Integrand size = 23, antiderivative size = 69 \[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {(a+2 b) \log (\cos (e+f x))}{b^2 f}-\frac {(a+b)^2 \log \left (b+a \cos ^2(e+f x)\right )}{2 a b^2 f}+\frac {\sec ^2(e+f x)}{2 b f} \] Output:
(a+2*b)*ln(cos(f*x+e))/b^2/f-1/2*(a+b)^2*ln(b+a*cos(f*x+e)^2)/a/b^2/f+1/2* sec(f*x+e)^2/b/f
Time = 0.20 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.43 \[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^2(e+f x) \left (2 a (a+2 b) \log (\cos (e+f x))-(a+b)^2 \log \left (a+b-a \sin ^2(e+f x)\right )+a b \sec ^2(e+f x)\right )}{4 a b^2 f \left (a+b \sec ^2(e+f x)\right )} \] Input:
Integrate[Tan[e + f*x]^5/(a + b*Sec[e + f*x]^2),x]
Output:
((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^2*(2*a*(a + 2*b)*Log[Cos[e + f*x]] - (a + b)^2*Log[a + b - a*Sin[e + f*x]^2] + a*b*Sec[e + f*x]^2))/(4* a*b^2*f*(a + b*Sec[e + f*x]^2))
Time = 0.30 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4626, 354, 99, 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.
\(\displaystyle \int \frac {\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (e+f x)^5}{a+b \sec (e+f x)^2}dx\) |
\(\Big \downarrow \) 4626 |
\(\displaystyle -\frac {\int \frac {\left (1-\cos ^2(e+f x)\right )^2 \sec ^3(e+f x)}{a \cos ^2(e+f x)+b}d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle -\frac {\int \frac {\left (1-\cos ^2(e+f x)\right )^2 \sec ^2(e+f x)}{a \cos ^2(e+f x)+b}d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {\int \left (\frac {(a+b)^2}{b^2 \left (a \cos ^2(e+f x)+b\right )}+\frac {\sec ^2(e+f x)}{b}+\frac {(-a-2 b) \sec (e+f x)}{b^2}\right )d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {(a+b)^2 \log \left (a \cos ^2(e+f x)+b\right )}{a b^2}-\frac {(a+2 b) \log \left (\cos ^2(e+f x)\right )}{b^2}-\frac {\sec (e+f x)}{b}}{2 f}\) |
Input:
Int[Tan[e + f*x]^5/(a + b*Sec[e + f*x]^2),x]
Output:
-1/2*(-(((a + 2*b)*Log[Cos[e + f*x]^2])/b^2) + ((a + b)^2*Log[b + a*Cos[e + f*x]^2])/(a*b^2) - Sec[e + f*x]/b)/f
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/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 = 1.24 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.97
method | result | size |
derivativedivides | \(\frac {-\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{2 b^{2} a}+\frac {\left (a +2 b \right ) \ln \left (\cos \left (f x +e \right )\right )}{b^{2}}+\frac {1}{2 b \cos \left (f x +e \right )^{2}}}{f}\) | \(67\) |
default | \(\frac {-\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{2 b^{2} a}+\frac {\left (a +2 b \right ) \ln \left (\cos \left (f x +e \right )\right )}{b^{2}}+\frac {1}{2 b \cos \left (f x +e \right )^{2}}}{f}\) | \(67\) |
risch | \(\frac {i x}{a}+\frac {2 i e}{a f}+\frac {2 \,{\mathrm e}^{2 i \left (f x +e \right )}}{f b \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) a}{b^{2} f}+\frac {2 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{b f}-\frac {a \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{2 b^{2} f}-\frac {\ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{b f}-\frac {\ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{2 a f}\) | \(207\) |
Input:
int(tan(f*x+e)^5/(a+b*sec(f*x+e)^2),x,method=_RETURNVERBOSE)
Output:
1/f*(-1/2*(a^2+2*a*b+b^2)/b^2/a*ln(b+a*cos(f*x+e)^2)+(a+2*b)/b^2*ln(cos(f* x+e))+1/2/b/cos(f*x+e)^2)
Time = 0.13 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.22 \[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {{\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} \log \left (a \cos \left (f x + e\right )^{2} + b\right ) - 2 \, {\left (a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{2} \log \left (-\cos \left (f x + e\right )\right ) - a b}{2 \, a b^{2} f \cos \left (f x + e\right )^{2}} \] Input:
integrate(tan(f*x+e)^5/(a+b*sec(f*x+e)^2),x, algorithm="fricas")
Output:
-1/2*((a^2 + 2*a*b + b^2)*cos(f*x + e)^2*log(a*cos(f*x + e)^2 + b) - 2*(a^ 2 + 2*a*b)*cos(f*x + e)^2*log(-cos(f*x + e)) - a*b)/(a*b^2*f*cos(f*x + e)^ 2)
