\(\int \frac {\cos ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx\) [558]

Optimal result

Integrand size = 21, antiderivative size = 152 \[ \int \frac {\cos ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {\left (a^4+6 a^2 b^2-3 b^4\right ) x}{2 \left (a^2+b^2\right )^3}+\frac {4 a b^3 \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right )^3 d}+\frac {b \left (a^2-3 b^2\right )}{2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))} \]

[Out]

1/2*(a^4+6*a^2*b^2-3*b^4)*x/(a^2+b^2)^3+4*a*b^3*ln(a*cos(d*x+c)+b*sin(d*x+c))/(a^2+b^2)^3/d+1/2*b*(a^2-3*b^2)/
(a^2+b^2)^2/d/(a+b*tan(d*x+c))+1/2*cos(d*x+c)^2*(b+a*tan(d*x+c))/(a^2+b^2)/d/(a+b*tan(d*x+c))

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3587, 755, 815, 649, 209, 266} \[ \int \frac {\cos ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {b \left (a^2-3 b^2\right )}{2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac {\cos ^2(c+d x) (a \tan (c+d x)+b)}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac {4 a b^3 \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}+\frac {x \left (a^4+6 a^2 b^2-3 b^4\right )}{2 \left (a^2+b^2\right )^3} \]

[In]

Int[Cos[c + d*x]^2/(a + b*Tan[c + d*x])^2,x]

[Out]

((a^4 + 6*a^2*b^2 - 3*b^4)*x)/(2*(a^2 + b^2)^3) + (4*a*b^3*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])/((a^2 + b^2)^
3*d) + (b*(a^2 - 3*b^2))/(2*(a^2 + b^2)^2*d*(a + b*Tan[c + d*x])) + (Cos[c + d*x]^2*(b + a*Tan[c + d*x]))/(2*(
a^2 + b^2)*d*(a + b*Tan[c + d*x]))

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 755

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(m + 1))*(a*e + c*d*x)*
((a + c*x^2)^(p + 1)/(2*a*(p + 1)*(c*d^2 + a*e^2))), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^
m*Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[
{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 3587

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(b*f), Subst
[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && NeQ[a^2 + b
^2, 0] && IntegerQ[m/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{(a+x)^2 \left (1+\frac {x^2}{b^2}\right )^2} \, dx,x,b \tan (c+d x)\right )}{b d} \\ & = \frac {\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac {b \text {Subst}\left (\int \frac {-3-\frac {a^2}{b^2}-\frac {2 a x}{b^2}}{(a+x)^2 \left (1+\frac {x^2}{b^2}\right )} \, dx,x,b \tan (c+d x)\right )}{2 \left (a^2+b^2\right ) d} \\ & = \frac {\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac {b \text {Subst}\left (\int \left (\frac {a^2-3 b^2}{\left (a^2+b^2\right ) (a+x)^2}-\frac {8 a b^2}{\left (a^2+b^2\right )^2 (a+x)}+\frac {-a^4-6 a^2 b^2+3 b^4+8 a b^2 x}{\left (a^2+b^2\right )^2 \left (b^2+x^2\right )}\right ) \, dx,x,b \tan (c+d x)\right )}{2 \left (a^2+b^2\right ) d} \\ & = \frac {4 a b^3 \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^3 d}+\frac {b \left (a^2-3 b^2\right )}{2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac {b \text {Subst}\left (\int \frac {-a^4-6 a^2 b^2+3 b^4+8 a b^2 x}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^3 d} \\ & = \frac {4 a b^3 \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^3 d}+\frac {b \left (a^2-3 b^2\right )}{2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac {\left (4 a b^3\right ) \text {Subst}\left (\int \frac {x}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{\left (a^2+b^2\right )^3 d}+\frac {\left (b \left (a^4+6 a^2 b^2-3 b^4\right )\right ) \text {Subst}\left (\int \frac {1}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^3 d} \\ & = \frac {\left (a^4+6 a^2 b^2-3 b^4\right ) x}{2 \left (a^2+b^2\right )^3}+\frac {4 a b^3 \log (\cos (c+d x))}{\left (a^2+b^2\right )^3 d}+\frac {4 a b^3 \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^3 d}+\frac {b \left (a^2-3 b^2\right )}{2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.41 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.00 \[ \int \frac {\cos ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {-\frac {a b \left (\left (-a+\sqrt {-b^2}\right ) \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )-2 \sqrt {-b^2} \log (a+b \tan (c+d x))+\left (a+\sqrt {-b^2}\right ) \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )\right )}{\sqrt {-b^2} \left (a^2+b^2\right )}+\frac {\cos ^2(c+d x) (b+a \tan (c+d x))}{a+b \tan (c+d x)}+\frac {b \left (a^2-3 b^2\right ) \left (\left (2 a+\frac {-a^2+b^2}{\sqrt {-b^2}}\right ) \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )-4 a \log (a+b \tan (c+d x))+\left (2 a+\frac {a^2-b^2}{\sqrt {-b^2}}\right ) \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )+\frac {2 \left (a^2+b^2\right )}{a+b \tan (c+d x)}\right )}{2 \left (a^2+b^2\right )^2}}{2 \left (a^2+b^2\right ) d} \]

