Integrand size = 21, antiderivative size = 164 \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {2 a \text {arctanh}(\sin (c+d x))}{b^3 d}+\frac {2 a^2 \left (2 a^2-3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^3 (a+b)^{3/2} d}+\frac {\left (2 a^2-b^2\right ) \tan (c+d x)}{b^2 \left (a^2-b^2\right ) d}-\frac {a^2 \sec (c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \]
-2*a*arctanh(sin(d*x+c))/b^3/d+2*a^2*(2*a^2-3*b^2)*arctanh((a-b)^(1/2)*tan (1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(3/2)/b^3/(a+b)^(3/2)/d+(2*a^2-b^2)*tan (d*x+c)/b^2/(a^2-b^2)/d-a^2*sec(d*x+c)*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d *x+c))
Time = 0.99 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.99 \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {-\frac {2 a^2 \left (2 a^2-3 b^2\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+2 a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a^3 b \sin (c+d x)}{(a-b) (a+b) (b+a \cos (c+d x))}+b \tan (c+d x)}{b^3 d} \]
((-2*a^2*(2*a^2 - 3*b^2)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^ 2]])/(a^2 - b^2)^(3/2) + 2*a*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 2* a*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (a^3*b*Sin[c + d*x])/((a - b) *(a + b)*(b + a*Cos[c + d*x])) + b*Tan[c + d*x])/(b^3*d)
Time = 1.09 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.12, 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, 4570, 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.
| \(\displaystyle \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4332 |
| \(\displaystyle -\frac {\int \frac {\sec (c+d x) \left (a^2-b \sec (c+d x) a-\left (2 a^2-b^2\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)}dx}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a^2-b \csc \left (c+d x+\frac {\pi }{2}\right ) a+\left (b^2-2 a^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4570 |
| \(\displaystyle -\frac {\frac {\int \frac {\sec (c+d x) \left (b a^2+2 \left (a^2-b^2\right ) \sec (c+d x) a\right )}{a+b \sec (c+d x)}dx}{b}-\frac {\left (2 a^2-b^2\right ) \tan (c+d x)}{b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b a^2+2 \left (a^2-b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) a\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {\left (2 a^2-b^2\right ) \tan (c+d x)}{b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4486 |
| \(\displaystyle -\frac {\frac {\frac {2 a \left (a^2-b^2\right ) \int \sec (c+d x)dx}{b}-a^2 \left (\frac {2 a^2}{b}-3 b\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{b}-\frac {\left (2 a^2-b^2\right ) \tan (c+d x)}{b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\frac {2 a \left (a^2-b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{b}-a^2 \left (\frac {2 a^2}{b}-3 b\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {\left (2 a^2-b^2\right ) \tan (c+d x)}{b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle -\frac {\frac {\frac {2 a \left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{b d}-a^2 \left (\frac {2 a^2}{b}-3 b\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {\left (2 a^2-b^2\right ) \tan (c+d x)}{b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle -\frac {\frac {\frac {2 a \left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {a^2 \left (\frac {2 a^2}{b}-3 b\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{b}}{b}-\frac {\left (2 a^2-b^2\right ) \tan (c+d x)}{b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\frac {2 a \left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {a^2 \left (\frac {2 a^2}{b}-3 b\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{b}}{b}-\frac {\left (2 a^2-b^2\right ) \tan (c+d x)}{b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle -\frac {\frac {\frac {2 a \left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a^2 \left (\frac {2 a^2}{b}-3 b\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}}{b}-\frac {\left (2 a^2-b^2\right ) \tan (c+d x)}{b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {\frac {\frac {2 a \left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a^2 \left (\frac {2 a^2}{b}-3 b\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b}}}{b}-\frac {\left (2 a^2-b^2\right ) \tan (c+d x)}{b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
-((a^2*Sec[c + d*x]*Tan[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Sec[c + d*x]))) - (((2*a*(a^2 - b^2)*ArcTanh[Sin[c + d*x]])/(b*d) - (2*a^2*((2*a^2)/b - 3* b)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[ a + b]*d))/b - ((2*a^2 - b^2)*Tan[c + d*x])/(b*d))/(b*(a^2 - b^2))
3.5.98.3.1 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[(-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]
Time = 0.67 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.27
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 a^{2} \left (\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (2 a^{2}-3 b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3}}+\frac {2 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{3}}-\frac {1}{b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{3}}}{d}\) | \(209\) |
| default | \(\frac {-\frac {2 a^{2} \left (\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (2 a^{2}-3 b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3}}+\frac {2 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{3}}-\frac {1}{b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{3}}}{d}\) | \(209\) |
| risch | \(\frac {2 i \left (-a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}-2 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}+2 b^{3} {\mathrm e}^{i \left (d x +c \right )}-2 a^{3}+a \,b^{2}\right )}{\left (-a^{2}+b^{2}\right ) d \,b^{2} \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{b^{3} d}-\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{b^{3} d}+\frac {2 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{3}}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}-\frac {2 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{3}}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}\) | \(537\) |
1/d*(-2*a^2/b^3*(a*b/(a^2-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*a^2-3*b^2)/(a-b)/(a+b)/((a-b)*(a+b))^(1/2)* arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2)))+2*a/b^3*ln(tan(1/2* d*x+1/2*c)-1)-1/b^2/(tan(1/2*d*x+1/2*c)-1)-1/b^2/(tan(1/2*d*x+1/2*c)+1)-2* a/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. 350 vs. \(2 (155) = 310\).
