Integrand size = 21, antiderivative size = 123 \[ \int \frac {\sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {(a+2 b) \log (1-\sin (c+d x))}{4 (a+b)^2 d}+\frac {(a-2 b) \log (1+\sin (c+d x))}{4 (a-b)^2 d}+\frac {b^3 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2 d}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 \left (a^2-b^2\right ) d} \]
-1/4*(a+2*b)*ln(1-sin(d*x+c))/(a+b)^2/d+1/4*(a-2*b)*ln(1+sin(d*x+c))/(a-b) ^2/d+b^3*ln(a+b*sin(d*x+c))/(a^2-b^2)^2/d-1/2*sec(d*x+c)^2*(b-a*sin(d*x+c) )/(a^2-b^2)/d
Time = 0.60 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.93 \[ \int \frac {\sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {(a+2 b) \log (1-\sin (c+d x))}{(a+b)^2}-\frac {(a-2 b) \log (1+\sin (c+d x))}{(a-b)^2}-\frac {4 b^3 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2}+\frac {1}{(a+b) (-1+\sin (c+d x))}+\frac {1}{(a-b) (1+\sin (c+d x))}}{4 d} \]
-1/4*(((a + 2*b)*Log[1 - Sin[c + d*x]])/(a + b)^2 - ((a - 2*b)*Log[1 + Sin [c + d*x]])/(a - b)^2 - (4*b^3*Log[a + b*Sin[c + d*x]])/(a^2 - b^2)^2 + 1/ ((a + b)*(-1 + Sin[c + d*x])) + 1/((a - b)*(1 + Sin[c + d*x])))/d
Time = 0.37 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.14, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (c+d x)^3 (a+b \sin (c+d x))}dx\) |
| \(\Big \downarrow \) 3147 |
| \(\displaystyle \frac {b^3 \int \frac {1}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 477 |
| \(\displaystyle \frac {\int \left (\frac {b^4}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac {b^2}{4 (a+b) (b-b \sin (c+d x))^2}+\frac {b^2}{4 (a-b) (\sin (c+d x) b+b)^2}+\frac {(a+2 b) b}{4 (a+b)^2 (b-b \sin (c+d x))}+\frac {(a-2 b) b}{4 (a-b)^2 (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))}{b d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {b^4 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2}+\frac {b^2}{4 (a+b) (b-b \sin (c+d x))}-\frac {b^2}{4 (a-b) (b \sin (c+d x)+b)}-\frac {b (a+2 b) \log (b-b \sin (c+d x))}{4 (a+b)^2}+\frac {b (a-2 b) \log (b \sin (c+d x)+b)}{4 (a-b)^2}}{b d}\) |
(-1/4*(b*(a + 2*b)*Log[b - b*Sin[c + d*x]])/(a + b)^2 + (b^4*Log[a + b*Sin [c + d*x]])/(a^2 - b^2)^2 + ((a - 2*b)*b*Log[b + b*Sin[c + d*x]])/(4*(a - b)^2) + b^2/(4*(a + b)*(b - b*Sin[c + d*x])) - b^2/(4*(a - b)*(b + b*Sin[c + d*x])))/(b*d)
3.5.29.3.1 Defintions of rubi rules used
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 = 1.04 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.98
| method | result | size |
| derivativedivides | \(\frac {\frac {b^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (a -2 b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}-\frac {1}{\left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-a -2 b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{2}}}{d}\) | \(121\) |
| default | \(\frac {\frac {b^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (a -2 b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}-\frac {1}{\left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-a -2 b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{2}}}{d}\) | \(121\) |
| parallelrisch | \(\frac {2 b^{3} \left (\cos \left (2 d x +2 c \right )+1\right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-\left (a +2 b \right ) \left (a -b \right )^{2} \left (\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (\left (a +b \right ) \left (a -2 b \right ) \left (\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 \left (a \sin \left (d x +c \right )+\frac {b \cos \left (2 d x +2 c \right )}{2}-\frac {b}{2}\right ) \left (a -b \right )\right ) \left (a +b \right )}{2 \left (a -b \right )^{2} \left (a +b \right )^{2} d \left (\cos \left (2 d x +2 c \right )+1\right )}\) | \(180\) |
| norman | \(\frac {\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (a^{2}-b^{2}\right )}+\frac {a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{2}-b^{2}\right )}-\frac {2 b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{2}-b^{2}\right )}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {b^{3} \ln \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {\left (a -2 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (a +2 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 \left (a^{2}+2 a b +b^{2}\right ) d}\) | \(221\) |
