3.1.59 \(\int \frac {\arctan (a+b x)}{c+\frac {d}{\sqrt {x}}} \, dx\) [59]

3.1.59.1 Optimal result
3.1.59.2 Mathematica [A] (verified)
3.1.59.3 Rubi [A] (verified)
3.1.59.4 Maple [C] (warning: unable to verify)
3.1.59.5 Fricas [F]
3.1.59.6 Sympy [F(-1)]
3.1.59.7 Maxima [F]
3.1.59.8 Giac [F(-2)]
3.1.59.9 Mupad [F(-1)]

3.1.59.1 Optimal result

Integrand size = 18, antiderivative size = 770 \[ \int \frac {\arctan (a+b x)}{c+\frac {d}{\sqrt {x}}} \, dx=-\frac {2 i \sqrt {i+a} d \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {i+a}}\right )}{\sqrt {b} c^2}+\frac {2 i \sqrt {i-a} d \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {i-a}}\right )}{\sqrt {b} c^2}-\frac {i d^2 \log \left (\frac {c \left (\sqrt {-i-a}-\sqrt {b} \sqrt {x}\right )}{\sqrt {-i-a} c+\sqrt {b} d}\right ) \log \left (d+c \sqrt {x}\right )}{c^3}+\frac {i d^2 \log \left (\frac {c \left (\sqrt {i-a}-\sqrt {b} \sqrt {x}\right )}{\sqrt {i-a} c+\sqrt {b} d}\right ) \log \left (d+c \sqrt {x}\right )}{c^3}-\frac {i d^2 \log \left (\frac {c \left (\sqrt {-i-a}+\sqrt {b} \sqrt {x}\right )}{\sqrt {-i-a} c-\sqrt {b} d}\right ) \log \left (d+c \sqrt {x}\right )}{c^3}+\frac {i d^2 \log \left (\frac {c \left (\sqrt {i-a}+\sqrt {b} \sqrt {x}\right )}{\sqrt {i-a} c-\sqrt {b} d}\right ) \log \left (d+c \sqrt {x}\right )}{c^3}-\frac {i d \sqrt {x} \log (1-i a-i b x)}{c^2}+\frac {i d^2 \log \left (d+c \sqrt {x}\right ) \log (1-i a-i b x)}{c^3}+\frac {i d \sqrt {x} \log (1+i a+i b x)}{c^2}-\frac {(1+i a+i b x) \log (1+i a+i b x)}{2 b c}-\frac {i d^2 \log \left (d+c \sqrt {x}\right ) \log (1+i a+i b x)}{c^3}-\frac {(1-i a-i b x) \log (-i (i+a+b x))}{2 b c}-\frac {i d^2 \operatorname {PolyLog}\left (2,-\frac {\sqrt {b} \left (d+c \sqrt {x}\right )}{\sqrt {-i-a} c-\sqrt {b} d}\right )}{c^3}+\frac {i d^2 \operatorname {PolyLog}\left (2,-\frac {\sqrt {b} \left (d+c \sqrt {x}\right )}{\sqrt {i-a} c-\sqrt {b} d}\right )}{c^3}-\frac {i d^2 \operatorname {PolyLog}\left (2,\frac {\sqrt {b} \left (d+c \sqrt {x}\right )}{\sqrt {-i-a} c+\sqrt {b} d}\right )}{c^3}+\frac {i d^2 \operatorname {PolyLog}\left (2,\frac {\sqrt {b} \left (d+c \sqrt {x}\right )}{\sqrt {i-a} c+\sqrt {b} d}\right )}{c^3} \]

