Integrand size = 22, antiderivative size = 706 \[ \int \frac {a+b \arctan (c+d x)}{e+f \sqrt {x}} \, dx=\frac {2 a \sqrt {x}}{f}+\frac {2 i b \sqrt {i+c} \arctan \left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {i+c}}\right )}{\sqrt {d} f}-\frac {2 i b \sqrt {i-c} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {i-c}}\right )}{\sqrt {d} f}+\frac {i b e \log \left (\frac {f \left (\sqrt {-i-c}-\sqrt {d} \sqrt {x}\right )}{\sqrt {d} e+\sqrt {-i-c} f}\right ) \log \left (e+f \sqrt {x}\right )}{f^2}-\frac {i b e \log \left (\frac {f \left (\sqrt {i-c}-\sqrt {d} \sqrt {x}\right )}{\sqrt {d} e+\sqrt {i-c} f}\right ) \log \left (e+f \sqrt {x}\right )}{f^2}+\frac {i b e \log \left (-\frac {f \left (\sqrt {-i-c}+\sqrt {d} \sqrt {x}\right )}{\sqrt {d} e-\sqrt {-i-c} f}\right ) \log \left (e+f \sqrt {x}\right )}{f^2}-\frac {i b e \log \left (-\frac {f \left (\sqrt {i-c}+\sqrt {d} \sqrt {x}\right )}{\sqrt {d} e-\sqrt {i-c} f}\right ) \log \left (e+f \sqrt {x}\right )}{f^2}+\frac {i b \sqrt {x} \log (1-i c-i d x)}{f}-\frac {e \log \left (e+f \sqrt {x}\right ) (a+i b \log (1-i c-i d x))}{f^2}-\frac {i b \sqrt {x} \log (1+i c+i d x)}{f}-\frac {e \log \left (e+f \sqrt {x}\right ) (a-i b \log (1+i c+i d x))}{f^2}+\frac {i b e \operatorname {PolyLog}\left (2,\frac {\sqrt {d} \left (e+f \sqrt {x}\right )}{\sqrt {d} e-\sqrt {-i-c} f}\right )}{f^2}+\frac {i b e \operatorname {PolyLog}\left (2,\frac {\sqrt {d} \left (e+f \sqrt {x}\right )}{\sqrt {d} e+\sqrt {-i-c} f}\right )}{f^2}-\frac {i b e \operatorname {PolyLog}\left (2,\frac {\sqrt {d} \left (e+f \sqrt {x}\right )}{\sqrt {d} e-\sqrt {i-c} f}\right )}{f^2}-\frac {i b e \operatorname {PolyLog}\left (2,\frac {\sqrt {d} \left (e+f \sqrt {x}\right )}{\sqrt {d} e+\sqrt {i-c} f}\right )}{f^2} \] Output:
2*a*x^(1/2)/f+2*I*b*(I+c)^(1/2)*arctan(d^(1/2)*x^(1/2)/(I+c)^(1/2))/d^(1/2 )/f-2*I*b*(I-c)^(1/2)*arctanh(d^(1/2)*x^(1/2)/(I-c)^(1/2))/d^(1/2)/f+I*b*e *ln(f*((-I-c)^(1/2)-d^(1/2)*x^(1/2))/(d^(1/2)*e+(-I-c)^(1/2)*f))*ln(e+f*x^ (1/2))/f^2-I*b*e*ln(f*((I-c)^(1/2)-d^(1/2)*x^(1/2))/(d^(1/2)*e+(I-c)^(1/2) *f))*ln(e+f*x^(1/2))/f^2+I*b*e*ln(-f*((-I-c)^(1/2)+d^(1/2)*x^(1/2))/(d^(1/ 2)*e-(-I-c)^(1/2)*f))*ln(e+f*x^(1/2))/f^2-I*b*e*ln(-f*((I-c)^(1/2)+d^(1/2) *x^(1/2))/(d^(1/2)*e-(I-c)^(1/2)*f))*ln(e+f*x^(1/2))/f^2+I*b*x^(1/2)*ln(1- I*c-I*d*x)/f-e*ln(e+f*x^(1/2))*(a+I*b*ln(1-I*c-I*d*x))/f^2-I*b*x^(1/2)*ln( 1+I*c+I*d*x)/f-e*ln(e+f*x^(1/2))*(a-I*b*ln(1+I*c+I*d*x))/f^2+I*b*e*polylog (2,d^(1/2)*(e+f*x^(1/2))/(d^(1/2)*e-(-I-c)^(1/2)*f))/f^2+I*b*e*polylog(2,d ^(1/2)*(e+f*x^(1/2))/(d^(1/2)*e+(-I-c)^(1/2)*f))/f^2-I*b*e*polylog(2,d^(1/ 2)*(e+f*x^(1/2))/(d^(1/2)*e-(I-c)^(1/2)*f))/f^2-I*b*e*polylog(2,d^(1/2)*(e +f*x^(1/2))/(d^(1/2)*e+(I-c)^(1/2)*f))/f^2
