\(\int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx\) [267]

Optimal result

Integrand size = 31, antiderivative size = 127 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {(A b-a B) x}{a^2+b^2}+\frac {(a A+b B) \log (\cos (c+d x))}{\left (a^2+b^2\right ) d}-\frac {a^3 (A b-a B) \log (a+b \tan (c+d x))}{b^3 \left (a^2+b^2\right ) d}+\frac {(A b-a B) \tan (c+d x)}{b^2 d}+\frac {B \tan ^2(c+d x)}{2 b d} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.07 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.09 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {-\frac {b (A+i B) \log (i-\tan (c+d x))}{a+i b}-\frac {b (A-i B) \log (i+\tan (c+d x))}{a-i b}+\frac {2 a^3 (-A b+a B) \log (a+b \tan (c+d x))}{b^2 \left (a^2+b^2\right )}+\frac {2 (A b-a B) \tan (c+d x)}{b}+B \tan ^2(c+d x)}{2 b d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.13, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 4090, 27, 3042, 4130, 3042, 4109, 3042, 3956, 4100, 16}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4090

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4130

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4109

\(\displaystyle \frac {B \tan ^2(c+d x)}{2 b d}-\frac {\frac {\frac {b^2 (a A+b B) \int \tan (c+d x)dx}{a^2+b^2}+\frac {a^3 (A b-a B) \int \frac {\tan ^2(c+d x)+1}{a+b \tan (c+d x)}dx}{a^2+b^2}+\frac {b^2 x (A b-a B)}{a^2+b^2}}{b}-\frac {(A b-a B) \tan (c+d x)}{b d}}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {B \tan ^2(c+d x)}{2 b d}-\frac {\frac {\frac {b^2 (a A+b B) \int \tan (c+d x)dx}{a^2+b^2}+\frac {a^3 (A b-a B) \int \frac {\tan (c+d x)^2+1}{a+b \tan (c+d x)}dx}{a^2+b^2}+\frac {b^2 x (A b-a B)}{a^2+b^2}}{b}-\frac {(A b-a B) \tan (c+d x)}{b d}}{b}\)

\(\Big \downarrow \) 3956

\(\displaystyle \frac {B \tan ^2(c+d x)}{2 b d}-\frac {\frac {\frac {a^3 (A b-a B) \int \frac {\tan (c+d x)^2+1}{a+b \tan (c+d x)}dx}{a^2+b^2}-\frac {b^2 (a A+b B) \log (\cos (c+d x))}{d \left (a^2+b^2\right )}+\frac {b^2 x (A b-a B)}{a^2+b^2}}{b}-\frac {(A b-a B) \tan (c+d x)}{b d}}{b}\)

\(\Big \downarrow \) 4100

\(\displaystyle \frac {B \tan ^2(c+d x)}{2 b d}-\frac {\frac {\frac {a^3 (A b-a B) \int \frac {1}{a+b \tan (c+d x)}d(b \tan (c+d x))}{b d \left (a^2+b^2\right )}-\frac {b^2 (a A+b B) \log (\cos (c+d x))}{d \left (a^2+b^2\right )}+\frac {b^2 x (A b-a B)}{a^2+b^2}}{b}-\frac {(A b-a B) \tan (c+d x)}{b d}}{b}\)

\(\Big \downarrow \) 16

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 16

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

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

rule 3956

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

rule 4090

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[b*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + 
 n))), x] + Simp[1/(d*(m + n))   Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Ta 
n[e + f*x])^n*Simp[a^2*A*d*(m + n) - b*B*(b*c*(m - 1) + a*d*(n + 1)) + d*(m 
 + n)*(2*a*A*b + B*(a^2 - b^2))*Tan[e + f*x] - (b*B*(b*c - a*d)*(m - 1) - b 
*(A*b + a*B)*d*(m + n))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, 
 f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2 
, 0] && GtQ[m, 1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 1] 
&& ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
 

rule 4100

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*tan[(e_.) + 
 (f_.)*(x_)]^2), x_Symbol] :> Simp[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 4109

