\(\int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx\) [276]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [C] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 31, antiderivative size = 157 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=-\frac {\left (a^2 A-A b^2+2 a b B\right ) x}{\left (a^2+b^2\right )^2}-\frac {\left (2 a A b-a^2 B+b^2 B\right ) \log (\cos (c+d x))}{\left (a^2+b^2\right )^2 d}-\frac {a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right ) \log (a+b \tan (c+d x))}{b^2 \left (a^2+b^2\right )^2 d}-\frac {a^2 (A b-a B)}{b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3685, 3707, 3698, 31, 3556} \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=-\frac {a^2 (A b-a B)}{b^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\left (a^2 (-B)+2 a A b+b^2 B\right ) \log (\cos (c+d x))}{d \left (a^2+b^2\right )^2}-\frac {x \left (a^2 A+2 a b B-A b^2\right )}{\left (a^2+b^2\right )^2}-\frac {a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right ) \log (a+b \tan (c+d x))}{b^2 d \left (a^2+b^2\right )^2} \]

[In]

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

[Out]

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

Rule 31

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

Rule 3556

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

Rule 3685

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.)
 + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(B*c - A*d))*(b*c - a*d)^2*((c + d*Tan[e + f*x])^(n + 1)/(f*d^2*(n +
 1)*(c^2 + d^2))), x] + Dist[1/(d*(c^2 + d^2)), Int[(c + d*Tan[e + f*x])^(n + 1)*Simp[B*(b*c - a*d)^2 + A*d*(a
^2*c - b^2*c + 2*a*b*d) + d*(B*(a^2*c - b^2*c + 2*a*b*d) + A*(2*a*b*c - a^2*d + b^2*d))*Tan[e + f*x] + b^2*B*(
c^2 + d^2)*Tan[e + f*x]^2, 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] && LtQ[n, -1]

Rule 3698

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[
A/(b*f), Subst[Int[(a + x)^m, x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && EqQ[A, C]

Rule 3707

Int[((A_) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2)/((a_.) + (b_.)*tan[(e_.) + (f_.)*
(x_)]), x_Symbol] :> Simp[(a*A + b*B - a*C)*(x/(a^2 + b^2)), x] + (Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 + b^2), I
nt[(1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*x]), x], x] - Dist[(A*b - a*B - b*C)/(a^2 + b^2), Int[Tan[e + f*x], x
], x]) /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && NeQ[a^2 + b^2, 0] && NeQ[A*b - a
*B - b*C, 0]

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.89 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.93 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=\frac {\frac {i (A+i B) \log (i-\tan (c+d x))}{(a+i b)^2}-\frac {i (A-i B) \log (i+\tan (c+d x))}{(a-i b)^2}+\frac {2 a \left (\left (-2 A b+a \left (3+\frac {a^2}{b^2}\right ) B\right ) \log (a+b \tan (c+d x))+\frac {a \left (a^2+b^2\right ) (-A b+a B)}{b^2 (a+b \tan (c+d x))}\right )}{\left (a^2+b^2\right )^2}}{2 d} \]

[In]

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

[Out]

((I*(A + I*B)*Log[I - Tan[c + d*x]])/(a + I*b)^2 - (I*(A - I*B)*Log[I + Tan[c + d*x]])/(a - I*b)^2 + (2*a*((-2
*A*b + a*(3 + a^2/b^2)*B)*Log[a + b*Tan[c + d*x]] + (a*(a^2 + b^2)*(-(A*b) + a*B))/(b^2*(a + b*Tan[c + d*x])))
)/(a^2 + b^2)^2)/(2*d)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.99

