Integrand size = 21, antiderivative size = 134 \[ \int \frac {\sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {(a+2 b) \log (1-\sin (c+d x))}{4 (a+b)^2 d}+\frac {(a-2 b) \log (1+\sin (c+d x))}{4 (a-b)^2 d}+\frac {b^3 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2 d}+\frac {1}{4 (a+b) d (1-\sin (c+d x))}-\frac {1}{4 (a-b) d (1+\sin (c+d x))} \] Output:
-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+1/4/(a+b)/d/(1-sin(d*x+c))-1/4/( a-b)/d/(1+sin(d*x+c))
Time = 0.96 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.85 \[ \int \frac {\sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {(a+2 b) \log (1-\sin (c+d x))}{(a+b)^2}-\frac {(a-2 b) \log (1+\sin (c+d x))}{(a-b)^2}-\frac {4 b^3 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2}+\frac {1}{(a+b) (-1+\sin (c+d x))}+\frac {1}{(a-b) (1+\sin (c+d x))}}{4 d} \] Input:
Integrate[Sec[c + d*x]^3/(a + b*Sin[c + d*x]),x]
Output:
-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 = 0.40 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.04, 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 |
| \(\displaystyle \frac {\frac {b^4 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2}+\frac {b^2}{4 (a+b) (b-b \sin (c+d x))}-\frac {b^2}{4 (a-b) (b \sin (c+d x)+b)}-\frac {b (a+2 b) \log (b-b \sin (c+d x))}{4 (a+b)^2}+\frac {b (a-2 b) \log (b \sin (c+d x)+b)}{4 (a-b)^2}}{b d}\) |
Input:
Int[Sec[c + d*x]^3/(a + b*Sin[c + d*x]),x]
Output:
(-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)
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 = 0.57 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.90
| 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 | \(\frac {2 b^{3} \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-\left (a +2 b \right ) \left (a -b \right )^{2} \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (a +b \right ) \left (\left (a +b \right ) \left (a -2 b \right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 \left (a -b \right ) \left (a \sin \left (d x +c \right )+\frac {b \cos \left (2 d x +2 c \right )}{2}-\frac {b}{2}\right )\right )}{2 \left (a -b \right )^{2} \left (a +b \right )^{2} d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(180\) |
| norman | \(\frac {\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (a^{2}-b^{2}\right )}+\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d \left (a^{2}-b^{2}\right )}-\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d \left (a^{2}-b^{2}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}+\frac {b^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}+\frac {\left (a -2 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (a +2 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 \left (a^{2}+2 a b +b^{2}\right ) d}\) | \(221\) |
| risch | \(-\frac {2 i b^{3} c}{d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}-\frac {i a c}{2 d \left (a^{2}-2 a b +b^{2}\right )}+\frac {i b c}{\left (a^{2}+2 a b +b^{2}\right ) d}-\frac {2 i b^{3} x}{a^{4}-2 b^{2} a^{2}+b^{4}}-\frac {i a x}{2 \left (a^{2}-2 a b +b^{2}\right )}+\frac {i a c}{2 \left (a^{2}+2 a b +b^{2}\right ) d}+\frac {i b c}{d \left (a^{2}-2 a b +b^{2}\right )}+\frac {i a x}{2 a^{2}+4 a b +2 b^{2}}+\frac {i b x}{a^{2}+2 a b +b^{2}}+\frac {i b x}{a^{2}-2 a b +b^{2}}+\frac {i a \,{\mathrm e}^{3 i \left (d x +c \right )}-i {\mathrm e}^{i \left (d x +c \right )} a +2 \,{\mathrm e}^{2 i \left (d x +c \right )} b}{d \left (-a^{2}+b^{2}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a}{2 \left (a^{2}+2 a b +b^{2}\right ) d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{\left (a^{2}+2 a b +b^{2}\right ) d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a}{2 d \left (a^{2}-2 a b +b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{d \left (a^{2}-2 a b +b^{2}\right )}+\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}\) | \(456\) |
Input:
int(sec(d*x+c)^3/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
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 = 0.12 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.14 \[ \int \frac {\sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {4 \, b^{3} \cos \left (d x + c\right )^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{3} - 3 \, a b^{2} + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, a^{2} b + 2 \, b^{3} + 2 \, {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \cos \left (d x + c\right )^{2}} \] Input:
integrate(sec(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="fricas")
Output:
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)
\[ \int \frac {\sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \] Input:
integrate(sec(d*x+c)**3/(a+b*sin(d*x+c)),x)
Output:
Integral(sec(c + d*x)**3/(a + b*sin(c + d*x)), x)
