\(\int \frac {\sec ^3(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) [1361]

Optimal result

Integrand size = 29, antiderivative size = 201 \[ \int \frac {\sec ^3(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a (a+3 b) \log (1-\sin (c+d x))}{16 (a+b)^3 d}-\frac {a (a-3 b) \log (1+\sin (c+d x))}{16 (a-b)^3 d}-\frac {a^2 b^3 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 d}-\frac {a (a-3 b)}{16 (a-b) (a+b)^2 d (1-\sin (c+d x))}+\frac {a (a+3 b)}{16 (a-b)^2 (a+b) d (1+\sin (c+d x))}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d} \] Output:

1/16*a*(a+3*b)*ln(1-sin(d*x+c))/(a+b)^3/d-1/16*a*(a-3*b)*ln(1+sin(d*x+c))/ 
(a-b)^3/d-a^2*b^3*ln(a+b*sin(d*x+c))/(a^2-b^2)^3/d-1/16*a*(a-3*b)/(a-b)/(a 
+b)^2/d/(1-sin(d*x+c))+1/16*a*(a+3*b)/(a-b)^2/(a+b)/d/(1+sin(d*x+c))-1/4*s 
ec(d*x+c)^4*(b-a*sin(d*x+c))/(a^2-b^2)/d
 

Mathematica [A] (verified)

Time = 1.50 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.81 \[ \int \frac {\sec ^3(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {a (a+3 b) \log (1-\sin (c+d x))}{(a+b)^3}-\frac {a (a-3 b) \log (1+\sin (c+d x))}{(a-b)^3}-\frac {16 a^2 b^3 \log (a+b \sin (c+d x))}{(a-b)^3 (a+b)^3}+\frac {1}{(a+b) (-1+\sin (c+d x))^2}+\frac {a-b}{(a+b)^2 (-1+\sin (c+d x))}-\frac {1}{(a-b) (1+\sin (c+d x))^2}+\frac {a+b}{(a-b)^2 (1+\sin (c+d x))}}{16 d} \] Input:

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

Output:

((a*(a + 3*b)*Log[1 - Sin[c + d*x]])/(a + b)^3 - (a*(a - 3*b)*Log[1 + Sin[ 
c + d*x]])/(a - b)^3 - (16*a^2*b^3*Log[a + b*Sin[c + d*x]])/((a - b)^3*(a 
+ b)^3) + 1/((a + b)*(-1 + Sin[c + d*x])^2) + (a - b)/((a + b)^2*(-1 + Sin 
[c + d*x])) - 1/((a - b)*(1 + Sin[c + d*x])^2) + (a + b)/((a - b)^2*(1 + S 
in[c + d*x])))/(16*d)
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.18, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {3042, 3316, 27, 601, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3316

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 601

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 663

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(b^3*(-1/4*(b^2/(a^2 - b^2) - (a*b*Sin[c + d*x])/(a^2 - b^2))/(b^2 - b^2*S 
in[c + d*x]^2)^2 - (a*(-1/4*((a - b)*(a + 3*b)*Log[b - b*Sin[c + d*x]])/(b 
^3*(a + b)^2) + (4*a*Log[a + b*Sin[c + d*x]])/(a^2 - b^2)^2 + ((a - 3*b)*( 
a + b)*Log[b + b*Sin[c + d*x]])/(4*(a - b)^2*b^3) + (a - 3*b)/(4*b^2*(a + 
b)*(b - b*Sin[c + d*x])) - (a + 3*b)/(4*(a - b)*b^2*(b + b*Sin[c + d*x]))) 
)/(4*(a^2 - b^2))))/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 601

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe 
ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol 
ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) 
*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[(c 
+ d*x)^n*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Qx)/(c + d*x)^n + (e* 
(2*p + 3))/(c + d*x)^n, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 1] 
 && LtQ[p, -1] && ILtQ[n, 0] && NeQ[b*c^2 + a*d^2, 0]
 

