\(\int (c+d x) \tan ^3(a+b x) \, dx\) [306]

Optimal result

Integrand size = 14, antiderivative size = 108 \[ \int (c+d x) \tan ^3(a+b x) \, dx=\frac {d x}{2 b}-\frac {i (c+d x)^2}{2 d}+\frac {(c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac {i d \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {d \tan (a+b x)}{2 b^2}+\frac {(c+d x) \tan ^2(a+b x)}{2 b} \] Output:

1/2*d*x/b-1/2*I*(d*x+c)^2/d+(d*x+c)*ln(1+exp(2*I*(b*x+a)))/b-1/2*I*d*polyl 
og(2,-exp(2*I*(b*x+a)))/b^2-1/2*d*tan(b*x+a)/b^2+1/2*(d*x+c)*tan(b*x+a)^2/ 
b
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(240\) vs. \(2(108)=216\).

Time = 6.16 (sec) , antiderivative size = 240, normalized size of antiderivative = 2.22 \[ \int (c+d x) \tan ^3(a+b x) \, dx=\frac {d x \sec ^2(a+b x)}{2 b}+\frac {c \left (2 \log (\cos (a+b x))+\sec ^2(a+b x)\right )}{2 b}+\frac {d \csc (a) \left (b^2 e^{-i \arctan (\cot (a))} x^2-\frac {\cot (a) \left (i b x (-\pi -2 \arctan (\cot (a)))-\pi \log \left (1+e^{-2 i b x}\right )-2 (b x-\arctan (\cot (a))) \log \left (1-e^{2 i (b x-\arctan (\cot (a)))}\right )+\pi \log (\cos (b x))-2 \arctan (\cot (a)) \log (\sin (b x-\arctan (\cot (a))))+i \operatorname {PolyLog}\left (2,e^{2 i (b x-\arctan (\cot (a)))}\right )\right )}{\sqrt {1+\cot ^2(a)}}\right ) \sec (a)}{2 b^2 \sqrt {\csc ^2(a) \left (\cos ^2(a)+\sin ^2(a)\right )}}-\frac {d \sec (a) \sec (a+b x) \sin (b x)}{2 b^2}-\frac {1}{2} d x^2 \tan (a) \] Input:

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

Output:

(d*x*Sec[a + b*x]^2)/(2*b) + (c*(2*Log[Cos[a + b*x]] + Sec[a + b*x]^2))/(2 
*b) + (d*Csc[a]*((b^2*x^2)/E^(I*ArcTan[Cot[a]]) - (Cot[a]*(I*b*x*(-Pi - 2* 
ArcTan[Cot[a]]) - Pi*Log[1 + E^((-2*I)*b*x)] - 2*(b*x - ArcTan[Cot[a]])*Lo 
g[1 - E^((2*I)*(b*x - ArcTan[Cot[a]]))] + Pi*Log[Cos[b*x]] - 2*ArcTan[Cot[ 
a]]*Log[Sin[b*x - ArcTan[Cot[a]]]] + I*PolyLog[2, E^((2*I)*(b*x - ArcTan[C 
ot[a]]))]))/Sqrt[1 + Cot[a]^2])*Sec[a])/(2*b^2*Sqrt[Csc[a]^2*(Cos[a]^2 + S 
in[a]^2)]) - (d*Sec[a]*Sec[a + b*x]*Sin[b*x])/(2*b^2) - (d*x^2*Tan[a])/2
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {3042, 4203, 3042, 3954, 24, 4202, 2620, 2715, 2838}

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.

\(\displaystyle \int (c+d x) \tan ^3(a+b x) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (c+d x) \tan (a+b x)^3dx\)

\(\Big \downarrow \) 4203

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3954

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 4202

\(\displaystyle 2 i \int \frac {e^{2 i (a+b x)} (c+d x)}{1+e^{2 i (a+b x)}}dx+\frac {(c+d x) \tan ^2(a+b x)}{2 b}-\frac {d \left (\frac {\tan (a+b x)}{b}-x\right )}{2 b}-\frac {i (c+d x)^2}{2 d}\)

\(\Big \downarrow \) 2620

\(\displaystyle 2 i \left (\frac {i d \int \log \left (1+e^{2 i (a+b x)}\right )dx}{2 b}-\frac {i (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )+\frac {(c+d x) \tan ^2(a+b x)}{2 b}-\frac {d \left (\frac {\tan (a+b x)}{b}-x\right )}{2 b}-\frac {i (c+d x)^2}{2 d}\)

\(\Big \downarrow \) 2715

\(\displaystyle 2 i \left (\frac {d \int e^{-2 i (a+b x)} \log \left (1+e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}-\frac {i (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )+\frac {(c+d x) \tan ^2(a+b x)}{2 b}-\frac {d \left (\frac {\tan (a+b x)}{b}-x\right )}{2 b}-\frac {i (c+d x)^2}{2 d}\)

