3.5.59 \(\int \frac {\sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx\) [459]

3.5.59.1 Optimal result
3.5.59.2 Mathematica [A] (verified)
3.5.59.3 Rubi [A] (verified)
3.5.59.4 Maple [A] (verified)
3.5.59.5 Fricas [B] (verification not implemented)
3.5.59.6 Sympy [F]
3.5.59.7 Maxima [F(-2)]
3.5.59.8 Giac [B] (verification not implemented)
3.5.59.9 Mupad [B] (verification not implemented)

3.5.59.1 Optimal result

Integrand size = 21, antiderivative size = 192 \[ \int \frac {\sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {3 b^2 \left (4 a^2+b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2} d}+\frac {b \sec (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}+\frac {5 a b \sec (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac {\sec (c+d x) \left (3 b \left (4 a^2+b^2\right )-a \left (2 a^2+13 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^3 d} \]

output
-3*b^2*(4*a^2+b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/(a^2-b 
^2)^(7/2)/d+1/2*b*sec(d*x+c)/(a^2-b^2)/d/(a+b*sin(d*x+c))^2+5/2*a*b*sec(d* 
x+c)/(a^2-b^2)^2/d/(a+b*sin(d*x+c))-1/2*sec(d*x+c)*(3*b*(4*a^2+b^2)-a*(2*a 
^2+13*b^2)*sin(d*x+c))/(a^2-b^2)^3/d
 
3.5.59.2 Mathematica [A] (verified)

Time = 2.29 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.01 \[ \int \frac {\sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {-\frac {6 b^2 \left (4 a^2+b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}+\sin \left (\frac {1}{2} (c+d x)\right ) \left (\frac {2}{(a+b)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2}{(a-b)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )+\frac {b^3 \cos (c+d x) \left (-8 a^2+b^2-7 a b \sin (c+d x)\right )}{(a-b)^3 (a+b)^3 (a+b \sin (c+d x))^2}}{2 d} \]

input
Integrate[Sec[c + d*x]^2/(a + b*Sin[c + d*x])^3,x]
 
output
((-6*b^2*(4*a^2 + b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/( 
a^2 - b^2)^(7/2) + Sin[(c + d*x)/2]*(2/((a + b)^3*(Cos[(c + d*x)/2] - Sin[ 
(c + d*x)/2])) + 2/((a - b)^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))) + (b 
^3*Cos[c + d*x]*(-8*a^2 + b^2 - 7*a*b*Sin[c + d*x]))/((a - b)^3*(a + b)^3* 
(a + b*Sin[c + d*x])^2))/(2*d)
 
3.5.59.3 Rubi [A] (verified)

Time = 0.91 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.16, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {3042, 3173, 25, 3042, 3343, 25, 3042, 3345, 27, 3042, 3139, 1083, 217}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\cos (c+d x)^2 (a+b \sin (c+d x))^3}dx\)

\(\Big \downarrow \) 3173

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3343

\(\displaystyle \frac {\frac {5 a b \sec (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {\int -\frac {\sec ^2(c+d x) \left (2 a^2-10 b \sin (c+d x) a+3 b^2\right )}{a+b \sin (c+d x)}dx}{a^2-b^2}}{2 \left (a^2-b^2\right )}+\frac {b \sec (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {\sec ^2(c+d x) \left (2 a^2-10 b \sin (c+d x) a+3 b^2\right )}{a+b \sin (c+d x)}dx}{a^2-b^2}+\frac {5 a b \sec (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}}{2 \left (a^2-b^2\right )}+\frac {b \sec (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {2 a^2-10 b \sin (c+d x) a+3 b^2}{\cos (c+d x)^2 (a+b \sin (c+d x))}dx}{a^2-b^2}+\frac {5 a b \sec (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}}{2 \left (a^2-b^2\right )}+\frac {b \sec (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}\)

