Integrand size = 21, antiderivative size = 157 \[ \int \frac {\sec ^5(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {a \left (2 a^2+b^2\right ) \text {arctanh}(\sin (c+d x))}{2 b^4 d}+\frac {2 a^4 \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^4 \sqrt {a+b} d}+\frac {\left (3 a^2+2 b^2\right ) \tan (c+d x)}{3 b^3 d}-\frac {a \sec (c+d x) \tan (c+d x)}{2 b^2 d}+\frac {\sec ^2(c+d x) \tan (c+d x)}{3 b d} \]
-1/2*a*(2*a^2+b^2)*arctanh(sin(d*x+c))/b^4/d+2*a^4*arctanh((a-b)^(1/2)*tan (1/2*d*x+1/2*c)/(a+b)^(1/2))/b^4/d/(a-b)^(1/2)/(a+b)^(1/2)+1/3*(3*a^2+2*b^ 2)*tan(d*x+c)/b^3/d-1/2*a*sec(d*x+c)*tan(d*x+c)/b^2/d+1/3*sec(d*x+c)^2*tan (d*x+c)/b/d
Time = 2.05 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.64 \[ \int \frac {\sec ^5(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {-\frac {24 a^4 \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {1}{2} \sec ^3(c+d x) \left (9 a \left (2 a^2+b^2\right ) \cos (c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+3 a \left (2 a^2+b^2\right ) \cos (3 (c+d x)) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+4 b \left (3 a^2+4 b^2-3 a b \cos (c+d x)+\left (3 a^2+2 b^2\right ) \cos (2 (c+d x))\right ) \sin (c+d x)\right )}{12 b^4 d} \]
((-24*a^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*a*(2*a^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*a* (2*a^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*b*(3*a^2 + 4*b^2 - 3*a*b*Co s[c + d*x] + (3*a^2 + 2*b^2)*Cos[2*(c + d*x)])*Sin[c + d*x]))/2)/(12*b^4*d )
Time = 1.28 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.11, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.762, Rules used = {3042, 4338, 3042, 4580, 25, 3042, 4570, 27, 3042, 4486, 3042, 4257, 4318, 3042, 3138, 221}
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.
| \(\displaystyle \int \frac {\sec ^5(c+d x)}{a+b \sec (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^5}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 4338 |
| \(\displaystyle \frac {\int \frac {\sec ^2(c+d x) \left (-3 a \sec ^2(c+d x)+2 b \sec (c+d x)+2 a\right )}{a+b \sec (c+d x)}dx}{3 b}+\frac {\tan (c+d x) \sec ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (-3 a \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 b \csc \left (c+d x+\frac {\pi }{2}\right )+2 a\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{3 b}+\frac {\tan (c+d x) \sec ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 4580 |
| \(\displaystyle \frac {\frac {\int -\frac {\sec (c+d x) \left (3 a^2-b \sec (c+d x) a-2 \left (3 a^2+2 b^2\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)}dx}{2 b}-\frac {3 a \tan (c+d x) \sec (c+d x)}{2 b d}}{3 b}+\frac {\tan (c+d x) \sec ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\int \frac {\sec (c+d x) \left (3 a^2-b \sec (c+d x) a-2 \left (3 a^2+2 b^2\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)}dx}{2 b}-\frac {3 a \tan (c+d x) \sec (c+d x)}{2 b d}}{3 b}+\frac {\tan (c+d x) \sec ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 a^2-b \csc \left (c+d x+\frac {\pi }{2}\right ) a-2 \left (3 a^2+2 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b}-\frac {3 a \tan (c+d x) \sec (c+d x)}{2 b d}}{3 b}+\frac {\tan (c+d x) \sec ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 4570 |
| \(\displaystyle \frac {-\frac {\frac {\int \frac {3 \sec (c+d x) \left (b a^2+\left (2 a^2+b^2\right ) \sec (c+d x) a\right )}{a+b \sec (c+d x)}dx}{b}-\frac {2 \left (3 a^2+2 b^2\right ) \tan (c+d x)}{b d}}{2 b}-\frac {3 a \tan (c+d x) \sec (c+d x)}{2 b d}}{3 b}+\frac {\tan (c+d x) \sec ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {\frac {3 \int \frac {\sec (c+d x) \left (b a^2+\left (2 a^2+b^2\right ) \sec (c+d x) a\right )}{a+b \sec (c+d x)}dx}{b}-\frac {2 \left (3 a^2+2 b^2\right ) \tan (c+d x)}{b d}}{2 b}-\frac {3 a \tan (c+d x) \sec (c+d x)}{2 b d}}{3 b}+\frac {\tan (c+d x) \sec ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {3 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b a^2+\left (2 a^2+b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) a\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {2 \left (3 a^2+2 b^2\right ) \tan (c+d x)}{b d}}{2 b}-\frac {3 a \tan (c+d x) \sec (c+d x)}{2 b d}}{3 b}+\frac {\tan (c+d x) \sec ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 4486 |
