Integrand size = 19, antiderivative size = 81 \[ \int \frac {1}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^3} \, dx=-\frac {1}{2 a c^3 \left (1+a^2 x^2\right )^2 \arctan (a x)^2}+\frac {2 x}{c^3 \left (1+a^2 x^2\right )^2 \arctan (a x)}-\frac {\operatorname {CosIntegral}(2 \arctan (a x))}{a c^3}-\frac {\operatorname {CosIntegral}(4 \arctan (a x))}{a c^3} \]
-1/2/a/c^3/(a^2*x^2+1)^2/arctan(a*x)^2+2*x/c^3/(a^2*x^2+1)^2/arctan(a*x)-C i(2*arctan(a*x))/a/c^3-Ci(4*arctan(a*x))/a/c^3
Time = 0.14 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.10 \[ \int \frac {1}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^3} \, dx=-\frac {1-4 a x \arctan (a x)+2 \left (1+a^2 x^2\right )^2 \arctan (a x)^2 \operatorname {CosIntegral}(2 \arctan (a x))+2 \left (1+a^2 x^2\right )^2 \arctan (a x)^2 \operatorname {CosIntegral}(4 \arctan (a x))}{2 a c^3 \left (1+a^2 x^2\right )^2 \arctan (a x)^2} \]
-1/2*(1 - 4*a*x*ArcTan[a*x] + 2*(1 + a^2*x^2)^2*ArcTan[a*x]^2*CosIntegral[ 2*ArcTan[a*x]] + 2*(1 + a^2*x^2)^2*ArcTan[a*x]^2*CosIntegral[4*ArcTan[a*x] ])/(a*c^3*(1 + a^2*x^2)^2*ArcTan[a*x]^2)
Time = 0.93 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.48, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {5437, 27, 5503, 5439, 3042, 3793, 2009, 5505, 4906, 2009}
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.
\(\displaystyle \int \frac {1}{\arctan (a x)^3 \left (a^2 c x^2+c\right )^3} \, dx\) |
\(\Big \downarrow \) 5437 |
\(\displaystyle -2 a \int \frac {x}{c^3 \left (a^2 x^2+1\right )^3 \arctan (a x)^2}dx-\frac {1}{2 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {2 a \int \frac {x}{\left (a^2 x^2+1\right )^3 \arctan (a x)^2}dx}{c^3}-\frac {1}{2 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^2}\) |
\(\Big \downarrow \) 5503 |
\(\displaystyle -\frac {2 a \left (\frac {\int \frac {1}{\left (a^2 x^2+1\right )^3 \arctan (a x)}dx}{a}-3 a \int \frac {x^2}{\left (a^2 x^2+1\right )^3 \arctan (a x)}dx-\frac {x}{a \left (a^2 x^2+1\right )^2 \arctan (a x)}\right )}{c^3}-\frac {1}{2 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^2}\) |
\(\Big \downarrow \) 5439 |
\(\displaystyle -\frac {2 a \left (-3 a \int \frac {x^2}{\left (a^2 x^2+1\right )^3 \arctan (a x)}dx+\frac {\int \frac {1}{\left (a^2 x^2+1\right )^2 \arctan (a x)}d\arctan (a x)}{a^2}-\frac {x}{a \left (a^2 x^2+1\right )^2 \arctan (a x)}\right )}{c^3}-\frac {1}{2 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 a \left (-3 a \int \frac {x^2}{\left (a^2 x^2+1\right )^3 \arctan (a x)}dx+\frac {\int \frac {\sin \left (\arctan (a x)+\frac {\pi }{2}\right )^4}{\arctan (a x)}d\arctan (a x)}{a^2}-\frac {x}{a \left (a^2 x^2+1\right )^2 \arctan (a x)}\right )}{c^3}-\frac {1}{2 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^2}\) |
\(\Big \downarrow \) 3793 |
\(\displaystyle -\frac {2 a \left (-3 a \int \frac {x^2}{\left (a^2 x^2+1\right )^3 \arctan (a x)}dx+\frac {\int \left (\frac {\cos (2 \arctan (a x))}{2 \arctan (a x)}+\frac {\cos (4 \arctan (a x))}{8 \arctan (a x)}+\frac {3}{8 \arctan (a x)}\right )d\arctan (a x)}{a^2}-\frac {x}{a \left (a^2 x^2+1\right )^2 \arctan (a x)}\right )}{c^3}-\frac {1}{2 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 a \left (-3 a \int \frac {x^2}{\left (a^2 x^2+1\right )^3 \arctan (a x)}dx+\frac {\frac {1}{2} \operatorname {CosIntegral}(2 \arctan (a x))+\frac {1}{8} \operatorname {CosIntegral}(4 \arctan (a x))+\frac {3}{8} \log (\arctan (a x))}{a^2}-\frac {x}{a \left (a^2 x^2+1\right )^2 \arctan (a x)}\right )}{c^3}-\frac {1}{2 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^2}\) |
\(\Big \downarrow \) 5505 |
