Integrand size = 21, antiderivative size = 188 \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{b^3 d}-\frac {a \left (2 a^4-5 a^2 b^2+6 b^4\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} b^3 (a+b)^{5/2} d}-\frac {a^2 \sec (c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {a^2 \left (2 a^2-5 b^2\right ) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))} \]
arctanh(sin(d*x+c))/b^3/d-a*(2*a^4-5*a^2*b^2+6*b^4)*arctanh((a-b)^(1/2)*ta n(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(5/2)/b^3/(a+b)^(5/2)/d-1/2*a^2*sec(d* x+c)*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^2-1/2*a^2*(2*a^2-5*b^2)*tan (d*x+c)/b^2/(a^2-b^2)^2/d/(a+b*sec(d*x+c))
Time = 1.10 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.03 \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {\frac {2 a \left (2 a^4-5 a^2 b^2+6 b^4\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-\frac {a^2 b \left (3 b \left (a^2-2 b^2\right )+a \left (2 a^2-5 b^2\right ) \cos (c+d x)\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (b+a \cos (c+d x))^2}}{2 b^3 d} \]
((2*a*(2*a^4 - 5*a^2*b^2 + 6*b^4)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt [a^2 - b^2]])/(a^2 - b^2)^(5/2) - 2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2 ]] + 2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - (a^2*b*(3*b*(a^2 - 2*b^2 ) + a*(2*a^2 - 5*b^2)*Cos[c + d*x])*Sin[c + d*x])/((a - b)^2*(a + b)^2*(b + a*Cos[c + d*x])^2))/(2*b^3*d)
Time = 1.22 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.24, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 4332, 3042, 4568, 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.
| \(\displaystyle \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 4332 |
| \(\displaystyle -\frac {\int \frac {\sec (c+d x) \left (a^2-2 b \sec (c+d x) a-2 \left (a^2-b^2\right ) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2}dx}{2 b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a^2-2 b \csc \left (c+d x+\frac {\pi }{2}\right ) a-2 \left (a^2-b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{2 b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 4568 |
| \(\displaystyle -\frac {\frac {a^2 \left (2 a^2-5 b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\int \frac {\sec (c+d x) \left (2 \sec (c+d x) \left (a^2-b^2\right )^2+a b \left (a^2-4 b^2\right )\right )}{a+b \sec (c+d x)}dx}{b \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {a^2 \left (2 a^2-5 b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (2 \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a^2-b^2\right )^2+a b \left (a^2-4 b^2\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 4486 |
| \(\displaystyle -\frac {\frac {a^2 \left (2 a^2-5 b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {2 \left (a^2-b^2\right )^2 \int \sec (c+d x)dx}{b}-\frac {a \left (2 a^4-5 a^2 b^2+6 b^4\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{b}}{b \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {a^2 \left (2 a^2-5 b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {2 \left (a^2-b^2\right )^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{b}-\frac {a \left (2 a^4-5 a^2 b^2+6 b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{b \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle -\frac {\frac {a^2 \left (2 a^2-5 b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {2 \left (a^2-b^2\right )^2 \text {arctanh}(\sin (c+d x))}{b d}-\frac {a \left (2 a^4-5 a^2 b^2+6 b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{b \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle -\frac {\frac {a^2 \left (2 a^2-5 b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {2 \left (a^2-b^2\right )^2 \text {arctanh}(\sin (c+d x))}{b d}-\frac {a \left (2 a^4-5 a^2 b^2+6 b^4\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{b^2}}{b \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {a^2 \left (2 a^2-5 b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {2 \left (a^2-b^2\right )^2 \text {arctanh}(\sin (c+d x))}{b d}-\frac {a \left (2 a^4-5 a^2 b^2+6 b^4\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{b^2}}{b \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle -\frac {\frac {a^2 \left (2 a^2-5 b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {2 \left (a^2-b^2\right )^2 \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a \left (2 a^4-5 a^2 b^2+6 b^4\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{b^2 d}}{b \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\frac {a^2 \left (2 a^2-5 b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {2 \left (a^2-b^2\right )^2 \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a \left (2 a^4-5 a^2 b^2+6 b^4\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}}{b \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}\) |
-1/2*(a^2*Sec[c + d*x]*Tan[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Sec[c + d*x]) ^2) - (-(((2*(a^2 - b^2)^2*ArcTanh[Sin[c + d*x]])/(b*d) - (2*a*(2*a^4 - 5* a^2*b^2 + 6*b^4)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqr t[a - b]*b*Sqrt[a + b]*d))/(b*(a^2 - b^2))) + (a^2*(2*a^2 - 5*b^2)*Tan[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])))/(2*b*(a^2 - b^2))
