3.1.58 \(\int \frac {\arctan (a+b x)}{c+d \sqrt {x}} \, dx\) [58]

3.1.58.1 Optimal result

Integrand size = 18, antiderivative size = 673 \[ \int \frac {\arctan (a+b x)}{c+d \sqrt {x}} \, dx=\frac {2 i \sqrt {i+a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {i+a}}\right )}{\sqrt {b} d}-\frac {2 i \sqrt {i-a} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {i-a}}\right )}{\sqrt {b} d}+\frac {i c \log \left (\frac {d \left (\sqrt {-i-a}-\sqrt {b} \sqrt {x}\right )}{\sqrt {b} c+\sqrt {-i-a} d}\right ) \log \left (c+d \sqrt {x}\right )}{d^2}-\frac {i c \log \left (\frac {d \left (\sqrt {i-a}-\sqrt {b} \sqrt {x}\right )}{\sqrt {b} c+\sqrt {i-a} d}\right ) \log \left (c+d \sqrt {x}\right )}{d^2}+\frac {i c \log \left (-\frac {d \left (\sqrt {-i-a}+\sqrt {b} \sqrt {x}\right )}{\sqrt {b} c-\sqrt {-i-a} d}\right ) \log \left (c+d \sqrt {x}\right )}{d^2}-\frac {i c \log \left (-\frac {d \left (\sqrt {i-a}+\sqrt {b} \sqrt {x}\right )}{\sqrt {b} c-\sqrt {i-a} d}\right ) \log \left (c+d \sqrt {x}\right )}{d^2}+\frac {i \sqrt {x} \log (1-i a-i b x)}{d}-\frac {i c \log \left (c+d \sqrt {x}\right ) \log (1-i a-i b x)}{d^2}-\frac {i \sqrt {x} \log (1+i a+i b x)}{d}+\frac {i c \log \left (c+d \sqrt {x}\right ) \log (1+i a+i b x)}{d^2}+\frac {i c \operatorname {PolyLog}\left (2,\frac {\sqrt {b} \left (c+d \sqrt {x}\right )}{\sqrt {b} c-\sqrt {-i-a} d}\right )}{d^2}+\frac {i c \operatorname {PolyLog}\left (2,\frac {\sqrt {b} \left (c+d \sqrt {x}\right )}{\sqrt {b} c+\sqrt {-i-a} d}\right )}{d^2}-\frac {i c \operatorname {PolyLog}\left (2,\frac {\sqrt {b} \left (c+d \sqrt {x}\right )}{\sqrt {b} c-\sqrt {i-a} d}\right )}{d^2}-\frac {i c \operatorname {PolyLog}\left (2,\frac {\sqrt {b} \left (c+d \sqrt {x}\right )}{\sqrt {b} c+\sqrt {i-a} d}\right )}{d^2} \]

output

-I*c*ln(1-I*a-I*b*x)*ln(c+d*x^(1/2))/d^2+I*c*ln(1+I*a+I*b*x)*ln(c+d*x^(1/2 
))/d^2+I*c*ln(c+d*x^(1/2))*ln(d*((-I-a)^(1/2)-b^(1/2)*x^(1/2))/(d*(-I-a)^( 
1/2)+c*b^(1/2)))/d^2-I*c*ln(c+d*x^(1/2))*ln(d*((I-a)^(1/2)-b^(1/2)*x^(1/2) 
)/(d*(I-a)^(1/2)+c*b^(1/2)))/d^2+I*c*ln(c+d*x^(1/2))*ln(-d*((-I-a)^(1/2)+b 
^(1/2)*x^(1/2))/(-d*(-I-a)^(1/2)+c*b^(1/2)))/d^2-I*c*ln(c+d*x^(1/2))*ln(-d 
*((I-a)^(1/2)+b^(1/2)*x^(1/2))/(-d*(I-a)^(1/2)+c*b^(1/2)))/d^2+I*c*polylog 
(2,b^(1/2)*(c+d*x^(1/2))/(-d*(-I-a)^(1/2)+c*b^(1/2)))/d^2+I*c*polylog(2,b^ 
(1/2)*(c+d*x^(1/2))/(d*(-I-a)^(1/2)+c*b^(1/2)))/d^2-I*c*polylog(2,b^(1/2)* 
(c+d*x^(1/2))/(-d*(I-a)^(1/2)+c*b^(1/2)))/d^2-I*c*polylog(2,b^(1/2)*(c+d*x 
^(1/2))/(d*(I-a)^(1/2)+c*b^(1/2)))/d^2-2*I*arctanh(b^(1/2)*x^(1/2)/(I-a)^( 
1/2))*(I-a)^(1/2)/d/b^(1/2)+2*I*arctan(b^(1/2)*x^(1/2)/(I+a)^(1/2))*(I+a)^ 
(1/2)/d/b^(1/2)+I*ln(1-I*a-I*b*x)*x^(1/2)/d-I*ln(1+I*a+I*b*x)*x^(1/2)/d
 