\(\Big \downarrow \) 3345

\(\displaystyle \frac {\frac {-\frac {\int \frac {3 b^2 \left (4 a^2+b^2\right )}{a+b \sin (c+d x)}dx}{a^2-b^2}-\frac {\sec (c+d x) \left (3 b \left (4 a^2+b^2\right )-a \left (2 a^2+13 b^2\right ) \sin (c+d x)\right )}{d \left (a^2-b^2\right )}}{a^2-b^2}+\frac {5 a b \sec (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}}{2 \left (a^2-b^2\right )}+\frac {b \sec (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-\frac {3 b^2 \left (4 a^2+b^2\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{a^2-b^2}-\frac {\sec (c+d x) \left (3 b \left (4 a^2+b^2\right )-a \left (2 a^2+13 b^2\right ) \sin (c+d x)\right )}{d \left (a^2-b^2\right )}}{a^2-b^2}+\frac {5 a b \sec (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}}{2 \left (a^2-b^2\right )}+\frac {b \sec (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {-\frac {3 b^2 \left (4 a^2+b^2\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{a^2-b^2}-\frac {\sec (c+d x) \left (3 b \left (4 a^2+b^2\right )-a \left (2 a^2+13 b^2\right ) \sin (c+d x)\right )}{d \left (a^2-b^2\right )}}{a^2-b^2}+\frac {5 a b \sec (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}}{2 \left (a^2-b^2\right )}+\frac {b \sec (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}\)

\(\Big \downarrow \) 3139

\(\displaystyle \frac {\frac {-\frac {6 b^2 \left (4 a^2+b^2\right ) \int \frac {1}{a \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tan \left (\frac {1}{2} (c+d x)\right )+a}d\tan \left (\frac {1}{2} (c+d x)\right )}{d \left (a^2-b^2\right )}-\frac {\sec (c+d x) \left (3 b \left (4 a^2+b^2\right )-a \left (2 a^2+13 b^2\right ) \sin (c+d x)\right )}{d \left (a^2-b^2\right )}}{a^2-b^2}+\frac {5 a b \sec (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}}{2 \left (a^2-b^2\right )}+\frac {b \sec (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {\frac {\frac {12 b^2 \left (4 a^2+b^2\right ) \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{d \left (a^2-b^2\right )}-\frac {\sec (c+d x) \left (3 b \left (4 a^2+b^2\right )-a \left (2 a^2+13 b^2\right ) \sin (c+d x)\right )}{d \left (a^2-b^2\right )}}{a^2-b^2}+\frac {5 a b \sec (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}}{2 \left (a^2-b^2\right )}+\frac {b \sec (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {-\frac {6 b^2 \left (4 a^2+b^2\right ) \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}-\frac {\sec (c+d x) \left (3 b \left (4 a^2+b^2\right )-a \left (2 a^2+13 b^2\right ) \sin (c+d x)\right )}{d \left (a^2-b^2\right )}}{a^2-b^2}+\frac {5 a b \sec (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}}{2 \left (a^2-b^2\right )}+\frac {b \sec (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}\)

input
Int[Sec[c + d*x]^2/(a + b*Sin[c + d*x])^3,x]
 
output
(b*Sec[c + d*x])/(2*(a^2 - b^2)*d*(a + b*Sin[c + d*x])^2) + ((5*a*b*Sec[c 
+ d*x])/((a^2 - b^2)*d*(a + b*Sin[c + d*x])) + ((-6*b^2*(4*a^2 + b^2)*ArcT 
an[(2*b + 2*a*Tan[(c + d*x)/2])/(2*Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d 
) - (Sec[c + d*x]*(3*b*(4*a^2 + b^2) - a*(2*a^2 + 13*b^2)*Sin[c + d*x]))/( 
(a^2 - b^2)*d))/(a^2 - b^2))/(2*(a^2 - b^2))
 

