Integrand size = 21, antiderivative size = 157 \[ \int \frac {\sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {2 b^4 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 \sqrt {a-b} \sqrt {a+b} d}-\frac {b \left (a^2+2 b^2\right ) \text {arctanh}(\sin (c+d x))}{2 a^4 d}+\frac {\left (2 a^2+3 b^2\right ) \tan (c+d x)}{3 a^3 d}-\frac {b \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac {\sec ^2(c+d x) \tan (c+d x)}{3 a d} \] Output:
2*b^4*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^4/(a-b)^(1/2)/( a+b)^(1/2)/d-1/2*b*(a^2+2*b^2)*arctanh(sin(d*x+c))/a^4/d+1/3*(2*a^2+3*b^2) *tan(d*x+c)/a^3/d-1/2*b*sec(d*x+c)*tan(d*x+c)/a^2/d+1/3*sec(d*x+c)^2*tan(d *x+c)/a/d
Time = 3.01 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.64 \[ \int \frac {\sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {-\frac {24 b^4 \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+\frac {1}{2} \sec ^3(c+d x) \left (9 b \left (a^2+2 b^2\right ) \cos (c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+3 b \left (a^2+2 b^2\right ) \cos (3 (c+d x)) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+4 a \left (4 a^2+3 b^2-3 a b \cos (c+d x)+\left (2 a^2+3 b^2\right ) \cos (2 (c+d x))\right ) \sin (c+d x)\right )}{12 a^4 d} \] Input:
Integrate[Sec[c + d*x]^4/(a + b*Cos[c + d*x]),x]
Output:
((-24*b^4*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + (Sec[c + d*x]^3*(9*b*(a^2 + 2*b^2)*Cos[c + d*x]*(Log[Cos[(c + d*x )/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 3*b *(a^2 + 2*b^2)*Cos[3*(c + d*x)]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 4*a*(4*a^2 + 3*b^2 - 3*a*b*C os[c + d*x] + (2*a^2 + 3*b^2)*Cos[2*(c + d*x)])*Sin[c + d*x]))/2)/(12*a^4* d)
Time = 1.15 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.11, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 3281, 25, 3042, 3534, 25, 3042, 3534, 27, 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.
| \(\displaystyle \int \frac {\sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 3281 |
| \(\displaystyle \frac {\int -\frac {\left (-2 b \cos ^2(c+d x)-2 a \cos (c+d x)+3 b\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)}dx}{3 a}+\frac {\tan (c+d x) \sec ^2(c+d x)}{3 a d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\int \frac {\left (-2 b \cos ^2(c+d x)-2 a \cos (c+d x)+3 b\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)}dx}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\int \frac {-2 b \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 a \sin \left (c+d x+\frac {\pi }{2}\right )+3 b}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 a}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {\int -\frac {\left (-3 b^2 \cos ^2(c+d x)+a b \cos (c+d x)+2 \left (2 a^2+3 b^2\right )\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 a}+\frac {3 b \tan (c+d x) \sec (c+d x)}{2 a d}}{3 a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\int \frac {\left (-3 b^2 \cos ^2(c+d x)+a b \cos (c+d x)+2 \left (2 a^2+3 b^2\right )\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\int \frac {-3 b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2+a b \sin \left (c+d x+\frac {\pi }{2}\right )+2 \left (2 a^2+3 b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {\int -\frac {3 \left (a \cos (c+d x) b^2+\left (a^2+2 b^2\right ) b\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}+\frac {2 \left (2 a^2+3 b^2\right ) \tan (c+d x)}{a d}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \left (2 a^2+3 b^2\right ) \tan (c+d x)}{a d}-\frac {3 \int \frac {\left (a \cos (c+d x) b^2+\left (a^2+2 b^2\right ) b\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \left (2 a^2+3 b^2\right ) \tan (c+d x)}{a d}-\frac {3 \int \frac {a \sin \left (c+d x+\frac {\pi }{2}\right ) b^2+\left (a^2+2 b^2\right ) b}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3480 |
| \(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \left (2 a^2+3 b^2\right ) \tan (c+d x)}{a d}-\frac {3 \left (\frac {b \left (a^2+2 b^2\right ) \int \sec (c+d x)dx}{a}-\frac {2 b^4 \int \frac {1}{a+b \cos (c+d x)}dx}{a}\right )}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \left (2 a^2+3 b^2\right ) \tan (c+d x)}{a d}-\frac {3 \left (\frac {b \left (a^2+2 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b^4 \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}\right )}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \left (2 a^2+3 b^2\right ) \tan (c+d x)}{a d}-\frac {3 \left (\frac {b \left (a^2+2 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {4 b^4 \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}\right )}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \left (2 a^2+3 b^2\right ) \tan (c+d x)}{a d}-\frac {3 \left (\frac {b \left (a^2+2 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {4 b^4 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \left (2 a^2+3 b^2\right ) \tan (c+d x)}{a d}-\frac {3 \left (\frac {b \left (a^2+2 b^2\right ) \text {arctanh}(\sin (c+d x))}{a d}-\frac {4 b^4 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}}{2 a}}{3 a}\) |
Input:
Int[Sec[c + d*x]^4/(a + b*Cos[c + d*x]),x]
Output:
(Sec[c + d*x]^2*Tan[c + d*x])/(3*a*d) - ((3*b*Sec[c + d*x]*Tan[c + d*x])/( 2*a*d) - ((-3*((-4*b^4*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]]) /(a*Sqrt[a - b]*Sqrt[a + b]*d) + (b*(a^2 + 2*b^2)*ArcTanh[Sin[c + d*x]])/( a*d)))/a + (2*(2*a^2 + 3*b^2)*Tan[c + d*x])/(a*d))/(2*a))/(3*a)
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/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 = 1.44 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.61
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{3 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a +b}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 a^{2}+a b +2 b^{2}}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {b \left (a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{4}}+\frac {2 b^{4} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{3 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {-a -b}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 a^{2}+a b +2 b^{2}}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {b \left (a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{4}}}{d}\) | \(252\) |
| default | \(\frac {-\frac {1}{3 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a +b}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 a^{2}+a b +2 b^{2}}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {b \left (a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{4}}+\frac {2 b^{4} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{3 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {-a -b}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 a^{2}+a b +2 b^{2}}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {b \left (a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{4}}}{d}\) | \(252\) |
| risch | \(\frac {i \left (3 a b \,{\mathrm e}^{5 i \left (d x +c \right )}+6 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+12 \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2}+12 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 a b \,{\mathrm e}^{i \left (d x +c \right )}+4 a^{2}+6 b^{2}\right )}{3 a^{3} d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a^{2} d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{4} d}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a^{2} d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{4} d}-\frac {b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,a^{4}}+\frac {b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,a^{4}}\) | \(343\) |
Input:
int(sec(d*x+c)^4/(a+cos(d*x+c)*b),x,method=_RETURNVERBOSE)
Output:
1/d*(-1/3/a/(tan(1/2*d*x+1/2*c)-1)^3-1/2*(a+b)/a^2/(tan(1/2*d*x+1/2*c)-1)^ 2-1/2*(2*a^2+a*b+2*b^2)/a^3/(tan(1/2*d*x+1/2*c)-1)+1/2*b*(a^2+2*b^2)/a^4*l n(tan(1/2*d*x+1/2*c)-1)+2*b^4/a^4/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2 *d*x+1/2*c)/((a-b)*(a+b))^(1/2))-1/3/a/(tan(1/2*d*x+1/2*c)+1)^3-1/2*(-a-b) /a^2/(tan(1/2*d*x+1/2*c)+1)^2-1/2*(2*a^2+a*b+2*b^2)/a^3/(tan(1/2*d*x+1/2*c )+1)-1/2*b*(a^2+2*b^2)/a^4*ln(tan(1/2*d*x+1/2*c)+1))
Time = 0.20 (sec) , antiderivative size = 535, normalized size of antiderivative = 3.41 \[ \int \frac {\sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\left [-\frac {6 \, \sqrt {-a^{2} + b^{2}} b^{4} \cos \left (d x + c\right )^{3} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + 3 \, {\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, a^{5} - 2 \, a^{3} b^{2} + 2 \, {\left (2 \, a^{5} + a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{6} - a^{4} b^{2}\right )} d \cos \left (d x + c\right )^{3}}, \frac {12 \, \sqrt {a^{2} - b^{2}} b^{4} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{3} - 3 \, {\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, a^{5} - 2 \, a^{3} b^{2} + 2 \, {\left (2 \, a^{5} + a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{6} - a^{4} b^{2}\right )} d \cos \left (d x + c\right )^{3}}\right ] \] Input:
integrate(sec(d*x+c)^4/(a+b*cos(d*x+c)),x, algorithm="fricas")
Output:
[-1/12*(6*sqrt(-a^2 + b^2)*b^4*cos(d*x + c)^3*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)) + 3*(a^4*b + a^2*b^3 - 2*b^5)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(a^4* b + a^2*b^3 - 2*b^5)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) - 2*(2*a^5 - 2* a^3*b^2 + 2*(2*a^5 + a^3*b^2 - 3*a*b^4)*cos(d*x + c)^2 - 3*(a^4*b - a^2*b^ 3)*cos(d*x + c))*sin(d*x + c))/((a^6 - a^4*b^2)*d*cos(d*x + c)^3), 1/12*(1 2*sqrt(a^2 - b^2)*b^4*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d* x + c)))*cos(d*x + c)^3 - 3*(a^4*b + a^2*b^3 - 2*b^5)*cos(d*x + c)^3*log(s in(d*x + c) + 1) + 3*(a^4*b + a^2*b^3 - 2*b^5)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*a^5 - 2*a^3*b^2 + 2*(2*a^5 + a^3*b^2 - 3*a*b^4)*cos(d*x + c)^2 - 3*(a^4*b - a^2*b^3)*cos(d*x + c))*sin(d*x + c))/((a^6 - a^4*b^2)* d*cos(d*x + c)^3)]
\[ \int \frac {\sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{a + b \cos {\left (c + d x \right )}}\, dx \] Input:
integrate(sec(d*x+c)**4/(a+b*cos(d*x+c)),x)
Output:
Integral(sec(c + d*x)**4/(a + b*cos(c + d*x)), x)
Exception generated. \[ \int \frac {\sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(sec(d*x+c)^4/(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
Leaf count of result is larger than twice the leaf count of optimal. 286 vs. \(2 (140) = 280\).
