\(\int \frac {\sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx\) [432]

Optimal result

Integrand size = 21, antiderivative size = 137 \[ \int \frac {\sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 b^4 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} d}-\frac {\sec ^3(c+d x) (b-a \sin (c+d x))}{3 \left (a^2-b^2\right ) d}+\frac {\sec (c+d x) \left (3 b^3+a \left (2 a^2-5 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d} \] Output:

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

Mathematica [A] (verified)

Time = 1.74 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.47 \[ \int \frac {\sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {24 b^4 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {\sec ^3(c+d x) \left (-4 a^2 b+10 b^3+\frac {3}{2} b \left (a^2-7 b^2\right ) \cos (c+d x)+6 b^3 \cos (2 (c+d x))+\frac {1}{2} a^2 b \cos (3 (c+d x))-\frac {7}{2} b^3 \cos (3 (c+d x))+6 a^3 \sin (c+d x)-9 a b^2 \sin (c+d x)+2 a^3 \sin (3 (c+d x))-5 a b^2 \sin (3 (c+d x))\right )}{(a-b)^2 (a+b)^2}}{12 d} \] Input:

Integrate[Sec[c + d*x]^4/(a + b*Sin[c + d*x]),x]
 

Output:

((24*b^4*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/ 
2) + (Sec[c + d*x]^3*(-4*a^2*b + 10*b^3 + (3*b*(a^2 - 7*b^2)*Cos[c + d*x]) 
/2 + 6*b^3*Cos[2*(c + d*x)] + (a^2*b*Cos[3*(c + d*x)])/2 - (7*b^3*Cos[3*(c 
 + d*x)])/2 + 6*a^3*Sin[c + d*x] - 9*a*b^2*Sin[c + d*x] + 2*a^3*Sin[3*(c + 
 d*x)] - 5*a*b^2*Sin[3*(c + d*x)]))/((a - b)^2*(a + b)^2))/(12*d)
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.14, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3175, 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.

\(\displaystyle \int \frac {\sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\cos (c+d x)^4 (a+b \sin (c+d x))}dx\)

\(\Big \downarrow \) 3175

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3345

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

Input:

Int[Sec[c + d*x]^4/(a + b*Sin[c + d*x]),x]
 

Output:

-1/3*(Sec[c + d*x]^3*(b - a*Sin[c + d*x]))/((a^2 - b^2)*d) + ((6*b^4*ArcTa 
n[(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^3 + a*(2*a^2 - 5*b^2)*Sin[c + d*x]))/((a^2 - b^2)*d) 
)/(3*(a^2 - b^2))
 

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 3175

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^ 
(m + 1)*((b - a*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* 
(a^2*(p + 2) - b^2*(m + p + 2) + a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; F 
reeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegersQ 
[2*m, 2*p]
 

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]
 

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.57

method	result	size
derivativedivides	\(\frac {-\frac {2}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3} \left (2 a -2 b \right )}+\frac {1}{\left (2 a -2 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 a -3 b}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {2 b^{4} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}}-\frac {2}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (2 a +2 b \right )}-\frac {1}{\left (2 a +2 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 a +3 b}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\)	\(215\)
default	\(\frac {-\frac {2}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3} \left (2 a -2 b \right )}+\frac {1}{\left (2 a -2 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 a -3 b}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {2 b^{4} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}}-\frac {2}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (2 a +2 b \right )}-\frac {1}{\left (2 a +2 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 a +3 b}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\)	\(215\)
risch	\(\frac {-2 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+2 b^{3} {\mathrm e}^{5 i \left (d x +c \right )}+4 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-8 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-\frac {8 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}}{3}+\frac {20 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}}{3}+\frac {4 i a^{3}}{3}-\frac {10 i a \,b^{2}}{3}+2 b^{3} {\mathrm e}^{i \left (d x +c \right )}}{d \left (-a^{2}+b^{2}\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}\)	\(310\)