3.5.59.3.1 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 217
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 1083
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3139
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre 
eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + 2*b*e*x + a 
*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ 
[a^2 - b^2, 0]
 

rule 3173
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + 
f*x])^(m + 1)/(f*g*(a^2 - b^2)*(m + 1))), x] + Simp[1/((a^2 - b^2)*(m + 1)) 
   Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1)*(a*(m + 1) - b*(m + 
p + 2)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b 
^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*p]
 

rule 3343
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c 
 - a*d))*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m + 1)/(f*g*(a^2 - 
 b^2)*(m + 1))), x] + Simp[1/((a^2 - b^2)*(m + 1))   Int[(g*Cos[e + f*x])^p 
*(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + p 
 + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ 
[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
 

rule 3345
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(g*Co 
s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c - a*d - (a*c - b*d)* 
Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Simp[1/(g^2*(a^2 - b^2)*(p + 
 1))   Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 
 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e + f*x], 
x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && Lt 
Q[p, -1] && IntegerQ[2*m]
 
3.5.59.4 Maple [A] (verified)

Time = 2.53 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.32

method result size
derivativedivides \(\frac {-\frac {1}{\left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 b^{2} \left (\frac {\frac {b^{2} \left (9 a^{2}-2 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {b \left (8 a^{4}+15 a^{2} b^{2}-2 b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{2}}+\frac {b^{2} \left (23 a^{2}-2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}+4 a^{2} b -\frac {b^{3}}{2}}{{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}^{2}}+\frac {3 \left (4 a^{2}+b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}-\frac {1}{\left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\) \(254\)
default \(\frac {-\frac {1}{\left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 b^{2} \left (\frac {\frac {b^{2} \left (9 a^{2}-2 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {b \left (8 a^{4}+15 a^{2} b^{2}-2 b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{2}}+\frac {b^{2} \left (23 a^{2}-2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}+4 a^{2} b -\frac {b^{3}}{2}}{{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}^{2}}+\frac {3 \left (4 a^{2}+b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}-\frac {1}{\left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\) \(254\)
risch \(\frac {i \left (-12 i a^{2} b^{3} {\mathrm e}^{5 i \left (d x +c \right )}-3 i b^{5} {\mathrm e}^{5 i \left (d x +c \right )}+16 i a^{4} b \,{\mathrm e}^{3 i \left (d x +c \right )}+12 i b^{3} a^{2} {\mathrm e}^{3 i \left (d x +c \right )}+2 i b^{5} {\mathrm e}^{3 i \left (d x +c \right )}+36 a^{3} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+9 a \,b^{4} {\mathrm e}^{4 i \left (d x +c \right )}+8 i a^{4} b \,{\mathrm e}^{i \left (d x +c \right )}+40 i a^{2} b^{3} {\mathrm e}^{i \left (d x +c \right )}-3 i b^{5} {\mathrm e}^{i \left (d x +c \right )}+8 a^{5} {\mathrm e}^{2 i \left (d x +c \right )}+18 a^{3} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+4 a \,b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-2 a^{3} b^{2}-13 a \,b^{4}\right )}{\left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right ) \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+i b \right )^{2} \left (a^{2}-b^{2}\right )^{3} d}-\frac {6 b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) a^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}-\frac {3 b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {6 b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) a^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {3 b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}\) \(618\)

input
int(sec(d*x+c)^2/(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
 
output
1/d*(-1/(a-b)^3/(tan(1/2*d*x+1/2*c)+1)-2*b^2/(a-b)^3/(a+b)^3*((1/2*b^2*(9* 
a^2-2*b^2)/a*tan(1/2*d*x+1/2*c)^3+1/2*b*(8*a^4+15*a^2*b^2-2*b^4)/a^2*tan(1 
/2*d*x+1/2*c)^2+1/2*b^2*(23*a^2-2*b^2)/a*tan(1/2*d*x+1/2*c)+4*a^2*b-1/2*b^ 
3)/(a*tan(1/2*d*x+1/2*c)^2+2*b*tan(1/2*d*x+1/2*c)+a)^2+3/2*(4*a^2+b^2)/(a^ 
2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2)))-1/( 
a+b)^3/(tan(1/2*d*x+1/2*c)-1))
 
3.5.59.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 405 vs. \(2 (180) = 360\).

