\(\int \frac {a+b \tan (c+d x)}{(b+a \tan (c+d x))^2} \, dx\) [316]

Optimal result

Integrand size = 23, antiderivative size = 101 \[ \int \frac {a+b \tan (c+d x)}{(b+a \tan (c+d x))^2} \, dx=-\frac {a \left (a^2-3 b^2\right ) x}{\left (a^2+b^2\right )^2}+\frac {b \left (3 a^2-b^2\right ) \log (b \cos (c+d x)+a \sin (c+d x))}{\left (a^2+b^2\right )^2 d}-\frac {a^2-b^2}{\left (a^2+b^2\right ) d (b+a \tan (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3610, 3612, 3611} \[ \int \frac {a+b \tan (c+d x)}{(b+a \tan (c+d x))^2} \, dx=-\frac {a^2-b^2}{d \left (a^2+b^2\right ) (a \tan (c+d x)+b)}+\frac {b \left (3 a^2-b^2\right ) \log (a \sin (c+d x)+b \cos (c+d x))}{d \left (a^2+b^2\right )^2}-\frac {a x \left (a^2-3 b^2\right )}{\left (a^2+b^2\right )^2} \]

[In]

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

[Out]

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

Rule 3610

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b
*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 + b^2))), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3611

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c/(b*f))
*Log[RemoveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rule 3612

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c +
b*d)*(x/(a^2 + b^2)), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.58 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.85 \[ \int \frac {a+b \tan (c+d x)}{(b+a \tan (c+d x))^2} \, dx=\frac {\frac {b (-((a+i b) \log (i-\tan (c+d x)))-(a-i b) \log (i+\tan (c+d x))+2 a \log (b+a \tan (c+d x)))}{a^2+b^2}+(a-b) (a+b) \left (\frac {i \log (i-\tan (c+d x))}{(a-i b)^2}-\frac {i \log (i+\tan (c+d x))}{(a+i b)^2}+\frac {2 a \left (2 b \log (b+a \tan (c+d x))-\frac {a^2+b^2}{b+a \tan (c+d x)}\right )}{\left (a^2+b^2\right )^2}\right )}{2 a d} \]

[In]

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

[Out]

((b*(-((a + I*b)*Log[I - Tan[c + d*x]]) - (a - I*b)*Log[I + Tan[c + d*x]] + 2*a*Log[b + a*Tan[c + d*x]]))/(a^2
+ b^2) + (a - b)*(a + b)*((I*Log[I - Tan[c + d*x]])/(a - I*b)^2 - (I*Log[I + Tan[c + d*x]])/(a + I*b)^2 + (2*
a*(2*b*Log[b + a*Tan[c + d*x]] - (a^2 + b^2)/(b + a*Tan[c + d*x])))/(a^2 + b^2)^2))/(2*a*d)