Input:

int(sec(d*x+c)^4/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

1/d*(-2/3/(tan(1/2*d*x+1/2*c)+1)^3/(2*a-2*b)+1/(2*a-2*b)/(tan(1/2*d*x+1/2* 
c)+1)^2-1/2*(2*a-3*b)/(a-b)^2/(tan(1/2*d*x+1/2*c)+1)+2*b^4/(a-b)^2/(a+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))- 
2/3/(tan(1/2*d*x+1/2*c)-1)^3/(2*a+2*b)-1/(2*a+2*b)/(tan(1/2*d*x+1/2*c)-1)^ 
2-1/2*(2*a+3*b)/(a+b)^2/(tan(1/2*d*x+1/2*c)-1))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 466, normalized size of antiderivative = 3.40 \[ \int \frac {\sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\left [-\frac {3 \, \sqrt {-a^{2} + b^{2}} b^{4} \cos \left (d x + c\right )^{3} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 2 \, a^{4} b - 4 \, a^{2} b^{3} + 2 \, b^{5} - 6 \, {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (2 \, a^{5} - 7 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}, -\frac {3 \, \sqrt {a^{2} - b^{2}} b^{4} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{3} + a^{4} b - 2 \, a^{2} b^{3} + b^{5} - 3 \, {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (2 \, a^{5} - 7 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}\right ] \] Input:

integrate(sec(d*x+c)^4/(a+b*sin(d*x+c)),x, algorithm="fricas")
 

Output:

[-1/6*(3*sqrt(-a^2 + b^2)*b^4*cos(d*x + c)^3*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*a^4*b - 4*a^2*b^3 + 2*b^5 - 6*(a^2*b^3 - b^5)*cos(d*x + 
c)^2 - 2*(a^5 - 2*a^3*b^2 + a*b^4 + (2*a^5 - 7*a^3*b^2 + 5*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)^ 
3), -1/3*(3*sqrt(a^2 - b^2)*b^4*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b 
^2)*cos(d*x + c)))*cos(d*x + c)^3 + a^4*b - 2*a^2*b^3 + b^5 - 3*(a^2*b^3 - 
 b^5)*cos(d*x + c)^2 - (a^5 - 2*a^3*b^2 + a*b^4 + (2*a^5 - 7*a^3*b^2 + 5*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)^3)]
 

Sympy [F]

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

integrate(sec(d*x+c)**4/(a+b*sin(d*x+c)),x)
 

Output:

Integral(sec(c + d*x)**4/(a + b*sin(c + d*x)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(sec(d*x+c)^4/(a+b*sin(d*x+c)),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
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 273 vs. \(2 (129) = 258\).

Time = 0.15 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.99 \[ \int \frac {\sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b^{4}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} - \frac {3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 6 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2} b + 4 \, b^{3}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}\right )}}{3 \, d} \] Input:

integrate(sec(d*x+c)^4/(a+b*sin(d*x+c)),x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 18.26 (sec) , antiderivative size = 387, normalized size of antiderivative = 2.82 \[ \int \frac {\sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2\,\left (a^2\,b-4\,b^3\right )}{3\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a\,b^2-a^3\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {4\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^4-2\,a^2\,b^2+b^4}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (4\,a\,b^2-a^3\right )}{3\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,a\,b^2-a^3\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2-2\,b^2\right )}{a^4-2\,a^2\,b^2+b^4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}+\frac {2\,b^4\,\mathrm {atan}\left (\frac {\frac {b^4\,\left (2\,a^4\,b-4\,a^2\,b^3+2\,b^5\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}+\frac {2\,a\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}}{2\,b^4}\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \] Input:

int(1/(cos(c + d*x)^4*(a + b*sin(c + d*x))),x)
 

Output:

