Integrand size = 16, antiderivative size = 66 \[ \int \frac {x}{a+b \sec \left (c+d x^2\right )} \, dx=\frac {x^2}{2 a}-\frac {b \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} \left (c+d x^2\right )\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} \sqrt {a+b} d} \]
1/2*x^2/a-b*arctanh((a-b)^(1/2)*tan(1/2*d*x^2+1/2*c)/(a+b)^(1/2))/a/d/(a-b )^(1/2)/(a+b)^(1/2)
Time = 0.34 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.02 \[ \int \frac {x}{a+b \sec \left (c+d x^2\right )} \, dx=\frac {\frac {c}{d}+x^2+\frac {2 b \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} \left (c+d x^2\right )\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d}}{2 a} \]
(c/d + x^2 + (2*b*ArcTanh[((-a + b)*Tan[(c + d*x^2)/2])/Sqrt[a^2 - b^2]])/ (Sqrt[a^2 - b^2]*d))/(2*a)
Time = 0.35 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {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.
\(\displaystyle \int \frac {x}{a+b \sec \left (c+d x^2\right )} \, dx\) |
\(\Big \downarrow \) 4692 |
\(\displaystyle \frac {1}{2} \int \frac {1}{a+b \sec \left (d x^2+c\right )}dx^2\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \int \frac {1}{a+b \csc \left (d x^2+c+\frac {\pi }{2}\right )}dx^2\) |
\(\Big \downarrow \) 4270 |
\(\displaystyle \frac {1}{2} \left (\frac {x^2}{a}-\frac {\int \frac {1}{\frac {a \cos \left (d x^2+c\right )}{b}+1}dx^2}{a}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {x^2}{a}-\frac {\int \frac {1}{\frac {a \sin \left (d x^2+c+\frac {\pi }{2}\right )}{b}+1}dx^2}{a}\right )\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {1}{2} \left (\frac {x^2}{a}-\frac {2 \int \frac {1}{\left (1-\frac {a}{b}\right ) x^4+\frac {a+b}{b}}d\tan \left (\frac {1}{2} \left (d x^2+c\right )\right )}{a d}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (\frac {x^2}{a}-\frac {2 b \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} \left (c+d x^2\right )\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )\) |
(x^2/a - (2*b*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x^2)/2])/Sqrt[a + b]])/(a*Sq rt[a - b]*Sqrt[a + b]*d))/2
3.1.20.3.1 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]
Time = 0.29 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.06
method | result | size |
derivativedivides | \(\frac {\frac {2 \arctan \left (\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right )}{a}-\frac {2 b \,\operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a \sqrt {\left (a -b \right ) \left (a +b \right )}}}{2 d}\) | \(70\) |
default | \(\frac {\frac {2 \arctan \left (\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right )}{a}-\frac {2 b \,\operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a \sqrt {\left (a -b \right ) \left (a +b \right )}}}{2 d}\) | \(70\) |
risch | \(\frac {x^{2}}{2 a}+\frac {b \ln \left ({\mathrm e}^{i \left (d \,x^{2}+c \right )}-\frac {i a^{2}-i b^{2}-b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{2 \sqrt {a^{2}-b^{2}}\, d a}-\frac {b \ln \left ({\mathrm e}^{i \left (d \,x^{2}+c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{2 \sqrt {a^{2}-b^{2}}\, d a}\) | \(160\) |
1/2/d*(2/a*arctan(tan(1/2*d*x^2+1/2*c))-2*b/a/((a-b)*(a+b))^(1/2)*arctanh( (a-b)*tan(1/2*d*x^2+1/2*c)/((a-b)*(a+b))^(1/2)))
Time = 0.29 (sec) , antiderivative size = 251, normalized size of antiderivative = 3.80 \[ \int \frac {x}{a+b \sec \left (c+d x^2\right )} \, dx=\left [\frac {2 \, {\left (a^{2} - b^{2}\right )} d x^{2} + \sqrt {a^{2} - b^{2}} b \log \left (\frac {2 \, a b \cos \left (d x^{2} + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x^{2} + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x^{2} + c\right ) + a\right )} \sin \left (d x^{2} + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x^{2} + c\right )^{2} + 2 \, a b \cos \left (d x^{2} + c\right ) + b^{2}}\right )}{4 \, {\left (a^{3} - a b^{2}\right )} d}, \frac {{\left (a^{2} - b^{2}\right )} d x^{2} - \sqrt {-a^{2} + b^{2}} b \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x^{2} + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x^{2} + c\right )}\right )}{2 \, {\left (a^{3} - a b^{2}\right )} d}\right ] \]
[1/4*(2*(a^2 - b^2)*d*x^2 + sqrt(a^2 - b^2)*b*log((2*a*b*cos(d*x^2 + c) - (a^2 - 2*b^2)*cos(d*x^2 + c)^2 - 2*sqrt(a^2 - b^2)*(b*cos(d*x^2 + c) + a)* sin(d*x^2 + c) + 2*a^2 - b^2)/(a^2*cos(d*x^2 + c)^2 + 2*a*b*cos(d*x^2 + c) + b^2)))/((a^3 - a*b^2)*d), 1/2*((a^2 - b^2)*d*x^2 - sqrt(-a^2 + b^2)*b*a rctan(-sqrt(-a^2 + b^2)*(b*cos(d*x^2 + c) + a)/((a^2 - b^2)*sin(d*x^2 + c) )))/((a^3 - a*b^2)*d)]
\[ \int \frac {x}{a+b \sec \left (c+d x^2\right )} \, dx=\int \frac {x}{a + b \sec {\left (c + d x^{2} \right )}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 7945 vs. \(2 (55) = 110\).