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+I*d^2*ln(d+c*x^(1/2))*ln(c*((I-a)^(1/2)+b^(1/2)*x^(1/2))/(c*(I-a)^(1/2 
)-d*b^(1/2)))/c^3-I*d^2*polylog(2,b^(1/2)*(d+c*x^(1/2))/(c*(-I-a)^(1/2)+d* 
b^(1/2)))/c^3-I*d^2*ln(d+c*x^(1/2))*ln(c*((-I-a)^(1/2)+b^(1/2)*x^(1/2))/(c 
*(-I-a)^(1/2)-d*b^(1/2)))/c^3+I*d^2*polylog(2,b^(1/2)*(d+c*x^(1/2))/(c*(I- 
a)^(1/2)+d*b^(1/2)))/c^3-I*d^2*ln(d+c*x^(1/2))*ln(c*((-I-a)^(1/2)-b^(1/2)* 
x^(1/2))/(c*(-I-a)^(1/2)+d*b^(1/2)))/c^3+I*d^2*ln(1-I*a-I*b*x)*ln(d+c*x^(1 
/2))/c^3+I*d*ln(1+I*a+I*b*x)*x^(1/2)/c^2+I*d^2*ln(d+c*x^(1/2))*ln(c*((I-a) 
^(1/2)-b^(1/2)*x^(1/2))/(c*(I-a)^(1/2)+d*b^(1/2)))/c^3-2*I*d*arctan(b^(1/2 
)*x^(1/2)/(I+a)^(1/2))*(I+a)^(1/2)/c^2/b^(1/2)-I*d^2*polylog(2,-b^(1/2)*(d 
+c*x^(1/2))/(c*(-I-a)^(1/2)-d*b^(1/2)))/c^3-I*d*ln(1-I*a-I*b*x)*x^(1/2)/c^ 
2-I*d^2*ln(1+I*a+I*b*x)*ln(d+c*x^(1/2))/c^3+I*d^2*polylog(2,-b^(1/2)*(d+c* 
x^(1/2))/(c*(I-a)^(1/2)-d*b^(1/2)))/c^3+2*I*d*arctanh(b^(1/2)*x^(1/2)/(I-a 
)^(1/2))*(I-a)^(1/2)/c^2/b^(1/2)
 
3.1.59.2 Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 770, normalized size of antiderivative = 1.00 \[ \int \frac {\arctan (a+b x)}{c+\frac {d}{\sqrt {x}}} \, dx=-\frac {i \left (4 \sqrt {i+a} \sqrt {b} c d \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {i+a}}\right )-4 \sqrt {i-a} \sqrt {b} c d \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {i-a}}\right )+2 b d^2 \log \left (\frac {c \left (\sqrt {-i-a}-\sqrt {b} \sqrt {x}\right )}{\sqrt {-i-a} c+\sqrt {b} d}\right ) \log \left (d+c \sqrt {x}\right )-2 b d^2 \log \left (\frac {c \left (\sqrt {i-a}-\sqrt {b} \sqrt {x}\right )}{\sqrt {i-a} c+\sqrt {b} d}\right ) \log \left (d+c \sqrt {x}\right )+2 b d^2 \log \left (\frac {c \left (\sqrt {-i-a}+\sqrt {b} \sqrt {x}\right )}{\sqrt {-i-a} c-\sqrt {b} d}\right ) \log \left (d+c \sqrt {x}\right )-2 b d^2 \log \left (\frac {c \left (\sqrt {i-a}+\sqrt {b} \sqrt {x}\right )}{\sqrt {i-a} c-\sqrt {b} d}\right ) \log \left (d+c \sqrt {x}\right )-i c^2 \log (1+i a+i b x)+a c^2 \log (1+i a+i b x)-2 b c d \sqrt {x} \log (1+i a+i b x)+b c^2 x \log (1+i a+i b x)+2 b d^2 \log \left (d+c \sqrt {x}\right ) \log (1+i a+i b x)-i c^2 \log (-i (i+a+b x))-a c^2 \log (-i (i+a+b x))+2 b c d \sqrt {x} \log (-i (i+a+b x))-b c^2 x \log (-i (i+a+b x))-2 b d^2 \log \left (d+c \sqrt {x}\right ) \log (-i (i+a+b x))+2 b d^2 \operatorname {PolyLog}\left (2,\frac {\sqrt {b} \left (d+c \sqrt {x}\right )}{-\sqrt {-i-a} c+\sqrt {b} d}\right )+2 b d^2 \operatorname {PolyLog}\left (2,\frac {\sqrt {b} \left (d+c \sqrt {x}\right )}{\sqrt {-i-a} c+\sqrt {b} d}\right )-2 b d^2 \operatorname {PolyLog}\left (2,\frac {\sqrt {b} \left (d+c \sqrt {x}\right )}{-\sqrt {i-a} c+\sqrt {b} d}\right )-2 b d^2 \operatorname {PolyLog}\left (2,\frac {\sqrt {b} \left (d+c \sqrt {x}\right )}{\sqrt {i-a} c+\sqrt {b} d}\right )\right )}{2 b c^3} \]

