\(\int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx\) [100]

Optimal result

Integrand size = 31, antiderivative size = 156 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {(A-3 B) \text {arctanh}(\sin (c+d x))}{a^3 d}-\frac {(7 A-27 B) \tan (c+d x)}{15 a^3 d}+\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(4 A-9 B) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(A-3 B) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )} \]

[Out]

(A-3*B)*arctanh(sin(d*x+c))/a^3/d-1/15*(7*A-27*B)*tan(d*x+c)/a^3/d+1/5*(A-B)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*se
c(d*x+c))^3+1/15*(4*A-9*B)*sec(d*x+c)^2*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^2-(A-3*B)*tan(d*x+c)/d/(a^3+a^3*sec(d*
x+c))

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4104, 4093, 3872, 3855, 3852, 8} \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {(A-3 B) \text {arctanh}(\sin (c+d x))}{a^3 d}-\frac {(7 A-27 B) \tan (c+d x)}{15 a^3 d}-\frac {(A-3 B) \tan (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {(4 A-9 B) \tan (c+d x) \sec ^2(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]

[In]

Int[(Sec[c + d*x]^4*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^3,x]

[Out]

((A - 3*B)*ArcTanh[Sin[c + d*x]])/(a^3*d) - ((7*A - 27*B)*Tan[c + d*x])/(15*a^3*d) + ((A - B)*Sec[c + d*x]^3*T
an[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + ((4*A - 9*B)*Sec[c + d*x]^2*Tan[c + d*x])/(15*a*d*(a + a*Sec[c + d
*x])^2) - ((A - 3*B)*Tan[c + d*x])/(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 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 4093

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

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 ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {\int \frac {\sec ^3(c+d x) (3 a (A-B)-a (A-6 B) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2} \\ & = \frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(4 A-9 B) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {\int \frac {\sec ^2(c+d x) \left (2 a^2 (4 A-9 B)-a^2 (7 A-27 B) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4} \\ & = \frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(4 A-9 B) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(A-3 B) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {\int \sec (c+d x) \left (-15 a^3 (A-3 B)+a^3 (7 A-27 B) \sec (c+d x)\right ) \, dx}{15 a^6} \\ & = \frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(4 A-9 B) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(A-3 B) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {(7 A-27 B) \int \sec ^2(c+d x) \, dx}{15 a^3}+\frac {(A-3 B) \int \sec (c+d x) \, dx}{a^3} \\ & = \frac {(A-3 B) \text {arctanh}(\sin (c+d x))}{a^3 d}+\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(4 A-9 B) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(A-3 B) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {(7 A-27 B) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d} \\ & = \frac {(A-3 B) \text {arctanh}(\sin (c+d x))}{a^3 d}-\frac {(7 A-27 B) \tan (c+d x)}{15 a^3 d}+\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(4 A-9 B) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(A-3 B) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.34 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.92 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (120 (A-3 B) \text {arctanh}(\sin (c+d x)) \cos ^5\left (\frac {1}{2} (c+d x)\right )-(51 A-201 B+(97 A-342 B) \cos (c+d x)+3 (17 A-57 B) \cos (2 (c+d x))+11 A \cos (3 (c+d x))-36 B \cos (3 (c+d x))) \sec (c+d x) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{15 a^3 d (1+\sec (c+d x))^3} \]

[In]

Integrate[(Sec[c + d*x]^4*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^3,x]

[Out]

(Cos[(c + d*x)/2]*Sec[c + d*x]^3*(120*(A - 3*B)*ArcTanh[Sin[c + d*x]]*Cos[(c + d*x)/2]^5 - (51*A - 201*B + (97
*A - 342*B)*Cos[c + d*x] + 3*(17*A - 57*B)*Cos[2*(c + d*x)] + 11*A*Cos[3*(c + d*x)] - 36*B*Cos[3*(c + d*x)])*S
ec[c + d*x]*Sin[(c + d*x)/2]))/(15*a^3*d*(1 + Sec[c + d*x])^3)

Maple [A] (verified)