Time = 28.61 (sec) , antiderivative size = 7945, normalized size of antiderivative = 120.38 \[ \int \frac {x}{a+b \sec \left (c+d x^2\right )} \, dx=\text {Too large to display} \]
1/2*(sqrt(-a^2 + b^2)*d*x^2 - b*arctan2(2*(4*(a^6 - a^4*b^2)*cos(d*x^2 + 2 *c)^4*cos(c)*sin(c) - 4*(a^6 - a^4*b^2)*cos(c)*sin(d*x^2 + 2*c)^4*sin(c) + 4*(3*(a^5*b - a^3*b^3)*cos(c)^2*sin(c) + (a^5*b - a^3*b^3)*sin(c)^3)*cos( d*x^2 + 2*c)^3 - 4*((a^5*b - a^3*b^3)*cos(c)^3 + 3*(a^5*b - a^3*b^3)*cos(c )*sin(c)^2 + ((a^6 - a^4*b^2)*cos(c)^2 - (a^6 - a^4*b^2)*sin(c)^2)*cos(d*x ^2 + 2*c))*sin(d*x^2 + 2*c)^3 - 4*((a^6 - 5*a^4*b^2 + 4*a^2*b^4)*cos(c)^3* sin(c) + (a^6 - 5*a^4*b^2 + 4*a^2*b^4)*cos(c)*sin(c)^3)*cos(d*x^2 + 2*c)^2 + 4*((a^6 - 5*a^4*b^2 + 4*a^2*b^4)*cos(c)^3*sin(c) + (a^6 - 5*a^4*b^2 + 4 *a^2*b^4)*cos(c)*sin(c)^3 - 3*((a^5*b - a^3*b^3)*cos(c)^2*sin(c) - (a^5*b - a^3*b^3)*sin(c)^3)*cos(d*x^2 + 2*c))*sin(d*x^2 + 2*c)^2 - 4*((a^5*b - 3* a^3*b^3 + 2*a*b^5)*cos(c)^4*sin(c) + 2*(a^5*b - 3*a^3*b^3 + 2*a*b^5)*cos(c )^2*sin(c)^3 + (a^5*b - 3*a^3*b^3 + 2*a*b^5)*sin(c)^5)*cos(d*x^2 + 2*c) + 4*((a^5*b - 3*a^3*b^3 + 2*a*b^5)*cos(c)^5 + 2*(a^5*b - 3*a^3*b^3 + 2*a*b^5 )*cos(c)^3*sin(c)^2 + (a^5*b - 3*a^3*b^3 + 2*a*b^5)*cos(c)*sin(c)^4 - ((a^ 6 - a^4*b^2)*cos(c)^2 - (a^6 - a^4*b^2)*sin(c)^2)*cos(d*x^2 + 2*c)^3 - 3*( (a^5*b - a^3*b^3)*cos(c)^3 - (a^5*b - a^3*b^3)*cos(c)*sin(c)^2)*cos(d*x^2 + 2*c)^2 + ((a^6 - 5*a^4*b^2 + 4*a^2*b^4)*cos(c)^4 - (a^6 - 5*a^4*b^2 + 4* a^2*b^4)*sin(c)^4)*cos(d*x^2 + 2*c))*sin(d*x^2 + 2*c) + (a^5*cos(c)*sin(d* x^2 + 2*c)^5 - a^5*cos(d*x^2 + 2*c)^5*sin(c) - 4*a^4*b*cos(d*x^2 + 2*c)^4* cos(c)*sin(c) - (a^5*cos(d*x^2 + 2*c)*sin(c) - 4*a^4*b*cos(c)*sin(c))*s...
