\(\int \sec (c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\) [987]

Optimal result

Integrand size = 29, antiderivative size = 81 \[ \int \sec (c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {4 a^3 (A+B) \log (1-\sin (c+d x))}{d}-\frac {3 a^3 (A+B) \sin (c+d x)}{d}-\frac {a^3 (A+B) \sin ^2(c+d x)}{2 d}-\frac {B (a+a \sin (c+d x))^3}{3 d} \] Output:

-4*a^3*(A+B)*ln(1-sin(d*x+c))/d-3*a^3*(A+B)*sin(d*x+c)/d-1/2*a^3*(A+B)*sin 
(d*x+c)^2/d-1/3*B*(a+a*sin(d*x+c))^3/d
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.84 \[ \int \sec (c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {a^3 \left (24 (A+B) \log (1-\sin (c+d x))+6 (3 A+4 B) \sin (c+d x)+3 (A+3 B) \sin ^2(c+d x)+2 B \sin ^3(c+d x)\right )}{6 d} \] Input:

Integrate[Sec[c + d*x]*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]
 

Output:

-1/6*(a^3*(24*(A + B)*Log[1 - Sin[c + d*x]] + 6*(3*A + 4*B)*Sin[c + d*x] + 
 3*(A + 3*B)*Sin[c + d*x]^2 + 2*B*Sin[c + d*x]^3))/d
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.91, 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.

\(\displaystyle \int \sec (c+d x) (a \sin (c+d x)+a)^3 (A+B \sin (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (c+d x)+a)^3 (A+B \sin (c+d x))}{\cos (c+d x)}dx\)

\(\Big \downarrow \) 3315

\(\displaystyle \frac {a \int \frac {(\sin (c+d x) a+a)^2 (a A+a B \sin (c+d x))}{a (a-a \sin (c+d x))}d(a \sin (c+d x))}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(\sin (c+d x) a+a)^2 (a A+a B \sin (c+d x))}{a-a \sin (c+d x)}d(a \sin (c+d x))}{d}\)

\(\Big \downarrow \) 86

\(\displaystyle \frac {\int \left (\frac {4 (A+B) a^3}{a-a \sin (c+d x)}-3 (A+B) a^2-(A+B) \sin (c+d x) a^2-B (\sin (c+d x) a+a)^2\right )d(a \sin (c+d x))}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {1}{2} a^3 (A+B) \sin ^2(c+d x)-3 a^3 (A+B) \sin (c+d x)-4 a^3 (A+B) \log (a-a \sin (c+d x))-\frac {1}{3} B (a \sin (c+d x)+a)^3}{d}\)

Input:

Int[Sec[c + d*x]*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]
 

Output:

(-4*a^3*(A + B)*Log[a - a*Sin[c + d*x]] - 3*a^3*(A + B)*Sin[c + d*x] - (a^ 
3*(A + B)*Sin[c + d*x]^2)/2 - (B*(a + a*Sin[c + d*x])^3)/3)/d
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 86

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]))))
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3315

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]
 