Time = 0.57 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.82 \[ \int \frac {\sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {\frac {12 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} b^{4}}{\sqrt {a^{2} - b^{2}} a^{4}} + \frac {3 \, {\left (a^{2} b + 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {3 \, {\left (a^{2} b + 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {2 \, {\left (6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a^{3}}}{6 \, d} \] Input:
integrate(sec(d*x+c)^4/(a+b*cos(d*x+c)),x, algorithm="giac")
Output:
-1/6*(12*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*ta n(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))*b^4/(sqrt(a ^2 - b^2)*a^4) + 3*(a^2*b + 2*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 3*(a^2*b + 2*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 + 2*(6*a^2*tan( 1/2*d*x + 1/2*c)^5 + 3*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*b^2*tan(1/2*d*x + 1/ 2*c)^5 - 4*a^2*tan(1/2*d*x + 1/2*c)^3 - 12*b^2*tan(1/2*d*x + 1/2*c)^3 + 6* a^2*tan(1/2*d*x + 1/2*c) - 3*a*b*tan(1/2*d*x + 1/2*c) + 6*b^2*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*a^3))/d
Time = 43.52 (sec) , antiderivative size = 991, normalized size of antiderivative = 6.31 \[ \int \frac {\sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \] Input:
int(1/(cos(c + d*x)^4*(a + b*cos(c + d*x))),x)
Output:
(a^5*(sin(c + d*x)/2 + sin(3*c + 3*d*x)/6) - a^4*((b*sin(2*c + 2*d*x))/4 + (b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/4 + (3* b*cos(c + d*x)*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/4) - a^2*((3* b^3*cos(c + d*x)*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/4 - (b^3*si n(2*c + 2*d*x))/4 + (b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos( 3*c + 3*d*x))/4) - a^3*((b^2*sin(c + d*x))/4 - (b^2*sin(3*c + 3*d*x))/12) - a*((b^4*sin(c + d*x))/4 + (b^4*sin(3*c + 3*d*x))/4) + (3*b^5*cos(c + d*x )*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 + (b^5*atanh(sin(c/2 + ( d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/2 + (3*b^4*atanh((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))/(cos(c/2 + (d*x)/2)*(a*b^2 - a^3 )*(a^7 - 3*a^3*b^4 + 2*a^5*b^2)))*cos(c + d*x)*(b^2 - a^2)^(1/2))/2 + (b^4 *atanh((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...
Time = 0.18 (sec) , antiderivative size = 623, normalized size of antiderivative = 3.97 \[ \int \frac {\sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^4/(a+b*cos(d*x+c)),x)
Output:
(12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**2*b**4 - 12*sqrt(a**2 - b**2)*ata n((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x )*b**4 + 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**4*b + 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b**3 - 6*co s(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**5 - 3*cos(c + d*x) *log(tan((c + d*x)/2) - 1)*a**4*b - 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**2*b**3 + 6*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b**5 - 3*cos(c + d *x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**4*b - 3*cos(c + d*x)*log( tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b**3 + 6*cos(c + d*x)*log(tan(( c + d*x)/2) + 1)*sin(c + d*x)**2*b**5 + 3*cos(c + d*x)*log(tan((c + d*x)/2 ) + 1)*a**4*b + 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**2*b**3 - 6*cos (c + d*x)*log(tan((c + d*x)/2) + 1)*b**5 + 3*cos(c + d*x)*sin(c + d*x)*a** 4*b - 3*cos(c + d*x)*sin(c + d*x)*a**2*b**3 + 4*sin(c + d*x)**3*a**5 + 2*s in(c + d*x)**3*a**3*b**2 - 6*sin(c + d*x)**3*a*b**4 - 6*sin(c + d*x)*a**5 + 6*sin(c + d*x)*a*b**4)/(6*cos(c + d*x)*a**4*d*(sin(c + d*x)**2*a**2 - si n(c + d*x)**2*b**2 - a**2 + b**2))