Time = 0.47 (sec) , antiderivative size = 760, normalized size of antiderivative = 4.63 \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\left [\frac {{\left ({\left (2 \, a^{5} - 3 \, a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, a^{4} b - 3 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \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 ) - 2 \, {\left ({\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left ({\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6} + {\left (2 \, a^{5} b - 3 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{5} b^{3} - 2 \, a^{3} b^{5} + a b^{7}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right )\right )}}, \frac {{\left ({\left (2 \, a^{5} - 3 \, a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, a^{4} b - 3 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \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 ) - {\left ({\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6} + {\left (2 \, a^{5} b - 3 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{{\left (a^{5} b^{3} - 2 \, a^{3} b^{5} + a b^{7}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right )}\right ] \]
[1/2*(((2*a^5 - 3*a^3*b^2)*cos(d*x + c)^2 + (2*a^4*b - 3*a^2*b^3)*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 - 2*a^4*b^2 + a^2 *b^4)*cos(d*x + c)^2 + (a^5*b - 2*a^3*b^3 + a*b^5)*cos(d*x + c))*log(sin(d *x + c) + 1) + 2*((a^6 - 2*a^4*b^2 + a^2*b^4)*cos(d*x + c)^2 + (a^5*b - 2* a^3*b^3 + a*b^5)*cos(d*x + c))*log(-sin(d*x + c) + 1) + 2*(a^4*b^2 - 2*a^2 *b^4 + b^6 + (2*a^5*b - 3*a^3*b^3 + a*b^5)*cos(d*x + c))*sin(d*x + c))/((a ^5*b^3 - 2*a^3*b^5 + a*b^7)*d*cos(d*x + c)^2 + (a^4*b^4 - 2*a^2*b^6 + b^8) *d*cos(d*x + c)), (((2*a^5 - 3*a^3*b^2)*cos(d*x + c)^2 + (2*a^4*b - 3*a^2* b^3)*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 - 2*a^4*b^2 + a^2*b^4)*cos(d*x + c)^2 + (a^5*b - 2*a^3*b^3 + a*b^5)*cos(d*x + c))*log(sin(d*x + c) + 1) + ((a^6 - 2*a^4*b^2 + a^2*b^4)*cos(d*x + c)^2 + (a^5*b - 2*a^3*b^3 + a*b^5 )*cos(d*x + c))*log(-sin(d*x + c) + 1) + (a^4*b^2 - 2*a^2*b^4 + b^6 + (2*a ^5*b - 3*a^3*b^3 + a*b^5)*cos(d*x + c))*sin(d*x + c))/((a^5*b^3 - 2*a^3*b^ 5 + a*b^7)*d*cos(d*x + c)^2 + (a^4*b^4 - 2*a^2*b^6 + b^8)*d*cos(d*x + c))]
\[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]
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. 331 vs. \(2 (155) = 310\).
