\(\int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\) [1537]

Optimal result

Integrand size = 31, antiderivative size = 122 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {(a+b) (a A-b (A+2 B)) \log (1-\sin (c+d x))}{4 d}+\frac {(a-b) (a A+b (A-2 B)) \log (1+\sin (c+d x))}{4 d}+\frac {(a+b)^2 (A+B)}{4 d (1-\sin (c+d x))}-\frac {(a-b)^2 (A-B)}{4 d (1+\sin (c+d x))} \] Output:

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

Mathematica [A] (verified)

Time = 1.55 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.43 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\left (a^2-b^2\right ) ((a+b) (a A-b (A+2 B)) \log (1-\sin (c+d x))-(a-b) (a A+b (A-2 B)) \log (1+\sin (c+d x)))-2 a^3 (-A b+a B) \sec ^2(c+d x)-2 \left (a^2-b^2\right ) \left (a^2 A+A b^2+2 a b B\right ) \sec (c+d x) \tan (c+d x)+\left (-6 a^3 A b+4 a A b^3+2 b^4 B\right ) \tan ^2(c+d x)}{4 \left (-a^2+b^2\right ) d} \] Input:

Integrate[Sec[c + d*x]^3*(a + b*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 3316, 27, 663, 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.

\(\displaystyle \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3316

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 663

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

\(\Big \downarrow \) 2009

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

Input:

Int[Sec[c + d*x]^3*(a + b*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]
 

Output:

(b^2*(-1/4*((a + b)*(a*A - b*(A + 2*B))*Log[b - b*Sin[c + d*x]])/b^2 + ((a 
 - b)*(a*A + b*(A - 2*B))*Log[b + b*Sin[c + d*x]])/(4*b^2) + ((a + b)^2*(A 
 + B))/(4*b*(b - b*Sin[c + d*x])) - ((a - b)^2*(A - B))/(4*b*(b + b*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 663

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_ 
)^2)^(p_.), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Simp[1/c^p   Int[ExpandI 
ntegrand[(d + e*x)^m*(f + g*x)^n*(-q + c*x)^p*(q + c*x)^p, x], x], x] /;  ! 
FractionalPowerFactorQ[q]] /; FreeQ[{a, c, d, e, f, g}, x] && ILtQ[p, -1] & 
& IntegersQ[m, n] && NiceSqrtQ[(-a)*c]
 

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 3316

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*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* 
Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) 
/2] && NeQ[a^2 - b^2, 0]
 

Maple [A] (verified)

Time = 1.51 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.53

method	result	size
derivativedivides	\(\frac {A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {a^{2} B}{2 \cos \left (d x +c \right )^{2}}+\frac {A a b}{\cos \left (d x +c \right )^{2}}+2 a b B \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+b^{2} A \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,b^{2} \left (\frac {\tan \left (d x +c \right )^{2}}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(187\)
default	\(\frac {A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {a^{2} B}{2 \cos \left (d x +c \right )^{2}}+\frac {A a b}{\cos \left (d x +c \right )^{2}}+2 a b B \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+b^{2} A \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,b^{2} \left (\frac {\tan \left (d x +c \right )^{2}}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(187\)
parallelrisch	\(\frac {-2 B \,b^{2} \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (a +b \right ) \left (\left (-A -2 B \right ) b +a A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (1+\cos \left (2 d x +2 c \right )\right ) \left (a -b \right ) \left (a A +b \left (A -2 B \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-2 A a b -a^{2} B -B \,b^{2}\right ) \cos \left (2 d x +2 c \right )+\left (2 A \,a^{2}+2 b^{2} A +4 a b B \right ) \sin \left (d x +c \right )+2 A a b +a^{2} B +B \,b^{2}}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(198\)
risch	\(-i B \,b^{2} x -\frac {2 i B \,b^{2} c}{d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (A \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 B a b \,{\mathrm e}^{2 i \left (d x +c \right )}-A \,a^{2}-b^{2} A +4 i A a b \,{\mathrm e}^{i \left (d x +c \right )}-2 a b B +2 i B \,a^{2} {\mathrm e}^{i \left (d x +c \right )}+2 i B \,b^{2} {\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,a^{2}}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2} A}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b B}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,b^{2}}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,a^{2}}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2} A}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b B}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,b^{2}}{d}\)	\(335\)
norman	\(\frac {\frac {\left (A \,a^{2}+b^{2} A +2 a b B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (A \,a^{2}+b^{2} A +2 a b B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}-\frac {4 A a b +2 a^{2} B +2 B \,b^{2}}{d}+\frac {4 \left (A \,a^{2}+b^{2} A +2 a b B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+\frac {6 \left (A \,a^{2}+b^{2} A +2 a b B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {4 \left (A \,a^{2}+b^{2} A +2 a b B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {\left (4 A a b +2 a^{2} B +2 B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}+\frac {2 \left (10 A a b +5 a^{2} B +5 B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {2 \left (10 A a b +5 a^{2} B +5 B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-\frac {\left (A \,a^{2}-b^{2} A -2 a b B -2 B \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (A \,a^{2}-b^{2} A -2 a b B +2 B \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}-\frac {B \,b^{2} \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d}\)	\(418\)