\(\Big \downarrow \) 2838

\(\displaystyle 2 i \left (-\frac {d \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{4 b^2}-\frac {i (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )+\frac {(c+d x) \tan ^2(a+b x)}{2 b}-\frac {d \left (\frac {\tan (a+b x)}{b}-x\right )}{2 b}-\frac {i (c+d x)^2}{2 d}\)

Input:

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

Output:

((-1/2*I)*(c + d*x)^2)/d + (2*I)*(((-1/2*I)*(c + d*x)*Log[1 + E^((2*I)*(a 
+ b*x))])/b - (d*PolyLog[2, -E^((2*I)*(a + b*x))])/(4*b^2)) + ((c + d*x)*T 
an[a + b*x]^2)/(2*b) - (d*(-x + Tan[a + b*x]/b))/(2*b)
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 3042

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

rule 3954

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d 
*x])^(n - 1)/(d*(n - 1))), x] - Simp[b^2   Int[(b*Tan[c + d*x])^(n - 2), x] 
, x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
 

rule 4202

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I 
*((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[2*I   Int[(c + d*x)^m*(E^(2*I*( 
e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] && IGt 
Q[m, 0]
 

rule 4203

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symb 
ol] :> Simp[b*(c + d*x)^m*((b*Tan[e + f*x])^(n - 1)/(f*(n - 1))), x] + (-Si 
mp[b*d*(m/(f*(n - 1)))   Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1), x] 
, x] - Simp[b^2   Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; Free 
Q[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 0]
 

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.69

method	result	size
risch	\(-\frac {i d \,x^{2}}{2}+i x c +\frac {2 b d x \,{\mathrm e}^{2 i \left (b x +a \right )}-i d \,{\mathrm e}^{2 i \left (b x +a \right )}+2 b c \,{\mathrm e}^{2 i \left (b x +a \right )}-i d}{b^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{2}}+\frac {c \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b}-\frac {2 c \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}-\frac {2 i d a x}{b}-\frac {i d \,a^{2}}{b^{2}}+\frac {d \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) x}{b}-\frac {i d \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{2}}+\frac {2 d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}\)	\(183\)

Input:

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

Output:

-1/2*I*d*x^2+I*x*c+(2*b*d*x*exp(2*I*(b*x+a))-I*d*exp(2*I*(b*x+a))+2*b*c*ex 
p(2*I*(b*x+a))-I*d)/b^2/(exp(2*I*(b*x+a))+1)^2+1/b*c*ln(exp(2*I*(b*x+a))+1 
)-2/b*c*ln(exp(I*(b*x+a)))-2*I/b*d*a*x-I/b^2*d*a^2+1/b*d*ln(exp(2*I*(b*x+a 
))+1)*x-1/2*I*d*polylog(2,-exp(2*I*(b*x+a)))/b^2+2/b^2*d*a*ln(exp(I*(b*x+a 
)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.56 \[ \int (c+d x) \tan ^3(a+b x) \, dx=\frac {2 \, b d x + 2 \, {\left (b d x + b c\right )} \tan \left (b x + a\right )^{2} + i \, d {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) - i \, d {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) + 2 \, {\left (b d x + b c\right )} \log \left (-\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 2 \, {\left (b d x + b c\right )} \log \left (-\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) - 2 \, d \tan \left (b x + a\right )}{4 \, b^{2}} \] Input:

integrate((d*x+c)*tan(b*x+a)^3,x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int (c+d x) \tan ^3(a+b x) \, dx=\int \left (c + d x\right ) \tan ^{3}{\left (a + b x \right )}\, dx \] Input:

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

Output:

Integral((c + d*x)*tan(a + b*x)**3, x)
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 517 vs. \(2 (89) = 178\).