input
Integrate[ArcTan[a + b*x]/(c + d/Sqrt[x]),x]
 
output
((-1/2*I)*(4*Sqrt[I + a]*Sqrt[b]*c*d*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[I + a]] 
 - 4*Sqrt[I - a]*Sqrt[b]*c*d*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[I - a]] + 2*b* 
d^2*Log[(c*(Sqrt[-I - a] - Sqrt[b]*Sqrt[x]))/(Sqrt[-I - a]*c + Sqrt[b]*d)] 
*Log[d + c*Sqrt[x]] - 2*b*d^2*Log[(c*(Sqrt[I - a] - Sqrt[b]*Sqrt[x]))/(Sqr 
t[I - a]*c + Sqrt[b]*d)]*Log[d + c*Sqrt[x]] + 2*b*d^2*Log[(c*(Sqrt[-I - a] 
 + Sqrt[b]*Sqrt[x]))/(Sqrt[-I - a]*c - Sqrt[b]*d)]*Log[d + c*Sqrt[x]] - 2* 
b*d^2*Log[(c*(Sqrt[I - a] + Sqrt[b]*Sqrt[x]))/(Sqrt[I - a]*c - Sqrt[b]*d)] 
*Log[d + c*Sqrt[x]] - I*c^2*Log[1 + I*a + I*b*x] + a*c^2*Log[1 + I*a + I*b 
*x] - 2*b*c*d*Sqrt[x]*Log[1 + I*a + I*b*x] + b*c^2*x*Log[1 + I*a + I*b*x] 
+ 2*b*d^2*Log[d + c*Sqrt[x]]*Log[1 + I*a + I*b*x] - I*c^2*Log[(-I)*(I + a 
+ b*x)] - a*c^2*Log[(-I)*(I + a + b*x)] + 2*b*c*d*Sqrt[x]*Log[(-I)*(I + a 
+ b*x)] - b*c^2*x*Log[(-I)*(I + a + b*x)] - 2*b*d^2*Log[d + c*Sqrt[x]]*Log 
[(-I)*(I + a + b*x)] + 2*b*d^2*PolyLog[2, (Sqrt[b]*(d + c*Sqrt[x]))/(-(Sqr 
t[-I - a]*c) + Sqrt[b]*d)] + 2*b*d^2*PolyLog[2, (Sqrt[b]*(d + c*Sqrt[x]))/ 
(Sqrt[-I - a]*c + Sqrt[b]*d)] - 2*b*d^2*PolyLog[2, (Sqrt[b]*(d + c*Sqrt[x] 
))/(-(Sqrt[I - a]*c) + Sqrt[b]*d)] - 2*b*d^2*PolyLog[2, (Sqrt[b]*(d + c*Sq 
rt[x]))/(Sqrt[I - a]*c + Sqrt[b]*d)]))/(b*c^3)
 
3.1.59.3 Rubi [A] (verified)

Time = 1.27 (sec) , antiderivative size = 792, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {5574, 2855, 2005, 2916, 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.

\(\displaystyle \int \frac {\arctan (a+b x)}{c+\frac {d}{\sqrt {x}}} \, dx\)

\(\Big \downarrow \) 5574

\(\displaystyle \frac {1}{2} i \int \frac {\log (-i a-i b x+1)}{c+\frac {d}{\sqrt {x}}}dx-\frac {1}{2} i \int \frac {\log (i a+i b x+1)}{c+\frac {d}{\sqrt {x}}}dx\)

