Integrand size = 19, antiderivative size = 57 \[ \int (a+a \sec (c+d x)) \tan ^3(c+d x) \, dx=\frac {a \log (\cos (c+d x))}{d}-\frac {a \sec (c+d x)}{d}+\frac {a \sec ^2(c+d x)}{2 d}+\frac {a \sec ^3(c+d x)}{3 d} \] Output:
a*ln(cos(d*x+c))/d-a*sec(d*x+c)/d+1/2*a*sec(d*x+c)^2/d+1/3*a*sec(d*x+c)^3/ d
Time = 0.15 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.96 \[ \int (a+a \sec (c+d x)) \tan ^3(c+d x) \, dx=-\frac {a \sec (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d}+\frac {a \left (2 \log (\cos (c+d x))+\sec ^2(c+d x)\right )}{2 d} \] Input:
Integrate[(a + a*Sec[c + d*x])*Tan[c + d*x]^3,x]
Output:
-((a*Sec[c + d*x])/d) + (a*Sec[c + d*x]^3)/(3*d) + (a*(2*Log[Cos[c + d*x]] + Sec[c + d*x]^2))/(2*d)
Time = 0.24 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.81, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3042, 25, 4367, 27, 84, 2009}
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.
\(\displaystyle \int \tan ^3(c+d x) (a \sec (c+d x)+a) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\cot \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )dx\) |
\(\Big \downarrow \) 4367 |
\(\displaystyle -\frac {\int a^3 (1-\cos (c+d x)) (\cos (c+d x)+1)^2 \sec ^4(c+d x)d\cos (c+d x)}{a^2 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a \int (1-\cos (c+d x)) (\cos (c+d x)+1)^2 \sec ^4(c+d x)d\cos (c+d x)}{d}\) |
\(\Big \downarrow \) 84 |
\(\displaystyle -\frac {a \int \left (\sec ^4(c+d x)+\sec ^3(c+d x)-\sec ^2(c+d x)-\sec (c+d x)\right )d\cos (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a \left (-\frac {1}{3} \sec ^3(c+d x)-\frac {1}{2} \sec ^2(c+d x)+\sec (c+d x)-\log (\cos (c+d x))\right )}{d}\) |
Input:
Int[(a + a*Sec[c + d*x])*Tan[c + d*x]^3,x]
Output:
-((a*(-Log[Cos[c + d*x]] + Sec[c + d*x] - Sec[c + d*x]^2/2 - Sec[c + d*x]^ 3/3))/d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] && !(ILtQ[n + p + 2, 0 ] && GtQ[n + 2*p, 0])
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _.), x_Symbol] :> Simp[1/(a^(m - n - 1)*b^n*d) Subst[Int[(a - b*x)^((m - 1)/2)*((a + b*x)^((m - 1)/2 + n)/x^(m + n)), x], x, Sin[c + d*x]], x] /; Fr eeQ[{a, b, c, d}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] && Integer Q[n]
Time = 0.14 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.77
method | result | size |
derivativedivides | \(\frac {a \left (\frac {\sec \left (d x +c \right )^{3}}{3}+\frac {\sec \left (d x +c \right )^{2}}{2}-\sec \left (d x +c \right )-\ln \left (\sec \left (d x +c \right )\right )\right )}{d}\) | \(44\) |
default | \(\frac {a \left (\frac {\sec \left (d x +c \right )^{3}}{3}+\frac {\sec \left (d x +c \right )^{2}}{2}-\sec \left (d x +c \right )-\ln \left (\sec \left (d x +c \right )\right )\right )}{d}\) | \(44\) |
parts | \(\frac {a \left (\frac {\tan \left (d x +c \right )^{2}}{2}-\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}+\frac {a \left (\frac {\sec \left (d x +c \right )^{3}}{3}-\sec \left (d x +c \right )\right )}{d}\) | \(55\) |
risch | \(-i a x -\frac {2 i a c}{d}-\frac {2 a \left (3 \,{\mathrm e}^{5 i \left (d x +c \right )}-3 \,{\mathrm e}^{4 i \left (d x +c \right )}+2 \,{\mathrm e}^{3 i \left (d x +c \right )}-3 \,{\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(107\) |