Time = 0.33 (sec) , antiderivative size = 894, normalized size of antiderivative = 4.66 \[ \int \frac {\sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\left [\frac {4 \, a^{6} b - 12 \, a^{4} b^{3} + 12 \, a^{2} b^{5} - 4 \, b^{7} + 2 \, {\left (4 \, a^{6} b + 10 \, a^{4} b^{3} - 17 \, a^{2} b^{5} + 3 \, b^{7}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left ({\left (4 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (4 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (4 \, a^{4} b^{2} + 5 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 2 \, {\left (2 \, a^{7} - 6 \, a^{5} b^{2} + 6 \, a^{3} b^{4} - 2 \, a b^{6} - {\left (2 \, a^{5} b^{2} + 11 \, a^{3} b^{4} - 13 \, a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{8} b^{2} - 4 \, a^{6} b^{4} + 6 \, a^{4} b^{6} - 4 \, a^{2} b^{8} + b^{10}\right )} d \cos \left (d x + c\right )^{3} - 2 \, {\left (a^{9} b - 4 \, a^{7} b^{3} + 6 \, a^{5} b^{5} - 4 \, a^{3} b^{7} + a b^{9}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (a^{10} - 3 \, a^{8} b^{2} + 2 \, a^{6} b^{4} + 2 \, a^{4} b^{6} - 3 \, a^{2} b^{8} + b^{10}\right )} d \cos \left (d x + c\right )\right )}}, \frac {2 \, a^{6} b - 6 \, a^{4} b^{3} + 6 \, a^{2} b^{5} - 2 \, b^{7} + {\left (4 \, a^{6} b + 10 \, a^{4} b^{3} - 17 \, a^{2} b^{5} + 3 \, b^{7}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left ({\left (4 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (4 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (4 \, a^{4} b^{2} + 5 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - {\left (2 \, a^{7} - 6 \, a^{5} b^{2} + 6 \, a^{3} b^{4} - 2 \, a b^{6} - {\left (2 \, a^{5} b^{2} + 11 \, a^{3} b^{4} - 13 \, a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{8} b^{2} - 4 \, a^{6} b^{4} + 6 \, a^{4} b^{6} - 4 \, a^{2} b^{8} + b^{10}\right )} d \cos \left (d x + c\right )^{3} - 2 \, {\left (a^{9} b - 4 \, a^{7} b^{3} + 6 \, a^{5} b^{5} - 4 \, a^{3} b^{7} + a b^{9}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (a^{10} - 3 \, a^{8} b^{2} + 2 \, a^{6} b^{4} + 2 \, a^{4} b^{6} - 3 \, a^{2} b^{8} + b^{10}\right )} d \cos \left (d x + c\right )\right )}}\right ] \]

input
integrate(sec(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="fricas")
 
