\(\int \frac {\sec ^3(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^4} \, dx\) [148]

Optimal result

Integrand size = 33, antiderivative size = 161 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {C \text {arctanh}(\sin (c+d x))}{a^4 d}-\frac {(8 A-55 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}+\frac {(16 A-215 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))}-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {2 (2 A-5 C) \sec ^2(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3} \]

[Out]

C*arctanh(sin(d*x+c))/a^4/d-1/105*(8*A-55*C)*tan(d*x+c)/a^4/d/(1+sec(d*x+c))^2+1/105*(16*A-215*C)*tan(d*x+c)/a
^4/d/(1+sec(d*x+c))-1/7*(A+C)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^4+2/35*(2*A-5*C)*sec(d*x+c)^2*tan(d*x
+c)/a/d/(a+a*sec(d*x+c))^3

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4170, 4104, 4093, 4083, 3855, 3879} \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {(16 A-215 C) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)}-\frac {(8 A-55 C) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}+\frac {C \text {arctanh}(\sin (c+d x))}{a^4 d}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}+\frac {2 (2 A-5 C) \tan (c+d x) \sec ^2(c+d x)}{35 a d (a \sec (c+d x)+a)^3} \]

[In]

Int[(Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^4,x]

[Out]

(C*ArcTanh[Sin[c + d*x]])/(a^4*d) - ((8*A - 55*C)*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])^2) + ((16*A - 21
5*C)*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])) - ((A + C)*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*(a + a*Sec[c +
d*x])^4) + (2*(2*A - 5*C)*Sec[c + d*x]^2*Tan[c + d*x])/(35*a*d*(a + a*Sec[c + d*x])^3)

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{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 4083

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

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]

Rule 4170

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

Rubi steps \begin{align*} \text {integral}& = -\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {\int \frac {\sec ^3(c+d x) (-a (4 A-3 C)-7 a C \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2} \\ & = -\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {2 (2 A-5 C) \sec ^2(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {\int \frac {\sec ^2(c+d x) \left (-4 a^2 (2 A-5 C)-35 a^2 C \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4} \\ & = -\frac {(8 A-55 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {2 (2 A-5 C) \sec ^2(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {\int \frac {\sec (c+d x) \left (2 a^3 (8 A-55 C)+105 a^3 C \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{105 a^6} \\ & = -\frac {(8 A-55 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {2 (2 A-5 C) \sec ^2(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {(16 A-215 C) \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{105 a^3}+\frac {C \int \sec (c+d x) \, dx}{a^4} \\ & = \frac {C \text {arctanh}(\sin (c+d x))}{a^4 d}-\frac {(8 A-55 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {2 (2 A-5 C) \sec ^2(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {(16 A-215 C) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.18 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.76 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=-\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (6720 C \cos ^7\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-\sec \left (\frac {c}{2}\right ) \left (70 (2 A-49 C) \sin \left (\frac {d x}{2}\right )-70 (2 A-31 C) \sin \left (c+\frac {d x}{2}\right )+168 A \sin \left (c+\frac {3 d x}{2}\right )-2625 C \sin \left (c+\frac {3 d x}{2}\right )+735 C \sin \left (2 c+\frac {3 d x}{2}\right )+56 A \sin \left (2 c+\frac {5 d x}{2}\right )-1015 C \sin \left (2 c+\frac {5 d x}{2}\right )+105 C \sin \left (3 c+\frac {5 d x}{2}\right )+8 A \sin \left (3 c+\frac {7 d x}{2}\right )-160 C \sin \left (3 c+\frac {7 d x}{2}\right )\right )\right )}{210 a^4 d (A+2 C+A \cos (2 (c+d x))) (1+\sec (c+d x))^4} \]

[In]

Integrate[(Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^4,x]

[Out]

-1/210*(Cos[(c + d*x)/2]*Sec[c + d*x]^2*(A + C*Sec[c + d*x]^2)*(6720*C*Cos[(c + d*x)/2]^7*(Log[Cos[(c + d*x)/2
] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Sec[c/2]*(70*(2*A - 49*C)*Sin[(d*x)/2] - 7
0*(2*A - 31*C)*Sin[c + (d*x)/2] + 168*A*Sin[c + (3*d*x)/2] - 2625*C*Sin[c + (3*d*x)/2] + 735*C*Sin[2*c + (3*d*
x)/2] + 56*A*Sin[2*c + (5*d*x)/2] - 1015*C*Sin[2*c + (5*d*x)/2] + 105*C*Sin[3*c + (5*d*x)/2] + 8*A*Sin[3*c + (
7*d*x)/2] - 160*C*Sin[3*c + (7*d*x)/2])))/(a^4*d*(A + 2*C + A*Cos[2*(c + d*x)])*(1 + Sec[c + d*x])^4)