\(\displaystyle -\frac {2 a \left (-\frac {3 \int \frac {a^2 x^2}{\left (a^2 x^2+1\right )^2 \arctan (a x)}d\arctan (a x)}{a^2}+\frac {\frac {1}{2} \operatorname {CosIntegral}(2 \arctan (a x))+\frac {1}{8} \operatorname {CosIntegral}(4 \arctan (a x))+\frac {3}{8} \log (\arctan (a x))}{a^2}-\frac {x}{a \left (a^2 x^2+1\right )^2 \arctan (a x)}\right )}{c^3}-\frac {1}{2 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^2}\) |
\(\Big \downarrow \) 4906 |
\(\displaystyle -\frac {2 a \left (-\frac {3 \int \left (\frac {1}{8 \arctan (a x)}-\frac {\cos (4 \arctan (a x))}{8 \arctan (a x)}\right )d\arctan (a x)}{a^2}+\frac {\frac {1}{2} \operatorname {CosIntegral}(2 \arctan (a x))+\frac {1}{8} \operatorname {CosIntegral}(4 \arctan (a x))+\frac {3}{8} \log (\arctan (a x))}{a^2}-\frac {x}{a \left (a^2 x^2+1\right )^2 \arctan (a x)}\right )}{c^3}-\frac {1}{2 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 a \left (-\frac {3 \left (\frac {1}{8} \log (\arctan (a x))-\frac {1}{8} \operatorname {CosIntegral}(4 \arctan (a x))\right )}{a^2}+\frac {\frac {1}{2} \operatorname {CosIntegral}(2 \arctan (a x))+\frac {1}{8} \operatorname {CosIntegral}(4 \arctan (a x))+\frac {3}{8} \log (\arctan (a x))}{a^2}-\frac {x}{a \left (a^2 x^2+1\right )^2 \arctan (a x)}\right )}{c^3}-\frac {1}{2 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^2}\) |
-1/2*1/(a*c^3*(1 + a^2*x^2)^2*ArcTan[a*x]^2) - (2*a*(-(x/(a*(1 + a^2*x^2)^ 2*ArcTan[a*x])) - (3*(-1/8*CosIntegral[4*ArcTan[a*x]] + Log[ArcTan[a*x]]/8 ))/a^2 + (CosIntegral[2*ArcTan[a*x]]/2 + CosIntegral[4*ArcTan[a*x]]/8 + (3 *Log[ArcTan[a*x]])/8)/a^2))/c^3
3.7.38.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f , m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b _.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin[a + b*x ]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IG tQ[p, 0]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_)^2)^(q_), x_S ymbol] :> Simp[(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])^(p + 1)/(b*c*d*(p + 1))), x] - Simp[2*c*((q + 1)/(b*(p + 1))) Int[x*(d + e*x^2)^q*(a + b*Arc Tan[c*x])^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && LtQ[q, -1] && LtQ[p, -1]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_), x_ Symbol] :> Simp[d^q/c Subst[Int[(a + b*x)^p/Cos[x]^(2*(q + 1)), x], x, Ar cTan[c*x]], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && ILtQ[2*( q + 1), 0] && (IntegerQ[q] || GtQ[d, 0])
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 2)^(q_), x_Symbol] :> Simp[x^m*(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])^(p + 1)/(b*c*d*(p + 1))), x] + (-Simp[c*((m + 2*q + 2)/(b*(p + 1))) Int[x^(m + 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p + 1), x], x] - Simp[m/(b*c*(p + 1)) Int[x^(m - 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p + 1), x], x]) /; F reeQ[{a, b, c, d, e, m}, x] && EqQ[e, c^2*d] && IntegerQ[m] && LtQ[q, -1] & & LtQ[p, -1] && NeQ[m + 2*q + 2, 0]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 2)^(q_), x_Symbol] :> Simp[d^q/c^(m + 1) Subst[Int[(a + b*x)^p*(Sin[x]^m/ Cos[x]^(m + 2*(q + 1))), x], x, ArcTan[c*x]], x] /; FreeQ[{a, b, c, d, e, p }, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && ILtQ[m + 2*q + 1, 0] && (IntegerQ[q ] || GtQ[d, 0])
Time = 11.20 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.10
method | result | size |
derivativedivides | \(-\frac {16 \,\operatorname {Ci}\left (4 \arctan \left (a x \right )\right ) \arctan \left (a x \right )^{2}+16 \,\operatorname {Ci}\left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )^{2}-4 \sin \left (4 \arctan \left (a x \right )\right ) \arctan \left (a x \right )-8 \sin \left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )+4 \cos \left (2 \arctan \left (a x \right )\right )+\cos \left (4 \arctan \left (a x \right )\right )+3}{16 a \,c^{3} \arctan \left (a x \right )^{2}}\) | \(89\) |
