Integrand size = 23, antiderivative size = 94 \[ \int \frac {\tan ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {a^2 \log (\cos (c+d x))}{(a+b)^3 d}+\frac {a^2 \log \left (a+b \sin ^2(c+d x)\right )}{2 (a+b)^3 d}-\frac {(2 a+b) \sec ^2(c+d x)}{2 (a+b)^2 d}+\frac {\sec ^4(c+d x)}{4 (a+b) d} \] Output:
-a^2*ln(cos(d*x+c))/(a+b)^3/d+1/2*a^2*ln(a+b*sin(d*x+c)^2)/(a+b)^3/d-1/2*( 2*a+b)*sec(d*x+c)^2/(a+b)^2/d+1/4*sec(d*x+c)^4/(a+b)/d
Time = 0.30 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.83 \[ \int \frac {\tan ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {2 a^2 \left (-2 \log (\cos (c+d x))+\log \left (a+b \sin ^2(c+d x)\right )\right )-2 \left (2 a^2+3 a b+b^2\right ) \sec ^2(c+d x)+(a+b)^2 \sec ^4(c+d x)}{4 (a+b)^3 d} \] Input:
Integrate[Tan[c + d*x]^5/(a + b*Sin[c + d*x]^2),x]
Output:
(2*a^2*(-2*Log[Cos[c + d*x]] + Log[a + b*Sin[c + d*x]^2]) - 2*(2*a^2 + 3*a *b + b^2)*Sec[c + d*x]^2 + (a + b)^2*Sec[c + d*x]^4)/(4*(a + b)^3*d)
Time = 0.31 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3673, 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.
\(\displaystyle \int \frac {\tan ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)^5}{a+b \sin (c+d x)^2}dx\) |
\(\Big \downarrow \) 3673 |
\(\displaystyle \frac {\int \frac {\sin ^4(c+d x)}{\left (1-\sin ^2(c+d x)\right )^3 \left (b \sin ^2(c+d x)+a\right )}d\sin ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (-\frac {a^2}{(a+b)^3 \left (\sin ^2(c+d x)-1\right )}+\frac {b a^2}{(a+b)^3 \left (b \sin ^2(c+d x)+a\right )}+\frac {-2 a-b}{(a+b)^2 \left (\sin ^2(c+d x)-1\right )^2}-\frac {1}{(a+b) \left (\sin ^2(c+d x)-1\right )^3}\right )d\sin ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {a^2 \log \left (1-\sin ^2(c+d x)\right )}{(a+b)^3}+\frac {a^2 \log \left (a+b \sin ^2(c+d x)\right )}{(a+b)^3}-\frac {2 a+b}{(a+b)^2 \left (1-\sin ^2(c+d x)\right )}+\frac {1}{2 (a+b) \left (1-\sin ^2(c+d x)\right )^2}}{2 d}\) |
Input:
Int[Tan[c + d*x]^5/(a + b*Sin[c + d*x]^2),x]
Output:
(-((a^2*Log[1 - Sin[c + d*x]^2])/(a + b)^3) + (a^2*Log[a + b*Sin[c + d*x]^ 2])/(a + b)^3 + 1/(2*(a + b)*(1 - Sin[c + d*x]^2)^2) - (2*a + b)/((a + b)^ 2*(1 - Sin[c + d*x]^2)))/(2*d)
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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^ (m_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x]^2, x]}, Simp[ff^((m + 1)/2)/(2*f) Subst[Int[x^((m - 1)/2)*((a + b*ff*x)^p/(1 - ff*x)^((m + 1 )/2)), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && Integ erQ[(m - 1)/2]
Time = 2.35 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.88
method | result | size |
derivativedivides | \(\frac {-\frac {2 a +b}{2 \left (a +b \right )^{2} \cos \left (d x +c \right )^{2}}+\frac {1}{4 \left (a +b \right ) \cos \left (d x +c \right )^{4}}-\frac {a^{2} \ln \left (\cos \left (d x +c \right )\right )}{\left (a +b \right )^{3}}+\frac {a^{2} \ln \left (a +b -b \cos \left (d x +c \right )^{2}\right )}{2 \left (a +b \right )^{3}}}{d}\) | \(83\) |
default | \(\frac {-\frac {2 a +b}{2 \left (a +b \right )^{2} \cos \left (d x +c \right )^{2}}+\frac {1}{4 \left (a +b \right ) \cos \left (d x +c \right )^{4}}-\frac {a^{2} \ln \left (\cos \left (d x +c \right )\right )}{\left (a +b \right )^{3}}+\frac {a^{2} \ln \left (a +b -b \cos \left (d x +c \right )^{2}\right )}{2 \left (a +b \right )^{3}}}{d}\) | \(83\) |
