Integrand size = 16, antiderivative size = 94 \[ \int \frac {x}{\left (a+b \tan \left (c+d x^2\right )\right )^2} \, dx=\frac {\left (a^2-b^2\right ) x^2}{2 \left (a^2+b^2\right )^2}+\frac {a b \log \left (a \cos \left (c+d x^2\right )+b \sin \left (c+d x^2\right )\right )}{\left (a^2+b^2\right )^2 d}-\frac {b}{2 \left (a^2+b^2\right ) d \left (a+b \tan \left (c+d x^2\right )\right )} \]
1/2*(a^2-b^2)*x^2/(a^2+b^2)^2+a*b*ln(a*cos(d*x^2+c)+b*sin(d*x^2+c))/(a^2+b ^2)^2/d-1/2*b/(a^2+b^2)/d/(a+b*tan(d*x^2+c))
Result contains complex when optimal does not.
Time = 0.99 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.21 \[ \int \frac {x}{\left (a+b \tan \left (c+d x^2\right )\right )^2} \, dx=\frac {-\frac {i \log \left (i-\tan \left (c+d x^2\right )\right )}{(a+i b)^2}+\frac {i \log \left (i+\tan \left (c+d x^2\right )\right )}{(a-i b)^2}+\frac {2 b \left (2 a \log \left (a+b \tan \left (c+d x^2\right )\right )-\frac {a^2+b^2}{a+b \tan \left (c+d x^2\right )}\right )}{\left (a^2+b^2\right )^2}}{4 d} \]
(((-I)*Log[I - Tan[c + d*x^2]])/(a + I*b)^2 + (I*Log[I + Tan[c + d*x^2]])/ (a - I*b)^2 + (2*b*(2*a*Log[a + b*Tan[c + d*x^2]] - (a^2 + b^2)/(a + b*Tan [c + d*x^2])))/(a^2 + b^2)^2)/(4*d)
Time = 0.52 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {4234, 3042, 3964, 3042, 4014, 3042, 4013}
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.
| \(\displaystyle \int \frac {x}{\left (a+b \tan \left (c+d x^2\right )\right )^2} \, dx\) |
| \(\Big \downarrow \) 4234 |
| \(\displaystyle \frac {1}{2} \int \frac {1}{\left (a+b \tan \left (d x^2+c\right )\right )^2}dx^2\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {1}{\left (a+b \tan \left (d x^2+c\right )\right )^2}dx^2\) |
| \(\Big \downarrow \) 3964 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int \frac {a-b \tan \left (d x^2+c\right )}{a+b \tan \left (d x^2+c\right )}dx^2}{a^2+b^2}-\frac {b}{d \left (a^2+b^2\right ) \left (a+b \tan \left (c+d x^2\right )\right )}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int \frac {a-b \tan \left (d x^2+c\right )}{a+b \tan \left (d x^2+c\right )}dx^2}{a^2+b^2}-\frac {b}{d \left (a^2+b^2\right ) \left (a+b \tan \left (c+d x^2\right )\right )}\right )\) |
| \(\Big \downarrow \) 4014 |
| \(\displaystyle \frac {1}{2} \left (\frac {\frac {2 a b \int \frac {b-a \tan \left (d x^2+c\right )}{a+b \tan \left (d x^2+c\right )}dx^2}{a^2+b^2}+\frac {x^2 \left (a^2-b^2\right )}{a^2+b^2}}{a^2+b^2}-\frac {b}{d \left (a^2+b^2\right ) \left (a+b \tan \left (c+d x^2\right )\right )}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {\frac {2 a b \int \frac {b-a \tan \left (d x^2+c\right )}{a+b \tan \left (d x^2+c\right )}dx^2}{a^2+b^2}+\frac {x^2 \left (a^2-b^2\right )}{a^2+b^2}}{a^2+b^2}-\frac {b}{d \left (a^2+b^2\right ) \left (a+b \tan \left (c+d x^2\right )\right )}\right )\) |
| \(\Big \downarrow \) 4013 |
| \(\displaystyle \frac {1}{2} \left (\frac {\frac {2 a b \log \left (a \cos \left (c+d x^2\right )+b \sin \left (c+d x^2\right )\right )}{d \left (a^2+b^2\right )}+\frac {x^2 \left (a^2-b^2\right )}{a^2+b^2}}{a^2+b^2}-\frac {b}{d \left (a^2+b^2\right ) \left (a+b \tan \left (c+d x^2\right )\right )}\right )\) |