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.69

method	result	size
parallelrisch	\(\frac {-56 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+56 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+7 \left (\frac {A}{5}+C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {7 \left (-A +11 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3}-7 A +105 C \right )}{56 a^{4} d}\)	\(111\)
derivativedivides	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} A}{7}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} C}{7}+8 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-8 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{8 d \,a^{4}}\)	\(147\)
default	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} A}{7}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} C}{7}+8 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-8 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{8 d \,a^{4}}\)	\(147\)
risch	\(-\frac {2 i \left (105 C \,{\mathrm e}^{6 i \left (d x +c \right )}+735 C \,{\mathrm e}^{5 i \left (d x +c \right )}-140 A \,{\mathrm e}^{4 i \left (d x +c \right )}+2170 C \,{\mathrm e}^{4 i \left (d x +c \right )}-140 A \,{\mathrm e}^{3 i \left (d x +c \right )}+3430 C \,{\mathrm e}^{3 i \left (d x +c \right )}-168 A \,{\mathrm e}^{2 i \left (d x +c \right )}+2625 C \,{\mathrm e}^{2 i \left (d x +c \right )}-56 A \,{\mathrm e}^{i \left (d x +c \right )}+1015 C \,{\mathrm e}^{i \left (d x +c \right )}-8 A +160 C \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{4} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{4} d}\)	\(194\)
norman	\(\frac {\frac {\left (A -15 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}-\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{56 a d}+\frac {\left (13 A -15 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{280 a d}+\frac {\left (29 A -55 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{840 a d}-\frac {\left (47 A -1465 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{280 a d}-\frac {\left (101 A +605 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{840 a d}+\frac {\left (-1145 C +67 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{120 a d}-\frac {\left (-169 C +11 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4} a^{3}}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4} d}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4} d}\)	\(263\)

[In]

int(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

1/56*(-56*C*ln(tan(1/2*d*x+1/2*c)-1)+56*C*ln(tan(1/2*d*x+1/2*c)+1)-tan(1/2*d*x+1/2*c)*((A+C)*tan(1/2*d*x+1/2*c
)^6+7*(1/5*A+C)*tan(1/2*d*x+1/2*c)^4+7/3*(-A+11*C)*tan(1/2*d*x+1/2*c)^2-7*A+105*C))/a^4/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.46 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {105 \, {\left (C \cos \left (d x + c\right )^{4} + 4 \, C \cos \left (d x + c\right )^{3} + 6 \, C \cos \left (d x + c\right )^{2} + 4 \, C \cos \left (d x + c\right ) + C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left (C \cos \left (d x + c\right )^{4} + 4 \, C \cos \left (d x + c\right )^{3} + 6 \, C \cos \left (d x + c\right )^{2} + 4 \, C \cos \left (d x + c\right ) + C\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (A - 20 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (32 \, A - 535 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (13 \, A - 155 \, C\right )} \cos \left (d x + c\right ) + 13 \, A - 260 \, C\right )} \sin \left (d x + c\right )}{210 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

[In]

integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

1/210*(105*(C*cos(d*x + c)^4 + 4*C*cos(d*x + c)^3 + 6*C*cos(d*x + c)^2 + 4*C*cos(d*x + c) + C)*log(sin(d*x + c
) + 1) - 105*(C*cos(d*x + c)^4 + 4*C*cos(d*x + c)^3 + 6*C*cos(d*x + c)^2 + 4*C*cos(d*x + c) + C)*log(-sin(d*x
+ c) + 1) + 2*(8*(A - 20*C)*cos(d*x + c)^3 + (32*A - 535*C)*cos(d*x + c)^2 + 4*(13*A - 155*C)*cos(d*x + c) + 1
3*A - 260*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*c
os(d*x + c) + a^4*d)

Sympy [F]

\[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {\int \frac {A \sec ^{3}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{5}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

[In]

integrate(sec(d*x+c)**3*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**4,x)

[Out]

(Integral(A*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)
+ Integral(C*sec(c + d*x)**5/(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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.42 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=-\frac {5 \, C {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} - \frac {A {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \]

[In]

integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/840*(5*C*((315*sin(d*x + c)/(cos(d*x + c) + 1) + 77*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5
/(cos(d*x + c) + 1)^5 + 3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 168*log(sin(d*x + c)/(cos(d*x + c) + 1) +
1)/a^4 + 168*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4) - A*(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)/d

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.13 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {\frac {840 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {840 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} - \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 21 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 105 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 35 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 385 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1575 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]

[In]

integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, algorithm="giac")

[Out]

1/840*(840*C*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 840*C*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 - (15*A*a^2
4*tan(1/2*d*x + 1/2*c)^7 + 15*C*a^24*tan(1/2*d*x + 1/2*c)^7 + 21*A*a^24*tan(1/2*d*x + 1/2*c)^5 + 105*C*a^24*ta
n(1/2*d*x + 1/2*c)^5 - 35*A*a^24*tan(1/2*d*x + 1/2*c)^3 + 385*C*a^24*tan(1/2*d*x + 1/2*c)^3 - 105*A*a^24*tan(1
/2*d*x + 1/2*c) + 1575*C*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d

Mupad [B] (verification not implemented)

Time = 14.94 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.97 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {2\,C\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A+C}{8\,a^4}+\frac {C}{a^4}-\frac {2\,A-6\,C}{8\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A+C}{24\,a^4}+\frac {C}{6\,a^4}-\frac {2\,A-6\,C}{24\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {A+C}{40\,a^4}+\frac {C}{10\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A+C\right )}{56\,a^4\,d} \]

[In]

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + a/cos(c + d*x))^4),x)

[Out]

(2*C*atanh(tan(c/2 + (d*x)/2)))/(a^4*d) - (tan(c/2 + (d*x)/2)*((A + C)/(8*a^4) + C/a^4 - (2*A - 6*C)/(8*a^4)))
/d - (tan(c/2 + (d*x)/2)^3*((A + C)/(24*a^4) + C/(6*a^4) - (2*A - 6*C)/(24*a^4)))/d - (tan(c/2 + (d*x)/2)^5*((
A + C)/(40*a^4) + C/(10*a^4)))/d - (tan(c/2 + (d*x)/2)^7*(A + C))/(56*a^4*d)