| risch | \(\frac {i b x}{a^{2}-2 a b +b^{2}}+\frac {i b c}{d \left (a^{2}-2 a b +b^{2}\right )}+\frac {i a x}{2 a^{2}+4 a b +2 b^{2}}+\frac {i a c}{2 \left (a^{2}+2 a b +b^{2}\right ) d}-\frac {2 i b^{3} x}{a^{4}-2 a^{2} b^{2}+b^{4}}-\frac {i a x}{2 \left (a^{2}-2 a b +b^{2}\right )}+\frac {i b x}{a^{2}+2 a b +b^{2}}-\frac {i a c}{2 d \left (a^{2}-2 a b +b^{2}\right )}+\frac {i b c}{\left (a^{2}+2 a b +b^{2}\right ) d}-\frac {2 i b^{3} c}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {i a \,{\mathrm e}^{3 i \left (d x +c \right )}-i a \,{\mathrm e}^{i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}}{d \left (-a^{2}+b^{2}\right ) \left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a}{2 d \left (a^{2}-2 a b +b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{d \left (a^{2}-2 a b +b^{2}\right )}-\frac {\ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right ) a}{2 \left (a^{2}+2 a b +b^{2}\right ) d}-\frac {\ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right ) b}{\left (a^{2}+2 a b +b^{2}\right ) d}+\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}\) | \(456\) |
1/d*(b^3/(a+b)^2/(a-b)^2*ln(a+b*sin(d*x+c))-1/(4*a-4*b)/(1+sin(d*x+c))+1/4 *(a-2*b)/(a-b)^2*ln(1+sin(d*x+c))-1/(4*a+4*b)/(sin(d*x+c)-1)+1/4/(a+b)^2*( -a-2*b)*ln(sin(d*x+c)-1))
Time = 0.34 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.24 \[ \int \frac {\sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {4 \, b^{3} \cos \left (d x + c\right )^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{3} - 3 \, a b^{2} + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, a^{2} b + 2 \, b^{3} + 2 \, {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \cos \left (d x + c\right )^{2}} \]
1/4*(4*b^3*cos(d*x + c)^2*log(b*sin(d*x + c) + a) + (a^3 - 3*a*b^2 - 2*b^3 )*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (a^3 - 3*a*b^2 + 2*b^3)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) - 2*a^2*b + 2*b^3 + 2*(a^3 - a*b^2)*sin(d*x + c))/((a^4 - 2*a^2*b^2 + b^4)*d*cos(d*x + c)^2)
\[ \int \frac {\sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]
Time = 0.18 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.13 \[ \int \frac {\sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {4 \, b^{3} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (a - 2 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {{\left (a + 2 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {2 \, {\left (a \sin \left (d x + c\right ) - b\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} - a^{2} + b^{2}}}{4 \, d} \]
1/4*(4*b^3*log(b*sin(d*x + c) + a)/(a^4 - 2*a^2*b^2 + b^4) + (a - 2*b)*log (sin(d*x + c) + 1)/(a^2 - 2*a*b + b^2) - (a + 2*b)*log(sin(d*x + c) - 1)/( a^2 + 2*a*b + b^2) - 2*(a*sin(d*x + c) - b)/((a^2 - b^2)*sin(d*x + c)^2 - a^2 + b^2))/d
Time = 0.32 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.44 \[ \int \frac {\sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {4 \, b^{4} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b - 2 \, a^{2} b^{3} + b^{5}} + \frac {{\left (a - 2 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {{\left (a + 2 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2} + 2 \, a b + b^{2}} + \frac {2 \, {\left (b^{3} \sin \left (d x + c\right )^{2} - a^{3} \sin \left (d x + c\right ) + a b^{2} \sin \left (d x + c\right ) + a^{2} b - 2 \, b^{3}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}}}{4 \, d} \]
1/4*(4*b^4*log(abs(b*sin(d*x + c) + a))/(a^4*b - 2*a^2*b^3 + b^5) + (a - 2 *b)*log(abs(sin(d*x + c) + 1))/(a^2 - 2*a*b + b^2) - (a + 2*b)*log(abs(sin (d*x + c) - 1))/(a^2 + 2*a*b + b^2) + 2*(b^3*sin(d*x + c)^2 - a^3*sin(d*x + c) + a*b^2*sin(d*x + c) + a^2*b - 2*b^3)/((a^4 - 2*a^2*b^2 + b^4)*(sin(d *x + c)^2 - 1)))/d
Time = 4.84 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.20 \[ \int \frac {\sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {b}{2\,\left (a^2-b^2\right )}-\frac {a\,\sin \left (c+d\,x\right )}{2\,\left (a^2-b^2\right )}}{d\,\left ({\sin \left (c+d\,x\right )}^2-1\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {b}{4\,{\left (a+b\right )}^2}+\frac {1}{4\,\left (a+b\right )}\right )}{d}+\frac {b^3\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{d\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (a-2\,b\right )}{4\,d\,{\left (a-b\right )}^2} \]