((((a^2 - b^2)*x^2)/(a^2 + b^2) + (2*a*b*Log[a*Cos[c + d*x^2] + b*Sin[c + d*x^2]])/((a^2 + b^2)*d))/(a^2 + b^2) - b/((a^2 + b^2)*d*(a + b*Tan[c + d* x^2])))/2
3.1.21.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + b*Tan[c + d*x])^(n + 1)/(d*(n + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2) Int[(a - b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -1]
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)* (x_)]), x_Symbol] :> Simp[(c/(b*f))*Log[RemoveContent[a*Cos[e + f*x] + b*Si n[e + f*x], x]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[(a*c + b*d)*(x/(a^2 + b^2)), x] + Simp[(b*c - a *d)/(a^2 + b^2) Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && N eQ[a*c + b*d, 0]
Int[(x_)^(m_.)*((a_.) + (b_.)*Tan[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol ] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Tan[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 = 0.15 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.13
| method | result | size |
| derivativedivides | \(\frac {-\frac {b}{\left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d \,x^{2}+c \right )\right )}+\frac {2 a b \ln \left (a +b \tan \left (d \,x^{2}+c \right )\right )}{\left (a^{2}+b^{2}\right )^{2}}+\frac {-a b \ln \left (1+\tan ^{2}\left (d \,x^{2}+c \right )\right )+\left (a^{2}-b^{2}\right ) \arctan \left (\tan \left (d \,x^{2}+c \right )\right )}{\left (a^{2}+b^{2}\right )^{2}}}{2 d}\) | \(106\) |
| default | \(\frac {-\frac {b}{\left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d \,x^{2}+c \right )\right )}+\frac {2 a b \ln \left (a +b \tan \left (d \,x^{2}+c \right )\right )}{\left (a^{2}+b^{2}\right )^{2}}+\frac {-a b \ln \left (1+\tan ^{2}\left (d \,x^{2}+c \right )\right )+\left (a^{2}-b^{2}\right ) \arctan \left (\tan \left (d \,x^{2}+c \right )\right )}{\left (a^{2}+b^{2}\right )^{2}}}{2 d}\) | \(106\) |
| norman | \(\frac {\frac {\left (a^{2}-b^{2}\right ) a \,x^{2}}{2 a^{4}+4 a^{2} b^{2}+2 b^{4}}+\frac {b \left (a^{2}-b^{2}\right ) x^{2} \tan \left (d \,x^{2}+c \right )}{2 a^{4}+4 a^{2} b^{2}+2 b^{4}}+\frac {b^{2} \tan \left (d \,x^{2}+c \right )}{2 a \left (a^{2}+b^{2}\right ) d}}{a +b \tan \left (d \,x^{2}+c \right )}+\frac {a b \ln \left (a +b \tan \left (d \,x^{2}+c \right )\right )}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}-\frac {a b \ln \left (1+\tan ^{2}\left (d \,x^{2}+c \right )\right )}{2 d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}\) | \(191\) |
| risch | \(-\frac {x^{2}}{2 \left (2 i a b -a^{2}+b^{2}\right )}-\frac {2 i a b \,x^{2}}{a^{4}+2 a^{2} b^{2}+b^{4}}-\frac {2 i a b c}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}-\frac {i b^{2}}{\left (-i a +b \right ) d \left (i a +b \right )^{2} \left ({\mathrm e}^{2 i \left (d \,x^{2}+c \right )} b +i a \,{\mathrm e}^{2 i \left (d \,x^{2}+c \right )}-b +i a \right )}+\frac {a b \ln \left ({\mathrm e}^{2 i \left (d \,x^{2}+c \right )}-\frac {i b +a}{i b -a}\right )}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}\) | \(191\) |