default | \(-\frac {16 \,\operatorname {Ci}\left (4 \arctan \left (a x \right )\right ) \arctan \left (a x \right )^{2}+16 \,\operatorname {Ci}\left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )^{2}-4 \sin \left (4 \arctan \left (a x \right )\right ) \arctan \left (a x \right )-8 \sin \left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )+4 \cos \left (2 \arctan \left (a x \right )\right )+\cos \left (4 \arctan \left (a x \right )\right )+3}{16 a \,c^{3} \arctan \left (a x \right )^{2}}\) | \(89\) |
-1/16/a/c^3*(16*Ci(4*arctan(a*x))*arctan(a*x)^2+16*Ci(2*arctan(a*x))*arcta n(a*x)^2-4*sin(4*arctan(a*x))*arctan(a*x)-8*sin(2*arctan(a*x))*arctan(a*x) +4*cos(2*arctan(a*x))+cos(4*arctan(a*x))+3)/arctan(a*x)^2
Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 297, normalized size of antiderivative = 3.67 \[ \int \frac {1}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^3} \, dx=-\frac {{\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} \operatorname {log\_integral}\left (\frac {a^{4} x^{4} + 4 i \, a^{3} x^{3} - 6 \, a^{2} x^{2} - 4 i \, a x + 1}{a^{4} x^{4} + 2 \, a^{2} x^{2} + 1}\right ) + {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} \operatorname {log\_integral}\left (\frac {a^{4} x^{4} - 4 i \, a^{3} x^{3} - 6 \, a^{2} x^{2} + 4 i \, a x + 1}{a^{4} x^{4} + 2 \, a^{2} x^{2} + 1}\right ) + {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} \operatorname {log\_integral}\left (-\frac {a^{2} x^{2} + 2 i \, a x - 1}{a^{2} x^{2} + 1}\right ) + {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} \operatorname {log\_integral}\left (-\frac {a^{2} x^{2} - 2 i \, a x - 1}{a^{2} x^{2} + 1}\right ) - 4 \, a x \arctan \left (a x\right ) + 1}{2 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{3} c^{3} x^{2} + a c^{3}\right )} \arctan \left (a x\right )^{2}} \]
-1/2*((a^4*x^4 + 2*a^2*x^2 + 1)*arctan(a*x)^2*log_integral((a^4*x^4 + 4*I* a^3*x^3 - 6*a^2*x^2 - 4*I*a*x + 1)/(a^4*x^4 + 2*a^2*x^2 + 1)) + (a^4*x^4 + 2*a^2*x^2 + 1)*arctan(a*x)^2*log_integral((a^4*x^4 - 4*I*a^3*x^3 - 6*a^2* x^2 + 4*I*a*x + 1)/(a^4*x^4 + 2*a^2*x^2 + 1)) + (a^4*x^4 + 2*a^2*x^2 + 1)* arctan(a*x)^2*log_integral(-(a^2*x^2 + 2*I*a*x - 1)/(a^2*x^2 + 1)) + (a^4* x^4 + 2*a^2*x^2 + 1)*arctan(a*x)^2*log_integral(-(a^2*x^2 - 2*I*a*x - 1)/( a^2*x^2 + 1)) - 4*a*x*arctan(a*x) + 1)/((a^5*c^3*x^4 + 2*a^3*c^3*x^2 + a*c ^3)*arctan(a*x)^2)
\[ \int \frac {1}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^3} \, dx=\frac {\int \frac {1}{a^{6} x^{6} \operatorname {atan}^{3}{\left (a x \right )} + 3 a^{4} x^{4} \operatorname {atan}^{3}{\left (a x \right )} + 3 a^{2} x^{2} \operatorname {atan}^{3}{\left (a x \right )} + \operatorname {atan}^{3}{\left (a x \right )}}\, dx}{c^{3}} \]
Integral(1/(a**6*x**6*atan(a*x)**3 + 3*a**4*x**4*atan(a*x)**3 + 3*a**2*x** 2*atan(a*x)**3 + atan(a*x)**3), x)/c**3
\[ \int \frac {1}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^3} \, dx=\int { \frac {1}{{\left (a^{2} c x^{2} + c\right )}^{3} \arctan \left (a x\right )^{3}} \,d x } \]
1/2*(2*(a^5*c^3*x^4 + 2*a^3*c^3*x^2 + a*c^3)*arctan(a*x)^2*integrate(2*(3* a^2*x^2 - 1)/((a^6*c^3*x^6 + 3*a^4*c^3*x^4 + 3*a^2*c^3*x^2 + c^3)*arctan(a *x)), x) + 4*a*x*arctan(a*x) - 1)/((a^5*c^3*x^4 + 2*a^3*c^3*x^2 + a*c^3)*a rctan(a*x)^2)
\[ \int \frac {1}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^3} \, dx=\int { \frac {1}{{\left (a^{2} c x^{2} + c\right )}^{3} \arctan \left (a x\right )^{3}} \,d x } \]
Timed out. \[ \int \frac {1}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^3} \, dx=\int \frac {1}{{\mathrm {atan}\left (a\,x\right )}^3\,{\left (c\,a^2\,x^2+c\right )}^3} \,d x \]