method result size
derivativedivides \(\frac {\frac {\frac {\left (2 A a b -B \,a^{2}+B \,b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (-A \,a^{2}+A \,b^{2}-2 B a b \right ) \arctan \left (\tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{2}}-\frac {a^{2} \left (A b -B a \right )}{b^{2} \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )}-\frac {a \left (2 A \,b^{3}-B \,a^{3}-3 B a \,b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{2} b^{2}}}{d}\) \(155\)
default \(\frac {\frac {\frac {\left (2 A a b -B \,a^{2}+B \,b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (-A \,a^{2}+A \,b^{2}-2 B a b \right ) \arctan \left (\tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{2}}-\frac {a^{2} \left (A b -B a \right )}{b^{2} \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )}-\frac {a \left (2 A \,b^{3}-B \,a^{3}-3 B a \,b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{2} b^{2}}}{d}\) \(155\)
norman \(\frac {-\frac {a \left (A \,a^{2}-A \,b^{2}+2 B a b \right ) x}{a^{4}+2 a^{2} b^{2}+b^{4}}-\frac {b \left (A \,a^{2}-A \,b^{2}+2 B a b \right ) x \tan \left (d x +c \right )}{a^{4}+2 a^{2} b^{2}+b^{4}}-\frac {\left (A a b -B \,a^{2}\right ) a}{d \,b^{2} \left (a^{2}+b^{2}\right )}}{a +b \tan \left (d x +c \right )}+\frac {\left (2 A a b -B \,a^{2}+B \,b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}-\frac {a \left (2 A \,b^{3}-B \,a^{3}-3 B a \,b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) d \,b^{2}}\) \(234\)
parallelrisch \(\frac {-2 A \,a^{2} b^{3}+2 B \,a^{5}-2 A x \tan \left (d x +c \right ) a^{2} b^{3} d -4 B x \tan \left (d x +c \right ) a \,b^{4} d +2 B \ln \left (a +b \tan \left (d x +c \right )\right ) a^{5}-2 A \,a^{4} b +2 B \,a^{3} b^{2}+B \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right ) b^{5}+2 A \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{2} b^{3}-4 A \ln \left (a +b \tan \left (d x +c \right )\right ) a^{2} b^{3}-B \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{3} b^{2}+B \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a \,b^{4}+6 B \ln \left (a +b \tan \left (d x +c \right )\right ) a^{3} b^{2}-4 A \ln \left (a +b \tan \left (d x +c \right )\right ) \tan \left (d x +c \right ) a \,b^{4}-B \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right ) a^{2} b^{3}+2 B \ln \left (a +b \tan \left (d x +c \right )\right ) \tan \left (d x +c \right ) a^{4} b +6 B \ln \left (a +b \tan \left (d x +c \right )\right ) \tan \left (d x +c \right ) a^{2} b^{3}+2 A x \tan \left (d x +c \right ) b^{5} d -2 A x \,a^{3} b^{2} d +2 A x a \,b^{4} d -4 B x \,a^{2} b^{3} d +2 A \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right ) a \,b^{4}}{2 \left (a +b \tan \left (d x +c \right )\right ) \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) d \,b^{2}}\) \(407\)
risch \(-\frac {2 i a^{4} B x}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) b^{2}}+\frac {x A}{2 i b a -a^{2}+b^{2}}+\frac {2 i B c}{d \,b^{2}}+\frac {4 i a b A c}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) d}+\frac {2 i a^{2} A}{\left (i b +a \right ) d \left (-i b +a \right )^{2} \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+a \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +a \right )}-\frac {i x B}{2 i b a -a^{2}+b^{2}}-\frac {2 i a^{4} B c}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) d \,b^{2}}-\frac {2 i a^{3} B}{\left (i b +a \right ) d b \left (-i b +a \right )^{2} \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+a \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +a \right )}-\frac {6 i a^{2} B x}{a^{4}+2 a^{2} b^{2}+b^{4}}+\frac {2 i B x}{b^{2}}+\frac {4 i a b A x}{a^{4}+2 a^{2} b^{2}+b^{4}}-\frac {6 i a^{2} B c}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) d}-\frac {2 a b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) A}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) d}+\frac {a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) B}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) d \,b^{2}}+\frac {3 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) B}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) d}-\frac {B \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d \,b^{2}}\) \(530\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 311 vs. \(2 (155) = 310\).