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] + (Simp[(A*b^2 - a*b*B + a^2*C)/(a^2 + b^2)   Int[( 
1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*x]), x], x] - Simp[(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]
 

rule 4130

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

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\frac {\frac {\frac {B \tan \left (d x +c \right )^{2} b}{2}+A \tan \left (d x +c \right ) b -B \tan \left (d x +c \right ) a}{b^{2}}+\frac {\frac {\left (-a A -B b \right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (-A b +B a \right ) \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}-\frac {a^{3} \left (A b -B a \right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{3} \left (a^{2}+b^{2}\right )}}{d}\)	\(127\)
default	\(\frac {\frac {\frac {B \tan \left (d x +c \right )^{2} b}{2}+A \tan \left (d x +c \right ) b -B \tan \left (d x +c \right ) a}{b^{2}}+\frac {\frac {\left (-a A -B b \right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (-A b +B a \right ) \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}-\frac {a^{3} \left (A b -B a \right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{3} \left (a^{2}+b^{2}\right )}}{d}\)	\(127\)
norman	\(\frac {\left (A b -B a \right ) \tan \left (d x +c \right )}{b^{2} d}-\frac {\left (A b -B a \right ) x}{a^{2}+b^{2}}+\frac {B \tan \left (d x +c \right )^{2}}{2 b d}-\frac {\left (a A +B b \right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d \left (a^{2}+b^{2}\right )}-\frac {a^{3} \left (A b -B a \right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{3} \left (a^{2}+b^{2}\right ) d}\)	\(131\)
parallelrisch	\(-\frac {2 A \,b^{4} d x -2 B a \,b^{3} d x -B \,a^{2} b^{2} \tan \left (d x +c \right )^{2}-B \,b^{4} \tan \left (d x +c \right )^{2}+A \ln \left (1+\tan \left (d x +c \right )^{2}\right ) a \,b^{3}+2 A \ln \left (a +b \tan \left (d x +c \right )\right ) a^{3} b -2 A \,a^{2} b^{2} \tan \left (d x +c \right )-2 A \,b^{4} \tan \left (d x +c \right )+B \ln \left (1+\tan \left (d x +c \right )^{2}\right ) b^{4}-2 B \ln \left (a +b \tan \left (d x +c \right )\right ) a^{4}+2 B \,a^{3} b \tan \left (d x +c \right )+2 B a \,b^{3} \tan \left (d x +c \right )}{2 \left (a^{2}+b^{2}\right ) b^{3} d}\)	\(188\)
risch	\(-\frac {x B}{i b -a}+\frac {2 i a^{3} A x}{b^{2} \left (a^{2}+b^{2}\right )}-\frac {2 i a^{4} B x}{b^{3} \left (a^{2}+b^{2}\right )}+\frac {2 i a^{3} A c}{b^{2} d \left (a^{2}+b^{2}\right )}-\frac {2 i A a c}{b^{2} d}+\frac {2 i B \,a^{2} c}{b^{3} d}-\frac {2 i a^{4} B c}{b^{3} d \left (a^{2}+b^{2}\right )}-\frac {i x A}{i b -a}-\frac {2 i A a x}{b^{2}}+\frac {2 i \left (-i B b \,{\mathrm e}^{2 i \left (d x +c \right )}+A b \,{\mathrm e}^{2 i \left (d x +c \right )}-B a \,{\mathrm e}^{2 i \left (d x +c \right )}+A b -B a \right )}{d \,b^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {2 i B \,a^{2} x}{b^{3}}-\frac {2 i B c}{b d}-\frac {2 i B x}{b}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) A a}{b^{2} d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B \,a^{2}}{b^{3} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B}{b d}-\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) A}{b^{2} d \left (a^{2}+b^{2}\right )}+\frac {a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) B}{b^{3} d \left (a^{2}+b^{2}\right )}\)	\(415\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.50 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {2 \, {\left (B a b^{3} - A b^{4}\right )} d x + {\left (B a^{2} b^{2} + B b^{4}\right )} \tan \left (d x + c\right )^{2} + {\left (B a^{4} - A a^{3} b\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - {\left (B a^{4} - A a^{3} b - A a b^{3} - B b^{4}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \, {\left (B a^{3} b - A a^{2} b^{2} + B a b^{3} - A b^{4}\right )} \tan \left (d x + c\right )}{2 \, {\left (a^{2} b^{3} + b^{5}\right )} d} \] Input:

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.81 (sec) , antiderivative size = 1297, normalized size of antiderivative = 10.21 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

