3.5.10 \(\int \sec (c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [410]

Optimal. Leaf size=92 \[ \frac {a (2 A+B+C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a (3 A+3 B+2 C) \tan (c+d x)}{3 d}+\frac {a (B+C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a C \sec ^2(c+d x) \tan (c+d x)}{3 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.09, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {4161, 4132, 3852, 8, 4131, 3855} \begin {gather*} \frac {a (3 A+3 B+2 C) \tan (c+d x)}{3 d}+\frac {a (2 A+B+C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a (B+C) \tan (c+d x) \sec (c+d x)}{2 d}+\frac {a C \tan (c+d x) \sec ^2(c+d x)}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 4131

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

Rule 4132

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 4161

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_)), x_Symbol] :> Simp[(-b)*C*Csc[e + f*x]*Cot[e + f*x]*((d*Csc[e + f
*x])^n/(f*(n + 2))), x] + Dist[1/(n + 2), Int[(d*Csc[e + f*x])^n*Simp[A*a*(n + 2) + (B*a*(n + 2) + b*(C*(n + 1
) + A*(n + 2)))*Csc[e + f*x] + (a*C + B*b)*(n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C
, n}, x] &&  !LtQ[n, -1]

Rubi steps

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

________________________________________________________________________________________

Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(485\) vs. \(2(92)=184\).
time = 4.43, size = 485, normalized size = 5.27 \begin {gather*} \frac {a \cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-6 (2 A+B+C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 (2 A+B+C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {2 C \sin \left (\frac {d x}{2}\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 )^3}+\frac {(3 B+4 C) \cos \left (\frac {c}{2}\right )-(3 B+2 C) \sin \left (\frac {c}{2}\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 )^2}+\frac {4 (3 A+3 B+2 C) \sin \left (\frac {d x}{2}\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 )}+\frac {2 C \sin \left (\frac {d x}{2}\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 )^3}-\frac {(3 B+4 C) \cos \left (\frac {c}{2}\right )+(3 B+2 C) \sin \left (\frac {c}{2}\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 )^2}+\frac {4 (3 A+3 B+2 C) \sin \left (\frac {d x}{2}\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 )}\right )}{6 d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x)))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.67, size = 131, normalized size = 1.42

method	result	size
derivativedivides	\(\frac {A a \tan \left (d x +c \right )+B a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-a C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+A a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B a \tan \left (d x +c \right )+a C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(131\)
default	\(\frac {A a \tan \left (d x +c \right )+B a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-a C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+A a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B a \tan \left (d x +c \right )+a C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(131\)
norman	\(\frac {-\frac {a \left (2 A +3 B +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 a \left (3 A +3 B +C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {a \left (C +B +2 A \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a \left (C +B +2 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a \left (C +B +2 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(140\)
risch	\(-\frac {i a \left (3 B \,{\mathrm e}^{5 i \left (d x +c \right )}+3 C \,{\mathrm e}^{5 i \left (d x +c \right )}-6 A \,{\mathrm e}^{4 i \left (d x +c \right )}-6 B \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A \,{\mathrm e}^{2 i \left (d x +c \right )}-12 B \,{\mathrm e}^{2 i \left (d x +c \right )}-12 C \,{\mathrm e}^{2 i \left (d x +c \right )}-3 B \,{\mathrm e}^{i \left (d x +c \right )}-3 C \,{\mathrm e}^{i \left (d x +c \right )}-6 A -6 B -4 C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}\)	\(259\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.29, size = 155, normalized size = 1.68 \begin {gather*} \frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a - 3 \, B a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, C a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 12 \, A a \tan \left (d x + c\right ) + 12 \, B a \tan \left (d x + c\right )}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a - 3*B*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c)
+ 1) + log(sin(d*x + c) - 1)) - 3*C*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d
*x + c) - 1)) + 12*A*a*log(sec(d*x + c) + tan(d*x + c)) + 12*A*a*tan(d*x + c) + 12*B*a*tan(d*x + c))/d

________________________________________________________________________________________

Fricas [A]
time = 2.38, size = 114, normalized size = 1.24 \begin {gather*} \frac {3 \, {\left (2 \, A + B + C\right )} a \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, A + B + C\right )} a \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (3 \, A + 3 \, B + 2 \, C\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (B + C\right )} a \cos \left (d x + c\right ) + 2 \, C a\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*(2*A + B + C)*a*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(2*A + B + C)*a*cos(d*x + c)^3*log(-sin(d*x +
c) + 1) + 2*(2*(3*A + 3*B + 2*C)*a*cos(d*x + c)^2 + 3*(B + C)*a*cos(d*x + c) + 2*C*a)*sin(d*x + c))/(d*cos(d*
x + c)^3)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 205 vs. \(2 (84) = 168\).
time = 0.53, size = 205, normalized size = 2.23 \begin {gather*} \frac {3 \, {\left (2 \, A a + B a + C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (2 \, A a + B a + C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, C a \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 )}^{3}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(2*A*a + B*a + C*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*A*a + B*a + C*a)*log(abs(tan(1/2*d*x + 1/
2*c) - 1)) - 2*(6*A*a*tan(1/2*d*x + 1/2*c)^5 + 3*B*a*tan(1/2*d*x + 1/2*c)^5 + 3*C*a*tan(1/2*d*x + 1/2*c)^5 - 1
2*A*a*tan(1/2*d*x + 1/2*c)^3 - 12*B*a*tan(1/2*d*x + 1/2*c)^3 - 4*C*a*tan(1/2*d*x + 1/2*c)^3 + 6*A*a*tan(1/2*d*
x + 1/2*c) + 9*B*a*tan(1/2*d*x + 1/2*c) + 9*C*a*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d

________________________________________________________________________________________

Mupad [B]
time = 5.80, size = 165, normalized size = 1.79 \begin {gather*} \frac {a\,\mathrm {atanh}\left (\frac {2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A+B+C\right )}{4\,A\,a+2\,B\,a+2\,C\,a}\right )\,\left (2\,A+B+C\right )}{d}-\frac {\left (2\,A\,a+B\,a+C\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-4\,A\,a-4\,B\,a-\frac {4\,C\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a+3\,B\,a+3\,C\,a\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*atanh((2*a*tan(c/2 + (d*x)/2)*(2*A + B + C))/(4*A*a + 2*B*a + 2*C*a))*(2*A + B + C))/d - (tan(c/2 + (d*x)/2
)*(2*A*a + 3*B*a + 3*C*a) + tan(c/2 + (d*x)/2)^5*(2*A*a + B*a + C*a) - tan(c/2 + (d*x)/2)^3*(4*A*a + 4*B*a + (
4*C*a)/3))/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 1))

________________________________________________________________________________________