3.1.58.2 Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 604, normalized size of antiderivative = 0.90 \[ \int \frac {\arctan (a+b x)}{c+d \sqrt {x}} \, dx=\frac {i \left (\frac {2 \sqrt {i+a} d \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {i+a}}\right )}{\sqrt {b}}-\frac {2 \sqrt {i-a} d \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {i-a}}\right )}{\sqrt {b}}+c \log \left (\frac {d \left (\sqrt {-i-a}-\sqrt {b} \sqrt {x}\right )}{\sqrt {b} c+\sqrt {-i-a} d}\right ) \log \left (c+d \sqrt {x}\right )-c \log \left (\frac {d \left (\sqrt {i-a}-\sqrt {b} \sqrt {x}\right )}{\sqrt {b} c+\sqrt {i-a} d}\right ) \log \left (c+d \sqrt {x}\right )+c \log \left (\frac {d \left (\sqrt {-i-a}+\sqrt {b} \sqrt {x}\right )}{-\sqrt {b} c+\sqrt {-i-a} d}\right ) \log \left (c+d \sqrt {x}\right )-c \log \left (\frac {d \left (\sqrt {i-a}+\sqrt {b} \sqrt {x}\right )}{-\sqrt {b} c+\sqrt {i-a} d}\right ) \log \left (c+d \sqrt {x}\right )-d \sqrt {x} \log (1+i a+i b x)+c \log \left (c+d \sqrt {x}\right ) \log (1+i a+i b x)+d \sqrt {x} \log (-i (i+a+b x))-c \log \left (c+d \sqrt {x}\right ) \log (-i (i+a+b x))+c \operatorname {PolyLog}\left (2,\frac {\sqrt {b} \left (c+d \sqrt {x}\right )}{\sqrt {b} c-\sqrt {-i-a} d}\right )+c \operatorname {PolyLog}\left (2,\frac {\sqrt {b} \left (c+d \sqrt {x}\right )}{\sqrt {b} c+\sqrt {-i-a} d}\right )-c \operatorname {PolyLog}\left (2,\frac {\sqrt {b} \left (c+d \sqrt {x}\right )}{\sqrt {b} c-\sqrt {i-a} d}\right )-c \operatorname {PolyLog}\left (2,\frac {\sqrt {b} \left (c+d \sqrt {x}\right )}{\sqrt {b} c+\sqrt {i-a} d}\right )\right )}{d^2} \]

input

Integrate[ArcTan[a + b*x]/(c + d*Sqrt[x]),x]
 

output

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

3.1.58.3 Rubi [A] (verified)

Time = 1.14 (sec) , antiderivative size = 665, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5574, 2855, 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.

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

\(\Big \downarrow \) 5574

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

\(\Big \downarrow \) 2855

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

\(\Big \downarrow \) 2916

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

\(\Big \downarrow \) 2009

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

input

Int[ArcTan[a + b*x]/(c + d*Sqrt[x]),x]
 

output

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

3.1.58.3.1 Defintions of rubi rules used

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]
 

3.1.58.4 Maple [C] (verified)

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

Time = 0.24 (sec) , antiderivative size = 364, normalized size of antiderivative = 0.54

