Integrand size = 13, antiderivative size = 230 \[ \int \log ^2(a+b \tan (c+d x)) \, dx=-\frac {i \log \left (\frac {b (i-\tan (c+d x))}{a+i b}\right ) \log ^2(a+b \tan (c+d x))}{2 d}+\frac {i \log \left (-\frac {b (i+\tan (c+d x))}{a-i b}\right ) \log ^2(a+b \tan (c+d x))}{2 d}+\frac {i \log (a+b \tan (c+d x)) \operatorname {PolyLog}\left (2,\frac {a+b \tan (c+d x)}{a-i b}\right )}{d}-\frac {i \log (a+b \tan (c+d x)) \operatorname {PolyLog}\left (2,\frac {a+b \tan (c+d x)}{a+i b}\right )}{d}-\frac {i \operatorname {PolyLog}\left (3,\frac {a+b \tan (c+d x)}{a-i b}\right )}{d}+\frac {i \operatorname {PolyLog}\left (3,\frac {a+b \tan (c+d x)}{a+i b}\right )}{d} \] Output:
-1/2*I*ln(b*(I-tan(d*x+c))/(a+I*b))*ln(a+b*tan(d*x+c))^2/d+1/2*I*ln(-b*(I+ tan(d*x+c))/(a-I*b))*ln(a+b*tan(d*x+c))^2/d+I*ln(a+b*tan(d*x+c))*polylog(2 ,(a+b*tan(d*x+c))/(a-I*b))/d-I*ln(a+b*tan(d*x+c))*polylog(2,(a+b*tan(d*x+c ))/(a+I*b))/d-I*polylog(3,(a+b*tan(d*x+c))/(a-I*b))/d+I*polylog(3,(a+b*tan (d*x+c))/(a+I*b))/d
Time = 0.35 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.88 \[ \int \log ^2(a+b \tan (c+d x)) \, dx=\frac {i \left (-\log \left (-\frac {b (-i+\tan (c+d x))}{a+i b}\right ) \log ^2(a+b \tan (c+d x))+\log \left (-\frac {b (i+\tan (c+d x))}{a-i b}\right ) \log ^2(a+b \tan (c+d x))+2 \log (a+b \tan (c+d x)) \operatorname {PolyLog}\left (2,\frac {a+b \tan (c+d x)}{a-i b}\right )-2 \log (a+b \tan (c+d x)) \operatorname {PolyLog}\left (2,\frac {a+b \tan (c+d x)}{a+i b}\right )-2 \operatorname {PolyLog}\left (3,\frac {a+b \tan (c+d x)}{a-i b}\right )+2 \operatorname {PolyLog}\left (3,\frac {a+b \tan (c+d x)}{a+i b}\right )\right )}{2 d} \] Input:
Integrate[Log[a + b*Tan[c + d*x]]^2,x]
Output:
((I/2)*(-(Log[-((b*(-I + Tan[c + d*x]))/(a + I*b))]*Log[a + b*Tan[c + d*x] ]^2) + Log[-((b*(I + Tan[c + d*x]))/(a - I*b))]*Log[a + b*Tan[c + d*x]]^2 + 2*Log[a + b*Tan[c + d*x]]*PolyLog[2, (a + b*Tan[c + d*x])/(a - I*b)] - 2 *Log[a + b*Tan[c + d*x]]*PolyLog[2, (a + b*Tan[c + d*x])/(a + I*b)] - 2*Po lyLog[3, (a + b*Tan[c + d*x])/(a - I*b)] + 2*PolyLog[3, (a + b*Tan[c + d*x ])/(a + I*b)]))/d
Time = 0.73 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {4853, 2856, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \log ^2(a+b \tan (c+d x)) \, dx\) |
\(\Big \downarrow \) 4853 |
\(\displaystyle \frac {\int \frac {\log ^2(a+b \tan (c+d x))}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 2856 |
\(\displaystyle \frac {\int \left (\frac {i \log ^2(a+b \tan (c+d x))}{2 (i-\tan (c+d x))}+\frac {i \log ^2(a+b \tan (c+d x))}{2 (\tan (c+d x)+i)}\right )d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-i \operatorname {PolyLog}\left (3,\frac {a+b \tan (c+d x)}{a-i b}\right )+i \operatorname {PolyLog}\left (3,\frac {a+b \tan (c+d x)}{a+i b}\right )+i \operatorname {PolyLog}\left (2,\frac {a+b \tan (c+d x)}{a-i b}\right ) \log (a+b \tan (c+d x))-i \operatorname {PolyLog}\left (2,\frac {a+b \tan (c+d x)}{a+i b}\right ) \log (a+b \tan (c+d x))-\frac {1}{2} i \log \left (\frac {b (-\tan (c+d x)+i)}{a+i b}\right ) \log ^2(a+b \tan (c+d x))+\frac {1}{2} i \log \left (-\frac {b (\tan (c+d x)+i)}{a-i b}\right ) \log ^2(a+b \tan (c+d x))}{d}\) |
