Integrand size = 16, antiderivative size = 933 \[ \int \frac {\arctan (a+b x)}{c+\frac {d}{x^3}} \, dx =\text {Too large to display} \] Output:
-1/2*(1+I*a+I*b*x)*ln(1+I*a+I*b*x)/b/c-1/2*(1-I*a-I*b*x)*ln(-I*(I+a+b*x))/ b/c-1/6*I*d^(1/3)*ln(1-I*a-I*b*x)*ln(-b*(d^(1/3)+c^(1/3)*x)/((I+a)*c^(1/3) -b*d^(1/3)))/c^(4/3)+1/6*I*d^(1/3)*ln(1+I*a+I*b*x)*ln(b*(d^(1/3)+c^(1/3)*x )/((I-a)*c^(1/3)+b*d^(1/3)))/c^(4/3)-1/6*(-1)^(1/6)*d^(1/3)*ln(1+I*a+I*b*x )*ln(-b*(d^(1/3)-(-1)^(1/3)*c^(1/3)*x)/((-1)^(1/3)*(I-a)*c^(1/3)-b*d^(1/3) ))/c^(4/3)+1/6*(-1)^(1/6)*d^(1/3)*ln(1-I*a-I*b*x)*ln(b*(d^(1/3)-(-1)^(1/3) *c^(1/3)*x)/((-1)^(1/3)*(I+a)*c^(1/3)+b*d^(1/3)))/c^(4/3)-1/6*(-1)^(5/6)*d ^(1/3)*ln(1+I*a+I*b*x)*ln(b*(d^(1/3)+(-1)^(2/3)*c^(1/3)*x)/((-1)^(2/3)*(I- a)*c^(1/3)+b*d^(1/3)))/c^(4/3)+1/6*(-1)^(5/6)*d^(1/3)*ln(1-I*a-I*b*x)*ln(b *(d^(1/3)+(-1)^(2/3)*c^(1/3)*x)/((-1)^(1/6)*(1-I*a)*c^(1/3)+b*d^(1/3)))/c^ (4/3)-1/6*(-1)^(1/6)*d^(1/3)*polylog(2,(-1)^(1/3)*c^(1/3)*(I-a-b*x)/((-1)^ (1/3)*(I-a)*c^(1/3)-b*d^(1/3)))/c^(4/3)-1/6*(-1)^(5/6)*d^(1/3)*polylog(2,( -1)^(1/6)*c^(1/3)*(I-a-b*x)/((-1)^(1/6)*(I-a)*c^(1/3)-I*b*d^(1/3)))/c^(4/3 )+1/6*I*d^(1/3)*polylog(2,c^(1/3)*(I-a-b*x)/((I-a)*c^(1/3)+b*d^(1/3)))/c^( 4/3)-1/6*I*d^(1/3)*polylog(2,c^(1/3)*(I+a+b*x)/((I+a)*c^(1/3)-b*d^(1/3)))/ c^(4/3)+1/6*(-1)^(5/6)*d^(1/3)*polylog(2,(-1)^(2/3)*c^(1/3)*(I+a+b*x)/((-1 )^(2/3)*(I+a)*c^(1/3)-b*d^(1/3)))/c^(4/3)+1/6*(-1)^(1/6)*d^(1/3)*polylog(2 ,(-1)^(1/3)*c^(1/3)*(I+a+b*x)/((-1)^(1/3)*(I+a)*c^(1/3)+b*d^(1/3)))/c^(4/3 )
Time = 0.87 (sec) , antiderivative size = 896, normalized size of antiderivative = 0.96 \[ \int \frac {\arctan (a+b x)}{c+\frac {d}{x^3}} \, dx =\text {Too large to display} \] Input:
Integrate[ArcTan[a + b*x]/(c + d/x^3),x]
Output:
((I/6)*((3*I)*c^(1/3)*Log[1 + I*a + I*b*x] - 3*a*c^(1/3)*Log[1 + I*a + I*b *x] - 3*b*c^(1/3)*x*Log[1 + I*a + I*b*x] + (3*I)*c^(1/3)*Log[(-I)*(I + a + b*x)] + 3*a*c^(1/3)*Log[(-I)*(I + a + b*x)] + 3*b*c^(1/3)*x*Log[(-I)*(I + a + b*x)] + b*d^(1/3)*Log[1 + I*a + I*b*x]*Log[(b*(d^(1/3) + c^(1/3)*x))/ (-((-I + a)*c^(1/3)) + b*d^(1/3))] - b*d^(1/3)*Log[(-I)*(I + a + b*x)]*Log [(b*(d^(1/3) + c^(1/3)*x))/(-((I + a)*c^(1/3)) + b*d^(1/3))] + (-1)^(2/3)* b*d^(1/3)*Log[1 + I*a + I*b*x]*Log[(b*(d^(1/3) - (-1)^(1/3)*c^(1/3)*x))/(( -1)^(1/3)*(-I + a)*c^(1/3) + b*d^(1/3))] - (-1)^(2/3)*b*d^(1/3)*Log[(-I)*( I + a + b*x)]*Log[(b*(d^(1/3) - (-1)^(1/3)*c^(1/3)*x))/((-1)^(1/3)*(I + a) *c^(1/3) + b*d^(1/3))] + (-1)^(1/3)*b*d^(1/3)*Log[(-I)*(I + a + b*x)]*Log[ (b*(d^(1/3) + (-1)^(2/3)*c^(1/3)*x))/((-1)^(1/6)*(1 - I*a)*c^(1/3) + b*d^( 1/3))] - (-1)^(1/3)*b*d^(1/3)*Log[1 + I*a + I*b*x]*Log[(b*(d^(1/3) + (-1)^ (2/3)*c^(1/3)*x))/(-((-1)^(2/3)*(-I + a)*c^(1/3)) + b*d^(1/3))] + b*d^(1/3 )*PolyLog[2, (c^(1/3)*(-I + a + b*x))/((-I + a)*c^(1/3) - b*d^(1/3))] - (- 1)^(1/3)*b*d^(1/3)*PolyLog[2, ((-1)^(1/6)*c^(1/3)*(-I + a + b*x))/((-1)^(1 /6)*(-I + a)*c^(1/3) + I*b*d^(1/3))] + (-1)^(2/3)*b*d^(1/3)*PolyLog[2, ((- 1)^(1/3)*c^(1/3)*(-I + a + b*x))/((-1)^(1/3)*(-I + a)*c^(1/3) + b*d^(1/3)) ] - b*d^(1/3)*PolyLog[2, (c^(1/3)*(I + a + b*x))/((I + a)*c^(1/3) - b*d^(1 /3))] + (-1)^(1/3)*b*d^(1/3)*PolyLog[2, ((-1)^(2/3)*c^(1/3)*(I + a + b*x)) /((-1)^(2/3)*(I + a)*c^(1/3) - b*d^(1/3))] - (-1)^(2/3)*b*d^(1/3)*PolyL...
Time = 1.79 (sec) , antiderivative size = 951, normalized size of antiderivative = 1.02, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {5574, 2856, 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.
\(\displaystyle \int \frac {\arctan (a+b x)}{c+\frac {d}{x^3}} \, dx\) |
\(\Big \downarrow \) 5574 |
\(\displaystyle \frac {1}{2} i \int \frac {\log (-i a-i b x+1)}{c+\frac {d}{x^3}}dx-\frac {1}{2} i \int \frac {\log (i a+i b x+1)}{c+\frac {d}{x^3}}dx\) |
\(\Big \downarrow \) 2856 |
\(\displaystyle \frac {1}{2} i \int \left (\frac {\log (-i a-i b x+1)}{c}-\frac {d \log (-i a-i b x+1)}{c \left (c x^3+d\right )}\right )dx-\frac {1}{2} i \int \left (\frac {\log (i a+i b x+1)}{c}-\frac {d \log (i a+i b x+1)}{c \left (c x^3+d\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} i \left (-\frac {x}{c}+\frac {i (-i a-i b x+1) \log (-i (a+b x+i))}{b c}-\frac {\sqrt [3]{d} \log (-i a-i b x+1) \log \left (-\frac {b \left (\sqrt [3]{c} x+\sqrt [3]{d}\right )}{(a+i) \sqrt [3]{c}-b \sqrt [3]{d}}\right )}{3 c^{4/3}}-\frac {(-1)^{2/3} \sqrt [3]{d} \log (-i a-i b x+1) \log \left (\frac {b \left (\sqrt [3]{d}-\sqrt [3]{-1} \sqrt [3]{c} x\right )}{\sqrt [3]{-1} \sqrt [3]{c} (a+i)+b \sqrt [3]{d}}\right )}{3 c^{4/3}}+\frac {\sqrt [3]{-1} \sqrt [3]{d} \log (-i a-i b x+1) \log \left (\frac {b \left ((-1)^{2/3} \sqrt [3]{c} x+\sqrt [3]{d}\right )}{\sqrt [6]{-1} \sqrt [3]{c} (1-i a)+b \sqrt [3]{d}}\right )}{3 c^{4/3}}-\frac {\sqrt [3]{d} \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{c} (a+b x+i)}{(a+i) \sqrt [3]{c}-b \sqrt [3]{d}}\right )}{3 c^{4/3}}+\frac {\sqrt [3]{-1} \sqrt [3]{d} \operatorname {PolyLog}\left (2,\frac {(-1)^{2/3} \sqrt [3]{c} (a+b x+i)}{(-1)^{2/3} (a+i) \sqrt [3]{c}-b \sqrt [3]{d}}\right )}{3 c^{4/3}}-\frac {(-1)^{2/3} \sqrt [3]{d} \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{-1} \sqrt [3]{c} (a+b x+i)}{\sqrt [3]{-1} \sqrt [3]{c} (a+i)+b \sqrt [3]{d}}\right )}{3 c^{4/3}}\right )-\frac {1}{2} i \left (-\frac {x}{c}-\frac {i (i a+i b x+1) \log (i a+i b x+1)}{b c}-\frac {\sqrt [3]{d} \log (i a+i b x+1) \log \left (\frac {b \left (\sqrt [3]{c} x+\sqrt [3]{d}\right )}{\sqrt [3]{c} (i-a)+b \sqrt [3]{d}}\right )}{3 c^{4/3}}-\frac {(-1)^{2/3} \sqrt [3]{d} \log (i a+i b x+1) \log \left (-\frac {b \left (\sqrt [3]{d}-\sqrt [3]{-1} \sqrt [3]{c} x\right )}{\sqrt [3]{-1} (i-a) \sqrt [3]{c}-b \sqrt [3]{d}}\right )}{3 c^{4/3}}+\frac {\sqrt [3]{-1} \sqrt [3]{d} \log (i a+i b x+1) \log \left (\frac {b \left ((-1)^{2/3} \sqrt [3]{c} x+\sqrt [3]{d}\right )}{(-1)^{2/3} \sqrt [3]{c} (i-a)+b \sqrt [3]{d}}\right )}{3 c^{4/3}}-\frac {(-1)^{2/3} \sqrt [3]{d} \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{-1} \sqrt [3]{c} (-a-b x+i)}{\sqrt [3]{-1} (i-a) \sqrt [3]{c}-b \sqrt [3]{d}}\right )}{3 c^{4/3}}+\frac {\sqrt [3]{-1} \sqrt [3]{d} \operatorname {PolyLog}\left (2,\frac {\sqrt [6]{-1} \sqrt [3]{c} (-a-b x+i)}{\sqrt [6]{-1} (i-a) \sqrt [3]{c}-i b \sqrt [3]{d}}\right )}{3 c^{4/3}}-\frac {\sqrt [3]{d} \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{c} (-a-b x+i)}{\sqrt [3]{c} (i-a)+b \sqrt [3]{d}}\right )}{3 c^{4/3}}\right )\) |
Input:
Int[ArcTan[a + b*x]/(c + d/x^3),x]
Output:
(-1/2*I)*(-(x/c) - (I*(1 + I*a + I*b*x)*Log[1 + I*a + I*b*x])/(b*c) - (d^( 1/3)*Log[1 + I*a + I*b*x]*Log[(b*(d^(1/3) + c^(1/3)*x))/((I - a)*c^(1/3) + b*d^(1/3))])/(3*c^(4/3)) - ((-1)^(2/3)*d^(1/3)*Log[1 + I*a + I*b*x]*Log[- ((b*(d^(1/3) - (-1)^(1/3)*c^(1/3)*x))/((-1)^(1/3)*(I - a)*c^(1/3) - b*d^(1 /3)))])/(3*c^(4/3)) + ((-1)^(1/3)*d^(1/3)*Log[1 + I*a + I*b*x]*Log[(b*(d^( 1/3) + (-1)^(2/3)*c^(1/3)*x))/((-1)^(2/3)*(I - a)*c^(1/3) + b*d^(1/3))])/( 3*c^(4/3)) - ((-1)^(2/3)*d^(1/3)*PolyLog[2, ((-1)^(1/3)*c^(1/3)*(I - a - b *x))/((-1)^(1/3)*(I - a)*c^(1/3) - b*d^(1/3))])/(3*c^(4/3)) + ((-1)^(1/3)* d^(1/3)*PolyLog[2, ((-1)^(1/6)*c^(1/3)*(I - a - b*x))/((-1)^(1/6)*(I - a)* c^(1/3) - I*b*d^(1/3))])/(3*c^(4/3)) - (d^(1/3)*PolyLog[2, (c^(1/3)*(I - a - b*x))/((I - a)*c^(1/3) + b*d^(1/3))])/(3*c^(4/3))) + (I/2)*(-(x/c) + (I *(1 - I*a - I*b*x)*Log[(-I)*(I + a + b*x)])/(b*c) - (d^(1/3)*Log[1 - I*a - I*b*x]*Log[-((b*(d^(1/3) + c^(1/3)*x))/((I + a)*c^(1/3) - b*d^(1/3)))])/( 3*c^(4/3)) - ((-1)^(2/3)*d^(1/3)*Log[1 - I*a - I*b*x]*Log[(b*(d^(1/3) - (- 1)^(1/3)*c^(1/3)*x))/((-1)^(1/3)*(I + a)*c^(1/3) + b*d^(1/3))])/(3*c^(4/3) ) + ((-1)^(1/3)*d^(1/3)*Log[1 - I*a - I*b*x]*Log[(b*(d^(1/3) + (-1)^(2/3)* c^(1/3)*x))/((-1)^(1/6)*(1 - I*a)*c^(1/3) + b*d^(1/3))])/(3*c^(4/3)) - (d^ (1/3)*PolyLog[2, (c^(1/3)*(I + a + b*x))/((I + a)*c^(1/3) - b*d^(1/3))])/( 3*c^(4/3)) + ((-1)^(1/3)*d^(1/3)*PolyLog[2, ((-1)^(2/3)*c^(1/3)*(I + a + b *x))/((-1)^(2/3)*(I + a)*c^(1/3) - b*d^(1/3))])/(3*c^(4/3)) - ((-1)^(2/...
