3.5.78 \(\int \frac {\sec (c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^4} \, dx\) [478]

Optimal. Leaf size=154 \[ -\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(8 A-B-6 C) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {(6 A+8 B+13 C) \tan (c+d x)}{105 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac {(6 A+8 B+13 C) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )} \]

[Out]

-1/7*(A-B+C)*sec(d*x+c)*tan(d*x+c)/d/(a+a*sec(d*x+c))^4+1/35*(8*A-B-6*C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^3+1/1
05*(6*A+8*B+13*C)*tan(d*x+c)/d/(a^2+a^2*sec(d*x+c))^2+1/105*(6*A+8*B+13*C)*tan(d*x+c)/d/(a^4+a^4*sec(d*x+c))

________________________________________________________________________________________

Rubi [A]
time = 0.19, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {4163, 4085, 3881, 3879} \begin {gather*} \frac {(6 A+8 B+13 C) \tan (c+d x)}{105 d \left (a^4 \sec (c+d x)+a^4\right )}+\frac {(6 A+8 B+13 C) \tan (c+d x)}{105 d \left (a^2 \sec (c+d x)+a^2\right )^2}+\frac {(8 A-B-6 C) \tan (c+d x)}{35 a d (a \sec (c+d x)+a)^3}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{7 d (a \sec (c+d x)+a)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/7*((A - B + C)*Sec[c + d*x]*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^4) + ((8*A - B - 6*C)*Tan[c + d*x])/(35*a
*d*(a + a*Sec[c + d*x])^3) + ((6*A + 8*B + 13*C)*Tan[c + d*x])/(105*d*(a^2 + a^2*Sec[c + d*x])^2) + ((6*A + 8*
B + 13*C)*Tan[c + d*x])/(105*d*(a^4 + a^4*Sec[c + d*x]))

Rule 3879

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[-Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 3881

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

Rule 4085

Int[csc[(e_.) + (f_.)*(x_)]*(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/(a*f*(2*m + 1))), x] + Dist[(a*B*m + A*b*
(m + 1))/(a*b*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, A, B, e, f}, x
] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] && LtQ[m, -2^(-1)]

Rule 4163

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

Rubi steps

\begin {align*} \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx &=-\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {\int \frac {\sec (c+d x) (a (6 A+B-C)-a (2 A-2 B-5 C) \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(8 A-B-6 C) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {(6 A+8 B+13 C) \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx}{35 a^2}\\ &=-\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(8 A-B-6 C) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {(6 A+8 B+13 C) \tan (c+d x)}{105 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac {(6 A+8 B+13 C) \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{105 a^3}\\ &=-\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(8 A-B-6 C) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {(6 A+8 B+13 C) \tan (c+d x)}{105 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac {(6 A+8 B+13 C) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.83, size = 231, normalized size = 1.50 \begin {gather*} \frac {\sec \left (\frac {c}{2}\right ) \sec ^7\left (\frac {1}{2} (c+d x)\right ) \left (70 (9 A+4 B+2 C) \sin \left (\frac {d x}{2}\right )-35 (18 A+5 B+4 C) \sin \left (c+\frac {d x}{2}\right )+441 A \sin \left (c+\frac {3 d x}{2}\right )+168 B \sin \left (c+\frac {3 d x}{2}\right )+168 C \sin \left (c+\frac {3 d x}{2}\right )-315 A \sin \left (2 c+\frac {3 d x}{2}\right )-105 B \sin \left (2 c+\frac {3 d x}{2}\right )+147 A \sin \left (2 c+\frac {5 d x}{2}\right )+91 B \sin \left (2 c+\frac {5 d x}{2}\right )+56 C \sin \left (2 c+\frac {5 d x}{2}\right )-105 A \sin \left (3 c+\frac {5 d x}{2}\right )+36 A \sin \left (3 c+\frac {7 d x}{2}\right )+13 B \sin \left (3 c+\frac {7 d x}{2}\right )+8 C \sin \left (3 c+\frac {7 d x}{2}\right )\right )}{6720 a^4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^7*(70*(9*A + 4*B + 2*C)*Sin[(d*x)/2] - 35*(18*A + 5*B + 4*C)*Sin[c + (d*x)/2] + 441
*A*Sin[c + (3*d*x)/2] + 168*B*Sin[c + (3*d*x)/2] + 168*C*Sin[c + (3*d*x)/2] - 315*A*Sin[2*c + (3*d*x)/2] - 105
*B*Sin[2*c + (3*d*x)/2] + 147*A*Sin[2*c + (5*d*x)/2] + 91*B*Sin[2*c + (5*d*x)/2] + 56*C*Sin[2*c + (5*d*x)/2] -
105*A*Sin[3*c + (5*d*x)/2] + 36*A*Sin[3*c + (7*d*x)/2] + 13*B*Sin[3*c + (7*d*x)/2] + 8*C*Sin[3*c + (7*d*x)/2]
))/(6720*a^4*d)