\(\Big \downarrow \) 2855

\(\displaystyle i \int \frac {\sqrt {x} \log (-i a-i b x+1)}{c+\frac {d}{\sqrt {x}}}d\sqrt {x}-i \int \frac {\sqrt {x} \log (i a+i b x+1)}{c+\frac {d}{\sqrt {x}}}d\sqrt {x}\)

\(\Big \downarrow \) 2005

\(\displaystyle i \int \frac {x \log (-i a-i b x+1)}{\sqrt {x} c+d}d\sqrt {x}-i \int \frac {x \log (i a+i b x+1)}{\sqrt {x} c+d}d\sqrt {x}\)

\(\Big \downarrow \) 2916

\(\displaystyle i \int \left (\frac {\log (-i a-i b x+1) d^2}{c^2 \left (\sqrt {x} c+d\right )}-\frac {\log (-i a-i b x+1) d}{c^2}+\frac {\sqrt {x} \log (-i a-i b x+1)}{c}\right )d\sqrt {x}-i \int \left (\frac {\log (i a+i b x+1) d^2}{c^2 \left (\sqrt {x} c+d\right )}-\frac {\log (i a+i b x+1) d}{c^2}+\frac {\sqrt {x} \log (i a+i b x+1)}{c}\right )d\sqrt {x}\)

\(\Big \downarrow \) 2009

\(\displaystyle i \left (-\frac {2 \sqrt {a+i} d \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+i}}\right )}{\sqrt {b} c^2}-\frac {d^2 \operatorname {PolyLog}\left (2,-\frac {\sqrt {b} \left (\sqrt {x} c+d\right )}{\sqrt {-a-i} c-\sqrt {b} d}\right )}{c^3}-\frac {d^2 \operatorname {PolyLog}\left (2,\frac {\sqrt {b} \left (\sqrt {x} c+d\right )}{\sqrt {-a-i} c+\sqrt {b} d}\right )}{c^3}-\frac {d^2 \log \left (c \sqrt {x}+d\right ) \log \left (\frac {c \left (-\sqrt {b} \sqrt {x}+\sqrt {-a-i}\right )}{\sqrt {b} d+\sqrt {-a-i} c}\right )}{c^3}-\frac {d^2 \log \left (c \sqrt {x}+d\right ) \log \left (\frac {c \left (\sqrt {b} \sqrt {x}+\sqrt {-a-i}\right )}{-\sqrt {b} d+\sqrt {-a-i} c}\right )}{c^3}+\frac {d^2 \log (-i a-i b x+1) \log \left (c \sqrt {x}+d\right )}{c^3}-\frac {d \sqrt {x} \log (-i a-i b x+1)}{c^2}+\frac {i (-i a-i b x+1) \log (-i (a+b x+i))}{2 b c}+\frac {2 d \sqrt {x}}{c^2}-\frac {x}{2 c}\right )-i \left (-\frac {2 \sqrt {-a+i} d \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {-a+i}}\right )}{\sqrt {b} c^2}-\frac {d^2 \operatorname {PolyLog}\left (2,-\frac {\sqrt {b} \left (\sqrt {x} c+d\right )}{\sqrt {i-a} c-\sqrt {b} d}\right )}{c^3}-\frac {d^2 \operatorname {PolyLog}\left (2,\frac {\sqrt {b} \left (\sqrt {x} c+d\right )}{\sqrt {i-a} c+\sqrt {b} d}\right )}{c^3}-\frac {d^2 \log \left (c \sqrt {x}+d\right ) \log \left (\frac {c \left (-\sqrt {b} \sqrt {x}+\sqrt {-a+i}\right )}{\sqrt {b} d+\sqrt {-a+i} c}\right )}{c^3}-\frac {d^2 \log \left (c \sqrt {x}+d\right ) \log \left (\frac {c \left (\sqrt {b} \sqrt {x}+\sqrt {-a+i}\right )}{-\sqrt {b} d+\sqrt {-a+i} c}\right )}{c^3}+\frac {d^2 \log (i a+i b x+1) \log \left (c \sqrt {x}+d\right )}{c^3}-\frac {d \sqrt {x} \log (i a+i b x+1)}{c^2}-\frac {i (i a+i b x+1) \log (i a+i b x+1)}{2 b c}+\frac {2 d \sqrt {x}}{c^2}-\frac {x}{2 c}\right )\)