Piecewise((zoo*x*(A + B*tan(c))*tan(c)**2, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)) 
, ((-A*log(tan(c + d*x)**2 + 1)/(2*d) + A*tan(c + d*x)**2/(2*d) + B*x + B* 
tan(c + d*x)**3/(3*d) - B*tan(c + d*x)/d)/a, Eq(b, 0)), (-3*A*d*x*tan(c + 
d*x)/(2*b*d*tan(c + d*x) - 2*I*b*d) + 3*I*A*d*x/(2*b*d*tan(c + d*x) - 2*I* 
b*d) + I*A*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*b*d*tan(c + d*x) - 2*I 
*b*d) + A*log(tan(c + d*x)**2 + 1)/(2*b*d*tan(c + d*x) - 2*I*b*d) + 2*A*ta 
n(c + d*x)**2/(2*b*d*tan(c + d*x) - 2*I*b*d) + 3*A/(2*b*d*tan(c + d*x) - 2 
*I*b*d) - 3*I*B*d*x*tan(c + d*x)/(2*b*d*tan(c + d*x) - 2*I*b*d) - 3*B*d*x/ 
(2*b*d*tan(c + d*x) - 2*I*b*d) - 2*B*log(tan(c + d*x)**2 + 1)*tan(c + d*x) 
/(2*b*d*tan(c + d*x) - 2*I*b*d) + 2*I*B*log(tan(c + d*x)**2 + 1)/(2*b*d*ta 
n(c + d*x) - 2*I*b*d) + B*tan(c + d*x)**3/(2*b*d*tan(c + d*x) - 2*I*b*d) + 
 I*B*tan(c + d*x)**2/(2*b*d*tan(c + d*x) - 2*I*b*d) + 3*I*B/(2*b*d*tan(c + 
 d*x) - 2*I*b*d), Eq(a, -I*b)), (-3*A*d*x*tan(c + d*x)/(2*b*d*tan(c + d*x) 
 + 2*I*b*d) - 3*I*A*d*x/(2*b*d*tan(c + d*x) + 2*I*b*d) - I*A*log(tan(c + d 
*x)**2 + 1)*tan(c + d*x)/(2*b*d*tan(c + d*x) + 2*I*b*d) + A*log(tan(c + d* 
x)**2 + 1)/(2*b*d*tan(c + d*x) + 2*I*b*d) + 2*A*tan(c + d*x)**2/(2*b*d*tan 
(c + d*x) + 2*I*b*d) + 3*A/(2*b*d*tan(c + d*x) + 2*I*b*d) + 3*I*B*d*x*tan( 
c + d*x)/(2*b*d*tan(c + d*x) + 2*I*b*d) - 3*B*d*x/(2*b*d*tan(c + d*x) + 2* 
I*b*d) - 2*B*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*b*d*tan(c + d*x) + 2 
*I*b*d) - 2*I*B*log(tan(c + d*x)**2 + 1)/(2*b*d*tan(c + d*x) + 2*I*b*d)...
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.02 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {\frac {2 \, {\left (B a - A b\right )} {\left (d x + c\right )}}{a^{2} + b^{2}} + \frac {2 \, {\left (B a^{4} - A a^{3} b\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} b^{3} + b^{5}} - \frac {{\left (A a + B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac {B b \tan \left (d x + c\right )^{2} - 2 \, {\left (B a - A b\right )} \tan \left (d x + c\right )}{b^{2}}}{2 \, d} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.15 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {{\left (B a - A b\right )} {\left (d x + c\right )}}{a^{2} d + b^{2} d} - \frac {{\left (A a + B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, {\left (a^{2} d + b^{2} d\right )}} + \frac {{\left (B a^{4} - A a^{3} b\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b^{3} d + b^{5} d} + \frac {B b d \tan \left (d x + c\right )^{2} - 2 \, B a d \tan \left (d x + c\right ) + 2 \, A b d \tan \left (d x + c\right )}{2 \, b^{2} d^{2}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 3.88 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.13 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {A}{b}-\frac {B\,a}{b^2}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-B+A\,1{}\mathrm {i}\right )}{2\,d\,\left (-b+a\,1{}\mathrm {i}\right )}+\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (B\,a^4-A\,a^3\,b\right )}{d\,\left (a^2\,b^3+b^5\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{2\,d\,\left (a-b\,1{}\mathrm {i}\right )}+\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,b\,d} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.21 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {-\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right )+\tan \left (d x +c \right )^{2}}{2 d} \] Input:

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

Output:

( - log(tan(c + d*x)**2 + 1) + tan(c + d*x)**2)/(2*d)