\(\int \tan ^3(c+d x) (a+b \tan (c+d x)) \, dx\) [414]

Optimal result

Integrand size = 19, antiderivative size = 60 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x)) \, dx=b x+\frac {a \log (\cos (c+d x))}{d}-\frac {b \tan (c+d x)}{d}+\frac {a \tan ^2(c+d x)}{2 d}+\frac {b \tan ^3(c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3609, 3606, 3556} \[ \int \tan ^3(c+d x) (a+b \tan (c+d x)) \, dx=\frac {a \tan ^2(c+d x)}{2 d}+\frac {a \log (\cos (c+d x))}{d}+\frac {b \tan ^3(c+d x)}{3 d}-\frac {b \tan (c+d x)}{d}+b x \]

[In]

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

[Out]

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

Rule 3556

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

Rule 3606

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c - b
*d)*x, x] + (Dist[b*c + a*d, Int[Tan[e + f*x], x], x] + Simp[b*d*(Tan[e + f*x]/f), x]) /; FreeQ[{a, b, c, d, e
, f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

Rule 3609

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*
((a + b*Tan[e + f*x])^m/(f*m)), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*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] && GtQ[m, 0]

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.12 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x)) \, dx=\frac {b \arctan (\tan (c+d x))}{d}-\frac {b \tan (c+d x)}{d}+\frac {b \tan ^3(c+d x)}{3 d}+\frac {a \left (2 \log (\cos (c+d x))+\tan ^2(c+d x)\right )}{2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.95

method	result	size
parallelrisch	\(-\frac {-2 b \left (\tan ^{3}\left (d x +c \right )\right )-6 b d x -3 \left (\tan ^{2}\left (d x +c \right )\right ) a +3 a \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+6 b \tan \left (d x +c \right )}{6 d}\)	\(57\)
derivativedivides	\(\frac {\frac {b \left (\tan ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\tan ^{2}\left (d x +c \right )\right ) a}{2}-b \tan \left (d x +c \right )-\frac {a \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+b \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(60\)
default	\(\frac {\frac {b \left (\tan ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\tan ^{2}\left (d x +c \right )\right ) a}{2}-b \tan \left (d x +c \right )-\frac {a \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+b \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(60\)
norman	\(b x +\frac {a \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {b \tan \left (d x +c \right )}{d}+\frac {b \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}-\frac {a \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(62\)
parts	\(\frac {a \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}+\frac {b \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(62\)
risch	\(b x -i a x -\frac {2 i a c}{d}-\frac {2 i \left (3 i a \,{\mathrm e}^{4 i \left (d x +c \right )}+6 b \,{\mathrm e}^{4 i \left (d x +c \right )}+3 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+6 b \,{\mathrm e}^{2 i \left (d x +c \right )}+4 b \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(108\)

[In]

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

[Out]

-1/6*(-2*b*tan(d*x+c)^3-6*b*d*x-3*tan(d*x+c)^2*a+3*a*ln(1+tan(d*x+c)^2)+6*b*tan(d*x+c))/d

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.97 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x)) \, dx=\frac {2 \, b \tan \left (d x + c\right )^{3} + 6 \, b d x + 3 \, a \tan \left (d x + c\right )^{2} + 3 \, a \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 6 \, b \tan \left (d x + c\right )}{6 \, d} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.17 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x)) \, dx=\begin {cases} - \frac {a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a \tan ^{2}{\left (c + d x \right )}}{2 d} + b x + \frac {b \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac {b \tan {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\left (c \right )}\right ) \tan ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-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, Ne(d, 0)), (x*(a + b*tan(c))*tan(c)**3, True))

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.98 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x)) \, dx=\frac {2 \, b \tan \left (d x + c\right )^{3} + 3 \, a \tan \left (d x + c\right )^{2} + 6 \, {\left (d x + c\right )} b - 3 \, a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 6 \, b \tan \left (d x + c\right )}{6 \, d} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 475 vs. \(2 (56) = 112\).

