Integrand size = 19, antiderivative size = 145 \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {\log (1-\sin (c+d x))}{2 (a+b)^3 d}+\frac {\log (1+\sin (c+d x))}{2 (a-b)^3 d}-\frac {b \left (3 a^2+b^2\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 d}+\frac {b}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}+\frac {2 a b}{\left (a^2-b^2\right )^2 d (a+b \sin (c+d x))} \] Output:
-1/2*ln(1-sin(d*x+c))/(a+b)^3/d+1/2*ln(1+sin(d*x+c))/(a-b)^3/d-b*(3*a^2+b^ 2)*ln(a+b*sin(d*x+c))/(a^2-b^2)^3/d+1/2*b/(a^2-b^2)/d/(a+b*sin(d*x+c))^2+2 *a*b/(a^2-b^2)^2/d/(a+b*sin(d*x+c))
Time = 0.63 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.93 \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {b \left (-\frac {\log (1-\sin (c+d x))}{b (a+b)^3}+\frac {\log (1+\sin (c+d x))}{(a-b)^3 b}-\frac {2 \left (3 a^2+b^2\right ) \log (a+b \sin (c+d x))}{(a-b)^3 (a+b)^3}+\frac {1}{\left (a^2-b^2\right ) (a+b \sin (c+d x))^2}+\frac {4 a}{(a-b)^2 (a+b)^2 (a+b \sin (c+d x))}\right )}{2 d} \] Input:
Integrate[Sec[c + d*x]/(a + b*Sin[c + d*x])^3,x]
Output:
(b*(-(Log[1 - Sin[c + d*x]]/(b*(a + b)^3)) + Log[1 + Sin[c + d*x]]/((a - b )^3*b) - (2*(3*a^2 + b^2)*Log[a + b*Sin[c + d*x]])/((a - b)^3*(a + b)^3) + 1/((a^2 - b^2)*(a + b*Sin[c + d*x])^2) + (4*a)/((a - b)^2*(a + b)^2*(a + b*Sin[c + d*x]))))/(2*d)
Time = 0.38 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3147, 477, 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.
| \(\displaystyle \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (c+d x) (a+b \sin (c+d x))^3}dx\) |
| \(\Big \downarrow \) 3147 |
| \(\displaystyle \frac {b \int \frac {1}{(a+b \sin (c+d x))^3 \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 477 |
| \(\displaystyle \frac {\int \left (-\frac {\left (3 a^2+b^2\right ) b^2}{\left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac {2 a b^2}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}-\frac {b^2}{\left (a^2-b^2\right ) (a+b \sin (c+d x))^3}+\frac {b}{2 (a+b)^3 (b-b \sin (c+d x))}+\frac {b}{2 (a-b)^3 (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))}{b d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {2 a b^2}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac {b^2}{2 \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}-\frac {b^2 \left (3 a^2+b^2\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3}-\frac {b \log (b-b \sin (c+d x))}{2 (a+b)^3}+\frac {b \log (b \sin (c+d x)+b)}{2 (a-b)^3}}{b d}\) |
Input:
Int[Sec[c + d*x]/(a + b*Sin[c + d*x])^3,x]
Output:
(-1/2*(b*Log[b - b*Sin[c + d*x]])/(a + b)^3 - (b^2*(3*a^2 + b^2)*Log[a + b *Sin[c + d*x]])/(a^2 - b^2)^3 + (b*Log[b + b*Sin[c + d*x]])/(2*(a - b)^3) + b^2/(2*(a^2 - b^2)*(a + b*Sin[c + d*x])^2) + (2*a*b^2)/((a^2 - b^2)^2*(a + b*Sin[c + d*x])))/(b*d)
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ a^p Int[ExpandIntegrand[(c + d*x)^n*(1 - Rt[-b/a, 2]*x)^p*(1 + Rt[-b/a, 2 ]*x)^p, x], x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[p, 0] && IntegerQ[n] & & NiceSqrtQ[-b/a] && !FractionalPowerFactorQ[Rt[-b/a, 2]]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 0.89 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.90
| method | result | size |
| derivativedivides | \(\frac {\frac {b}{2 \left (a -b \right ) \left (a +b \right ) \left (a +b \sin \left (d x +c \right )\right )^{2}}+\frac {2 a b}{\left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )}-\frac {b \left (3 a^{2}+b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 \left (a -b \right )^{3}}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{3}}}{d}\) | \(130\) |