output
[1/4*(4*a^6*b - 12*a^4*b^3 + 12*a^2*b^5 - 4*b^7 + 2*(4*a^6*b + 10*a^4*b^3 
- 17*a^2*b^5 + 3*b^7)*cos(d*x + c)^2 + 3*((4*a^2*b^4 + b^6)*cos(d*x + c)^3 
 - 2*(4*a^3*b^3 + a*b^5)*cos(d*x + c)*sin(d*x + c) - (4*a^4*b^2 + 5*a^2*b^ 
4 + b^6)*cos(d*x + c))*sqrt(-a^2 + b^2)*log(((2*a^2 - b^2)*cos(d*x + c)^2 
- 2*a*b*sin(d*x + c) - a^2 - b^2 + 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos( 
d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 
 - b^2)) - 2*(2*a^7 - 6*a^5*b^2 + 6*a^3*b^4 - 2*a*b^6 - (2*a^5*b^2 + 11*a^ 
3*b^4 - 13*a*b^6)*cos(d*x + c)^2)*sin(d*x + c))/((a^8*b^2 - 4*a^6*b^4 + 6* 
a^4*b^6 - 4*a^2*b^8 + b^10)*d*cos(d*x + c)^3 - 2*(a^9*b - 4*a^7*b^3 + 6*a^ 
5*b^5 - 4*a^3*b^7 + a*b^9)*d*cos(d*x + c)*sin(d*x + c) - (a^10 - 3*a^8*b^2 
 + 2*a^6*b^4 + 2*a^4*b^6 - 3*a^2*b^8 + b^10)*d*cos(d*x + c)), 1/2*(2*a^6*b 
 - 6*a^4*b^3 + 6*a^2*b^5 - 2*b^7 + (4*a^6*b + 10*a^4*b^3 - 17*a^2*b^5 + 3* 
b^7)*cos(d*x + c)^2 + 3*((4*a^2*b^4 + b^6)*cos(d*x + c)^3 - 2*(4*a^3*b^3 + 
 a*b^5)*cos(d*x + c)*sin(d*x + c) - (4*a^4*b^2 + 5*a^2*b^4 + b^6)*cos(d*x 
+ c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d* 
x + c))) - (2*a^7 - 6*a^5*b^2 + 6*a^3*b^4 - 2*a*b^6 - (2*a^5*b^2 + 11*a^3* 
b^4 - 13*a*b^6)*cos(d*x + c)^2)*sin(d*x + c))/((a^8*b^2 - 4*a^6*b^4 + 6*a^ 
4*b^6 - 4*a^2*b^8 + b^10)*d*cos(d*x + c)^3 - 2*(a^9*b - 4*a^7*b^3 + 6*a^5* 
b^5 - 4*a^3*b^7 + a*b^9)*d*cos(d*x + c)*sin(d*x + c) - (a^10 - 3*a^8*b^2 + 
 2*a^6*b^4 + 2*a^4*b^6 - 3*a^2*b^8 + b^10)*d*cos(d*x + c))]
 
3.5.59.6 Sympy [F]

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

input
integrate(sec(d*x+c)**2/(a+b*sin(d*x+c))**3,x)
 
output
Integral(sec(c + d*x)**2/(a + b*sin(c + d*x))**3, x)
 
3.5.59.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {\sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]

input
integrate(sec(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="maxima")
 
output
Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 
3.5.59.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 385 vs. \(2 (180) = 360\).

Time = 0.54 (sec) , antiderivative size = 385, normalized size of antiderivative = 2.01 \[ \int \frac {\sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {\frac {3 \, {\left (4 \, a^{2} b^{2} + b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {2 \, {\left (a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{2} b - b^{3}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}} + \frac {9 \, a^{3} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, a^{4} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, a^{2} b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 23 \, a^{3} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, a^{4} b^{3} - a^{2} b^{5}}{{\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{2}}}{d} \]

input
integrate(sec(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="giac")
 
output
-(3*(4*a^2*b^2 + b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a 
*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 
 - b^6)*sqrt(a^2 - b^2)) + 2*(a^3*tan(1/2*d*x + 1/2*c) + 3*a*b^2*tan(1/2*d 
*x + 1/2*c) - 3*a^2*b - b^3)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(tan(1/2 
*d*x + 1/2*c)^2 - 1)) + (9*a^3*b^4*tan(1/2*d*x + 1/2*c)^3 - 2*a*b^6*tan(1/ 
2*d*x + 1/2*c)^3 + 8*a^4*b^3*tan(1/2*d*x + 1/2*c)^2 + 15*a^2*b^5*tan(1/2*d 
*x + 1/2*c)^2 - 2*b^7*tan(1/2*d*x + 1/2*c)^2 + 23*a^3*b^4*tan(1/2*d*x + 1/ 
2*c) - 2*a*b^6*tan(1/2*d*x + 1/2*c) + 8*a^4*b^3 - a^2*b^5)/((a^8 - 3*a^6*b 
^2 + 3*a^4*b^4 - a^2*b^6)*(a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/ 
2*c) + a)^2))/d
 