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_. )*(x_)^(r_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x) ^n])^p, (f + g*x^r)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, r}, x] && I GtQ[p, 0] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[r] && NeQ[r, 1]))
Int[ArcTan[(a_) + (b_.)*(x_)]/((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Simp[ I/2 Int[Log[1 - I*a - I*b*x]/(c + d*x^n), x], x] - Simp[I/2 Int[Log[1 + I*a + I*b*x]/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d}, x] && RationalQ[n]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.09 (sec) , antiderivative size = 511, normalized size of antiderivative = 0.55
method | result | size |
risch | \(\frac {i \ln \left (-b x i-a i+1\right ) x}{2 c}+\frac {i \ln \left (-b x i-a i+1\right ) a}{2 b c}-\frac {\ln \left (-b x i-a i+1\right )}{2 b c}+\frac {1}{b c}+\frac {i b^{2} d \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{3}+\left (3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1, \operatorname {index} =1\right ) a c -3 c \right ) \textit {\_Z}^{2}+\left (-6 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1, \operatorname {index} =1\right ) a c -3 a^{2} c +3 c \right ) \textit {\_Z} -\operatorname {RootOf}\left (\textit {\_Z}^{2}+1, \operatorname {index} =1\right ) a^{3} c +\operatorname {RootOf}\left (\textit {\_Z}^{2}+1, \operatorname {index} =1\right ) b^{3} d +3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1, \operatorname {index} =1\right ) a c +3 a^{2} c -c \right )}{\sum }\frac {\ln \left (-b x i-a i+1\right ) \ln \left (\frac {b x i+a i+\textit {\_R1} -1}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {b x i+a i+\textit {\_R1} -1}{\textit {\_R1}}\right )}{2 \textit {\_R1} a i+\textit {\_R1}^{2}-a^{2}-2 a i-2 \textit {\_R1} +1}\right )}{6 c^{2}}-\frac {i \ln \left (b x i+a i+1\right ) x}{2 c}-\frac {i \ln \left (b x i+a i+1\right ) a}{2 b c}-\frac {\ln \left (b x i+a i+1\right )}{2 b c}-\frac {i b^{2} d \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{3}+\left (-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1, \operatorname {index} =1\right ) a c -3 c \right ) \textit {\_Z}^{2}+\left (6 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1, \operatorname {index} =1\right ) a c -3 a^{2} c +3 c \right ) \textit {\_Z} +\operatorname {RootOf}\left (\textit {\_Z}^{2}+1, \operatorname {index} =1\right ) a^{3} c -\operatorname {RootOf}\left (\textit {\_Z}^{2}+1, \operatorname {index} =1\right ) b^{3} d -3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1, \operatorname {index} =1\right ) a c +3 a^{2} c -c \right )}{\sum }\frac {\ln \left (b x i+a i+1\right ) \ln \left (\frac {-b x i-a i+\textit {\_R1} -1}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-b x i-a i+\textit {\_R1} -1}{\textit {\_R1}}\right )}{-2 \textit {\_R1} a i+\textit {\_R1}^{2}-a^{2}+2 a i-2 \textit {\_R1} +1}\right )}{6 c^{2}}\) | \(511\) |
