Integrand size = 27, antiderivative size = 117 \[ \int \frac {\sec ^2(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {b \log (1-\sin (c+d x))}{4 (a+b)^2 d}+\frac {b \log (1+\sin (c+d x))}{4 (a-b)^2 d}-\frac {a b^2 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2 d}+\frac {\sec ^2(c+d x) (a-b \sin (c+d x))}{2 \left (a^2-b^2\right ) d} \]
-1/4*b*ln(1-sin(d*x+c))/(a+b)^2/d+1/4*b*ln(1+sin(d*x+c))/(a-b)^2/d-a*b^2*l n(a+b*sin(d*x+c))/(a^2-b^2)^2/d+1/2*sec(d*x+c)^2*(a-b*sin(d*x+c))/(a^2-b^2 )/d
Time = 0.45 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.92 \[ \int \frac {\sec ^2(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-\frac {b \log (1-\sin (c+d x))}{(a+b)^2}+\frac {b \log (1+\sin (c+d x))}{(a-b)^2}-\frac {4 a b^2 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2}-\frac {1}{(a+b) (-1+\sin (c+d x))}+\frac {1}{(a-b) (1+\sin (c+d x))}}{4 d} \]
(-((b*Log[1 - Sin[c + d*x]])/(a + b)^2) + (b*Log[1 + Sin[c + d*x]])/(a - b )^2 - (4*a*b^2*Log[a + b*Sin[c + d*x]])/(a^2 - b^2)^2 - 1/((a + b)*(-1 + S in[c + d*x])) + 1/((a - b)*(1 + Sin[c + d*x])))/(4*d)
Time = 0.35 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.29, 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, 593, 25, 657, 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.
| \(\displaystyle \int \frac {\tan (c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)}{\cos (c+d x)^3 (a+b \sin (c+d x))}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^3 \int \frac {\sin (c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^2 \int \frac {b \sin (c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 593 |
| \(\displaystyle \frac {b^2 \left (\frac {a-b \sin (c+d x)}{2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int -\frac {a-b \sin (c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 \left (a^2-b^2\right )}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b^2 \left (\frac {\int \frac {a-b \sin (c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 \left (a^2-b^2\right )}+\frac {a-b \sin (c+d x)}{2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{d}\) |
| \(\Big \downarrow \) 657 |
| \(\displaystyle \frac {b^2 \left (\frac {\int \left (-\frac {2 a}{(a-b) (a+b) (a+b \sin (c+d x))}+\frac {a-b}{2 b (a+b) (b-b \sin (c+d x))}+\frac {a+b}{2 (a-b) b (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))}{2 \left (a^2-b^2\right )}+\frac {a-b \sin (c+d x)}{2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^2 \left (\frac {a-b \sin (c+d x)}{2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}+\frac {-\frac {2 a \log (a+b \sin (c+d x))}{a^2-b^2}-\frac {(a-b) \log (b-b \sin (c+d x))}{2 b (a+b)}+\frac {(a+b) \log (b \sin (c+d x)+b)}{2 b (a-b)}}{2 \left (a^2-b^2\right )}\right )}{d}\) |
(b^2*((-1/2*((a - b)*Log[b - b*Sin[c + d*x]])/(b*(a + b)) - (2*a*Log[a + b *Sin[c + d*x]])/(a^2 - b^2) + ((a + b)*Log[b + b*Sin[c + d*x]])/(2*(a - b) *b))/(2*(a^2 - b^2)) + (a - b*Sin[c + d*x])/(2*(a^2 - b^2)*(b^2 - b^2*Sin[ c + d*x]^2))))/d
3.14.48.3.1 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_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c + d*x)^(n + 1)*(c - d*x)*((a + b*x^2)^(p + 1)/(2*(p + 1)*(b*c^2 + a*d^2))), x] - Simp[d/(2*(p + 1)*(b*c^2 + a*d^2)) Int[(c + d*x)^n*(a + b* x^2)^(p + 1)*(c*n - d*(n + 2*p + 4)*x), x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[p, -1] && NeQ[b*c^2 + a*d^2, 0]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
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 = 0.49 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.96
| method | result | size |
| derivativedivides | \(\frac {-\frac {b^{2} a \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {1}{\left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}-\frac {b \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{2}}+\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {b \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}}{d}\) | \(112\) |
| default | \(\frac {-\frac {b^{2} a \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {1}{\left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}-\frac {b \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{2}}+\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {b \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}}{d}\) | \(112\) |
| parallelrisch | \(\frac {-2 b^{2} a \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-b \left (a -b \right )^{2} \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (-b \left (1+\cos \left (2 d x +2 c \right )\right ) \left (a +b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (a -b \right ) \left (a \cos \left (2 d x +2 c \right )+2 b \sin \left (d x +c \right )-a \right )\right ) \left (a +b \right )}{2 \left (a -b \right )^{2} \left (a +b \right )^{2} d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(174\) |
