Integrand size = 21, antiderivative size = 117 \[ \int \frac {\sec ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{b^2 d}-\frac {2 a \left (a^2-2 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^2 (a+b)^{3/2} d}-\frac {a^2 \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \]
arctanh(sin(d*x+c))/b^2/d-2*a*(a^2-2*b^2)*arctanh((a-b)^(1/2)*tan(1/2*d*x+ 1/2*c)/(a+b)^(1/2))/(a-b)^(3/2)/b^2/(a+b)^(3/2)/d-a^2*tan(d*x+c)/b/(a^2-b^ 2)/d/(a+b*sec(d*x+c))
Time = 0.47 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.25 \[ \int \frac {\sec ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {\frac {2 a \left (a^2-2 b^2\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a^2 b \sin (c+d x)}{(-a+b) (a+b) (b+a \cos (c+d x))}}{b^2 d} \]
((2*a*(a^2 - 2*b^2)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/ (a^2 - b^2)^(3/2) - Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (a^2*b*Sin[c + d*x])/((-a + b)*(a + b)*(b + a*Cos[c + d*x])))/(b^2*d)
Time = 0.75 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.21, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3042, 4326, 25, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4326 |
| \(\displaystyle -\frac {\int -\frac {\sec (c+d x) \left (a b+\left (a^2-b^2\right ) \sec (c+d x)\right )}{a+b \sec (c+d x)}dx}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\sec (c+d x) \left (a b+\left (a^2-b^2\right ) \sec (c+d x)\right )}{a+b \sec (c+d x)}dx}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a b+\left (a^2-b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4486 |
| \(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \int \sec (c+d x)dx}{b}-\frac {a \left (a^2-2 b^2\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{b}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{b}-\frac {a \left (a^2-2 b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {a \left (a^2-2 b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {a \left (a^2-2 b^2\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{b^2}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {a \left (a^2-2 b^2\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{b^2}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a \left (a^2-2 b^2\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{b^2 d}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a \left (a^2-2 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
(((a^2 - b^2)*ArcTanh[Sin[c + d*x]])/(b*d) - (2*a*(a^2 - 2*b^2)*ArcTanh[(S qrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]*d))/ (b*(a^2 - b^2)) - (a^2*Tan[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Sec[c + d*x]) )
3.5.99.3.1 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_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-a^2)*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[a*b*(m + 1) - (a^2 + b^2*(m + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
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]
Time = 0.52 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.42
| method | result | size |
| derivativedivides | \(\frac {\frac {2 a \left (\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (a^{2}-2 b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{2}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{2}}}{d}\) | \(166\) |
| default | \(\frac {\frac {2 a \left (\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (a^{2}-2 b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{2}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{2}}}{d}\) | \(166\) |
| risch | \(-\frac {2 i a \left (b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}{\left (a^{2}-b^{2}\right ) d b \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}-\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}+\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{b^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{b^{2} d}\) | \(437\) |
1/d*(2*a/b^2*(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)-(a^2-2*b^2)/(a-b)/(a+b)/((a-b)*(a+b))^(1/2)*arcta nh((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2)))+1/b^2*ln(tan(1/2*d*x+1/2 *c)+1)-1/b^2*ln(tan(1/2*d*x+1/2*c)-1))
Leaf count of result is larger than twice the leaf count of optimal. 270 vs. \(2 (108) = 216\).