\[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\int \frac {\tan ^{5}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \] Input:
integrate(tan(f*x+e)**5/(a+b*sec(f*x+e)**2),x)
Output:
Integral(tan(e + f*x)**5/(a + b*sec(e + f*x)**2), x)
Time = 0.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.17 \[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {{\left (a + 2 \, b\right )} \log \left (\sin \left (f x + e\right )^{2} - 1\right )}{b^{2}} - \frac {1}{b \sin \left (f x + e\right )^{2} - b} - \frac {{\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a b^{2}}}{2 \, f} \] Input:
integrate(tan(f*x+e)^5/(a+b*sec(f*x+e)^2),x, algorithm="maxima")
Output:
1/2*((a + 2*b)*log(sin(f*x + e)^2 - 1)/b^2 - 1/(b*sin(f*x + e)^2 - b) - (a ^2 + 2*a*b + b^2)*log(a*sin(f*x + e)^2 - a - b)/(a*b^2))/f
Time = 0.18 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.06 \[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {{\left (a + 2 \, b\right )} \log \left ({\left | \cos \left (f x + e\right ) \right |}\right )}{b^{2} f} - \frac {{\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left ({\left | a \cos \left (f x + e\right )^{2} + b \right |}\right )}{2 \, a b^{2} f} + \frac {1}{2 \, b f \cos \left (f x + e\right )^{2}} \] Input:
integrate(tan(f*x+e)^5/(a+b*sec(f*x+e)^2),x, algorithm="giac")
Output:
(a + 2*b)*log(abs(cos(f*x + e)))/(b^2*f) - 1/2*(a^2 + 2*a*b + b^2)*log(abs (a*cos(f*x + e)^2 + b))/(a*b^2*f) + 1/2/(b*f*cos(f*x + e)^2)
Time = 16.23 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.49 \[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,a\,f}-\frac {\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}{b\,f}-\frac {\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}{2\,a\,f}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2}{2\,b\,f}-\frac {a\,\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}{2\,b^2\,f} \] Input:
int(tan(e + f*x)^5/(a + b/cos(e + f*x)^2),x)
Output:
log(tan(e + f*x)^2 + 1)/(2*a*f) - log(a + b + b*tan(e + f*x)^2)/(b*f) - lo g(a + b + b*tan(e + f*x)^2)/(2*a*f) + tan(e + f*x)^2/(2*b*f) - (a*log(a + b + b*tan(e + f*x)^2))/(2*b^2*f)
Time = 0.17 (sec) , antiderivative size = 766, normalized size of antiderivative = 11.10 \[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx =\text {Too large to display} \] Input:
int(tan(f*x+e)^5/(a+b*sec(f*x+e)^2),x)
Output:
(2*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**2*b**2 - 2*log(tan((e + f*x) /2)**2 + 1)*b**2 + 2*log(tan((e + f*x)/2) - 1)*sin(e + f*x)**2*a**2 + 4*lo g(tan((e + f*x)/2) - 1)*sin(e + f*x)**2*a*b - 2*log(tan((e + f*x)/2) - 1)* a**2 - 4*log(tan((e + f*x)/2) - 1)*a*b + 2*log(tan((e + f*x)/2) + 1)*sin(e + f*x)**2*a**2 + 4*log(tan((e + f*x)/2) + 1)*sin(e + f*x)**2*a*b - 2*log( tan((e + f*x)/2) + 1)*a**2 - 4*log(tan((e + f*x)/2) + 1)*a*b - log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan((e + f*x)/2))*sin(e + f*x)**2*a**2 - 2*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2* sqrt(a)*tan((e + f*x)/2))*sin(e + f*x)**2*a*b - log(sqrt(a + b)*tan((e + f *x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan((e + f*x)/2))*sin(e + f*x)**2*b**2 + log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan((e + f*x)/2))*a**2 + 2*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sq rt(a)*tan((e + f*x)/2))*a*b + log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan((e + f*x)/2))*b**2 - log(sqrt(a + b)*tan((e + f*x)/2 )**2 + sqrt(a + b) + 2*sqrt(a)*tan((e + f*x)/2))*sin(e + f*x)**2*a**2 - 2* log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) + 2*sqrt(a)*tan((e + f*x )/2))*sin(e + f*x)**2*a*b - log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) + 2*sqrt(a)*tan((e + f*x)/2))*sin(e + f*x)**2*b**2 + log(sqrt(a + b)*t an((e + f*x)/2)**2 + sqrt(a + b) + 2*sqrt(a)*tan((e + f*x)/2))*a**2 + 2*lo g(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) + 2*sqrt(a)*tan((e + f*...