Integrand size = 21, antiderivative size = 167 \[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {(a+3 b) \log (1-\sin (c+d x))}{4 (a+b)^3 d}+\frac {(a-3 b) \log (1+\sin (c+d x))}{4 (a-b)^3 d}+\frac {4 a b^3 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 d}+\frac {1}{4 (a+b)^2 d (1-\sin (c+d x))}-\frac {1}{4 (a-b)^2 d (1+\sin (c+d x))}-\frac {b^3}{\left (a^2-b^2\right )^2 d (a+b \sin (c+d x))} \] Output:
-1/4*(a+3*b)*ln(1-sin(d*x+c))/(a+b)^3/d+1/4*(a-3*b)*ln(1+sin(d*x+c))/(a-b) ^3/d+4*a*b^3*ln(a+b*sin(d*x+c))/(a^2-b^2)^3/d+1/4/(a+b)^2/d/(1-sin(d*x+c)) -1/4/(a-b)^2/d/(1+sin(d*x+c))-b^3/(a^2-b^2)^2/d/(a+b*sin(d*x+c))
Time = 2.16 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.33 \[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {a ((a-b) \log (1-\sin (c+d x))-(a+b) \log (1+\sin (c+d x))+2 b \log (a+b \sin (c+d x)))}{(a-b) (a+b)}+\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{a+b \sin (c+d x)}-b \left (-a^2-3 b^2\right ) \left (-\frac {\log (1-\sin (c+d x))}{2 b (a+b)^2}+\frac {\log (1+\sin (c+d x))}{2 (a-b)^2 b}-\frac {2 a \log (a+b \sin (c+d x))}{(a-b)^2 (a+b)^2}+\frac {1}{\left (a^2-b^2\right ) (a+b \sin (c+d x))}\right )}{2 \left (-a^2+b^2\right ) d} \] Input:
Integrate[Sec[c + d*x]^3/(a + b*Sin[c + d*x])^2,x]
Output:
((a*((a - b)*Log[1 - Sin[c + d*x]] - (a + b)*Log[1 + Sin[c + d*x]] + 2*b*L og[a + b*Sin[c + d*x]]))/((a - b)*(a + b)) + (Sec[c + d*x]^2*(b - a*Sin[c + d*x]))/(a + b*Sin[c + d*x]) - b*(-a^2 - 3*b^2)*(-1/2*Log[1 - Sin[c + d*x ]]/(b*(a + b)^2) + Log[1 + Sin[c + d*x]]/(2*(a - b)^2*b) - (2*a*Log[a + b* Sin[c + d*x]])/((a - b)^2*(a + b)^2) + 1/((a^2 - b^2)*(a + b*Sin[c + d*x]) )))/(2*(-a^2 + b^2)*d)
Time = 0.42 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (c+d x)^3 (a+b \sin (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3147 |
| \(\displaystyle \frac {b^3 \int \frac {1}{(a+b \sin (c+d x))^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 477 |
| \(\displaystyle \frac {\int \left (\frac {4 a b^4}{\left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac {b^4}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac {b^2}{4 (a+b)^2 (b-b \sin (c+d x))^2}+\frac {b^2}{4 (a-b)^2 (\sin (c+d x) b+b)^2}+\frac {(a+3 b) b}{4 (a+b)^3 (b-b \sin (c+d x))}+\frac {(a-3 b) b}{4 (a-b)^3 (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))}{b d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {b^4}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac {4 a b^4 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3}+\frac {b^2}{4 (a+b)^2 (b-b \sin (c+d x))}-\frac {b^2}{4 (a-b)^2 (b \sin (c+d x)+b)}-\frac {b (a+3 b) \log (b-b \sin (c+d x))}{4 (a+b)^3}+\frac {b (a-3 b) \log (b \sin (c+d x)+b)}{4 (a-b)^3}}{b d}\) |
Input:
Int[Sec[c + d*x]^3/(a + b*Sin[c + d*x])^2,x]
Output:
(-1/4*(b*(a + 3*b)*Log[b - b*Sin[c + d*x]])/(a + b)^3 + (4*a*b^4*Log[a + b *Sin[c + d*x]])/(a^2 - b^2)^3 + ((a - 3*b)*b*Log[b + b*Sin[c + d*x]])/(4*( a - b)^3) + b^2/(4*(a + b)^2*(b - b*Sin[c + d*x])) - b^4/((a^2 - b^2)^2*(a + b*Sin[c + d*x])) - b^2/(4*(a - b)^2*(b + 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 = 1.65 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.87
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{4 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-a -3 b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{3}}-\frac {b^{3}}{\left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )}+\frac {4 a \,b^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}-\frac {1}{4 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (a -3 b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{3}}}{d}\) | \(146\) |
| default | \(\frac {-\frac {1}{4 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-a -3 b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{3}}-\frac {b^{3}}{\left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )}+\frac {4 a \,b^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}-\frac {1}{4 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (a -3 b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{3}}}{d}\) | \(146\) |