| parallelrisch | \(-\frac {-x^{2} \tan \left (d \,x^{2}+c \right ) a^{2} b^{2} d +x^{2} \tan \left (d \,x^{2}+c \right ) b^{4} d -x^{2} a^{3} b d +x^{2} a \,b^{3} d +\ln \left (1+\tan ^{2}\left (d \,x^{2}+c \right )\right ) \tan \left (d \,x^{2}+c \right ) a \,b^{3}-2 \ln \left (a +b \tan \left (d \,x^{2}+c \right )\right ) \tan \left (d \,x^{2}+c \right ) a \,b^{3}+\ln \left (1+\tan ^{2}\left (d \,x^{2}+c \right )\right ) a^{2} b^{2}-2 \ln \left (a +b \tan \left (d \,x^{2}+c \right )\right ) a^{2} b^{2}+a^{2} b^{2}+b^{4}}{2 \left (a +b \tan \left (d \,x^{2}+c \right )\right ) \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) b d}\) | \(200\) |
1/2/d*(-b/(a^2+b^2)/(a+b*tan(d*x^2+c))+2*a*b/(a^2+b^2)^2*ln(a+b*tan(d*x^2+ c))+1/(a^2+b^2)^2*(-a*b*ln(1+tan(d*x^2+c)^2)+(a^2-b^2)*arctan(tan(d*x^2+c) )))
Time = 0.25 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.80 \[ \int \frac {x}{\left (a+b \tan \left (c+d x^2\right )\right )^2} \, dx=\frac {{\left (a^{3} - a b^{2}\right )} d x^{2} - b^{3} + {\left (a b^{2} \tan \left (d x^{2} + c\right ) + a^{2} b\right )} \log \left (\frac {b^{2} \tan \left (d x^{2} + c\right )^{2} + 2 \, a b \tan \left (d x^{2} + c\right ) + a^{2}}{\tan \left (d x^{2} + c\right )^{2} + 1}\right ) + {\left ({\left (a^{2} b - b^{3}\right )} d x^{2} + a b^{2}\right )} \tan \left (d x^{2} + c\right )}{2 \, {\left ({\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d \tan \left (d x^{2} + c\right ) + {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}} \]
1/2*((a^3 - a*b^2)*d*x^2 - b^3 + (a*b^2*tan(d*x^2 + c) + a^2*b)*log((b^2*t an(d*x^2 + c)^2 + 2*a*b*tan(d*x^2 + c) + a^2)/(tan(d*x^2 + c)^2 + 1)) + (( a^2*b - b^3)*d*x^2 + a*b^2)*tan(d*x^2 + c))/((a^4*b + 2*a^2*b^3 + b^5)*d*t an(d*x^2 + c) + (a^5 + 2*a^3*b^2 + a*b^4)*d)
Result contains complex when optimal does not.
Time = 0.78 (sec) , antiderivative size = 1584, normalized size of antiderivative = 16.85 \[ \int \frac {x}{\left (a+b \tan \left (c+d x^2\right )\right )^2} \, dx=\text {Too large to display} \]
Piecewise((zoo*x**2/tan(c)**2, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (x**2/(2*a **2), Eq(b, 0)), (-(atan(tan(c + d*x**2)) + pi*floor((c + d*x**2 - pi/2)/p i))*tan(c + d*x**2)**2/(8*b**2*d*tan(c + d*x**2)**2 - 16*I*b**2*d*tan(c + d*x**2) - 8*b**2*d) + 2*I*(atan(tan(c + d*x**2)) + pi*floor((c + d*x**2 - pi/2)/pi))*tan(c + d*x**2)/(8*b**2*d*tan(c + d*x**2)**2 - 16*I*b**2*d*tan( c + d*x**2) - 8*b**2*d) + (atan(tan(c + d*x**2)) + pi*floor((c + d*x**2 - pi/2)/pi))/(8*b**2*d*tan(c + d*x**2)**2 - 16*I*b**2*d*tan(c + d*x**2) - 8* b**2*d) - tan(c + d*x**2)/(8*b**2*d*tan(c + d*x**2)**2 - 16*I*b**2*d*tan(c + d*x**2) - 8*b**2*d) + 2*I/(8*b**2*d*tan(c + d*x**2)**2 - 16*I*b**2*d*ta n(c + d*x**2) - 8*b**2*d), Eq(a, -I*b)), (-(atan(tan(c + d*x**2)) + pi*flo or((c + d*x**2 - pi/2)/pi))*tan(c + d*x**2)**2/(8*b**2*d*tan(c + d*x**2)** 2 + 16*I*b**2*d*tan(c + d*x**2) - 8*b**2*d) - 2*I*(atan(tan(c + d*x**2)) + pi*floor((c + d*x**2 - pi/2)/pi))*tan(c + d*x**2)/(8*b**2*d*tan(c + d*x** 2)**2 + 16*I*b**2*d*tan(c + d*x**2) - 8*b**2*d) + (atan(tan(c + d*x**2)) + pi*floor((c + d*x**2 - pi/2)/pi))/(8*b**2*d*tan(c + d*x**2)**2 + 16*I*b** 2*d*tan(c + d*x**2) - 8*b**2*d) - tan(c + d*x**2)/(8*b**2*d*tan(c + d*x**2 )**2 + 16*I*b**2*d*tan(c + d*x**2) - 8*b**2*d) - 2*I/(8*b**2*d*tan(c + d*x **2)**2 + 16*I*b**2*d*tan(c + d*x**2) - 8*b**2*d), Eq(a, I*b)), (x**2/(2*( a + b*tan(c))**2), Eq(d, 0)), (a**3*d*x**2/(2*a**5*d + 2*a**4*b*d*tan(c + d*x**2) + 4*a**3*b**2*d + 4*a**2*b**3*d*tan(c + d*x**2) + 2*a*b**4*d + ...
