Integrand size = 31, antiderivative size = 81 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^2 (A-B) \text {arctanh}(\sin (c+d x))}{4 d}+\frac {a^4 (A+B)}{4 d (a-a \sin (c+d x))^2}+\frac {a^5 (A-B)}{4 d \left (a^3-a^3 \sin (c+d x)\right )} \] Output:
1/4*a^2*(A-B)*arctanh(sin(d*x+c))/d+1/4*a^4*(A+B)/d/(a-a*sin(d*x+c))^2+1/4 *a^5*(A-B)/d/(a^3-a^3*sin(d*x+c))
Time = 0.16 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.93 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^5 \left (\frac {(A-B) \text {arctanh}(\sin (c+d x))}{4 a^3}+\frac {A+B}{4 a (a-a \sin (c+d x))^2}+\frac {A-B}{4 a^2 (a-a \sin (c+d x))}\right )}{d} \] Input:
Integrate[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]
Output:
(a^5*(((A - B)*ArcTanh[Sin[c + d*x]])/(4*a^3) + (A + B)/(4*a*(a - a*Sin[c + d*x])^2) + (A - B)/(4*a^2*(a - a*Sin[c + d*x]))))/d
Time = 0.32 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.89, 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.
| \(\displaystyle \int \sec ^5(c+d x) (a \sin (c+d x)+a)^2 (A+B \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^2 (A+B \sin (c+d x))}{\cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {a^5 \int \frac {a A+a B \sin (c+d x)}{a (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a^4 \int \frac {a A+a B \sin (c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {a^4 \int \left (\frac {A-B}{4 a \left (a^2-a^2 \sin ^2(c+d x)\right )}+\frac {A-B}{4 a (a-a \sin (c+d x))^2}+\frac {A+B}{2 (a-a \sin (c+d x))^3}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^4 \left (\frac {(A-B) \text {arctanh}(\sin (c+d x))}{4 a^2}+\frac {A-B}{4 a (a-a \sin (c+d x))}+\frac {A+B}{4 (a-a \sin (c+d x))^2}\right )}{d}\) |
Input:
Int[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]
Output:
(a^4*(((A - B)*ArcTanh[Sin[c + d*x]])/(4*a^2) + (A + B)/(4*(a - a*Sin[c + d*x])^2) + (A - B)/(4*a*(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 = 0.95 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.67
| method | result | size |
| parallelrisch | \(-\frac {\left (\left (-3+\cos \left (2 d x +2 c \right )+4 \sin \left (d x +c \right )\right ) \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (-3+\cos \left (2 d x +2 c \right )+4 \sin \left (d x +c \right )\right ) \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 A \cos \left (2 d x +2 c \right )+\left (2 B +6 A \right ) \sin \left (d x +c \right )-2 A \right ) a^{2}}{4 d \left (-3+\cos \left (2 d x +2 c \right )+4 \sin \left (d x +c \right )\right )}\) | \(135\) |
| risch | \(\frac {i a^{2} {\mathrm e}^{i \left (d x +c \right )} \left (4 i A \,{\mathrm e}^{i \left (d x +c \right )}-A \,{\mathrm e}^{2 i \left (d x +c \right )}+B \,{\mathrm e}^{2 i \left (d x +c \right )}+A -B \right )}{2 d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4}}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{4 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{4 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{4 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{4 d}\) | \(167\) |
| derivativedivides | \(\frac {A \,a^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a^{2} B \sin \left (d x +c \right )^{4}}{4 \cos \left (d x +c \right )^{4}}+\frac {A \,a^{2}}{2 \cos \left (d x +c \right )^{4}}+2 a^{2} B \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+A \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a^{2} B}{4 \cos \left (d x +c \right )^{4}}}{d}\) | \(238\) |
| default | \(\frac {A \,a^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a^{2} B \sin \left (d x +c \right )^{4}}{4 \cos \left (d x +c \right )^{4}}+\frac {A \,a^{2}}{2 \cos \left (d x +c \right )^{4}}+2 a^{2} B \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+A \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a^{2} B}{4 \cos \left (d x +c \right )^{4}}}{d}\) | \(238\) |
| norman | \(\frac {\frac {\left (4 A \,a^{2}+2 a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {\left (4 A \,a^{2}+2 a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{d}+\frac {\left (12 A \,a^{2}+10 a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {\left (12 A \,a^{2}+10 a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}+\frac {\left (16 A \,a^{2}+20 a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}+\frac {\left (16 A \,a^{2}+20 a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}+\frac {a^{2} \left (7 A +5 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+\frac {a^{2} \left (7 A +5 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}+\frac {a^{2} \left (3 A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {a^{2} \left (3 A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{2 d}+\frac {2 a^{2} \left (9 A +11 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {a^{2} \left (29 A +31 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 d}+\frac {a^{2} \left (29 A +31 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{2 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-\frac {a^{2} \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}+\frac {a^{2} \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d}\) | \(426\) |
Input:
int(sec(d*x+c)^5*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x,method=_RETURNVERBO SE)
Output:
-1/4*((-3+cos(2*d*x+2*c)+4*sin(d*x+c))*(A-B)*ln(tan(1/2*d*x+1/2*c)-1)-(-3+ cos(2*d*x+2*c)+4*sin(d*x+c))*(A-B)*ln(tan(1/2*d*x+1/2*c)+1)+2*A*cos(2*d*x+ 2*c)+(2*B+6*A)*sin(d*x+c)-2*A)*a^2/d/(-3+cos(2*d*x+2*c)+4*sin(d*x+c))
Leaf count of result is larger than twice the leaf count of optimal. 161 vs. \(2 (77) = 154\).
