Integrand size = 19, antiderivative size = 55 \[ \int (a+b \sec (c+d x)) \tan ^3(c+d x) \, dx=\frac {a \log (\cos (c+d x))}{d}-\frac {2 b \sec (c+d x)}{3 d}+\frac {(3 a+2 b \sec (c+d x)) \tan ^2(c+d x)}{6 d} \] Output:
a*ln(cos(d*x+c))/d-2/3*b*sec(d*x+c)/d+1/6*(3*a+2*b*sec(d*x+c))*tan(d*x+c)^ 2/d
Time = 0.11 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00 \[ \int (a+b \sec (c+d x)) \tan ^3(c+d x) \, dx=-\frac {b \sec (c+d x)}{d}+\frac {b \sec ^3(c+d x)}{3 d}+\frac {a \left (2 \log (\cos (c+d x))+\sec ^2(c+d x)\right )}{2 d} \] Input:
Integrate[(a + b*Sec[c + d*x])*Tan[c + d*x]^3,x]
Output:
-((b*Sec[c + d*x])/d) + (b*Sec[c + d*x]^3)/(3*d) + (a*(2*Log[Cos[c + d*x]] + Sec[c + d*x]^2))/(2*d)
Time = 0.42 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.07, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.632, Rules used = {3042, 25, 4369, 25, 3042, 25, 4372, 25, 3042, 3086, 24, 3956}
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.
\(\displaystyle \int \tan ^3(c+d x) (a+b \sec (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\cot \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \left (a+b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )dx\) |
\(\Big \downarrow \) 4369 |
\(\displaystyle \frac {1}{3} \int -((3 a+2 b \sec (c+d x)) \tan (c+d x))dx+\frac {\tan ^2(c+d x) (3 a+2 b \sec (c+d x))}{6 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\tan ^2(c+d x) (3 a+2 b \sec (c+d x))}{6 d}-\frac {1}{3} \int (3 a+2 b \sec (c+d x)) \tan (c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan ^2(c+d x) (3 a+2 b \sec (c+d x))}{6 d}-\frac {1}{3} \int -\cot \left (c+d x+\frac {\pi }{2}\right ) \left (3 a+2 b \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{3} \int \cot \left (\frac {1}{2} (2 c+\pi )+d x\right ) \left (3 a+2 b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )dx+\frac {\tan ^2(c+d x) (3 a+2 b \sec (c+d x))}{6 d}\) |
\(\Big \downarrow \) 4372 |
\(\displaystyle \frac {1}{3} (3 a \int -\tan (c+d x)dx+2 b \int -\sec (c+d x) \tan (c+d x)dx)+\frac {\tan ^2(c+d x) (3 a+2 b \sec (c+d x))}{6 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{3} (-3 a \int \tan (c+d x)dx-2 b \int \sec (c+d x) \tan (c+d x)dx)+\frac {\tan ^2(c+d x) (3 a+2 b \sec (c+d x))}{6 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} (-3 a \int \tan (c+d x)dx-2 b \int \sec (c+d x) \tan (c+d x)dx)+\frac {\tan ^2(c+d x) (3 a+2 b \sec (c+d x))}{6 d}\) |
\(\Big \downarrow \) 3086 |
\(\displaystyle \frac {1}{3} \left (-3 a \int \tan (c+d x)dx-\frac {2 b \int 1d\sec (c+d x)}{d}\right )+\frac {\tan ^2(c+d x) (3 a+2 b \sec (c+d x))}{6 d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {1}{3} \left (-3 a \int \tan (c+d x)dx-\frac {2 b \sec (c+d x)}{d}\right )+\frac {\tan ^2(c+d x) (3 a+2 b \sec (c+d x))}{6 d}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {\tan ^2(c+d x) (3 a+2 b \sec (c+d x))}{6 d}+\frac {1}{3} \left (\frac {3 a \log (\cos (c+d x))}{d}-\frac {2 b \sec (c+d x)}{d}\right )\) |
Input:
Int[(a + b*Sec[c + d*x])*Tan[c + d*x]^3,x]
Output:
((3*a*Log[Cos[c + d*x]])/d - (2*b*Sec[c + d*x])/d)/3 + ((3*a + 2*b*Sec[c + d*x])*Tan[c + d*x]^2)/(6*d)
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[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( a_)), x_Symbol] :> Simp[(-e)*(e*Cot[c + d*x])^(m - 1)*((a*m + b*(m - 1)*Csc [c + d*x])/(d*m*(m - 1))), x] - Simp[e^2/m Int[(e*Cot[c + d*x])^(m - 2)*( a*m + b*(m - 1)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[m , 1]
Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(e*Cot[c + d*x])^m, x], x] + Simp[b Int[ (e*Cot[c + d*x])^m*Csc[c + d*x], x], x] /; FreeQ[{a, b, c, d, e, m}, x]
Time = 0.24 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.85
method | result | size |
derivativedivides | \(\frac {\frac {b \sec \left (d x +c \right )^{3}}{3}+\frac {a \sec \left (d x +c \right )^{2}}{2}-b \sec \left (d x +c \right )-a \ln \left (\sec \left (d x +c \right )\right )}{d}\) | \(47\) |
default | \(\frac {\frac {b \sec \left (d x +c \right )^{3}}{3}+\frac {a \sec \left (d x +c \right )^{2}}{2}-b \sec \left (d x +c \right )-a \ln \left (\sec \left (d x +c \right )\right )}{d}\) | \(47\) |
