Integrand size = 28, antiderivative size = 108 \[ \int \frac {\sec ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\log (1-\cos (c+d x))}{2 (a+b) d}+\frac {a \log (\cos (c+d x))}{b^2 d}+\frac {\log (1+\cos (c+d x))}{2 (a-b) d}-\frac {a^3 \log (b+a \cos (c+d x))}{b^2 \left (a^2-b^2\right ) d}+\frac {\sec (c+d x)}{b d} \] Output:
1/2*ln(1-cos(d*x+c))/(a+b)/d+a*ln(cos(d*x+c))/b^2/d+1/2*ln(1+cos(d*x+c))/( a-b)/d-a^3*ln(b+a*cos(d*x+c))/b^2/(a^2-b^2)/d+sec(d*x+c)/b/d
Time = 0.39 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.85 \[ \int \frac {\sec ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\frac {\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a-b}+\frac {a \log (\cos (c+d x))}{b^2}+\frac {a^3 \log (b+a \cos (c+d x))}{-a^2 b^2+b^4}+\frac {\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a+b}+\frac {\sec (c+d x)}{b}}{d} \] Input:
Integrate[Sec[c + d*x]^3/(a*Sin[c + d*x] + b*Tan[c + d*x]),x]
Output:
(Log[Cos[(c + d*x)/2]]/(a - b) + (a*Log[Cos[c + d*x]])/b^2 + (a^3*Log[b + a*Cos[c + d*x]])/(-(a^2*b^2) + b^4) + Log[Sin[(c + d*x)/2]]/(a + b) + Sec[ c + d*x]/b)/d
Time = 0.52 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 4897, 3042, 3316, 27, 615, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (c+d x)^3}{a \sin (c+d x)+b \tan (c+d x)}dx\) |
| \(\Big \downarrow \) 4897 |
| \(\displaystyle \int \frac {\csc (c+d x) \sec ^2(c+d x)}{a \cos (c+d x)+b}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin \left (c+d x-\frac {\pi }{2}\right )^2 \cos \left (c+d x-\frac {\pi }{2}\right ) \left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle -\frac {a \int \frac {\sec ^2(c+d x)}{(b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a^3 \int \frac {\sec ^2(c+d x)}{a^2 (b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))}{d}\) |
| \(\Big \downarrow \) 615 |
| \(\displaystyle -\frac {a^3 \int \left (\frac {\sec ^2(c+d x)}{a^4 b}-\frac {\sec (c+d x)}{a^3 b^2}+\frac {1}{2 a^3 (a+b) (a-a \cos (c+d x))}-\frac {1}{2 a^3 (a-b) (\cos (c+d x) a+a)}-\frac {1}{b^2 (b-a) (a+b) (b+a \cos (c+d x))}\right )d(a \cos (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^3 \left (-\frac {\sec (c+d x)}{a^3 b}-\frac {\log (a-a \cos (c+d x))}{2 a^3 (a+b)}-\frac {\log (a \cos (c+d x)+a)}{2 a^3 (a-b)}-\frac {\log (a \cos (c+d x))}{a^2 b^2}+\frac {\log (a \cos (c+d x)+b)}{b^2 \left (a^2-b^2\right )}\right )}{d}\) |
Input:
Int[Sec[c + d*x]^3/(a*Sin[c + d*x] + b*Tan[c + d*x]),x]
Output:
-((a^3*(-(Log[a*Cos[c + d*x]]/(a^2*b^2)) - Log[a - a*Cos[c + d*x]]/(2*a^3* (a + b)) - Log[a + a*Cos[c + d*x]]/(2*a^3*(a - b)) + Log[b + a*Cos[c + d*x ]]/(b^2*(a^2 - b^2)) - Sec[c + d*x]/(a^3*b)))/d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
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 = 2.30 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.92
| method | result | size |
| derivativedivides | \(\frac {\frac {a \ln \left (\cos \left (d x +c \right )\right )}{b^{2}}+\frac {1}{b \cos \left (d x +c \right )}+\frac {\ln \left (1+\cos \left (d x +c \right )\right )}{2 a -2 b}+\frac {\ln \left (-1+\cos \left (d x +c \right )\right )}{2 a +2 b}-\frac {a^{3} \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right ) b^{2}}}{d}\) | \(99\) |
| default | \(\frac {\frac {a \ln \left (\cos \left (d x +c \right )\right )}{b^{2}}+\frac {1}{b \cos \left (d x +c \right )}+\frac {\ln \left (1+\cos \left (d x +c \right )\right )}{2 a -2 b}+\frac {\ln \left (-1+\cos \left (d x +c \right )\right )}{2 a +2 b}-\frac {a^{3} \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right ) b^{2}}}{d}\) | \(99\) |
| risch | \(-\frac {i x}{a -b}-\frac {i c}{d \left (a -b \right )}-\frac {i x}{a +b}-\frac {i c}{d \left (a +b \right )}-\frac {2 i a x}{b^{2}}-\frac {2 i a c}{b^{2} d}+\frac {2 i a^{3} x}{b^{2} \left (a^{2}-b^{2}\right )}+\frac {2 i a^{3} c}{b^{2} d \left (a^{2}-b^{2}\right )}+\frac {2 \,{\mathrm e}^{i \left (d x +c \right )}}{d b \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \left (a -b \right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \left (a +b \right )}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b^{2} d}-\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{b^{2} d \left (a^{2}-b^{2}\right )}\) | \(255\) |
Input:
int(sec(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(a/b^2*ln(cos(d*x+c))+1/b/cos(d*x+c)+1/(2*a-2*b)*ln(1+cos(d*x+c))+1/(2 *a+2*b)*ln(-1+cos(d*x+c))-a^3/(a+b)/(a-b)/b^2*ln(b+a*cos(d*x+c)))