Time = 0.24 (sec) , antiderivative size = 517, normalized size of antiderivative = 4.79 \[ \int (c+d x) \tan ^3(a+b x) \, dx=-\frac {b^{2} d x^{2} + 2 \, b^{2} c x - 2 \, {\left (b d x + b c + {\left (b d x + b c\right )} \cos \left (4 \, b x + 4 \, a\right ) + 2 \, {\left (b d x + b c\right )} \cos \left (2 \, b x + 2 \, a\right ) - {\left (-i \, b d x - i \, b c\right )} \sin \left (4 \, b x + 4 \, a\right ) - 2 \, {\left (-i \, b d x - i \, b c\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + {\left (b^{2} d x^{2} + 2 \, b^{2} c x\right )} \cos \left (4 \, b x + 4 \, a\right ) + 2 \, {\left (b^{2} d x^{2} + 2 i \, b c + 2 \, {\left (b^{2} c + i \, b d\right )} x + d\right )} \cos \left (2 \, b x + 2 \, a\right ) + {\left (d \cos \left (4 \, b x + 4 \, a\right ) + 2 \, d \cos \left (2 \, b x + 2 \, a\right ) + i \, d \sin \left (4 \, b x + 4 \, a\right ) + 2 i \, d \sin \left (2 \, b x + 2 \, a\right ) + d\right )} {\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) - {\left (-i \, b d x - i \, b c + {\left (-i \, b d x - i \, b c\right )} \cos \left (4 \, b x + 4 \, a\right ) - 2 \, {\left (i \, b d x + i \, b c\right )} \cos \left (2 \, b x + 2 \, a\right ) + {\left (b d x + b c\right )} \sin \left (4 \, b x + 4 \, a\right ) + 2 \, {\left (b d x + b c\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - {\left (-i \, b^{2} d x^{2} - 2 i \, b^{2} c x\right )} \sin \left (4 \, b x + 4 \, a\right ) + 2 \, {\left (i \, b^{2} d x^{2} - 2 \, b c + 2 \, {\left (i \, b^{2} c - b d\right )} x + i \, d\right )} \sin \left (2 \, b x + 2 \, a\right ) + 2 \, d}{-2 i \, b^{2} \cos \left (4 \, b x + 4 \, a\right ) - 4 i \, b^{2} \cos \left (2 \, b x + 2 \, a\right ) + 2 \, b^{2} \sin \left (4 \, b x + 4 \, a\right ) + 4 \, b^{2} \sin \left (2 \, b x + 2 \, a\right ) - 2 i \, b^{2}} \] Input:

integrate((d*x+c)*tan(b*x+a)^3,x, algorithm="maxima")
 

Output:

-(b^2*d*x^2 + 2*b^2*c*x - 2*(b*d*x + b*c + (b*d*x + b*c)*cos(4*b*x + 4*a) 
+ 2*(b*d*x + b*c)*cos(2*b*x + 2*a) - (-I*b*d*x - I*b*c)*sin(4*b*x + 4*a) - 
 2*(-I*b*d*x - I*b*c)*sin(2*b*x + 2*a))*arctan2(sin(2*b*x + 2*a), cos(2*b* 
x + 2*a) + 1) + (b^2*d*x^2 + 2*b^2*c*x)*cos(4*b*x + 4*a) + 2*(b^2*d*x^2 + 
2*I*b*c + 2*(b^2*c + I*b*d)*x + d)*cos(2*b*x + 2*a) + (d*cos(4*b*x + 4*a) 
+ 2*d*cos(2*b*x + 2*a) + I*d*sin(4*b*x + 4*a) + 2*I*d*sin(2*b*x + 2*a) + d 
)*dilog(-e^(2*I*b*x + 2*I*a)) - (-I*b*d*x - I*b*c + (-I*b*d*x - I*b*c)*cos 
(4*b*x + 4*a) - 2*(I*b*d*x + I*b*c)*cos(2*b*x + 2*a) + (b*d*x + b*c)*sin(4 
*b*x + 4*a) + 2*(b*d*x + b*c)*sin(2*b*x + 2*a))*log(cos(2*b*x + 2*a)^2 + s 
in(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) - (-I*b^2*d*x^2 - 2*I*b^2*c*x) 
*sin(4*b*x + 4*a) + 2*(I*b^2*d*x^2 - 2*b*c + 2*(I*b^2*c - b*d)*x + I*d)*si 
n(2*b*x + 2*a) + 2*d)/(-2*I*b^2*cos(4*b*x + 4*a) - 4*I*b^2*cos(2*b*x + 2*a 
) + 2*b^2*sin(4*b*x + 4*a) + 4*b^2*sin(2*b*x + 2*a) - 2*I*b^2)
 

Giac [F]

\[ \int (c+d x) \tan ^3(a+b x) \, dx=\int { {\left (d x + c\right )} \tan \left (b x + a\right )^{3} \,d x } \] Input:

integrate((d*x+c)*tan(b*x+a)^3,x, algorithm="giac")
 

Output:

integrate((d*x + c)*tan(b*x + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int (c+d x) \tan ^3(a+b x) \, dx=\int {\mathrm {tan}\left (a+b\,x\right )}^3\,\left (c+d\,x\right ) \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int (c+d x) \tan ^3(a+b x) \, dx=\frac {2 \left (\int \tan \left (b x +a \right )^{3} x d x \right ) b d -\mathrm {log}\left (\tan \left (b x +a \right )^{2}+1\right ) c +\tan \left (b x +a \right )^{2} c}{2 b} \] Input:

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

Output:

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