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.24

method	result	size
derivativedivides	\(\frac {-\frac {a^{2}-b^{2}}{\left (a^{2}+b^{2}\right ) \left (b +a \tan \left (d x +c \right )\right )}+\frac {b \left (3 a^{2}-b^{2}\right ) \ln \left (b +a \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{2}}+\frac {\frac {\left (-3 a^{2} b +b^{3}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (-a^{3}+3 a \,b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{2}}}{d}\)	\(125\)
default	\(\frac {-\frac {a^{2}-b^{2}}{\left (a^{2}+b^{2}\right ) \left (b +a \tan \left (d x +c \right )\right )}+\frac {b \left (3 a^{2}-b^{2}\right ) \ln \left (b +a \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{2}}+\frac {\frac {\left (-3 a^{2} b +b^{3}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (-a^{3}+3 a \,b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{2}}}{d}\)	\(125\)
norman	\(\frac {\frac {\left (a^{2}-b^{2}\right ) a \tan \left (d x +c \right )}{b d \left (a^{2}+b^{2}\right )}-\frac {a^{2} \left (a^{2}-3 b^{2}\right ) x \tan \left (d x +c \right )}{a^{4}+2 a^{2} b^{2}+b^{4}}-\frac {b a \left (a^{2}-3 b^{2}\right ) x}{a^{4}+2 a^{2} b^{2}+b^{4}}}{b +a \tan \left (d x +c \right )}+\frac {b \left (3 a^{2}-b^{2}\right ) \ln \left (b +a \tan \left (d x +c \right )\right )}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}-\frac {b \left (3 a^{2}-b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}\)	\(206\)
parallelrisch	\(-\frac {2 x \tan \left (d x +c \right ) a^{4} b d -6 x \tan \left (d x +c \right ) a^{2} b^{3} d +3 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right ) a^{3} b^{2}-\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right ) a \,b^{4}-6 \ln \left (b +a \tan \left (d x +c \right )\right ) \tan \left (d x +c \right ) a^{3} b^{2}+2 \ln \left (b +a \tan \left (d x +c \right )\right ) \tan \left (d x +c \right ) a \,b^{4}+2 x \,a^{3} b^{2} d -6 x a \,b^{4} d +3 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{2} b^{3}-\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) b^{5}-6 \ln \left (b +a \tan \left (d x +c \right )\right ) a^{2} b^{3}+2 \ln \left (b +a \tan \left (d x +c \right )\right ) b^{5}-2 \tan \left (d x +c \right ) a^{5}+2 \tan \left (d x +c \right ) a \,b^{4}}{2 \left (b +a \tan \left (d x +c \right )\right ) \left (a^{2}+b^{2}\right )^{2} b d}\)	\(268\)
risch	\(\frac {i x b}{2 i b a +a^{2}-b^{2}}-\frac {x a}{2 i b a +a^{2}-b^{2}}-\frac {6 i b \,a^{2} x}{a^{4}+2 a^{2} b^{2}+b^{4}}+\frac {2 i b^{3} x}{a^{4}+2 a^{2} b^{2}+b^{4}}-\frac {6 i b \,a^{2} c}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}+\frac {2 i b^{3} c}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}-\frac {2 i a^{3}}{\left (-i b +a \right ) d \left (i b +a \right )^{2} \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+i b \,{\mathrm e}^{2 i \left (d x +c \right )}-a +i b \right )}+\frac {2 i a \,b^{2}}{\left (-i b +a \right ) d \left (i b +a \right )^{2} \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+i b \,{\mathrm e}^{2 i \left (d x +c \right )}-a +i b \right )}+\frac {3 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {i b -a}{i b +a}\right ) a^{2}}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}-\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {i b -a}{i b +a}\right )}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}\)	\(373\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.89 \[ \int \frac {a+b \tan (c+d x)}{(b+a \tan (c+d x))^2} \, dx=-\frac {2 \, a^{4} - 2 \, a^{2} b^{2} + 2 \, {\left (a^{3} b - 3 \, a b^{3}\right )} d x - {\left (3 \, a^{2} b^{2} - b^{4} + {\left (3 \, a^{3} b - a b^{3}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {a^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + b^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \, {\left (a^{3} b - a b^{3} - {\left (a^{4} - 3 \, a^{2} b^{2}\right )} d x\right )} \tan \left (d x + c\right )}{2 \, {\left ({\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d \tan \left (d x + c\right ) + {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d\right )}} \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.75 (sec) , antiderivative size = 1348, normalized size of antiderivative = 13.35 \[ \int \frac {a+b \tan (c+d x)}{(b+a \tan (c+d x))^2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((zoo*x*tan(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (log(tan(c + d*x)**2 + 1)/(2*b*d), Eq(a, 0)), (I/(2*
