3.883 \(\int \cos ^3(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=196 \[ \frac{a \sin (c+d x) \left (a^2 (2 A+3 C)+6 a b B+3 A b^2\right )}{3 d}+\frac{1}{2} x \left (3 a^2 b (A+2 C)+a^3 B+6 a b^2 B+2 A b^3\right )-\frac{b^2 \tan (c+d x) (3 a B+5 A b-6 b C)}{6 d}+\frac{(a B+A b) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d}+\frac{A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^3}{3 d}+\frac{b^2 (3 a C+b B) \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

((2*A*b^3 + a^3*B + 6*a*b^2*B + 3*a^2*b*(A + 2*C))*x)/2 + (b^2*(b*B + 3*a*C)*ArcTanh[Sin[c + d*x]])/d + (a*(3*
A*b^2 + 6*a*b*B + a^2*(2*A + 3*C))*Sin[c + d*x])/(3*d) + ((A*b + a*B)*Cos[c + d*x]*(a + b*Sec[c + d*x])^2*Sin[
c + d*x])/(2*d) + (A*Cos[c + d*x]^2*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(3*d) - (b^2*(5*A*b + 3*a*B - 6*b*C)*
Tan[c + d*x])/(6*d)

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Rubi [A]  time = 0.60091, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {4094, 4076, 4047, 8, 4045, 3770} \[ \frac{a \sin (c+d x) \left (a^2 (2 A+3 C)+6 a b B+3 A b^2\right )}{3 d}+\frac{1}{2} x \left (3 a^2 b (A+2 C)+a^3 B+6 a b^2 B+2 A b^3\right )-\frac{b^2 \tan (c+d x) (3 a B+5 A b-6 b C)}{6 d}+\frac{(a B+A b) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d}+\frac{A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^3}{3 d}+\frac{b^2 (3 a C+b B) \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

((2*A*b^3 + a^3*B + 6*a*b^2*B + 3*a^2*b*(A + 2*C))*x)/2 + (b^2*(b*B + 3*a*C)*ArcTanh[Sin[c + d*x]])/d + (a*(3*
A*b^2 + 6*a*b*B + a^2*(2*A + 3*C))*Sin[c + d*x])/(3*d) + ((A*b + a*B)*Cos[c + d*x]*(a + b*Sec[c + d*x])^2*Sin[
c + d*x])/(2*d) + (A*Cos[c + d*x]^2*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(3*d) - (b^2*(5*A*b + 3*a*B - 6*b*C)*
Tan[c + d*x])/(6*d)

Rule 4094

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/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*
m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Csc[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]

Rule 4076

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]

Rule 4047

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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 4045