((2*(a^2*b - 4*b^3))/(3*(a^4 + b^4 - 2*a^2*b^2)) + (2*tan(c/2 + (d*x)/2)*( 
2*a*b^2 - a^3))/(a^4 + b^4 - 2*a^2*b^2) + (4*b^3*tan(c/2 + (d*x)/2)^2)/(a^ 
4 + b^4 - 2*a^2*b^2) - (4*tan(c/2 + (d*x)/2)^3*(4*a*b^2 - a^3))/(3*(a^4 + 
b^4 - 2*a^2*b^2)) + (2*tan(c/2 + (d*x)/2)^5*(2*a*b^2 - a^3))/(a^4 + b^4 - 
2*a^2*b^2) + (2*b*tan(c/2 + (d*x)/2)^4*(a^2 - 2*b^2))/(a^4 + b^4 - 2*a^2*b 
^2))/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x) 
/2)^6 - 1)) + (2*b^4*atan(((b^4*(2*a^4*b + 2*b^5 - 4*a^2*b^3))/((a + b)^(5 
/2)*(a - b)^(5/2)) + (2*a*b^4*tan(c/2 + (d*x)/2)*(a^4 + b^4 - 2*a^2*b^2))/ 
((a + b)^(5/2)*(a - b)^(5/2)))/(2*b^4)))/(d*(a + b)^(5/2)*(a - b)^(5/2))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 395, normalized size of antiderivative = 2.88 \[ \int \frac {\sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {6 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{4}-6 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \cos \left (d x +c \right ) b^{4}+\cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{4} b -\cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{2} b^{3}-\cos \left (d x +c \right ) a^{4} b +\cos \left (d x +c \right ) a^{2} b^{3}+2 \sin \left (d x +c \right )^{3} a^{5}-7 \sin \left (d x +c \right )^{3} a^{3} b^{2}+5 \sin \left (d x +c \right )^{3} a \,b^{4}+3 \sin \left (d x +c \right )^{2} a^{2} b^{3}-3 \sin \left (d x +c \right )^{2} b^{5}-3 \sin \left (d x +c \right ) a^{5}+9 \sin \left (d x +c \right ) a^{3} b^{2}-6 \sin \left (d x +c \right ) a \,b^{4}+a^{4} b -5 a^{2} b^{3}+4 b^{5}}{3 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2} a^{6}-3 \sin \left (d x +c \right )^{2} a^{4} b^{2}+3 \sin \left (d x +c \right )^{2} a^{2} b^{4}-\sin \left (d x +c \right )^{2} b^{6}-a^{6}+3 a^{4} b^{2}-3 a^{2} b^{4}+b^{6}\right )} \] Input:

int(sec(d*x+c)^4/(a+b*sin(d*x+c)),x)
 

Output:

(6*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*cos( 
c + d*x)*sin(c + d*x)**2*b**4 - 6*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2) 
*a + b)/sqrt(a**2 - b**2))*cos(c + d*x)*b**4 + cos(c + d*x)*sin(c + d*x)** 
2*a**4*b - cos(c + d*x)*sin(c + d*x)**2*a**2*b**3 - cos(c + d*x)*a**4*b + 
cos(c + d*x)*a**2*b**3 + 2*sin(c + d*x)**3*a**5 - 7*sin(c + d*x)**3*a**3*b 
**2 + 5*sin(c + d*x)**3*a*b**4 + 3*sin(c + d*x)**2*a**2*b**3 - 3*sin(c + d 
*x)**2*b**5 - 3*sin(c + d*x)*a**5 + 9*sin(c + d*x)*a**3*b**2 - 6*sin(c + d 
*x)*a*b**4 + a**4*b - 5*a**2*b**3 + 4*b**5)/(3*cos(c + d*x)*d*(sin(c + d*x 
)**2*a**6 - 3*sin(c + d*x)**2*a**4*b**2 + 3*sin(c + d*x)**2*a**2*b**4 - si 
n(c + d*x)**2*b**6 - a**6 + 3*a**4*b**2 - 3*a**2*b**4 + b**6))