Integrand size = 21, antiderivative size = 119 \[ \int \frac {\sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {2 b^3 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 \sqrt {a-b} \sqrt {a+b} d}+\frac {\left (a^2+2 b^2\right ) \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {b \tan (c+d x)}{a^2 d}+\frac {\sec (c+d x) \tan (c+d x)}{2 a d} \] Output:
-2*b^3*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^3/(a-b)^(1/2)/ (a+b)^(1/2)/d+1/2*(a^2+2*b^2)*arctanh(sin(d*x+c))/a^3/d-b*tan(d*x+c)/a^2/d +1/2*sec(d*x+c)*tan(d*x+c)/a/d
Time = 1.60 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.00 \[ \int \frac {\sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\frac {8 b^3 \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-2 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-4 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {a^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-4 a b \tan (c+d x)}{4 a^3 d} \] Input:
Integrate[Sec[c + d*x]^3/(a + b*Cos[c + d*x]),x]
Output:
((8*b^3*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] - 2*a^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 4*b^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*a^2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/ 2]] + 4*b^2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + a^2/(Cos[(c + d*x)/ 2] - Sin[(c + d*x)/2])^2 - a^2/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - 4 *a*b*Tan[c + d*x])/(4*a^3*d)
Time = 0.82 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.09, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 3281, 25, 3042, 3534, 25, 3042, 3480, 3042, 3138, 218, 4257}
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.
| \(\displaystyle \int \frac {\sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 3281 |
| \(\displaystyle \frac {\int -\frac {\left (-b \cos ^2(c+d x)-a \cos (c+d x)+2 b\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 a}+\frac {\tan (c+d x) \sec (c+d x)}{2 a d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\int \frac {\left (-b \cos ^2(c+d x)-a \cos (c+d x)+2 b\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\int \frac {-b \sin \left (c+d x+\frac {\pi }{2}\right )^2-a \sin \left (c+d x+\frac {\pi }{2}\right )+2 b}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {\int -\frac {\left (a^2+b \cos (c+d x) a+2 b^2\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}+\frac {2 b \tan (c+d x)}{a d}}{2 a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 b \tan (c+d x)}{a d}-\frac {\int \frac {\left (a^2+b \cos (c+d x) a+2 b^2\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 b \tan (c+d x)}{a d}-\frac {\int \frac {a^2+b \sin \left (c+d x+\frac {\pi }{2}\right ) a+2 b^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{2 a}\) |
| \(\Big \downarrow \) 3480 |
| \(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 b \tan (c+d x)}{a d}-\frac {\frac {\left (a^2+2 b^2\right ) \int \sec (c+d x)dx}{a}-\frac {2 b^3 \int \frac {1}{a+b \cos (c+d x)}dx}{a}}{a}}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 b \tan (c+d x)}{a d}-\frac {\frac {\left (a^2+2 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b^3 \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a}}{2 a}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 b \tan (c+d x)}{a d}-\frac {\frac {\left (a^2+2 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {4 b^3 \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}}{a}}{2 a}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 b \tan (c+d x)}{a d}-\frac {\frac {\left (a^2+2 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {4 b^3 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}}{2 a}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 b \tan (c+d x)}{a d}-\frac {\frac {\left (a^2+2 b^2\right ) \text {arctanh}(\sin (c+d x))}{a d}-\frac {4 b^3 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}}{2 a}\) |
Input:
Int[Sec[c + d*x]^3/(a + b*Cos[c + d*x]),x]
Output:
(Sec[c + d*x]*Tan[c + d*x])/(2*a*d) - (-(((-4*b^3*ArcTan[(Sqrt[a - b]*Tan[ (c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d) + ((a^2 + 2*b^2) *ArcTanh[Sin[c + d*x]])/(a*d))/a) + (2*b*Tan[c + d*x])/(a*d))/(2*a)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*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[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2 ))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n + 3)*Si n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[ 2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2* n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0])))