risch | \(-\frac {2 \left (2 \,{\mathrm e}^{6 i \left (d x +c \right )} a +{\mathrm e}^{6 i \left (d x +c \right )} b +2 \,{\mathrm e}^{4 i \left (d x +c \right )} a +2 a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{2 i \left (d x +c \right )}\right )}{d \left (a +b \right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}+\frac {a^{2} \ln \left ({\mathrm e}^{4 i \left (d x +c \right )}-\frac {2 \left (2 a +b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{b}+1\right )}{2 d \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}\) | \(185\) |
Input:
int(tan(d*x+c)^5/(a+b*sin(d*x+c)^2),x,method=_RETURNVERBOSE)
Output:
1/d*(-1/2*(2*a+b)/(a+b)^2/cos(d*x+c)^2+1/4/(a+b)/cos(d*x+c)^4-a^2/(a+b)^3* ln(cos(d*x+c))+1/2*a^2/(a+b)^3*ln(a+b-b*cos(d*x+c)^2))
Time = 0.16 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.26 \[ \int \frac {\tan ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {2 \, a^{2} \cos \left (d x + c\right )^{4} \log \left (-b \cos \left (d x + c\right )^{2} + a + b\right ) - 4 \, a^{2} \cos \left (d x + c\right )^{4} \log \left (-\cos \left (d x + c\right )\right ) - 2 \, {\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}{4 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d \cos \left (d x + c\right )^{4}} \] Input:
integrate(tan(d*x+c)^5/(a+b*sin(d*x+c)^2),x, algorithm="fricas")
Output:
1/4*(2*a^2*cos(d*x + c)^4*log(-b*cos(d*x + c)^2 + a + b) - 4*a^2*cos(d*x + c)^4*log(-cos(d*x + c)) - 2*(2*a^2 + 3*a*b + b^2)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cos(d*x + c)^4)
\[ \int \frac {\tan ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\int \frac {\tan ^{5}{\left (c + d x \right )}}{a + b \sin ^{2}{\left (c + d x \right )}}\, dx \] Input:
integrate(tan(d*x+c)**5/(a+b*sin(d*x+c)**2),x)
Output:
Integral(tan(c + d*x)**5/(a + b*sin(c + d*x)**2), x)
Time = 0.04 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.69 \[ \int \frac {\tan ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {\frac {2 \, a^{2} \log \left (b \sin \left (d x + c\right )^{2} + a\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {2 \, a^{2} \log \left (\sin \left (d x + c\right )^{2} - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (2 \, a + b\right )} \sin \left (d x + c\right )^{2} - 3 \, a - b}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sin \left (d x + c\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sin \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}}{4 \, d} \] Input:
integrate(tan(d*x+c)^5/(a+b*sin(d*x+c)^2),x, algorithm="maxima")
Output:
1/4*(2*a^2*log(b*sin(d*x + c)^2 + a)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 2*a ^2*log(sin(d*x + c)^2 - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + (2*(2*a + b)* sin(d*x + c)^2 - 3*a - b)/((a^2 + 2*a*b + b^2)*sin(d*x + c)^4 - 2*(a^2 + 2 *a*b + b^2)*sin(d*x + c)^2 + a^2 + 2*a*b + b^2))/d