| default | \(\frac {\frac {b}{2 \left (a -b \right ) \left (a +b \right ) \left (a +b \sin \left (d x +c \right )\right )^{2}}+\frac {2 a b}{\left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )}-\frac {b \left (3 a^{2}+b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 \left (a -b \right )^{3}}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{3}}}{d}\) | \(130\) |
| parallelrisch | \(\frac {-3 \left (a^{2}+\frac {b^{2}}{3}\right ) a^{2} \left (\frac {b^{2}}{2}-\frac {b^{2} \cos \left (2 d x +2 c \right )}{2}+2 a b \sin \left (d x +c \right )+a^{2}\right ) b \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-\left (a -b \right )^{3} a^{2} \left (\frac {b^{2}}{2}-\frac {b^{2} \cos \left (2 d x +2 c \right )}{2}+2 a b \sin \left (d x +c \right )+a^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (a +b \right ) \left (\left (a +b \right )^{2} a^{2} \left (\frac {b^{2}}{2}-\frac {b^{2} \cos \left (2 d x +2 c \right )}{2}+2 a b \sin \left (d x +c \right )+a^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-3 \left (a -b \right ) \left (\frac {\left (-5 a^{2} b +b^{3}\right ) \cos \left (2 d x +2 c \right )}{12}+\left (a^{3}-\frac {1}{3} a \,b^{2}\right ) \sin \left (d x +c \right )+\frac {5 a^{2} b}{12}-\frac {b^{3}}{12}\right ) b^{2}\right )}{d \left (a -b \right )^{3} \left (a +b \right )^{3} a^{2} \left (\frac {b^{2}}{2}-\frac {b^{2} \cos \left (2 d x +2 c \right )}{2}+2 a b \sin \left (d x +c \right )+a^{2}\right )}\) | \(307\) |
| norman | \(\frac {\frac {6 b^{2} a^{2}-2 b^{4}}{4 d b \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}+\frac {\left (6 b^{2} a^{2}-2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4 d b \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}+\frac {\left (6 a^{3} b^{2}-10 a \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 d a b \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}-\frac {\left (3 a^{2}+b^{2}\right ) b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) d}\) | \(323\) |
| risch | \(\frac {i x}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}+\frac {i c}{\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}-\frac {i x}{a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}}-\frac {i c}{d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {6 i b \,a^{2} x}{a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}+\frac {6 i b \,a^{2} c}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) d}+\frac {2 i b^{3} x}{a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}+\frac {2 i b^{3} c}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) d}+\frac {2 b \left (2 i a b \,{\mathrm e}^{3 i \left (d x +c \right )}-2 i a b \,{\mathrm e}^{i \left (d x +c \right )}-5 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+b^{2} {\mathrm e}^{2 i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )} b -b +2 i {\mathrm e}^{i \left (d x +c \right )} a \right )^{2} \left (-a^{2}+b^{2}\right )^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {3 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right ) a^{2}}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) d}-\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) d}\) | \(565\) |
Input:
int(sec(d*x+c)/(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d*(1/2*b/(a-b)/(a+b)/(a+b*sin(d*x+c))^2+2*a*b/(a+b)^2/(a-b)^2/(a+b*sin(d *x+c))-b*(3*a^2+b^2)/(a+b)^3/(a-b)^3*ln(a+b*sin(d*x+c))+1/2/(a-b)^3*ln(1+s in(d*x+c))-1/2/(a+b)^3*ln(sin(d*x+c)-1))
Leaf count of result is larger than twice the leaf count of optimal. 462 vs. \(2 (139) = 278\).
