Integrand size = 22, antiderivative size = 87 \[ \int \frac {(e x)^{-1+n}}{a+b \sec \left (c+d x^n\right )} \, dx=\frac {(e x)^n}{a e n}-\frac {2 b x^{-n} (e x)^n \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} \left (c+d x^n\right )\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} \sqrt {a+b} d e n} \]
(e*x)^n/a/e/n-2*b*(e*x)^n*arctanh((a-b)^(1/2)*tan(1/2*c+1/2*d*x^n)/(a+b)^( 1/2))/a/d/e/n/(x^n)/(a-b)^(1/2)/(a+b)^(1/2)
Time = 0.85 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.92 \[ \int \frac {(e x)^{-1+n}}{a+b \sec \left (c+d x^n\right )} \, dx=\frac {(e x)^n \left (d+c x^{-n}+\frac {2 b x^{-n} \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} \left (c+d x^n\right )\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}\right )}{a d e n} \]
((e*x)^n*(d + c/x^n + (2*b*ArcTanh[((-a + b)*Tan[(c + d*x^n)/2])/Sqrt[a^2 - b^2]])/(Sqrt[a^2 - b^2]*x^n)))/(a*d*e*n)
Time = 0.43 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.92, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {4696, 4692, 3042, 4270, 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.
| \(\displaystyle \int \frac {(e x)^{n-1}}{a+b \sec \left (c+d x^n\right )} \, dx\) |
| \(\Big \downarrow \) 4696 |
| \(\displaystyle \frac {x^{-n} (e x)^n \int \frac {x^{n-1}}{a+b \sec \left (d x^n+c\right )}dx}{e}\) |
| \(\Big \downarrow \) 4692 |
| \(\displaystyle \frac {x^{-n} (e x)^n \int \frac {1}{a+b \sec \left (d x^n+c\right )}dx^n}{e n}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {x^{-n} (e x)^n \int \frac {1}{a+b \csc \left (d x^n+c+\frac {\pi }{2}\right )}dx^n}{e n}\) |
| \(\Big \downarrow \) 4270 |
| \(\displaystyle \frac {x^{-n} (e x)^n \left (\frac {x^n}{a}-\frac {\int \frac {1}{\frac {a \cos \left (d x^n+c\right )}{b}+1}dx^n}{a}\right )}{e n}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {x^{-n} (e x)^n \left (\frac {x^n}{a}-\frac {\int \frac {1}{\frac {a \sin \left (d x^n+c+\frac {\pi }{2}\right )}{b}+1}dx^n}{a}\right )}{e n}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {x^{-n} (e x)^n \left (\frac {x^n}{a}-\frac {2 \int \frac {1}{\left (1-\frac {a}{b}\right ) x^{2 n}+\frac {a+b}{b}}d\tan \left (\frac {1}{2} \left (d x^n+c\right )\right )}{a d}\right )}{e n}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {x^{-n} (e x)^n \left (\frac {x^n}{a}-\frac {2 b \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} \left (c+d x^n\right )\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{e n}\) |
((e*x)^n*(x^n/a - (2*b*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x^n)/2])/Sqrt[a + b ]])/(a*Sqrt[a - b]*Sqrt[a + b]*d)))/(e*n*x^n)
3.1.78.3.1 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_)]*(b_.) + (a_))^(-1), x_Symbol] :> Simp[x/a, x] - Simp[1/a Int[1/(1 + (a/b)*Sin[c + d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[(x_)^(m_.)*((a_.) + (b_.)*Sec[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol ] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sec[c + d*x])^ p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[(m + 1)/n], 0] && IntegerQ[p]