3.6.7.3.1 Defintions of rubi rules used
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_))^(m_), x_Symbol] :> Simp[(-a^2)*d^3*Cot[e + f*x]*(a + b*Csc[e + f*x])^( m + 1)*((d*Csc[e + f*x])^(n - 3)/(b*f*(m + 1)*(a^2 - b^2))), x] + Simp[d^3/ (b*(m + 1)*(a^2 - b^2)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x]) ^(n - 3)*Simp[a^2*(n - 3) + a*b*(m + 1)*Csc[e + f*x] - (a^2*(n - 2) + b^2*( m + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && (IGtQ[n, 3] || (IntegersQ[n + 1/2, 2*m] && GtQ[n , 2]))
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[(-(A*b^2 - a*b*B + a^2*C))*Cot[e + f*x]*((a + b*Csc[e + f*x] )^(m + 1)/(b*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^ 2, 0]
Time = 0.92 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.36
| method | result | size |
| derivativedivides | \(\frac {-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{3}}+\frac {2 a \left (\frac {\frac {\left (2 a^{2}-a b -6 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (2 a^{2}+a b -6 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{2}}-\frac {\left (2 a^{4}-5 a^{2} b^{2}+6 b^{4}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{3}}}{d}\) | \(255\) |
| default | \(\frac {-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{3}}+\frac {2 a \left (\frac {\frac {\left (2 a^{2}-a b -6 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (2 a^{2}+a b -6 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{2}}-\frac {\left (2 a^{4}-5 a^{2} b^{2}+6 b^{4}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{3}}}{d}\) | \(255\) |
| risch | \(\frac {i a \left (-a^{3} b \,{\mathrm e}^{3 i \left (d x +c \right )}+4 a \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-2 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}+a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+10 b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-7 a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}+16 b^{3} a \,{\mathrm e}^{i \left (d x +c \right )}-2 a^{4}+5 a^{2} b^{2}\right )}{\left (-a^{2}+b^{2}\right )^{2} d \,b^{2} \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )^{2}}+\frac {a^{5} \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}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,b^{3}}-\frac {5 a^{3} \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 )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d b}+\frac {3 a b \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}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {a^{5} \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}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,b^{3}}+\frac {5 a^{3} \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 )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d b}-\frac {3 a b \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}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{b^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{b^{3} d}\) | \(722\) |
1/d*(-1/b^3*ln(tan(1/2*d*x+1/2*c)-1)+2*a/b^3*((1/2*(2*a^2-a*b-6*b^2)*a*b/( a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3-1/2*(2*a^2+a*b-6*b^2)*a*b/(a+b)/ (a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c))/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/ 2*c)^2*b-a-b)^2-1/2*(2*a^4-5*a^2*b^2+6*b^4)/(a^4-2*a^2*b^2+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/b^3*ln( tan(1/2*d*x+1/2*c)+1))
Leaf count of result is larger than twice the leaf count of optimal. 547 vs. \(2 (175) = 350\).
Time = 0.84 (sec) , antiderivative size = 1153, normalized size of antiderivative = 6.13 \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\text {Too large to display} \]
[1/4*((2*a^5*b^2 - 5*a^3*b^4 + 6*a*b^6 + (2*a^7 - 5*a^5*b^2 + 6*a^3*b^4)*c os(d*x + c)^2 + 2*(2*a^6*b - 5*a^4*b^3 + 6*a^2*b^5)*cos(d*x + c))*sqrt(a^2 - b^2)*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)) + 2*(a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^ 8 + (a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*cos(d*x + c)^2 + 2*(a^7*b - 3* a^5*b^3 + 3*a^3*b^5 - a*b^7)*cos(d*x + c))*log(sin(d*x + c) + 1) - 2*(a^6* b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8 + (a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6 )*cos(d*x + c)^2 + 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*cos(d*x + c)) *log(-sin(d*x + c) + 1) - 2*(3*a^6*b^2 - 9*a^4*b^4 + 6*a^2*b^6 + (2*a^7*b - 7*a^5*b^3 + 5*a^3*b^5)*cos(d*x + c))*sin(d*x + c))/((a^8*b^3 - 3*a^6*b^5 + 3*a^4*b^7 - a^2*b^9)*d*cos(d*x + c)^2 + 2*(a^7*b^4 - 3*a^5*b^6 + 3*a^3* b^8 - a*b^10)*d*cos(d*x + c) + (a^6*b^5 - 3*a^4*b^7 + 3*a^2*b^9 - b^11)*d) , -1/2*((2*a^5*b^2 - 5*a^3*b^4 + 6*a*b^6 + (2*a^7 - 5*a^5*b^2 + 6*a^3*b^4) *cos(d*x + c)^2 + 2*(2*a^6*b - 5*a^4*b^3 + 6*a^2*b^5)*cos(d*x + c))*sqrt(- a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin( d*x + c))) - (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8 + (a^8 - 3*a^6*b^2 + 3 *a^4*b^4 - a^2*b^6)*cos(d*x + c)^2 + 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a* b^7)*cos(d*x + c))*log(sin(d*x + c) + 1) + (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^ 6 - b^8 + (a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*cos(d*x + c)^2 + 2*(a...
