Integrand size = 25, antiderivative size = 82 \[ \int \frac {\sec (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 a b \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}+\frac {\sec (c+d x) (a-b \sin (c+d x))}{\left (a^2-b^2\right ) d} \] Output:
2*a*b*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/(a^2-b^2)^(3/2)/d+s ec(d*x+c)*(a-b*sin(d*x+c))/(a^2-b^2)/d
Time = 0.46 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.84 \[ \int \frac {\sec (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 a b \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right ) \cos (c+d x)+\sqrt {a^2-b^2} (a-a \cos (c+d x)-b \sin (c+d x))}{(a-b) (a+b) \sqrt {a^2-b^2} d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \] Input:
Integrate[(Sec[c + d*x]*Tan[c + d*x])/(a + b*Sin[c + d*x]),x]
Output:
(2*a*b*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]]*Cos[c + d*x] + Sqr t[a^2 - b^2]*(a - a*Cos[c + d*x] - b*Sin[c + d*x]))/((a - b)*(a + b)*Sqrt[ a^2 - b^2]*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin [(c + d*x)/2]))
Time = 0.40 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3345, 25, 27, 3042, 3139, 1083, 217}
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.
\(\displaystyle \int \frac {\tan (c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)}{\cos (c+d x)^2 (a+b \sin (c+d x))}dx\) |
\(\Big \downarrow \) 3345 |
\(\displaystyle \frac {\sec (c+d x) (a-b \sin (c+d x))}{d \left (a^2-b^2\right )}-\frac {\int -\frac {a b}{a+b \sin (c+d x)}dx}{a^2-b^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {a b}{a+b \sin (c+d x)}dx}{a^2-b^2}+\frac {\sec (c+d x) (a-b \sin (c+d x))}{d \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a b \int \frac {1}{a+b \sin (c+d x)}dx}{a^2-b^2}+\frac {\sec (c+d x) (a-b \sin (c+d x))}{d \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a b \int \frac {1}{a+b \sin (c+d x)}dx}{a^2-b^2}+\frac {\sec (c+d x) (a-b \sin (c+d x))}{d \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3139 |
\(\displaystyle \frac {2 a b \int \frac {1}{a \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tan \left (\frac {1}{2} (c+d x)\right )+a}d\tan \left (\frac {1}{2} (c+d x)\right )}{d \left (a^2-b^2\right )}+\frac {\sec (c+d x) (a-b \sin (c+d x))}{d \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {\sec (c+d x) (a-b \sin (c+d x))}{d \left (a^2-b^2\right )}-\frac {4 a b \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{d \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {2 a b \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}+\frac {\sec (c+d x) (a-b \sin (c+d x))}{d \left (a^2-b^2\right )}\) |
Input:
Int[(Sec[c + d*x]*Tan[c + d*x])/(a + b*Sin[c + d*x]),x]
Output:
(2*a*b*ArcTan[(2*b + 2*a*Tan[(c + d*x)/2])/(2*Sqrt[a^2 - b^2])])/((a^2 - b ^2)^(3/2)*d) + (Sec[c + d*x]*(a - b*Sin[c + d*x]))/((a^2 - b^2)*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ [a^2 - b^2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(g*Co s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c - a*d - (a*c - b*d)* Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Simp[1/(g^2*(a^2 - b^2)*(p + 1)) Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && Lt Q[p, -1] && IntegerQ[2*m]
Time = 0.51 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.35
method | result | size |
derivativedivides | \(\frac {\frac {2 a b \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {a^{2}-b^{2}}}+\frac {4}{\left (4 a -4 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {4}{\left (4 a +4 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\) | \(111\) |
default | \(\frac {\frac {2 a b \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {a^{2}-b^{2}}}+\frac {4}{\left (4 a -4 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {4}{\left (4 a +4 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\) | \(111\) |
risch | \(\frac {2 i \left (i a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{d \left (-a^{2}+b^{2}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {i b a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {i b a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}\) | \(209\) |