| norman | \(\frac {-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (a^{2}-b^{2}\right )}-\frac {b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{2}-b^{2}\right )}+\frac {2 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a^{2}-b^{2}\right ) d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \left (a^{2}-2 a b +b^{2}\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 \left (a^{2}+2 a b +b^{2}\right ) d}-\frac {a \,b^{2} \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}\) | \(217\) |
| risch | \(\frac {i b x}{2 a^{2}+4 a b +2 b^{2}}+\frac {i b c}{2 \left (a^{2}+2 a b +b^{2}\right ) d}-\frac {i b x}{2 \left (a^{2}-2 a b +b^{2}\right )}-\frac {i b c}{2 d \left (a^{2}-2 a b +b^{2}\right )}+\frac {2 i a \,b^{2} x}{a^{4}-2 a^{2} b^{2}+b^{4}}+\frac {2 i a \,b^{2} c}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {i \left (-2 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{3 i \left (d x +c \right )}-b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{\left (a^{2}-b^{2}\right ) d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{2 \left (a^{2}+2 a b +b^{2}\right ) d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{2 \left (a^{2}-2 a b +b^{2}\right ) d}-\frac {a \,b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}\) | \(318\) |
1/d*(-b^2*a/(a+b)^2/(a-b)^2*ln(a+b*sin(d*x+c))-1/(4*a+4*b)/(sin(d*x+c)-1)- 1/4*b/(a+b)^2*ln(sin(d*x+c)-1)+1/(4*a-4*b)/(1+sin(d*x+c))+1/4*b/(a-b)^2*ln (1+sin(d*x+c)))
Time = 0.41 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.32 \[ \int \frac {\sec ^2(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {4 \, a b^{2} \cos \left (d x + c\right )^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, a^{3} + 2 \, a b^{2} + 2 \, {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \cos \left (d x + c\right )^{2}} \]
-1/4*(4*a*b^2*cos(d*x + c)^2*log(b*sin(d*x + c) + a) - (a^2*b + 2*a*b^2 + b^3)*cos(d*x + c)^2*log(sin(d*x + c) + 1) + (a^2*b - 2*a*b^2 + b^3)*cos(d* x + c)^2*log(-sin(d*x + c) + 1) - 2*a^3 + 2*a*b^2 + 2*(a^2*b - b^3)*sin(d* x + c))/((a^4 - 2*a^2*b^2 + b^4)*d*cos(d*x + c)^2)
\[ \int \frac {\sec ^2(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\sin {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.13 \[ \int \frac {\sec ^2(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {4 \, a b^{2} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {b \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {b \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {2 \, {\left (b \sin \left (d x + c\right ) - a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} - a^{2} + b^{2}}}{4 \, d} \]
-1/4*(4*a*b^2*log(b*sin(d*x + c) + a)/(a^4 - 2*a^2*b^2 + b^4) - b*log(sin( d*x + c) + 1)/(a^2 - 2*a*b + b^2) + b*log(sin(d*x + c) - 1)/(a^2 + 2*a*b + b^2) - 2*(b*sin(d*x + c) - a)/((a^2 - b^2)*sin(d*x + c)^2 - a^2 + b^2))/d
Time = 0.44 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.45 \[ \int \frac {\sec ^2(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {4 \, a b^{3} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b - 2 \, a^{2} b^{3} + b^{5}} - \frac {b \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {b \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2} + 2 \, a b + b^{2}} + \frac {2 \, {\left (a b^{2} \sin \left (d x + c\right )^{2} - a^{2} b \sin \left (d x + c\right ) + b^{3} \sin \left (d x + c\right ) + a^{3} - 2 \, a b^{2}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}}}{4 \, d} \]
-1/4*(4*a*b^3*log(abs(b*sin(d*x + c) + a))/(a^4*b - 2*a^2*b^3 + b^5) - b*l og(abs(sin(d*x + c) + 1))/(a^2 - 2*a*b + b^2) + b*log(abs(sin(d*x + c) - 1 ))/(a^2 + 2*a*b + b^2) + 2*(a*b^2*sin(d*x + c)^2 - a^2*b*sin(d*x + c) + b^ 3*sin(d*x + c) + a^3 - 2*a*b^2)/((a^4 - 2*a^2*b^2 + b^4)*(sin(d*x + c)^2 - 1)))/d
Time = 12.54 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.78 \[ \int \frac {\sec ^2(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{2\,d\,{\left (a-b\right )}^2}-\frac {\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^2-b^2}-\frac {2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^2-b^2}+\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{a^2-b^2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{2\,d\,{\left (a+b\right )}^2}-\frac {a\,b^2\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (a^4-2\,a^2\,b^2+b^4\right )} \]
(b*log(tan(c/2 + (d*x)/2) + 1))/(2*d*(a - b)^2) - ((b*tan(c/2 + (d*x)/2))/ (a^2 - b^2) - (2*a*tan(c/2 + (d*x)/2)^2)/(a^2 - b^2) + (b*tan(c/2 + (d*x)/ 2)^3)/(a^2 - b^2))/(d*(tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^2 + 1)) - (b*log(tan(c/2 + (d*x)/2) - 1))/(2*d*(a + b)^2) - (a*b^2*log(a + 2*b*ta n(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2))/(d*(a^4 + b^4 - 2*a^2*b^2))