| \(\displaystyle \frac {-\frac {\frac {3 \left (\frac {a \left (2 a^2+b^2\right ) \int \sec (c+d x)dx}{b}-\frac {2 a^4 \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{b}\right )}{b}-\frac {2 \left (3 a^2+2 b^2\right ) \tan (c+d x)}{b d}}{2 b}-\frac {3 a \tan (c+d x) \sec (c+d x)}{2 b d}}{3 b}+\frac {\tan (c+d x) \sec ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {3 \left (\frac {a \left (2 a^2+b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{b}-\frac {2 a^4 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\right )}{b}-\frac {2 \left (3 a^2+2 b^2\right ) \tan (c+d x)}{b d}}{2 b}-\frac {3 a \tan (c+d x) \sec (c+d x)}{2 b d}}{3 b}+\frac {\tan (c+d x) \sec ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {-\frac {\frac {3 \left (\frac {a \left (2 a^2+b^2\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a^4 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\right )}{b}-\frac {2 \left (3 a^2+2 b^2\right ) \tan (c+d x)}{b d}}{2 b}-\frac {3 a \tan (c+d x) \sec (c+d x)}{2 b d}}{3 b}+\frac {\tan (c+d x) \sec ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle \frac {-\frac {\frac {3 \left (\frac {a \left (2 a^2+b^2\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a^4 \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{b^2}\right )}{b}-\frac {2 \left (3 a^2+2 b^2\right ) \tan (c+d x)}{b d}}{2 b}-\frac {3 a \tan (c+d x) \sec (c+d x)}{2 b d}}{3 b}+\frac {\tan (c+d x) \sec ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {3 \left (\frac {a \left (2 a^2+b^2\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a^4 \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{b^2}\right )}{b}-\frac {2 \left (3 a^2+2 b^2\right ) \tan (c+d x)}{b d}}{2 b}-\frac {3 a \tan (c+d x) \sec (c+d x)}{2 b d}}{3 b}+\frac {\tan (c+d x) \sec ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {-\frac {\frac {3 \left (\frac {a \left (2 a^2+b^2\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {4 a^4 \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{b^2 d}\right )}{b}-\frac {2 \left (3 a^2+2 b^2\right ) \tan (c+d x)}{b d}}{2 b}-\frac {3 a \tan (c+d x) \sec (c+d x)}{2 b d}}{3 b}+\frac {\tan (c+d x) \sec ^2(c+d x)}{3 b d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {-\frac {\frac {3 \left (\frac {a \left (2 a^2+b^2\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {4 a^4 \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}\right )}{b}-\frac {2 \left (3 a^2+2 b^2\right ) \tan (c+d x)}{b d}}{2 b}-\frac {3 a \tan (c+d x) \sec (c+d x)}{2 b d}}{3 b}+\frac {\tan (c+d x) \sec ^2(c+d x)}{3 b d}\) |
(Sec[c + d*x]^2*Tan[c + d*x])/(3*b*d) + ((-3*a*Sec[c + d*x]*Tan[c + d*x])/ (2*b*d) - ((3*((a*(2*a^2 + b^2)*ArcTanh[Sin[c + d*x]])/(b*d) - (4*a^4*ArcT anh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b ]*d)))/b - (2*(3*a^2 + 2*b^2)*Tan[c + d*x])/(b*d))/(2*b))/(3*b)
3.5.87.3.1 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/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[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[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[1/b Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_)), x_Symbol] :> Simp[(-d^3)*Cot[e + f*x]*((d*Csc[e + f*x])^(n - 3)/(b*f* (n - 2))), x] + Simp[d^3/(b*(n - 2)) Int[(d*Csc[e + f*x])^(n - 3)*(Simp[a *(n - 3) + b*(n - 3)*Csc[e + f*x] - a*(n - 2)*Csc[e + f*x]^2, x]/(a + b*Csc [e + f*x])), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && Gt Q[n, 3]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[( e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[B/b Int[Csc[e + f*x], x], x] + Simp[(A*b - a*B)/b Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x ] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]
Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e _.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_S ymbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2) )), x] + Simp[1/(b*(m + 2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[ b*A*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[ (e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x _Symbol] :> Simp[(-C)*Csc[e + f*x]*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[a*C + b*(C*(m + 2) + A*(m + 3))*Csc[e + f*x] - (2*a*C - b*B* (m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] & & NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
Time = 0.71 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.61
| method | result | size |
| derivativedivides | \(\frac {\frac {2 a^{4} \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{3 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {-a -b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 a^{2}+a b +2 b^{2}}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {a \left (2 a^{2}+b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{4}}-\frac {1}{3 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a +b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 a^{2}+a b +2 b^{2}}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {a \left (2 a^{2}+b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{4}}}{d}\) | \(252\) |
| default | \(\frac {\frac {2 a^{4} \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{3 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {-a -b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 a^{2}+a b +2 b^{2}}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {a \left (2 a^{2}+b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{4}}-\frac {1}{3 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a +b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 a^{2}+a b +2 b^{2}}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {a \left (2 a^{2}+b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{4}}}{d}\) | \(252\) |
| risch | \(\frac {i \left (3 a b \,{\mathrm e}^{5 i \left (d x +c \right )}+6 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+12 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+12 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 a b \,{\mathrm e}^{i \left (d x +c \right )}+6 a^{2}+4 b^{2}\right )}{3 b^{3} d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, d \,b^{4}}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d \,b^{4}}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{b^{4} d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 b^{2} d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{b^{4} d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 b^{2} d}\) | \(341\) |
1/d*(2*a^4/b^4/((a-b)*(a+b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a-b) *(a+b))^(1/2))-1/3/b/(tan(1/2*d*x+1/2*c)+1)^3-1/2*(-a-b)/b^2/(tan(1/2*d*x+ 1/2*c)+1)^2-1/2*(2*a^2+a*b+2*b^2)/b^3/(tan(1/2*d*x+1/2*c)+1)-1/2*a*(2*a^2+ b^2)/b^4*ln(tan(1/2*d*x+1/2*c)+1)-1/3/b/(tan(1/2*d*x+1/2*c)-1)^3-1/2*(a+b) /b^2/(tan(1/2*d*x+1/2*c)-1)^2-1/2*(2*a^2+a*b+2*b^2)/b^3/(tan(1/2*d*x+1/2*c )-1)+1/2*a*(2*a^2+b^2)/b^4*ln(tan(1/2*d*x+1/2*c)-1))
Time = 0.39 (sec) , antiderivative size = 557, normalized size of antiderivative = 3.55 \[ \int \frac {\sec ^5(c+d x)}{a+b \sec (c+d x)} \, dx=\left [\frac {6 \, \sqrt {a^{2} - b^{2}} a^{4} \cos \left (d x + c\right )^{3} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - 3 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, a^{2} b^{3} - 2 \, b^{5} + 2 \, {\left (3 \, a^{4} b - a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}, \frac {12 \, \sqrt {-a^{2} + b^{2}} a^{4} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{3} - 3 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, a^{2} b^{3} - 2 \, b^{5} + 2 \, {\left (3 \, a^{4} b - a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}\right ] \]