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

Rule 3770

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

Rubi steps

\begin{align*} \int \cos ^3(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos ^2(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac{1}{3} \int \cos ^2(c+d x) (a+b \sec (c+d x))^2 \left (3 (A b+a B)+(2 a A+3 b B+3 a C) \sec (c+d x)-b (A-3 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{(A b+a B) \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{A \cos ^2(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac{1}{6} \int \cos (c+d x) (a+b \sec (c+d x)) \left (2 \left (3 A b^2+6 a b B+\frac{1}{2} a^2 (4 A+6 C)\right )+\left (3 a^2 B+6 b^2 B+a b (5 A+12 C)\right ) \sec (c+d x)-b (5 A b+3 a B-6 b C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{(A b+a B) \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{A \cos ^2(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}-\frac{b^2 (5 A b+3 a B-6 b C) \tan (c+d x)}{6 d}+\frac{1}{6} \int \cos (c+d x) \left (2 a \left (3 A b^2+6 a b B+a^2 (2 A+3 C)\right )+3 \left (2 A b^3+a^3 B+6 a b^2 B+3 a^2 b (A+2 C)\right ) \sec (c+d x)+6 b^2 (b B+3 a C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{(A b+a B) \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{A \cos ^2(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}-\frac{b^2 (5 A b+3 a B-6 b C) \tan (c+d x)}{6 d}+\frac{1}{6} \int \cos (c+d x) \left (2 a \left (3 A b^2+6 a b B+a^2 (2 A+3 C)\right )+6 b^2 (b B+3 a C) \sec ^2(c+d x)\right ) \, dx+\frac{1}{2} \left (2 A b^3+a^3 B+6 a b^2 B+3 a^2 b (A+2 C)\right ) \int 1 \, dx\\ &=\frac{1}{2} \left (2 A b^3+a^3 B+6 a b^2 B+3 a^2 b (A+2 C)\right ) x+\frac{a \left (3 A b^2+6 a b B+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 d}+\frac{(A b+a B) \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{A \cos ^2(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}-\frac{b^2 (5 A b+3 a B-6 b C) \tan (c+d x)}{6 d}+\left (b^2 (b B+3 a C)\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} \left (2 A b^3+a^3 B+6 a b^2 B+3 a^2 b (A+2 C)\right ) x+\frac{b^2 (b B+3 a C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a \left (3 A b^2+6 a b B+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 d}+\frac{(A b+a B) \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{A \cos ^2(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}-\frac{b^2 (5 A b+3 a B-6 b C) \tan (c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 1.35415, size = 263, normalized size = 1.34 \[ \frac{6 (c+d x) \left (3 a^2 b (A+2 C)+a^3 B+6 a b^2 B+2 A b^3\right )+3 a \sin (c+d x) \left (a^2 (3 A+4 C)+12 a b B+12 A b^2\right )+3 a^2 (a B+3 A b) \sin (2 (c+d x))+a^3 A \sin (3 (c+d x))-12 b^2 (3 a C+b B) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+12 b^2 (3 a C+b B) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{12 b^3 C \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{12 b^3 C \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(6*(2*A*b^3 + a^3*B + 6*a*b^2*B + 3*a^2*b*(A + 2*C))*(c + d*x) - 12*b^2*(b*B + 3*a*C)*Log[Cos[(c + d*x)/2] - S
in[(c + d*x)/2]] + 12*b^2*(b*B + 3*a*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (12*b^3*C*Sin[(c + d*x)/2])
/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + (12*b^3*C*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 3
*a*(12*A*b^2 + 12*a*b*B + a^2*(3*A + 4*C))*Sin[c + d*x] + 3*a^2*(3*A*b + a*B)*Sin[2*(c + d*x)] + a^3*A*Sin[3*(
c + d*x)])/(12*d)

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Maple [A]  time = 0.076, size = 278, normalized size = 1.4 \begin{align*}{\frac{A \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ){a}^{3}}{3\,d}}+{\frac{2\,A{a}^{3}\sin \left ( dx+c \right ) }{3\,d}}+{\frac{B{a}^{3}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{3}Bx}{2}}+{\frac{B{a}^{3}c}{2\,d}}+{\frac{{a}^{3}C\sin \left ( dx+c \right ) }{d}}+{\frac{3\,A{a}^{2}b\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2\,d}}+{\frac{3\,{a}^{2}Abx}{2}}+{\frac{3\,A{a}^{2}bc}{2\,d}}+3\,{\frac{B{a}^{2}b\sin \left ( dx+c \right ) }{d}}+3\,{a}^{2}bCx+3\,{\frac{C{a}^{2}bc}{d}}+3\,{\frac{Aa{b}^{2}\sin \left ( dx+c \right ) }{d}}+3\,Ba{b}^{2}x+3\,{\frac{Ba{b}^{2}c}{d}}+3\,{\frac{Ca{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+A{b}^{3}x+{\frac{A{b}^{3}c}{d}}+{\frac{B{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{C{b}^{3}\tan \left ( dx+c \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