a*d*tan(c + d*x)**2 - 4*I*a*d*tan(c + d*x) - 2*a*d), Eq(b, -I*a)), (-I/(2*a*d*tan(c + d*x)**2 + 4*I*a*d*tan(c
+ d*x) - 2*a*d), Eq(b, I*a)), (x*(a + b*tan(c))/(a*tan(c) + b)**2, Eq(d, 0)), (-2*a**4*d*x*tan(c + d*x)/(2*a**
5*d*tan(c + d*x) + 2*a**4*b*d + 4*a**3*b**2*d*tan(c + d*x) + 4*a**2*b**3*d + 2*a*b**4*d*tan(c + d*x) + 2*b**5*
d) - 2*a**4/(2*a**5*d*tan(c + d*x) + 2*a**4*b*d + 4*a**3*b**2*d*tan(c + d*x) + 4*a**2*b**3*d + 2*a*b**4*d*tan(
c + d*x) + 2*b**5*d) - 2*a**3*b*d*x/(2*a**5*d*tan(c + d*x) + 2*a**4*b*d + 4*a**3*b**2*d*tan(c + d*x) + 4*a**2*
b**3*d + 2*a*b**4*d*tan(c + d*x) + 2*b**5*d) + 6*a**3*b*log(tan(c + d*x) + b/a)*tan(c + d*x)/(2*a**5*d*tan(c +
d*x) + 2*a**4*b*d + 4*a**3*b**2*d*tan(c + d*x) + 4*a**2*b**3*d + 2*a*b**4*d*tan(c + d*x) + 2*b**5*d) - 3*a**3
*b*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*a**5*d*tan(c + d*x) + 2*a**4*b*d + 4*a**3*b**2*d*tan(c + d*x) + 4*
a**2*b**3*d + 2*a*b**4*d*tan(c + d*x) + 2*b**5*d) + 6*a**2*b**2*d*x*tan(c + d*x)/(2*a**5*d*tan(c + d*x) + 2*a*
*4*b*d + 4*a**3*b**2*d*tan(c + d*x) + 4*a**2*b**3*d + 2*a*b**4*d*tan(c + d*x) + 2*b**5*d) + 6*a**2*b**2*log(ta
n(c + d*x) + b/a)/(2*a**5*d*tan(c + d*x) + 2*a**4*b*d + 4*a**3*b**2*d*tan(c + d*x) + 4*a**2*b**3*d + 2*a*b**4*
d*tan(c + d*x) + 2*b**5*d) - 3*a**2*b**2*log(tan(c + d*x)**2 + 1)/(2*a**5*d*tan(c + d*x) + 2*a**4*b*d + 4*a**3
*b**2*d*tan(c + d*x) + 4*a**2*b**3*d + 2*a*b**4*d*tan(c + d*x) + 2*b**5*d) + 6*a*b**3*d*x/(2*a**5*d*tan(c + d*
x) + 2*a**4*b*d + 4*a**3*b**2*d*tan(c + d*x) + 4*a**2*b**3*d + 2*a*b**4*d*tan(c + d*x) + 2*b**5*d) - 2*a*b**3*
log(tan(c + d*x) + b/a)*tan(c + d*x)/(2*a**5*d*tan(c + d*x) + 2*a**4*b*d + 4*a**3*b**2*d*tan(c + d*x) + 4*a**2
*b**3*d + 2*a*b**4*d*tan(c + d*x) + 2*b**5*d) + a*b**3*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*a**5*d*tan(c +
d*x) + 2*a**4*b*d + 4*a**3*b**2*d*tan(c + d*x) + 4*a**2*b**3*d + 2*a*b**4*d*tan(c + d*x) + 2*b**5*d) - 2*b**4
*log(tan(c + d*x) + b/a)/(2*a**5*d*tan(c + d*x) + 2*a**4*b*d + 4*a**3*b**2*d*tan(c + d*x) + 4*a**2*b**3*d + 2*
a*b**4*d*tan(c + d*x) + 2*b**5*d) + b**4*log(tan(c + d*x)**2 + 1)/(2*a**5*d*tan(c + d*x) + 2*a**4*b*d + 4*a**3
*b**2*d*tan(c + d*x) + 4*a**2*b**3*d + 2*a*b**4*d*tan(c + d*x) + 2*b**5*d) + 2*b**4/(2*a**5*d*tan(c + d*x) + 2
*a**4*b*d + 4*a**3*b**2*d*tan(c + d*x) + 4*a**2*b**3*d + 2*a*b**4*d*tan(c + d*x) + 2*b**5*d), True))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.59 \[ \int \frac {a+b \tan (c+d x)}{(b+a \tan (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (a^{3} - 3 \, a b^{2}\right )} {\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, {\left (3 \, a^{2} b - b^{3}\right )} \log \left (a \tan \left (d x + c\right ) + b\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {2 \, {\left (a^{2} - b^{2}\right )}}{a^{2} b + b^{3} + {\left (a^{3} + a b^{2}\right )} \tan \left (d x + c\right )}}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.97 \[ \int \frac {a+b \tan (c+d x)}{(b+a \tan (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (a^{3} - 3 \, a b^{2}\right )} {\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, {\left (3 \, a^{3} b - a b^{3}\right )} \log \left ({\left | a \tan \left (d x + c\right ) + b \right |}\right )}{a^{5} + 2 \, a^{3} b^{2} + a b^{4}} + \frac {2 \, {\left (3 \, a^{3} b \tan \left (d x + c\right ) - a b^{3} \tan \left (d x + c\right ) + a^{4} + 3 \, a^{2} b^{2} - 2 \, b^{4}\right )}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a \tan \left (d x + c\right ) + b\right )}}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 7.81 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.50 \[ \int \frac {a+b \tan (c+d x)}{(b+a \tan (c+d x))^2} \, dx=\frac {b\,\ln \left (b+a\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (3\,a^2-b^2\right )}{d\,{\left (a^2+b^2\right )}^2}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (a-b\,1{}\mathrm {i}\right )}{2\,d\,\left (-a^2\,1{}\mathrm {i}+2\,a\,b+b^2\,1{}\mathrm {i}\right )}-\frac {a^2-b^2}{d\,\left (a^2+b^2\right )\,\left (b+a\,\mathrm {tan}\left (c+d\,x\right )\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (a+b\,1{}\mathrm {i}\right )}{2\,d\,\left (a^2\,1{}\mathrm {i}+2\,a\,b-b^2\,1{}\mathrm {i}\right )} \]

[In]

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

[Out]

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