Input:
Int[Log[a + b*Tan[c + d*x]]^2,x]
Output:
((-1/2*I)*Log[(b*(I - Tan[c + d*x]))/(a + I*b)]*Log[a + b*Tan[c + d*x]]^2 + (I/2)*Log[-((b*(I + Tan[c + d*x]))/(a - I*b))]*Log[a + b*Tan[c + d*x]]^2 + I*Log[a + b*Tan[c + d*x]]*PolyLog[2, (a + b*Tan[c + d*x])/(a - I*b)] - I*Log[a + b*Tan[c + d*x]]*PolyLog[2, (a + b*Tan[c + d*x])/(a + I*b)] - I*P olyLog[3, (a + b*Tan[c + d*x])/(a - I*b)] + I*PolyLog[3, (a + b*Tan[c + d* x])/(a + I*b)])/d
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_. )*(x_)^(r_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x) ^n])^p, (f + g*x^r)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, r}, x] && I GtQ[p, 0] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[r] && NeQ[r, 1]))
Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, Simp[With[{d = FreeFa ctors[Tan[v], x]}, d/Coefficient[v, x, 1] Subst[Int[SubstFor[1/(1 + d^2*x ^2), Tan[v]/d, u, x], x], x, Tan[v]/d]], x] /; !FalseQ[v] && FunctionOfQ[N onfreeFactors[Tan[v], x], u, x, True] && TryPureTanSubst[ActivateTrig[u], x ]]
\[\int \ln \left (a +b \tan \left (d x +c \right )\right )^{2}d x\]
Input:
int(ln(a+b*tan(d*x+c))^2,x)
Output:
int(ln(a+b*tan(d*x+c))^2,x)
\[ \int \log ^2(a+b \tan (c+d x)) \, dx=\int { \log \left (b \tan \left (d x + c\right ) + a\right )^{2} \,d x } \] Input:
integrate(log(a+b*tan(d*x+c))^2,x, algorithm="fricas")
Output:
integral(log(b*tan(d*x + c) + a)^2, x)
\[ \int \log ^2(a+b \tan (c+d x)) \, dx=\int \log {\left (a + b \tan {\left (c + d x \right )} \right )}^{2}\, dx \] Input:
integrate(ln(a+b*tan(d*x+c))**2,x)
Output:
Integral(log(a + b*tan(c + d*x))**2, x)
Exception generated. \[ \int \log ^2(a+b \tan (c+d x)) \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(log(a+b*tan(d*x+c))^2,x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: BINDING-STACK overflow at size 10240. Stack can probably be resized.Proceed with caution.
\[ \int \log ^2(a+b \tan (c+d x)) \, dx=\int { \log \left (b \tan \left (d x + c\right ) + a\right )^{2} \,d x } \] Input:
integrate(log(a+b*tan(d*x+c))^2,x, algorithm="giac")
Output:
integrate(log(b*tan(d*x + c) + a)^2, x)
Timed out. \[ \int \log ^2(a+b \tan (c+d x)) \, dx=\int {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^2 \,d x \] Input:
int(log(a + b*tan(c + d*x))^2,x)
Output:
int(log(a + b*tan(c + d*x))^2, x)
\[ \int \log ^2(a+b \tan (c+d x)) \, dx=\int \mathrm {log}\left (\tan \left (d x +c \right ) b +a \right )^{2}d x \] Input:
int(log(a+b*tan(d*x+c))^2,x)
Output:
int(log(tan(c + d*x)*b + a)**2,x)