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A*b - a*B)/(b*c - a*d) Int[1/(a + b*Sin[e + f*x]), x], x] + Simp[(B*c - A*d)/ (b*c - a*d) Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int [(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A *b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ [n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) | | EqQ[a, 0])))
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 1.35 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.61
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-a -2 b}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3}}+\frac {1}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-a -2 b}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-a^{2}-2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3}}-\frac {2 b^{3} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(192\) |
| default | \(\frac {-\frac {1}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-a -2 b}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3}}+\frac {1}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-a -2 b}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-a^{2}-2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3}}-\frac {2 b^{3} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(192\) |
| risch | \(-\frac {i \left ({\mathrm e}^{3 i \left (d x +c \right )} a +2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-a \,{\mathrm e}^{i \left (d x +c \right )}+2 b \right )}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{a^{3} d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,a^{3}}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,a^{3}}\) | \(301\) |
Input:
int(sec(d*x+c)^3/(a+cos(d*x+c)*b),x,method=_RETURNVERBOSE)
Output:
1/d*(-1/2/a/(tan(1/2*d*x+1/2*c)+1)^2-1/2*(-a-2*b)/a^2/(tan(1/2*d*x+1/2*c)+ 1)+1/2*(a^2+2*b^2)/a^3*ln(tan(1/2*d*x+1/2*c)+1)+1/2/a/(tan(1/2*d*x+1/2*c)- 1)^2-1/2*(-a-2*b)/a^2/(tan(1/2*d*x+1/2*c)-1)+1/2/a^3*(-a^2-2*b^2)*ln(tan(1 /2*d*x+1/2*c)-1)-2*b^3/a^3/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/ 2*c)/((a-b)*(a+b))^(1/2)))
Time = 0.20 (sec) , antiderivative size = 459, normalized size of antiderivative = 3.86 \[ \int \frac {\sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\left [-\frac {2 \, \sqrt {-a^{2} + b^{2}} b^{3} \cos \left (d x + c\right )^{2} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - {\left (a^{4} + a^{2} b^{2} - 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{4} + a^{2} b^{2} - 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{4} - a^{2} b^{2} - 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{2}}, -\frac {4 \, \sqrt {a^{2} - b^{2}} b^{3} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{2} - {\left (a^{4} + a^{2} b^{2} - 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{4} + a^{2} b^{2} - 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{4} - a^{2} b^{2} - 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{2}}\right ] \] Input:
integrate(sec(d*x+c)^3/(a+b*cos(d*x+c)),x, algorithm="fricas")
Output:
[-1/4*(2*sqrt(-a^2 + b^2)*b^3*cos(d*x + c)^2*log((2*a*b*cos(d*x + c) + (2* a^2 - b^2)*cos(d*x + c)^2 - 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d* x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) - ( a^4 + a^2*b^2 - 2*b^4)*cos(d*x + c)^2*log(sin(d*x + c) + 1) + (a^4 + a^2*b ^2 - 2*b^4)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) - 2*(a^4 - a^2*b^2 - 2*( a^3*b - a*b^3)*cos(d*x + c))*sin(d*x + c))/((a^5 - a^3*b^2)*d*cos(d*x + c) ^2), -1/4*(4*sqrt(a^2 - b^2)*b^3*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c)))*cos(d*x + c)^2 - (a^4 + a^2*b^2 - 2*b^4)*cos(d*x + c)^ 2*log(sin(d*x + c) + 1) + (a^4 + a^2*b^2 - 2*b^4)*cos(d*x + c)^2*log(-sin( d*x + c) + 1) - 2*(a^4 - a^2*b^2 - 2*(a^3*b - a*b^3)*cos(d*x + c))*sin(d*x + c))/((a^5 - a^3*b^2)*d*cos(d*x + c)^2)]
\[ \int \frac {\sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{a + b \cos {\left (c + d x \right )}}\, dx \] Input:
integrate(sec(d*x+c)**3/(a+b*cos(d*x+c)),x)
Output:
Integral(sec(c + d*x)**3/(a + b*cos(c + d*x)), x)
Exception generated. \[ \int \frac {\sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(sec(d*x+c)^3/(a+b*cos(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
Time = 0.59 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.77 \[ \int \frac {\sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\frac {4 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} b^{3}}{\sqrt {a^{2} - b^{2}} a^{3}} + \frac {{\left (a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {{\left (a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {2 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{2}}}{2 \, d} \] Input:
integrate(sec(d*x+c)^3/(a+b*cos(d*x+c)),x, algorithm="giac")
Output:
1/2*(4*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan( 1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))*b^3/(sqrt(a^2 - b^2)*a^3) + (a^2 + 2*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - (a^2 + 2*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 + 2*(a*tan(1/2*d*x + 1/2* c)^3 + 2*b*tan(1/2*d*x + 1/2*c)^3 + a*tan(1/2*d*x + 1/2*c) - 2*b*tan(1/2*d *x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^2))/d
Time = 42.54 (sec) , antiderivative size = 1087, normalized size of antiderivative = 9.13 \[ \int \frac {\sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \] Input:
int(1/(cos(c + d*x)^3*(a + b*cos(c + d*x))),x)
Output:
(a*(sin(c + d*x)/2 + atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))/2 + (ata nh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x))/2))/(d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) + ((b^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 - (b^2*sin(c + d*x))/2 + (b^2*atanh(sin(c/2 + (d*x)/2)/cos( c/2 + (d*x)/2))*cos(2*c + 2*d*x))/2)/(a*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (b*sin(2*c + 2*d*x))/(2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2) ) - (b^3*atan(((a^9*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 8*b^7*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2) - 8*b^9*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 8*a^2*b^7*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 3*a^4*b^5*sin(c/2 + (d*x )/2)*(b^2 - a^2)^(1/2) - 3*a^5*b^4*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - 2*a^6*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 2*a^7*b^2*sin(c/2 + (d*x) /2)*(b^2 - a^2)^(1/2) - a^8*b*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))*1i)/(c os(c/2 + (d*x)/2)*(a*b^2 - a^3)*(a^7 - 3*a^3*b^4 + 2*a^5*b^2)))*1i)/(a^3*d *(b^2 - a^2)^(1/2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (b^4*atanh(sin(c/2 + (d*x )/2)/cos(c/2 + (d*x)/2)))/(a^3*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) + (b^3*sin(2*c + 2*d*x))/(2*a^2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (b^3*atan(((a^9*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 8*b^7*sin(c/2 + (d *x)/2)*(b^2 - a^2)^(3/2) - 8*b^9*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 8* a^2*b^7*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 3*a^4*b^5*sin(c/2 + (d*x)/2 )*(b^2 - a^2)^(1/2) - 3*a^5*b^4*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - ...
Time = 0.17 (sec) , antiderivative size = 479, normalized size of antiderivative = 4.03 \[ \int \frac {\sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {-4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d x +c \right )^{2} b^{3}+4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) b^{3}+2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} b -2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{3}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{4}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{2} 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} b^{4}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{4}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} b^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{4}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{4}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{2} 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} b^{4}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{4}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} b^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{4}-\sin \left (d x +c \right ) a^{4}+\sin \left (d x +c \right ) a^{2} b^{2}}{2 a^{3} d \left (\sin \left (d x +c \right )^{2} a^{2}-\sin \left (d x +c \right )^{2} b^{2}-a^{2}+b^{2}\right )} \] Input:
int(sec(d*x+c)^3/(a+b*cos(d*x+c)),x)
Output:
( - 4*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqr t(a**2 - b**2))*sin(c + d*x)**2*b**3 + 4*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*b**3 + 2*cos(c + d*x)*s in(c + d*x)*a**3*b - 2*cos(c + d*x)*sin(c + d*x)*a*b**3 - log(tan((c + d*x )/2) - 1)*sin(c + d*x)**2*a**4 - log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 *a**2*b**2 + 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**4 + log(tan((c + d*x)/2) - 1)*a**4 + log(tan((c + d*x)/2) - 1)*a**2*b**2 - 2*log(tan((c + d*x)/2) - 1)*b**4 + log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**4 + log (tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b**2 - 2*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b**4 - log(tan((c + d*x)/2) + 1)*a**4 - log(tan((c + d*x)/2) + 1)*a**2*b**2 + 2*log(tan((c + d*x)/2) + 1)*b**4 - sin(c + d*x)*a **4 + sin(c + d*x)*a**2*b**2)/(2*a**3*d*(sin(c + d*x)**2*a**2 - sin(c + d* x)**2*b**2 - a**2 + b**2))