Integrand size = 31, antiderivative size = 97 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {(3 A+B) \text {arctanh}(\sin (c+d x))}{8 a d}+\frac {A+B}{8 d (a-a \sin (c+d x))}-\frac {A}{4 d (a+a \sin (c+d x))}-\frac {a^3 (A-B)}{8 d \left (a^2+a^2 \sin (c+d x)\right )^2} \] Output:
1/8*(3*A+B)*arctanh(sin(d*x+c))/a/d+1/8*(A+B)/d/(a-a*sin(d*x+c))-1/4*A/d/( a+a*sin(d*x+c))-1/8*a^3*(A-B)/d/(a^2+a^2*sin(d*x+c))^2
Time = 0.31 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.77 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\frac {(3 A+B) \text {arctanh}(\sin (c+d x))}{a}+\frac {-A+B}{a (1+\sin (c+d x))^2}+\frac {A+B}{a-a \sin (c+d x)}-\frac {2 A}{a+a \sin (c+d x)}}{8 d} \] Input:
Integrate[(Sec[c + d*x]^3*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x]),x]
Output:
(((3*A + B)*ArcTanh[Sin[c + d*x]])/a + (-A + B)/(a*(1 + Sin[c + d*x])^2) + (A + B)/(a - a*Sin[c + d*x]) - (2*A)/(a + a*Sin[c + d*x]))/(8*d)
Time = 0.35 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 3315, 27, 86, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin (c+d x)}{\cos (c+d x)^3 (a \sin (c+d x)+a)}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {a^3 \int \frac {a A+a B \sin (c+d x)}{a (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^3}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a^2 \int \frac {a A+a B \sin (c+d x)}{(a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^3}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {a^2 \int \left (\frac {A}{4 a^2 (\sin (c+d x) a+a)^2}+\frac {3 A+B}{8 a^2 \left (a^2-a^2 \sin ^2(c+d x)\right )}+\frac {A+B}{8 a^2 (a-a \sin (c+d x))^2}+\frac {A-B}{4 a (\sin (c+d x) a+a)^3}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^2 \left (\frac {(3 A+B) \text {arctanh}(\sin (c+d x))}{8 a^3}+\frac {A+B}{8 a^2 (a-a \sin (c+d x))}-\frac {A}{4 a^2 (a \sin (c+d x)+a)}-\frac {A-B}{8 a (a \sin (c+d x)+a)^2}\right )}{d}\) |
Input:
Int[(Sec[c + d*x]^3*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x]),x]
Output:
(a^2*(((3*A + B)*ArcTanh[Sin[c + d*x]])/(8*a^3) + (A + B)/(8*a^2*(a - a*Si n[c + d*x])) - (A - B)/(8*a*(a + a*Sin[c + d*x])^2) - A/(4*a^2*(a + a*Sin[ c + d*x]))))/d
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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 + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
Time = 0.90 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.97
| method | result | size |
| derivativedivides | \(\frac {-\frac {A}{4 \left (1+\sin \left (d x +c \right )\right )}-\frac {\frac {A}{4}-\frac {B}{4}}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}+\left (\frac {3 A}{16}+\frac {B}{16}\right ) \ln \left (1+\sin \left (d x +c \right )\right )-\frac {\frac {A}{8}+\frac {B}{8}}{\sin \left (d x +c \right )-1}+\left (-\frac {3 A}{16}-\frac {B}{16}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{d a}\) | \(94\) |
| default | \(\frac {-\frac {A}{4 \left (1+\sin \left (d x +c \right )\right )}-\frac {\frac {A}{4}-\frac {B}{4}}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}+\left (\frac {3 A}{16}+\frac {B}{16}\right ) \ln \left (1+\sin \left (d x +c \right )\right )-\frac {\frac {A}{8}+\frac {B}{8}}{\sin \left (d x +c \right )-1}+\left (-\frac {3 A}{16}-\frac {B}{16}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{d a}\) | \(94\) |
| parallelrisch | \(\frac {-3 \left (\frac {\sin \left (3 d x +3 c \right )}{2}+\frac {\sin \left (d x +c \right )}{2}+\cos \left (2 d x +2 c \right )+1\right ) \left (A +\frac {B}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 \left (\frac {\sin \left (3 d x +3 c \right )}{2}+\frac {\sin \left (d x +c \right )}{2}+\cos \left (2 d x +2 c \right )+1\right ) \left (A +\frac {B}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-A -3 B \right ) \cos \left (2 d x +2 c \right )+\left (A -B \right ) \sin \left (3 d x +3 c \right )+\left (7 A +B \right ) \sin \left (d x +c \right )+A +3 B}{4 a d \left (\sin \left (3 d x +3 c \right )+\sin \left (d x +c \right )+2 \cos \left (2 d x +2 c \right )+2\right )}\) | \(186\) |