derivativedivides | \(\frac {\frac {\arctan \left (b x +a \right ) \left (b x +a \right )}{c}+\frac {\arctan \left (b x +a \right ) \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{3}-3 a c \,\textit {\_Z}^{2}+3 a^{2} c \textit {\_Z} -a^{3} c +b^{3} d \right )}{\sum }\frac {\ln \left (b x -\textit {\_R} +a \right )}{-\textit {\_R}^{2}+2 \textit {\_R} a -a^{2}}\right ) d \,b^{3}}{3 c^{2}}-\frac {d \,b^{3} \left (\arctan \left (b x +a \right ) \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{3}-3 a c \,\textit {\_Z}^{2}+3 a^{2} c \textit {\_Z} -a^{3} c +b^{3} d \right )}{\sum }\frac {\ln \left (b x -\textit {\_R} +a \right )}{-\textit {\_R}^{2}+2 \textit {\_R} a -a^{2}}\right )-3 c \left (-\frac {2 \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\left (a^{3} c +3 i a^{2} c -b^{3} d -3 a c -i c \right ) \textit {\_Z}^{6}+\left (3 a^{3} c +3 i a^{2} c -3 b^{3} d +3 a c +3 i c \right ) \textit {\_Z}^{4}+\left (3 a^{3} c -3 i a^{2} c -3 b^{3} d +3 a c -3 i c \right ) \textit {\_Z}^{2}-3 i a^{2} c +a^{3} c -b^{3} d +i c -3 a c \right )}{\sum }\frac {i \arctan \left (b x +a \right ) \ln \left (\frac {\textit {\_R1} -\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {\textit {\_R1} -\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}}{\textit {\_R1}}\right )}{a^{3} c \,\textit {\_R1}^{4}+3 i a^{2} c \,\textit {\_R1}^{4}-b^{3} d \,\textit {\_R1}^{4}-3 a c \,\textit {\_R1}^{4}-i c \,\textit {\_R1}^{4}+2 a^{3} c \,\textit {\_R1}^{2}+2 i a^{2} c \,\textit {\_R1}^{2}-2 b^{3} d \,\textit {\_R1}^{2}+2 a c \,\textit {\_R1}^{2}+2 i c \,\textit {\_R1}^{2}+a^{3} c -i a^{2} c -b^{3} d +a c -i c}\right )}{3}-\frac {2 \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\left (a^{3} c +3 i a^{2} c -b^{3} d -3 a c -i c \right ) \textit {\_Z}^{6}+\left (3 a^{3} c +3 i a^{2} c -3 b^{3} d +3 a c +3 i c \right ) \textit {\_Z}^{4}+\left (3 a^{3} c -3 i a^{2} c -3 b^{3} d +3 a c -3 i c \right ) \textit {\_Z}^{2}-3 i a^{2} c +a^{3} c -b^{3} d +i c -3 a c \right )}{\sum }\frac {\textit {\_R1}^{2} \left (i \arctan \left (b x +a \right ) \ln \left (\frac {\textit {\_R1} -\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {\textit {\_R1} -\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}}{\textit {\_R1}}\right )\right )}{a^{3} c \,\textit {\_R1}^{4}+3 i a^{2} c \,\textit {\_R1}^{4}-b^{3} d \,\textit {\_R1}^{4}-3 a c \,\textit {\_R1}^{4}-i c \,\textit {\_R1}^{4}+2 a^{3} c \,\textit {\_R1}^{2}+2 i a^{2} c \,\textit {\_R1}^{2}-2 b^{3} d \,\textit {\_R1}^{2}+2 a c \,\textit {\_R1}^{2}+2 i c \,\textit {\_R1}^{2}+a^{3} c -i a^{2} c -b^{3} d +a c -i c}\right )}{3}\right )\right )}{3 c^{2}}-\frac {\ln \left (1+\left (b x +a \right )^{2}\right )}{2 c}}{b}\) | \(819\) |