Time = 0.08 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.99 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {2 \, {\left (A - B\right )} a^{2} \sin \left (d x + c\right ) - 4 \, A a^{2} + {\left ({\left (A - B\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (A - B\right )} a^{2} \sin \left (d x + c\right ) - 2 \, {\left (A - B\right )} a^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (A - B\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (A - B\right )} a^{2} \sin \left (d x + c\right ) - 2 \, {\left (A - B\right )} a^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{8 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \] Input:
integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="f ricas")
Output:
1/8*(2*(A - B)*a^2*sin(d*x + c) - 4*A*a^2 + ((A - B)*a^2*cos(d*x + c)^2 + 2*(A - B)*a^2*sin(d*x + c) - 2*(A - B)*a^2)*log(sin(d*x + c) + 1) - ((A - B)*a^2*cos(d*x + c)^2 + 2*(A - B)*a^2*sin(d*x + c) - 2*(A - B)*a^2)*log(-s in(d*x + c) + 1))/(d*cos(d*x + c)^2 + 2*d*sin(d*x + c) - 2*d)
Timed out. \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**5*(a+a*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.07 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {{\left (A - B\right )} a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A - B\right )} a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left ({\left (A - B\right )} a^{2} \sin \left (d x + c\right ) - 2 \, A a^{2}\right )}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{8 \, d} \] Input:
integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="m axima")
Output:
1/8*((A - B)*a^2*log(sin(d*x + c) + 1) - (A - B)*a^2*log(sin(d*x + c) - 1) - 2*((A - B)*a^2*sin(d*x + c) - 2*A*a^2)/(sin(d*x + c)^2 - 2*sin(d*x + c) + 1))/d
Time = 0.17 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.20 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {{\left (A a^{2} - B a^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{8 \, d} - \frac {{\left (A a^{2} - B a^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{8 \, d} + \frac {2 \, A a^{2} - {\left (A a^{2} - B a^{2}\right )} \sin \left (d x + c\right )}{4 \, d {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} \] Input:
integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="g iac")
Output:
1/8*(A*a^2 - B*a^2)*log(abs(sin(d*x + c) + 1))/d - 1/8*(A*a^2 - B*a^2)*log (abs(sin(d*x + c) - 1))/d + 1/4*(2*A*a^2 - (A*a^2 - B*a^2)*sin(d*x + c))/( d*(sin(d*x + c) - 1)^2)
Time = 0.14 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.90 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\frac {A\,a^2}{2}-\sin \left (c+d\,x\right )\,\left (\frac {A\,a^2}{4}-\frac {B\,a^2}{4}\right )}{d\,\left ({\sin \left (c+d\,x\right )}^2-2\,\sin \left (c+d\,x\right )+1\right )}+\frac {a^2\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (A-B\right )}{4\,d} \] Input:
int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x))^2)/cos(c + d*x)^5,x)
Output:
((A*a^2)/2 - sin(c + d*x)*((A*a^2)/4 - (B*a^2)/4))/(d*(sin(c + d*x)^2 - 2* sin(c + d*x) + 1)) + (a^2*atanh(sin(c + d*x))*(A - B))/(4*d)
Time = 0.17 (sec) , antiderivative size = 290, normalized size of antiderivative = 3.58 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^{2} \left (-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b +4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) a -4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) b -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b -4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) a +4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) b +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b -\sin \left (d x +c \right )^{2} a +\sin \left (d x +c \right )^{2} b +3 a +b \right )}{8 d \left (\sin \left (d x +c \right )^{2}-2 \sin \left (d x +c \right )+1\right )} \] Input:
int(sec(d*x+c)^5*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)
Output:
(a**2*( - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a + 2*log(tan((c + d *x)/2) - 1)*sin(c + d*x)**2*b + 4*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a - 4*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*b - 2*log(tan((c + d*x)/2) - 1 )*a + 2*log(tan((c + d*x)/2) - 1)*b + 2*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a - 2*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b - 4*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*a + 4*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*b + 2*log(tan((c + d*x)/2) + 1)*a - 2*log(tan((c + d*x)/2) + 1)*b - sin(c + d*x)**2*a + sin(c + d*x)**2*b + 3*a + b))/(8*d*(sin(c + d*x)**2 - 2*sin(c + d*x) + 1))