\(\int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx\) [1492]

Optimal result

Integrand size = 29, antiderivative size = 111 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {\left (a^2-3 b^2\right ) \text {arctanh}(\sin (c+d x))}{8 d}-\frac {(a+b) (a+3 b)}{16 d (1-\sin (c+d x))}+\frac {(a-3 b) (a-b)}{16 d (1+\sin (c+d x))}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d} \] Output:

-1/8*(a^2-3*b^2)*arctanh(sin(d*x+c))/d-1/16*(a+b)*(a+3*b)/d/(1-sin(d*x+c)) 
+1/16*(a-3*b)*(a-b)/d/(1+sin(d*x+c))+1/4*sec(d*x+c)^3*(a+b*sin(d*x+c))^2*t 
an(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.88 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.77 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {\left (a^2-3 b^2\right ) \text {arctanh}(\sin (c+d x))+\frac {1}{4} \sec ^4(c+d x) \left (16 a b \cos (2 (c+d x))+2 \left (-3 a^2+b^2+\left (a^2+5 b^2\right ) \cos (2 (c+d x))\right ) \sin (c+d x)\right )}{8 d} \] Input:

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

Output:

-1/8*((a^2 - 3*b^2)*ArcTanh[Sin[c + d*x]] + (Sec[c + d*x]^4*(16*a*b*Cos[2* 
(c + d*x)] + 2*(-3*a^2 + b^2 + (a^2 + 5*b^2)*Cos[2*(c + d*x)])*Sin[c + d*x 
]))/4)/d
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.22, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {3042, 3316, 27, 531, 25, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)^2 (a+b \sin (c+d x))^2}{\cos (c+d x)^5}dx\)

\(\Big \downarrow \) 3316

\(\displaystyle \frac {b^5 \int \frac {\sin ^2(c+d x) (a+b \sin (c+d x))^2}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 531

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {b^3 \left (\frac {b \sin (c+d x) (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {1}{4} \int \frac {(a+b \sin (c+d x)) (a+3 b \sin (c+d x))}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))\right )}{d}\)

\(\Big \downarrow \) 663

\(\displaystyle \frac {b^3 \left (\frac {b \sin (c+d x) (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {1}{4} \int \left (\frac {(a-3 b) (a-b)}{4 b^2 (\sin (c+d x) b+b)^2}+\frac {a^2-3 b^2}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}+\frac {(a+b) (a+3 b)}{4 b^2 (b-b \sin (c+d x))^2}\right )d(b \sin (c+d x))\right )}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b^3 \left (\frac {1}{4} \left (-\frac {\left (a^2-3 b^2\right ) \text {arctanh}(\sin (c+d x))}{2 b^3}+\frac {(a-3 b) (a-b)}{4 b^2 (b \sin (c+d x)+b)}-\frac {(a+b) (a+3 b)}{4 b^2 (b-b \sin (c+d x))}\right )+\frac {b \sin (c+d x) (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\)

Input:

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

Output:

(b^3*((b*Sin[c + d*x]*(a + b*Sin[c + d*x])^2)/(4*(b^2 - b^2*Sin[c + d*x]^2 
)^2) + (-1/2*((a^2 - 3*b^2)*ArcTanh[Sin[c + d*x]])/b^3 - ((a + b)*(a + 3*b 
))/(4*b^2*(b - b*Sin[c + d*x])) + ((a - 3*b)*(a - b))/(4*b^2*(b + b*Sin[c 
+ d*x])))/4))/d
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 531

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symb 
ol] :> With[{Qx = PolynomialQuotient[x^m, a + b*x^2, x], e = Coeff[Polynomi 
alRemainder[x^m, a + b*x^2, x], x, 0], f = Coeff[PolynomialRemainder[x^m, a 
 + b*x^2, x], x, 1]}, Simp[(c + d*x)^n*(a*f - b*e*x)*((a + b*x^2)^(p + 1)/( 
2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1))   Int[(c + d*x)^(n - 1)*(a + b 
*x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*(c + d*x)*Qx - a*d*f*n + b*c*e*(2*p 
 + 3) + b*d*e*(n + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGt 
Q[n, 0] && IGtQ[m, 0] && LtQ[p, -1] && GtQ[n, 1] && IntegerQ[2*p]
 

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 = 2.63 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.50

method	result	size
derivativedivides	\(\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a b \sin \left (d x +c \right )^{4}}{2 \cos \left (d x +c \right )^{4}}+b^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(166\)
default	\(\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a b \sin \left (d x +c \right )^{4}}{2 \cos \left (d x +c \right )^{4}}+b^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(166\)
risch	\(\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+5 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-7 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-3 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+16 i a b \,{\mathrm e}^{5 i \left (d x +c \right )}+7 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+3 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-a^{2}-5 b^{2}+16 i a b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{8 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{8 d}\)	\(252\)

Input:

int(sec(d*x+c)^3*(a+b*sin(d*x+c))^2*tan(d*x+c)^2,x,method=_RETURNVERBOSE)
 

Output:

1/d*(a^2*(1/4*sin(d*x+c)^3/cos(d*x+c)^4+1/8*sin(d*x+c)^3/cos(d*x+c)^2+1/8* 
sin(d*x+c)-1/8*ln(sec(d*x+c)+tan(d*x+c)))+1/2*a*b*sin(d*x+c)^4/cos(d*x+c)^ 
4+b^2*(1/4*sin(d*x+c)^5/cos(d*x+c)^4-1/8*sin(d*x+c)^5/cos(d*x+c)^2-1/8*sin 
(d*x+c)^3-3/8*sin(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c))))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.12 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {{\left (a^{2} - 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} - 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 16 \, a b \cos \left (d x + c\right )^{2} - 8 \, a b + 2 \, {\left ({\left (a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \] Input:

integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))^2*tan(d*x+c)^2,x, algorithm="frica 
s")
 

Output:

-1/16*((a^2 - 3*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (a^2 - 3*b^2)* 
cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 16*a*b*cos(d*x + c)^2 - 8*a*b + 2* 
((a^2 + 5*b^2)*cos(d*x + c)^2 - 2*a^2 - 2*b^2)*sin(d*x + c))/(d*cos(d*x + 
c)^4)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.08 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {{\left (a^{2} - 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} - 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (8 \, a b \sin \left (d x + c\right )^{2} + {\left (a^{2} + 5 \, b^{2}\right )} \sin \left (d x + c\right )^{3} - 4 \, a b + {\left (a^{2} - 3 \, b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \] Input:

integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))^2*tan(d*x+c)^2,x, algorithm="maxim 
a")
 

Output:

-1/16*((a^2 - 3*b^2)*log(sin(d*x + c) + 1) - (a^2 - 3*b^2)*log(sin(d*x + c 
) - 1) - 2*(8*a*b*sin(d*x + c)^2 + (a^2 + 5*b^2)*sin(d*x + c)^3 - 4*a*b + 
(a^2 - 3*b^2)*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d
 

Giac [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.16 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {{\left (a^{2} - 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, d} + \frac {{\left (a^{2} - 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, d} + \frac {a^{2} \sin \left (d x + c\right )^{3} + 5 \, b^{2} \sin \left (d x + c\right )^{3} + 8 \, a b \sin \left (d x + c\right )^{2} + a^{2} \sin \left (d x + c\right ) - 3 \, b^{2} \sin \left (d x + c\right ) - 4 \, a b}{8 \, {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2} d} \] Input:

integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))^2*tan(d*x+c)^2,x, algorithm="giac" 
)
 

Output:

-1/16*(a^2 - 3*b^2)*log(abs(sin(d*x + c) + 1))/d + 1/16*(a^2 - 3*b^2)*log( 
abs(sin(d*x + c) - 1))/d + 1/8*(a^2*sin(d*x + c)^3 + 5*b^2*sin(d*x + c)^3 
+ 8*a*b*sin(d*x + c)^2 + a^2*sin(d*x + c) - 3*b^2*sin(d*x + c) - 4*a*b)/(( 
sin(d*x + c)^2 - 1)^2*d)
 

Mupad [B] (verification not implemented)

Time = 21.96 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.72 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {\left (\frac {a^2}{4}-\frac {3\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {7\,a^2}{4}+\frac {11\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (\frac {7\,a^2}{4}+\frac {11\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {a^2}{4}-\frac {3\,b^2}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {a^2}{4}-\frac {3\,b^2}{4}\right )}{d} \] Input:

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

Output:

(tan(c/2 + (d*x)/2)^7*(a^2/4 - (3*b^2)/4) + tan(c/2 + (d*x)/2)^3*((7*a^2)/ 
4 + (11*b^2)/4) + tan(c/2 + (d*x)/2)^5*((7*a^2)/4 + (11*b^2)/4) + tan(c/2 
+ (d*x)/2)*(a^2/4 - (3*b^2)/4) + 8*a*b*tan(c/2 + (d*x)/2)^4)/(d*(6*tan(c/2 
 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 
+ (d*x)/2)^8 + 1)) - (atanh(tan(c/2 + (d*x)/2))*(a^2/4 - (3*b^2)/4))/d
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 352, normalized size of antiderivative = 3.17 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a^{2}-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} b^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{2}+6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2}-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a^{2}+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} b^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{2}-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2}+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{2}+4 \sin \left (d x +c \right )^{4} a b +\sin \left (d x +c \right )^{3} a^{2}+5 \sin \left (d x +c \right )^{3} b^{2}+\sin \left (d x +c \right ) a^{2}-3 \sin \left (d x +c \right ) b^{2}}{8 d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:

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

Output:

(log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**2 - 3*log(tan((c + d*x)/2) - 
 1)*sin(c + d*x)**4*b**2 - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a** 
2 + 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**2 + log(tan((c + d*x)/2 
) - 1)*a**2 - 3*log(tan((c + d*x)/2) - 1)*b**2 - log(tan((c + d*x)/2) + 1) 
*sin(c + d*x)**4*a**2 + 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b**2 + 
 2*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2 - 6*log(tan((c + d*x)/2) 
 + 1)*sin(c + d*x)**2*b**2 - log(tan((c + d*x)/2) + 1)*a**2 + 3*log(tan((c 
 + d*x)/2) + 1)*b**2 + 4*sin(c + d*x)**4*a*b + sin(c + d*x)**3*a**2 + 5*si 
n(c + d*x)**3*b**2 + sin(c + d*x)*a**2 - 3*sin(c + d*x)*b**2)/(8*d*(sin(c 
+ d*x)**4 - 2*sin(c + d*x)**2 + 1))