Integrand size = 31, antiderivative size = 73 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^2 (A-2 B) \sec (c+d x)}{3 d}+\frac {(A+B) \sec ^3(c+d x) (a+a \sin (c+d x))^2}{3 d}+\frac {a^2 (A-2 B) \tan (c+d x)}{3 d} \] Output:
1/3*a^2*(A-2*B)*sec(d*x+c)/d+1/3*(A+B)*sec(d*x+c)^3*(a+a*sin(d*x+c))^2/d+1 /3*a^2*(A-2*B)*tan(d*x+c)/d
Time = 0.18 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.66 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {a^2 \sec (c+d x) (-2 A+B+(A-2 B) \sin (c+d x)) (\sec (c+d x)+\tan (c+d x))^2}{3 d} \] Input:
Integrate[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]
Output:
-1/3*(a^2*Sec[c + d*x]*(-2*A + B + (A - 2*B)*Sin[c + d*x])*(Sec[c + d*x] + Tan[c + d*x])^2)/d
Time = 0.43 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.88, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {3042, 3334, 3042, 3148, 3042, 4254, 24}
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.
| \(\displaystyle \int \sec ^4(c+d x) (a \sin (c+d x)+a)^2 (A+B \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^2 (A+B \sin (c+d x))}{\cos (c+d x)^4}dx\) |
| \(\Big \downarrow \) 3334 |
| \(\displaystyle \frac {1}{3} a (A-2 B) \int \sec ^2(c+d x) (\sin (c+d x) a+a)dx+\frac {(A+B) \sec ^3(c+d x) (a \sin (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} a (A-2 B) \int \frac {\sin (c+d x) a+a}{\cos (c+d x)^2}dx+\frac {(A+B) \sec ^3(c+d x) (a \sin (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 3148 |
| \(\displaystyle \frac {1}{3} a (A-2 B) \left (a \int \sec ^2(c+d x)dx+\frac {a \sec (c+d x)}{d}\right )+\frac {(A+B) \sec ^3(c+d x) (a \sin (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} a (A-2 B) \left (a \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {a \sec (c+d x)}{d}\right )+\frac {(A+B) \sec ^3(c+d x) (a \sin (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{3} a (A-2 B) \left (\frac {a \sec (c+d x)}{d}-\frac {a \int 1d(-\tan (c+d x))}{d}\right )+\frac {(A+B) \sec ^3(c+d x) (a \sin (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {(A+B) \sec ^3(c+d x) (a \sin (c+d x)+a)^2}{3 d}+\frac {1}{3} a (A-2 B) \left (\frac {a \tan (c+d x)}{d}+\frac {a \sec (c+d x)}{d}\right )\) |
Input:
Int[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]
Output:
((A + B)*Sec[c + d*x]^3*(a + a*Sin[c + d*x])^2)/(3*d) + (a*(A - 2*B)*((a*S ec[c + d*x])/d + (a*Tan[c + d*x])/d))/3
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Simp[a Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b* c + a*d))*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(p + 1))) , x] + Simp[b*((a*d*m + b*c*(m + p + 1))/(a*g^2*(p + 1))) Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Time = 0.99 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.78
| method | result | size |
| parallelrisch | \(-\frac {2 \left (A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {2 A}{3}-\frac {B}{3}\right ) a^{2}}{d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\) | \(57\) |
| risch | \(-\frac {2 \left (-3 i B \,a^{2} {\mathrm e}^{i \left (d x +c \right )}+3 B \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+A \,a^{2}-2 a^{2} B +3 i A \,a^{2} {\mathrm e}^{i \left (d x +c \right )}\right )}{3 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} d}\) | \(79\) |
| derivativedivides | \(\frac {\frac {A \,a^{2} \sin \left (d x +c \right )^{3}}{3 \cos \left (d x +c \right )^{3}}+a^{2} B \left (\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}\right )+\frac {2 A \,a^{2}}{3 \cos \left (d x +c \right )^{3}}+\frac {2 a^{2} B \sin \left (d x +c \right )^{3}}{3 \cos \left (d x +c \right )^{3}}-A \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+\frac {a^{2} B}{3 \cos \left (d x +c \right )^{3}}}{d}\) | \(162\) |
| default | \(\frac {\frac {A \,a^{2} \sin \left (d x +c \right )^{3}}{3 \cos \left (d x +c \right )^{3}}+a^{2} B \left (\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}\right )+\frac {2 A \,a^{2}}{3 \cos \left (d x +c \right )^{3}}+\frac {2 a^{2} B \sin \left (d x +c \right )^{3}}{3 \cos \left (d x +c \right )^{3}}-A \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+\frac {a^{2} B}{3 \cos \left (d x +c \right )^{3}}}{d}\) | \(162\) |
| norman | \(\frac {-\frac {4 A \,a^{2}-2 a^{2} B}{3 d}-\frac {\left (8 A \,a^{2}+12 a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}-\frac {\left (12 A \,a^{2}+10 a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}-\frac {2 A \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 A \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}-\frac {2 a^{2} \left (2 A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}-\frac {2 a^{2} \left (2 A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}-\frac {4 a^{2} \left (3 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {4 a^{2} \left (3 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {4 a^{2} \left (10 A +13 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d}-\frac {2 a^{2} \left (11 A +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {2 a^{2} \left (11 A +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}\) | \(325\) |