Time = 0.79 (sec) , antiderivative size = 475, normalized size of antiderivative = 7.92 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x)) \, dx=\frac {6 \, b d x \tan \left (d x\right )^{3} \tan \left (c\right )^{3} + 3 \, a \log \left (\frac {4 \, {\left (\tan \left (d x\right )^{2} \tan \left (c\right )^{2} - 2 \, \tan \left (d x\right ) \tan \left (c\right ) + 1\right )}}{\tan \left (d x\right )^{2} \tan \left (c\right )^{2} + \tan \left (d x\right )^{2} + \tan \left (c\right )^{2} + 1}\right ) \tan \left (d x\right )^{3} \tan \left (c\right )^{3} - 18 \, b d x \tan \left (d x\right )^{2} \tan \left (c\right )^{2} + 3 \, a \tan \left (d x\right )^{3} \tan \left (c\right )^{3} - 9 \, a \log \left (\frac {4 \, {\left (\tan \left (d x\right )^{2} \tan \left (c\right )^{2} - 2 \, \tan \left (d x\right ) \tan \left (c\right ) + 1\right )}}{\tan \left (d x\right )^{2} \tan \left (c\right )^{2} + \tan \left (d x\right )^{2} + \tan \left (c\right )^{2} + 1}\right ) \tan \left (d x\right )^{2} \tan \left (c\right )^{2} + 6 \, b \tan \left (d x\right )^{3} \tan \left (c\right )^{2} + 6 \, b \tan \left (d x\right )^{2} \tan \left (c\right )^{3} + 18 \, b d x \tan \left (d x\right ) \tan \left (c\right ) + 3 \, a \tan \left (d x\right )^{3} \tan \left (c\right ) - 3 \, a \tan \left (d x\right )^{2} \tan \left (c\right )^{2} + 3 \, a \tan \left (d x\right ) \tan \left (c\right )^{3} - 2 \, b \tan \left (d x\right )^{3} + 9 \, a \log \left (\frac {4 \, {\left (\tan \left (d x\right )^{2} \tan \left (c\right )^{2} - 2 \, \tan \left (d x\right ) \tan \left (c\right ) + 1\right )}}{\tan \left (d x\right )^{2} \tan \left (c\right )^{2} + \tan \left (d x\right )^{2} + \tan \left (c\right )^{2} + 1}\right ) \tan \left (d x\right ) \tan \left (c\right ) - 18 \, b \tan \left (d x\right )^{2} \tan \left (c\right ) - 18 \, b \tan \left (d x\right ) \tan \left (c\right )^{2} - 2 \, b \tan \left (c\right )^{3} - 6 \, b d x - 3 \, a \tan \left (d x\right )^{2} + 3 \, a \tan \left (d x\right ) \tan \left (c\right ) - 3 \, a \tan \left (c\right )^{2} - 3 \, a \log \left (\frac {4 \, {\left (\tan \left (d x\right )^{2} \tan \left (c\right )^{2} - 2 \, \tan \left (d x\right ) \tan \left (c\right ) + 1\right )}}{\tan \left (d x\right )^{2} \tan \left (c\right )^{2} + \tan \left (d x\right )^{2} + \tan \left (c\right )^{2} + 1}\right ) + 6 \, b \tan \left (d x\right ) + 6 \, b \tan \left (c\right ) - 3 \, a}{6 \, {\left (d \tan \left (d x\right )^{3} \tan \left (c\right )^{3} - 3 \, d \tan \left (d x\right )^{2} \tan \left (c\right )^{2} + 3 \, d \tan \left (d x\right ) \tan \left (c\right ) - d\right )}} \]

[In]

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

[Out]

1/6*(6*b*d*x*tan(d*x)^3*tan(c)^3 + 3*a*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^
2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^3*tan(c)^3 - 18*b*d*x*tan(d*x)^2*tan(c)^2 + 3*a*tan(d*x)^3*tan(c)^3 -
9*a*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*ta
n(d*x)^2*tan(c)^2 + 6*b*tan(d*x)^3*tan(c)^2 + 6*b*tan(d*x)^2*tan(c)^3 + 18*b*d*x*tan(d*x)*tan(c) + 3*a*tan(d*x
)^3*tan(c) - 3*a*tan(d*x)^2*tan(c)^2 + 3*a*tan(d*x)*tan(c)^3 - 2*b*tan(d*x)^3 + 9*a*log(4*(tan(d*x)^2*tan(c)^2
- 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)*tan(c) - 18*b*tan(d*x)^2
*tan(c) - 18*b*tan(d*x)*tan(c)^2 - 2*b*tan(c)^3 - 6*b*d*x - 3*a*tan(d*x)^2 + 3*a*tan(d*x)*tan(c) - 3*a*tan(c)^
2 - 3*a*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))
+ 6*b*tan(d*x) + 6*b*tan(c) - 3*a)/(d*tan(d*x)^3*tan(c)^3 - 3*d*tan(d*x)^2*tan(c)^2 + 3*d*tan(d*x)*tan(c) - d
)

Mupad [B] (verification not implemented)

Time = 4.82 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.90 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x)) \, dx=\frac {\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}-\frac {a\,\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}{2}-b\,\mathrm {tan}\left (c+d\,x\right )+\frac {b\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3}+b\,d\,x}{d} \]

[In]

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

[Out]

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