\(\int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx\) [285]

Optimal result

Integrand size = 29, antiderivative size = 179 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx=\frac {\left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right ) x}{\left (a^2+b^2\right )^3}-\frac {\left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right )^3 d}+\frac {a (A b-a B)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {a^2 A-A b^2+2 a b B}{\left (a^2+b^2\right )^2 d (a+b \tan (c+d x))} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.58 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.05 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx=\frac {\frac {(A+i B) \log (i-\tan (c+d x))}{(a+i b)^3}+\frac {(A-i B) \log (i+\tan (c+d x))}{(a-i b)^3}-\frac {2 \left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right ) \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^3}+\frac {a (A b-a B)}{b \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac {2 \left (a^2 A-A b^2+2 a b B\right )}{\left (a^2+b^2\right )^2 (a+b \tan (c+d x))}}{2 d} \] Input:

Integrate[(Tan[c + d*x]*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^3,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {3042, 4074, 3042, 4012, 3042, 4014, 3042, 4013}

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.

\(\displaystyle \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3}dx\)

\(\Big \downarrow \) 4074

\(\displaystyle \frac {\int \frac {A b-a B+(a A+b B) \tan (c+d x)}{(a+b \tan (c+d x))^2}dx}{a^2+b^2}+\frac {a (A b-a B)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {A b-a B+(a A+b B) \tan (c+d x)}{(a+b \tan (c+d x))^2}dx}{a^2+b^2}+\frac {a (A b-a B)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 4012

\(\displaystyle \frac {\frac {\int \frac {-B a^2+2 A b a+b^2 B+\left (A a^2+2 b B a-A b^2\right ) \tan (c+d x)}{a+b \tan (c+d x)}dx}{a^2+b^2}+\frac {a^2 A+2 a b B-A b^2}{d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{a^2+b^2}+\frac {a (A b-a B)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {-B a^2+2 A b a+b^2 B+\left (A a^2+2 b B a-A b^2\right ) \tan (c+d x)}{a+b \tan (c+d x)}dx}{a^2+b^2}+\frac {a^2 A+2 a b B-A b^2}{d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{a^2+b^2}+\frac {a (A b-a B)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 4014

\(\displaystyle \frac {\frac {\frac {x \left (a^3 (-B)+3 a^2 A b+3 a b^2 B-A b^3\right )}{a^2+b^2}-\frac {\left (a^3 A+3 a^2 b B-3 a A b^2-b^3 B\right ) \int \frac {b-a \tan (c+d x)}{a+b \tan (c+d x)}dx}{a^2+b^2}}{a^2+b^2}+\frac {a^2 A+2 a b B-A b^2}{d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{a^2+b^2}+\frac {a (A b-a B)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {x \left (a^3 (-B)+3 a^2 A b+3 a b^2 B-A b^3\right )}{a^2+b^2}-\frac {\left (a^3 A+3 a^2 b B-3 a A b^2-b^3 B\right ) \int \frac {b-a \tan (c+d x)}{a+b \tan (c+d x)}dx}{a^2+b^2}}{a^2+b^2}+\frac {a^2 A+2 a b B-A b^2}{d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{a^2+b^2}+\frac {a (A b-a B)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 4013

\(\displaystyle \frac {a (A b-a B)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac {\frac {a^2 A+2 a b B-A b^2}{d \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac {\frac {x \left (a^3 (-B)+3 a^2 A b+3 a b^2 B-A b^3\right )}{a^2+b^2}-\frac {\left (a^3 A+3 a^2 b B-3 a A b^2-b^3 B\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )}}{a^2+b^2}}{a^2+b^2}\)

Input:

Int[(Tan[c + d*x]*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^3,x]
 

Output:

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

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4012

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] + Simp[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 4013

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*Si 
n[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 4014

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] + Simp[(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] && N 
eQ[a*c + b*d, 0]
 