input
Int[ArcTan[a + b*x]/(c + d/Sqrt[x]),x]
 
output
I*((2*d*Sqrt[x])/c^2 - x/(2*c) - (2*Sqrt[I + a]*d*ArcTan[(Sqrt[b]*Sqrt[x]) 
/Sqrt[I + a]])/(Sqrt[b]*c^2) - (d^2*Log[(c*(Sqrt[-I - a] - Sqrt[b]*Sqrt[x] 
))/(Sqrt[-I - a]*c + Sqrt[b]*d)]*Log[d + c*Sqrt[x]])/c^3 - (d^2*Log[(c*(Sq 
rt[-I - a] + Sqrt[b]*Sqrt[x]))/(Sqrt[-I - a]*c - Sqrt[b]*d)]*Log[d + c*Sqr 
t[x]])/c^3 - (d*Sqrt[x]*Log[1 - I*a - I*b*x])/c^2 + (d^2*Log[d + c*Sqrt[x] 
]*Log[1 - I*a - I*b*x])/c^3 + ((I/2)*(1 - I*a - I*b*x)*Log[(-I)*(I + a + b 
*x)])/(b*c) - (d^2*PolyLog[2, -((Sqrt[b]*(d + c*Sqrt[x]))/(Sqrt[-I - a]*c 
- Sqrt[b]*d))])/c^3 - (d^2*PolyLog[2, (Sqrt[b]*(d + c*Sqrt[x]))/(Sqrt[-I - 
 a]*c + Sqrt[b]*d)])/c^3) - I*((2*d*Sqrt[x])/c^2 - x/(2*c) - (2*Sqrt[I - a 
]*d*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[I - a]])/(Sqrt[b]*c^2) - (d^2*Log[(c*(S 
qrt[I - a] - Sqrt[b]*Sqrt[x]))/(Sqrt[I - a]*c + Sqrt[b]*d)]*Log[d + c*Sqrt 
[x]])/c^3 - (d^2*Log[(c*(Sqrt[I - a] + Sqrt[b]*Sqrt[x]))/(Sqrt[I - a]*c - 
Sqrt[b]*d)]*Log[d + c*Sqrt[x]])/c^3 - (d*Sqrt[x]*Log[1 + I*a + I*b*x])/c^2 
 - ((I/2)*(1 + I*a + I*b*x)*Log[1 + I*a + I*b*x])/(b*c) + (d^2*Log[d + c*S 
qrt[x]]*Log[1 + I*a + I*b*x])/c^3 - (d^2*PolyLog[2, -((Sqrt[b]*(d + c*Sqrt 
[x]))/(Sqrt[I - a]*c - Sqrt[b]*d))])/c^3 - (d^2*PolyLog[2, (Sqrt[b]*(d + c 
*Sqrt[x]))/(Sqrt[I - a]*c + Sqrt[b]*d)])/c^3)
 

3.1.59.3.1 Defintions of rubi rules used

rule 2005
Int[(Fx_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[x^(m 
+ n*p)*(b + a/x^n)^p*Fx, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && Neg 
Q[n]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2855
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_ 
.)*(x_)^(r_))^(q_.), x_Symbol] :> With[{k = Denominator[r]}, Simp[k   Subst 
[Int[x^(k - 1)*(f + g*x^(k*r))^q*(a + b*Log[c*(d + e*x^k)^n])^p, x], x, x^( 
1/k)], x]] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x] && FractionQ[r] && I 
GtQ[p, 0]
 

rule 2916
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m 
_.)*((f_.) + (g_.)*(x_))^(r_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log 
[c*(d + e*x^n)^p])^q, x^m*(f + g*x)^r, x], x] /; FreeQ[{a, b, c, d, e, f, g 
, n, p, q}, x] && IntegerQ[m] && IntegerQ[r]
 

rule 5574
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]
 