| parallelrisch | \(\frac {8 \left (\frac {b \sin \left (3 d x +3 c \right )}{2}+\frac {b \sin \left (d x +c \right )}{2}+\cos \left (2 d x +2 c \right ) a +a \right ) a^{2} b^{3} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-\left (a -b \right )^{3} \left (\frac {b \sin \left (3 d x +3 c \right )}{2}+\frac {b \sin \left (d x +c \right )}{2}+\cos \left (2 d x +2 c \right ) a +a \right ) a \left (a +3 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (a +b \right ) \left (\left (a -3 b \right ) \left (\frac {b \sin \left (3 d x +3 c \right )}{2}+\frac {b \sin \left (d x +c \right )}{2}+\cos \left (2 d x +2 c \right ) a +a \right ) \left (a +b \right )^{2} a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 \left (a -b \right ) \left (\frac {3 \left (-a^{3} b +a \,b^{3}\right ) \cos \left (2 d x +2 c \right )}{2}+\frac {\left (-b^{2} a^{2}+3 b^{4}\right ) \sin \left (3 d x +3 c \right )}{2}+\left (a^{4}-\frac {3}{2} b^{2} a^{2}+\frac {3}{2} b^{4}\right ) \sin \left (d x +c \right )-\frac {5 a^{3} b}{2}+\frac {5 a \,b^{3}}{2}\right )\right )}{2 d \left (a -b \right )^{3} \left (\frac {b \sin \left (3 d x +3 c \right )}{2}+\frac {b \sin \left (d x +c \right )}{2}+\cos \left (2 d x +2 c \right ) a +a \right ) \left (a +b \right )^{3} a}\) | \(343\) |
| norman | \(\frac {\frac {4 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d \left (a^{2}-b^{2}\right )}+\frac {-a^{4}-b^{2} a^{2}-2 b^{4}}{2 b d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}-\frac {\left (-a^{4}+3 b^{2} a^{2}-6 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 d b \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}-\frac {\left (-a^{4}+3 b^{2} a^{2}-6 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2 d b \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}+\frac {\left (-a^{4}-b^{2} a^{2}-2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2 b d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}+\frac {\left (a -3 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {\left (a +3 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}+\frac {4 a \,b^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}\) | \(437\) |
| risch | \(\frac {3 i b x}{2 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {i a c}{2 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}+\frac {i a x}{2 a^{3}+6 a^{2} b +6 a \,b^{2}+2 b^{3}}+\frac {3 i b c}{2 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {3 i b x}{2 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {i a c}{2 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {8 i a \,b^{3} x}{a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}-\frac {i \left (-i a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}-3 i b^{3} {\mathrm e}^{5 i \left (d x +c \right )}-6 i a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}-2 i b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+2 \,{\mathrm e}^{4 i \left (d x +c \right )} a^{3}-2 a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-i a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}-3 i b^{3} {\mathrm e}^{i \left (d x +c \right )}-2 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+2 \,{\mathrm e}^{2 i \left (d x +c \right )} a \,b^{2}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \left (-i {\mathrm e}^{2 i \left (d x +c \right )} b +i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right ) d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}-\frac {i a x}{2 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {3 i b c}{2 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}-\frac {8 i a \,b^{3} c}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a}{2 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{2 \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 ) a}{2 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{2 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {4 a \,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 )}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}\) | \(751\) |
Input:
int(sec(d*x+c)^3/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(-1/4/(a+b)^2/(sin(d*x+c)-1)+1/4/(a+b)^3*(-a-3*b)*ln(sin(d*x+c)-1)-b^3 /(a+b)^2/(a-b)^2/(a+b*sin(d*x+c))+4*a*b^3/(a+b)^3/(a-b)^3*ln(a+b*sin(d*x+c ))-1/4/(a-b)^2/(1+sin(d*x+c))+1/4*(a-3*b)/(a-b)^3*ln(1+sin(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 381 vs. \(2 (157) = 314\).