Input:
int((a+a*sec(d*x+c))*tan(d*x+c)^3,x,method=_RETURNVERBOSE)
Output:
a/d*(1/3*sec(d*x+c)^3+1/2*sec(d*x+c)^2-sec(d*x+c)-ln(sec(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00 \[ \int (a+a \sec (c+d x)) \tan ^3(c+d x) \, dx=\frac {6 \, a \cos \left (d x + c\right )^{3} \log \left (-\cos \left (d x + c\right )\right ) - 6 \, a \cos \left (d x + c\right )^{2} + 3 \, a \cos \left (d x + c\right ) + 2 \, a}{6 \, d \cos \left (d x + c\right )^{3}} \] Input:
integrate((a+a*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*a*cos(d*x + c)^2 + 3*a*cos( d*x + c) + 2*a)/(d*cos(d*x + c)^3)
Time = 0.21 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.33 \[ \int (a+a \sec (c+d x)) \tan ^3(c+d x) \, dx=\begin {cases} - \frac {a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a \tan ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{3 d} + \frac {a \tan ^{2}{\left (c + d x \right )}}{2 d} - \frac {2 a \sec {\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \sec {\left (c \right )} + a\right ) \tan ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate((a+a*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*sec(c + d *x)/(3*d) + a*tan(c + d*x)**2/(2*d) - 2*a*sec(c + d*x)/(3*d), Ne(d, 0)), ( x*(a*sec(c) + a)*tan(c)**3, True))
Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.88 \[ \int (a+a \sec (c+d x)) \tan ^3(c+d x) \, dx=\frac {6 \, a \log \left (\cos \left (d x + c\right )\right ) - \frac {6 \, a \cos \left (d x + c\right )^{2} - 3 \, a \cos \left (d x + c\right ) - 2 \, a}{\cos \left (d x + c\right )^{3}}}{6 \, d} \] Input:
integrate((a+a*sec(d*x+c))*tan(d*x+c)^3,x, algorithm="maxima")
Output:
1/6*(6*a*log(cos(d*x + c)) - (6*a*cos(d*x + c)^2 - 3*a*cos(d*x + c) - 2*a) /cos(d*x + c)^3)/d
Time = 0.21 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.89 \[ \int (a+a \sec (c+d x)) \tan ^3(c+d x) \, dx=\frac {6 \, a \log \left ({\left | \cos \left (d x + c\right ) \right |}\right ) - \frac {6 \, a \cos \left (d x + c\right )^{2} - 3 \, a \cos \left (d x + c\right ) - 2 \, a}{\cos \left (d x + c\right )^{3}}}{6 \, d} \] Input:
integrate((a+a*sec(d*x+c))*tan(d*x+c)^3,x, algorithm="giac")
Output:
1/6*(6*a*log(abs(cos(d*x + c))) - (6*a*cos(d*x + c)^2 - 3*a*cos(d*x + c) - 2*a)/cos(d*x + c)^3)/d
Time = 13.00 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.68 \[ \int (a+a \sec (c+d x)) \tan ^3(c+d x) \, dx=\frac {2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {4\,a}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {2\,a\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d} \] Input:
int(tan(c + d*x)^3*(a + a/cos(c + d*x)),x)
Output:
((4*a)/3 - 6*a*tan(c/2 + (d*x)/2)^2 + 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 = 0.15 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.95 \[ \int (a+a \sec (c+d x)) \tan ^3(c+d x) \, dx=\frac {a \left (-3 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right )+2 \sec \left (d x +c \right ) \tan \left (d x +c \right )^{2}-4 \sec \left (d x +c \right )+3 \tan \left (d x +c \right )^{2}\right )}{6 d} \] Input:
int((a+a*sec(d*x+c))*tan(d*x+c)^3,x)
Output:
(a*( - 3*log(tan(c + d*x)**2 + 1) + 2*sec(c + d*x)*tan(c + d*x)**2 - 4*sec (c + d*x) + 3*tan(c + d*x)**2))/(6*d)