default | \(\frac {\frac {\arctan \left (b x +a \right ) \left (b x +a \right )}{c}+\frac {\arctan \left (b x +a \right ) \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{3}-3 a c \,\textit {\_Z}^{2}+3 a^{2} c \textit {\_Z} -a^{3} c +b^{3} d \right )}{\sum }\frac {\ln \left (b x -\textit {\_R} +a \right )}{-\textit {\_R}^{2}+2 \textit {\_R} a -a^{2}}\right ) d \,b^{3}}{3 c^{2}}-\frac {d \,b^{3} \left (\arctan \left (b x +a \right ) \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{3}-3 a c \,\textit {\_Z}^{2}+3 a^{2} c \textit {\_Z} -a^{3} c +b^{3} d \right )}{\sum }\frac {\ln \left (b x -\textit {\_R} +a \right )}{-\textit {\_R}^{2}+2 \textit {\_R} a -a^{2}}\right )-3 c \left (-\frac {2 \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\left (a^{3} c +3 i a^{2} c -b^{3} d -3 a c -i c \right ) \textit {\_Z}^{6}+\left (3 a^{3} c +3 i a^{2} c -3 b^{3} d +3 a c +3 i c \right ) \textit {\_Z}^{4}+\left (3 a^{3} c -3 i a^{2} c -3 b^{3} d +3 a c -3 i c \right ) \textit {\_Z}^{2}-3 i a^{2} c +a^{3} c -b^{3} d +i c -3 a c \right )}{\sum }\frac {i \arctan \left (b x +a \right ) \ln \left (\frac {\textit {\_R1} -\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {\textit {\_R1} -\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}}{\textit {\_R1}}\right )}{a^{3} c \,\textit {\_R1}^{4}+3 i a^{2} c \,\textit {\_R1}^{4}-b^{3} d \,\textit {\_R1}^{4}-3 a c \,\textit {\_R1}^{4}-i c \,\textit {\_R1}^{4}+2 a^{3} c \,\textit {\_R1}^{2}+2 i a^{2} c \,\textit {\_R1}^{2}-2 b^{3} d \,\textit {\_R1}^{2}+2 a c \,\textit {\_R1}^{2}+2 i c \,\textit {\_R1}^{2}+a^{3} c -i a^{2} c -b^{3} d +a c -i c}\right )}{3}-\frac {2 \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\left (a^{3} c +3 i a^{2} c -b^{3} d -3 a c -i c \right ) \textit {\_Z}^{6}+\left (3 a^{3} c +3 i a^{2} c -3 b^{3} d +3 a c +3 i c \right ) \textit {\_Z}^{4}+\left (3 a^{3} c -3 i a^{2} c -3 b^{3} d +3 a c -3 i c \right ) \textit {\_Z}^{2}-3 i a^{2} c +a^{3} c -b^{3} d +i c -3 a c \right )}{\sum }\frac {\textit {\_R1}^{2} \left (i \arctan \left (b x +a \right ) \ln \left (\frac {\textit {\_R1} -\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {\textit {\_R1} -\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}}{\textit {\_R1}}\right )\right )}{a^{3} c \,\textit {\_R1}^{4}+3 i a^{2} c \,\textit {\_R1}^{4}-b^{3} d \,\textit {\_R1}^{4}-3 a c \,\textit {\_R1}^{4}-i c \,\textit {\_R1}^{4}+2 a^{3} c \,\textit {\_R1}^{2}+2 i a^{2} c \,\textit {\_R1}^{2}-2 b^{3} d \,\textit {\_R1}^{2}+2 a c \,\textit {\_R1}^{2}+2 i c \,\textit {\_R1}^{2}+a^{3} c -i a^{2} c -b^{3} d +a c -i c}\right )}{3}\right )\right )}{3 c^{2}}-\frac {\ln \left (1+\left (b x +a \right )^{2}\right )}{2 c}}{b}\) | \(819\) |
