3.569 \(\int \frac{(A+C \cos ^2(c+d x)) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx\)

Optimal. Leaf size=184 \[ \frac{2 b^2 \left (a^2 C+A b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d \sqrt{a-b} \sqrt{a+b}}+\frac{\left (a^2 (2 A+3 C)+3 A b^2\right ) \tan (c+d x)}{3 a^3 d}-\frac{b \left (a^2 (A+2 C)+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac{A b \tan (c+d x) \sec (c+d x)}{2 a^2 d}+\frac{A \tan (c+d x) \sec ^2(c+d x)}{3 a d} \]

[Out]

(2*b^2*(A*b^2 + a^2*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^4*Sqrt[a - b]*Sqrt[a + b]*d) - (
b*(2*A*b^2 + a^2*(A + 2*C))*ArcTanh[Sin[c + d*x]])/(2*a^4*d) + ((3*A*b^2 + a^2*(2*A + 3*C))*Tan[c + d*x])/(3*a
^3*d) - (A*b*Sec[c + d*x]*Tan[c + d*x])/(2*a^2*d) + (A*Sec[c + d*x]^2*Tan[c + d*x])/(3*a*d)

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Rubi [A]  time = 0.740514, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3056, 3055, 3001, 3770, 2659, 205} \[ \frac{2 b^2 \left (a^2 C+A b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d \sqrt{a-b} \sqrt{a+b}}+\frac{\left (a^2 (2 A+3 C)+3 A b^2\right ) \tan (c+d x)}{3 a^3 d}-\frac{b \left (a^2 (A+2 C)+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac{A b \tan (c+d x) \sec (c+d x)}{2 a^2 d}+\frac{A \tan (c+d x) \sec ^2(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b^2*(A*b^2 + a^2*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^4*Sqrt[a - b]*Sqrt[a + b]*d) - (
b*(2*A*b^2 + a^2*(A + 2*C))*ArcTanh[Sin[c + d*x]])/(2*a^4*d) + ((3*A*b^2 + a^2*(2*A + 3*C))*Tan[c + d*x])/(3*a
^3*d) - (A*b*Sec[c + d*x]*Tan[c + d*x])/(2*a^2*d) + (A*Sec[c + d*x]^2*Tan[c + d*x])/(3*a*d)

Rule 3056

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c +
d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3770

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

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [B]  time = 2.88109, size = 413, normalized size = 2.24 \[ \frac{\frac{4 a \left (a^2 (2 A+3 C)+3 A b^2\right ) \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{4 a \left (a^2 (2 A+3 C)+3 A b^2\right ) \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}-\frac{24 b^2 \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}+6 b \left (a^2 (A+2 C)+2 A b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-6 b \left (a^2 (A+2 C)+2 A b^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{a^2 A (a-3 b)}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{a^2 A (a-3 b)}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{2 a^3 A \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{2 a^3 A \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}}{12 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-24*b^2*(A*b^2 + a^2*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + 6*b*(2*A*b^
2 + a^2*(A + 2*C))*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 6*b*(2*A*b^2 + a^2*(A + 2*C))*Log[Cos[(c + d*x)/
2] + Sin[(c + d*x)/2]] + (a^2*A*(a - 3*b))/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (2*a^3*A*Sin[(c + d*x)/2]
)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3 + (4*a*(3*A*b^2 + a^2*(2*A + 3*C))*Sin[(c + d*x)/2])/(Cos[(c + d*x)/
2] - Sin[(c + d*x)/2]) + (2*a^3*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 - (a^2*A*(a - 3*b)
)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (4*a*(3*A*b^2 + a^2*(2*A + 3*C))*Sin[(c + d*x)/2])/(Cos[(c + d*x)/
2] + Sin[(c + d*x)/2]))/(12*a^4*d)