Input:
int(sec(d*x+c)^4*(a+a*sin(d*x+c))^2*(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)+2/3*A-1/3*B)*a^2/d/(t an(1/2*d*x+1/2*c)-1)^3
Time = 0.07 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.64 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {{\left (A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right ) + {\left (A + B\right )} a^{2} - {\left ({\left (A - 2 \, B\right )} a^{2} \cos \left (d x + c\right ) - {\left (A + B\right )} a^{2}\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \] Input:
integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="f ricas")
Output:
-1/3*((A - 2*B)*a^2*cos(d*x + c)^2 + (2*A - B)*a^2*cos(d*x + c) + (A + B)* a^2 - ((A - 2*B)*a^2*cos(d*x + c) - (A + B)*a^2)*sin(d*x + c))/(d*cos(d*x + c)^2 - d*cos(d*x + c) + (d*cos(d*x + c) + 2*d)*sin(d*x + c) - 2*d)
\[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=a^{2} \left (\int A \sec ^{4}{\left (c + d x \right )}\, dx + \int 2 A \sin {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int A \sin ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int B \sin {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int 2 B \sin ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int B \sin ^{3}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(sec(d*x+c)**4*(a+a*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)
Output:
a**2*(Integral(A*sec(c + d*x)**4, x) + Integral(2*A*sin(c + d*x)*sec(c + d *x)**4, x) + Integral(A*sin(c + d*x)**2*sec(c + d*x)**4, x) + Integral(B*s in(c + d*x)*sec(c + d*x)**4, x) + Integral(2*B*sin(c + d*x)**2*sec(c + d*x )**4, x) + Integral(B*sin(c + d*x)**3*sec(c + d*x)**4, x))
Time = 0.05 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.48 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {A a^{2} \tan \left (d x + c\right )^{3} + 2 \, B a^{2} \tan \left (d x + c\right )^{3} + {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} - \frac {{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} B a^{2}}{\cos \left (d x + c\right )^{3}} + \frac {2 \, A a^{2}}{\cos \left (d x + c\right )^{3}} + \frac {B a^{2}}{\cos \left (d x + c\right )^{3}}}{3 \, d} \] Input:
integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="m axima")
Output:
1/3*(A*a^2*tan(d*x + c)^3 + 2*B*a^2*tan(d*x + c)^3 + (tan(d*x + c)^3 + 3*t an(d*x + c))*A*a^2 - (3*cos(d*x + c)^2 - 1)*B*a^2/cos(d*x + c)^3 + 2*A*a^2 /cos(d*x + c)^3 + B*a^2/cos(d*x + c)^3)/d
Time = 0.18 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.07 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {2 \, {\left (3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, A a^{2} - B a^{2}\right )}}{3 \, d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} \] Input:
integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="g iac")
Output:
-2/3*(3*A*a^2*tan(1/2*d*x + 1/2*c)^2 - 3*A*a^2*tan(1/2*d*x + 1/2*c) + 3*B* a^2*tan(1/2*d*x + 1/2*c) + 2*A*a^2 - B*a^2)/(d*(tan(1/2*d*x + 1/2*c) - 1)^ 3)
Time = 34.57 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.05 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {\sqrt {2}\,a^2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {B}{2}-\frac {5\,A}{2}+\frac {A\,\cos \left (c+d\,x\right )}{2}+\frac {B\,\cos \left (c+d\,x\right )}{2}+\frac {3\,A\,\sin \left (c+d\,x\right )}{2}-\frac {3\,B\,\sin \left (c+d\,x\right )}{2}\right )}{6\,d\,{\cos \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d\,x}{2}\right )}^3} \] Input:
int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x))^2)/cos(c + d*x)^4,x)
Output:
-(2^(1/2)*a^2*cos(c/2 + (d*x)/2)*(B/2 - (5*A)/2 + (A*cos(c + d*x))/2 + (B* cos(c + d*x))/2 + (3*A*sin(c + d*x))/2 - (3*B*sin(c + d*x))/2))/(6*d*cos(c /2 + pi/4 + (d*x)/2)^3)
Time = 0.18 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.07 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {2 a^{2} \left (-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a -3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a +b \right )}{3 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )} \] Input:
int(sec(d*x+c)^4*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)
Output:
(2*a**2*( - tan((c + d*x)/2)**3*a - 3*tan((c + d*x)/2)*b - a + b))/(3*d*(t an((c + d*x)/2)**3 - 3*tan((c + d*x)/2)**2 + 3*tan((c + d*x)/2) - 1))