Leaf count of result is larger than twice the leaf count of optimal. 556 vs. \(2 (90) = 180\).
Time = 0.37 (sec) , antiderivative size = 556, normalized size of antiderivative = 5.91 \[ \int \frac {x}{\left (a+b \tan \left (c+d x^2\right )\right )^2} \, dx=\frac {{\left (a^{4} - b^{4}\right )} d x^{2} \cos \left (2 \, d x^{2} + 2 \, c\right )^{2} + {\left (a^{4} - b^{4}\right )} d x^{2} \sin \left (2 \, d x^{2} + 2 \, c\right )^{2} + {\left (a^{4} - b^{4}\right )} d x^{2} - 2 \, {\left (2 \, a b^{3} - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d x^{2}\right )} \cos \left (2 \, d x^{2} + 2 \, c\right ) + {\left (4 \, a^{2} b^{2} \sin \left (2 \, d x^{2} + 2 \, c\right ) + a^{3} b + a b^{3} + {\left (a^{3} b + a b^{3}\right )} \cos \left (2 \, d x^{2} + 2 \, c\right )^{2} + {\left (a^{3} b + a b^{3}\right )} \sin \left (2 \, d x^{2} + 2 \, c\right )^{2} + 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (2 \, d x^{2} + 2 \, c\right )\right )} \log \left (\frac {{\left (a^{2} + b^{2}\right )} \cos \left (2 \, d x^{2} + 2 \, c\right )^{2} + 4 \, a b \sin \left (2 \, d x^{2} + 2 \, c\right ) + {\left (a^{2} + b^{2}\right )} \sin \left (2 \, d x^{2} + 2 \, c\right )^{2} + a^{2} + b^{2} + 2 \, {\left (a^{2} - b^{2}\right )} \cos \left (2 \, d x^{2} + 2 \, c\right )}{{\left (a^{2} + b^{2}\right )} \cos \left (2 \, c\right )^{2} + {\left (a^{2} + b^{2}\right )} \sin \left (2 \, c\right )^{2}}\right ) + 2 \, {\left (a^{2} b^{2} - b^{4} + 2 \, {\left (a^{3} b - a b^{3}\right )} d x^{2}\right )} \sin \left (2 \, d x^{2} + 2 \, c\right )}{2 \, {\left ({\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} d \cos \left (2 \, d x^{2} + 2 \, c\right )^{2} + {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} d \sin \left (2 \, d x^{2} + 2 \, c\right )^{2} + 2 \, {\left (a^{6} + a^{4} b^{2} - a^{2} b^{4} - b^{6}\right )} d \cos \left (2 \, d x^{2} + 2 \, c\right ) + 4 \, {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} d \sin \left (2 \, d x^{2} + 2 \, c\right ) + {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} d\right )}} \]
