Integrand size = 27, antiderivative size = 93 \[ \int \frac {\sin ^2(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\log (1-\sin (c+d x))}{2 (a+b) d}-\frac {\log (1+\sin (c+d x))}{2 (a-b) d}+\frac {a^3 \log (a+b \sin (c+d x))}{b^2 \left (a^2-b^2\right ) d}-\frac {\sin (c+d x)}{b d} \]
-1/2*ln(1-sin(d*x+c))/(a+b)/d-1/2*ln(1+sin(d*x+c))/(a-b)/d+a^3*ln(a+b*sin( d*x+c))/b^2/(a^2-b^2)/d-sin(d*x+c)/b/d
Time = 0.16 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.89 \[ \int \frac {\sin ^2(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {\log (1-\sin (c+d x))}{a+b}+\frac {\log (1+\sin (c+d x))}{a-b}-\frac {2 a^3 \log (a+b \sin (c+d x))}{b^2 \left (a^2-b^2\right )}+\frac {2 \sin (c+d x)}{b}}{2 d} \]
-1/2*(Log[1 - Sin[c + d*x]]/(a + b) + Log[1 + Sin[c + d*x]]/(a - b) - (2*a ^3*Log[a + b*Sin[c + d*x]])/(b^2*(a^2 - b^2)) + (2*Sin[c + d*x])/b)/d
Time = 0.39 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {3042, 3316, 27, 604, 25, 2160, 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.
\(\displaystyle \int \frac {\sin ^2(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^3}{\cos (c+d x) (a+b \sin (c+d x))}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle \frac {b \int \frac {\sin ^3(c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {b^3 \sin ^3(c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{b^2 d}\) |
\(\Big \downarrow \) 604 |
\(\displaystyle \frac {-\int -\frac {\sin (c+d x) b^3-a \sin ^2(c+d x) b^2+a b^2}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))-a-b \sin (c+d x)}{b^2 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {\sin (c+d x) b^3-a \sin ^2(c+d x) b^2+a b^2}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))-a-b \sin (c+d x)}{b^2 d}\) |
\(\Big \downarrow \) 2160 |
\(\displaystyle \frac {\int \left (\frac {a^3}{(a-b) (a+b) (a+b \sin (c+d x))}+\frac {b^2}{2 (a+b) (b-b \sin (c+d x))}-\frac {b^2}{2 (a-b) (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))-a-b \sin (c+d x)}{b^2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {a^3 \log (a+b \sin (c+d x))}{a^2-b^2}-\frac {b^2 \log (b-b \sin (c+d x))}{2 (a+b)}-\frac {b^2 \log (b \sin (c+d x)+b)}{2 (a-b)}-a-b \sin (c+d x)}{b^2 d}\) |
(-a - (b^2*Log[b - b*Sin[c + d*x]])/(2*(a + b)) + (a^3*Log[a + b*Sin[c + d *x]])/(a^2 - b^2) - (b^2*Log[b + b*Sin[c + d*x]])/(2*(a - b)) - b*Sin[c + d*x])/(b^2*d)
3.14.32.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[(c + d*x)^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*d^(m - 1)*(m + n + 2*p + 1))), x] + Simp[1/(b*d^m*(m + n + 2*p + 1)) Int[(c + d*x)^n*(a + b* x^2)^p*ExpandToSum[b*d^m*(m + n + 2*p + 1)*x^m - b*(m + n + 2*p + 1)*(c + d *x)^m - (c + d*x)^(m - 2)*(a*d^2*(m + n - 1) - b*c^2*(m + n + 2*p + 1) - 2* b*c*d*(m + n + p)*x), x], x], x] /; FreeQ[{a, b, c, d, n, p}, x] && IGtQ[m, 1] && NeQ[m + n + 2*p + 1, 0] && IntegerQ[2*p]
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Time = 0.38 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.94
method | result | size |
derivativedivides | \(\frac {-\frac {\sin \left (d x +c \right )}{b}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}+\frac {a^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{2} \left (a +b \right ) \left (a -b \right )}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}}{d}\) | \(87\) |
default | \(\frac {-\frac {\sin \left (d x +c \right )}{b}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}+\frac {a^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{2} \left (a +b \right ) \left (a -b \right )}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}}{d}\) | \(87\) |
norman | \(\frac {-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b d}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {a^{3} \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{b^{2} d \left (a^{2}-b^{2}\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \left (a -b \right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{\left (a +b \right ) d}-\frac {a \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2} d}\) | \(173\) |
parallelrisch | \(\frac {-\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a \,b^{2}+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{3}-\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a \,b^{2}-\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{3}-a^{3} \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \,b^{2}+a^{3} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-\sin \left (d x +c \right ) a^{2} b +b^{3} \sin \left (d x +c \right )}{\left (a^{2}-b^{2}\right ) b^{2} d}\) | \(176\) |
risch | \(\frac {i a x}{b^{2}}+\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 b d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 b d}+\frac {i x}{a -b}+\frac {i c}{d \left (a -b \right )}+\frac {i x}{a +b}+\frac {i c}{d \left (a +b \right )}-\frac {2 i a^{3} x}{b^{2} \left (a^{2}-b^{2}\right )}-\frac {2 i a^{3} c}{b^{2} d \left (a^{2}-b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \left (a -b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d \left (a +b \right )}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{2} d \left (a^{2}-b^{2}\right )}\) | \(234\) |
1/d*(-sin(d*x+c)/b-1/(2*a+2*b)*ln(sin(d*x+c)-1)+1/b^2*a^3/(a+b)/(a-b)*ln(a +b*sin(d*x+c))-1/(2*a-2*b)*ln(1+sin(d*x+c)))
Time = 0.39 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.08 \[ \int \frac {\sin ^2(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 \, a^{3} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (a b^{2} + b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a b^{2} - b^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} d} \]
1/2*(2*a^3*log(b*sin(d*x + c) + a) - (a*b^2 + b^3)*log(sin(d*x + c) + 1) - (a*b^2 - b^3)*log(-sin(d*x + c) + 1) - 2*(a^2*b - b^3)*sin(d*x + c))/((a^ 2*b^2 - b^4)*d)
Timed out. \[ \int \frac {\sin ^2(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.88 \[ \int \frac {\sin ^2(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, a^{3} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{2} b^{2} - b^{4}} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a - b} - \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a + b} - \frac {2 \, \sin \left (d x + c\right )}{b}}{2 \, d} \]
1/2*(2*a^3*log(b*sin(d*x + c) + a)/(a^2*b^2 - b^4) - log(sin(d*x + c) + 1) /(a - b) - log(sin(d*x + c) - 1)/(a + b) - 2*sin(d*x + c)/b)/d
Time = 0.34 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.91 \[ \int \frac {\sin ^2(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, a^{3} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{2} b^{2} - b^{4}} - \frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a - b} - \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a + b} - \frac {2 \, \sin \left (d x + c\right )}{b}}{2 \, d} \]
1/2*(2*a^3*log(abs(b*sin(d*x + c) + a))/(a^2*b^2 - b^4) - log(abs(sin(d*x + c) + 1))/(a - b) - log(abs(sin(d*x + c) - 1))/(a + b) - 2*sin(d*x + c)/b )/d
Time = 12.15 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.44 \[ \int \frac {\sin ^2(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{d\,\left (a+b\right )}-\frac {\sin \left (c+d\,x\right )}{b\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{d\,\left (a-b\right )}-\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{b^2\,d}-\frac {a^3\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (b^4-a^2\,b^2\right )} \]