-1/2*((6*A - 13*B)*ArcTanh[Sin[c + d*x]])/(a^3*d) + (8*(9*A - 19*B)*Tan[c + d*x])/(15*a^3*d) - ((6*A - 13*B)*S
ec[c + d*x]*Tan[c + d*x])/(2*a^3*d) + ((A - B)*Sec[c + d*x]^4*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + ((6
*A - 11*B)*Sec[c + d*x]^3*Tan[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) + (4*(9*A - 19*B)*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 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 {(A-B) \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 (A-B)-a (2 A-7 B) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2} \\ & = \frac {(A-B) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(6 A-11 B) \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 (3 a^2 (6 A-11 B)-a^2 (18 A-43 B) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4} \\ & = \frac {(A-B) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(6 A-11 B) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {4 (9 A-19 B) \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 (8 a^3 (9 A-19 B)-15 a^3 (6 A-13 B) \sec (c+d x)\right ) \, dx}{15 a^6} \\ & = \frac {(A-B) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(6 A-11 B) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {4 (9 A-19 B) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {(8 (9 A-19 B)) \int \sec ^2(c+d x) \, dx}{15 a^3}-\frac {(6 A-13 B) \int \sec ^3(c+d x) \, dx}{a^3} \\ & = -\frac {(6 A-13 B) \sec (c+d x) \tan (c+d x)}{2 a^3 d}+\frac {(A-B) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(6 A-11 B) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {4 (9 A-19 B) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {(6 A-13 B) \int \sec (c+d x) \, dx}{2 a^3}-\frac {(8 (9 A-19 B)) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d} \\ & = -\frac {(6 A-13 B) \text {arctanh}(\sin (c+d x))}{2 a^3 d}+\frac {8 (9 A-19 B) \tan (c+d x)}{15 a^3 d}-\frac {(6 A-13 B) \sec (c+d x) \tan (c+d x)}{2 a^3 d}+\frac {(A-B) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(6 A-11 B) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {4 (9 A-19 B) \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 = 2.79 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.76 \[ \int \frac {\sec ^5(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\sec ^3(c+d x) \left (-480 (6 A-13 B) \text {arctanh}(\sin (c+d x)) \cos ^6\left (\frac {1}{2} (c+d x)\right )+(684 A-1354 B+3 (382 A-777 B) \cos (c+d x)+54 (14 A-29 B) \cos (2 (c+d x))+342 A \cos (3 (c+d x))-717 B \cos (3 (c+d x))+72 A \cos (4 (c+d x))-152 B \cos (4 (c+d x))) \sec (c+d x) \tan (c+d x)\right )}{120 a^3 d (1+\sec (c+d x))^3} \]

[In]

Integrate[(Sec[c + d*x]^5*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^3,x]

[Out]

(Sec[c + d*x]^3*(-480*(6*A - 13*B)*ArcTanh[Sin[c + d*x]]*Cos[(c + d*x)/2]^6 + (684*A - 1354*B + 3*(382*A - 777
*B)*Cos[c + d*x] + 54*(14*A - 29*B)*Cos[2*(c + d*x)] + 342*A*Cos[3*(c + d*x)] - 717*B*Cos[3*(c + d*x)] + 72*A*
Cos[4*(c + d*x)] - 152*B*Cos[4*(c + d*x)])*Sec[c + d*x]*Tan[c + d*x]))/(120*a^3*d*(1 + Sec[c + d*x])^3)

Maple [A] (verified)

Time = 0.99 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.87

method	result	size
parallelrisch	\(\frac {720 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A -\frac {13 B}{6}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-720 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A -\frac {13 B}{6}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+72 \left (\left (\frac {21 A}{2}-\frac {87 B}{4}\right ) \cos \left (2 d x +2 c \right )+\left (\frac {19 A}{4}-\frac {239 B}{24}\right ) \cos \left (3 d x +3 c \right )+\left (A -\frac {19 B}{9}\right ) \cos \left (4 d x +4 c \right )+\left (\frac {191 A}{12}-\frac {259 B}{8}\right ) \cos \left (d x +c \right )+\frac {19 A}{2}-\frac {677 B}{36}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{240 d \,a^{3} \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(175\)
derivativedivides	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A -\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\frac {-14 B +4 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (26 B -12 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 B}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\left (-26 B +12 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {-14 B +4 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {2 B}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}}{4 d \,a^{3}}\)	\(206\)
default	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A -\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\frac {-14 B +4 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (26 B -12 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 B}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\left (-26 B +12 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {-14 B +4 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {2 B}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}}{4 d \,a^{3}}\)	\(206\)
norman	\(\frac {\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{20 a d}+\frac {\left (3 A -5 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{12 a d}-\frac {25 \left (9 A -19 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 a d}-\frac {\left (25 A -51 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (27 A -59 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{12 a d}+\frac {\left (345 A -721 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}+\frac {\left (549 A -1165 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{12 a d}-\frac {\left (3123 A -6613 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} a^{2}}+\frac {\left (6 A -13 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3} d}-\frac {\left (6 A -13 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3} d}\)	\(280\)
risch	\(\frac {i \left (90 A \,{\mathrm e}^{8 i \left (d x +c \right )}-195 B \,{\mathrm e}^{8 i \left (d x +c \right )}+450 A \,{\mathrm e}^{7 i \left (d x +c \right )}-975 B \,{\mathrm e}^{7 i \left (d x +c \right )}+1050 A \,{\mathrm e}^{6 i \left (d x +c \right )}-2275 B \,{\mathrm e}^{6 i \left (d x +c \right )}+1650 A \,{\mathrm e}^{5 i \left (d x +c \right )}-3575 B \,{\mathrm e}^{5 i \left (d x +c \right )}+2094 A \,{\mathrm e}^{4 i \left (d x +c \right )}-4329 B \,{\mathrm e}^{4 i \left (d x +c \right )}+1830 A \,{\mathrm e}^{3 i \left (d x +c \right )}-3805 B \,{\mathrm e}^{3 i \left (d x +c \right )}+1278 A \,{\mathrm e}^{2 i \left (d x +c \right )}-2673 B \,{\mathrm e}^{2 i \left (d x +c \right )}+630 \,{\mathrm e}^{i \left (d x +c \right )} A -1325 B \,{\mathrm e}^{i \left (d x +c \right )}+144 A -304 B \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{a^{3} d}-\frac {13 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 a^{3} d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{a^{3} d}+\frac {13 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 a^{3} d}\)	\(324\)