3.5.59.9 Mupad [B] (verification not implemented)

Time = 8.45 (sec) , antiderivative size = 650, normalized size of antiderivative = 3.39 \[ \int \frac {\sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,a^5-2\,a^3\,b^2+15\,a\,b^4\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {6\,a^4\,b+10\,a^2\,b^3-b^5}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,a^6+6\,a^4\,b^2+9\,a^2\,b^4-2\,b^6\right )}{a\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^6\,b+2\,a^4\,b^3+12\,a^2\,b^5-b^7\right )}{a^2\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (2\,a^6\,b+30\,a^4\,b^3+15\,a^2\,b^5-2\,b^7\right )}{a^2\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^6-18\,a^4\,b^2-31\,a^2\,b^4+2\,b^6\right )}{a\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-a^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^2+4\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2+4\,b^2\right )+4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {3\,b^2\,\mathrm {atan}\left (\frac {\frac {3\,b^2\,\left (4\,a^2+b^2\right )\,\left (2\,a^6\,b-6\,a^4\,b^3+6\,a^2\,b^5-2\,b^7\right )}{2\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}+\frac {3\,a\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,a^2+b^2\right )\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}{{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}}{12\,a^2\,b^2+3\,b^4}\right )\,\left (4\,a^2+b^2\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \]

input
int(1/(cos(c + d*x)^2*(a + b*sin(c + d*x))^3),x)
 
output
- ((2*tan(c/2 + (d*x)/2)^3*(15*a*b^4 + 2*a^5 - 2*a^3*b^2))/(a^6 - b^6 + 3* 
a^2*b^4 - 3*a^4*b^2) - (6*a^4*b - b^5 + 10*a^2*b^3)/(a^6 - b^6 + 3*a^2*b^4 
 - 3*a^4*b^2) + (tan(c/2 + (d*x)/2)^5*(2*a^6 - 2*b^6 + 9*a^2*b^4 + 6*a^4*b 
^2))/(a*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (2*tan(c/2 + (d*x)/2)^2*(2* 
a^6*b - b^7 + 12*a^2*b^5 + 2*a^4*b^3))/(a^2*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4 
*b^2)) + (tan(c/2 + (d*x)/2)^4*(2*a^6*b - 2*b^7 + 15*a^2*b^5 + 30*a^4*b^3) 
)/(a^2*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (tan(c/2 + (d*x)/2)*(2*a^6 + 
 2*b^6 - 31*a^2*b^4 - 18*a^4*b^2))/(a*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) 
)/(d*(a^2*tan(c/2 + (d*x)/2)^6 - a^2 - tan(c/2 + (d*x)/2)^2*(a^2 + 4*b^2) 
+ tan(c/2 + (d*x)/2)^4*(a^2 + 4*b^2) + 4*a*b*tan(c/2 + (d*x)/2)^5 - 4*a*b* 
tan(c/2 + (d*x)/2))) - (3*b^2*atan(((3*b^2*(4*a^2 + b^2)*(2*a^6*b - 2*b^7 
+ 6*a^2*b^5 - 6*a^4*b^3))/(2*(a + b)^(7/2)*(a - b)^(7/2)) + (3*a*b^2*tan(c 
/2 + (d*x)/2)*(4*a^2 + b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2))/((a + b)^ 
(7/2)*(a - b)^(7/2)))/(3*b^4 + 12*a^2*b^2))*(4*a^2 + b^2))/(d*(a + b)^(7/2 
)*(a - b)^(7/2))