| risch | \(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (6 i A \,{\mathrm e}^{3 i \left (d x +c \right )}+3 A \,{\mathrm e}^{4 i \left (d x +c \right )}+2 i B \,{\mathrm e}^{3 i \left (d x +c \right )}+B \,{\mathrm e}^{4 i \left (d x +c \right )}-6 i A \,{\mathrm e}^{i \left (d x +c \right )}+2 A \,{\mathrm e}^{2 i \left (d x +c \right )}-2 i B \,{\mathrm e}^{i \left (d x +c \right )}-10 B \,{\mathrm e}^{2 i \left (d x +c \right )}+3 A +B \right )}{4 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{4} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{2} d a}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{8 d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{8 d a}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{8 d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{8 d a}\) | \(252\) |
| norman | \(\frac {\frac {\left (A +3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d a}+\frac {\left (A +3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 d a}+\frac {\left (A +3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2 d a}+\frac {\left (7 A +5 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 d a}+\frac {\left (7 A +5 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d a}+\frac {\left (5 A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d a}+\frac {\left (5 A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d a}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {\left (3 A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d a}+\frac {\left (3 A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d a}\) | \(268\) |
Input:
int(sec(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE )
Output:
1/d/a*(-1/4*A/(1+sin(d*x+c))-1/2*(1/4*A-1/4*B)/(1+sin(d*x+c))^2+(3/16*A+1/ 16*B)*ln(1+sin(d*x+c))-(1/8*A+1/8*B)/(sin(d*x+c)-1)+(-3/16*A-1/16*B)*ln(si n(d*x+c)-1))
Time = 0.09 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.66 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=-\frac {2 \, {\left (3 \, A + B\right )} \cos \left (d x + c\right )^{2} - {\left ({\left (3 \, A + B\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (3 \, A + B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (3 \, A + B\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (3 \, A + B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (3 \, A + B\right )} \sin \left (d x + c\right ) - 2 \, A - 6 \, B}{16 \, {\left (a d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate(sec(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="fri cas")
Output:
-1/16*(2*(3*A + B)*cos(d*x + c)^2 - ((3*A + B)*cos(d*x + c)^2*sin(d*x + c) + (3*A + B)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + ((3*A + B)*cos(d*x + c)^2*sin(d*x + c) + (3*A + B)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*( 3*A + B)*sin(d*x + c) - 2*A - 6*B)/(a*d*cos(d*x + c)^2*sin(d*x + c) + a*d* cos(d*x + c)^2)
\[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {A \sec ^{3}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sin {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(sec(d*x+c)**3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x)
Output:
(Integral(A*sec(c + d*x)**3/(sin(c + d*x) + 1), x) + Integral(B*sin(c + d* x)*sec(c + d*x)**3/(sin(c + d*x) + 1), x))/a
Time = 0.03 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.16 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\frac {{\left (3 \, A + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac {{\left (3 \, A + B\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a} - \frac {2 \, {\left ({\left (3 \, A + B\right )} \sin \left (d x + c\right )^{2} + {\left (3 \, A + B\right )} \sin \left (d x + c\right ) - 2 \, A + 2 \, B\right )}}{a \sin \left (d x + c\right )^{3} + a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - a}}{16 \, d} \] Input:
integrate(sec(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="max ima")