Time = 0.64 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.62 \[ \int \frac {\tan ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {a^{2} b \log \left ({\left | b \cos \left (d x + c\right )^{2} - a - b \right |}\right )}{2 \, {\left (a^{3} b d + 3 \, a^{2} b^{2} d + 3 \, a b^{3} d + b^{4} d\right )}} - \frac {a^{2} \log \left ({\left | \cos \left (d x + c\right ) \right |}\right )}{a^{3} d + 3 \, a^{2} b d + 3 \, a b^{2} d + b^{3} d} - \frac {2 \, {\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 2 \, a b - b^{2}}{4 \, {\left (a + b\right )}^{3} d \cos \left (d x + c\right )^{4}} \] Input:
integrate(tan(d*x+c)^5/(a+b*sin(d*x+c)^2),x, algorithm="giac")
Output:
1/2*a^2*b*log(abs(b*cos(d*x + c)^2 - a - b))/(a^3*b*d + 3*a^2*b^2*d + 3*a* b^3*d + b^4*d) - a^2*log(abs(cos(d*x + c)))/(a^3*d + 3*a^2*b*d + 3*a*b^2*d + b^3*d) - 1/4*(2*(2*a^2 + 3*a*b + b^2)*cos(d*x + c)^2 - a^2 - 2*a*b - b^ 2)/((a + b)^3*d*cos(d*x + c)^4)
Time = 34.77 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.96 \[ \int \frac {\tan ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {a^2\,\left (\frac {\ln \left (\left (a+b\right )\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\right )}{2}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2}{2}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^4}{4}\right )+\frac {b^2\,{\mathrm {tan}\left (c+d\,x\right )}^4}{4}-a\,b\,\left (\frac {{\mathrm {tan}\left (c+d\,x\right )}^2}{2}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^4}{2}\right )}{d\,{\left (a+b\right )}^3} \] Input:
int(tan(c + d*x)^5/(a + b*sin(c + d*x)^2),x)
Output:
(a^2*(log(a + tan(c + d*x)^2*(a + b))/2 - tan(c + d*x)^2/2 + tan(c + d*x)^ 4/4) + (b^2*tan(c + d*x)^4)/4 - a*b*(tan(c + d*x)^2/2 - tan(c + d*x)^4/2)) /(d*(a + b)^3)
Time = 0.25 (sec) , antiderivative size = 689, normalized size of antiderivative = 7.33 \[ \int \frac {\tan ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx =\text {Too large to display} \] Input:
int(tan(d*x+c)^5/(a+b*sin(d*x+c)^2),x)
Output:
(2*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + sqrt(a)*tan((c + d*x)/2) )*sin(c + d*x)**4*a**2 - 4*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + sqrt(a)*tan((c + d*x)/2))*sin(c + d*x)**2*a**2 + 2*log( - sqrt(2*sqrt(b)*s qrt(a + b) - a - 2*b) + sqrt(a)*tan((c + d*x)/2))*a**2 + 2*log(sqrt(2*sqrt (b)*sqrt(a + b) - a - 2*b) + sqrt(a)*tan((c + d*x)/2))*sin(c + d*x)**4*a** 2 - 4*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + sqrt(a)*tan((c + d*x)/2) )*sin(c + d*x)**2*a**2 + 2*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + sqr t(a)*tan((c + d*x)/2))*a**2 - 4*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4* a**2 + 8*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2 - 4*log(tan((c + d *x)/2) - 1)*a**2 - 4*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**2 + 8*lo g(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2 - 4*log(tan((c + d*x)/2) + 1) *a**2 + 2*log(2*sqrt(b)*sqrt(a + b) + tan((c + d*x)/2)**2*a + a + 2*b)*sin (c + d*x)**4*a**2 - 4*log(2*sqrt(b)*sqrt(a + b) + tan((c + d*x)/2)**2*a + a + 2*b)*sin(c + d*x)**2*a**2 + 2*log(2*sqrt(b)*sqrt(a + b) + tan((c + d*x )/2)**2*a + a + 2*b)*a**2 + 3*sin(c + d*x)**4*a**2 + 4*sin(c + d*x)**4*a*b + sin(c + d*x)**4*b**2 - 2*sin(c + d*x)**2*a**2 - 2*sin(c + d*x)**2*a*b)/ (4*d*(sin(c + d*x)**4*a**3 + 3*sin(c + d*x)**4*a**2*b + 3*sin(c + d*x)**4* a*b**2 + sin(c + d*x)**4*b**3 - 2*sin(c + d*x)**2*a**3 - 6*sin(c + d*x)**2 *a**2*b - 6*sin(c + d*x)**2*a*b**2 - 2*sin(c + d*x)**2*b**3 + a**3 + 3*a** 2*b + 3*a*b**2 + b**3))