1/3/d*A*cos(d*x+c)^2*sin(d*x+c)*a^3+2/3*a^3*A*sin(d*x+c)/d+1/2/d*B*a^3*sin(d*x+c)*cos(d*x+c)+1/2*a^3*B*x+1/2/d
*B*a^3*c+a^3*C*sin(d*x+c)/d+3/2/d*A*a^2*b*sin(d*x+c)*cos(d*x+c)+3/2*a^2*A*b*x+3/2/d*A*a^2*b*c+3/d*B*a^2*b*sin(
d*x+c)+3*a^2*b*C*x+3/d*C*a^2*b*c+3/d*A*a*b^2*sin(d*x+c)+3*B*a*b^2*x+3/d*B*a*b^2*c+3/d*C*a*b^2*ln(sec(d*x+c)+ta
n(d*x+c))+A*b^3*x+1/d*A*b^3*c+1/d*B*b^3*ln(sec(d*x+c)+tan(d*x+c))+1/d*C*b^3*tan(d*x+c)

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Maxima [A]  time = 1.0333, size = 292, normalized size = 1.49 \begin{align*} -\frac{4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} - 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 9 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} b - 36 \,{\left (d x + c\right )} C a^{2} b - 36 \,{\left (d x + c\right )} B a b^{2} - 12 \,{\left (d x + c\right )} A b^{3} - 18 \, C a b^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 6 \, B b^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, C a^{3} \sin \left (d x + c\right ) - 36 \, B a^{2} b \sin \left (d x + c\right ) - 36 \, A a b^{2} \sin \left (d x + c\right ) - 12 \, C b^{3} \tan \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^3 - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3 - 9*(2*d*x + 2*c +
sin(2*d*x + 2*c))*A*a^2*b - 36*(d*x + c)*C*a^2*b - 36*(d*x + c)*B*a*b^2 - 12*(d*x + c)*A*b^3 - 18*C*a*b^2*(lo
g(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 6*B*b^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 12*C*
a^3*sin(d*x + c) - 36*B*a^2*b*sin(d*x + c) - 36*A*a*b^2*sin(d*x + c) - 12*C*b^3*tan(d*x + c))/d

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Fricas [A]  time = 0.572554, size = 486, normalized size = 2.48 \begin{align*} \frac{3 \,{\left (B a^{3} + 3 \,{\left (A + 2 \, C\right )} a^{2} b + 6 \, B a b^{2} + 2 \, A b^{3}\right )} d x \cos \left (d x + c\right ) + 3 \,{\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (2 \, A a^{3} \cos \left (d x + c\right )^{3} + 6 \, C b^{3} + 3 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left ({\left (2 \, A + 3 \, C\right )} a^{3} + 9 \, B a^{2} b + 9 \, A a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.3365, size = 564, normalized size = 2.88 \begin{align*} -\frac{\frac{12 \, C b^{3} \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} - 3 \,{\left (B a^{3} + 3 \, A a^{2} b + 6 \, C a^{2} b + 6 \, B a b^{2} + 2 \, A b^{3}\right )}{\left (d x + c\right )} - 6 \,{\left (3 \, C a b^{2} + B b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) + 6 \,{\left (3 \, C a b^{2} + B b^{3}\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^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 9 \, A a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 18 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 18 \, A a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 4 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 36 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 36 \, A a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, A a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 18 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 18 \, A a b^{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 )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(12*C*b^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - 3*(B*a^3 + 3*A*a^2*b + 6*C*a^2*b + 6*B*a*b^
2 + 2*A*b^3)*(d*x + c) - 6*(3*C*a*b^2 + B*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 6*(3*C*a*b^2 + B*b^3)*log(
abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*A*a^3*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 6*C*a^3*
tan(1/2*d*x + 1/2*c)^5 - 9*A*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 18*B*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 18*A*a*b^2*tan
(1/2*d*x + 1/2*c)^5 + 4*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 12*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 36*B*a^2*b*tan(1/2*d*
x + 1/2*c)^3 + 36*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^3*tan(1/2*d*x + 1/2*c) + 3*B*a^3*tan(1/2*d*x + 1/2*c)
+ 6*C*a^3*tan(1/2*d*x + 1/2*c) + 9*A*a^2*b*tan(1/2*d*x + 1/2*c) + 18*B*a^2*b*tan(1/2*d*x + 1/2*c) + 18*A*a*b^
2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d