Time = 0.18 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.36 \[ \int \frac {\sec ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {2 \, a^{3} \cos \left (d x + c\right ) \log \left (a \cos \left (d x + c\right ) + b\right ) - 2 \, a^{2} b + 2 \, b^{3} - 2 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) - {\left (a b^{2} + b^{3}\right )} \cos \left (d x + c\right ) \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (a b^{2} - b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} d \cos \left (d x + c\right )} \] Input:
integrate(sec(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="fricas")
Output:
-1/2*(2*a^3*cos(d*x + c)*log(a*cos(d*x + c) + b) - 2*a^2*b + 2*b^3 - 2*(a^ 3 - a*b^2)*cos(d*x + c)*log(-cos(d*x + c)) - (a*b^2 + b^3)*cos(d*x + c)*lo g(1/2*cos(d*x + c) + 1/2) - (a*b^2 - b^3)*cos(d*x + c)*log(-1/2*cos(d*x + c) + 1/2))/((a^2*b^2 - b^4)*d*cos(d*x + c))
\[ \int \frac {\sec ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}}\, dx \] Input:
integrate(sec(d*x+c)**3/(a*sin(d*x+c)+b*tan(d*x+c)),x)
Output:
Integral(sec(c + d*x)**3/(a*sin(c + d*x) + b*tan(c + d*x)), x)
Time = 0.04 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.46 \[ \int \frac {\sec ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {\frac {a^{3} \log \left (a + b - \frac {{\left (a - b\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{2} b^{2} - b^{4}} - \frac {a \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{b^{2}} - \frac {a \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{b^{2}} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a + b} - \frac {2}{b - \frac {b \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}}{d} \] Input:
integrate(sec(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="maxima")
Output:
-(a^3*log(a + b - (a - b)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/(a^2*b^2 - b^4) - a*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/b^2 - a*log(sin(d*x + c) /(cos(d*x + c) + 1) - 1)/b^2 - log(sin(d*x + c)/(cos(d*x + c) + 1))/(a + b ) - 2/(b - b*sin(d*x + c)^2/(cos(d*x + c) + 1)^2))/d
Time = 0.15 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.02 \[ \int \frac {\sec ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {a^{4} \log \left ({\left | a \cos \left (d x + c\right ) + b \right |}\right )}{a^{3} b^{2} d - a b^{4} d} + \frac {\log \left ({\left | \cos \left (d x + c\right ) + 1 \right |}\right )}{2 \, {\left (a d - b d\right )}} + \frac {\log \left ({\left | \cos \left (d x + c\right ) - 1 \right |}\right )}{2 \, {\left (a d + b d\right )}} + \frac {a \log \left ({\left | \cos \left (d x + c\right ) \right |}\right )}{b^{2} d} + \frac {1}{b d \cos \left (d x + c\right )} \] Input:
integrate(sec(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="giac")
Output:
-a^4*log(abs(a*cos(d*x + c) + b))/(a^3*b^2*d - a*b^4*d) + 1/2*log(abs(cos( d*x + c) + 1))/(a*d - b*d) + 1/2*log(abs(cos(d*x + c) - 1))/(a*d + b*d) + a*log(abs(cos(d*x + c)))/(b^2*d) + 1/(b*d*cos(d*x + c))
Time = 17.61 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.09 \[ \int \frac {\sec ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d\,\left (a+b\right )}-\frac {2}{b\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}+\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}{b^2\,d}-\frac {a^3\,\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{b^2\,d\,\left (a^2-b^2\right )} \] Input:
int(1/(cos(c + d*x)^3*(a*sin(c + d*x) + b*tan(c + d*x))),x)
Output:
log(tan(c/2 + (d*x)/2))/(d*(a + b)) - 2/(b*d*(tan(c/2 + (d*x)/2)^2 - 1)) + (a*log(tan(c/2 + (d*x)/2)^2 - 1))/(b^2*d) - (a^3*log(a + b - a*tan(c/2 + (d*x)/2)^2 + b*tan(c/2 + (d*x)/2)^2))/(b^2*d*(a^2 - b^2))
Time = 0.17 (sec) , antiderivative size = 239, normalized size of antiderivative = 2.21 \[ \int \frac {\sec ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{3}-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a \,b^{2}+\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{3}-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a \,b^{2}-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right ) a^{3}+\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \,b^{2}-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}-\cos \left (d x +c \right ) a^{2} b +\cos \left (d x +c \right ) b^{3}+a^{2} b -b^{3}}{\cos \left (d x +c \right ) b^{2} d \left (a^{2}-b^{2}\right )} \] Input:
int(sec(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c)),x)
Output:
(cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**3 - cos(c + d*x)*log(tan((c + d *x)/2) - 1)*a*b**2 + cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**3 - cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a*b**2 - cos(c + d*x)*log(tan((c + d*x)/2) **2*a - tan((c + d*x)/2)**2*b - a - b)*a**3 + cos(c + d*x)*log(tan((c + d* x)/2))*a*b**2 - cos(c + d*x)*log(tan((c + d*x)/2))*b**3 - cos(c + d*x)*a** 2*b + cos(c + d*x)*b**3 + a**2*b - b**3)/(cos(c + d*x)*b**2*d*(a**2 - b**2 ))