Time = 0.31 (sec) , antiderivative size = 331, normalized size of antiderivative = 2.02 \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {2 \, {\left (\frac {{\left (2 \, a^{4} - 3 \, a^{2} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{2} b^{3} - b^{5}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {2 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )} {\left (a^{2} b^{2} - b^{4}\right )}} - \frac {a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{3}} + \frac {a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{3}}\right )}}{d} \]
2*((2*a^4 - 3*a^2*b^2)*(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^2*b^3 - b^5)*sqrt(-a^2 + b^2)) - (2*a^3*tan(1/2*d*x + 1/2*c)^3 - a ^2*b*tan(1/2*d*x + 1/2*c)^3 - a*b^2*tan(1/2*d*x + 1/2*c)^3 + b^3*tan(1/2*d *x + 1/2*c)^3 - 2*a^3*tan(1/2*d*x + 1/2*c) - a^2*b*tan(1/2*d*x + 1/2*c) + a*b^2*tan(1/2*d*x + 1/2*c) + b^3*tan(1/2*d*x + 1/2*c))/((a*tan(1/2*d*x + 1 /2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b)*( a^2*b^2 - b^4)) - a*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^3 + a*log(abs(tan (1/2*d*x + 1/2*c) - 1))/b^3)/d
Time = 19.21 (sec) , antiderivative size = 3159, normalized size of antiderivative = 19.26 \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\text {Too large to display} \]
((2*tan(c/2 + (d*x)/2)^3*(a*b^2 + a^2*b - 2*a^3 - b^3))/(b^2*(a + b)*(a - b)) - (2*tan(c/2 + (d*x)/2)*(a*b^2 - a^2*b - 2*a^3 + b^3))/(b^2*(a + b)*(a - b)))/(d*(a + b + tan(c/2 + (d*x)/2)^4*(a - b) - 2*a*tan(c/2 + (d*x)/2)^ 2)) + (a*atan(((a*((32*tan(c/2 + (d*x)/2)*(8*a^8 - 8*a^7*b + 4*a^2*b^6 - 8 *a^3*b^5 + 5*a^4*b^4 + 16*a^5*b^3 - 16*a^6*b^2))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4) - (2*a*((32*(2*a*b^11 - 3*a^2*b^10 - 3*a^3*b^9 + 5*a^4*b^8 + a^5* b^7 - 2*a^6*b^6))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) - (64*a*tan(c/2 + (d*x )/2)*(2*a*b^11 - 2*a^2*b^10 - 4*a^3*b^9 + 4*a^4*b^8 + 2*a^5*b^7 - 2*a^6*b^ 6))/(b^3*(a*b^6 + b^7 - a^2*b^5 - a^3*b^4))))/b^3)*2i)/b^3 + (a*((32*tan(c /2 + (d*x)/2)*(8*a^8 - 8*a^7*b + 4*a^2*b^6 - 8*a^3*b^5 + 5*a^4*b^4 + 16*a^ 5*b^3 - 16*a^6*b^2))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4) + (2*a*((32*(2*a*b^ 11 - 3*a^2*b^10 - 3*a^3*b^9 + 5*a^4*b^8 + a^5*b^7 - 2*a^6*b^6))/(a*b^8 + b ^9 - a^2*b^7 - a^3*b^6) + (64*a*tan(c/2 + (d*x)/2)*(2*a*b^11 - 2*a^2*b^10 - 4*a^3*b^9 + 4*a^4*b^8 + 2*a^5*b^7 - 2*a^6*b^6))/(b^3*(a*b^6 + b^7 - a^2* b^5 - a^3*b^4))))/b^3)*2i)/b^3)/((64*(8*a^8 - 4*a^7*b + 12*a^4*b^4 + 6*a^5 *b^3 - 20*a^6*b^2))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) - (2*a*((32*tan(c/2 + (d*x)/2)*(8*a^8 - 8*a^7*b + 4*a^2*b^6 - 8*a^3*b^5 + 5*a^4*b^4 + 16*a^5*b ^3 - 16*a^6*b^2))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4) - (2*a*((32*(2*a*b^11 - 3*a^2*b^10 - 3*a^3*b^9 + 5*a^4*b^8 + a^5*b^7 - 2*a^6*b^6))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) - (64*a*tan(c/2 + (d*x)/2)*(2*a*b^11 - 2*a^2*b^10 ...