Time = 0.20 (sec) , antiderivative size = 462, normalized size of antiderivative = 3.19 \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {5 \, a^{4} b - 6 \, a^{2} b^{3} + b^{5} - 2 \, {\left (3 \, a^{4} b + 4 \, a^{2} b^{3} + b^{5} - {\left (3 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left (a^{5} + 3 \, a^{4} b + 4 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5} - {\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{4} b + 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{5} - 3 \, a^{4} b + 4 \, a^{3} b^{2} - 4 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5} - {\left (a^{3} b^{2} - 3 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{4} b - 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} - a b^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, {\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} d \sin \left (d x + c\right ) - {\left (a^{8} - 2 \, a^{6} b^{2} + 2 \, a^{2} b^{6} - b^{8}\right )} d\right )}} \] Input:
integrate(sec(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="fricas")
Output:
-1/2*(5*a^4*b - 6*a^2*b^3 + b^5 - 2*(3*a^4*b + 4*a^2*b^3 + b^5 - (3*a^2*b^ 3 + b^5)*cos(d*x + c)^2 + 2*(3*a^3*b^2 + a*b^4)*sin(d*x + c))*log(b*sin(d* x + c) + a) + (a^5 + 3*a^4*b + 4*a^3*b^2 + 4*a^2*b^3 + 3*a*b^4 + b^5 - (a^ 3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*cos(d*x + c)^2 + 2*(a^4*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*sin(d*x + c))*log(sin(d*x + c) + 1) - (a^5 - 3*a^4*b + 4*a^3*b^2 - 4*a^2*b^3 + 3*a*b^4 - b^5 - (a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*cos(d*x + c)^2 + 2*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*sin(d*x + c))*log(-sin(d*x + c) + 1) + 4*(a^3*b^2 - a*b^4)*sin(d*x + c))/((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*cos(d*x + c)^2 - 2*(a^7*b - 3*a^5*b^3 + 3* a^3*b^5 - a*b^7)*d*sin(d*x + c) - (a^8 - 2*a^6*b^2 + 2*a^2*b^6 - b^8)*d)
\[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\int \frac {\sec {\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{3}}\, dx \] Input:
integrate(sec(d*x+c)/(a+b*sin(d*x+c))**3,x)
Output:
Integral(sec(c + d*x)/(a + b*sin(c + d*x))**3, x)
Time = 0.04 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.54 \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {\frac {2 \, {\left (3 \, a^{2} b + b^{3}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac {4 \, a b^{2} \sin \left (d x + c\right ) + 5 \, a^{2} b - b^{3}}{a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4} + {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \sin \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \sin \left (d x + c\right )} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}}}{2 \, d} \] Input:
integrate(sec(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="maxima")
Output:
-1/2*(2*(3*a^2*b + b^3)*log(b*sin(d*x + c) + a)/(a^6 - 3*a^4*b^2 + 3*a^2*b ^4 - b^6) - (4*a*b^2*sin(d*x + c) + 5*a^2*b - b^3)/(a^6 - 2*a^4*b^2 + a^2* b^4 + (a^4*b^2 - 2*a^2*b^4 + b^6)*sin(d*x + c)^2 + 2*(a^5*b - 2*a^3*b^3 + a*b^5)*sin(d*x + c)) - log(sin(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^ 3) + log(sin(d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3))/d
Time = 0.13 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.46 \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {{\left (3 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b d - 3 \, a^{4} b^{3} d + 3 \, a^{2} b^{5} d - b^{7} d} - \frac {\log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{2 \, {\left (a^{3} d + 3 \, a^{2} b d + 3 \, a b^{2} d + b^{3} d\right )}} + \frac {\log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{2 \, {\left (a^{3} d - 3 \, a^{2} b d + 3 \, a b^{2} d - b^{3} d\right )}} + \frac {5 \, a^{4} b - 6 \, a^{2} b^{3} + b^{5} + 4 \, {\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )}{2 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{2} {\left (a + b\right )}^{3} {\left (a - b\right )}^{3} d} \] Input:
integrate(sec(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="giac")
Output:
-(3*a^2*b^2 + b^4)*log(abs(b*sin(d*x + c) + a))/(a^6*b*d - 3*a^4*b^3*d + 3 *a^2*b^5*d - b^7*d) - 1/2*log(abs(-sin(d*x + c) + 1))/(a^3*d + 3*a^2*b*d + 3*a*b^2*d + b^3*d) + 1/2*log(abs(-sin(d*x + c) - 1))/(a^3*d - 3*a^2*b*d + 3*a*b^2*d - b^3*d) + 1/2*(5*a^4*b - 6*a^2*b^3 + b^5 + 4*(a^3*b^2 - a*b^4) *sin(d*x + c))/((b*sin(d*x + c) + a)^2*(a + b)^3*(a - b)^3*d)
Time = 16.60 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.17 \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (\frac {1}{2\,{\left (a+b\right )}^3}-\frac {1}{2\,{\left (a-b\right )}^3}\right )}{d}+\frac {\frac {5\,a^2\,b-b^3}{2\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {2\,a\,b^2\,\sin \left (c+d\,x\right )}{a^4-2\,a^2\,b^2+b^4}}{d\,\left (a^2+2\,a\,b\,\sin \left (c+d\,x\right )+b^2\,{\sin \left (c+d\,x\right )}^2\right )}+\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )}{2\,d\,{\left (a-b\right )}^3}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )}{2\,d\,{\left (a+b\right )}^3} \] Input:
int(1/(cos(c + d*x)*(a + b*sin(c + d*x))^3),x)
Output:
(log(a + b*sin(c + d*x))*(1/(2*(a + b)^3) - 1/(2*(a - b)^3)))/d + ((5*a^2* b - b^3)/(2*(a^4 + b^4 - 2*a^2*b^2)) + (2*a*b^2*sin(c + d*x))/(a^4 + b^4 - 2*a^2*b^2))/(d*(a^2 + b^2*sin(c + d*x)^2 + 2*a*b*sin(c + d*x))) + log(sin (c + d*x) + 1)/(2*d*(a - b)^3) - log(sin(c + d*x) - 1)/(2*d*(a + b)^3)
Time = 0.18 (sec) , antiderivative size = 995, normalized size of antiderivative = 6.86 \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)/(a+b*sin(d*x+c))^3,x)
Output:
( - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3*b**2 + 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b**3 - 6*log(tan((c + d*x)/2) - 1)*sin( c + d*x)**2*a*b**4 + 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**5 - 4* log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a**4*b + 12*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a**3*b**2 - 12*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a**2 *b**3 + 4*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a*b**4 - 2*log(tan((c + d *x)/2) - 1)*a**5 + 6*log(tan((c + d*x)/2) - 1)*a**4*b - 6*log(tan((c + d*x )/2) - 1)*a**3*b**2 + 2*log(tan((c + d*x)/2) - 1)*a**2*b**3 + 2*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**3*b**2 + 6*log(tan((c + d*x)/2) + 1)*si n(c + d*x)**2*a**2*b**3 + 6*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b* *4 + 2*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b**5 + 4*log(tan((c + d*x )/2) + 1)*sin(c + d*x)*a**4*b + 12*log(tan((c + d*x)/2) + 1)*sin(c + d*x)* a**3*b**2 + 12*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*a**2*b**3 + 4*log(ta n((c + d*x)/2) + 1)*sin(c + d*x)*a*b**4 + 2*log(tan((c + d*x)/2) + 1)*a**5 + 6*log(tan((c + d*x)/2) + 1)*a**4*b + 6*log(tan((c + d*x)/2) + 1)*a**3*b **2 + 2*log(tan((c + d*x)/2) + 1)*a**2*b**3 - 6*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)**2*a**2*b**3 - 2*log(tan((c + d*x )/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)**2*b**5 - 12*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)*a**3*b**2 - 4*log (tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)*a*b**4 ...