Input:
int(arctan(b*x+a)/(c+d/x^3),x,method=_RETURNVERBOSE)
Output:
1/2*I/c*ln(1-I*a-I*b*x)*x+1/2*I/b/c*ln(1-I*a-I*b*x)*a-1/2/b/c*ln(1-I*a-I*b *x)+1/b/c+1/6*I*b^2*d/c^2*sum(1/(1+2*I*a*_R1-2*I*a+_R1^2-a^2-2*_R1)*(ln(1- I*a-I*b*x)*ln((_R1+I*b*x+I*a-1)/_R1)+dilog((_R1+I*b*x+I*a-1)/_R1)),_R1=Roo tOf(c*_Z^3+(3*RootOf(_Z^2+1,index=1)*a*c-3*c)*_Z^2+(-6*RootOf(_Z^2+1,index =1)*a*c-3*a^2*c+3*c)*_Z-RootOf(_Z^2+1,index=1)*a^3*c+RootOf(_Z^2+1,index=1 )*b^3*d+3*RootOf(_Z^2+1,index=1)*a*c+3*a^2*c-c))-1/2*I/c*ln(1+I*a+I*b*x)*x -1/2*I/b/c*ln(1+I*a+I*b*x)*a-1/2/b/c*ln(1+I*a+I*b*x)-1/6*I*b^2*d/c^2*sum(1 /(1-2*I*a*_R1+2*I*a+_R1^2-a^2-2*_R1)*(ln(1+I*a+I*b*x)*ln((_R1-I*b*x-I*a-1) /_R1)+dilog((_R1-I*b*x-I*a-1)/_R1)),_R1=RootOf(c*_Z^3+(-3*RootOf(_Z^2+1,in dex=1)*a*c-3*c)*_Z^2+(6*RootOf(_Z^2+1,index=1)*a*c-3*a^2*c+3*c)*_Z+RootOf( _Z^2+1,index=1)*a^3*c-RootOf(_Z^2+1,index=1)*b^3*d-3*RootOf(_Z^2+1,index=1 )*a*c+3*a^2*c-c))
\[ \int \frac {\arctan (a+b x)}{c+\frac {d}{x^3}} \, dx=\int { \frac {\arctan \left (b x + a\right )}{c + \frac {d}{x^{3}}} \,d x } \] Input:
integrate(arctan(b*x+a)/(c+d/x^3),x, algorithm="fricas")
Output:
integral(x^3*arctan(b*x + a)/(c*x^3 + d), x)
Timed out. \[ \int \frac {\arctan (a+b x)}{c+\frac {d}{x^3}} \, dx=\text {Timed out} \] Input:
integrate(atan(b*x+a)/(c+d/x**3),x)
Output:
Timed out
\[ \int \frac {\arctan (a+b x)}{c+\frac {d}{x^3}} \, dx=\int { \frac {\arctan \left (b x + a\right )}{c + \frac {d}{x^{3}}} \,d x } \] Input:
integrate(arctan(b*x+a)/(c+d/x^3),x, algorithm="maxima")
Output:
integrate(arctan(b*x + a)/(c + d/x^3), x)
\[ \int \frac {\arctan (a+b x)}{c+\frac {d}{x^3}} \, dx=\int { \frac {\arctan \left (b x + a\right )}{c + \frac {d}{x^{3}}} \,d x } \] Input:
integrate(arctan(b*x+a)/(c+d/x^3),x, algorithm="giac")
Output:
integrate(arctan(b*x + a)/(c + d/x^3), x)
Timed out. \[ \int \frac {\arctan (a+b x)}{c+\frac {d}{x^3}} \, dx=\int \frac {\mathrm {atan}\left (a+b\,x\right )}{c+\frac {d}{x^3}} \,d x \] Input:
int(atan(a + b*x)/(c + d/x^3),x)
Output:
int(atan(a + b*x)/(c + d/x^3), x)
\[ \int \frac {\arctan (a+b x)}{c+\frac {d}{x^3}} \, dx=\int \frac {\mathit {atan} \left (b x +a \right ) x^{3}}{c \,x^{3}+d}d x \] Input:
int(atan(b*x+a)/(c+d/x^3),x)
Output:
int((atan(a + b*x)*x**3)/(c*x**3 + d),x)