Leaf count of result is larger than twice the leaf count of optimal. 278 vs. \(2 (55) = 110\).
Time = 0.29 (sec) , antiderivative size = 278, normalized size of antiderivative = 4.21 \[ \int \frac {x}{a+b \sec \left (c+d x^2\right )} \, dx=\frac {{\left (\sqrt {-a^{2} + b^{2}} {\left (a - 2 \, b\right )} d {\left | -a + b \right |} - \sqrt {-a^{2} + b^{2}} {\left | a \right |} {\left | -a + b \right |} {\left | d \right |}\right )} {\left (\pi \left \lfloor \frac {d x^{2} + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x^{2} + \frac {1}{2} \, c\right )}{\sqrt {-\frac {b d + \sqrt {b^{2} d^{2} + {\left (a d + b d\right )} {\left (a d - b d\right )}}}{a d - b d}}}\right )\right )}}{2 \, {\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} a^{2} d^{2} + {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} d {\left | a \right |} {\left | d \right |}\right )}} + \frac {{\left (a d - 2 \, b d + {\left | a \right |} {\left | d \right |}\right )} {\left (\pi \left \lfloor \frac {d x^{2} + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x^{2} + \frac {1}{2} \, c\right )}{\sqrt {-\frac {b d - \sqrt {b^{2} d^{2} + {\left (a d + b d\right )} {\left (a d - b d\right )}}}{a d - b d}}}\right )\right )}}{2 \, {\left (a^{2} d^{2} - b d {\left | a \right |} {\left | d \right |}\right )}} \]
1/2*(sqrt(-a^2 + b^2)*(a - 2*b)*d*abs(-a + b) - sqrt(-a^2 + b^2)*abs(a)*ab s(-a + b)*abs(d))*(pi*floor(1/2*(d*x^2 + c)/pi + 1/2) + arctan(tan(1/2*d*x ^2 + 1/2*c)/sqrt(-(b*d + sqrt(b^2*d^2 + (a*d + b*d)*(a*d - b*d)))/(a*d - b *d))))/((a^2 - 2*a*b + b^2)*a^2*d^2 + (a^2*b - 2*a*b^2 + b^3)*d*abs(a)*abs (d)) + 1/2*(a*d - 2*b*d + abs(a)*abs(d))*(pi*floor(1/2*(d*x^2 + c)/pi + 1/ 2) + arctan(tan(1/2*d*x^2 + 1/2*c)/sqrt(-(b*d - sqrt(b^2*d^2 + (a*d + b*d) *(a*d - b*d)))/(a*d - b*d))))/(a^2*d^2 - b*d*abs(a)*abs(d))
Time = 15.09 (sec) , antiderivative size = 157, normalized size of antiderivative = 2.38 \[ \int \frac {x}{a+b \sec \left (c+d x^2\right )} \, dx=\frac {x^2}{2\,a}+\frac {b\,\ln \left (2\,b\,x\,{\mathrm {e}}^{d\,x^2\,1{}\mathrm {i}}\,{\mathrm {e}}^{c\,1{}\mathrm {i}}-\frac {b\,x\,\left (a+b\,{\mathrm {e}}^{d\,x^2\,1{}\mathrm {i}}\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\right )\,2{}\mathrm {i}}{\sqrt {a+b}\,\sqrt {a-b}}\right )}{2\,a\,d\,\sqrt {a+b}\,\sqrt {a-b}}-\frac {b\,\ln \left (2\,b\,x\,{\mathrm {e}}^{d\,x^2\,1{}\mathrm {i}}\,{\mathrm {e}}^{c\,1{}\mathrm {i}}+\frac {b\,x\,\left (a+b\,{\mathrm {e}}^{d\,x^2\,1{}\mathrm {i}}\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\right )\,2{}\mathrm {i}}{\sqrt {a+b}\,\sqrt {a-b}}\right )}{2\,a\,d\,\sqrt {a+b}\,\sqrt {a-b}} \]
x^2/(2*a) + (b*log(2*b*x*exp(d*x^2*1i)*exp(c*1i) - (b*x*(a + b*exp(d*x^2*1 i)*exp(c*1i))*2i)/((a + b)^(1/2)*(a - b)^(1/2))))/(2*a*d*(a + b)^(1/2)*(a - b)^(1/2)) - (b*log(2*b*x*exp(d*x^2*1i)*exp(c*1i) + (b*x*(a + b*exp(d*x^2 *1i)*exp(c*1i))*2i)/((a + b)^(1/2)*(a - b)^(1/2))))/(2*a*d*(a + b)^(1/2)*( a - b)^(1/2))