Time = 0.97 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.93

method	result	size
parallelrisch	\(\frac {-120 \cos \left (d x +c \right ) \left (A -3 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+120 \cos \left (d x +c \right ) \left (A -3 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-97 \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\frac {3 \left (17 A -57 B \right ) \cos \left (2 d x +2 c \right )}{97}+\frac {\left (11 A -36 B \right ) \cos \left (3 d x +3 c \right )}{97}+\left (A -\frac {342 B}{97}\right ) \cos \left (d x +c \right )+\frac {51 A}{97}-\frac {201 B}{97}\right )}{120 a^{3} d \cos \left (d x +c \right )}\)	\(145\)
derivativedivides	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\left (12 B -4 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-12 B +4 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{4 d \,a^{3}}\)	\(162\)
default	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\left (12 B -4 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-12 B +4 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{4 d \,a^{3}}\)	\(162\)
norman	\(\frac {-\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{20 a d}-\frac {\left (4 A -9 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{30 a d}-\frac {\left (7 A -25 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {5 \left (8 A -27 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 a d}+\frac {\left (26 A -81 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{5 a d}-\frac {\left (43 A -153 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{60 a d}-\frac {\left (553 A -1773 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} a^{2}}+\frac {\left (A -3 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3} d}-\frac {\left (A -3 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3} d}\)	\(249\)
risch	\(-\frac {2 i \left (15 A \,{\mathrm e}^{6 i \left (d x +c \right )}-45 B \,{\mathrm e}^{6 i \left (d x +c \right )}+75 A \,{\mathrm e}^{5 i \left (d x +c \right )}-225 B \,{\mathrm e}^{5 i \left (d x +c \right )}+160 A \,{\mathrm e}^{4 i \left (d x +c \right )}-480 B \,{\mathrm e}^{4 i \left (d x +c \right )}+170 A \,{\mathrm e}^{3 i \left (d x +c \right )}-600 B \,{\mathrm e}^{3 i \left (d x +c \right )}+167 A \,{\mathrm e}^{2 i \left (d x +c \right )}-567 B \,{\mathrm e}^{2 i \left (d x +c \right )}+95 \,{\mathrm e}^{i \left (d x +c \right )} A -315 B \,{\mathrm e}^{i \left (d x +c \right )}+22 A -72 B \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{a^{3} d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{a^{3} d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{3} d}\)	\(275\)

[In]

int(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/120*(-120*cos(d*x+c)*(A-3*B)*ln(tan(1/2*d*x+1/2*c)-1)+120*cos(d*x+c)*(A-3*B)*ln(tan(1/2*d*x+1/2*c)+1)-97*sec
(1/2*d*x+1/2*c)^4*tan(1/2*d*x+1/2*c)*(3/97*(17*A-57*B)*cos(2*d*x+2*c)+1/97*(11*A-36*B)*cos(3*d*x+3*c)+(A-342/9
7*B)*cos(d*x+c)+51/97*A-201/97*B))/a^3/d/cos(d*x+c)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.64 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {15 \, {\left ({\left (A - 3 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (A - 3 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (A - 3 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (A - 3 \, B\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (A - 3 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (A - 3 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (A - 3 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (A - 3 \, B\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, {\left (11 \, A - 36 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (17 \, A - 57 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (32 \, A - 117 \, B\right )} \cos \left (d x + c\right ) - 15 \, B\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \]

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/30*(15*((A - 3*B)*cos(d*x + c)^4 + 3*(A - 3*B)*cos(d*x + c)^3 + 3*(A - 3*B)*cos(d*x + c)^2 + (A - 3*B)*cos(d
*x + c))*log(sin(d*x + c) + 1) - 15*((A - 3*B)*cos(d*x + c)^4 + 3*(A - 3*B)*cos(d*x + c)^3 + 3*(A - 3*B)*cos(d
*x + c)^2 + (A - 3*B)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(2*(11*A - 36*B)*cos(d*x + c)^3 + 3*(17*A - 57*
B)*cos(d*x + c)^2 + (32*A - 117*B)*cos(d*x + c) - 15*B)*sin(d*x + c))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x
+ c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))

Sympy [F]

\[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {A \sec ^{4}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{5}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

[In]

integrate(sec(d*x+c)**4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**3,x)

[Out]

(Integral(A*sec(c + d*x)**4/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(B*sec(c
+ d*x)**5/(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 = 286, normalized size of antiderivative = 1.83 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {3 \, B {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - A {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )}}{60 \, d} \]

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(3*B*(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) - A*((105*s
in(d*x + c)/(cos(d*x + c) + 1) + 20*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 - 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 = 0.33 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.19 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {60 \, {\left (A - 3 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, {\left (A - 3 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {120 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 20 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 255 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(60*(A - 3*B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 60*(A - 3*B)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^
3 - 120*B*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a^3) - (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^1
2*tan(1/2*d*x + 1/2*c)^5 + 20*A*a^12*tan(1/2*d*x + 1/2*c)^3 - 30*B*a^12*tan(1/2*d*x + 1/2*c)^3 + 105*A*a^12*ta
n(1/2*d*x + 1/2*c) - 255*B*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d

Mupad [B] (verification not implemented)

Time = 13.74 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.08 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A-3\,B\right )}{a^3\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-B\right )}{4\,a^3}-\frac {3\,B}{2\,a^3}+\frac {2\,A-4\,B}{2\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B\right )}{20\,a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B}{6\,a^3}+\frac {2\,A-4\,B}{12\,a^3}\right )}{d}-\frac {2\,B\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3\right )} \]

[In]

int((A + B/cos(c + d*x))/(cos(c + d*x)^4*(a + a/cos(c + d*x))^3),x)

[Out]

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