Integrand size = 19, antiderivative size = 61 \[ \int \sec (c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {\left (2 a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {2 a b \sec (c+d x)}{d}+\frac {b^2 \sec (c+d x) \tan (c+d x)}{2 d} \] Output:
1/2*(2*a^2-b^2)*arctanh(sin(d*x+c))/d+2*a*b*sec(d*x+c)/d+1/2*b^2*sec(d*x+c )*tan(d*x+c)/d
Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.10 \[ \int \sec (c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {a^2 \coth ^{-1}(\sin (c+d x))}{d}-\frac {b^2 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {2 a b \sec (c+d x)}{d}+\frac {b^2 \sec (c+d x) \tan (c+d x)}{2 d} \] Input:
Integrate[Sec[c + d*x]*(a + b*Tan[c + d*x])^2,x]
Output:
(a^2*ArcCoth[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)
Time = 0.36 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {3042, 3991, 27, 3042, 3086, 24, 4159, 298, 219}
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.
\(\displaystyle \int \sec (c+d x) (a+b \tan (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (c+d x) (a+b \tan (c+d x))^2dx\) |
\(\Big \downarrow \) 3991 |
\(\displaystyle \int \sec (c+d x) \left (a^2+b^2 \tan ^2(c+d x)\right )dx+\int 2 a b \sec (c+d x) \tan (c+d x)dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \sec (c+d x) \left (a^2+b^2 \tan ^2(c+d x)\right )dx+2 a b \int \sec (c+d x) \tan (c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (c+d x) \left (a^2+b^2 \tan (c+d x)^2\right )dx+2 a b \int \sec (c+d x) \tan (c+d x)dx\) |
\(\Big \downarrow \) 3086 |
\(\displaystyle \int \sec (c+d x) \left (a^2+b^2 \tan (c+d x)^2\right )dx+\frac {2 a b \int 1d\sec (c+d x)}{d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \int \sec (c+d x) \left (a^2+b^2 \tan (c+d x)^2\right )dx+\frac {2 a b \sec (c+d x)}{d}\) |
\(\Big \downarrow \) 4159 |
\(\displaystyle \frac {\int \frac {a^2-\left (a^2-b^2\right ) \sin ^2(c+d x)}{\left (1-\sin ^2(c+d x)\right )^2}d\sin (c+d x)}{d}+\frac {2 a b \sec (c+d x)}{d}\) |
\(\Big \downarrow \) 298 |
\(\displaystyle \frac {\frac {1}{2} \left (2 a^2-b^2\right ) \int \frac {1}{1-\sin ^2(c+d x)}d\sin (c+d x)+\frac {b^2 \sin (c+d x)}{2 \left (1-\sin ^2(c+d x)\right )}}{d}+\frac {2 a b \sec (c+d x)}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {1}{2} \left (2 a^2-b^2\right ) \text {arctanh}(\sin (c+d x))+\frac {b^2 \sin (c+d x)}{2 \left (1-\sin ^2(c+d x)\right )}}{d}+\frac {2 a b \sec (c+d x)}{d}\) |
Input:
Int[Sec[c + d*x]*(a + b*Tan[c + d*x])^2,x]
Output:
(2*a*b*Sec[c + d*x])/d + (((2*a^2 - b^2)*ArcTanh[Sin[c + d*x]])/2 + (b^2*S in[c + d*x])/(2*(1 - Sin[c + d*x]^2)))/d
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_.), x_Symbol] :> Simp[a/f Subst[Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2 ), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2 ] && !(IntegerQ[m/2] && LtQ[0, m, n + 1])
Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Module[{k}, Int[Sec[e + f*x]^m*Sum[Binomial[n, 2*k]*a^(n - 2*k)*b^(2*k)*Tan[e + f*x]^(2*k), {k, 0, n/2}], x] + Int[Sec[e + f*x]^m*Tan [e + f*x]*Sum[Binomial[n, 2*k + 1]*a^(n - 2*k - 1)*b^(2*k + 1)*Tan[e + f*x] ^(2*k), {k, 0, (n - 1)/2}], x]] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 + b^2, 0] && IntegerQ[(m - 1)/2] && IGtQ[n, 0]
Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_ ))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2 *x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f} , x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2] && IntegerQ[p]
Time = 1.11 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.36
method | result | size |
derivativedivides | \(\frac {b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {2 a b}{\cos \left (d x +c \right )}+a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(83\) |
default | \(\frac {b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {2 a b}{\cos \left (d x +c \right )}+a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(83\) |
risch | \(\frac {b \,{\mathrm e}^{i \left (d x +c \right )} \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+4 a \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +4 a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{2 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{2 d}\) | \(151\) |
Input:
int(sec(d*x+c)*(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
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)+t an(d*x+c)))+2*a*b/cos(d*x+c)+a^2*ln(sec(d*x+c)+tan(d*x+c)))
Time = 0.10 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.57 \[ \int \sec (c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 8 \, a b \cos \left (d x + c\right ) + 2 \, b^{2} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate(sec(d*x+c)*(a+b*tan(d*x+c))^2,x, algorithm="fricas")
Output:
1/4*((2*a^2 - b^2)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (2*a^2 - b^2)*co s(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)
\[ \int \sec (c+d x) (a+b \tan (c+d x))^2 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{2} \sec {\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)*(a+b*tan(d*x+c))**2,x)
Output:
Integral((a + b*tan(c + d*x))**2*sec(c + d*x), x)
Time = 0.06 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.34 \[ \int \sec (c+d x) (a+b \tan (c+d x))^2 \, dx=-\frac {b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - \frac {8 \, a b}{\cos \left (d x + c\right )}}{4 \, d} \] Input:
integrate(sec(d*x+c)*(a+b*tan(d*x+c))^2,x, algorithm="maxima")
Output:
-1/4*(b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + log(sin(d*x + c) + 1) - l og(sin(d*x + c) - 1)) - 4*a^2*log(sec(d*x + c) + tan(d*x + c)) - 8*a*b/cos (d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 122 vs. \(2 (57) = 114\).
Time = 0.24 (sec) , antiderivative size = 122, normalized size of antiderivative = 2.00 \[ \int \sec (c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {{\left (2 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, a b\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \] Input:
integrate(sec(d*x+c)*(a+b*tan(d*x+c))^2,x, algorithm="giac")
Output:
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
Time = 1.18 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.74 \[ \int \sec (c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+4\,a\,b}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,a^2-b^2\right )}{d} \] Input:
int((a + b*tan(c + d*x))^2/cos(c + d*x),x)
Output:
(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
Time = 0.20 (sec) , antiderivative size = 221, normalized size of antiderivative = 3.62 \[ \int \sec (c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {-4 \cos \left (d x +c \right ) a b -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{2}-4 \sin \left (d x +c \right )^{2} a b -\sin \left (d x +c \right ) b^{2}+4 a b}{2 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)*(a+b*tan(d*x+c))^2,x)
Output:
( - 4*cos(c + d*x)*a*b - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2 + log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**2 + 2*log(tan((c + d*x)/2) - 1)*a**2 - log(tan((c + d*x)/2) - 1)*b**2 + 2*log(tan((c + d*x)/2) + 1)*s in(c + d*x)**2*a**2 - log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b**2 - 2*l og(tan((c + d*x)/2) + 1)*a**2 + log(tan((c + d*x)/2) + 1)*b**2 - 4*sin(c + d*x)**2*a*b - sin(c + d*x)*b**2 + 4*a*b)/(2*d*(sin(c + d*x)**2 - 1))