Integrand size = 21, antiderivative size = 137 \[ \int \frac {\sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 b^4 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} d}-\frac {\sec ^3(c+d x) (b-a \sin (c+d x))}{3 \left (a^2-b^2\right ) d}+\frac {\sec (c+d x) \left (3 b^3+a \left (2 a^2-5 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d} \]
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
Time = 1.14 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.47 \[ \int \frac {\sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {24 b^4 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {\sec ^3(c+d x) \left (-4 a^2 b+10 b^3+\frac {3}{2} b \left (a^2-7 b^2\right ) \cos (c+d x)+6 b^3 \cos (2 (c+d x))+\frac {1}{2} a^2 b \cos (3 (c+d x))-\frac {7}{2} b^3 \cos (3 (c+d x))+6 a^3 \sin (c+d x)-9 a b^2 \sin (c+d x)+2 a^3 \sin (3 (c+d x))-5 a b^2 \sin (3 (c+d x))\right )}{(a-b)^2 (a+b)^2}}{12 d} \]
((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)
Time = 0.63 (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 |
\(\displaystyle \frac {\frac {6 b^4 \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}+\frac {\sec (c+d x) \left (a \left (2 a^2-5 b^2\right ) \sin (c+d x)+3 b^3\right )}{d \left (a^2-b^2\right )}}{3 \left (a^2-b^2\right )}-\frac {\sec ^3(c+d x) (b-a \sin (c+d x))}{3 d \left (a^2-b^2\right )}\) |
-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))
3.5.35.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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]
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]
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]
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]
Time = 1.05 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.57
method | result | size |
derivativedivides | \(\frac {\frac {2 b^{4} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}}-\frac {2}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (2 a +2 b \right )}-\frac {1}{\left (2 a +2 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 a +3 b}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3} \left (2 a -2 b \right )}+\frac {1}{\left (2 a -2 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 a -3 b}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\) | \(215\) |
default | \(\frac {\frac {2 b^{4} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}}-\frac {2}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (2 a +2 b \right )}-\frac {1}{\left (2 a +2 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 a +3 b}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3} \left (2 a -2 b \right )}+\frac {1}{\left (2 a -2 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 a -3 b}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\) | \(215\) |
risch | \(\frac {-2 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+2 b^{3} {\mathrm e}^{5 i \left (d x +c \right )}+4 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-8 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-\frac {8 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}}{3}+\frac {20 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}}{3}+\frac {4 i a^{3}}{3}-\frac {10 i a \,b^{2}}{3}+2 b^{3} {\mathrm e}^{i \left (d x +c \right )}}{d \left (-a^{2}+b^{2}\right )^{2} \left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right )^{3}}-\frac {b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}\) | \(310\) |
1/d*(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)-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))
Time = 0.30 (sec) , antiderivative size = 466, normalized size of antiderivative = 3.40 \[ \int \frac {\sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\left [-\frac {3 \, \sqrt {-a^{2} + b^{2}} b^{4} \cos \left (d x + c\right )^{3} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 2 \, a^{4} b - 4 \, a^{2} b^{3} + 2 \, b^{5} - 6 \, {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (2 \, a^{5} - 7 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}, -\frac {3 \, \sqrt {a^{2} - b^{2}} b^{4} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{3} + a^{4} b - 2 \, a^{2} b^{3} + b^{5} - 3 \, {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (2 \, a^{5} - 7 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}\right ] \]
[-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)]
\[ \int \frac {\sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {\sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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
Leaf count of result is larger than twice the leaf count of optimal. 273 vs. \(2 (129) = 258\).
Time = 0.33 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.99 \[ \int \frac {\sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b^{4}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} - \frac {3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 6 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2} b + 4 \, b^{3}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}\right )}}{3 \, d} \]
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
Time = 7.61 (sec) , antiderivative size = 387, normalized size of antiderivative = 2.82 \[ \int \frac {\sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2\,\left (a^2\,b-4\,b^3\right )}{3\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a\,b^2-a^3\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {4\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^4-2\,a^2\,b^2+b^4}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (4\,a\,b^2-a^3\right )}{3\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,a\,b^2-a^3\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2-2\,b^2\right )}{a^4-2\,a^2\,b^2+b^4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}+\frac {2\,b^4\,\mathrm {atan}\left (\frac {\frac {b^4\,\left (2\,a^4\,b-4\,a^2\,b^3+2\,b^5\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}+\frac {2\,a\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}}{2\,b^4}\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \]
((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))