Integrand size = 23, antiderivative size = 64 \[ \int \frac {\tan ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {a \log (\cos (c+d x))}{(a+b)^2 d}-\frac {a \log \left (a+b \sin ^2(c+d x)\right )}{2 (a+b)^2 d}+\frac {\sec ^2(c+d x)}{2 (a+b) d} \] Output:
a*ln(cos(d*x+c))/(a+b)^2/d-1/2*a*ln(a+b*sin(d*x+c)^2)/(a+b)^2/d+1/2*sec(d* x+c)^2/(a+b)/d
Time = 0.12 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.81 \[ \int \frac {\tan ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {a \left (2 \log (\cos (c+d x))-\log \left (a+b \sin ^2(c+d x)\right )\right )+(a+b) \sec ^2(c+d x)}{2 (a+b)^2 d} \] Input:
Integrate[Tan[c + d*x]^3/(a + b*Sin[c + d*x]^2),x]
Output:
(a*(2*Log[Cos[c + d*x]] - Log[a + b*Sin[c + d*x]^2]) + (a + b)*Sec[c + d*x ]^2)/(2*(a + b)^2*d)
Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3673, 86, 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.
\(\displaystyle \int \frac {\tan ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)^3}{a+b \sin (c+d x)^2}dx\) |
\(\Big \downarrow \) 3673 |
\(\displaystyle \frac {\int \frac {\sin ^2(c+d x)}{\left (1-\sin ^2(c+d x)\right )^2 \left (b \sin ^2(c+d x)+a\right )}d\sin ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\int \left (\frac {a}{(a+b)^2 \left (\sin ^2(c+d x)-1\right )}-\frac {b a}{(a+b)^2 \left (b \sin ^2(c+d x)+a\right )}+\frac {1}{(a+b) \left (\sin ^2(c+d x)-1\right )^2}\right )d\sin ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{(a+b) \left (1-\sin ^2(c+d x)\right )}+\frac {a \log \left (1-\sin ^2(c+d x)\right )}{(a+b)^2}-\frac {a \log \left (a+b \sin ^2(c+d x)\right )}{(a+b)^2}}{2 d}\) |
Input:
Int[Tan[c + d*x]^3/(a + b*Sin[c + d*x]^2),x]
Output:
((a*Log[1 - Sin[c + d*x]^2])/(a + b)^2 - (a*Log[a + b*Sin[c + d*x]^2])/(a + b)^2 + 1/((a + b)*(1 - Sin[c + d*x]^2)))/(2*d)
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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 = 1.24 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.91
method | result | size |
derivativedivides | \(\frac {\frac {a \ln \left (\cos \left (d x +c \right )\right )}{\left (a +b \right )^{2}}+\frac {1}{2 \left (a +b \right ) \cos \left (d x +c \right )^{2}}-\frac {a \ln \left (a +b -b \cos \left (d x +c \right )^{2}\right )}{2 \left (a +b \right )^{2}}}{d}\) | \(58\) |
default | \(\frac {\frac {a \ln \left (\cos \left (d x +c \right )\right )}{\left (a +b \right )^{2}}+\frac {1}{2 \left (a +b \right ) \cos \left (d x +c \right )^{2}}-\frac {a \ln \left (a +b -b \cos \left (d x +c \right )^{2}\right )}{2 \left (a +b \right )^{2}}}{d}\) | \(58\) |
risch | \(\frac {2 \,{\mathrm e}^{2 i \left (d x +c \right )}}{d \left (a +b \right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d \left (a^{2}+2 a b +b^{2}\right )}-\frac {a \ln \left ({\mathrm e}^{4 i \left (d x +c \right )}-\frac {2 \left (2 a +b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{b}+1\right )}{2 d \left (a^{2}+2 a b +b^{2}\right )}\) | \(114\) |
Input:
int(tan(d*x+c)^3/(a+b*sin(d*x+c)^2),x,method=_RETURNVERBOSE)
Output:
1/d*(a/(a+b)^2*ln(cos(d*x+c))+1/2/(a+b)/cos(d*x+c)^2-1/2*a/(a+b)^2*ln(a+b- b*cos(d*x+c)^2))
Time = 0.12 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.22 \[ \int \frac {\tan ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {a \cos \left (d x + c\right )^{2} \log \left (-b \cos \left (d x + c\right )^{2} + a + b\right ) - 2 \, a \cos \left (d x + c\right )^{2} \log \left (-\cos \left (d x + c\right )\right ) - a - b}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d \cos \left (d x + c\right )^{2}} \] Input:
integrate(tan(d*x+c)^3/(a+b*sin(d*x+c)^2),x, algorithm="fricas")
Output:
-1/2*(a*cos(d*x + c)^2*log(-b*cos(d*x + c)^2 + a + b) - 2*a*cos(d*x + c)^2 *log(-cos(d*x + c)) - a - b)/((a^2 + 2*a*b + b^2)*d*cos(d*x + c)^2)
\[ \int \frac {\tan ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\int \frac {\tan ^{3}{\left (c + d x \right )}}{a + b \sin ^{2}{\left (c + d x \right )}}\, dx \] Input:
integrate(tan(d*x+c)**3/(a+b*sin(d*x+c)**2),x)
Output:
Integral(tan(c + d*x)**3/(a + b*sin(c + d*x)**2), x)
Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.28 \[ \int \frac {\tan ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {\frac {a \log \left (b \sin \left (d x + c\right )^{2} + a\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {a \log \left (\sin \left (d x + c\right )^{2} - 1\right )}{a^{2} + 2 \, a b + b^{2}} + \frac {1}{{\left (a + b\right )} \sin \left (d x + c\right )^{2} - a - b}}{2 \, d} \] Input:
integrate(tan(d*x+c)^3/(a+b*sin(d*x+c)^2),x, algorithm="maxima")
Output:
-1/2*(a*log(b*sin(d*x + c)^2 + a)/(a^2 + 2*a*b + b^2) - a*log(sin(d*x + c) ^2 - 1)/(a^2 + 2*a*b + b^2) + 1/((a + b)*sin(d*x + c)^2 - a - b))/d
Time = 0.59 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.42 \[ \int \frac {\tan ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {a b \log \left ({\left | b \cos \left (d x + c\right )^{2} - a - b \right |}\right )}{2 \, {\left (a^{2} b d + 2 \, a b^{2} d + b^{3} d\right )}} + \frac {a \log \left ({\left | \cos \left (d x + c\right ) \right |}\right )}{a^{2} d + 2 \, a b d + b^{2} d} + \frac {1}{2 \, {\left (a + b\right )} d \cos \left (d x + c\right )^{2}} \] Input:
integrate(tan(d*x+c)^3/(a+b*sin(d*x+c)^2),x, algorithm="giac")
Output:
-1/2*a*b*log(abs(b*cos(d*x + c)^2 - a - b))/(a^2*b*d + 2*a*b^2*d + b^3*d) + a*log(abs(cos(d*x + c)))/(a^2*d + 2*a*b*d + b^2*d) + 1/2/((a + b)*d*cos( d*x + c)^2)
Time = 34.65 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.81 \[ \int \frac {\tan ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {a\,\left (\frac {\ln \left (\left (a+b\right )\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\right )}{2}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2}{2}\right )-\frac {b\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}}{d\,{\left (a+b\right )}^2} \] Input:
int(tan(c + d*x)^3/(a + b*sin(c + d*x)^2),x)
Output:
-(a*(log(a + tan(c + d*x)^2*(a + b))/2 - tan(c + d*x)^2/2) - (b*tan(c + d* x)^2)/2)/(d*(a + b)^2)
Time = 0.20 (sec) , antiderivative size = 374, normalized size of antiderivative = 5.84 \[ \int \frac {\tan ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {-\mathrm {log}\left (-\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+\sqrt {a}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a +\mathrm {log}\left (-\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+\sqrt {a}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -\mathrm {log}\left (\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+\sqrt {a}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a +\mathrm {log}\left (\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+\sqrt {a}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -\mathrm {log}\left (2 \sqrt {b}\, \sqrt {a +b}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +a +2 b \right ) \sin \left (d x +c \right )^{2} a +\mathrm {log}\left (2 \sqrt {b}\, \sqrt {a +b}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +a +2 b \right ) a -\sin \left (d x +c \right )^{2} a -\sin \left (d x +c \right )^{2} b}{2 d \left (\sin \left (d x +c \right )^{2} a^{2}+2 \sin \left (d x +c \right )^{2} a b +\sin \left (d x +c \right )^{2} b^{2}-a^{2}-2 a b -b^{2}\right )} \] Input:
int(tan(d*x+c)^3/(a+b*sin(d*x+c)^2),x)
Output:
( - log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + sqrt(a)*tan((c + d*x)/2 ))*sin(c + d*x)**2*a + log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + sqrt (a)*tan((c + d*x)/2))*a - log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + sqrt (a)*tan((c + d*x)/2))*sin(c + d*x)**2*a + log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + sqrt(a)*tan((c + d*x)/2))*a + 2*log(tan((c + d*x)/2) - 1)*sin( c + d*x)**2*a - 2*log(tan((c + d*x)/2) - 1)*a + 2*log(tan((c + d*x)/2) + 1 )*sin(c + d*x)**2*a - 2*log(tan((c + d*x)/2) + 1)*a - log(2*sqrt(b)*sqrt(a + b) + tan((c + d*x)/2)**2*a + a + 2*b)*sin(c + d*x)**2*a + log(2*sqrt(b) *sqrt(a + b) + tan((c + d*x)/2)**2*a + a + 2*b)*a - sin(c + d*x)**2*a - si n(c + d*x)**2*b)/(2*d*(sin(c + d*x)**2*a**2 + 2*sin(c + d*x)**2*a*b + sin( c + d*x)**2*b**2 - a**2 - 2*a*b - b**2))