Integrand size = 29, antiderivative size = 51 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a (A-B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a^3 (A+B)}{2 d \left (a^2-a^2 \sin (c+d x)\right )} \] Output:
1/2*a*(A-B)*arctanh(sin(d*x+c))/d+1/2*a^3*(A+B)/d/(a^2-a^2*sin(d*x+c))
Time = 1.48 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.73 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a \left ((A-B) \text {arctanh}(\sin (c+d x))-\frac {A+B}{-1+\sin (c+d x)}\right )}{2 d} \] Input:
Integrate[Sec[c + d*x]^3*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]
Output:
(a*((A - B)*ArcTanh[Sin[c + d*x]] - (A + B)/(-1 + Sin[c + d*x])))/(2*d)
Time = 0.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, 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.
\(\displaystyle \int \sec ^3(c+d x) (a \sin (c+d x)+a) (A+B \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (c+d x)+a) (A+B \sin (c+d x))}{\cos (c+d x)^3}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {a^3 \int \frac {a A+a B \sin (c+d x)}{a (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^2 \int \frac {a A+a B \sin (c+d x)}{(a-a \sin (c+d x))^2 (\sin (c+d x) a+a)}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {a^2 \int \left (\frac {A-B}{2 \left (a^2-a^2 \sin ^2(c+d x)\right )}+\frac {A+B}{2 (a-a \sin (c+d x))^2}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 \left (\frac {(A-B) \text {arctanh}(\sin (c+d x))}{2 a}+\frac {A+B}{2 (a-a \sin (c+d x))}\right )}{d}\) |
Input:
Int[Sec[c + d*x]^3*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]
Output:
(a^2*(((A - B)*ArcTanh[Sin[c + d*x]])/(2*a) + (A + B)/(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 = 0.72 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.61
method | result | size |
parallelrisch | \(\frac {\left (-\left (\sin \left (d x +c \right )-1\right ) \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (\sin \left (d x +c \right )-1\right ) \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\left (A +B \right ) \sin \left (d x +c \right )\right ) a}{2 d \left (\sin \left (d x +c \right )-1\right )}\) | \(82\) |
derivativedivides | \(\frac {\frac {a A}{2 \cos \left (d x +c \right )^{2}}+a B \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {a B}{2 \cos \left (d x +c \right )^{2}}}{d}\) | \(110\) |
default | \(\frac {\frac {a A}{2 \cos \left (d x +c \right )^{2}}+a B \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {a B}{2 \cos \left (d x +c \right )^{2}}}{d}\) | \(110\) |
risch | \(-\frac {i a \,{\mathrm e}^{i \left (d x +c \right )} \left (A +B \right )}{d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{2}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}\) | \(115\) |
norman | \(\frac {\frac {a \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {2 \left (2 a A +2 a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {2 a \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {3 a \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+\frac {3 a \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {2 a \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-\frac {a \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(221\) |
Input:
int(sec(d*x+c)^3*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x,method=_RETURNVERBOSE )
Output:
1/2*(-(sin(d*x+c)-1)*(A-B)*ln(tan(1/2*d*x+1/2*c)-1)+(sin(d*x+c)-1)*(A-B)*l n(tan(1/2*d*x+1/2*c)+1)-(A+B)*sin(d*x+c))*a/d/(sin(d*x+c)-1)
Time = 0.08 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.76 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {2 \, {\left (A + B\right )} a - {\left ({\left (A - B\right )} a \sin \left (d x + c\right ) - {\left (A - B\right )} a\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (A - B\right )} a \sin \left (d x + c\right ) - {\left (A - B\right )} a\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{4 \, {\left (d \sin \left (d x + c\right ) - d\right )}} \] Input:
integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="fri cas")
Output:
-1/4*(2*(A + B)*a - ((A - B)*a*sin(d*x + c) - (A - B)*a)*log(sin(d*x + c) + 1) + ((A - B)*a*sin(d*x + c) - (A - B)*a)*log(-sin(d*x + c) + 1))/(d*sin (d*x + c) - d)
\[ \int \sec ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=a \left (\int A \sec ^{3}{\left (c + d x \right )}\, dx + \int A \sin {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int B \sin {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int B \sin ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(sec(d*x+c)**3*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)
Output:
a*(Integral(A*sec(c + d*x)**3, x) + Integral(A*sin(c + d*x)*sec(c + d*x)** 3, x) + Integral(B*sin(c + d*x)*sec(c + d*x)**3, x) + Integral(B*sin(c + d *x)**2*sec(c + d*x)**3, x))
Time = 0.04 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.08 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {{\left (A - B\right )} a \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A - B\right )} a \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (A + B\right )} a}{\sin \left (d x + c\right ) - 1}}{4 \, d} \] Input:
integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="max ima")
Output:
1/4*((A - B)*a*log(sin(d*x + c) + 1) - (A - B)*a*log(sin(d*x + c) - 1) - 2 *(A + B)*a/(sin(d*x + c) - 1))/d
Time = 0.17 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.35 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {{\left (A a - B a\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{4 \, d} - \frac {{\left (A a - B a\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{4 \, d} - \frac {A a + B a}{2 \, d {\left (\sin \left (d x + c\right ) - 1\right )}} \] Input:
integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="gia c")
Output:
1/4*(A*a - B*a)*log(abs(sin(d*x + c) + 1))/d - 1/4*(A*a - B*a)*log(abs(sin (d*x + c) - 1))/d - 1/2*(A*a + B*a)/(d*(sin(d*x + c) - 1))
Time = 33.80 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.84 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (A-B\right )}{2\,d}-\frac {\frac {A\,a}{2}+\frac {B\,a}{2}}{d\,\left (\sin \left (c+d\,x\right )-1\right )} \] Input:
int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x)))/cos(c + d*x)^3,x)
Output:
(a*atanh(sin(c + d*x))*(A - B))/(2*d) - ((A*a)/2 + (B*a)/2)/(d*(sin(c + d* x) - 1))
Time = 0.18 (sec) , antiderivative size = 163, normalized size of antiderivative = 3.20 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a \left (-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) a +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) b +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) a -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) b -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b -a -b \right )}{2 d \left (\sin \left (d x +c \right )-1\right )} \] Input:
int(sec(d*x+c)^3*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)
Output:
(a*( - log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a + log(tan((c + d*x)/2) - 1 )*sin(c + d*x)*b + log(tan((c + d*x)/2) - 1)*a - log(tan((c + d*x)/2) - 1) *b + log(tan((c + d*x)/2) + 1)*sin(c + d*x)*a - log(tan((c + d*x)/2) + 1)* sin(c + d*x)*b - log(tan((c + d*x)/2) + 1)*a + log(tan((c + d*x)/2) + 1)*b - a - b))/(2*d*(sin(c + d*x) - 1))