[In]

int(sec(d*x+c)^5*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/240*(720*(1+cos(2*d*x+2*c))*(A-13/6*B)*ln(tan(1/2*d*x+1/2*c)-1)-720*(1+cos(2*d*x+2*c))*(A-13/6*B)*ln(tan(1/2
*d*x+1/2*c)+1)+72*((21/2*A-87/4*B)*cos(2*d*x+2*c)+(19/4*A-239/24*B)*cos(3*d*x+3*c)+(A-19/9*B)*cos(4*d*x+4*c)+(
191/12*A-259/8*B)*cos(d*x+c)+19/2*A-677/36*B)*sec(1/2*d*x+1/2*c)^4*tan(1/2*d*x+1/2*c))/d/a^3/(1+cos(2*d*x+2*c)
)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.46 \[ \int \frac {\sec ^5(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {15 \, {\left ({\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (16 \, {\left (9 \, A - 19 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (114 \, A - 239 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (234 \, A - 479 \, B\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right ) + 15 \, B\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

integrate(sec(d*x+c)^5*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/60*(15*((6*A - 13*B)*cos(d*x + c)^5 + 3*(6*A - 13*B)*cos(d*x + c)^4 + 3*(6*A - 13*B)*cos(d*x + c)^3 + (6*A
- 13*B)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 15*((6*A - 13*B)*cos(d*x + c)^5 + 3*(6*A - 13*B)*cos(d*x + c)^
4 + 3*(6*A - 13*B)*cos(d*x + c)^3 + (6*A - 13*B)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(16*(9*A - 19*B)*c
os(d*x + c)^4 + 3*(114*A - 239*B)*cos(d*x + c)^3 + (234*A - 479*B)*cos(d*x + c)^2 + 15*(2*A - 3*B)*cos(d*x + c
) + 15*B)*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]

\[ \int \frac {\sec ^5(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {A \sec ^{5}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{6}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

[In]

integrate(sec(d*x+c)**5*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**3,x)

[Out]

(Integral(A*sec(c + d*x)**5/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(B*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 = 0.22 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.87 \[ \int \frac {\sec ^5(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {B {\left (\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} - \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - 3 \, A {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )}}{60 \, d} \]

[In]

integrate(sec(d*x+c)^5*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*(B*(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)
+ 40*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) - 3*A*(40*sin(d*x + c)/((a^3 -
a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (85*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*
x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 60*log(sin(d*x + c)/(cos(d*x + c) +
1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3))/d

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.15 \[ \int \frac {\sec ^5(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {30 \, {\left (6 \, A - 13 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {30 \, {\left (6 \, A - 13 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {60 \, {\left (2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 30 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 255 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 465 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

[In]

integrate(sec(d*x+c)^5*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

-1/60*(30*(6*A - 13*B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 30*(6*A - 13*B)*log(abs(tan(1/2*d*x + 1/2*c) -
1))/a^3 + 60*(2*A*tan(1/2*d*x + 1/2*c)^3 - 7*B*tan(1/2*d*x + 1/2*c)^3 - 2*A*tan(1/2*d*x + 1/2*c) + 5*B*tan(1/
2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^3) - (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^12*tan(1/2*d*x
+ 1/2*c)^5 + 30*A*a^12*tan(1/2*d*x + 1/2*c)^3 - 40*B*a^12*tan(1/2*d*x + 1/2*c)^3 + 255*A*a^12*tan(1/2*d*x + 1
/2*c) - 465*B*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d

Mupad [B] (verification not implemented)

Time = 13.65 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.07 \[ \int \frac {\sec ^5(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-B\right )}{2\,a^3}+\frac {3\,\left (3\,A-5\,B\right )}{4\,a^3}+\frac {2\,A-10\,B}{4\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,A-7\,B\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A-5\,B\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B}{4\,a^3}+\frac {3\,A-5\,B}{12\,a^3}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B\right )}{20\,a^3\,d}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (6\,A-13\,B\right )}{a^3\,d} \]

[In]

int((A + B/cos(c + d*x))/(cos(c + d*x)^5*(a + a/cos(c + d*x))^3),x)

[Out]

(tan(c/2 + (d*x)/2)*((3*(A - B))/(2*a^3) + (3*(3*A - 5*B))/(4*a^3) + (2*A - 10*B)/(4*a^3)))/d - (tan(c/2 + (d*
x)/2)^3*(2*A - 7*B) - tan(c/2 + (d*x)/2)*(2*A - 5*B))/(d*(a^3*tan(c/2 + (d*x)/2)^4 - 2*a^3*tan(c/2 + (d*x)/2)^
2 + a^3)) + (tan(c/2 + (d*x)/2)^3*((A - B)/(4*a^3) + (3*A - 5*B)/(12*a^3)))/d + (tan(c/2 + (d*x)/2)^5*(A - B))
/(20*a^3*d) - (atanh(tan(c/2 + (d*x)/2))*(6*A - 13*B))/(a^3*d)