Input:

int(sec(d*x+c)^3*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x,method=_RETURNVERBO 
SE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.11 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {{\left (A a^{2} - 2 \, B a b - {\left (A - 2 \, B\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a^{2} - 2 \, B a b - {\left (A + 2 \, B\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, B a^{2} + 4 \, A a b + 2 \, B b^{2} + 2 \, {\left (A a^{2} + 2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \] Input:

integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="f 
ricas")
 

Output:

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

Sympy [F]

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

integrate(sec(d*x+c)**3*(a+b*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {{\left (A a^{2} - 2 \, B a b - {\left (A - 2 \, B\right )} b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a^{2} - 2 \, B a b - {\left (A + 2 \, B\right )} b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (B a^{2} + 2 \, A a b + B b^{2} + {\left (A a^{2} + 2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \] Input:

integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="m 
axima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.16 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {{\left (A a^{2} - 2 \, B a b - A b^{2} + 2 \, B b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{4 \, d} - \frac {{\left (A a^{2} - 2 \, B a b - A b^{2} - 2 \, B b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{4 \, d} - \frac {B a^{2} + 2 \, A a b + B b^{2} + {\left (A a^{2} + 2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )}{2 \, d {\left (\sin \left (d x + c\right ) + 1\right )} {\left (\sin \left (d x + c\right ) - 1\right )}} \] Input:

integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="g 
iac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 33.84 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.97 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (a-b\right )\,\left (A\,a+A\,b-2\,B\,b\right )}{4\,d}-\frac {\sin \left (c+d\,x\right )\,\left (\frac {A\,a^2}{2}+B\,a\,b+\frac {A\,b^2}{2}\right )+\frac {B\,a^2}{2}+\frac {B\,b^2}{2}+A\,a\,b}{d\,\left ({\sin \left (c+d\,x\right )}^2-1\right )}+\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (a+b\right )\,\left (A\,b-A\,a+2\,B\,b\right )}{4\,d} \] Input:

int(((A + B*sin(c + d*x))*(a + b*sin(c + d*x))^2)/cos(c + d*x)^3,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 368, normalized size of antiderivative = 3.02 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2} b^{3}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) b^{3}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{3}+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a \,b^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b^{3}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{3}-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a \,b^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{3}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{3}-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a \,b^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b^{3}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{3}+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a \,b^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{3}-3 \sin \left (d x +c \right )^{2} a^{2} b -\sin \left (d x +c \right )^{2} b^{3}-\sin \left (d x +c \right ) a^{3}-3 \sin \left (d x +c \right ) a \,b^{2}}{2 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:

int(sec(d*x+c)^3*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)
 

Output:

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