1/2*((a^4 - b^4)*d*x^2*cos(2*d*x^2 + 2*c)^2 + (a^4 - b^4)*d*x^2*sin(2*d*x^ 2 + 2*c)^2 + (a^4 - b^4)*d*x^2 - 2*(2*a*b^3 - (a^4 - 2*a^2*b^2 + b^4)*d*x^ 2)*cos(2*d*x^2 + 2*c) + (4*a^2*b^2*sin(2*d*x^2 + 2*c) + a^3*b + a*b^3 + (a ^3*b + a*b^3)*cos(2*d*x^2 + 2*c)^2 + (a^3*b + a*b^3)*sin(2*d*x^2 + 2*c)^2 + 2*(a^3*b - a*b^3)*cos(2*d*x^2 + 2*c))*log(((a^2 + b^2)*cos(2*d*x^2 + 2*c )^2 + 4*a*b*sin(2*d*x^2 + 2*c) + (a^2 + b^2)*sin(2*d*x^2 + 2*c)^2 + a^2 + b^2 + 2*(a^2 - b^2)*cos(2*d*x^2 + 2*c))/((a^2 + b^2)*cos(2*c)^2 + (a^2 + b ^2)*sin(2*c)^2)) + 2*(a^2*b^2 - b^4 + 2*(a^3*b - a*b^3)*d*x^2)*sin(2*d*x^2 + 2*c))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d*cos(2*d*x^2 + 2*c)^2 + (a^ 6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d*sin(2*d*x^2 + 2*c)^2 + 2*(a^6 + a^4*b^2 - a^2*b^4 - b^6)*d*cos(2*d*x^2 + 2*c) + 4*(a^5*b + 2*a^3*b^3 + a*b^5)*d*s in(2*d*x^2 + 2*c) + (a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d)
Time = 0.36 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.69 \[ \int \frac {x}{\left (a+b \tan \left (c+d x^2\right )\right )^2} \, dx=\frac {a b^{2} \log \left ({\left | b \tan \left (d x^{2} + c\right ) + a \right |}\right )}{a^{4} b d + 2 \, a^{2} b^{3} d + b^{5} d} - \frac {a b \log \left (\tan \left (d x^{2} + c\right )^{2} + 1\right )}{2 \, {\left (a^{4} d + 2 \, a^{2} b^{2} d + b^{4} d\right )}} + \frac {{\left (d x^{2} + c\right )} {\left (a^{2} - b^{2}\right )}}{2 \, {\left (a^{4} d + 2 \, a^{2} b^{2} d + b^{4} d\right )}} - \frac {a^{2} b + b^{3}}{2 \, {\left (a^{2} + b^{2}\right )}^{2} {\left (b \tan \left (d x^{2} + c\right ) + a\right )} d} \]
a*b^2*log(abs(b*tan(d*x^2 + c) + a))/(a^4*b*d + 2*a^2*b^3*d + b^5*d) - 1/2 *a*b*log(tan(d*x^2 + c)^2 + 1)/(a^4*d + 2*a^2*b^2*d + b^4*d) + 1/2*(d*x^2 + c)*(a^2 - b^2)/(a^4*d + 2*a^2*b^2*d + b^4*d) - 1/2*(a^2*b + b^3)/((a^2 + b^2)^2*(b*tan(d*x^2 + c) + a)*d)
Time = 3.86 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.84 \[ \int \frac {x}{\left (a+b \tan \left (c+d x^2\right )\right )^2} \, dx=\frac {\frac {x^2\,\mathrm {tan}\left (d\,x^2+c\right )\,\left (\frac {a^2\,b}{2}-\frac {b^3}{2}\right )}{{\left (a^2+b^2\right )}^2}-\frac {x^2\,\left (\frac {a\,b^2}{2}-\frac {a^3}{2}\right )}{{\left (a^2+b^2\right )}^2}+\frac {b^2\,\mathrm {tan}\left (d\,x^2+c\right )}{2\,a\,d\,\left (a^2+b^2\right )}}{a+b\,\mathrm {tan}\left (d\,x^2+c\right )}-\frac {a\,b\,\ln \left ({\mathrm {tan}\left (d\,x^2+c\right )}^2+1\right )}{2\,\left (d\,a^4+2\,d\,a^2\,b^2+d\,b^4\right )}+\frac {a\,b\,\ln \left (a+b\,\mathrm {tan}\left (d\,x^2+c\right )\right )}{d\,{\left (a^2+b^2\right )}^2} \]