parts | \(\frac {a \left (\frac {\tan \left (d x +c \right )^{2}}{2}-\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}+\frac {b \left (\frac {\sec \left (d x +c \right )^{3}}{3}-\sec \left (d x +c \right )\right )}{d}\) | \(55\) |
risch | \(-i a x -\frac {2 i a c}{d}-\frac {2 \left (3 b \,{\mathrm e}^{5 i \left (d x +c \right )}-3 a \,{\mathrm e}^{4 i \left (d x +c \right )}+2 b \,{\mathrm e}^{3 i \left (d x +c \right )}-3 a \,{\mathrm e}^{2 i \left (d x +c \right )}+3 b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(111\) |
Input:
int((a+b*sec(d*x+c))*tan(d*x+c)^3,x,method=_RETURNVERBOSE)
Output:
1/d*(1/3*b*sec(d*x+c)^3+1/2*a*sec(d*x+c)^2-b*sec(d*x+c)-a*ln(sec(d*x+c)))
Time = 0.13 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.04 \[ \int (a+b \sec (c+d x)) \tan ^3(c+d x) \, dx=\frac {6 \, a \cos \left (d x + c\right )^{3} \log \left (-\cos \left (d x + c\right )\right ) - 6 \, b \cos \left (d x + c\right )^{2} + 3 \, a \cos \left (d x + c\right ) + 2 \, b}{6 \, d \cos \left (d x + c\right )^{3}} \] Input:
integrate((a+b*sec(d*x+c))*tan(d*x+c)^3,x, algorithm="fricas")
Output:
1/6*(6*a*cos(d*x + c)^3*log(-cos(d*x + c)) - 6*b*cos(d*x + c)^2 + 3*a*cos( d*x + c) + 2*b)/(d*cos(d*x + c)^3)
Time = 0.25 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.38 \[ \int (a+b \sec (c+d x)) \tan ^3(c+d x) \, dx=\begin {cases} - \frac {a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a \tan ^{2}{\left (c + d x \right )}}{2 d} + \frac {b \tan ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{3 d} - \frac {2 b \sec {\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a + b \sec {\left (c \right )}\right ) \tan ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate((a+b*sec(d*x+c))*tan(d*x+c)**3,x)
Output:
Piecewise((-a*log(tan(c + d*x)**2 + 1)/(2*d) + a*tan(c + d*x)**2/(2*d) + b *tan(c + d*x)**2*sec(c + d*x)/(3*d) - 2*b*sec(c + d*x)/(3*d), Ne(d, 0)), ( x*(a + b*sec(c))*tan(c)**3, True))
Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.91 \[ \int (a+b \sec (c+d x)) \tan ^3(c+d x) \, dx=\frac {6 \, a \log \left (\cos \left (d x + c\right )\right ) - \frac {6 \, b \cos \left (d x + c\right )^{2} - 3 \, a \cos \left (d x + c\right ) - 2 \, b}{\cos \left (d x + c\right )^{3}}}{6 \, d} \] Input:
integrate((a+b*sec(d*x+c))*tan(d*x+c)^3,x, algorithm="maxima")
Output:
1/6*(6*a*log(cos(d*x + c)) - (6*b*cos(d*x + c)^2 - 3*a*cos(d*x + c) - 2*b) /cos(d*x + c)^3)/d
Time = 0.17 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int (a+b \sec (c+d x)) \tan ^3(c+d x) \, dx=\frac {6 \, a \log \left ({\left | \cos \left (d x + c\right ) \right |}\right ) - \frac {6 \, b \cos \left (d x + c\right )^{2} - 3 \, a \cos \left (d x + c\right ) - 2 \, b}{\cos \left (d x + c\right )^{3}}}{6 \, d} \] Input:
integrate((a+b*sec(d*x+c))*tan(d*x+c)^3,x, algorithm="giac")
Output:
1/6*(6*a*log(abs(cos(d*x + c))) - (6*b*cos(d*x + c)^2 - 3*a*cos(d*x + c) - 2*b)/cos(d*x + c)^3)/d
Time = 11.77 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.85 \[ \int (a+b \sec (c+d x)) \tan ^3(c+d x) \, dx=\frac {2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (-2\,a-4\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {4\,b}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {2\,a\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d} \] Input:
int(tan(c + d*x)^3*(a + b/cos(c + d*x)),x)
Output:
((4*b)/3 - tan(c/2 + (d*x)/2)^2*(2*a + 4*b) + 2*a*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)) - (2*a*atanh(tan(c/2 + (d*x)/2)^2))/d
Time = 0.16 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.04 \[ \int (a+b \sec (c+d x)) \tan ^3(c+d x) \, dx=\frac {-3 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) a +2 \sec \left (d x +c \right ) \tan \left (d x +c \right )^{2} b -4 \sec \left (d x +c \right ) b +3 \tan \left (d x +c \right )^{2} a}{6 d} \] Input:
int((a+b*sec(d*x+c))*tan(d*x+c)^3,x)
Output:
( - 3*log(tan(c + d*x)**2 + 1)*a + 2*sec(c + d*x)*tan(c + d*x)**2*b - 4*se c(c + d*x)*b + 3*tan(c + d*x)**2*a)/(6*d)