Integrand size = 19, antiderivative size = 52 \[ \int \sec ^3(c+d x) (a+b \tan (c+d x)) \, dx=\frac {a \text {arctanh}(\sin (c+d x))}{2 d}+\frac {b \sec ^3(c+d x)}{3 d}+\frac {a \sec (c+d x) \tan (c+d x)}{2 d} \] Output:
1/2*a*arctanh(sin(d*x+c))/d+1/3*b*sec(d*x+c)^3/d+1/2*a*sec(d*x+c)*tan(d*x+ c)/d
Time = 0.01 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00 \[ \int \sec ^3(c+d x) (a+b \tan (c+d x)) \, dx=\frac {a \text {arctanh}(\sin (c+d x))}{2 d}+\frac {b \sec ^3(c+d x)}{3 d}+\frac {a \sec (c+d x) \tan (c+d x)}{2 d} \] Input:
Integrate[Sec[c + d*x]^3*(a + b*Tan[c + d*x]),x]
Output:
(a*ArcTanh[Sin[c + d*x]])/(2*d) + (b*Sec[c + d*x]^3)/(3*d) + (a*Sec[c + d* x]*Tan[c + d*x])/(2*d)
Time = 0.33 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3042, 3967, 3042, 4255, 3042, 4257}
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.
\(\displaystyle \int \sec ^3(c+d x) (a+b \tan (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (c+d x)^3 (a+b \tan (c+d x))dx\) |
\(\Big \downarrow \) 3967 |
\(\displaystyle a \int \sec ^3(c+d x)dx+\frac {b \sec ^3(c+d x)}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {b \sec ^3(c+d x)}{3 d}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle a \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {b \sec ^3(c+d x)}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {b \sec ^3(c+d x)}{3 d}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle a \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {b \sec ^3(c+d x)}{3 d}\) |
Input:
Int[Sec[c + d*x]^3*(a + b*Tan[c + d*x]),x]
Output:
(b*Sec[c + d*x]^3)/(3*d) + a*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]* Tan[c + d*x])/(2*d))
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] + Simp[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])
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 2.39 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.96
method | result | size |
derivativedivides | \(\frac {\frac {b}{3 \cos \left (d x +c \right )^{3}}+a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(50\) |
default | \(\frac {\frac {b}{3 \cos \left (d x +c \right )^{3}}+a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(50\) |
risch | \(\frac {-3 i a \,{\mathrm e}^{5 i \left (d x +c \right )}+8 b \,{\mathrm e}^{3 i \left (d x +c \right )}+3 i a \,{\mathrm e}^{i \left (d x +c \right )}}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}\) | \(97\) |
Input:
int(sec(d*x+c)^3*(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(1/3*b/cos(d*x+c)^3+a*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan (d*x+c))))
Time = 0.09 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.42 \[ \int \sec ^3(c+d x) (a+b \tan (c+d x)) \, dx=\frac {3 \, a \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 6 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 4 \, b}{12 \, d \cos \left (d x + c\right )^{3}} \] Input:
integrate(sec(d*x+c)^3*(a+b*tan(d*x+c)),x, algorithm="fricas")
Output:
1/12*(3*a*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*a*cos(d*x + c)^3*log(-s in(d*x + c) + 1) + 6*a*cos(d*x + c)*sin(d*x + c) + 4*b)/(d*cos(d*x + c)^3)
\[ \int \sec ^3(c+d x) (a+b \tan (c+d x)) \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)**3*(a+b*tan(d*x+c)),x)
Output:
Integral((a + b*tan(c + d*x))*sec(c + d*x)**3, x)
Time = 0.05 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.17 \[ \int \sec ^3(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {3 \, a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - \frac {4 \, b}{\cos \left (d x + c\right )^{3}}}{12 \, d} \] Input:
integrate(sec(d*x+c)^3*(a+b*tan(d*x+c)),x, algorithm="maxima")
Output:
-1/12*(3*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 4*b/cos(d*x + c)^3)/d
Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (46) = 92\).
Time = 0.21 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.90 \[ \int \sec ^3(c+d x) (a+b \tan (c+d x)) \, dx=\frac {3 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \] Input:
integrate(sec(d*x+c)^3*(a+b*tan(d*x+c)),x, algorithm="giac")
Output:
1/6*(3*a*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*a*log(abs(tan(1/2*d*x + 1/ 2*c) - 1)) + 2*(3*a*tan(1/2*d*x + 1/2*c)^5 - 6*b*tan(1/2*d*x + 1/2*c)^4 - 3*a*tan(1/2*d*x + 1/2*c) - 2*b)/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d
Time = 2.61 (sec) , antiderivative size = 105, normalized size of antiderivative = 2.02 \[ \int \sec ^3(c+d x) (a+b \tan (c+d x)) \, dx=\frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {2\,b}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int((a + b*tan(c + d*x))/cos(c + d*x)^3,x)
Output:
(a*atanh(tan(c/2 + (d*x)/2)))/d - ((2*b)/3 + a*tan(c/2 + (d*x)/2) - a*tan( c/2 + (d*x)/2)^5 + 2*b*tan(c/2 + (d*x)/2)^4)/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 1))
Time = 0.15 (sec) , antiderivative size = 170, normalized size of antiderivative = 3.27 \[ \int \sec ^3(c+d x) (a+b \tan (c+d x)) \, dx=\frac {-3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a -3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b -3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a +2 \cos \left (d x +c \right ) b -2 b}{6 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)^3*(a+b*tan(d*x+c)),x)
Output:
( - 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a + 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a + 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a - 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a - 2*cos( c + d*x)*sin(c + d*x)**2*b - 3*cos(c + d*x)*sin(c + d*x)*a + 2*cos(c + d*x )*b - 2*b)/(6*cos(c + d*x)*d*(sin(c + d*x)**2 - 1))