________________________________________________________________________________________

Maple [A]
time = 0.64, size = 108, normalized size = 0.70

method	result	size
derivativedivides	\(\frac {\frac {\left (-A +B -C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {\left (3 A -B -C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (-3 A -B +C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\)	\(108\)
default	\(\frac {\frac {\left (-A +B -C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {\left (3 A -B -C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (-3 A -B +C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\)	\(108\)
norman	\(\frac {-\frac {\left (A -B +C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 a d}+\frac {\left (A +B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}-\frac {\left (9 A +7 B +5 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 a d}+\frac {\left (27 A +11 B +C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}+\frac {\left (31 A -17 B +3 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{280 a d}-\frac {\left (123 A -11 B -31 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{420 a d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} a^{3}}\)	\(181\)
risch	\(\frac {2 i \left (105 A \,{\mathrm e}^{6 i \left (d x +c \right )}+315 A \,{\mathrm e}^{5 i \left (d x +c \right )}+105 B \,{\mathrm e}^{5 i \left (d x +c \right )}+630 A \,{\mathrm e}^{4 i \left (d x +c \right )}+175 B \,{\mathrm e}^{4 i \left (d x +c \right )}+140 C \,{\mathrm e}^{4 i \left (d x +c \right )}+630 A \,{\mathrm e}^{3 i \left (d x +c \right )}+280 B \,{\mathrm e}^{3 i \left (d x +c \right )}+140 C \,{\mathrm e}^{3 i \left (d x +c \right )}+441 A \,{\mathrm e}^{2 i \left (d x +c \right )}+168 B \,{\mathrm e}^{2 i \left (d x +c \right )}+168 C \,{\mathrm e}^{2 i \left (d x +c \right )}+147 \,{\mathrm e}^{i \left (d x +c \right )} A +91 B \,{\mathrm e}^{i \left (d x +c \right )}+56 C \,{\mathrm e}^{i \left (d x +c \right )}+36 A +13 B +8 C \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}\)	\(213\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.29, size = 259, normalized size = 1.68 \begin {gather*} \frac {\frac {C {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} + \frac {B {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} + \frac {3 \, A {\left (\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/840*(C*(105*sin(d*x + c)/(cos(d*x + c) + 1) + 35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 21*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5 - 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 + B*(105*sin(d*x + c)/(cos(d*x + c) + 1) - 35*
sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 15*sin(d*x + c)^7/(cos(d*x + c)
+ 1)^7)/a^4 + 3*A*(35*sin(d*x + c)/(cos(d*x + c) + 1) - 35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x +
c)^5/(cos(d*x + c) + 1)^5 - 5*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4)/d

________________________________________________________________________________________

Fricas [A]
time = 1.93, size = 135, normalized size = 0.88 \begin {gather*} \frac {{\left ({\left (36 \, A + 13 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (39 \, A + 52 \, B + 32 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (6 \, A + 8 \, B + 13 \, C\right )} \cos \left (d x + c\right ) + 6 \, A + 8 \, B + 13 \, C\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*((36*A + 13*B + 8*C)*cos(d*x + c)^3 + (39*A + 52*B + 32*C)*cos(d*x + c)^2 + 4*(6*A + 8*B + 13*C)*cos(d*x
+ c) + 6*A + 8*B + 13*C)*sin(d*x + c)/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2
+ 4*a^4*d*cos(d*x + c) + a^4*d)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {A \sec {\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{2}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{3}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(A*sec(c + d*x)/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x) +
Integral(B*sec(c + d*x)**2/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x)
+ Integral(C*sec(c + d*x)**3/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x
))/a**4

________________________________________________________________________________________

Giac [A]
time = 0.51, size = 171, normalized size = 1.11 \begin {gather*} -\frac {15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 63 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 21 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 21 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 105 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 35 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 35 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 105 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{840 \, a^{4} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/840*(15*A*tan(1/2*d*x + 1/2*c)^7 - 15*B*tan(1/2*d*x + 1/2*c)^7 + 15*C*tan(1/2*d*x + 1/2*c)^7 - 63*A*tan(1/2
*d*x + 1/2*c)^5 + 21*B*tan(1/2*d*x + 1/2*c)^5 + 21*C*tan(1/2*d*x + 1/2*c)^5 + 105*A*tan(1/2*d*x + 1/2*c)^3 + 3
5*B*tan(1/2*d*x + 1/2*c)^3 - 35*C*tan(1/2*d*x + 1/2*c)^3 - 105*A*tan(1/2*d*x + 1/2*c) - 105*B*tan(1/2*d*x + 1/
2*c) - 105*C*tan(1/2*d*x + 1/2*c))/(a^4*d)

________________________________________________________________________________________

Mupad [B]
time = 3.31, size = 99, normalized size = 0.64 \begin {gather*} \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+B+C\right )}{8\,a^4\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A-B+C\right )}{56\,a^4\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (B-3\,A+C\right )}{40\,a^4\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,A+B-C\right )}{24\,a^4\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(A + B + C))/(8*a^4*d) - (tan(c/2 + (d*x)/2)^7*(A - B + C))/(56*a^4*d) - (tan(c/2 + (d*x)/
2)^5*(B - 3*A + C))/(40*a^4*d) - (tan(c/2 + (d*x)/2)^3*(3*A + B - C))/(24*a^4*d)

________________________________________________________________________________________