3.5.22 \(\int \cos ^2(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [422]

Optimal. Leaf size=128 \[ \frac {1}{2} a^2 (3 A+4 B+2 C) x+\frac {a^2 (B+2 C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a^2 (3 A+2 B-2 C) \sin (c+d x)}{2 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {(A-2 C) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.20, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {4171, 4103, 4081, 3855} \begin {gather*} \frac {a^2 (3 A+2 B-2 C) \sin (c+d x)}{2 d}+\frac {1}{2} a^2 x (3 A+4 B+2 C)-\frac {(A-2 C) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{2 d}+\frac {a^2 (B+2 C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^2}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3855

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

Rule 4081

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.)
+ (A_)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Dist[1/(d*n), Int[(d*Csc[e + f*x
])^(n + 1)*Simp[n*(B*a + A*b) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B},
x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]

Rule 4103

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

Rule 4171

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*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*
Csc[e + f*x])^n/(f*n)), x] - Dist[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m -
b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 -
b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \cos (c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x))^2 (2 a (A+B)-a (A-2 C) \sec (c+d x)) \, dx}{2 a}\\ &=\frac {A \cos (c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {(A-2 C) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x)) \left (a^2 (3 A+2 B-2 C)+2 a^2 (B+2 C) \sec (c+d x)\right ) \, dx}{2 a}\\ &=\frac {a^2 (3 A+2 B-2 C) \sin (c+d x)}{2 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {(A-2 C) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{2 d}-\frac {\int \left (-a^3 (3 A+4 B+2 C)-2 a^3 (B+2 C) \sec (c+d x)\right ) \, dx}{2 a}\\ &=\frac {1}{2} a^2 (3 A+4 B+2 C) x+\frac {a^2 (3 A+2 B-2 C) \sin (c+d x)}{2 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {(A-2 C) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{2 d}+\left (a^2 (B+2 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} a^2 (3 A+4 B+2 C) x+\frac {a^2 (B+2 C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a^2 (3 A+2 B-2 C) \sin (c+d x)}{2 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {(A-2 C) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{2 d}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(329\) vs. \(2(128)=256\).
time = 3.84, size = 329, normalized size = 2.57 \begin {gather*} \frac {a^2 \cos ^2(c+d x) (1+\cos (c+d x))^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (2 (3 A+4 B+2 C) x-\frac {4 (B+2 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {4 (B+2 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {4 (2 A+B) \cos (d x) \sin (c)}{d}+\frac {A \cos (2 d x) \sin (2 c)}{d}+\frac {4 (2 A+B) \cos (c) \sin (d x)}{d}+\frac {A \cos (2 c) \sin (2 d x)}{d}+\frac {4 C \sin \left (\frac {d x}{2}\right )}{d \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 {4 C \sin \left (\frac {d x}{2}\right )}{d \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 )}{8 (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x)))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Cos[c + d*x]^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(2*(3*A +
4*B + 2*C)*x - (4*(B + 2*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (4*(B + 2*C)*Log[Cos[(c + d*x)/2] +
Sin[(c + d*x)/2]])/d + (4*(2*A + B)*Cos[d*x]*Sin[c])/d + (A*Cos[2*d*x]*Sin[2*c])/d + (4*(2*A + B)*Cos[c]*Sin[d
*x])/d + (A*Cos[2*c]*Sin[2*d*x])/d + (4*C*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c +
d*x)/2])) + (4*C*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/(8*(A + 2*C +
2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]))

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Maple [A]
time = 0.67, size = 137, normalized size = 1.07

method	result	size
derivativedivides	\(\frac {a^{2} A \left (d x +c \right )+a^{2} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{2} C \tan \left (d x +c \right )+2 a^{2} A \sin \left (d x +c \right )+2 a^{2} B \left (d x +c \right )+2 a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{2} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{2} B \sin \left (d x +c \right )+a^{2} C \left (d x +c \right )}{d}\)	\(137\)
default	\(\frac {a^{2} A \left (d x +c \right )+a^{2} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{2} C \tan \left (d x +c \right )+2 a^{2} A \sin \left (d x +c \right )+2 a^{2} B \left (d x +c \right )+2 a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{2} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{2} B \sin \left (d x +c \right )+a^{2} C \left (d x +c \right )}{d}\)	\(137\)
risch	\(\frac {3 a^{2} A x}{2}+2 a^{2} B x +a^{2} x C -\frac {i a^{2} A \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {i a^{2} A \,{\mathrm e}^{i \left (d x +c \right )}}{d}-\frac {i a^{2} B \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a^{2} A \,{\mathrm e}^{-i \left (d x +c \right )}}{d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{2} B}{2 d}+\frac {i a^{2} A \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i a^{2} C}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}-\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}\)	\(246\)
norman	\(\frac {\left (-\frac {3}{2} a^{2} A -2 a^{2} B -a^{2} C \right ) x +\left (-\frac {3}{2} a^{2} A -2 a^{2} B -a^{2} C \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {3}{2} a^{2} A +2 a^{2} B +a^{2} C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {3}{2} a^{2} A +2 a^{2} B +a^{2} C \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-3 a^{2} A -4 a^{2} B -2 a^{2} C \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 a^{2} A +4 a^{2} B +2 a^{2} C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {a^{2} \left (3 A +2 B -2 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{2} \left (A +B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a^{2} \left (3 A +B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a^{2} \left (3 A -2 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2} \left (5 A +2 B +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {a^{2} \left (B +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{2} \left (B +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(389\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 151, normalized size = 1.18 \begin {gather*} \frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} + 4 \, {\left (d x + c\right )} A a^{2} + 8 \, {\left (d x + c\right )} B a^{2} + 4 \, {\left (d x + c\right )} C a^{2} + 2 \, B a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, A a^{2} \sin \left (d x + c\right ) + 4 \, B a^{2} \sin \left (d x + c\right ) + 4 \, C a^{2} \tan \left (d x + c\right )}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2 + 4*(d*x + c)*A*a^2 + 8*(d*x + c)*B*a^2 + 4*(d*x + c)*C*a^2 + 2*B*
a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*C*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1))
+ 8*A*a^2*sin(d*x + c) + 4*B*a^2*sin(d*x + c) + 4*C*a^2*tan(d*x + c))/d

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Fricas [A]
time = 2.16, size = 130, normalized size = 1.02 \begin {gather*} \frac {{\left (3 \, A + 4 \, B + 2 \, C\right )} a^{2} d x \cos \left (d x + c\right ) + {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (A a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + 2 \, C a^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.57, size = 198, normalized size = 1.55 \begin {gather*} -\frac {\frac {4 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - {\left (3 \, A a^{2} + 4 \, B a^{2} + 2 \, C a^{2}\right )} {\left (d x + c\right )} - 2 \, {\left (B a^{2} + 2 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 2 \, {\left (B a^{2} + 2 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B a^{2} \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}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.55, size = 232, normalized size = 1.81 \begin {gather*} \frac {3\,A\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+4\,B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}+2\,C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,4{}\mathrm {i}}{d}+\frac {A\,a^2\,\sin \left (2\,c+2\,d\,x\right )+\frac {A\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{8}+\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {A\,a^2\,\sin \left (c+d\,x\right )}{8}+C\,a^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*A*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 4*B*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - B
*a^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i + 2*C*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))
- C*a^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*4i)/d + (A*a^2*sin(2*c + 2*d*x) + (A*a^2*sin(3*c + 3
*d*x))/8 + (B*a^2*sin(2*c + 2*d*x))/2 + (A*a^2*sin(c + d*x))/8 + C*a^2*sin(c + d*x))/(d*cos(c + d*x))

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