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Maple [B]  time = 0.072, size = 554, normalized size = 3. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*b^4/a^4/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A+2/d*b^2/a^2/((a+b)*(a-b
))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C-1/3/a/d*A/(tan(1/2*d*x+1/2*c)-1)^3-1/a/d*A/(ta
n(1/2*d*x+1/2*c)-1)-1/2/d*A/a^2/(tan(1/2*d*x+1/2*c)-1)*b-1/d/a^3/(tan(1/2*d*x+1/2*c)-1)*A*b^2-1/a/d/(tan(1/2*d
*x+1/2*c)-1)*C-1/2/a/d*A/(tan(1/2*d*x+1/2*c)-1)^2-1/2/d*A/a^2/(tan(1/2*d*x+1/2*c)-1)^2*b+1/2/d*A*b/a^2*ln(tan(
1/2*d*x+1/2*c)-1)+1/d*b^3/a^4*ln(tan(1/2*d*x+1/2*c)-1)*A+1/d*b/a^2*ln(tan(1/2*d*x+1/2*c)-1)*C-1/3/a/d*A/(tan(1
/2*d*x+1/2*c)+1)^3-1/a/d*A/(tan(1/2*d*x+1/2*c)+1)-1/2/d*A/a^2/(tan(1/2*d*x+1/2*c)+1)*b-1/d/a^3/(tan(1/2*d*x+1/
2*c)+1)*A*b^2-1/a/d/(tan(1/2*d*x+1/2*c)+1)*C+1/2/a/d*A/(tan(1/2*d*x+1/2*c)+1)^2+1/2/d*A/a^2/(tan(1/2*d*x+1/2*c
)+1)^2*b-1/2/d*A*b/a^2*ln(tan(1/2*d*x+1/2*c)+1)-1/d*b^3/a^4*ln(tan(1/2*d*x+1/2*c)+1)*A-1/d*b/a^2*ln(tan(1/2*d*
x+1/2*c)+1)*C

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 16.4688, size = 1472, normalized size = 8. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/12*(6*(C*a^2*b^2 + A*b^4)*sqrt(-a^2 + b^2)*cos(d*x + c)^3*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x
+ c)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d
*x + c) + a^2)) + 3*((A + 2*C)*a^4*b + (A - 2*C)*a^2*b^3 - 2*A*b^5)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(
(A + 2*C)*a^4*b + (A - 2*C)*a^2*b^3 - 2*A*b^5)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) - 2*(2*A*a^5 - 2*A*a^3*b^
2 + 2*((2*A + 3*C)*a^5 + (A - 3*C)*a^3*b^2 - 3*A*a*b^4)*cos(d*x + c)^2 - 3*(A*a^4*b - A*a^2*b^3)*cos(d*x + c))
*sin(d*x + c))/((a^6 - a^4*b^2)*d*cos(d*x + c)^3), 1/12*(12*(C*a^2*b^2 + A*b^4)*sqrt(a^2 - b^2)*arctan(-(a*cos
(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c)))*cos(d*x + c)^3 - 3*((A + 2*C)*a^4*b + (A - 2*C)*a^2*b^3 - 2*A*b
^5)*cos(d*x + c)^3*log(sin(d*x + c) + 1) + 3*((A + 2*C)*a^4*b + (A - 2*C)*a^2*b^3 - 2*A*b^5)*cos(d*x + c)^3*lo
g(-sin(d*x + c) + 1) + 2*(2*A*a^5 - 2*A*a^3*b^2 + 2*((2*A + 3*C)*a^5 + (A - 3*C)*a^3*b^2 - 3*A*a*b^4)*cos(d*x
+ c)^2 - 3*(A*a^4*b - A*a^2*b^3)*cos(d*x + c))*sin(d*x + c))/((a^6 - a^4*b^2)*d*cos(d*x + c)^3)]

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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((A+C*cos(d*x+c)**2)*sec(d*x+c)**4/(a+b*cos(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 1.41986, size = 502, normalized size = 2.73 \begin{align*} -\frac{\frac{3 \,{\left (A a^{2} b + 2 \, C a^{2} b + 2 \, A b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac{3 \,{\left (A a^{2} b + 2 \, C a^{2} b + 2 \, A b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac{12 \,{\left (C a^{2} b^{2} + A b^{4}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{\sqrt{a^{2} - b^{2}} a^{4}} + \frac{2 \,{\left (6 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 4 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, 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} a^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(A*a^2*b + 2*C*a^2*b + 2*A*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 3*(A*a^2*b + 2*C*a^2*b + 2*A*
b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 + 12*(C*a^2*b^2 + A*b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*
a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*a^4) +
2*(6*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 6*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 3*A*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^2
*tan(1/2*d*x + 1/2*c)^5 - 4*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 12*C*a^2*tan(1/2*d*x + 1/2*c)^3 - 12*A*b^2*tan(1/2*
d*x + 1/2*c)^3 + 6*A*a^2*tan(1/2*d*x + 1/2*c) + 6*C*a^2*tan(1/2*d*x + 1/2*c) - 3*A*a*b*tan(1/2*d*x + 1/2*c) +
6*A*b^2*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*a^3))/d