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

Optimal result

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

[Out]

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*polylog(2,-exp(2*I*(b*x+a)))/b^2-1/2*d*ta
n(b*x+a)/b^2+1/2*(d*x+c)*tan(b*x+a)^2/b

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3801, 3554, 8, 3800, 2221, 2317, 2438} \[ \int (c+d x) \tan ^3(a+b x) \, dx=-\frac {i d \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {d \tan (a+b x)}{2 b^2}+\frac {(c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {(c+d x) \tan ^2(a+b x)}{2 b}+\frac {d x}{2 b}-\frac {i (c+d x)^2}{2 d} \]

[In]

Int[(c + d*x)*Tan[a + b*x]^3,x]

[Out]

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

Rule 8

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

Rule 2221

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]
- Dist[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 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[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 2438

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 3554

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

Rule 3800

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I*((c + d*x)^(m + 1)/(d*(m + 1))), x
] - Dist[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] &&
IGtQ[m, 0]

Rule 3801

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

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

Mathematica [B] (verified)

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

Time = 6.26 (sec) , antiderivative size = 240, normalized size of antiderivative = 2.22 \[ \int (c+d x) \tan ^3(a+b x) \, dx=\frac {d x \sec ^2(a+b x)}{2 b}+\frac {d \csc (a) \left (b^2 e^{-i \arctan (\cot (a))} x^2-\frac {\cot (a) \left (i b x (-\pi -2 \arctan (\cot (a)))-\pi \log \left (1+e^{-2 i b x}\right )-2 (b x-\arctan (\cot (a))) \log \left (1-e^{2 i (b x-\arctan (\cot (a)))}\right )+\pi \log (\cos (b x))-2 \arctan (\cot (a)) \log (\sin (b x-\arctan (\cot (a))))+i \operatorname {PolyLog}\left (2,e^{2 i (b x-\arctan (\cot (a)))}\right )\right )}{\sqrt {1+\cot ^2(a)}}\right ) \sec (a)}{2 b^2 \sqrt {\csc ^2(a) \left (\cos ^2(a)+\sin ^2(a)\right )}}-\frac {d \sec (a) \sec (a+b x) \sin (b x)}{2 b^2}-\frac {1}{2} d x^2 \tan (a)+\frac {c \left (2 \log (\cos (a+b x))+\tan ^2(a+b x)\right )}{2 b} \]

[In]

Integrate[(c + d*x)*Tan[a + b*x]^3,x]

[Out]

(d*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]])*Log[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[Cot[a]]))]))
/Sqrt[1 + Cot[a]^2])*Sec[a])/(2*b^2*Sqrt[Csc[a]^2*(Cos[a]^2 + Sin[a]^2)]) - (d*Sec[a]*Sec[a + b*x]*Sin[b*x])/(
2*b^2) - (d*x^2*Tan[a])/2 + (c*(2*Log[Cos[a + b*x]] + Tan[a + b*x]^2))/(2*b)

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.69

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

[In]

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

[Out]

-1/2*I*d*x^2+I*c*x+(2*b*d*x*exp(2*I*(b*x+a))-I*d*exp(2*I*(b*x+a))+2*b*c*exp(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*x*a-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)

none

Time = 0.26 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.56 \[ \int (c+d x) \tan ^3(a+b x) \, dx=\frac {2 \, b d x + 2 \, {\left (b d x + b c\right )} \tan \left (b x + a\right )^{2} + i \, d {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) - i \, d {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) + 2 \, {\left (b d x + b c\right )} \log \left (-\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 2 \, {\left (b d x + b c\right )} \log \left (-\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) - 2 \, d \tan \left (b x + a\right )}{4 \, b^{2}} \]

[In]

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

[Out]

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]

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

[In]

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

[Out]

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 = 0.44 (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}} \]

[In]

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

[Out]

-(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 + sin(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)*sin(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]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(tan(a + b*x)^3*(c + d*x),x)

[Out]

int(tan(a + b*x)^3*(c + d*x), x)