Output:
1/16*((3*A + B)*log(sin(d*x + c) + 1)/a - (3*A + B)*log(sin(d*x + c) - 1)/ a - 2*((3*A + B)*sin(d*x + c)^2 + (3*A + B)*sin(d*x + c) - 2*A + 2*B)/(a*s in(d*x + c)^3 + a*sin(d*x + c)^2 - a*sin(d*x + c) - a))/d
Time = 0.15 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.11 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {{\left (3 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, a d} - \frac {{\left (3 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, a d} - \frac {{\left (3 \, A + B\right )} \sin \left (d x + c\right )^{2} + {\left (3 \, A + B\right )} \sin \left (d x + c\right ) - 2 \, A + 2 \, B}{8 \, a d {\left (\sin \left (d x + c\right ) + 1\right )}^{2} {\left (\sin \left (d x + c\right ) - 1\right )}} \] Input:
integrate(sec(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="gia c")
Output:
1/16*(3*A + B)*log(abs(sin(d*x + c) + 1))/(a*d) - 1/16*(3*A + B)*log(abs(s in(d*x + c) - 1))/(a*d) - 1/8*((3*A + B)*sin(d*x + c)^2 + (3*A + B)*sin(d* x + c) - 2*A + 2*B)/(a*d*(sin(d*x + c) + 1)^2*(sin(d*x + c) - 1))
Time = 0.14 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.99 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\left (\frac {3\,A}{8}+\frac {B}{8}\right )\,{\sin \left (c+d\,x\right )}^2+\left (\frac {3\,A}{8}+\frac {B}{8}\right )\,\sin \left (c+d\,x\right )-\frac {A}{4}+\frac {B}{4}}{d\,\left (-a\,{\sin \left (c+d\,x\right )}^3-a\,{\sin \left (c+d\,x\right )}^2+a\,\sin \left (c+d\,x\right )+a\right )}+\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (3\,A+B\right )}{8\,a\,d} \] Input:
int((A + B*sin(c + d*x))/(cos(c + d*x)^3*(a + a*sin(c + d*x))),x)
Output:
(B/4 - A/4 + sin(c + d*x)*((3*A)/8 + B/8) + sin(c + d*x)^2*((3*A)/8 + B/8) )/(d*(a + a*sin(c + d*x) - a*sin(c + d*x)^2 - a*sin(c + d*x)^3)) + (atanh( sin(c + d*x))*(3*A + B))/(8*a*d)
Time = 0.18 (sec) , antiderivative size = 411, normalized size of antiderivative = 4.24 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3} a -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3} 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 -\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 ) \sin \left (d x +c \right ) a +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) b +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +\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 )^{3} a +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3} 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 +\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 ) \sin \left (d x +c \right ) a -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) b -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b -3 \sin \left (d x +c \right )^{3} a -\sin \left (d x +c \right )^{3} b -6 \sin \left (d x +c \right )^{2} a -2 \sin \left (d x +c \right )^{2} b +5 a -b}{8 a d \left (\sin \left (d x +c \right )^{3}+\sin \left (d x +c \right )^{2}-\sin \left (d x +c \right )-1\right )} \] Input:
int(sec(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x)
Output:
( - 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*a - log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*b - 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a - lo g(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b + 3*log(tan((c + d*x)/2) - 1)*si n(c + d*x)*a + log(tan((c + d*x)/2) - 1)*sin(c + d*x)*b + 3*log(tan((c + d *x)/2) - 1)*a + log(tan((c + d*x)/2) - 1)*b + 3*log(tan((c + d*x)/2) + 1)* sin(c + d*x)**3*a + log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3*b + 3*log(ta n((c + d*x)/2) + 1)*sin(c + d*x)**2*a + log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b - 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*a - log(tan((c + d*x) /2) + 1)*sin(c + d*x)*b - 3*log(tan((c + d*x)/2) + 1)*a - log(tan((c + d*x )/2) + 1)*b - 3*sin(c + d*x)**3*a - sin(c + d*x)**3*b - 6*sin(c + d*x)**2* a - 2*sin(c + d*x)**2*b + 5*a - b)/(8*a*d*(sin(c + d*x)**3 + sin(c + d*x)* *2 - sin(c + d*x) - 1))