method	result	size
derivativedivides	\(\frac {2 \arctan \left (b x +a \right ) \sqrt {x}}{d}-\frac {2 \arctan \left (b x +a \right ) c \ln \left (c +d \sqrt {x}\right )}{d^{2}}-\frac {4 b \left (\frac {d^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b^{2} \textit {\_Z}^{4}-4 b^{2} c \,\textit {\_Z}^{3}+\left (2 a b \,d^{2}+6 b^{2} c^{2}\right ) \textit {\_Z}^{2}+\left (-4 a b c \,d^{2}-4 b^{2} c^{3}\right ) \textit {\_Z} +a^{2} d^{4}+2 a b \,c^{2} d^{2}+b^{2} c^{4}+d^{4}\right )}{\sum }\frac {\left (\textit {\_R}^{2}-2 \textit {\_R} c +c^{2}\right ) \ln \left (d \sqrt {x}-\textit {\_R} +c \right )}{b \,\textit {\_R}^{3}-3 \textit {\_R}^{2} b c +\textit {\_R} a \,d^{2}+3 \textit {\_R} b \,c^{2}-a c \,d^{2}-b \,c^{3}}\right )}{4 b}-\frac {c \,d^{2} \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (b^{2} \textit {\_Z}^{4}-4 b^{2} c \,\textit {\_Z}^{3}+\left (2 a b \,d^{2}+6 b^{2} c^{2}\right ) \textit {\_Z}^{2}+\left (-4 a b c \,d^{2}-4 b^{2} c^{3}\right ) \textit {\_Z} +a^{2} d^{4}+2 a b \,c^{2} d^{2}+b^{2} c^{4}+d^{4}\right )}{\sum }\frac {\ln \left (c +d \sqrt {x}\right ) \ln \left (\frac {-d \sqrt {x}+\textit {\_R1} -c}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-d \sqrt {x}+\textit {\_R1} -c}{\textit {\_R1}}\right )}{\textit {\_R1}^{2} b -2 \textit {\_R1} b c +a \,d^{2}+b \,c^{2}}\right )}{4 b}\right )}{d^{2}}\)	\(364\)
default	\(\frac {2 \arctan \left (b x +a \right ) \sqrt {x}}{d}-\frac {2 \arctan \left (b x +a \right ) c \ln \left (c +d \sqrt {x}\right )}{d^{2}}-\frac {4 b \left (\frac {d^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b^{2} \textit {\_Z}^{4}-4 b^{2} c \,\textit {\_Z}^{3}+\left (2 a b \,d^{2}+6 b^{2} c^{2}\right ) \textit {\_Z}^{2}+\left (-4 a b c \,d^{2}-4 b^{2} c^{3}\right ) \textit {\_Z} +a^{2} d^{4}+2 a b \,c^{2} d^{2}+b^{2} c^{4}+d^{4}\right )}{\sum }\frac {\left (\textit {\_R}^{2}-2 \textit {\_R} c +c^{2}\right ) \ln \left (d \sqrt {x}-\textit {\_R} +c \right )}{b \,\textit {\_R}^{3}-3 \textit {\_R}^{2} b c +\textit {\_R} a \,d^{2}+3 \textit {\_R} b \,c^{2}-a c \,d^{2}-b \,c^{3}}\right )}{4 b}-\frac {c \,d^{2} \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (b^{2} \textit {\_Z}^{4}-4 b^{2} c \,\textit {\_Z}^{3}+\left (2 a b \,d^{2}+6 b^{2} c^{2}\right ) \textit {\_Z}^{2}+\left (-4 a b c \,d^{2}-4 b^{2} c^{3}\right ) \textit {\_Z} +a^{2} d^{4}+2 a b \,c^{2} d^{2}+b^{2} c^{4}+d^{4}\right )}{\sum }\frac {\ln \left (c +d \sqrt {x}\right ) \ln \left (\frac {-d \sqrt {x}+\textit {\_R1} -c}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-d \sqrt {x}+\textit {\_R1} -c}{\textit {\_R1}}\right )}{\textit {\_R1}^{2} b -2 \textit {\_R1} b c +a \,d^{2}+b \,c^{2}}\right )}{4 b}\right )}{d^{2}}\)	\(364\)

input

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

output

2*arctan(b*x+a)/d*x^(1/2)-2*arctan(b*x+a)*c/d^2*ln(c+d*x^(1/2))-4*b/d^2*(1 
/4*d^2/b*sum((_R^2-2*_R*c+c^2)/(_R^3*b-3*_R^2*b*c+_R*a*d^2+3*_R*b*c^2-a*c* 
d^2-b*c^3)*ln(d*x^(1/2)-_R+c),_R=RootOf(b^2*_Z^4-4*b^2*c*_Z^3+(2*a*b*d^2+6 
*b^2*c^2)*_Z^2+(-4*a*b*c*d^2-4*b^2*c^3)*_Z+a^2*d^4+2*a*b*c^2*d^2+b^2*c^4+d 
^4))-1/4*c*d^2/b*sum(1/(_R1^2*b-2*_R1*b*c+a*d^2+b*c^2)*(ln(c+d*x^(1/2))*ln 
((-d*x^(1/2)+_R1-c)/_R1)+dilog((-d*x^(1/2)+_R1-c)/_R1)),_R1=RootOf(b^2*_Z^ 
4-4*b^2*c*_Z^3+(2*a*b*d^2+6*b^2*c^2)*_Z^2+(-4*a*b*c*d^2-4*b^2*c^3)*_Z+a^2* 
d^4+2*a*b*c^2*d^2+b^2*c^4+d^4)))
 

3.1.58.5 Fricas [F]

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

input

integrate(arctan(b*x+a)/(c+d*x^(1/2)),x, algorithm="fricas")
 

output

integral((d*sqrt(x)*arctan(b*x + a) - c*arctan(b*x + a))/(d^2*x - c^2), x)
 

3.1.58.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.1.58.7 Maxima [F]

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

input

integrate(arctan(b*x+a)/(c+d*x^(1/2)),x, algorithm="maxima")
 

output

integrate(arctan(b*x + a)/(d*sqrt(x) + c), x)
 

3.1.58.8 Giac [F(-2)]

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

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
 

3.1.58.9 Mupad [F(-1)]

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

input

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

output

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