[In]

Integrate[Cos[c + d*x]^2/(a + b*Tan[c + d*x])^2,x]

[Out]

(-((a*b*((-a + Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Tan[c + d*x]] - 2*Sqrt[-b^2]*Log[a + b*Tan[c + d*x]] + (a + Sqrt
[-b^2])*Log[Sqrt[-b^2] + b*Tan[c + d*x]]))/(Sqrt[-b^2]*(a^2 + b^2))) + (Cos[c + d*x]^2*(b + a*Tan[c + d*x]))/(
a + b*Tan[c + d*x]) + (b*(a^2 - 3*b^2)*((2*a + (-a^2 + b^2)/Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Tan[c + d*x]] - 4*a
*Log[a + b*Tan[c + d*x]] + (2*a + (a^2 - b^2)/Sqrt[-b^2])*Log[Sqrt[-b^2] + b*Tan[c + d*x]] + (2*(a^2 + b^2))/(
a + b*Tan[c + d*x])))/(2*(a^2 + b^2)^2))/(2*(a^2 + b^2)*d)

Maple [A] (verified)

Time = 4.17 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.01

method	result	size
derivativedivides	\(\frac {-\frac {b^{3}}{\left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}+\frac {4 b^{3} a \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{3}}+\frac {\frac {\left (\frac {a^{4}}{2}-\frac {b^{4}}{2}\right ) \tan \left (d x +c \right )+a^{3} b +a \,b^{3}}{1+\tan ^{2}\left (d x +c \right )}-2 a \,b^{3} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+\frac {\left (a^{4}+6 a^{2} b^{2}-3 b^{4}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{2}}{\left (a^{2}+b^{2}\right )^{3}}}{d}\)	\(154\)
default	\(\frac {-\frac {b^{3}}{\left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}+\frac {4 b^{3} a \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{3}}+\frac {\frac {\left (\frac {a^{4}}{2}-\frac {b^{4}}{2}\right ) \tan \left (d x +c \right )+a^{3} b +a \,b^{3}}{1+\tan ^{2}\left (d x +c \right )}-2 a \,b^{3} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+\frac {\left (a^{4}+6 a^{2} b^{2}-3 b^{4}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{2}}{\left (a^{2}+b^{2}\right )^{3}}}{d}\)	\(154\)
risch	\(\frac {3 i x b}{6 i b \,a^{2}-2 i b^{3}-2 a^{3}+6 a \,b^{2}}-\frac {x a}{6 i b \,a^{2}-2 i b^{3}-2 a^{3}+6 a \,b^{2}}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 \left (-2 i a b +a^{2}-b^{2}\right ) d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 \left (2 i a b +a^{2}-b^{2}\right ) d}-\frac {8 i a \,b^{3} x}{a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}}-\frac {8 i a \,b^{3} c}{d \left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right )}-\frac {2 i b^{4}}{\left (-i a +b \right )^{2} d \left (i a +b \right )^{3} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right )}+\frac {4 a \,b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{d \left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right )}\)	\(318\)

[In]

int(cos(d*x+c)^2/(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(-b^3/(a^2+b^2)^2/(a+b*tan(d*x+c))+4*b^3/(a^2+b^2)^3*a*ln(a+b*tan(d*x+c))+1/(a^2+b^2)^3*(((1/2*a^4-1/2*b^4
)*tan(d*x+c)+a^3*b+a*b^3)/(1+tan(d*x+c)^2)-2*a*b^3*ln(1+tan(d*x+c)^2)+1/2*(a^4+6*a^2*b^2-3*b^4)*arctan(tan(d*x
+c))))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.84 \[ \int \frac {\cos ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{2} b^{3} + 3 \, b^{5} - {\left (a^{5} + 6 \, a^{3} b^{2} - 3 \, a b^{4}\right )} d x\right )} \cos \left (d x + c\right ) + 4 \, {\left (a^{2} b^{3} \cos \left (d x + c\right ) + a b^{4} \sin \left (d x + c\right )\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - {\left (a^{3} b^{2} - a b^{4} - {\left (a^{4} b + 6 \, a^{2} b^{3} - 3 \, b^{5}\right )} d x - {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} d \sin \left (d x + c\right )\right )}} \]