Time = 0.31 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.98 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=\frac {2 \, B a^{3} b^{2} - 2 \, A a^{2} b^{3} - 2 \, {\left (A a^{3} b^{2} + 2 \, B a^{2} b^{3} - A a b^{4}\right )} d x + {\left (B a^{5} + 3 \, B a^{3} b^{2} - 2 \, A a^{2} b^{3} + {\left (B a^{4} b + 3 \, B a^{2} b^{3} - 2 \, A a b^{4}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - {\left (B a^{5} + 2 \, B a^{3} b^{2} + B a b^{4} + {\left (B a^{4} b + 2 \, B a^{2} b^{3} + B b^{5}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \, {\left (B a^{4} b - A a^{3} b^{2} + {\left (A a^{2} b^{3} + 2 \, B a b^{4} - A b^{5}\right )} d x\right )} \tan \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7}\right )} d \tan \left (d x + c\right ) + {\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} d\right )}} \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.05 (sec) , antiderivative size = 3485, normalized size of antiderivative = 22.20 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((zoo*x*(A + B*tan(c)), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), ((-A*x + A*tan(c + d*x)/d - B*log(tan(c + d*
x)**2 + 1)/(2*d) + B*tan(c + d*x)**2/(2*d))/a**2, Eq(b, 0)), (A*d*x*tan(c + d*x)**2/(4*b**2*d*tan(c + d*x)**2
- 8*I*b**2*d*tan(c + d*x) - 4*b**2*d) - 2*I*A*d*x*tan(c + d*x)/(4*b**2*d*tan(c + d*x)**2 - 8*I*b**2*d*tan(c +
d*x) - 4*b**2*d) - A*d*x/(4*b**2*d*tan(c + d*x)**2 - 8*I*b**2*d*tan(c + d*x) - 4*b**2*d) - 3*A*tan(c + d*x)/(4
*b**2*d*tan(c + d*x)**2 - 8*I*b**2*d*tan(c + d*x) - 4*b**2*d) + 2*I*A/(4*b**2*d*tan(c + d*x)**2 - 8*I*b**2*d*t
an(c + d*x) - 4*b**2*d) + 3*I*B*d*x*tan(c + d*x)**2/(4*b**2*d*tan(c + d*x)**2 - 8*I*b**2*d*tan(c + d*x) - 4*b*
*2*d) + 6*B*d*x*tan(c + d*x)/(4*b**2*d*tan(c + d*x)**2 - 8*I*b**2*d*tan(c + d*x) - 4*b**2*d) - 3*I*B*d*x/(4*b*
*2*d*tan(c + d*x)**2 - 8*I*b**2*d*tan(c + d*x) - 4*b**2*d) + 2*B*log(tan(c + d*x)**2 + 1)*tan(c + d*x)**2/(4*b
**2*d*tan(c + d*x)**2 - 8*I*b**2*d*tan(c + d*x) - 4*b**2*d) - 4*I*B*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(4*b
**2*d*tan(c + d*x)**2 - 8*I*b**2*d*tan(c + d*x) - 4*b**2*d) - 2*B*log(tan(c + d*x)**2 + 1)/(4*b**2*d*tan(c + d
*x)**2 - 8*I*b**2*d*tan(c + d*x) - 4*b**2*d) - 5*I*B*tan(c + d*x)/(4*b**2*d*tan(c + d*x)**2 - 8*I*b**2*d*tan(c
 + d*x) - 4*b**2*d) - 4*B/(4*b**2*d*tan(c + d*x)**2 - 8*I*b**2*d*tan(c + d*x) - 4*b**2*d), Eq(a, -I*b)), (A*d*
x*tan(c + d*x)**2/(4*b**2*d*tan(c + d*x)**2 + 8*I*b**2*d*tan(c + d*x) - 4*b**2*d) + 2*I*A*d*x*tan(c + d*x)/(4*
b**2*d*tan(c + d*x)**2 + 8*I*b**2*d*tan(c + d*x) - 4*b**2*d) - A*d*x/(4*b**2*d*tan(c + d*x)**2 + 8*I*b**2*d*ta
n(c + d*x) - 4*b**2*d) - 3*A*tan(c + d*x)/(4*b**2*d*tan(c + d*x)**2 + 8*I*b**2*d*tan(c + d*x) - 4*b**2*d) - 2*
I*A/(4*b**2*d*tan(c + d*x)**2 + 8*I*b**2*d*tan(c + d*x) - 4*b**2*d) - 3*I*B*d*x*tan(c + d*x)**2/(4*b**2*d*tan(
c + d*x)**2 + 8*I*b**2*d*tan(c + d*x) - 4*b**2*d) + 6*B*d*x*tan(c + d*x)/(4*b**2*d*tan(c + d*x)**2 + 8*I*b**2*
d*tan(c + d*x) - 4*b**2*d) + 3*I*B*d*x/(4*b**2*d*tan(c + d*x)**2 + 8*I*b**2*d*tan(c + d*x) - 4*b**2*d) + 2*B*l
og(tan(c + d*x)**2 + 1)*tan(c + d*x)**2/(4*b**2*d*tan(c + d*x)**2 + 8*I*b**2*d*tan(c + d*x) - 4*b**2*d) + 4*I*
B*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(4*b**2*d*tan(c + d*x)**2 + 8*I*b**2*d*tan(c + d*x) - 4*b**2*d) - 2*B*
log(tan(c + d*x)**2 + 1)/(4*b**2*d*tan(c + d*x)**2 + 8*I*b**2*d*tan(c + d*x) - 4*b**2*d) + 5*I*B*tan(c + d*x)/
(4*b**2*d*tan(c + d*x)**2 + 8*I*b**2*d*tan(c + d*x) - 4*b**2*d) - 4*B/(4*b**2*d*tan(c + d*x)**2 + 8*I*b**2*d*t
an(c + d*x) - 4*b**2*d), Eq(a, I*b)), (x*(A + B*tan(c))*tan(c)**2/(a + b*tan(c))**2, Eq(d, 0)), (-2*A*a**4*b/(
2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*
d*tan(c + d*x)) - 2*A*a**3*b**2*d*x/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*
d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) - 2*A*a**2*b**3*d*x*tan(c + d*x)/(2*a**5*b**2*d + 2*a**4*
b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) - 4*A*a
**2*b**3*log(a/b + tan(c + d*x))/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*t
an(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) + 2*A*a**2*b**3*log(tan(c + d*x)**2 + 1)/(2*a**5*b**2*d + 2*