Input:
int(sec(d*x+c)*tan(d*x+c)/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(2*a*b/(a-b)/(a+b)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+ 2*b)/(a^2-b^2)^(1/2))+4/(4*a-4*b)/(tan(1/2*d*x+1/2*c)+1)-4/(4*a+4*b)/(tan( 1/2*d*x+1/2*c)-1))
Time = 0.10 (sec) , antiderivative size = 308, normalized size of antiderivative = 3.76 \[ \int \frac {\sec (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\left [\frac {\sqrt {-a^{2} + b^{2}} a b \cos \left (d x + c\right ) \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 2 \, a^{3} - 2 \, a b^{2} - 2 \, {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \cos \left (d x + c\right )}, -\frac {\sqrt {a^{2} - b^{2}} a b \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - a^{3} + a b^{2} + {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \cos \left (d x + c\right )}\right ] \] Input:
integrate(sec(d*x+c)*tan(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="fricas")
Output:
[1/2*(sqrt(-a^2 + b^2)*a*b*cos(d*x + c)*log(-((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 - 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos (d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^ 2 - b^2)) + 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)), -(sqrt(a^2 - b^2)*a*b*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c)))*cos(d*x + c) - a^3 + a*b^2 + (a^2*b - b^3)*sin(d*x + c))/((a^4 - 2*a^2*b^2 + b^4)*d*cos(d*x + c))]
\[ \int \frac {\sec (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\tan {\left (c + d x \right )} \sec {\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \] Input:
integrate(sec(d*x+c)*tan(d*x+c)/(a+b*sin(d*x+c)),x)
Output:
Integral(tan(c + d*x)*sec(c + d*x)/(a + b*sin(c + d*x)), x)
Exception generated. \[ \int \frac {\sec (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(sec(d*x+c)*tan(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.29 \[ \int \frac {\sec (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a b}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}} + \frac {b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a}{{\left (a^{2} - b^{2}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}}\right )}}{d} \] Input:
integrate(sec(d*x+c)*tan(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
2*((pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2* c) + b)/sqrt(a^2 - b^2)))*a*b/(a^2 - b^2)^(3/2) + (b*tan(1/2*d*x + 1/2*c) - a)/((a^2 - b^2)*(tan(1/2*d*x + 1/2*c)^2 - 1)))/d
Time = 20.07 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.84 \[ \int \frac {\sec (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2\,a\,b\,\mathrm {atan}\left (\frac {\frac {a\,b\,\left (2\,a^2\,b-2\,b^3\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}+\frac {2\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}}{2\,a\,b}\right )}{d\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}-\frac {\frac {2\,a}{a^2-b^2}-\frac {2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^2-b^2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int(tan(c + d*x)/(cos(c + d*x)*(a + b*sin(c + d*x))),x)
Output:
(2*a*b*atan(((a*b*(2*a^2*b - 2*b^3))/((a + b)^(3/2)*(a - b)^(3/2)) + (2*a^ 2*b*tan(c/2 + (d*x)/2)*(a^2 - b^2))/((a + b)^(3/2)*(a - b)^(3/2)))/(2*a*b) ))/(d*(a + b)^(3/2)*(a - b)^(3/2)) - ((2*a)/(a^2 - b^2) - (2*b*tan(c/2 + ( d*x)/2))/(a^2 - b^2))/(d*(tan(c/2 + (d*x)/2)^2 - 1))
Time = 0.18 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.59 \[ \int \frac {\sec (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \cos \left (d x +c \right ) a b -\cos \left (d x +c \right ) a^{3}+\cos \left (d x +c \right ) a \,b^{2}-\sin \left (d x +c \right ) a^{2} b +\sin \left (d x +c \right ) b^{3}+a^{3}-a \,b^{2}}{\cos \left (d x +c \right ) d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )} \] Input:
int(sec(d*x+c)*tan(d*x+c)/(a+b*sin(d*x+c)),x)
Output:
(2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*cos( c + d*x)*a*b - cos(c + d*x)*a**3 + cos(c + d*x)*a*b**2 - sin(c + d*x)*a**2 *b + sin(c + d*x)*b**3 + a**3 - a*b**2)/(cos(c + d*x)*d*(a**4 - 2*a**2*b** 2 + b**4))