[In]

integrate(cos(d*x+c)^2/(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/2*((a^4*b + 2*a^2*b^3 + b^5)*cos(d*x + c)^3 - (a^2*b^3 + 3*b^5 - (a^5 + 6*a^3*b^2 - 3*a*b^4)*d*x)*cos(d*x +
c) + 4*(a^2*b^3*cos(d*x + c) + a*b^4*sin(d*x + c))*log(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x +
c)^2 + b^2) - (a^3*b^2 - a*b^4 - (a^4*b + 6*a^2*b^3 - 3*b^5)*d*x - (a^5 + 2*a^3*b^2 + a*b^4)*cos(d*x + c)^2)*
sin(d*x + c))/((a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*d*cos(d*x + c) + (a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*d*
sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**2/(a+b*tan(d*x+c))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.86 \[ \int \frac {\cos ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {\frac {8 \, a b^{3} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {4 \, a b^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {{\left (a^{4} + 6 \, a^{2} b^{2} - 3 \, b^{4}\right )} {\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {2 \, a^{2} b - 2 \, b^{3} + {\left (a^{2} b - 3 \, b^{3}\right )} \tan \left (d x + c\right )^{2} + {\left (a^{3} + a b^{2}\right )} \tan \left (d x + c\right )}{a^{5} + 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \tan \left (d x + c\right )^{3} + {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \tan \left (d x + c\right )^{2} + {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \tan \left (d x + c\right )}}{2 \, d} \]

[In]

integrate(cos(d*x+c)^2/(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

1/2*(8*a*b^3*log(b*tan(d*x + c) + a)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - 4*a*b^3*log(tan(d*x + c)^2 + 1)/(a^
6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + (a^4 + 6*a^2*b^2 - 3*b^4)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + (
2*a^2*b - 2*b^3 + (a^2*b - 3*b^3)*tan(d*x + c)^2 + (a^3 + a*b^2)*tan(d*x + c))/(a^5 + 2*a^3*b^2 + a*b^4 + (a^4
*b + 2*a^2*b^3 + b^5)*tan(d*x + c)^3 + (a^5 + 2*a^3*b^2 + a*b^4)*tan(d*x + c)^2 + (a^4*b + 2*a^2*b^3 + b^5)*ta
n(d*x + c)))/d

Giac [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.64 \[ \int \frac {\cos ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {\frac {8 \, a b^{4} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}} - \frac {4 \, a b^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {{\left (a^{4} + 6 \, a^{2} b^{2} - 3 \, b^{4}\right )} {\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {a^{2} b \tan \left (d x + c\right )^{2} - 3 \, b^{3} \tan \left (d x + c\right )^{2} + a^{3} \tan \left (d x + c\right ) + a b^{2} \tan \left (d x + c\right ) + 2 \, a^{2} b - 2 \, b^{3}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (b \tan \left (d x + c\right )^{3} + a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right ) + a\right )}}}{2 \, d} \]

[In]

integrate(cos(d*x+c)^2/(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

1/2*(8*a*b^4*log(abs(b*tan(d*x + c) + a))/(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7) - 4*a*b^3*log(tan(d*x + c)^2 +
1)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + (a^4 + 6*a^2*b^2 - 3*b^4)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b
^6) + (a^2*b*tan(d*x + c)^2 - 3*b^3*tan(d*x + c)^2 + a^3*tan(d*x + c) + a*b^2*tan(d*x + c) + 2*a^2*b - 2*b^3)/
((a^4 + 2*a^2*b^2 + b^4)*(b*tan(d*x + c)^3 + a*tan(d*x + c)^2 + b*tan(d*x + c) + a)))/d

Mupad [B] (verification not implemented)

Time = 5.13 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.62 \[ \int \frac {\cos ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {\frac {a^2\,b-b^3}{{\left (a^2+b^2\right )}^2}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (a^2\,b-3\,b^3\right )}{2\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {a\,\mathrm {tan}\left (c+d\,x\right )}{2\,\left (a^2+b^2\right )}}{d\,\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^3+a\,{\mathrm {tan}\left (c+d\,x\right )}^2+b\,\mathrm {tan}\left (c+d\,x\right )+a\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-3\,b+a\,1{}\mathrm {i}\right )}{4\,d\,\left (-a^3-a^2\,b\,3{}\mathrm {i}+3\,a\,b^2+b^3\,1{}\mathrm {i}\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (a-b\,3{}\mathrm {i}\right )}{4\,d\,\left (-a^3\,1{}\mathrm {i}-3\,a^2\,b+a\,b^2\,3{}\mathrm {i}+b^3\right )}+\frac {4\,a\,b^3\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}{d\,{\left (a^2+b^2\right )}^3} \]

[In]

int(cos(c + d*x)^2/(a + b*tan(c + d*x))^2,x)

[Out]

((a^2*b - b^3)/(a^2 + b^2)^2 + (tan(c + d*x)^2*(a^2*b - 3*b^3))/(2*(a^4 + b^4 + 2*a^2*b^2)) + (a*tan(c + d*x))
/(2*(a^2 + b^2)))/(d*(a + b*tan(c + d*x) + a*tan(c + d*x)^2 + b*tan(c + d*x)^3)) + (log(tan(c + d*x) - 1i)*(a*
1i - 3*b))/(4*d*(3*a*b^2 - a^2*b*3i - a^3 + b^3*1i)) + (log(tan(c + d*x) + 1i)*(a - b*3i))/(4*d*(a*b^2*3i - 3*
a^2*b - a^3*1i + b^3)) + (4*a*b^3*log(a + b*tan(c + d*x)))/(d*(a^2 + b^2)^3)