rule 4074

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

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.19

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (A \,a^{3}-3 A a \,b^{2}+3 B \,a^{2} b -B \,b^{3}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (3 A \,a^{2} b -A \,b^{3}-B \,a^{3}+3 B a \,b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{3}}+\frac {a \left (A b -B a \right )}{2 \left (a^{2}+b^{2}\right ) b \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {A \,a^{2}-A \,b^{2}+2 B a b}{\left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}-\frac {\left (A \,a^{3}-3 A a \,b^{2}+3 B \,a^{2} b -B \,b^{3}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{3}}}{d}\)	\(213\)
default	\(\frac {\frac {\frac {\left (A \,a^{3}-3 A a \,b^{2}+3 B \,a^{2} b -B \,b^{3}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (3 A \,a^{2} b -A \,b^{3}-B \,a^{3}+3 B a \,b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{3}}+\frac {a \left (A b -B a \right )}{2 \left (a^{2}+b^{2}\right ) b \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {A \,a^{2}-A \,b^{2}+2 B a b}{\left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}-\frac {\left (A \,a^{3}-3 A a \,b^{2}+3 B \,a^{2} b -B \,b^{3}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{3}}}{d}\)	\(213\)
norman	\(\frac {\frac {\left (A \,a^{2} b^{2}-A \,b^{4}+2 B a \,b^{3}\right ) \tan \left (d x +c \right )}{d b \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}+\frac {\left (3 A \,a^{2} b -A \,b^{3}-B \,a^{3}+3 B a \,b^{2}\right ) a^{2} x}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (a^{2}+b^{2}\right )}+\frac {b^{2} \left (3 A \,a^{2} b -A \,b^{3}-B \,a^{3}+3 B a \,b^{2}\right ) x \tan \left (d x +c \right )^{2}}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (a^{2}+b^{2}\right )}+\frac {a \left (3 A \,a^{2} b^{2}-A \,b^{4}-B \,a^{3} b +3 B a \,b^{3}\right )}{2 b^{2} \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) d}+\frac {2 b \left (3 A \,a^{2} b -A \,b^{3}-B \,a^{3}+3 B a \,b^{2}\right ) a x \tan \left (d x +c \right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (a^{2}+b^{2}\right )}}{\left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {\left (A \,a^{3}-3 A a \,b^{2}+3 B \,a^{2} b -B \,b^{3}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d \left (a^{6}+3 b^{2} a^{4}+3 b^{4} a^{2}+b^{6}\right )}-\frac {\left (A \,a^{3}-3 A a \,b^{2}+3 B \,a^{2} b -B \,b^{3}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{d \left (a^{6}+3 b^{2} a^{4}+3 b^{4} a^{2}+b^{6}\right )}\)	\(446\)
risch	\(\frac {x B}{3 i a^{2} b -i b^{3}-a^{3}+3 a \,b^{2}}+\frac {i x A}{3 i a^{2} b -i b^{3}-a^{3}+3 a \,b^{2}}+\frac {2 i a^{3} A x}{a^{6}+3 b^{2} a^{4}+3 b^{4} a^{2}+b^{6}}-\frac {6 i a \,b^{2} A x}{a^{6}+3 b^{2} a^{4}+3 b^{4} a^{2}+b^{6}}+\frac {6 i a^{2} b B x}{a^{6}+3 b^{2} a^{4}+3 b^{4} a^{2}+b^{6}}-\frac {2 i B \,b^{3} x}{a^{6}+3 b^{2} a^{4}+3 b^{4} a^{2}+b^{6}}+\frac {2 i a^{3} A c}{d \left (a^{6}+3 b^{2} a^{4}+3 b^{4} a^{2}+b^{6}\right )}-\frac {6 i a \,b^{2} A c}{d \left (a^{6}+3 b^{2} a^{4}+3 b^{4} a^{2}+b^{6}\right )}+\frac {6 i a^{2} b B c}{d \left (a^{6}+3 b^{2} a^{4}+3 b^{4} a^{2}+b^{6}\right )}-\frac {2 i B \,b^{3} c}{d \left (a^{6}+3 b^{2} a^{4}+3 b^{4} a^{2}+b^{6}\right )}-\frac {2 i \left (-A \,a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-i B \,a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 A \,a^{2} b^{2}-B \,a^{3} b +i A a \,b^{3}-2 i B \,a^{2} b^{2}+A \,b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-2 B a \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-2 i A \,a^{3} b +i B \,a^{4}-A \,b^{4}+2 B a \,b^{3}-2 i A \,a^{3} b \,{\mathrm e}^{2 i \left (d x +c \right )}+i B \,a^{4} {\mathrm e}^{2 i \left (d x +c \right )}\right )}{\left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right )^{2} \left (-i a +b \right )^{2} d \left (i a +b \right )^{3}}-\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) A}{d \left (a^{6}+3 b^{2} a^{4}+3 b^{4} a^{2}+b^{6}\right )}+\frac {3 a \,b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) A}{d \left (a^{6}+3 b^{2} a^{4}+3 b^{4} a^{2}+b^{6}\right )}-\frac {3 a^{2} b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) B}{d \left (a^{6}+3 b^{2} a^{4}+3 b^{4} a^{2}+b^{6}\right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) B \,b^{3}}{d \left (a^{6}+3 b^{2} a^{4}+3 b^{4} a^{2}+b^{6}\right )}\)	\(816\)
parallelrisch	\(\frac {-2 A \ln \left (a +b \tan \left (d x +c \right )\right ) \tan \left (d x +c \right )^{2} a^{3} b^{4}+6 A \ln \left (a +b \tan \left (d x +c \right )\right ) \tan \left (d x +c \right )^{2} a \,b^{6}-6 B \ln \left (a +b \tan \left (d x +c \right )\right ) \tan \left (d x +c \right )^{2} a^{2} b^{5}+6 A x \tan \left (d x +c \right )^{2} a^{2} b^{5} d -2 B x \tan \left (d x +c \right )^{2} a^{3} b^{4} d +6 B x \tan \left (d x +c \right )^{2} a \,b^{6} d +4 B a \,b^{6} \tan \left (d x +c \right )+2 B \ln \left (a +b \tan \left (d x +c \right )\right ) \tan \left (d x +c \right )^{2} b^{7}+A \ln \left (1+\tan \left (d x +c \right )^{2}\right ) a^{5} b^{2}-3 A \ln \left (1+\tan \left (d x +c \right )^{2}\right ) a^{3} b^{4}-2 A \ln \left (a +b \tan \left (d x +c \right )\right ) a^{5} b^{2}+3 B \ln \left (1+\tan \left (d x +c \right )^{2}\right ) a^{4} b^{3}-B \ln \left (1+\tan \left (d x +c \right )^{2}\right ) a^{2} b^{5}-B \ln \left (1+\tan \left (d x +c \right )^{2}\right ) \tan \left (d x +c \right )^{2} b^{7}-A a \,b^{6}+2 B \,a^{4} b^{3}+3 B \,a^{2} b^{5}-2 A \,b^{7} \tan \left (d x +c \right )-2 B x \,a^{5} b^{2} d -B \,a^{6} b +6 A \ln \left (a +b \tan \left (d x +c \right )\right ) a^{3} b^{4}+A \ln \left (1+\tan \left (d x +c \right )^{2}\right ) \tan \left (d x +c \right )^{2} a^{3} b^{4}-3 A \ln \left (1+\tan \left (d x +c \right )^{2}\right ) \tan \left (d x +c \right )^{2} a \,b^{6}+3 B \ln \left (1+\tan \left (d x +c \right )^{2}\right ) \tan \left (d x +c \right )^{2} a^{2} b^{5}+2 A \ln \left (1+\tan \left (d x +c \right )^{2}\right ) \tan \left (d x +c \right ) a^{4} b^{3}-6 A \ln \left (1+\tan \left (d x +c \right )^{2}\right ) \tan \left (d x +c \right ) a^{2} b^{5}-4 A \ln \left (a +b \tan \left (d x +c \right )\right ) \tan \left (d x +c \right ) a^{4} b^{3}+6 B \ln \left (1+\tan \left (d x +c \right )^{2}\right ) \tan \left (d x +c \right ) a^{3} b^{4}-2 B \ln \left (1+\tan \left (d x +c \right )^{2}\right ) \tan \left (d x +c \right ) a \,b^{6}-2 A x \tan \left (d x +c \right )^{2} b^{7} d -6 B \ln \left (a +b \tan \left (d x +c \right )\right ) a^{4} b^{3}+2 B \ln \left (a +b \tan \left (d x +c \right )\right ) a^{2} b^{5}+12 A x \tan \left (d x +c \right ) a^{3} b^{4} d -4 A x \tan \left (d x +c \right ) a \,b^{6} d +12 B x \tan \left (d x +c \right ) a^{2} b^{5} d +3 A \,a^{5} b^{2}+2 A \,a^{3} b^{4}+2 A \tan \left (d x +c \right ) a^{4} b^{3}+4 B \tan \left (d x +c \right ) a^{3} b^{4}+6 A x \,a^{4} b^{3} d -2 A x \,a^{2} b^{5} d +6 B x \,a^{3} b^{4} d +12 A \ln \left (a +b \tan \left (d x +c \right )\right ) a^{2} b^{5} \tan \left (d x +c \right )-12 B \ln \left (a +b \tan \left (d x +c \right )\right ) a^{3} b^{4} \tan \left (d x +c \right )+4 B \ln \left (a +b \tan \left (d x +c \right )\right ) a \,b^{6} \tan \left (d x +c \right )-4 B x \,a^{4} b^{3} d \tan \left (d x +c \right )}{2 \left (a +b \tan \left (d x +c \right )\right )^{2} \left (a^{6}+3 b^{2} a^{4}+3 b^{4} a^{2}+b^{6}\right ) b^{2} d}\)	\(909\)

