Integrand size = 21, antiderivative size = 92 \[ \int \left (a+b \sec ^3(e+f x)\right ) \tan ^5(e+f x) \, dx=-\frac {a \log (\cos (e+f x))}{f}-\frac {a \sec ^2(e+f x)}{f}+\frac {b \sec ^3(e+f x)}{3 f}+\frac {a \sec ^4(e+f x)}{4 f}-\frac {2 b \sec ^5(e+f x)}{5 f}+\frac {b \sec ^7(e+f x)}{7 f} \] Output:
-a*ln(cos(f*x+e))/f-a*sec(f*x+e)^2/f+1/3*b*sec(f*x+e)^3/f+1/4*a*sec(f*x+e) ^4/f-2/5*b*sec(f*x+e)^5/f+1/7*b*sec(f*x+e)^7/f
Time = 0.20 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00 \[ \int \left (a+b \sec ^3(e+f x)\right ) \tan ^5(e+f x) \, dx=-\frac {a \log (\cos (e+f x))}{f}-\frac {a \sec ^2(e+f x)}{f}+\frac {b \sec ^3(e+f x)}{3 f}+\frac {a \sec ^4(e+f x)}{4 f}-\frac {2 b \sec ^5(e+f x)}{5 f}+\frac {b \sec ^7(e+f x)}{7 f} \] Input:
Integrate[(a + b*Sec[e + f*x]^3)*Tan[e + f*x]^5,x]
Output:
-((a*Log[Cos[e + f*x]])/f) - (a*Sec[e + f*x]^2)/f + (b*Sec[e + f*x]^3)/(3* f) + (a*Sec[e + f*x]^4)/(4*f) - (2*b*Sec[e + f*x]^5)/(5*f) + (b*Sec[e + f* x]^7)/(7*f)
Time = 0.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.84, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 4626, 2333, 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.
\(\displaystyle \int \tan ^5(e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (e+f x)^5 \left (a+b \sec (e+f x)^3\right )dx\) |
\(\Big \downarrow \) 4626 |
\(\displaystyle -\frac {\int \left (1-\cos ^2(e+f x)\right )^2 \left (a \cos ^3(e+f x)+b\right ) \sec ^8(e+f x)d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 2333 |
\(\displaystyle -\frac {\int \left (b \sec ^8(e+f x)-2 b \sec ^6(e+f x)+a \sec ^5(e+f x)+b \sec ^4(e+f x)-2 a \sec ^3(e+f x)+a \sec (e+f x)\right )d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {1}{4} a \sec ^4(e+f x)+a \sec ^2(e+f x)+a \log (\cos (e+f x))-\frac {1}{7} b \sec ^7(e+f x)+\frac {2}{5} b \sec ^5(e+f x)-\frac {1}{3} b \sec ^3(e+f x)}{f}\) |
Input:
Int[(a + b*Sec[e + f*x]^3)*Tan[e + f*x]^5,x]
Output:
-((a*Log[Cos[e + f*x]] + a*Sec[e + f*x]^2 - (b*Sec[e + f*x]^3)/3 - (a*Sec[ e + f*x]^4)/4 + (2*b*Sec[e + f*x]^5)/5 - (b*Sec[e + f*x]^7)/7)/f)
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_ )]^(m_.), x_Symbol] :> Module[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-(f *ff^(m + n*p - 1))^(-1) Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*((b + a*(ff* x)^n)^p/x^(m + n*p)), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n} , x] && IntegerQ[(m - 1)/2] && IntegerQ[n] && IntegerQ[p]
Time = 1.91 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.76
method | result | size |
derivativedivides | \(\frac {\frac {b \sec \left (f x +e \right )^{7}}{7}-\frac {2 b \sec \left (f x +e \right )^{5}}{5}+\frac {a \sec \left (f x +e \right )^{4}}{4}+\frac {b \sec \left (f x +e \right )^{3}}{3}-a \sec \left (f x +e \right )^{2}+a \ln \left (\sec \left (f x +e \right )\right )}{f}\) | \(70\) |
default | \(\frac {\frac {b \sec \left (f x +e \right )^{7}}{7}-\frac {2 b \sec \left (f x +e \right )^{5}}{5}+\frac {a \sec \left (f x +e \right )^{4}}{4}+\frac {b \sec \left (f x +e \right )^{3}}{3}-a \sec \left (f x +e \right )^{2}+a \ln \left (\sec \left (f x +e \right )\right )}{f}\) | \(70\) |