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 = 2.52 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {-a +b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {a \left (a +3 b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}-\frac {a^{2} b^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}-\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {-a -b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}-\frac {a \left (a -3 b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}}{d}\)	\(173\)
default	\(\frac {\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {-a +b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {a \left (a +3 b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}-\frac {a^{2} b^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}-\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {-a -b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}-\frac {a \left (a -3 b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}}{d}\)	\(173\)
risch	\(-\frac {i a^{2} x}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {2 i a^{2} b^{3} x}{a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}+\frac {i a^{2} c}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {i a^{2} x}{8 a^{3}-24 a^{2} b +24 a \,b^{2}-8 b^{3}}-\frac {3 i a b c}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {3 i a b c}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {3 i a b x}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {2 i a^{2} b^{3} c}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}-\frac {3 i a b x}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {i a^{2} c}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {i a^{3} {\mathrm e}^{7 i \left (d x +c \right )}+3 i a \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}-7 i a^{3} {\mathrm e}^{5 i \left (d x +c \right )}+11 i a \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+8 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+7 i a^{3} {\mathrm e}^{3 i \left (d x +c \right )}-11 i a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+16 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-i a^{3} {\mathrm e}^{i \left (d x +c \right )}-3 i a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}+8 \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2} b}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2}}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2}}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {a^{2} b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}\)	\(759\)

Input:

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

Output:

1/d*(1/2/(8*a+8*b)/(sin(d*x+c)-1)^2-1/16*(-a+b)/(a+b)^2/(sin(d*x+c)-1)+1/1 
6*a*(a+3*b)/(a+b)^3*ln(sin(d*x+c)-1)-a^2*b^3/(a+b)^3/(a-b)^3*ln(a+b*sin(d* 
x+c))-1/2/(8*a-8*b)/(1+sin(d*x+c))^2-1/16*(-a-b)/(a-b)^2/(1+sin(d*x+c))-1/ 
16*a*(a-3*b)/(a-b)^3*ln(1+sin(d*x+c)))
 

Fricas [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.28 \[ \int \frac {\sec ^3(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {16 \, a^{2} b^{3} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left (a^{5} - 6 \, a^{3} b^{2} - 8 \, a^{2} b^{3} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{5} - 6 \, a^{3} b^{2} + 8 \, a^{2} b^{3} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a^{4} b - 8 \, a^{2} b^{3} + 4 \, b^{5} - 8 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{5} - 4 \, a^{3} b^{2} + 2 \, a b^{4} - {\left (a^{5} + 2 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{4}} \] Input:

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

Output:

-1/16*(16*a^2*b^3*cos(d*x + c)^4*log(b*sin(d*x + c) + a) + (a^5 - 6*a^3*b^ 
2 - 8*a^2*b^3 - 3*a*b^4)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (a^5 - 6*a 
^3*b^2 + 8*a^2*b^3 - 3*a*b^4)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 4*a^ 
4*b - 8*a^2*b^3 + 4*b^5 - 8*(a^4*b - a^2*b^3)*cos(d*x + c)^2 - 2*(2*a^5 - 
4*a^3*b^2 + 2*a*b^4 - (a^5 + 2*a^3*b^2 - 3*a*b^4)*cos(d*x + c)^2)*sin(d*x 
+ c))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d*cos(d*x + c)^4)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.32 \[ \int \frac {\sec ^3(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {16 \, a^{2} b^{3} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac {{\left (a^{2} - 3 \, a b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (a^{2} + 3 \, a b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (4 \, a^{2} b \sin \left (d x + c\right )^{2} - {\left (a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )^{3} - 2 \, a^{2} b - 2 \, b^{3} - {\left (a^{3} - 5 \, a b^{2}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}}}{16 \, d} \] Input:

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

Output:

-1/16*(16*a^2*b^3*log(b*sin(d*x + c) + a)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b 
^6) + (a^2 - 3*a*b)*log(sin(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) 
- (a^2 + 3*a*b)*log(sin(d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + 2* 
(4*a^2*b*sin(d*x + c)^2 - (a^3 + 3*a*b^2)*sin(d*x + c)^3 - 2*a^2*b - 2*b^3 
 - (a^3 - 5*a*b^2)*sin(d*x + c))/((a^4 - 2*a^2*b^2 + b^4)*sin(d*x + c)^4 + 
 a^4 - 2*a^2*b^2 + b^4 - 2*(a^4 - 2*a^2*b^2 + b^4)*sin(d*x + c)^2))/d
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.38 \[ \int \frac {\sec ^3(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a^{2} b^{4} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b d - 3 \, a^{4} b^{3} d + 3 \, a^{2} b^{5} d - b^{7} d} + \frac {{\left (a^{2} + 3 \, a b\right )} \log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, {\left (a^{3} d + 3 \, a^{2} b d + 3 \, a b^{2} d + b^{3} d\right )}} - \frac {{\left (a^{2} - 3 \, a b\right )} \log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, {\left (a^{3} d - 3 \, a^{2} b d + 3 \, a b^{2} d - b^{3} d\right )}} + \frac {2 \, a^{4} b - 2 \, b^{5} + {\left (a^{5} + 2 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \sin \left (d x + c\right )^{3} - 4 \, {\left (a^{4} b - a^{2} b^{3}\right )} \sin \left (d x + c\right )^{2} + {\left (a^{5} - 6 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \sin \left (d x + c\right )}{8 \, {\left (a + b\right )}^{3} {\left (a - b\right )}^{3} d {\left (\sin \left (d x + c\right ) + 1\right )}^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} \] Input:

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

Output:

-a^2*b^4*log(abs(b*sin(d*x + c) + a))/(a^6*b*d - 3*a^4*b^3*d + 3*a^2*b^5*d 
 - b^7*d) + 1/16*(a^2 + 3*a*b)*log(abs(-sin(d*x + c) + 1))/(a^3*d + 3*a^2* 
b*d + 3*a*b^2*d + b^3*d) - 1/16*(a^2 - 3*a*b)*log(abs(-sin(d*x + c) - 1))/ 
(a^3*d - 3*a^2*b*d + 3*a*b^2*d - b^3*d) + 1/8*(2*a^4*b - 2*b^5 + (a^5 + 2* 
a^3*b^2 - 3*a*b^4)*sin(d*x + c)^3 - 4*(a^4*b - a^2*b^3)*sin(d*x + c)^2 + ( 
a^5 - 6*a^3*b^2 + 5*a*b^4)*sin(d*x + c))/((a + b)^3*(a - b)^3*d*(sin(d*x + 
 c) + 1)^2*(sin(d*x + c) - 1)^2)
 

Mupad [B] (verification not implemented)

Time = 21.38 (sec) , antiderivative size = 498, normalized size of antiderivative = 2.48 \[ \int \frac {\sec ^3(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {b^2}{4\,{\left (a-b\right )}^3}+\frac {b}{8\,{\left (a-b\right )}^2}-\frac {1}{8\,\left (a-b\right )}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {b}{8\,{\left (a+b\right )}^2}+\frac {1}{8\,\left (a+b\right )}-\frac {b^2}{4\,{\left (a+b\right )}^3}\right )}{d}-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,a\,b^2-7\,a^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {2\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{a^4-2\,a^2\,b^2+b^4}-\frac {2\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^4-2\,a^2\,b^2+b^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (5\,a\,b^2-a^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (3\,a\,b^2-7\,a^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {4\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{a^4-2\,a^2\,b^2+b^4}-\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-5\,b^2\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {a^2\,b^3\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 1011, normalized size of antiderivative = 5.03 \[ \int \frac {\sec ^3(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

(log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**5 - 6*log(tan((c + d*x)/2) - 
 1)*sin(c + d*x)**4*a**3*b**2 + 8*log(tan((c + d*x)/2) - 1)*sin(c + d*x)** 
4*a**2*b**3 - 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a*b**4 - 2*log(t 
an((c + d*x)/2) - 1)*sin(c + d*x)**2*a**5 + 12*log(tan((c + d*x)/2) - 1)*s 
in(c + d*x)**2*a**3*b**2 - 16*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a* 
*2*b**3 + 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**4 + log(tan((c 
+ d*x)/2) - 1)*a**5 - 6*log(tan((c + d*x)/2) - 1)*a**3*b**2 + 8*log(tan((c 
 + d*x)/2) - 1)*a**2*b**3 - 3*log(tan((c + d*x)/2) - 1)*a*b**4 - log(tan(( 
c + d*x)/2) + 1)*sin(c + d*x)**4*a**5 + 6*log(tan((c + d*x)/2) + 1)*sin(c 
+ d*x)**4*a**3*b**2 + 8*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**2*b** 
3 + 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a*b**4 + 2*log(tan((c + d* 
x)/2) + 1)*sin(c + d*x)**2*a**5 - 12*log(tan((c + d*x)/2) + 1)*sin(c + d*x 
)**2*a**3*b**2 - 16*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b**3 - 
6*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b**4 - log(tan((c + d*x)/2) 
+ 1)*a**5 + 6*log(tan((c + d*x)/2) + 1)*a**3*b**2 + 8*log(tan((c + d*x)/2) 
 + 1)*a**2*b**3 + 3*log(tan((c + d*x)/2) + 1)*a*b**4 - 8*log(tan((c + d*x) 
/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)**4*a**2*b**3 + 16*log(ta 
n((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)**2*a**2*b**3 
- 8*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*a**2*b**3 - 2*si 
n(c + d*x)**4*a**4*b + 4*sin(c + d*x)**4*a**2*b**3 - 2*sin(c + d*x)**4*...