Time = 0.51 (sec) , antiderivative size = 596, normalized size of antiderivative = 5.09 \[ \int \frac {\sec ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\left [\frac {{\left (a^{3} b - 2 \, a b^{3} + {\left (a^{4} - 2 \, a^{2} b^{2}\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 ) + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5} + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5} + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{4} b - a^{2} b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} d \cos \left (d x + c\right ) + {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} d\right )}}, -\frac {2 \, {\left (a^{3} b - 2 \, a b^{3} + {\left (a^{4} - 2 \, a^{2} b^{2}\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 (a^{4} b - 2 \, a^{2} b^{3} + b^{5} + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5} + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (a^{4} b - a^{2} b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} d \cos \left (d x + c\right ) + {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} d\right )}}\right ] \]
[1/2*((a^3*b - 2*a*b^3 + (a^4 - 2*a^2*b^2)*cos(d*x + c))*sqrt(a^2 - b^2)*l og((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)) + (a^4*b - 2*a^2*b^3 + b^5 + (a^5 - 2*a^3*b^2 + a* b^4)*cos(d*x + c))*log(sin(d*x + c) + 1) - (a^4*b - 2*a^2*b^3 + b^5 + (a^5 - 2*a^3*b^2 + a*b^4)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(a^4*b - a^ 2*b^3)*sin(d*x + c))/((a^5*b^2 - 2*a^3*b^4 + a*b^6)*d*cos(d*x + c) + (a^4* b^3 - 2*a^2*b^5 + b^7)*d), -1/2*(2*(a^3*b - 2*a*b^3 + (a^4 - 2*a^2*b^2)*co s(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^4*b - 2*a^2*b^3 + b^5 + (a^5 - 2*a^3*b^2 + a*b^4)*cos(d*x + c))*log(sin(d*x + c) + 1) + (a^4*b - 2*a^2*b^3 + b^5 + (a^5 - 2*a^3*b^2 + a*b^4)*cos(d*x + c))*log(-sin(d*x + c) + 1) + 2*(a^4*b - a^2*b^3)*sin(d*x + c))/((a^5*b^2 - 2*a^3*b^4 + a*b^6)*d*cos(d*x + c) + (a^4*b^3 - 2*a^2*b^5 + b^7)*d)]
\[ \int \frac {\sec ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {\sec ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]
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
Time = 0.32 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.74 \[ \int \frac {\sec ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {\frac {2 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{2} b - b^{3}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}} - \frac {2 \, {\left (a^{3} - 2 \, a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{2} b^{2} - b^{4}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{2}} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{2}}}{d} \]
(2*a^2*tan(1/2*d*x + 1/2*c)/((a^2*b - b^3)*(a*tan(1/2*d*x + 1/2*c)^2 - b*t an(1/2*d*x + 1/2*c)^2 - a - b)) - 2*(a^3 - 2*a*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^2 - b^4)*sqrt(-a^2 + b^2)) + log( abs(tan(1/2*d*x + 1/2*c) + 1))/b^2 - log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^ 2)/d
Time = 19.07 (sec) , antiderivative size = 2848, normalized size of antiderivative = 24.34 \[ \int \frac {\sec ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\text {Too large to display} \]
- (atan((((((32*(2*a*b^8 - b^9 + a^2*b^7 - 3*a^3*b^6 + a^5*b^4))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) - (32*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4 *a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^2)))/b^2 - (32*tan(c/2 + (d*x)/2)*(2*a^6 - 2*a^5*b - 2*a*b^5 + b^ 6 + 3*a^2*b^4 + 4*a^3*b^3 - 5*a^4*b^2))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2)) *1i)/b^2 - ((((32*(2*a*b^8 - b^9 + a^2*b^7 - 3*a^3*b^6 + a^5*b^4))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) + (32*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/(b^2*(a*b^4 + b^5 - a^2*b ^3 - a^3*b^2)))/b^2 + (32*tan(c/2 + (d*x)/2)*(2*a^6 - 2*a^5*b - 2*a*b^5 + b^6 + 3*a^2*b^4 + 4*a^3*b^3 - 5*a^4*b^2))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2 ))*1i)/b^2)/((((32*(2*a*b^8 - b^9 + a^2*b^7 - 3*a^3*b^6 + a^5*b^4))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) - (32*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/(b^2*(a*b^4 + b^5 - a^2* b^3 - a^3*b^2)))/b^2 - (32*tan(c/2 + (d*x)/2)*(2*a^6 - 2*a^5*b - 2*a*b^5 + b^6 + 3*a^2*b^4 + 4*a^3*b^3 - 5*a^4*b^2))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^ 2))/b^2 - (64*(2*a*b^4 - a^4*b + a^5 + 2*a^2*b^3 - 3*a^3*b^2))/(a*b^5 + b^ 6 - a^2*b^4 - a^3*b^3) + (((32*(2*a*b^8 - b^9 + a^2*b^7 - 3*a^3*b^6 + a^5* b^4))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) + (32*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^2)))/b^2 + (32*tan(c/2 + (d*x)/2)*(2*a^6 - 2*a^...