3.1.59.4 Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.21 (sec) , antiderivative size = 388, normalized size of antiderivative = 0.50

method result size
derivativedivides \(\frac {\arctan \left (b x +a \right ) x}{c}-\frac {2 \arctan \left (b x +a \right ) d \sqrt {x}}{c^{2}}+\frac {2 \arctan \left (b x +a \right ) d^{2} \ln \left (d +c \sqrt {x}\right )}{c^{3}}-\frac {4 b \left (-\frac {c \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b^{2} \textit {\_Z}^{4}-4 b^{2} d \,\textit {\_Z}^{3}+\left (2 a \,c^{2} b +6 b^{2} d^{2}\right ) \textit {\_Z}^{2}+\left (-4 a b \,c^{2} d -4 b^{2} d^{3}\right ) \textit {\_Z} +a^{2} c^{4}+2 a b \,c^{2} d^{2}+b^{2} d^{4}+c^{4}\right )}{\sum }\frac {\left (-\textit {\_R}^{3}+5 \textit {\_R}^{2} d -7 \textit {\_R} \,d^{2}+3 d^{3}\right ) \ln \left (c \sqrt {x}-\textit {\_R} +d \right )}{b \,\textit {\_R}^{3}-3 \textit {\_R}^{2} b d +\textit {\_R} a \,c^{2}+3 \textit {\_R} b \,d^{2}-a \,c^{2} d -b \,d^{3}}\right )}{8 b}+\frac {c \,d^{2} \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (b^{2} \textit {\_Z}^{4}-4 b^{2} d \,\textit {\_Z}^{3}+\left (2 a \,c^{2} b +6 b^{2} d^{2}\right ) \textit {\_Z}^{2}+\left (-4 a b \,c^{2} d -4 b^{2} d^{3}\right ) \textit {\_Z} +a^{2} c^{4}+2 a b \,c^{2} d^{2}+b^{2} d^{4}+c^{4}\right )}{\sum }\frac {\ln \left (d +c \sqrt {x}\right ) \ln \left (\frac {-c \sqrt {x}+\textit {\_R1} -d}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-c \sqrt {x}+\textit {\_R1} -d}{\textit {\_R1}}\right )}{\textit {\_R1}^{2} b -2 \textit {\_R1} b d +a \,c^{2}+b \,d^{2}}\right )}{4 b}\right )}{c^{2}}\) \(388\)
default \(\frac {\arctan \left (b x +a \right ) x}{c}-\frac {2 \arctan \left (b x +a \right ) d \sqrt {x}}{c^{2}}+\frac {2 \arctan \left (b x +a \right ) d^{2} \ln \left (d +c \sqrt {x}\right )}{c^{3}}-\frac {4 b \left (-\frac {c \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b^{2} \textit {\_Z}^{4}-4 b^{2} d \,\textit {\_Z}^{3}+\left (2 a \,c^{2} b +6 b^{2} d^{2}\right ) \textit {\_Z}^{2}+\left (-4 a b \,c^{2} d -4 b^{2} d^{3}\right ) \textit {\_Z} +a^{2} c^{4}+2 a b \,c^{2} d^{2}+b^{2} d^{4}+c^{4}\right )}{\sum }\frac {\left (-\textit {\_R}^{3}+5 \textit {\_R}^{2} d -7 \textit {\_R} \,d^{2}+3 d^{3}\right ) \ln \left (c \sqrt {x}-\textit {\_R} +d \right )}{b \,\textit {\_R}^{3}-3 \textit {\_R}^{2} b d +\textit {\_R} a \,c^{2}+3 \textit {\_R} b \,d^{2}-a \,c^{2} d -b \,d^{3}}\right )}{8 b}+\frac {c \,d^{2} \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (b^{2} \textit {\_Z}^{4}-4 b^{2} d \,\textit {\_Z}^{3}+\left (2 a \,c^{2} b +6 b^{2} d^{2}\right ) \textit {\_Z}^{2}+\left (-4 a b \,c^{2} d -4 b^{2} d^{3}\right ) \textit {\_Z} +a^{2} c^{4}+2 a b \,c^{2} d^{2}+b^{2} d^{4}+c^{4}\right )}{\sum }\frac {\ln \left (d +c \sqrt {x}\right ) \ln \left (\frac {-c \sqrt {x}+\textit {\_R1} -d}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-c \sqrt {x}+\textit {\_R1} -d}{\textit {\_R1}}\right )}{\textit {\_R1}^{2} b -2 \textit {\_R1} b d +a \,c^{2}+b \,d^{2}}\right )}{4 b}\right )}{c^{2}}\) \(388\)

