Integrand size = 31, antiderivative size = 43 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^2 B \log (1-\sin (c+d x))}{d}+\frac {a^3 (A+B)}{d (a-a \sin (c+d x))} \] Output:
a^2*B*ln(1-sin(d*x+c))/d+a^3*(A+B)/d/(a-a*sin(d*x+c))
Time = 0.08 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.95 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^3 \left (\frac {B \log (1-\sin (c+d x))}{a}+\frac {A+B}{a-a \sin (c+d x)}\right )}{d} \] Input:
Integrate[Sec[c + d*x]^3*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]
Output:
(a^3*((B*Log[1 - Sin[c + d*x]])/a + (A + B)/(a - a*Sin[c + d*x])))/d
Time = 0.29 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.93, 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, 49, 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.
\(\displaystyle \int \sec ^3(c+d x) (a \sin (c+d x)+a)^2 (A+B \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (c+d x)+a)^2 (A+B \sin (c+d x))}{\cos (c+d x)^3}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {a^3 \int \frac {a A+a B \sin (c+d x)}{a (a-a \sin (c+d x))^2}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^2 \int \frac {a A+a B \sin (c+d x)}{(a-a \sin (c+d x))^2}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {a^2 \int \left (\frac {a (A+B)}{(a-a \sin (c+d x))^2}-\frac {B}{a-a \sin (c+d x)}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 \left (\frac {a (A+B)}{a-a \sin (c+d x)}+B \log (a-a \sin (c+d x))\right )}{d}\) |
Input:
Int[Sec[c + d*x]^3*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]
Output:
(a^2*(B*Log[a - a*Sin[c + d*x]] + (a*(A + B))/(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_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
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 = 0.72 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.74
method | result | size |
parallelrisch | \(-\frac {\left (\left (\sin \left (d x +c \right )-1\right ) B \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-2 B \left (\sin \left (d x +c \right )-1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (A +B \right ) \sin \left (d x +c \right )\right ) a^{2}}{d \left (\sin \left (d x +c \right )-1\right )}\) | \(75\) |
risch | \(-i x \,a^{2} B -\frac {2 i a^{2} B c}{d}-\frac {2 i a^{2} {\mathrm e}^{i \left (d x +c \right )} \left (A +B \right )}{d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{2}}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}\) | \(78\) |
derivativedivides | \(\frac {A \,a^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{2} B \left (\frac {\tan \left (d x +c \right )^{2}}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )+\frac {A \,a^{2}}{\cos \left (d x +c \right )^{2}}+2 a^{2} B \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {a^{2} B}{2 \cos \left (d x +c \right )^{2}}}{d}\) | \(189\) |
default | \(\frac {A \,a^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{2} B \left (\frac {\tan \left (d x +c \right )^{2}}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )+\frac {A \,a^{2}}{\cos \left (d x +c \right )^{2}}+2 a^{2} B \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {a^{2} B}{2 \cos \left (d x +c \right )^{2}}}{d}\) | \(189\) |
norman | \(\frac {-\frac {4 A \,a^{2}+4 a^{2} B}{d}-\frac {\left (4 A \,a^{2}+4 a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}+\frac {2 \left (10 A \,a^{2}+10 a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {2 \left (10 A \,a^{2}+10 a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}+\frac {2 a^{2} \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {8 a^{2} \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+\frac {12 a^{2} \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {8 a^{2} \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {2 a^{2} \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\frac {2 a^{2} B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {a^{2} B \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d}\) | \(291\) |
Input:
int(sec(d*x+c)^3*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x,method=_RETURNVERBO SE)
Output:
-((sin(d*x+c)-1)*B*ln(sec(1/2*d*x+1/2*c)^2)-2*B*(sin(d*x+c)-1)*ln(tan(1/2* d*x+1/2*c)-1)+(A+B)*sin(d*x+c))*a^2/d/(sin(d*x+c)-1)
Time = 0.09 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.28 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {{\left (A + B\right )} a^{2} - {\left (B a^{2} \sin \left (d x + c\right ) - B a^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{d \sin \left (d x + c\right ) - d} \] Input:
integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="f ricas")
Output:
-((A + B)*a^2 - (B*a^2*sin(d*x + c) - B*a^2)*log(-sin(d*x + c) + 1))/(d*si n(d*x + c) - d)
\[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=a^{2} \left (\int A \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 A \sin {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int A \sin ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int B \sin {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 B \sin ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int B \sin ^{3}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(sec(d*x+c)**3*(a+a*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)
Output:
a**2*(Integral(A*sec(c + d*x)**3, x) + Integral(2*A*sin(c + d*x)*sec(c + d *x)**3, x) + Integral(A*sin(c + d*x)**2*sec(c + d*x)**3, x) + Integral(B*s in(c + d*x)*sec(c + d*x)**3, x) + Integral(2*B*sin(c + d*x)**2*sec(c + d*x )**3, x) + Integral(B*sin(c + d*x)**3*sec(c + d*x)**3, x))
Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {B a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {{\left (A + B\right )} a^{2}}{\sin \left (d x + c\right ) - 1}}{d} \] Input:
integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="m axima")
Output:
(B*a^2*log(sin(d*x + c) - 1) - (A + B)*a^2/(sin(d*x + c) - 1))/d
Time = 0.18 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.05 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {B a^{2} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{d} - \frac {A a^{2} + B a^{2}}{d {\left (\sin \left (d x + c\right ) - 1\right )}} \] Input:
integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="g iac")
Output:
B*a^2*log(abs(sin(d*x + c) - 1))/d - (A*a^2 + B*a^2)/(d*(sin(d*x + c) - 1) )
Time = 33.79 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.02 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {B\,a^2\,\ln \left (\sin \left (c+d\,x\right )-1\right )}{d}-\frac {A\,a^2+B\,a^2}{d\,\left (\sin \left (c+d\,x\right )-1\right )} \] Input:
int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x))^2)/cos(c + d*x)^3,x)
Output:
(B*a^2*log(sin(c + d*x) - 1))/d - (A*a^2 + B*a^2)/(d*(sin(c + d*x) - 1))
Time = 0.17 (sec) , antiderivative size = 99, normalized size of antiderivative = 2.30 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^{2} \left (-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right ) b +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) b +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) b -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b -a -b \right )}{d \left (\sin \left (d x +c \right )-1\right )} \] Input:
int(sec(d*x+c)^3*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)
Output:
(a**2*( - log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)*b + log(tan((c + d*x)/ 2)**2 + 1)*b + 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*b - 2*log(tan((c + d*x)/2) - 1)*b - a - b))/(d*(sin(c + d*x) - 1))