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.19

method	result	size
parallelrisch	\(\frac {4 \left (\left (A +B \right ) \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+2 \left (-A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {\left (A +3 B \right ) \cos \left (2 d x +2 c \right )}{16}+\frac {B \sin \left (3 d x +3 c \right )}{48}+\frac {\left (-3 A -\frac {17 B}{4}\right ) \sin \left (d x +c \right )}{4}-\frac {A}{16}-\frac {3 B}{16}\right ) a^{3}}{d}\)	\(96\)
derivativedivides	\(\frac {a^{3} A \left (-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+a^{3} B \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+3 a^{3} A \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+3 a^{3} B \left (-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )-3 a^{3} A \ln \left (\cos \left (d x +c \right )\right )+3 a^{3} B \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a^{3} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-a^{3} B \ln \left (\cos \left (d x +c \right )\right )}{d}\)	\(198\)
default	\(\frac {a^{3} A \left (-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+a^{3} B \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+3 a^{3} A \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+3 a^{3} B \left (-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )-3 a^{3} A \ln \left (\cos \left (d x +c \right )\right )+3 a^{3} B \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a^{3} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-a^{3} B \ln \left (\cos \left (d x +c \right )\right )}{d}\)	\(198\)
risch	\(4 i x \,a^{3} A +4 i x \,a^{3} B +\frac {3 i {\mathrm e}^{i \left (d x +c \right )} a^{3} A}{2 d}+\frac {17 i {\mathrm e}^{i \left (d x +c \right )} a^{3} B}{8 d}-\frac {3 i a^{3} {\mathrm e}^{-i \left (d x +c \right )} A}{2 d}-\frac {17 i a^{3} {\mathrm e}^{-i \left (d x +c \right )} B}{8 d}+\frac {8 i a^{3} A c}{d}+\frac {8 i a^{3} B c}{d}-\frac {8 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {8 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}+\frac {\sin \left (3 d x +3 c \right ) a^{3} B}{12 d}+\frac {a^{3} \cos \left (2 d x +2 c \right ) A}{4 d}+\frac {3 a^{3} \cos \left (2 d x +2 c \right ) B}{4 d}\)	\(214\)
norman	\(\frac {-\frac {2 \left (a^{3} A +3 a^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}-\frac {2 \left (a^{3} A +3 a^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}-\frac {2 \left (2 a^{3} A +6 a^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}-\frac {2 a^{3} \left (3 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a^{3} \left (3 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {2 a^{3} \left (27 A +40 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {2 a^{3} \left (27 A +40 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}-\frac {8 a^{3} \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {4 a^{3} \left (A +B \right ) \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d}\)	\(254\)

Input:

int(sec(d*x+c)*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x,method=_RETURNVERBOSE 
)
 

Output:

4*((A+B)*ln(sec(1/2*d*x+1/2*c)^2)+2*(-A-B)*ln(tan(1/2*d*x+1/2*c)-1)+1/16*( 
A+3*B)*cos(2*d*x+2*c)+1/48*B*sin(3*d*x+3*c)+1/4*(-3*A-17/4*B)*sin(d*x+c)-1 
/16*A-3/16*B)*a^3/d
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.95 \[ \int \sec (c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {3 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} - 24 \, {\left (A + B\right )} a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (B a^{3} \cos \left (d x + c\right )^{2} - {\left (9 \, A + 13 \, B\right )} a^{3}\right )} \sin \left (d x + c\right )}{6 \, d} \] Input:

integrate(sec(d*x+c)*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="fri 
cas")
 

Output:

1/6*(3*(A + 3*B)*a^3*cos(d*x + c)^2 - 24*(A + B)*a^3*log(-sin(d*x + c) + 1 
) + 2*(B*a^3*cos(d*x + c)^2 - (9*A + 13*B)*a^3)*sin(d*x + c))/d
 

Sympy [F]

\[ \int \sec (c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=a^{3} \left (\int A \sec {\left (c + d x \right )}\, dx + \int 3 A \sin {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 A \sin ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int A \sin ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \sin {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 B \sin ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 B \sin ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \sin ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx\right ) \] Input:

integrate(sec(d*x+c)*(a+a*sin(d*x+c))**3*(A+B*sin(d*x+c)),x)
 

Output:

a**3*(Integral(A*sec(c + d*x), x) + Integral(3*A*sin(c + d*x)*sec(c + d*x) 
, x) + Integral(3*A*sin(c + d*x)**2*sec(c + d*x), x) + Integral(A*sin(c + 
d*x)**3*sec(c + d*x), x) + Integral(B*sin(c + d*x)*sec(c + d*x), x) + Inte 
gral(3*B*sin(c + d*x)**2*sec(c + d*x), x) + Integral(3*B*sin(c + d*x)**3*s 
ec(c + d*x), x) + Integral(B*sin(c + d*x)**4*sec(c + d*x), x))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.90 \[ \int \sec (c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {2 \, B a^{3} \sin \left (d x + c\right )^{3} + 3 \, {\left (A + 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{2} + 24 \, {\left (A + B\right )} a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, {\left (3 \, A + 4 \, B\right )} a^{3} \sin \left (d x + c\right )}{6 \, d} \] Input:

integrate(sec(d*x+c)*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="max 
ima")
 

