Integrand size = 31, antiderivative size = 47 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {a^5 (a A+a B \sin (c+d x))^2}{2 (A+B) d \left (a^2-a^2 \sin (c+d x)\right )^2} \] Output:
1/2*a^5*(a*A+a*B*sin(d*x+c))^2/(A+B)/d/(a^2-a^2*sin(d*x+c))^2
Time = 0.07 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.79 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {a^3 (A+B \sin (c+d x))^2}{2 (A+B) d (-1+\sin (c+d x))^2} \] Input:
Integrate[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]
Output:
(a^3*(A + B*Sin[c + d*x])^2)/(2*(A + B)*d*(-1 + Sin[c + d*x])^2)
Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3042, 3315, 27, 48}
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.
\(\displaystyle \int \sec ^5(c+d x) (a \sin (c+d x)+a)^3 (A+B \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (c+d x)+a)^3 (A+B \sin (c+d x))}{\cos (c+d x)^5}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {a^5 \int \frac {a A+a B \sin (c+d x)}{a (a-a \sin (c+d x))^3}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^4 \int \frac {a A+a B \sin (c+d x)}{(a-a \sin (c+d x))^3}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {a^3 (a A+a B \sin (c+d x))^2}{2 d (A+B) (a-a \sin (c+d x))^2}\) |
Input:
Int[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]
Output:
(a^3*(a*A + a*B*Sin[c + d*x])^2)/(2*(A + B)*d*(a - a*Sin[c + d*x])^2)
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] :> Simp [(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
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 = 0.66 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.30
method | result | size |
parallelrisch | \(\frac {2 \left (A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3}}{d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}\) | \(61\) |
risch | \(-\frac {2 \left (A \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+i B \,a^{3} {\mathrm e}^{i \left (d x +c \right )}-B \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-i B \,a^{3} {\mathrm e}^{3 i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d}\) | \(82\) |
derivativedivides | \(\frac {\frac {a^{3} A \sin \left (d x +c \right )^{4}}{4 \cos \left (d x +c \right )^{4}}+a^{3} B \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{3} A \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {3 a^{3} B \sin \left (d x +c \right )^{4}}{4 \cos \left (d x +c \right )^{4}}+\frac {3 a^{3} A}{4 \cos \left (d x +c \right )^{4}}+3 a^{3} B \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{3} A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a^{3} B}{4 \cos \left (d x +c \right )^{4}}}{d}\) | \(337\) |
default | \(\frac {\frac {a^{3} A \sin \left (d x +c \right )^{4}}{4 \cos \left (d x +c \right )^{4}}+a^{3} B \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{3} A \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {3 a^{3} B \sin \left (d x +c \right )^{4}}{4 \cos \left (d x +c \right )^{4}}+\frac {3 a^{3} A}{4 \cos \left (d x +c \right )^{4}}+3 a^{3} B \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{3} A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a^{3} B}{4 \cos \left (d x +c \right )^{4}}}{d}\) | \(337\) |
norman | \(\frac {\frac {4 \left (7 a^{3} A +5 a^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {4 \left (7 a^{3} A +5 a^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{d}+\frac {2 \left (36 a^{3} A +44 a^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}+\frac {2 a^{3} A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a^{3} A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{d}+\frac {2 a^{3} \left (3 A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {2 a^{3} \left (3 A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{d}+\frac {2 a^{3} \left (7 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+\frac {2 a^{3} \left (7 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d}+\frac {10 a^{3} \left (7 A +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {10 a^{3} \left (7 A +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+\frac {2 a^{3} \left (21 A +20 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {2 a^{3} \left (21 A +20 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}+\frac {2 a^{3} \left (29 A +31 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}+\frac {2 a^{3} \left (29 A +31 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}\) | \(414\) |
Input:
int(sec(d*x+c)^5*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x,method=_RETURNVERBO SE)
Output:
2*(A*tan(1/2*d*x+1/2*c)^2+(-A+B)*tan(1/2*d*x+1/2*c)+A)*tan(1/2*d*x+1/2*c)* a^3/d/(tan(1/2*d*x+1/2*c)-1)^4
Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.04 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {2 \, B a^{3} \sin \left (d x + c\right ) + {\left (A - B\right )} a^{3}}{2 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \] Input:
integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="f ricas")
Output:
-1/2*(2*B*a^3*sin(d*x + c) + (A - B)*a^3)/(d*cos(d*x + c)^2 + 2*d*sin(d*x + c) - 2*d)
Timed out. \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**5*(a+a*sin(d*x+c))**3*(A+B*sin(d*x+c)),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {2 \, B a^{3} \sin \left (d x + c\right ) + {\left (A - B\right )} a^{3}}{2 \, {\left (\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )} d} \] Input:
integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="m axima")
Output:
1/2*(2*B*a^3*sin(d*x + c) + (A - B)*a^3)/((sin(d*x + c)^2 - 2*sin(d*x + c) + 1)*d)
Time = 0.16 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.83 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {2 \, B a^{3} \sin \left (d x + c\right ) + A a^{3} - B a^{3}}{2 \, d {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} \] Input:
integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="g iac")
Output:
1/2*(2*B*a^3*sin(d*x + c) + A*a^3 - B*a^3)/(d*(sin(d*x + c) - 1)^2)
Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.77 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {\frac {a^3\,\left (A-B\right )}{2}+B\,a^3\,\sin \left (c+d\,x\right )}{d\,{\left (\sin \left (c+d\,x\right )-1\right )}^2} \] Input:
int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x))^3)/cos(c + d*x)^5,x)
Output:
((a^3*(A - B))/2 + B*a^3*sin(c + d*x))/(d*(sin(c + d*x) - 1)^2)
Time = 0.16 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.85 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {a^{3} \left (\sin \left (d x +c \right )^{2} b +a \right )}{2 d \left (\sin \left (d x +c \right )^{2}-2 \sin \left (d x +c \right )+1\right )} \] Input:
int(sec(d*x+c)^5*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x)
Output:
(a**3*(sin(c + d*x)**2*b + a))/(2*d*(sin(c + d*x)**2 - 2*sin(c + d*x) + 1) )