parts | \(\frac {a \left (\frac {\tan \left (f x +e \right )^{4}}{4}-\frac {\tan \left (f x +e \right )^{2}}{2}+\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2}\right )}{f}+\frac {b \left (\frac {\sec \left (f x +e \right )^{7}}{7}-\frac {2 \sec \left (f x +e \right )^{5}}{5}+\frac {\sec \left (f x +e \right )^{3}}{3}\right )}{f}\) | \(77\) |
risch | \(i a x +\frac {2 i a e}{f}-\frac {4 \left (105 a \,{\mathrm e}^{12 i \left (f x +e \right )}-70 b \,{\mathrm e}^{11 i \left (f x +e \right )}+420 a \,{\mathrm e}^{10 i \left (f x +e \right )}+56 b \,{\mathrm e}^{9 i \left (f x +e \right )}+735 a \,{\mathrm e}^{8 i \left (f x +e \right )}-228 b \,{\mathrm e}^{7 i \left (f x +e \right )}+735 a \,{\mathrm e}^{6 i \left (f x +e \right )}+56 b \,{\mathrm e}^{5 i \left (f x +e \right )}+420 \,{\mathrm e}^{4 i \left (f x +e \right )} a -70 b \,{\mathrm e}^{3 i \left (f x +e \right )}+105 \,{\mathrm e}^{2 i \left (f x +e \right )} a \right )}{105 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{7}}-\frac {a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{f}\) | \(184\) |
Input:
int((a+b*sec(f*x+e)^3)*tan(f*x+e)^5,x,method=_RETURNVERBOSE)
Output:
1/f*(1/7*b*sec(f*x+e)^7-2/5*b*sec(f*x+e)^5+1/4*a*sec(f*x+e)^4+1/3*b*sec(f* x+e)^3-a*sec(f*x+e)^2+a*ln(sec(f*x+e)))
Time = 0.09 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.88 \[ \int \left (a+b \sec ^3(e+f x)\right ) \tan ^5(e+f x) \, dx=-\frac {420 \, a \cos \left (f x + e\right )^{7} \log \left (-\cos \left (f x + e\right )\right ) + 420 \, a \cos \left (f x + e\right )^{5} - 140 \, b \cos \left (f x + e\right )^{4} - 105 \, a \cos \left (f x + e\right )^{3} + 168 \, b \cos \left (f x + e\right )^{2} - 60 \, b}{420 \, f \cos \left (f x + e\right )^{7}} \] Input:
integrate((a+b*sec(f*x+e)^3)*tan(f*x+e)^5,x, algorithm="fricas")
Output:
-1/420*(420*a*cos(f*x + e)^7*log(-cos(f*x + e)) + 420*a*cos(f*x + e)^5 - 1 40*b*cos(f*x + e)^4 - 105*a*cos(f*x + e)^3 + 168*b*cos(f*x + e)^2 - 60*b)/ (f*cos(f*x + e)^7)
Time = 0.79 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.29 \[ \int \left (a+b \sec ^3(e+f x)\right ) \tan ^5(e+f x) \, dx=\begin {cases} \frac {a \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {a \tan ^{4}{\left (e + f x \right )}}{4 f} - \frac {a \tan ^{2}{\left (e + f x \right )}}{2 f} + \frac {b \tan ^{4}{\left (e + f x \right )} \sec ^{3}{\left (e + f x \right )}}{7 f} - \frac {4 b \tan ^{2}{\left (e + f x \right )} \sec ^{3}{\left (e + f x \right )}}{35 f} + \frac {8 b \sec ^{3}{\left (e + f x \right )}}{105 f} & \text {for}\: f \neq 0 \\x \left (a + b \sec ^{3}{\left (e \right )}\right ) \tan ^{5}{\left (e \right )} & \text {otherwise} \end {cases} \] Input:
integrate((a+b*sec(f*x+e)**3)*tan(f*x+e)**5,x)
Output:
Piecewise((a*log(tan(e + f*x)**2 + 1)/(2*f) + a*tan(e + f*x)**4/(4*f) - a* tan(e + f*x)**2/(2*f) + b*tan(e + f*x)**4*sec(e + f*x)**3/(7*f) - 4*b*tan( e + f*x)**2*sec(e + f*x)**3/(35*f) + 8*b*sec(e + f*x)**3/(105*f), Ne(f, 0) ), (x*(a + b*sec(e)**3)*tan(e)**5, True))
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.79 \[ \int \left (a+b \sec ^3(e+f x)\right ) \tan ^5(e+f x) \, dx=-\frac {420 \, a \log \left (\cos \left (f x + e\right )\right ) + \frac {420 \, a \cos \left (f x + e\right )^{5} - 140 \, b \cos \left (f x + e\right )^{4} - 105 \, a \cos \left (f x + e\right )^{3} + 168 \, b \cos \left (f x + e\right )^{2} - 60 \, b}{\cos \left (f x + e\right )^{7}}}{420 \, f} \] Input:
integrate((a+b*sec(f*x+e)^3)*tan(f*x+e)^5,x, algorithm="maxima")
Output:
-1/420*(420*a*log(cos(f*x + e)) + (420*a*cos(f*x + e)^5 - 140*b*cos(f*x + e)^4 - 105*a*cos(f*x + e)^3 + 168*b*cos(f*x + e)^2 - 60*b)/cos(f*x + e)^7) /f
Time = 0.29 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.80 \[ \int \left (a+b \sec ^3(e+f x)\right ) \tan ^5(e+f x) \, dx=-\frac {420 \, a \log \left ({\left | \cos \left (f x + e\right ) \right |}\right ) + \frac {420 \, a \cos \left (f x + e\right )^{5} - 140 \, b \cos \left (f x + e\right )^{4} - 105 \, a \cos \left (f x + e\right )^{3} + 168 \, b \cos \left (f x + e\right )^{2} - 60 \, b}{\cos \left (f x + e\right )^{7}}}{420 \, f} \] Input:
integrate((a+b*sec(f*x+e)^3)*tan(f*x+e)^5,x, algorithm="giac")
Output:
-1/420*(420*a*log(abs(cos(f*x + e))) + (420*a*cos(f*x + e)^5 - 140*b*cos(f *x + e)^4 - 105*a*cos(f*x + e)^3 + 168*b*cos(f*x + e)^2 - 60*b)/cos(f*x + e)^7)/f
Time = 18.51 (sec) , antiderivative size = 227, normalized size of antiderivative = 2.47 \[ \int \left (a+b \sec ^3(e+f x)\right ) \tan ^5(e+f x) \, dx=\frac {2\,a\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\right )}{f}-\frac {2\,a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{12}-14\,a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}+\left (32\,a+\frac {32\,b}{3}\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+\left (\frac {16\,b}{3}-32\,a\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+\left (14\,a+\frac {16\,b}{5}\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+\left (-2\,a-\frac {16\,b}{15}\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+\frac {16\,b}{105}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{14}-7\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{12}+21\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}-35\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+35\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6-21\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+7\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )} \] Input:
int(tan(e + f*x)^5*(a + b/cos(e + f*x)^3),x)
Output:
(2*a*atanh(tan(e/2 + (f*x)/2)^2))/f - ((16*b)/105 - tan(e/2 + (f*x)/2)^2*( 2*a + (16*b)/15) + tan(e/2 + (f*x)/2)^4*(14*a + (16*b)/5) - tan(e/2 + (f*x )/2)^6*(32*a - (16*b)/3) + tan(e/2 + (f*x)/2)^8*(32*a + (32*b)/3) - 14*a*t an(e/2 + (f*x)/2)^10 + 2*a*tan(e/2 + (f*x)/2)^12)/(f*(7*tan(e/2 + (f*x)/2) ^2 - 21*tan(e/2 + (f*x)/2)^4 + 35*tan(e/2 + (f*x)/2)^6 - 35*tan(e/2 + (f*x )/2)^8 + 21*tan(e/2 + (f*x)/2)^10 - 7*tan(e/2 + (f*x)/2)^12 + tan(e/2 + (f *x)/2)^14 - 1))
Time = 0.18 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.99 \[ \int \left (a+b \sec ^3(e+f x)\right ) \tan ^5(e+f x) \, dx=\frac {210 \,\mathrm {log}\left (\tan \left (f x +e \right )^{2}+1\right ) a +60 \sec \left (f x +e \right )^{3} \tan \left (f x +e \right )^{4} b -48 \sec \left (f x +e \right )^{3} \tan \left (f x +e \right )^{2} b +32 \sec \left (f x +e \right )^{3} b +105 \tan \left (f x +e \right )^{4} a -210 \tan \left (f x +e \right )^{2} a}{420 f} \] Input:
int((a+b*sec(f*x+e)^3)*tan(f*x+e)^5,x)
Output:
(210*log(tan(e + f*x)**2 + 1)*a + 60*sec(e + f*x)**3*tan(e + f*x)**4*b - 4 8*sec(e + f*x)**3*tan(e + f*x)**2*b + 32*sec(e + f*x)**3*b + 105*tan(e + f *x)**4*a - 210*tan(e + f*x)**2*a)/(420*f)