\(\int \sec ^3(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\) [988]

Optimal result

Integrand size = 31, antiderivative size = 61 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {a^3 (A+3 B) \log (1-\sin (c+d x))}{d}+\frac {2 a^3 (A+B)}{d (1-\sin (c+d x))}+\frac {a^3 B \sin (c+d x)}{d} \] Output:

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.79 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {a^3 \left ((A+3 B) \log (1-\sin (c+d x))-\frac {2 (A+B)}{-1+\sin (c+d x)}+B \sin (c+d x)\right )}{d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 57, 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, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3315

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 86

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(a^2*(a*(A + 3*B)*Log[a - a*Sin[c + d*x]] + a*B*Sin[c + d*x] + (2*a^2*(A + 
 B))/(a - a*Sin[c + d*x])))/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 = 0.87 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.67

method	result	size
parallelrisch	\(-\frac {\left (\left (\sin \left (d x +c \right )-1\right ) \left (A +3 B \right ) \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-2 \left (\sin \left (d x +c \right )-1\right ) \left (A +3 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {B \cos \left (2 d x +2 c \right )}{2}+\sin \left (d x +c \right ) \left (2 A +3 B \right )-\frac {B}{2}\right ) a^{3}}{d \left (\sin \left (d x +c \right )-1\right )}\)	\(102\)
risch	\(-i x \,a^{3} A -3 i x \,a^{3} B -\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{3} B}{2 d}+\frac {i a^{3} {\mathrm e}^{-i \left (d x +c \right )} B}{2 d}-\frac {2 i a^{3} A c}{d}-\frac {6 i a^{3} B c}{d}-\frac {4 i a^{3} {\mathrm e}^{i \left (d x +c \right )} \left (A +B \right )}{d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{2}}+\frac {2 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}+\frac {6 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}\)	\(157\)
derivativedivides	\(\frac {a^{3} A \left (\frac {\tan \left (d x +c \right )^{2}}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )+a^{3} B \left (\frac {\sin \left (d x +c \right )^{5}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{3}}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{3} A \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{3} B \left (\frac {\tan \left (d x +c \right )^{2}}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )+\frac {3 a^{3} A}{2 \cos \left (d x +c \right )^{2}}+3 a^{3} B \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{3} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {a^{3} B}{2 \cos \left (d x +c \right )^{2}}}{d}\)	\(273\)
default	\(\frac {a^{3} A \left (\frac {\tan \left (d x +c \right )^{2}}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )+a^{3} B \left (\frac {\sin \left (d x +c \right )^{5}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{3}}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{3} A \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{3} B \left (\frac {\tan \left (d x +c \right )^{2}}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )+\frac {3 a^{3} A}{2 \cos \left (d x +c \right )^{2}}+3 a^{3} B \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{3} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {a^{3} B}{2 \cos \left (d x +c \right )^{2}}}{d}\)	\(273\)
norman	\(\frac {\frac {48 a^{3} A +48 a^{3} B}{4 d}+\frac {\left (48 a^{3} A +48 a^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{4 d}+\frac {\left (64 a^{3} A +64 a^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 d}+\frac {\left (64 a^{3} A +64 a^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{2 d}+\frac {\left (80 a^{3} A +80 a^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4 d}+\frac {\left (80 a^{3} A +80 a^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{4 d}+\frac {2 a^{3} \left (2 A +3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a^{3} \left (2 A +3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}+\frac {4 a^{3} \left (10 A +9 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {4 a^{3} \left (10 A +9 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {2 a^{3} \left (10 A +11 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+\frac {2 a^{3} \left (10 A +11 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {2 a^{3} \left (A +3 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {a^{3} \left (A +3 B \right ) \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d}\)	\(403\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.46 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {B a^{3} \cos \left (d x + c\right )^{2} + B a^{3} \sin \left (d x + c\right ) + {\left (2 \, A + B\right )} a^{3} - {\left ({\left (A + 3 \, B\right )} a^{3} \sin \left (d x + c\right ) - {\left (A + 3 \, B\right )} a^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{d \sin \left (d x + c\right ) - d} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.85 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {{\left (A + 3 \, B\right )} a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) + B a^{3} \sin \left (d x + c\right ) - \frac {2 \, {\left (A + B\right )} a^{3}}{\sin \left (d x + c\right ) - 1}}{d} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.10 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {B a^{3} \sin \left (d x + c\right )}{d} + \frac {{\left (A a^{3} + 3 \, B a^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{d} - \frac {2 \, {\left (A a^{3} + B a^{3}\right )}}{d {\left (\sin \left (d x + c\right ) - 1\right )}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.03 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (A\,a^3+3\,B\,a^3\right )-\frac {2\,A\,a^3+2\,B\,a^3}{\sin \left (c+d\,x\right )-1}+B\,a^3\,\sin \left (c+d\,x\right )}{d} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 185, normalized size of antiderivative = 3.03 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {a^{3} \left (-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right ) a -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right ) b +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) b +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) a +6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) b -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a -6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +\sin \left (d x +c \right )^{2} b -2 a -3 b \right )}{d \left (\sin \left (d x +c \right )-1\right )} \] Input:

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

Output:

(a**3*( - log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)*a - 3*log(tan((c + d*x 
)/2)**2 + 1)*sin(c + d*x)*b + log(tan((c + d*x)/2)**2 + 1)*a + 3*log(tan(( 
c + d*x)/2)**2 + 1)*b + 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a + 6*log 
(tan((c + d*x)/2) - 1)*sin(c + d*x)*b - 2*log(tan((c + d*x)/2) - 1)*a - 6* 
log(tan((c + d*x)/2) - 1)*b + sin(c + d*x)**2*b - 2*a - 3*b))/(d*(sin(c + 
d*x) - 1))