Time = 0.04 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.04 \[ \int \frac {\sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {4 \, b^{3} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (a - 2 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {{\left (a + 2 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {2 \, {\left (a \sin \left (d x + c\right ) - b\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} - a^{2} + b^{2}}}{4 \, d} \] Input:
integrate(sec(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="maxima")
Output:
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 = 0.13 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.34 \[ \int \frac {\sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b^{4} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b d - 2 \, a^{2} b^{3} d + b^{5} d} - \frac {{\left (a + 2 \, b\right )} \log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{4 \, {\left (a^{2} d + 2 \, a b d + b^{2} d\right )}} + \frac {{\left (a - 2 \, b\right )} \log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{4 \, {\left (a^{2} d - 2 \, a b d + b^{2} d\right )}} + \frac {a^{2} b - b^{3} - {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (a + b\right )}^{2} {\left (a - b\right )}^{2} d {\left (\sin \left (d x + c\right ) + 1\right )} {\left (\sin \left (d x + c\right ) - 1\right )}} \] Input:
integrate(sec(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
b^4*log(abs(b*sin(d*x + c) + a))/(a^4*b*d - 2*a^2*b^3*d + b^5*d) - 1/4*(a + 2*b)*log(abs(-sin(d*x + c) + 1))/(a^2*d + 2*a*b*d + b^2*d) + 1/4*(a - 2* b)*log(abs(-sin(d*x + c) - 1))/(a^2*d - 2*a*b*d + b^2*d) + 1/2*(a^2*b - b^ 3 - (a^3 - a*b^2)*sin(d*x + c))/((a + b)^2*(a - b)^2*d*(sin(d*x + c) + 1)* (sin(d*x + c) - 1))
Time = 15.79 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.10 \[ \int \frac {\sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {b}{2\,\left (a^2-b^2\right )}-\frac {a\,\sin \left (c+d\,x\right )}{2\,\left (a^2-b^2\right )}}{d\,\left ({\sin \left (c+d\,x\right )}^2-1\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {b}{4\,{\left (a+b\right )}^2}+\frac {1}{4\,\left (a+b\right )}\right )}{d}+\frac {b^3\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{d\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (a-2\,b\right )}{4\,d\,{\left (a-b\right )}^2} \] Input:
int(1/(cos(c + d*x)^3*(a + b*sin(c + d*x))),x)
Output:
(b/(2*(a^2 - b^2)) - (a*sin(c + d*x))/(2*(a^2 - b^2)))/(d*(sin(c + d*x)^2 - 1)) - (log(sin(c + d*x) - 1)*(b/(4*(a + b)^2) + 1/(4*(a + b))))/d + (b^3 *log(a + b*sin(c + d*x)))/(d*(a^4 + b^4 - 2*a^2*b^2)) + (log(sin(c + d*x) + 1)*(a - 2*b))/(4*d*(a - b)^2)
Time = 0.19 (sec) , antiderivative size = 454, normalized size of antiderivative = 3.39 \[ \int \frac {\sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{3}+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a \,b^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b^{3}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{3}-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a \,b^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{3}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{3}-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a \,b^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b^{3}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{3}+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a \,b^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{3}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) \sin \left (d x +c \right )^{2} b^{3}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) b^{3}-\sin \left (d x +c \right )^{2} a^{2} b +\sin \left (d x +c \right )^{2} b^{3}-\sin \left (d x +c \right ) a^{3}+\sin \left (d x +c \right ) a \,b^{2}+2 a^{2} b -2 b^{3}}{2 d \left (\sin \left (d x +c \right )^{2} a^{4}-2 \sin \left (d x +c \right )^{2} a^{2} b^{2}+\sin \left (d x +c \right )^{2} b^{4}-a^{4}+2 a^{2} b^{2}-b^{4}\right )} \] Input:
int(sec(d*x+c)^3/(a+b*sin(d*x+c)),x)
Output:
( - log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3 + 3*log(tan((c + d*x)/2 ) - 1)*sin(c + d*x)**2*a*b**2 - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)** 2*b**3 + log(tan((c + d*x)/2) - 1)*a**3 - 3*log(tan((c + d*x)/2) - 1)*a*b* *2 + 2*log(tan((c + d*x)/2) - 1)*b**3 + log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**3 - 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b**2 - 2*log( tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b**3 - log(tan((c + d*x)/2) + 1)*a** 3 + 3*log(tan((c + d*x)/2) + 1)*a*b**2 + 2*log(tan((c + d*x)/2) + 1)*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**3 - 2*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*b**3 - sin( c + d*x)**2*a**2*b + sin(c + d*x)**2*b**3 - sin(c + d*x)*a**3 + sin(c + d* x)*a*b**2 + 2*a**2*b - 2*b**3)/(2*d*(sin(c + d*x)**2*a**4 - 2*sin(c + d*x) **2*a**2*b**2 + sin(c + d*x)**2*b**4 - a**4 + 2*a**2*b**2 - b**4))