\(\int \sec (c+d x) (a+b \tan (c+d x))^2 \, dx\) [526]

Optimal result

Integrand size = 19, antiderivative size = 65 \[ \int \sec (c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {\left (2 a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {3 a b \sec (c+d x)}{2 d}+\frac {b \sec (c+d x) (a+b \tan (c+d x))}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 65, 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 = {3589, 3567, 3855} \[ \int \sec (c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {\left (2 a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {3 a b \sec (c+d x)}{2 d}+\frac {b \sec (c+d x) (a+b \tan (c+d x))}{2 d} \]

[In]

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

[Out]

((2*a^2 - b^2)*ArcTanh[Sin[c + d*x]])/(2*d) + (3*a*b*Sec[c + d*x])/(2*d) + (b*Sec[c + d*x]*(a + b*Tan[c + d*x]
))/(2*d)

Rule 3567

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*((d*Sec[
e + f*x])^m/(f*m)), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3589

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[b*(d*Sec
[e + f*x])^m*((a + b*Tan[e + f*x])/(f*(m + 1))), x] + Dist[1/(m + 1), Int[(d*Sec[e + f*x])^m*(a^2*(m + 1) - b^
2 + a*b*(m + 2)*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 + b^2, 0] && NeQ[m, -1]

Rule 3855

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

Rubi steps \begin{align*} \text {integral}& = \frac {b \sec (c+d x) (a+b \tan (c+d x))}{2 d}+\frac {1}{2} \int \sec (c+d x) \left (2 a^2-b^2+3 a b \tan (c+d x)\right ) \, dx \\ & = \frac {3 a b \sec (c+d x)}{2 d}+\frac {b \sec (c+d x) (a+b \tan (c+d x))}{2 d}+\frac {1}{2} \left (2 a^2-b^2\right ) \int \sec (c+d x) \, dx \\ & = \frac {\left (2 a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {3 a b \sec (c+d x)}{2 d}+\frac {b \sec (c+d x) (a+b \tan (c+d x))}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.03 \[ \int \sec (c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {a^2 \text {arctanh}(\sin (c+d x))}{d}-\frac {b^2 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {2 a b \sec (c+d x)}{d}+\frac {b^2 \sec (c+d x) \tan (c+d x)}{2 d} \]

[In]

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

[Out]

(a^2*ArcTanh[Sin[c + d*x]])/d - (b^2*ArcTanh[Sin[c + d*x]])/(2*d) + (2*a*b*Sec[c + d*x])/d + (b^2*Sec[c + d*x]
*Tan[c + d*x])/(2*d)

Maple [A] (verified)

Time = 1.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.28

method	result	size
derivativedivides	\(\frac {b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {2 a b}{\cos \left (d x +c \right )}+a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(83\)
default	\(\frac {b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {2 a b}{\cos \left (d x +c \right )}+a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(83\)
risch	\(\frac {b \,{\mathrm e}^{i \left (d x +c \right )} \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+4 a \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +4 a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{2 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{2 d}\)	\(151\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.48 \[ \int \sec (c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 8 \, a b \cos \left (d x + c\right ) + 2 \, b^{2} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

1/4*((2*a^2 - b^2)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (2*a^2 - b^2)*cos(d*x + c)^2*log(-sin(d*x + c) + 1)
+ 8*a*b*cos(d*x + c) + 2*b^2*sin(d*x + c))/(d*cos(d*x + c)^2)

Sympy [F]

\[ \int \sec (c+d x) (a+b \tan (c+d x))^2 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{2} \sec {\left (c + d x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.26 \[ \int \sec (c+d x) (a+b \tan (c+d x))^2 \, dx=-\frac {b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - \frac {8 \, a b}{\cos \left (d x + c\right )}}{4 \, d} \]

[In]

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

[Out]

-1/4*(b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 4*a^2*log(se
c(d*x + c) + tan(d*x + c)) - 8*a*b/cos(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 122 vs. \(2 (59) = 118\).

Time = 0.51 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.88 \[ \int \sec (c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {{\left (2 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, a b\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.65 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.63 \[ \int \sec (c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+4\,a\,b}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,a^2-b^2\right )}{d} \]

[In]

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

[Out]

(4*a*b + b^2*tan(c/2 + (d*x)/2)^3 + b^2*tan(c/2 + (d*x)/2) - 4*a*b*tan(c/2 + (d*x)/2)^2)/(d*(tan(c/2 + (d*x)/2
)^4 - 2*tan(c/2 + (d*x)/2)^2 + 1)) + (atanh(tan(c/2 + (d*x)/2))*(2*a^2 - b^2))/d