Output:

-1/6*(2*B*a^3*sin(d*x + c)^3 + 3*(A + 3*B)*a^3*sin(d*x + c)^2 + 24*(A + B) 
*a^3*log(sin(d*x + c) - 1) + 6*(3*A + 4*B)*a^3*sin(d*x + c))/d
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.41 \[ \int \sec (c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {4 \, {\left (A a^{3} + B a^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{d} - \frac {2 \, B a^{3} d^{2} \sin \left (d x + c\right )^{3} + 3 \, A a^{3} d^{2} \sin \left (d x + c\right )^{2} + 9 \, B a^{3} d^{2} \sin \left (d x + c\right )^{2} + 18 \, A a^{3} d^{2} \sin \left (d x + c\right ) + 24 \, B a^{3} d^{2} \sin \left (d x + c\right )}{6 \, d^{3}} \] Input:

integrate(sec(d*x+c)*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="gia 
c")
 

Output:

-4*(A*a^3 + B*a^3)*log(abs(sin(d*x + c) - 1))/d - 1/6*(2*B*a^3*d^2*sin(d*x 
 + c)^3 + 3*A*a^3*d^2*sin(d*x + c)^2 + 9*B*a^3*d^2*sin(d*x + c)^2 + 18*A*a 
^3*d^2*sin(d*x + c) + 24*B*a^3*d^2*sin(d*x + c))/d^3
 

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.23 \[ \int \sec (c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {{\sin \left (c+d\,x\right )}^2\,\left (\frac {a^3\,\left (A+2\,B\right )}{2}+\frac {B\,a^3}{2}\right )+\sin \left (c+d\,x\right )\,\left (a^3\,\left (A+2\,B\right )+a^3\,\left (2\,A+B\right )+B\,a^3\right )+\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (4\,A\,a^3+4\,B\,a^3\right )+\frac {B\,a^3\,{\sin \left (c+d\,x\right )}^3}{3}}{d} \] Input:

int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x))^3)/cos(c + d*x),x)
 

Output:

-(sin(c + d*x)^2*((a^3*(A + 2*B))/2 + (B*a^3)/2) + sin(c + d*x)*(a^3*(A + 
2*B) + a^3*(2*A + B) + B*a^3) + log(sin(c + d*x) - 1)*(4*A*a^3 + 4*B*a^3) 
+ (B*a^3*sin(c + d*x)^3)/3)/d
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.60 \[ \int \sec (c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {a^{3} \left (24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a +24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) b -48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a -48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b -2 \sin \left (d x +c \right )^{3} b -3 \sin \left (d x +c \right )^{2} a -9 \sin \left (d x +c \right )^{2} b -18 \sin \left (d x +c \right ) a -24 \sin \left (d x +c \right ) b +4 a +12 b \right )}{6 d} \] Input:

int(sec(d*x+c)*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x)
 

Output:

(a**3*(24*log(tan((c + d*x)/2)**2 + 1)*a + 24*log(tan((c + d*x)/2)**2 + 1) 
*b - 48*log(tan((c + d*x)/2) - 1)*a - 48*log(tan((c + d*x)/2) - 1)*b - 2*s 
in(c + d*x)**3*b - 3*sin(c + d*x)**2*a - 9*sin(c + d*x)**2*b - 18*sin(c + 
d*x)*a - 24*sin(c + d*x)*b + 4*a + 12*b))/(6*d)