a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) -
2*A*a**2*b**3/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b
**6*d + 2*b**7*d*tan(c + d*x)) + 2*A*a*b**4*d*x/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d +
4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) - 4*A*a*b**4*log(a/b + tan(c + d*x))*tan(c +
d*x)/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2
*b**7*d*tan(c + d*x)) + 2*A*a*b**4*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c
+ d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) + 2*A*b**5*d*x*tan(c
 + d*x)/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d
+ 2*b**7*d*tan(c + d*x)) + 2*B*a**5*log(a/b + tan(c + d*x))/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a*
*3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) + 2*B*a**5/(2*a**5*b**2*d + 2*a**
4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) + 2*B
*a**4*b*log(a/b + tan(c + d*x))*tan(c + d*x)/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a
**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) + 6*B*a**3*b**2*log(a/b + tan(c + d*x))/(2*a**5*
b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c
 + d*x)) - B*a**3*b**2*log(tan(c + d*x)**2 + 1)/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d +
4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) + 2*B*a**3*b**2/(2*a**5*b**2*d + 2*a**4*b**3*
d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) - 4*B*a**2*b
**3*d*x/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d
+ 2*b**7*d*tan(c + d*x)) + 6*B*a**2*b**3*log(a/b + tan(c + d*x))*tan(c + d*x)/(2*a**5*b**2*d + 2*a**4*b**3*d*t
an(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) - B*a**2*b**3*l
og(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5
*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) - 4*B*a*b**4*d*x*tan(c + d*x)/(2*a**5*b**2*d + 2*a**4*b*
*3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) + B*a*b**
4*log(tan(c + d*x)**2 + 1)/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c +
 d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) + B*b**5*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*a**5*b**2*d + 2*
a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)), T
rue))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.25 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (A a^{2} + 2 \, B a b - A b^{2}\right )} {\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, {\left (B a^{4} + 3 \, B a^{2} b^{2} - 2 \, A a b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}} + \frac {{\left (B a^{2} - 2 \, A a b - B b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, {\left (B a^{3} - A a^{2} b\right )}}{a^{3} b^{2} + a b^{4} + {\left (a^{2} b^{3} + b^{5}\right )} \tan \left (d x + c\right )}}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.58 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.55 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (A a^{2} + 2 \, B a b - A b^{2}\right )} {\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (B a^{2} - 2 \, A a b - B b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, {\left (B a^{4} + 3 \, B a^{2} b^{2} - 2 \, A a b^{3}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}} + \frac {2 \, {\left (B a^{4} \tan \left (d x + c\right ) + 3 \, B a^{2} b^{2} \tan \left (d x + c\right ) - 2 \, A a b^{3} \tan \left (d x + c\right ) + A a^{4} + 2 \, B a^{3} b - A a^{2} b^{2}\right )}}{{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 7.84 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.05 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{2\,d\,\left (-a^2+a\,b\,2{}\mathrm {i}+b^2\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (A+B\,1{}\mathrm {i}\right )}{2\,d\,\left (-a^2\,1{}\mathrm {i}+2\,a\,b+b^2\,1{}\mathrm {i}\right )}-\frac {a^2\,\left (A\,b-B\,a\right )}{b^2\,d\,\left (a^2+b^2\right )\,\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}+\frac {a\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (B\,a^3+3\,B\,a\,b^2-2\,A\,b^3\right )}{b^2\,d\,{\left (a^2+b^2\right )}^2} \]

[In]

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

[Out]

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