\[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{3}}\, dx \]
Exception generated. \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]
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
Time = 0.36 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.85 \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {\frac {{\left (2 \, a^{5} - 5 \, a^{3} b^{2} + 6 \, a b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {2 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}^{2}} + \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{3}} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{3}}}{d} \]
((2*a^5 - 5*a^3*b^2 + 6*a*b^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)))/((a^4*b^3 - 2*a^2*b^5 + b^7)*sqrt(-a^2 + b^2)) + (2*a^5*tan(1/2* d*x + 1/2*c)^3 - 3*a^4*b*tan(1/2*d*x + 1/2*c)^3 - 5*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 - 2*a^5*tan(1/2*d*x + 1/2*c) - 3*a^4*b*tan(1/2*d*x + 1/2*c) + 5*a^3*b^2*tan(1/2*d*x + 1/2*c) + 6*a^2*b^3 *tan(1/2*d*x + 1/2*c))/((a^4*b^2 - 2*a^2*b^4 + b^6)*(a*tan(1/2*d*x + 1/2*c )^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)^2) + log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^3 - log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^3)/d
Time = 22.43 (sec) , antiderivative size = 5078, normalized size of antiderivative = 27.01 \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\text {Too large to display} \]
- (atan((((((8*(12*a*b^14 - 4*b^15 + 8*a^2*b^13 - 34*a^3*b^12 - 6*a^4*b^11 + 36*a^5*b^10 + 4*a^6*b^9 - 18*a^7*b^8 - 2*a^8*b^7 + 4*a^9*b^6))/(a*b^12 + b^13 - 3*a^2*b^11 - 3*a^3*b^10 + 3*a^4*b^9 + 3*a^5*b^8 - a^6*b^7 - a^7*b ^6) - (8*tan(c/2 + (d*x)/2)*(8*a*b^15 - 8*a^2*b^14 - 32*a^3*b^13 + 32*a^4* b^12 + 48*a^5*b^11 - 48*a^6*b^10 - 32*a^7*b^9 + 32*a^8*b^8 + 8*a^9*b^7 - 8 *a^10*b^6))/(b^3*(a*b^10 + b^11 - 3*a^2*b^9 - 3*a^3*b^8 + 3*a^4*b^7 + 3*a^ 5*b^6 - a^6*b^5 - a^7*b^4)))/b^3 - (8*tan(c/2 + (d*x)/2)*(8*a^10 - 8*a^9*b - 8*a*b^9 + 4*b^10 + 24*a^2*b^8 + 32*a^3*b^7 - 52*a^4*b^6 - 48*a^5*b^5 + 57*a^6*b^4 + 32*a^7*b^3 - 32*a^8*b^2))/(a*b^10 + b^11 - 3*a^2*b^9 - 3*a^3* b^8 + 3*a^4*b^7 + 3*a^5*b^6 - a^6*b^5 - a^7*b^4))*1i)/b^3 - ((((8*(12*a*b^ 14 - 4*b^15 + 8*a^2*b^13 - 34*a^3*b^12 - 6*a^4*b^11 + 36*a^5*b^10 + 4*a^6* b^9 - 18*a^7*b^8 - 2*a^8*b^7 + 4*a^9*b^6))/(a*b^12 + b^13 - 3*a^2*b^11 - 3 *a^3*b^10 + 3*a^4*b^9 + 3*a^5*b^8 - a^6*b^7 - a^7*b^6) + (8*tan(c/2 + (d*x )/2)*(8*a*b^15 - 8*a^2*b^14 - 32*a^3*b^13 + 32*a^4*b^12 + 48*a^5*b^11 - 48 *a^6*b^10 - 32*a^7*b^9 + 32*a^8*b^8 + 8*a^9*b^7 - 8*a^10*b^6))/(b^3*(a*b^1 0 + b^11 - 3*a^2*b^9 - 3*a^3*b^8 + 3*a^4*b^7 + 3*a^5*b^6 - a^6*b^5 - a^7*b ^4)))/b^3 + (8*tan(c/2 + (d*x)/2)*(8*a^10 - 8*a^9*b - 8*a*b^9 + 4*b^10 + 2 4*a^2*b^8 + 32*a^3*b^7 - 52*a^4*b^6 - 48*a^5*b^5 + 57*a^6*b^4 + 32*a^7*b^3 - 32*a^8*b^2))/(a*b^10 + b^11 - 3*a^2*b^9 - 3*a^3*b^8 + 3*a^4*b^7 + 3*a^5 *b^6 - a^6*b^5 - a^7*b^4))*1i)/b^3)/((((8*(12*a*b^14 - 4*b^15 + 8*a^2*b...