Time = 22.83 (sec) , antiderivative size = 631, normalized size of antiderivative = 0.89 \[ \int \frac {a+b \arctan (c+d x)}{e+f \sqrt {x}} \, dx=\frac {2 a \left (f \sqrt {x}-e \log \left (e+f \sqrt {x}\right )\right )+i b \left (\frac {2 \sqrt {i+c} f \arctan \left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {i+c}}\right )}{\sqrt {d}}-\frac {2 \sqrt {i-c} f \text {arctanh}\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {i-c}}\right )}{\sqrt {d}}+e \log \left (\frac {f \left (\sqrt {-i-c}-\sqrt {d} \sqrt {x}\right )}{\sqrt {d} e+\sqrt {-i-c} f}\right ) \log \left (e+f \sqrt {x}\right )-e \log \left (\frac {f \left (\sqrt {i-c}-\sqrt {d} \sqrt {x}\right )}{\sqrt {d} e+\sqrt {i-c} f}\right ) \log \left (e+f \sqrt {x}\right )+e \log \left (\frac {f \left (\sqrt {-i-c}+\sqrt {d} \sqrt {x}\right )}{-\sqrt {d} e+\sqrt {-i-c} f}\right ) \log \left (e+f \sqrt {x}\right )-e \log \left (\frac {f \left (\sqrt {i-c}+\sqrt {d} \sqrt {x}\right )}{-\sqrt {d} e+\sqrt {i-c} f}\right ) \log \left (e+f \sqrt {x}\right )-f \sqrt {x} \log (1+i c+i d x)+e \log \left (e+f \sqrt {x}\right ) \log (1+i c+i d x)+f \sqrt {x} \log (-i (i+c+d x))-e \log \left (e+f \sqrt {x}\right ) \log (-i (i+c+d x))+e \operatorname {PolyLog}\left (2,\frac {\sqrt {d} \left (e+f \sqrt {x}\right )}{\sqrt {d} e-\sqrt {-i-c} f}\right )+e \operatorname {PolyLog}\left (2,\frac {\sqrt {d} \left (e+f \sqrt {x}\right )}{\sqrt {d} e+\sqrt {-i-c} f}\right )-e \operatorname {PolyLog}\left (2,\frac {\sqrt {d} \left (e+f \sqrt {x}\right )}{\sqrt {d} e-\sqrt {i-c} f}\right )-e \operatorname {PolyLog}\left (2,\frac {\sqrt {d} \left (e+f \sqrt {x}\right )}{\sqrt {d} e+\sqrt {i-c} f}\right )\right )}{f^2} \] Input:
Integrate[(a + b*ArcTan[c + d*x])/(e + f*Sqrt[x]),x]
Output:
(2*a*(f*Sqrt[x] - e*Log[e + f*Sqrt[x]]) + I*b*((2*Sqrt[I + c]*f*ArcTan[(Sq rt[d]*Sqrt[x])/Sqrt[I + c]])/Sqrt[d] - (2*Sqrt[I - c]*f*ArcTanh[(Sqrt[d]*S qrt[x])/Sqrt[I - c]])/Sqrt[d] + e*Log[(f*(Sqrt[-I - c] - Sqrt[d]*Sqrt[x])) /(Sqrt[d]*e + Sqrt[-I - c]*f)]*Log[e + f*Sqrt[x]] - e*Log[(f*(Sqrt[I - c] - Sqrt[d]*Sqrt[x]))/(Sqrt[d]*e + Sqrt[I - c]*f)]*Log[e + f*Sqrt[x]] + e*Lo g[(f*(Sqrt[-I - c] + Sqrt[d]*Sqrt[x]))/(-(Sqrt[d]*e) + Sqrt[-I - c]*f)]*Lo g[e + f*Sqrt[x]] - e*Log[(f*(Sqrt[I - c] + Sqrt[d]*Sqrt[x]))/(-(Sqrt[d]*e) + Sqrt[I - c]*f)]*Log[e + f*Sqrt[x]] - f*Sqrt[x]*Log[1 + I*c + I*d*x] + e *Log[e + f*Sqrt[x]]*Log[1 + I*c + I*d*x] + f*Sqrt[x]*Log[(-I)*(I + c + d*x )] - e*Log[e + f*Sqrt[x]]*Log[(-I)*(I + c + d*x)] + e*PolyLog[2, (Sqrt[d]* (e + f*Sqrt[x]))/(Sqrt[d]*e - Sqrt[-I - c]*f)] + e*PolyLog[2, (Sqrt[d]*(e + f*Sqrt[x]))/(Sqrt[d]*e + Sqrt[-I - c]*f)] - e*PolyLog[2, (Sqrt[d]*(e + f *Sqrt[x]))/(Sqrt[d]*e - Sqrt[I - c]*f)] - e*PolyLog[2, (Sqrt[d]*(e + f*Sqr t[x]))/(Sqrt[d]*e + Sqrt[I - c]*f)]))/f^2
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.