Input:

int(tan(d*x+c)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE 
)
 

Output:

1/d*(1/(a^2+b^2)^3*(1/2*(A*a^3-3*A*a*b^2+3*B*a^2*b-B*b^3)*ln(1+tan(d*x+c)^ 
2)+(3*A*a^2*b-A*b^3-B*a^3+3*B*a*b^2)*arctan(tan(d*x+c)))+1/2*a*(A*b-B*a)/( 
a^2+b^2)/b/(a+b*tan(d*x+c))^2+(A*a^2-A*b^2+2*B*a*b)/(a^2+b^2)^2/(a+b*tan(d 
*x+c))-(A*a^3-3*A*a*b^2+3*B*a^2*b-B*b^3)/(a^2+b^2)^3*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. 488 vs. \(2 (176) = 352\).

Time = 0.12 (sec) , antiderivative size = 488, normalized size of antiderivative = 2.73 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx=-\frac {3 \, B a^{4} b - 5 \, A a^{3} b^{2} - 3 \, B a^{2} b^{3} + A a b^{4} + 2 \, {\left (B a^{5} - 3 \, A a^{4} b - 3 \, B a^{3} b^{2} + A a^{2} b^{3}\right )} d x - {\left (B a^{4} b - 3 \, A a^{3} b^{2} - 5 \, B a^{2} b^{3} + 3 \, A a b^{4} - 2 \, {\left (B a^{3} b^{2} - 3 \, A a^{2} b^{3} - 3 \, B a b^{4} + A b^{5}\right )} d x\right )} \tan \left (d x + c\right )^{2} + {\left (A a^{5} + 3 \, B a^{4} b - 3 \, A a^{3} b^{2} - B a^{2} b^{3} + {\left (A a^{3} b^{2} + 3 \, B a^{2} b^{3} - 3 \, A a b^{4} - B b^{5}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (A a^{4} b + 3 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3} - B a b^{4}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \, {\left (B a^{5} - 2 \, A a^{4} b - 3 \, B a^{3} b^{2} + 3 \, A a^{2} b^{3} + 2 \, B a b^{4} - A b^{5} - 2 \, {\left (B a^{4} b - 3 \, A a^{3} b^{2} - 3 \, B a^{2} b^{3} + A a b^{4}\right )} d x\right )} \tan \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} d \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} d \tan \left (d x + c\right ) + {\left (a^{8} + 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} + a^{2} b^{6}\right )} d\right )}} \] Input:

integrate(tan(d*x+c)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x, algorithm="fri 
cas")
 

Output:

-1/2*(3*B*a^4*b - 5*A*a^3*b^2 - 3*B*a^2*b^3 + A*a*b^4 + 2*(B*a^5 - 3*A*a^4 
*b - 3*B*a^3*b^2 + A*a^2*b^3)*d*x - (B*a^4*b - 3*A*a^3*b^2 - 5*B*a^2*b^3 + 
 3*A*a*b^4 - 2*(B*a^3*b^2 - 3*A*a^2*b^3 - 3*B*a*b^4 + A*b^5)*d*x)*tan(d*x 
+ c)^2 + (A*a^5 + 3*B*a^4*b - 3*A*a^3*b^2 - B*a^2*b^3 + (A*a^3*b^2 + 3*B*a 
^2*b^3 - 3*A*a*b^4 - B*b^5)*tan(d*x + c)^2 + 2*(A*a^4*b + 3*B*a^3*b^2 - 3* 
A*a^2*b^3 - B*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)) - 2*(B*a^5 - 2*A*a^4*b - 3*B*a^3*b^2 + 
3*A*a^2*b^3 + 2*B*a*b^4 - A*b^5 - 2*(B*a^4*b - 3*A*a^3*b^2 - 3*B*a^2*b^3 + 
 A*a*b^4)*d*x)*tan(d*x + c))/((a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*d*ta 
n(d*x + c)^2 + 2*(a^7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*d*tan(d*x + c) + 
(a^8 + 3*a^6*b^2 + 3*a^4*b^4 + a^2*b^6)*d)
 

Sympy [F(-2)]

Exception generated. \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx=\text {Exception raised: AttributeError} \] Input:

integrate(tan(d*x+c)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))**3,x)
 

Output:

Exception raised: AttributeError >> 'NoneType' object has no attribute 'pr 
imitive'
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.84 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx=-\frac {\frac {2 \, {\left (B a^{3} - 3 \, A a^{2} b - 3 \, B a b^{2} + A b^{3}\right )} {\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {2 \, {\left (A a^{3} + 3 \, B a^{2} b - 3 \, A a b^{2} - B b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {{\left (A a^{3} + 3 \, B a^{2} b - 3 \, A a b^{2} - B b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {B a^{4} - 3 \, A a^{3} b - 3 \, B a^{2} b^{2} + A a b^{3} - 2 \, {\left (A a^{2} b^{2} + 2 \, B a b^{3} - A b^{4}\right )} \tan \left (d x + c\right )}{a^{6} b + 2 \, a^{4} b^{3} + a^{2} b^{5} + {\left (a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} \tan \left (d x + c\right )}}{2 \, d} \] Input:

integrate(tan(d*x+c)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x, algorithm="max 
ima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.80 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx=-\frac {{\left (B a^{3} - 3 \, A a^{2} b - 3 \, B a b^{2} + A b^{3}\right )} {\left (d x + c\right )}}{a^{6} d + 3 \, a^{4} b^{2} d + 3 \, a^{2} b^{4} d + b^{6} d} + \frac {{\left (A a^{3} + 3 \, B a^{2} b - 3 \, A a b^{2} - B b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, {\left (a^{6} d + 3 \, a^{4} b^{2} d + 3 \, a^{2} b^{4} d + b^{6} d\right )}} - \frac {{\left (A a^{3} b + 3 \, B a^{2} b^{2} - 3 \, A a b^{3} - B b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b d + 3 \, a^{4} b^{3} d + 3 \, a^{2} b^{5} d + b^{7} d} - \frac {B a^{6} - 3 \, A a^{5} b - 2 \, B a^{4} b^{2} - 2 \, A a^{3} b^{3} - 3 \, B a^{2} b^{4} + A a b^{5} - 2 \, {\left (A a^{4} b^{2} + 2 \, B a^{3} b^{3} + 2 \, B a b^{5} - A b^{6}\right )} \tan \left (d x + c\right )}{2 \, {\left (a^{2} + b^{2}\right )}^{3} {\left (b \tan \left (d x + c\right ) + a\right )}^{2} b d} \] Input:

integrate(tan(d*x+c)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x, algorithm="gia 
c")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 3.80 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.58 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx=\frac {\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (A\,a^2\,b+2\,B\,a\,b^2-A\,b^3\right )}{a^4+2\,a^2\,b^2+b^4}-\frac {B\,a^4-3\,A\,a^3\,b-3\,B\,a^2\,b^2+A\,a\,b^3}{2\,b\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{d\,\left (a^2+2\,a\,b\,\mathrm {tan}\left (c+d\,x\right )+b^2\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )}-\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (\frac {A\,a+3\,B\,b}{{\left (a^2+b^2\right )}^2}-\frac {4\,b^2\,\left (A\,a+B\,b\right )}{{\left (a^2+b^2\right )}^3}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-B+A\,1{}\mathrm {i}\right )}{2\,d\,\left (-a^3\,1{}\mathrm {i}+3\,a^2\,b+a\,b^2\,3{}\mathrm {i}-b^3\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{2\,d\,\left (-a^3+a^2\,b\,3{}\mathrm {i}+3\,a\,b^2-b^3\,1{}\mathrm {i}\right )} \] Input:

int((tan(c + d*x)*(A + B*tan(c + d*x)))/(a + b*tan(c + d*x))^3,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.45 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx=\frac {\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) \tan \left (d x +c \right ) a^{2} b -\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) \tan \left (d x +c \right ) b^{3}+\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) a^{3}-\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) a \,b^{2}-2 \,\mathrm {log}\left (a +\tan \left (d x +c \right ) b \right ) \tan \left (d x +c \right ) a^{2} b +2 \,\mathrm {log}\left (a +\tan \left (d x +c \right ) b \right ) \tan \left (d x +c \right ) b^{3}-2 \,\mathrm {log}\left (a +\tan \left (d x +c \right ) b \right ) a^{3}+2 \,\mathrm {log}\left (a +\tan \left (d x +c \right ) b \right ) a \,b^{2}-2 \tan \left (d x +c \right ) a^{2} b +4 \tan \left (d x +c \right ) a \,b^{2} d x -2 \tan \left (d x +c \right ) b^{3}+4 a^{2} b d x}{2 d \left (\tan \left (d x +c \right ) a^{4} b +2 \tan \left (d x +c \right ) a^{2} b^{3}+\tan \left (d x +c \right ) b^{5}+a^{5}+2 a^{3} b^{2}+a \,b^{4}\right )} \] Input:

int(tan(d*x+c)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x)
 

Output:

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