\(\int \tan (c+d x) (a+b \tan (c+d x))^2 \, dx\) [427]

Optimal result

Integrand size = 19, antiderivative size = 58 \[ \int \tan (c+d x) (a+b \tan (c+d x))^2 \, dx=-2 a b x-\frac {\left (a^2-b^2\right ) \log (\cos (c+d x))}{d}+\frac {a b \tan (c+d x)}{d}+\frac {(a+b \tan (c+d x))^2}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3609, 3606, 3556} \[ \int \tan (c+d x) (a+b \tan (c+d x))^2 \, dx=-\frac {\left (a^2-b^2\right ) \log (\cos (c+d x))}{d}+\frac {(a+b \tan (c+d x))^2}{2 d}+\frac {a b \tan (c+d x)}{d}-2 a b x \]

[In]

Int[Tan[c + d*x]*(a + b*Tan[c + d*x])^2,x]

[Out]

-2*a*b*x - ((a^2 - b^2)*Log[Cos[c + d*x]])/d + (a*b*Tan[c + d*x])/d + (a + b*Tan[c + d*x])^2/(2*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 {(a+b \tan (c+d x))^2}{2 d}+\int (-b+a \tan (c+d x)) (a+b \tan (c+d x)) \, dx \\ & = -2 a b x+\frac {a b \tan (c+d x)}{d}+\frac {(a+b \tan (c+d x))^2}{2 d}+\left (a^2-b^2\right ) \int \tan (c+d x) \, dx \\ & = -2 a b x-\frac {\left (a^2-b^2\right ) \log (\cos (c+d x))}{d}+\frac {a b \tan (c+d x)}{d}+\frac {(a+b \tan (c+d x))^2}{2 d} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.79 \[ \int \tan (c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {a^2 \log (i-\tan (c+d x))}{2 d}+\frac {i a b \log (i-\tan (c+d x))}{d}-\frac {b^2 \log (i-\tan (c+d x))}{2 d}+\frac {a^2 \log (i+\tan (c+d x))}{2 d}-\frac {i a b \log (i+\tan (c+d x))}{d}-\frac {b^2 \log (i+\tan (c+d x))}{2 d}+\frac {2 a b \tan (c+d x)}{d}+\frac {b^2 \tan ^2(c+d x)}{2 d} \]

[In]

Integrate[Tan[c + d*x]*(a + b*Tan[c + d*x])^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.05

method	result	size
norman	\(-2 a b x +\frac {b^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {2 a b \tan \left (d x +c \right )}{d}+\frac {\left (a^{2}-b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(61\)
derivativedivides	\(\frac {\frac {b^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{2}+2 a b \tan \left (d x +c \right )+\frac {\left (a^{2}-b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}-2 a b \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(62\)
default	\(\frac {\frac {b^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{2}+2 a b \tan \left (d x +c \right )+\frac {\left (a^{2}-b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}-2 a b \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(62\)
parallelrisch	\(\frac {-4 a b d x +b^{2} \left (\tan ^{2}\left (d x +c \right )\right )+\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{2}-\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) b^{2}+4 a b \tan \left (d x +c \right )}{2 d}\)	\(66\)
parts	\(\frac {b^{2} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}+\frac {a^{2} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {2 a b \left (\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(75\)
risch	\(-2 a b x +i a^{2} x -i b^{2} x +\frac {2 i a^{2} c}{d}-\frac {2 i b^{2} c}{d}+\frac {2 i b \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{2}}{d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) b^{2}}{d}\)	\(129\)

[In]

int(tan(d*x+c)*(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00 \[ \int \tan (c+d x) (a+b \tan (c+d x))^2 \, dx=-\frac {4 \, a b d x - b^{2} \tan \left (d x + c\right )^{2} - 4 \, a b \tan \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right )}{2 \, d} \]

[In]

integrate(tan(d*x+c)*(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.47 \[ \int \tan (c+d x) (a+b \tan (c+d x))^2 \, dx=\begin {cases} \frac {a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - 2 a b x + \frac {2 a b \tan {\left (c + d x \right )}}{d} - \frac {b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {b^{2} \tan ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\left (c \right )}\right )^{2} \tan {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(tan(d*x+c)*(a+b*tan(d*x+c))**2,x)

[Out]

Piecewise((a**2*log(tan(c + d*x)**2 + 1)/(2*d) - 2*a*b*x + 2*a*b*tan(c + d*x)/d - b**2*log(tan(c + d*x)**2 + 1
)/(2*d) + b**2*tan(c + d*x)**2/(2*d), Ne(d, 0)), (x*(a + b*tan(c))**2*tan(c), True))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00 \[ \int \tan (c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {b^{2} \tan \left (d x + c\right )^{2} - 4 \, {\left (d x + c\right )} a b + 4 \, a b \tan \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} \]

[In]

integrate(tan(d*x+c)*(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 494 vs. \(2 (56) = 112\).

Time = 0.62 (sec) , antiderivative size = 494, normalized size of antiderivative = 8.52 \[ \int \tan (c+d x) (a+b \tan (c+d x))^2 \, dx=-\frac {4 \, a b d x \tan \left (d x\right )^{2} \tan \left (c\right )^{2} + a^{2} \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} - b^{2} \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} - 8 \, a b d x \tan \left (d x\right ) \tan \left (c\right ) - b^{2} \tan \left (d x\right )^{2} \tan \left (c\right )^{2} - 2 \, a^{2} \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 ) + 2 \, b^{2} \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 ) + 4 \, a b \tan \left (d x\right )^{2} \tan \left (c\right ) + 4 \, a b \tan \left (d x\right ) \tan \left (c\right )^{2} + 4 \, a b d x - b^{2} \tan \left (d x\right )^{2} - b^{2} \tan \left (c\right )^{2} + a^{2} \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 ) - b^{2} \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 ) - 4 \, a b \tan \left (d x\right ) - 4 \, a b \tan \left (c\right ) - b^{2}}{2 \, {\left (d \tan \left (d x\right )^{2} \tan \left (c\right )^{2} - 2 \, d \tan \left (d x\right ) \tan \left (c\right ) + d\right )}} \]

[In]

integrate(tan(d*x+c)*(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/2*(4*a*b*d*x*tan(d*x)^2*tan(c)^2 + a^2*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)^2*tan(c)^2 - b^2*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)^2*tan(c)^2 - 8*a*b*d*x*tan(d*x)*tan(c) - b^2*ta
n(d*x)^2*tan(c)^2 - 2*a^2*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) + 2*b^2*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) + 4*a*b*tan(d*x)^2*tan(c) + 4*a*b*tan(d*x)*tan(c)^2 + 4*a*
b*d*x - b^2*tan(d*x)^2 - b^2*tan(c)^2 + a^2*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*ta
n(c)^2 + tan(d*x)^2 + tan(c)^2 + 1)) - b^2*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)) - 4*a*b*tan(d*x) - 4*a*b*tan(c) - b^2)/(d*tan(d*x)^2*tan(c)^2 - 2*d*tan(d*
x)*tan(c) + d)

Mupad [B] (verification not implemented)

Time = 5.48 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.98 \[ \int \tan (c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,\left (\frac {a^2}{2}-\frac {b^2}{2}\right )+\frac {b^2\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}+2\,a\,b\,\mathrm {tan}\left (c+d\,x\right )-2\,a\,b\,d\,x}{d} \]

[In]

int(tan(c + d*x)*(a + b*tan(c + d*x))^2,x)

[Out]

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