\(\int \frac {\sec ^2(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^2} \, dx\) [460]

Optimal result

Integrand size = 41, antiderivative size = 112 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {(B-2 C) \text {arctanh}(\sin (c+d x))}{a^2 d}+\frac {(A-B+4 C) \tan (c+d x)}{3 a^2 d}-\frac {(B-2 C) \tan (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac {(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {4169, 4093, 3872, 3855, 3852, 8} \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {(A-B+4 C) \tan (c+d x)}{3 a^2 d}+\frac {(B-2 C) \text {arctanh}(\sin (c+d x))}{a^2 d}-\frac {(B-2 C) \tan (c+d x)}{a^2 d (\sec (c+d x)+1)}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[In]

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

[Out]

((B - 2*C)*ArcTanh[Sin[c + d*x]])/(a^2*d) + ((A - B + 4*C)*Tan[c + d*x])/(3*a^2*d) - ((B - 2*C)*Tan[c + d*x])/
(a^2*d*(1 + Sec[c + d*x])) - ((A - B + C)*Sec[c + d*x]^2*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 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 4169

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-(a*A - b*B + a*C))*Cot[e + f*x]*(a + b*C
sc[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[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(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, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(312\) vs. \(2(112)=224\).

Time = 3.12 (sec) , antiderivative size = 312, normalized size of antiderivative = 2.79 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {4 \cos \left (\frac {1}{2} (c+d x)\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left ((A-B+C) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+2 (A-4 B+7 C) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+\cos ^3\left (\frac {1}{2} (c+d x)\right ) \left (-6 (B-2 C) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\frac {6 C \sin (d x)}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )+(A-B+C) \cos \left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )\right )}{3 a^2 d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) (1+\sec (c+d x))^2} \]

[In]

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

[Out]

(4*Cos[(c + d*x)/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((A - B + C)*Sec[c/2]*Sin[(d*x)/2] + 2*(A - 4*B +
7*C)*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] + Cos[(c + d*x)/2]^3*(-6*(B - 2*C)*(Log[Cos[(c + d*x)/2] - Sin[(
c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + (6*C*Sin[d*x])/((Cos[c/2] - Sin[c/2])*(Cos[c/2] + S
in[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))) + (A - B + C)*Cos[(c +
d*x)/2]*Tan[c/2]))/(3*a^2*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*(1 + Sec[c + d*x])^2)

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.20

method	result	size
parallelrisch	\(\frac {-3 \cos \left (d x +c \right ) \left (B -2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 \cos \left (d x +c \right ) \left (B -2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (\frac {A}{4}-B +\frac {5 C}{2}\right ) \cos \left (2 d x +2 c \right )+\left (A -\frac {5 B}{2}+7 C \right ) \cos \left (d x +c \right )+\frac {A}{4}-B +4 C \right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 d \,a^{2} \cos \left (d x +c \right )}\)	\(134\)
derivativedivides	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\left (-4 C +2 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (-2 B +4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {2 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}}{2 d \,a^{2}}\)	\(159\)
default	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\left (-4 C +2 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (-2 B +4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {2 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}}{2 d \,a^{2}}\)	\(159\)
norman	\(\frac {-\frac {\left (B -2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}-\frac {\left (A -4 B +9 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a d}+\frac {\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{6 a d}+\frac {\left (-13 B +34 C +4 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}-\frac {\left (-3 B +9 C +A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3} a}+\frac {\left (B -2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2} d}-\frac {\left (B -2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2} d}\)	\(201\)
risch	\(-\frac {2 i \left (3 B \,{\mathrm e}^{4 i \left (d x +c \right )}-6 C \,{\mathrm e}^{4 i \left (d x +c \right )}-3 A \,{\mathrm e}^{3 i \left (d x +c \right )}+9 B \,{\mathrm e}^{3 i \left (d x +c \right )}-18 C \,{\mathrm e}^{3 i \left (d x +c \right )}-A \,{\mathrm e}^{2 i \left (d x +c \right )}+7 B \,{\mathrm e}^{2 i \left (d x +c \right )}-22 C \,{\mathrm e}^{2 i \left (d x +c \right )}-3 A \,{\mathrm e}^{i \left (d x +c \right )}+9 B \,{\mathrm e}^{i \left (d x +c \right )}-24 C \,{\mathrm e}^{i \left (d x +c \right )}-A +4 B -10 C \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{2} d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a^{2} d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{2} d}\)	\(266\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.77 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {3 \, {\left ({\left (B - 2 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (B - 2 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (B - 2 \, C\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (B - 2 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (B - 2 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (B - 2 \, C\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left ({\left (A - 4 \, B + 10 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, A - 5 \, B + 14 \, C\right )} \cos \left (d x + c\right ) + 3 \, C\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \]

[In]

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

[Out]

1/6*(3*((B - 2*C)*cos(d*x + c)^3 + 2*(B - 2*C)*cos(d*x + c)^2 + (B - 2*C)*cos(d*x + c))*log(sin(d*x + c) + 1)
- 3*((B - 2*C)*cos(d*x + c)^3 + 2*(B - 2*C)*cos(d*x + c)^2 + (B - 2*C)*cos(d*x + c))*log(-sin(d*x + c) + 1) +
2*((A - 4*B + 10*C)*cos(d*x + c)^2 + (2*A - 5*B + 14*C)*cos(d*x + c) + 3*C)*sin(d*x + c))/(a^2*d*cos(d*x + c)^
3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))

Sympy [F]

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

[In]

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

[Out]

(Integral(A*sec(c + d*x)**2/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)**3/(sec(c + d
*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**4/(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. 287 vs. \(2 (108) = 216\).

Time = 0.26 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.56 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {C {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} - \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - B {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} + \frac {A {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, d} \]

[In]

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

[Out]

1/6*(C*((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))) - B*((9*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^
3/(cos(d*x + c) + 1)^3)/a^2 - 6*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 6*log(sin(d*x + c)/(cos(d*x + c
) + 1) - 1)/a^2) + A*(3*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2)/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.62 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {\frac {6 \, {\left (B - 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {6 \, {\left (B - 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} - \frac {12 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{2}} + \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

[In]

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

[Out]

1/6*(6*(B - 2*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 6*(B - 2*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 -
12*C*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*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 + C*a^4*tan(1/2*d*x + 1/2*c)^3 + 3*A*a^4*tan(1/2*d*x + 1/2*c) - 9*B*a^4*tan(1/2*d*x + 1/2*c) +
15*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

Mupad [B] (verification not implemented)

Time = 16.07 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.09 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A-B+C}{a^2}-\frac {A+B-3\,C}{2\,a^2}\right )}{d}+\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (B-2\,C\right )}{a^2\,d}-\frac {2\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B+C\right )}{6\,a^2\,d} \]

[In]

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

[Out]

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