Integrand size = 21, antiderivative size = 85 \[ \int \frac {\sec ^3(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {a \text {arctanh}(\sin (c+d x))}{b^2 d}+\frac {2 a^2 \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^2 \sqrt {a+b} d}+\frac {\tan (c+d x)}{b d} \]
-a*arctanh(sin(d*x+c))/b^2/d+2*a^2*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/ (a+b)^(1/2))/b^2/d/(a-b)^(1/2)/(a+b)^(1/2)+tan(d*x+c)/b/d
Time = 0.50 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.35 \[ \int \frac {\sec ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {-\frac {2 a^2 \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+a \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+b \tan (c+d x)}{b^2 d} \]
((-2*a^2*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + a*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + b*Tan[c + d*x])/(b^2*d)
Time = 0.58 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 4277, 3042, 4276, 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.
\(\displaystyle \int \frac {\sec ^3(c+d x)}{a+b \sec (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 4277 |
\(\displaystyle \frac {\tan (c+d x)}{b d}-\frac {a \int \frac {\sec ^2(c+d x)}{a+b \sec (c+d x)}dx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan (c+d x)}{b d}-\frac {a \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\) |
\(\Big \downarrow \) 4276 |
\(\displaystyle \frac {\tan (c+d x)}{b d}-\frac {a \left (\frac {\int \sec (c+d x)dx}{b}-\frac {a \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{b}\right )}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan (c+d x)}{b d}-\frac {a \left (\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{b}-\frac {a \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\right )}{b}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\tan (c+d x)}{b d}-\frac {a \left (\frac {\text {arctanh}(\sin (c+d x))}{b d}-\frac {a \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\right )}{b}\) |
\(\Big \downarrow \) 4318 |
\(\displaystyle \frac {\tan (c+d x)}{b d}-\frac {a \left (\frac {\text {arctanh}(\sin (c+d x))}{b d}-\frac {a \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{b^2}\right )}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan (c+d x)}{b d}-\frac {a \left (\frac {\text {arctanh}(\sin (c+d x))}{b d}-\frac {a \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{b^2}\right )}{b}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {\tan (c+d x)}{b d}-\frac {a \left (\frac {\text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{b^2 d}\right )}{b}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\tan (c+d x)}{b d}-\frac {a \left (\frac {\text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}\right )}{b}\) |
-((a*(ArcTanh[Sin[c + d*x]]/(b*d) - (2*a*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x )/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]*d)))/b) + Tan[c + d*x]/(b*d )
3.5.89.3.1 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_)]^2/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Sym bol] :> Simp[1/b Int[Csc[e + f*x], x], x] - Simp[a/b Int[Csc[e + f*x]/( a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x]
Int[csc[(e_.) + (f_.)*(x_)]^3/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Sym bol] :> Simp[-Cot[e + f*x]/(b*f), x] - Simp[a/b Int[Csc[e + f*x]^2/(a + b *Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, 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]
Time = 0.43 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.45
method | result | size |
derivativedivides | \(\frac {-\frac {1}{b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{2}}-\frac {1}{b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{2}}+\frac {2 a^{2} \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(123\) |
default | \(\frac {-\frac {1}{b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{2}}-\frac {1}{b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{2}}+\frac {2 a^{2} \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(123\) |
risch | \(\frac {2 i}{d b \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, d \,b^{2}}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d \,b^{2}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{b^{2} d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{b^{2} d}\) | \(214\) |
1/d*(-1/b/(tan(1/2*d*x+1/2*c)+1)-a/b^2*ln(tan(1/2*d*x+1/2*c)+1)-1/b/(tan(1 /2*d*x+1/2*c)-1)+a/b^2*ln(tan(1/2*d*x+1/2*c)-1)+2*a^2/b^2/((a-b)*(a+b))^(1 /2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (76) = 152\).
Time = 0.31 (sec) , antiderivative size = 392, normalized size of antiderivative = 4.61 \[ \int \frac {\sec ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\left [\frac {\sqrt {a^{2} - b^{2}} a^{2} \cos \left (d x + c\right ) \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} d \cos \left (d x + c\right )}, \frac {2 \, \sqrt {-a^{2} + b^{2}} a^{2} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} d \cos \left (d x + c\right )}\right ] \]
[1/2*(sqrt(a^2 - b^2)*a^2*cos(d*x + c)*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)) - (a^3 - a *b^2)*cos(d*x + c)*log(sin(d*x + c) + 1) + (a^3 - a*b^2)*cos(d*x + c)*log( -sin(d*x + c) + 1) + 2*(a^2*b - b^3)*sin(d*x + c))/((a^2*b^2 - b^4)*d*cos( d*x + c)), 1/2*(2*sqrt(-a^2 + b^2)*a^2*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c)))*cos(d*x + c) - (a^3 - a*b^2)*cos(d* x + c)*log(sin(d*x + c) + 1) + (a^3 - a*b^2)*cos(d*x + c)*log(-sin(d*x + c ) + 1) + 2*(a^2*b - b^3)*sin(d*x + c))/((a^2*b^2 - b^4)*d*cos(d*x + c))]
\[ \int \frac {\sec ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {\sec ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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
Time = 0.34 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.79 \[ \int \frac {\sec ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )} a^{2}}{\sqrt {-a^{2} + b^{2}} b^{2}} - \frac {a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{2}} + \frac {a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{2}} - \frac {2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} b}}{d} \]
(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/(sqrt(-a^2 + b^2)*b^2) - a*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^2 + a*log(abs(tan(1/2* d*x + 1/2*c) - 1))/b^2 - 2*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*b))/d
Time = 14.05 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.40 \[ \int \frac {\sec ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\mathrm {tan}\left (c+d\,x\right )}{b\,d}-\frac {2\,a\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^2\,d}-\frac {a^2\,\mathrm {atan}\left (\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}}\right )\,2{}\mathrm {i}}{b^2\,d\,\sqrt {a^2-b^2}} \]