Int[((e_)*(x_))^(m_.)*((a_.) + (b_.)*Sec[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x _Symbol] :> Simp[e^IntPart[m]*((e*x)^FracPart[m]/x^FracPart[m]) Int[x^m*( a + b*Sec[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.50 (sec) , antiderivative size = 314, normalized size of antiderivative = 3.61
| method | result | size |
| risch | \(\frac {x \,{\mathrm e}^{\frac {\left (-1+n \right ) \left (-i \pi \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )+i \pi \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e x \right )^{2}+i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )^{2}-i \pi \operatorname {csgn}\left (i e x \right )^{3}+2 \ln \left (x \right )+2 \ln \left (e \right )\right )}{2}}}{a n}+\frac {2 i \arctan \left (\frac {2 a \,{\mathrm e}^{i \left (d \,x^{n}+2 c \right )}+2 \,{\mathrm e}^{i c} b}{2 \sqrt {a^{2} {\mathrm e}^{2 i c}-{\mathrm e}^{2 i c} b^{2}}}\right ) e^{n} b \,{\mathrm e}^{\frac {i \left (-\pi n \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )+\pi n \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e x \right )^{2}+\pi n \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )^{2}-\pi n \operatorname {csgn}\left (i e x \right )^{3}+\pi \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )-\pi \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e x \right )^{2}-\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )^{2}+\pi \operatorname {csgn}\left (i e x \right )^{3}+2 c \right )}{2}}}{\sqrt {a^{2} {\mathrm e}^{2 i c}-{\mathrm e}^{2 i c} b^{2}}\, d e n a}\) | \(314\) |
1/a/n*x*exp(1/2*(-1+n)*(-I*Pi*csgn(I*e)*csgn(I*x)*csgn(I*e*x)+I*Pi*csgn(I* e)*csgn(I*e*x)^2+I*Pi*csgn(I*x)*csgn(I*e*x)^2-I*Pi*csgn(I*e*x)^3+2*ln(x)+2 *ln(e)))+2*I*arctan(1/2*(2*a*exp(I*(d*x^n+2*c))+2*exp(I*c)*b)/(a^2*exp(2*I *c)-exp(2*I*c)*b^2)^(1/2))/(a^2*exp(2*I*c)-exp(2*I*c)*b^2)^(1/2)/d/e*e^n/n /a*b*exp(1/2*I*(-Pi*n*csgn(I*e)*csgn(I*x)*csgn(I*e*x)+Pi*n*csgn(I*e)*csgn( I*e*x)^2+Pi*n*csgn(I*x)*csgn(I*e*x)^2-Pi*n*csgn(I*e*x)^3+Pi*csgn(I*e)*csgn (I*x)*csgn(I*e*x)-Pi*csgn(I*e)*csgn(I*e*x)^2-Pi*csgn(I*x)*csgn(I*e*x)^2+Pi *csgn(I*e*x)^3+2*c))
Time = 0.32 (sec) , antiderivative size = 300, normalized size of antiderivative = 3.45 \[ \int \frac {(e x)^{-1+n}}{a+b \sec \left (c+d x^n\right )} \, dx=\left [\frac {2 \, {\left (a^{2} - b^{2}\right )} d e^{n - 1} x^{n} + \sqrt {a^{2} - b^{2}} b e^{n - 1} \log \left (\frac {2 \, a b \cos \left (d x^{n} + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x^{n} + c\right )^{2} + 2 \, a^{2} - b^{2} - 2 \, {\left (\sqrt {a^{2} - b^{2}} b \cos \left (d x^{n} + c\right ) + \sqrt {a^{2} - b^{2}} a\right )} \sin \left (d x^{n} + c\right )}{a^{2} \cos \left (d x^{n} + c\right )^{2} + 2 \, a b \cos \left (d x^{n} + c\right ) + b^{2}}\right )}{2 \, {\left (a^{3} - a b^{2}\right )} d n}, \frac {{\left (a^{2} - b^{2}\right )} d e^{n - 1} x^{n} - \sqrt {-a^{2} + b^{2}} b e^{n - 1} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} b \cos \left (d x^{n} + c\right ) + \sqrt {-a^{2} + b^{2}} a}{{\left (a^{2} - b^{2}\right )} \sin \left (d x^{n} + c\right )}\right )}{{\left (a^{3} - a b^{2}\right )} d n}\right ] \]
