\(\int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx\) [91]

Optimal result

Integrand size = 31, antiderivative size = 156 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=-\frac {(4 A-7 B) \text {arctanh}(\sin (c+d x))}{2 a^2 d}+\frac {2 (5 A-8 B) \tan (c+d x)}{3 a^2 d}-\frac {(4 A-7 B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac {(5 A-8 B) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2} \]

[Out]

-1/2*(4*A-7*B)*arctanh(sin(d*x+c))/a^2/d+2/3*(5*A-8*B)*tan(d*x+c)/a^2/d-1/2*(4*A-7*B)*sec(d*x+c)*tan(d*x+c)/a^
2/d+1/3*(5*A-8*B)*sec(d*x+c)^2*tan(d*x+c)/a^2/d/(1+sec(d*x+c))+1/3*(A-B)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*
x+c))^2

Rubi [A] (verified)

Time = 0.38 (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, 3872, 3852, 8, 3853, 3855} \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=-\frac {(4 A-7 B) \text {arctanh}(\sin (c+d x))}{2 a^2 d}+\frac {2 (5 A-8 B) \tan (c+d x)}{3 a^2 d}+\frac {(5 A-8 B) \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d (\sec (c+d x)+1)}-\frac {(4 A-7 B) \tan (c+d x) \sec (c+d x)}{2 a^2 d}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[In]

Int[(Sec[c + d*x]^4*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^2,x]

[Out]

-1/2*((4*A - 7*B)*ArcTanh[Sin[c + d*x]])/(a^2*d) + (2*(5*A - 8*B)*Tan[c + d*x])/(3*a^2*d) - ((4*A - 7*B)*Sec[c
+ d*x]*Tan[c + d*x])/(2*a^2*d) + ((5*A - 8*B)*Sec[c + d*x]^2*Tan[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) + ((A
- B)*Sec[c + d*x]^3*Tan[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

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 ^3(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\int \frac {\sec ^3(c+d x) (3 a (A-B)-a (2 A-5 B) \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2} \\ & = \frac {(5 A-8 B) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\int \sec ^2(c+d x) \left (2 a^2 (5 A-8 B)-3 a^2 (4 A-7 B) \sec (c+d x)\right ) \, dx}{3 a^4} \\ & = \frac {(5 A-8 B) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {(2 (5 A-8 B)) \int \sec ^2(c+d x) \, dx}{3 a^2}-\frac {(4 A-7 B) \int \sec ^3(c+d x) \, dx}{a^2} \\ & = -\frac {(4 A-7 B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac {(5 A-8 B) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(4 A-7 B) \int \sec (c+d x) \, dx}{2 a^2}-\frac {(2 (5 A-8 B)) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 a^2 d} \\ & = -\frac {(4 A-7 B) \text {arctanh}(\sin (c+d x))}{2 a^2 d}+\frac {2 (5 A-8 B) \tan (c+d x)}{3 a^2 d}-\frac {(4 A-7 B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac {(5 A-8 B) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.50 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.83 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\frac {\sec ^2(c+d x) \left (-24 (4 A-7 B) \text {arctanh}(\sin (c+d x)) \cos ^4\left (\frac {1}{2} (c+d x)\right )+(28 A-37 B+6 (7 A-10 B) \cos (c+d x)+(28 A-43 B) \cos (2 (c+d x))+10 A \cos (3 (c+d x))-16 B \cos (3 (c+d x))) \sec (c+d x) \tan (c+d x)\right )}{12 a^2 d (1+\sec (c+d x))^2} \]

[In]

Integrate[(Sec[c + d*x]^4*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^2,x]

[Out]

(Sec[c + d*x]^2*(-24*(4*A - 7*B)*ArcTanh[Sin[c + d*x]]*Cos[(c + d*x)/2]^4 + (28*A - 37*B + 6*(7*A - 10*B)*Cos[
c + d*x] + (28*A - 43*B)*Cos[2*(c + d*x)] + 10*A*Cos[3*(c + d*x)] - 16*B*Cos[3*(c + d*x)])*Sec[c + d*x]*Tan[c
+ d*x]))/(12*a^2*d*(1 + Sec[c + d*x])^2)

Maple [A] (verified)