input
int(arctan(b*x+a)/(c+d/x^(1/2)),x,method=_RETURNVERBOSE)
 
output
arctan(b*x+a)*x/c-2*arctan(b*x+a)/c^2*d*x^(1/2)+2*arctan(b*x+a)*d^2/c^3*ln 
(d+c*x^(1/2))-4*b/c^2*(-1/8*c/b*sum((-_R^3+5*_R^2*d-7*_R*d^2+3*d^3)/(_R^3* 
b-3*_R^2*b*d+_R*a*c^2+3*_R*b*d^2-a*c^2*d-b*d^3)*ln(c*x^(1/2)-_R+d),_R=Root 
Of(b^2*_Z^4-4*b^2*d*_Z^3+(2*a*b*c^2+6*b^2*d^2)*_Z^2+(-4*a*b*c^2*d-4*b^2*d^ 
3)*_Z+a^2*c^4+2*a*b*c^2*d^2+b^2*d^4+c^4))+1/4*c*d^2/b*sum(1/(_R1^2*b-2*_R1 
*b*d+a*c^2+b*d^2)*(ln(d+c*x^(1/2))*ln((-c*x^(1/2)+_R1-d)/_R1)+dilog((-c*x^ 
(1/2)+_R1-d)/_R1)),_R1=RootOf(b^2*_Z^4-4*b^2*d*_Z^3+(2*a*b*c^2+6*b^2*d^2)* 
_Z^2+(-4*a*b*c^2*d-4*b^2*d^3)*_Z+a^2*c^4+2*a*b*c^2*d^2+b^2*d^4+c^4)))
 
3.1.59.5 Fricas [F]

\[ \int \frac {\arctan (a+b x)}{c+\frac {d}{\sqrt {x}}} \, dx=\int { \frac {\arctan \left (b x + a\right )}{c + \frac {d}{\sqrt {x}}} \,d x } \]

input
integrate(arctan(b*x+a)/(c+d/x^(1/2)),x, algorithm="fricas")
 
output
integral((c*x*arctan(b*x + a) - d*sqrt(x)*arctan(b*x + a))/(c^2*x - d^2), 
x)
 
3.1.59.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\arctan (a+b x)}{c+\frac {d}{\sqrt {x}}} \, dx=\text {Timed out} \]

input
integrate(atan(b*x+a)/(c+d/x**(1/2)),x)
 
output
Timed out
 
3.1.59.7 Maxima [F]

\[ \int \frac {\arctan (a+b x)}{c+\frac {d}{\sqrt {x}}} \, dx=\int { \frac {\arctan \left (b x + a\right )}{c + \frac {d}{\sqrt {x}}} \,d x } \]

input
integrate(arctan(b*x+a)/(c+d/x^(1/2)),x, algorithm="maxima")
 
output
integrate(arctan(b*x + a)/(c + d/sqrt(x)), x)
 
3.1.59.8 Giac [F(-2)]

Exception generated. \[ \int \frac {\arctan (a+b x)}{c+\frac {d}{\sqrt {x}}} \, dx=\text {Exception raised: TypeError} \]

input
integrate(arctan(b*x+a)/(c+d/x^(1/2)),x, algorithm="giac")
 
output
Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:The choice was done assuming 0=[0]W 
arning, replacing 0 by -24, a substitution variable should perhaps be purg 
ed.Warnin
 
3.1.59.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\arctan (a+b x)}{c+\frac {d}{\sqrt {x}}} \, dx=\int \frac {\mathrm {atan}\left (a+b\,x\right )}{c+\frac {d}{\sqrt {x}}} \,d x \]

input
int(atan(a + b*x)/(c + d/x^(1/2)),x)
 
output
int(atan(a + b*x)/(c + d/x^(1/2)), x)