Time = 0.16 (sec) , antiderivative size = 381, normalized size of antiderivative = 2.28 \[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {2 \, a^{4} b - 4 \, a^{2} b^{3} + 2 \, b^{5} + 2 \, {\left (a^{4} b + 2 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{2} - 16 \, {\left (a b^{4} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + a^{2} b^{3} \cos \left (d x + c\right )^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left ({\left (a^{4} b - 6 \, a^{2} b^{3} - 8 \, a b^{4} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (a^{5} - 6 \, a^{3} b^{2} - 8 \, a^{2} b^{3} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (a^{4} b - 6 \, a^{2} b^{3} + 8 \, a b^{4} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (a^{5} - 6 \, a^{3} b^{2} + 8 \, a^{2} b^{3} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate(sec(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="fricas")
Output:
-1/4*(2*a^4*b - 4*a^2*b^3 + 2*b^5 + 2*(a^4*b + 2*a^2*b^3 - 3*b^5)*cos(d*x + c)^2 - 16*(a*b^4*cos(d*x + c)^2*sin(d*x + c) + a^2*b^3*cos(d*x + c)^2)*l og(b*sin(d*x + c) + a) - ((a^4*b - 6*a^2*b^3 - 8*a*b^4 - 3*b^5)*cos(d*x + c)^2*sin(d*x + c) + (a^5 - 6*a^3*b^2 - 8*a^2*b^3 - 3*a*b^4)*cos(d*x + c)^2 )*log(sin(d*x + c) + 1) + ((a^4*b - 6*a^2*b^3 + 8*a*b^4 - 3*b^5)*cos(d*x + c)^2*sin(d*x + c) + (a^5 - 6*a^3*b^2 + 8*a^2*b^3 - 3*a*b^4)*cos(d*x + c)^ 2)*log(-sin(d*x + c) + 1) - 2*(a^5 - 2*a^3*b^2 + a*b^4)*sin(d*x + c))/((a^ 6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*d*cos(d*x + c)^2*sin(d*x + c) + (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c)^2)
\[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \] Input:
integrate(sec(d*x+c)**3/(a+b*sin(d*x+c))**2,x)
Output:
Integral(sec(c + d*x)**3/(a + b*sin(c + d*x))**2, x)
Time = 0.04 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.65 \[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {16 \, a b^{3} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac {{\left (a - 3 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (a + 3 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {2 \, {\left (2 \, a^{2} b + 2 \, b^{3} - {\left (a^{2} b + 3 \, b^{3}\right )} \sin \left (d x + c\right )^{2} - {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )\right )}}{a^{5} - 2 \, a^{3} b^{2} + a b^{4} - {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )^{3} - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )^{2} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )}}{4 \, d} \] Input:
integrate(sec(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="maxima")
Output:
1/4*(16*a*b^3*log(b*sin(d*x + c) + a)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) + (a - 3*b)*log(sin(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (a + 3 *b)*log(sin(d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 2*(2*a^2*b + 2 *b^3 - (a^2*b + 3*b^3)*sin(d*x + c)^2 - (a^3 - a*b^2)*sin(d*x + c))/(a^5 - 2*a^3*b^2 + a*b^4 - (a^4*b - 2*a^2*b^3 + b^5)*sin(d*x + c)^3 - (a^5 - 2*a ^3*b^2 + a*b^4)*sin(d*x + c)^2 + (a^4*b - 2*a^2*b^3 + b^5)*sin(d*x + c)))/ d
Time = 0.13 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.56 \[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {4 \, a b^{4} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b d - 3 \, a^{4} b^{3} d + 3 \, a^{2} b^{5} d - b^{7} d} + \frac {{\left (a - 3 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{4 \, {\left (a^{3} d - 3 \, a^{2} b d + 3 \, a b^{2} d - b^{3} d\right )}} - \frac {{\left (a + 3 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{4 \, {\left (a^{3} d + 3 \, a^{2} b d + 3 \, a b^{2} d + b^{3} d\right )}} - \frac {a^{2} b \sin \left (d x + c\right )^{2} + 3 \, b^{3} \sin \left (d x + c\right )^{2} + a^{3} \sin \left (d x + c\right ) - a b^{2} \sin \left (d x + c\right ) - 2 \, a^{2} b - 2 \, b^{3}}{2 \, {\left (a^{4} d - 2 \, a^{2} b^{2} d + b^{4} d\right )} {\left (b \sin \left (d x + c\right )^{3} + a \sin \left (d x + c\right )^{2} - b \sin \left (d x + c\right ) - a\right )}} \] Input:
integrate(sec(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="giac")
Output:
4*a*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) + 1/4*(a - 3*b)*log(abs(sin(d*x + c) + 1))/(a^3*d - 3*a^2*b*d + 3 *a*b^2*d - b^3*d) - 1/4*(a + 3*b)*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*(a^2*b*sin(d*x + c)^2 + 3*b^3*sin(d*x + c )^2 + a^3*sin(d*x + c) - a*b^2*sin(d*x + c) - 2*a^2*b - 2*b^3)/((a^4*d - 2 *a^2*b^2*d + b^4*d)*(b*sin(d*x + c)^3 + a*sin(d*x + c)^2 - b*sin(d*x + c) - a))
Time = 16.64 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.36 \[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {{\sin \left (c+d\,x\right )}^2\,\left (a^2\,b+3\,b^3\right )}{2\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {a^2\,b+b^3}{{\left (a^2-b^2\right )}^2}+\frac {a\,\sin \left (c+d\,x\right )}{2\,\left (a^2-b^2\right )}}{d\,\left (-b\,{\sin \left (c+d\,x\right )}^3-a\,{\sin \left (c+d\,x\right )}^2+b\,\sin \left (c+d\,x\right )+a\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {b}{2\,{\left (a+b\right )}^3}+\frac {1}{4\,{\left (a+b\right )}^2}\right )}{d}+\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (a-3\,b\right )}{4\,d\,{\left (a-b\right )}^3}+\frac {4\,a\,b^3\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{d\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )} \] Input:
int(1/(cos(c + d*x)^3*(a + b*sin(c + d*x))^2),x)
Output:
((sin(c + d*x)^2*(a^2*b + 3*b^3))/(2*(a^4 + b^4 - 2*a^2*b^2)) - (a^2*b + b ^3)/(a^2 - b^2)^2 + (a*sin(c + d*x))/(2*(a^2 - b^2)))/(d*(a + b*sin(c + d* x) - a*sin(c + d*x)^2 - b*sin(c + d*x)^3)) - (log(sin(c + d*x) - 1)*(b/(2* (a + b)^3) + 1/(4*(a + b)^2)))/d + (log(sin(c + d*x) + 1)*(a - 3*b))/(4*d* (a - b)^3) + (4*a*b^3*log(a + b*sin(c + d*x)))/(d*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2))
Time = 0.19 (sec) , antiderivative size = 1258, normalized size of antiderivative = 7.53 \[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^3/(a+b*sin(d*x+c))^2,x)
Output:
( - log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*a**4*b**2 + 6*log(tan((c + d *x)/2) - 1)*sin(c + d*x)**3*a**2*b**4 - 8*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*a*b**5 + 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*b**6 - log( tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**5*b + 6*log(tan((c + d*x)/2) - 1) *sin(c + d*x)**2*a**3*b**3 - 8*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a **2*b**4 + 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**5 + log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a**4*b**2 - 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a**2*b**4 + 8*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a*b**5 - 3*lo g(tan((c + d*x)/2) - 1)*sin(c + d*x)*b**6 + log(tan((c + d*x)/2) - 1)*a**5 *b - 6*log(tan((c + d*x)/2) - 1)*a**3*b**3 + 8*log(tan((c + d*x)/2) - 1)*a **2*b**4 - 3*log(tan((c + d*x)/2) - 1)*a*b**5 + log(tan((c + d*x)/2) + 1)* sin(c + d*x)**3*a**4*b**2 - 6*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3*a* *2*b**4 - 8*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3*a*b**5 - 3*log(tan(( c + d*x)/2) + 1)*sin(c + d*x)**3*b**6 + log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**5*b - 6*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**3*b**3 - 8 *log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b**4 - 3*log(tan((c + d*x) /2) + 1)*sin(c + d*x)**2*a*b**5 - log(tan((c + d*x)/2) + 1)*sin(c + d*x)*a **4*b**2 + 6*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*a**2*b**4 + 8*log(tan( (c + d*x)/2) + 1)*sin(c + d*x)*a*b**5 + 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*b**6 - log(tan((c + d*x)/2) + 1)*a**5*b + 6*log(tan((c + d*x)/2)...