[1/12*(6*sqrt(a^2 - b^2)*a^4*cos(d*x + c)^3*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 + 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) - 3*( 2*a^5 - a^3*b^2 - a*b^4)*cos(d*x + c)^3*log(sin(d*x + c) + 1) + 3*(2*a^5 - a^3*b^2 - a*b^4)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*a^2*b^3 - 2 *b^5 + 2*(3*a^4*b - a^2*b^3 - 2*b^5)*cos(d*x + c)^2 - 3*(a^3*b^2 - a*b^4)* cos(d*x + c))*sin(d*x + c))/((a^2*b^4 - b^6)*d*cos(d*x + c)^3), 1/12*(12*s qrt(-a^2 + b^2)*a^4*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c)))*cos(d*x + c)^3 - 3*(2*a^5 - a^3*b^2 - a*b^4)*cos(d*x + c)^3*log(sin(d*x + c) + 1) + 3*(2*a^5 - a^3*b^2 - a*b^4)*cos(d*x + c)^3*l og(-sin(d*x + c) + 1) + 2*(2*a^2*b^3 - 2*b^5 + 2*(3*a^4*b - a^2*b^3 - 2*b^ 5)*cos(d*x + c)^2 - 3*(a^3*b^2 - a*b^4)*cos(d*x + c))*sin(d*x + c))/((a^2* b^4 - b^6)*d*cos(d*x + c)^3)]
\[ \int \frac {\sec ^5(c+d x)}{a+b \sec (c+d x)} \, dx=\int \frac {\sec ^{5}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {\sec ^5(c+d x)}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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*a^2-4*b^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 = 0.33 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.82 \[ \int \frac {\sec ^5(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\frac {12 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )} a^{4}}{\sqrt {-a^{2} + b^{2}} b^{4}} - \frac {3 \, {\left (2 \, a^{3} + a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{4}} + \frac {3 \, {\left (2 \, a^{3} + a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{4}} - \frac {2 \, {\left (6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} b^{3}}}{6 \, d} \]
1/6*(12*(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)))*a^4/(sqrt(- a^2 + b^2)*b^4) - 3*(2*a^3 + a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^4 + 3*(2*a^3 + a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^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 - 12*a^2*tan(1/2*d*x + 1/2*c)^3 - 4*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*b^3))/d
Time = 15.23 (sec) , antiderivative size = 1021, normalized size of antiderivative = 6.50 \[ \int \frac {\sec ^5(c+d x)}{a+b \sec (c+d x)} \, dx=\text {Too large to display} \]
-((9*a^4*cos(c + d*x)*atanh((8*a^6*sin(c/2 + (d*x)/2)*(a^2 - b^2) - 8*a^8* sin(c/2 + (d*x)/2) + b^6*sin(c/2 + (d*x)/2)*(a^2 - b^2) + 8*a^7*b*sin(c/2 + (d*x)/2) - 2*a*b^5*sin(c/2 + (d*x)/2)*(a^2 - b^2) - 8*a^5*b*sin(c/2 + (d *x)/2)*(a^2 - b^2) + 5*a^2*b^4*sin(c/2 + (d*x)/2)*(a^2 - b^2) - 8*a^3*b^3* sin(c/2 + (d*x)/2)*(a^2 - b^2) + 8*a^4*b^2*sin(c/2 + (d*x)/2)*(a^2 - b^2)) /(b*cos(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2)*(4*a^4*(a^2 - b^2) + b^4*(a^2 - b ^2) + 2*a^5*b - 4*a^6 + 2*a^3*b^3 + 4*a^2*b^2*(a^2 - b^2) - a*b^3*(a^2 - b ^2) - 2*a^3*b*(a^2 - b^2)))))/2 - (3*b^3*sin(c + d*x)*(a^2 - b^2)^(1/2))/2 - (b^3*sin(3*c + 3*d*x)*(a^2 - b^2)^(1/2))/2 + (3*a^4*cos(3*c + 3*d*x)*at anh((8*a^6*sin(c/2 + (d*x)/2)*(a^2 - b^2) - 8*a^8*sin(c/2 + (d*x)/2) + b^6 *sin(c/2 + (d*x)/2)*(a^2 - b^2) + 8*a^7*b*sin(c/2 + (d*x)/2) - 2*a*b^5*sin (c/2 + (d*x)/2)*(a^2 - b^2) - 8*a^5*b*sin(c/2 + (d*x)/2)*(a^2 - b^2) + 5*a ^2*b^4*sin(c/2 + (d*x)/2)*(a^2 - b^2) - 8*a^3*b^3*sin(c/2 + (d*x)/2)*(a^2 - b^2) + 8*a^4*b^2*sin(c/2 + (d*x)/2)*(a^2 - b^2))/(b*cos(c/2 + (d*x)/2)*( a^2 - b^2)^(1/2)*(4*a^4*(a^2 - b^2) + b^4*(a^2 - b^2) + 2*a^5*b - 4*a^6 + 2*a^3*b^3 + 4*a^2*b^2*(a^2 - b^2) - a*b^3*(a^2 - b^2) - 2*a^3*b*(a^2 - b^2 )))))/2 + (3*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3* d*x)*(a^2 - b^2)^(1/2))/2 - (3*a^2*b*sin(c + d*x)*(a^2 - b^2)^(1/2))/4 + ( 9*a^3*cos(c + d*x)*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*(a^2 - b^2 )^(1/2))/2 + (3*a*b^2*sin(2*c + 2*d*x)*(a^2 - b^2)^(1/2))/4 - (3*a^2*b*...