\(\int \frac {\arctan (a+b x)}{c+\frac {d}{\sqrt {x}}} \, dx\) [61]

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:

I*d^2*ln(c*((I-a)^(1/2)-b^(1/2)*x^(1/2))/((I-a)^(1/2)*c+b^(1/2)*d))*ln(d+c 
*x^(1/2))/c^3-2*I*(I+a)^(1/2)*d*arctan(b^(1/2)*x^(1/2)/(I+a)^(1/2))/b^(1/2 
)/c^2+I*d^2*polylog(2,b^(1/2)*(d+c*x^(1/2))/((I-a)^(1/2)*c+b^(1/2)*d))/c^3 
+I*d^2*polylog(2,-b^(1/2)*(d+c*x^(1/2))/((I-a)^(1/2)*c-b^(1/2)*d))/c^3-I*d 
^2*ln(d+c*x^(1/2))*ln(1+I*a+I*b*x)/c^3+I*d*x^(1/2)*ln(1+I*a+I*b*x)/c^2-I*d 
^2*polylog(2,b^(1/2)*(d+c*x^(1/2))/((-I-a)^(1/2)*c+b^(1/2)*d))/c^3-I*d^2*l 
n(c*((-I-a)^(1/2)-b^(1/2)*x^(1/2))/((-I-a)^(1/2)*c+b^(1/2)*d))*ln(d+c*x^(1 
/2))/c^3-I*d*x^(1/2)*ln(1-I*a-I*b*x)/c^2-1/2*(1+I*a+I*b*x)*ln(1+I*a+I*b*x) 
/b/c+I*d^2*ln(c*((I-a)^(1/2)+b^(1/2)*x^(1/2))/((I-a)^(1/2)*c-b^(1/2)*d))*l 
n(d+c*x^(1/2))/c^3-1/2*(1-I*a-I*b*x)*ln(-I*(I+a+b*x))/b/c-I*d^2*polylog(2, 
-b^(1/2)*(d+c*x^(1/2))/((-I-a)^(1/2)*c-b^(1/2)*d))/c^3+2*I*(I-a)^(1/2)*d*a 
rctanh(b^(1/2)*x^(1/2)/(I-a)^(1/2))/b^(1/2)/c^2+I*d^2*ln(d+c*x^(1/2))*ln(1 
-I*a-I*b*x)/c^3-I*d^2*ln(c*((-I-a)^(1/2)+b^(1/2)*x^(1/2))/((-I-a)^(1/2)*c- 
b^(1/2)*d))*ln(d+c*x^(1/2))/c^3
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.33 (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)
 

Rubi [A] (verified)

Time = 1.38 (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)
 

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]
 

Maple [C] (warning: unable to verify)

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

Time = 0.17 (sec) , antiderivative size = 388, normalized size of antiderivative = 0.50

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

Input:

int(arctan(b*x+a)/(c+d/x^(1/2)),x,method=_RETURNVERBOSE)
 

Output:

arctan(b*x+a)/c*x-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)))
 

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)
                                                                                    
                                                                                    
 

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
 

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)
 

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:Warning, need to choose a branch fo 
r the root of a polynomial with parameters. This might be wrong.The choice 
 was done
 

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)
 

Reduce [F]

\[ \int \frac {\arctan (a+b x)}{c+\frac {d}{\sqrt {x}}} \, dx=\left (\int \frac {\mathit {atan} \left (b x +a \right ) x}{c^{2} x -d^{2}}d x \right ) c -\left (\int \frac {\sqrt {x}\, \mathit {atan} \left (b x +a \right )}{c^{2} x -d^{2}}d x \right ) d \] Input:

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

Output:

int((atan(a + b*x)*x)/(c**2*x - d**2),x)*c - int((sqrt(x)*atan(a + b*x))/( 
c**2*x - d**2),x)*d