[1/2*(2*(a^2 - b^2)*d*e^(n - 1)*x^n + sqrt(a^2 - b^2)*b*e^(n - 1)*log((2*a *b*cos(d*x^n + c) - (a^2 - 2*b^2)*cos(d*x^n + c)^2 + 2*a^2 - b^2 - 2*(sqrt (a^2 - b^2)*b*cos(d*x^n + c) + sqrt(a^2 - b^2)*a)*sin(d*x^n + c))/(a^2*cos (d*x^n + c)^2 + 2*a*b*cos(d*x^n + c) + b^2)))/((a^3 - a*b^2)*d*n), ((a^2 - b^2)*d*e^(n - 1)*x^n - sqrt(-a^2 + b^2)*b*e^(n - 1)*arctan(-(sqrt(-a^2 + b^2)*b*cos(d*x^n + c) + sqrt(-a^2 + b^2)*a)/((a^2 - b^2)*sin(d*x^n + c)))) /((a^3 - a*b^2)*d*n)]
\[ \int \frac {(e x)^{-1+n}}{a+b \sec \left (c+d x^n\right )} \, dx=\int \frac {\left (e x\right )^{n - 1}}{a + b \sec {\left (c + d x^{n} \right )}}\, dx \]
\[ \int \frac {(e x)^{-1+n}}{a+b \sec \left (c+d x^n\right )} \, dx=\int { \frac {\left (e x\right )^{n - 1}}{b \sec \left (d x^{n} + c\right ) + a} \,d x } \]
-(2*a*b*e^(n + 1)*n*integrate((a*x^n*cos(2*d*x^n + 2*c)*cos(d*x^n + c) + 2 *b*x^n*cos(d*x^n + c)^2 + a*x^n*sin(2*d*x^n + 2*c)*sin(d*x^n + c) + 2*b*x^ n*sin(d*x^n + c)^2 + a*x^n*cos(d*x^n + c))/(a^3*e*x*cos(2*d*x^n + 2*c)^2 + 4*a*b^2*e*x*cos(d*x^n + c)^2 + a^3*e*x*sin(2*d*x^n + 2*c)^2 + 4*a^2*b*e*x *sin(2*d*x^n + 2*c)*sin(d*x^n + c) + 4*a*b^2*e*x*sin(d*x^n + c)^2 + 4*a^2* b*e*x*cos(d*x^n + c) + a^3*e*x + 2*(2*a^2*b*e*x*cos(d*x^n + c) + a^3*e*x)* cos(2*d*x^n + 2*c)), x) - e^n*x^n)/(a*e*n)
\[ \int \frac {(e x)^{-1+n}}{a+b \sec \left (c+d x^n\right )} \, dx=\int { \frac {\left (e x\right )^{n - 1}}{b \sec \left (d x^{n} + c\right ) + a} \,d x } \]
Time = 15.28 (sec) , antiderivative size = 223, normalized size of antiderivative = 2.56 \[ \int \frac {(e x)^{-1+n}}{a+b \sec \left (c+d x^n\right )} \, dx=\frac {x\,{\left (e\,x\right )}^{n-1}}{a\,n}+\frac {b\,x\,\ln \left (2\,b\,x\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\,{\mathrm {e}}^{d\,x^n\,1{}\mathrm {i}}\,{\left (e\,x\right )}^{n-1}-\frac {b\,x\,\left (a+b\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\,{\mathrm {e}}^{d\,x^n\,1{}\mathrm {i}}\right )\,{\left (e\,x\right )}^{n-1}\,2{}\mathrm {i}}{\sqrt {a+b}\,\sqrt {a-b}}\right )\,{\left (e\,x\right )}^{n-1}}{a\,d\,n\,x^n\,\sqrt {a+b}\,\sqrt {a-b}}-\frac {b\,x\,\ln \left (2\,b\,x\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\,{\mathrm {e}}^{d\,x^n\,1{}\mathrm {i}}\,{\left (e\,x\right )}^{n-1}+\frac {b\,x\,\left (a+b\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\,{\mathrm {e}}^{d\,x^n\,1{}\mathrm {i}}\right )\,{\left (e\,x\right )}^{n-1}\,2{}\mathrm {i}}{\sqrt {a+b}\,\sqrt {a-b}}\right )\,{\left (e\,x\right )}^{n-1}}{a\,d\,n\,x^n\,\sqrt {a+b}\,\sqrt {a-b}} \]
(x*(e*x)^(n - 1))/(a*n) + (b*x*log(2*b*x*exp(c*1i)*exp(d*x^n*1i)*(e*x)^(n - 1) - (b*x*(a + b*exp(c*1i)*exp(d*x^n*1i))*(e*x)^(n - 1)*2i)/((a + b)^(1/ 2)*(a - b)^(1/2)))*(e*x)^(n - 1))/(a*d*n*x^n*(a + b)^(1/2)*(a - b)^(1/2)) - (b*x*log(2*b*x*exp(c*1i)*exp(d*x^n*1i)*(e*x)^(n - 1) + (b*x*(a + b*exp(c *1i)*exp(d*x^n*1i))*(e*x)^(n - 1)*2i)/((a + b)^(1/2)*(a - b)^(1/2)))*(e*x) ^(n - 1))/(a*d*n*x^n*(a + b)^(1/2)*(a - b)^(1/2))