Time = 0.93 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.03

method	result	size
parallelrisch	\(\frac {4 \left (A -\frac {7 B}{4}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-4 \left (A -\frac {7 B}{4}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+7 \left (\frac {\left (2 A -\frac {43 B}{14}\right ) \cos \left (2 d x +2 c \right )}{3}+\frac {\left (5 A -8 B \right ) \cos \left (3 d x +3 c \right )}{21}+\left (A -\frac {10 B}{7}\right ) \cos \left (d x +c \right )+\frac {2 A}{3}-\frac {37 B}{42}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2} \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(160\)
derivativedivides	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\frac {2 A -5 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (4 A -7 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {B}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 A -5 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (7 B -4 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {B}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}{2 d \,a^{2}}\)	\(177\)
default	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\frac {2 A -5 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (4 A -7 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {B}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 A -5 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (7 B -4 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {B}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}{2 d \,a^{2}}\)	\(177\)
norman	\(\frac {\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{6 a d}+\frac {\left (9 A -13 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {\left (11 A -18 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}+\frac {\left (11 A -17 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{6 a d}+\frac {\left (61 A -100 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 a d}-\frac {\left (95 A -149 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 a d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} a}+\frac {\left (4 A -7 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{2} d}-\frac {\left (4 A -7 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{2} d}\)	\(228\)
risch	\(\frac {i \left (12 A \,{\mathrm e}^{6 i \left (d x +c \right )}-21 B \,{\mathrm e}^{6 i \left (d x +c \right )}+36 A \,{\mathrm e}^{5 i \left (d x +c \right )}-63 B \,{\mathrm e}^{5 i \left (d x +c \right )}+56 A \,{\mathrm e}^{4 i \left (d x +c \right )}-98 B \,{\mathrm e}^{4 i \left (d x +c \right )}+84 A \,{\mathrm e}^{3 i \left (d x +c \right )}-126 B \,{\mathrm e}^{3 i \left (d x +c \right )}+64 A \,{\mathrm e}^{2 i \left (d x +c \right )}-97 B \,{\mathrm e}^{2 i \left (d x +c \right )}+48 \,{\mathrm e}^{i \left (d x +c \right )} A -75 B \,{\mathrm e}^{i \left (d x +c \right )}+20 A -32 B \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{a^{2} d}-\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 a^{2} d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{a^{2} d}+\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 a^{2} d}\)	\(276\)

[In]

int(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/2*(4*(A-7/4*B)*(1+cos(2*d*x+2*c))*ln(tan(1/2*d*x+1/2*c)-1)-4*(A-7/4*B)*(1+cos(2*d*x+2*c))*ln(tan(1/2*d*x+1/2
*c)+1)+7*(1/3*(2*A-43/14*B)*cos(2*d*x+2*c)+1/21*(5*A-8*B)*cos(3*d*x+3*c)+(A-10/7*B)*cos(d*x+c)+2/3*A-37/42*B)*
sec(1/2*d*x+1/2*c)^2*tan(1/2*d*x+1/2*c))/d/a^2/(1+cos(2*d*x+2*c))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.46 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=-\frac {3 \, {\left ({\left (4 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (4 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (4 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (4 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (4 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (4 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (5 \, A - 8 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (28 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{2} + 6 \, {\left (A - B\right )} \cos \left (d x + c\right ) + 3 \, B\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/12*(3*((4*A - 7*B)*cos(d*x + c)^4 + 2*(4*A - 7*B)*cos(d*x + c)^3 + (4*A - 7*B)*cos(d*x + c)^2)*log(sin(d*x
+ c) + 1) - 3*((4*A - 7*B)*cos(d*x + c)^4 + 2*(4*A - 7*B)*cos(d*x + c)^3 + (4*A - 7*B)*cos(d*x + c)^2)*log(-si
n(d*x + c) + 1) - 2*(4*(5*A - 8*B)*cos(d*x + c)^3 + (28*A - 43*B)*cos(d*x + c)^2 + 6*(A - B)*cos(d*x + c) + 3*
B)*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)

Sympy [F]

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

[In]

integrate(sec(d*x+c)**4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**2,x)

[Out]

(Integral(A*sec(c + d*x)**4/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)**5/(sec(c + d
*x)**2 + 2*sec(c + d*x) + 1), x))/a**2

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 336 vs. \(2 (146) = 292\).

Time = 0.22 (sec) , antiderivative size = 336, normalized size of antiderivative = 2.15 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=-\frac {B {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} - \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} - A {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} - \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{6 \, d} \]

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/6*(B*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) - 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 - 2*a^2*sin(d*x +
c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) + s
in(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 21*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 21*log(sin(d*x + c
)/(cos(d*x + c) + 1) - 1)/a^2) - A*((15*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)
/a^2 - 12*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 12*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2 + 12*
sin(d*x + c)/((a^2 - a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1))))/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.27 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=-\frac {\frac {3 \, {\left (4 \, A - 7 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {3 \, {\left (4 \, A - 7 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {6 \, {\left (2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{2}} - \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 21 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-1/6*(3*(4*A - 7*B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 3*(4*A - 7*B)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/
a^2 + 6*(2*A*tan(1/2*d*x + 1/2*c)^3 - 5*B*tan(1/2*d*x + 1/2*c)^3 - 2*A*tan(1/2*d*x + 1/2*c) + 3*B*tan(1/2*d*x
+ 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^2) - (A*a^4*tan(1/2*d*x + 1/2*c)^3 - B*a^4*tan(1/2*d*x + 1/2*c)^3
+ 15*A*a^4*tan(1/2*d*x + 1/2*c) - 21*B*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

Mupad [B] (verification not implemented)

Time = 13.60 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.06 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-B\right )}{2\,a^2}+\frac {2\,A-4\,B}{2\,a^2}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,A-5\,B\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A-3\,B\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B\right )}{6\,a^2\,d}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,A-7\,B\right )}{a^2\,d} \]

[In]

int((A + B/cos(c + d*x))/(cos(c + d*x)^4*(a + a/cos(c + d*x))^2),x)

[Out]

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