| \(\displaystyle \int \frac {a+b \arctan (c+d x)}{e+f \sqrt {x}} \, dx\) |
| \(\Big \downarrow \) 7267 |
| \(\displaystyle 2 \int \frac {\sqrt {x} (a+b \arctan (c+d x))}{e+f \sqrt {x}}d\sqrt {x}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 2 \int \left (\frac {\sqrt {x} a}{e+f \sqrt {x}}+\frac {b \sqrt {x} \arctan (c+d x)}{e+f \sqrt {x}}\right )d\sqrt {x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 \left (-\frac {b e \int \frac {\arctan (c+d x)}{e+f \sqrt {x}}d\sqrt {x}}{f}-\frac {a e \log \left (e+f \sqrt {x}\right )}{f^2}+\frac {a \sqrt {x}}{f}+\frac {b \arctan \left (\frac {\sqrt {\sqrt {c^2+1}-c}-\sqrt {2} \sqrt {d} \sqrt {x}}{\sqrt {\sqrt {c^2+1}+c}}\right )}{\sqrt {2} \sqrt {\sqrt {c^2+1}+c} \sqrt {d} f}-\frac {b \arctan \left (\frac {\sqrt {\sqrt {c^2+1}-c}+\sqrt {2} \sqrt {d} \sqrt {x}}{\sqrt {\sqrt {c^2+1}+c}}\right )}{\sqrt {2} \sqrt {\sqrt {c^2+1}+c} \sqrt {d} f}+\frac {b \sqrt {x} \arctan (c+d x)}{f}-\frac {b \log \left (-\sqrt {2} \sqrt {\sqrt {c^2+1}-c} \sqrt {d} \sqrt {x}+\sqrt {c^2+1}+d x\right )}{2 \sqrt {2} \sqrt {\sqrt {c^2+1}-c} \sqrt {d} f}+\frac {b \log \left (\sqrt {2} \sqrt {\sqrt {c^2+1}-c} \sqrt {d} \sqrt {x}+\sqrt {c^2+1}+d x\right )}{2 \sqrt {2} \sqrt {\sqrt {c^2+1}-c} \sqrt {d} f}\right )\) |
Input:
Int[(a + b*ArcTan[c + d*x])/(e + f*Sqrt[x]),x]
Output:
$Aborted
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.27 (sec) , antiderivative size = 372, normalized size of antiderivative = 0.53
| method | result | size |
| derivativedivides | \(\frac {2 a \sqrt {x}}{f}-\frac {2 a e \ln \left (e +f \sqrt {x}\right )}{f^{2}}+\frac {2 b \arctan \left (d x +c \right ) \sqrt {x}}{f}-\frac {2 b \arctan \left (d x +c \right ) e \ln \left (e +f \sqrt {x}\right )}{f^{2}}-b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (d^{2} \textit {\_Z}^{4}-4 d^{2} e \,\textit {\_Z}^{3}+\left (2 c d \,f^{2}+6 d^{2} e^{2}\right ) \textit {\_Z}^{2}+\left (-4 c d e \,f^{2}-4 d^{2} e^{3}\right ) \textit {\_Z} +c^{2} f^{4}+2 c d \,e^{2} f^{2}+d^{2} e^{4}+f^{4}\right )}{\sum }\frac {\left (\textit {\_R}^{2}-2 \textit {\_R} e +e^{2}\right ) \ln \left (f \sqrt {x}-\textit {\_R} +e \right )}{d \,\textit {\_R}^{3}-3 \textit {\_R}^{2} d e +\textit {\_R} c \,f^{2}+3 \textit {\_R} d \,e^{2}-c e \,f^{2}-d \,e^{3}}\right )+b e \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (d^{2} \textit {\_Z}^{4}-4 d^{2} e \,\textit {\_Z}^{3}+\left (2 c d \,f^{2}+6 d^{2} e^{2}\right ) \textit {\_Z}^{2}+\left (-4 c d e \,f^{2}-4 d^{2} e^{3}\right ) \textit {\_Z} +c^{2} f^{4}+2 c d \,e^{2} f^{2}+d^{2} e^{4}+f^{4}\right )}{\sum }\frac {\ln \left (e +f \sqrt {x}\right ) \ln \left (\frac {-f \sqrt {x}+\textit {\_R1} -e}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-f \sqrt {x}+\textit {\_R1} -e}{\textit {\_R1}}\right )}{\textit {\_R1}^{2} d -2 \textit {\_R1} d e +c \,f^{2}+d \,e^{2}}\right )\) | \(372\) |
| default | \(\frac {2 a \sqrt {x}}{f}-\frac {2 a e \ln \left (e +f \sqrt {x}\right )}{f^{2}}+\frac {2 b \arctan \left (d x +c \right ) \sqrt {x}}{f}-\frac {2 b \arctan \left (d x +c \right ) e \ln \left (e +f \sqrt {x}\right )}{f^{2}}-b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (d^{2} \textit {\_Z}^{4}-4 d^{2} e \,\textit {\_Z}^{3}+\left (2 c d \,f^{2}+6 d^{2} e^{2}\right ) \textit {\_Z}^{2}+\left (-4 c d e \,f^{2}-4 d^{2} e^{3}\right ) \textit {\_Z} +c^{2} f^{4}+2 c d \,e^{2} f^{2}+d^{2} e^{4}+f^{4}\right )}{\sum }\frac {\left (\textit {\_R}^{2}-2 \textit {\_R} e +e^{2}\right ) \ln \left (f \sqrt {x}-\textit {\_R} +e \right )}{d \,\textit {\_R}^{3}-3 \textit {\_R}^{2} d e +\textit {\_R} c \,f^{2}+3 \textit {\_R} d \,e^{2}-c e \,f^{2}-d \,e^{3}}\right )+b e \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (d^{2} \textit {\_Z}^{4}-4 d^{2} e \,\textit {\_Z}^{3}+\left (2 c d \,f^{2}+6 d^{2} e^{2}\right ) \textit {\_Z}^{2}+\left (-4 c d e \,f^{2}-4 d^{2} e^{3}\right ) \textit {\_Z} +c^{2} f^{4}+2 c d \,e^{2} f^{2}+d^{2} e^{4}+f^{4}\right )}{\sum }\frac {\ln \left (e +f \sqrt {x}\right ) \ln \left (\frac {-f \sqrt {x}+\textit {\_R1} -e}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-f \sqrt {x}+\textit {\_R1} -e}{\textit {\_R1}}\right )}{\textit {\_R1}^{2} d -2 \textit {\_R1} d e +c \,f^{2}+d \,e^{2}}\right )\) | \(372\) |
| parts | \(a \left (\frac {2 \sqrt {x}}{f}+\frac {e \ln \left (f \sqrt {x}-e \right )}{f^{2}}-\frac {e \ln \left (e +f \sqrt {x}\right )}{f^{2}}-\frac {e \ln \left (f^{2} x -e^{2}\right )}{f^{2}}\right )+\frac {2 b \arctan \left (d x +c \right ) \sqrt {x}}{f}-\frac {2 b \arctan \left (d x +c \right ) e \ln \left (e +f \sqrt {x}\right )}{f^{2}}+b e \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (d^{2} \textit {\_Z}^{4}-4 d^{2} e \,\textit {\_Z}^{3}+\left (2 c d \,f^{2}+6 d^{2} e^{2}\right ) \textit {\_Z}^{2}+\left (-4 c d e \,f^{2}-4 d^{2} e^{3}\right ) \textit {\_Z} +c^{2} f^{4}+2 c d \,e^{2} f^{2}+d^{2} e^{4}+f^{4}\right )}{\sum }\frac {\ln \left (e +f \sqrt {x}\right ) \ln \left (\frac {-f \sqrt {x}+\textit {\_R1} -e}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-f \sqrt {x}+\textit {\_R1} -e}{\textit {\_R1}}\right )}{\textit {\_R1}^{2} d -2 \textit {\_R1} d e +c \,f^{2}+d \,e^{2}}\right )-b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (d^{2} \textit {\_Z}^{4}-4 d^{2} e \,\textit {\_Z}^{3}+\left (2 c d \,f^{2}+6 d^{2} e^{2}\right ) \textit {\_Z}^{2}+\left (-4 c d e \,f^{2}-4 d^{2} e^{3}\right ) \textit {\_Z} +c^{2} f^{4}+2 c d \,e^{2} f^{2}+d^{2} e^{4}+f^{4}\right )}{\sum }\frac {\left (\textit {\_R}^{2}-2 \textit {\_R} e +e^{2}\right ) \ln \left (f \sqrt {x}-\textit {\_R} +e \right )}{d \,\textit {\_R}^{3}-3 \textit {\_R}^{2} d e +\textit {\_R} c \,f^{2}+3 \textit {\_R} d \,e^{2}-c e \,f^{2}-d \,e^{3}}\right )\) | \(406\) |
Input:
int((a+b*arctan(d*x+c))/(e+f*x^(1/2)),x,method=_RETURNVERBOSE)
Output:
2*a*x^(1/2)/f-2*a*e/f^2*ln(e+f*x^(1/2))+2*b*arctan(d*x+c)/f*x^(1/2)-2*b*ar ctan(d*x+c)*e/f^2*ln(e+f*x^(1/2))-b*sum((_R^2-2*_R*e+e^2)/(_R^3*d-3*_R^2*d *e+_R*c*f^2+3*_R*d*e^2-c*e*f^2-d*e^3)*ln(f*x^(1/2)-_R+e),_R=RootOf(d^2*_Z^ 4-4*d^2*e*_Z^3+(2*c*d*f^2+6*d^2*e^2)*_Z^2+(-4*c*d*e*f^2-4*d^2*e^3)*_Z+c^2* f^4+2*c*d*e^2*f^2+d^2*e^4+f^4))+b*e*sum(1/(_R1^2*d-2*_R1*d*e+c*f^2+d*e^2)* (ln(e+f*x^(1/2))*ln((-f*x^(1/2)+_R1-e)/_R1)+dilog((-f*x^(1/2)+_R1-e)/_R1)) ,_R1=RootOf(d^2*_Z^4-4*d^2*e*_Z^3+(2*c*d*f^2+6*d^2*e^2)*_Z^2+(-4*c*d*e*f^2 -4*d^2*e^3)*_Z+c^2*f^4+2*c*d*e^2*f^2+d^2*e^4+f^4))
\[ \int \frac {a+b \arctan (c+d x)}{e+f \sqrt {x}} \, dx=\int { \frac {b \arctan \left (d x + c\right ) + a}{f \sqrt {x} + e} \,d x } \] Input:
integrate((a+b*arctan(d*x+c))/(e+f*x^(1/2)),x, algorithm="fricas")
Output:
integral(-(b*e*arctan(d*x + c) + a*e - (b*f*arctan(d*x + c) + a*f)*sqrt(x) )/(f^2*x - e^2), x)
Timed out. \[ \int \frac {a+b \arctan (c+d x)}{e+f \sqrt {x}} \, dx=\text {Timed out} \] Input:
integrate((a+b*atan(d*x+c))/(e+f*x**(1/2)),x)
Output:
Timed out
\[ \int \frac {a+b \arctan (c+d x)}{e+f \sqrt {x}} \, dx=\int { \frac {b \arctan \left (d x + c\right ) + a}{f \sqrt {x} + e} \,d x } \] Input:
integrate((a+b*arctan(d*x+c))/(e+f*x^(1/2)),x, algorithm="maxima")
Output:
2*(b*f^2*integrate(1/2*arctan(d*x + c)/(f*sqrt(x) + e), x) - a*e*log(f*sqr t(x) + e) + a*f*sqrt(x))/f^2
\[ \int \frac {a+b \arctan (c+d x)}{e+f \sqrt {x}} \, dx=\int { \frac {b \arctan \left (d x + c\right ) + a}{f \sqrt {x} + e} \,d x } \] Input:
integrate((a+b*arctan(d*x+c))/(e+f*x^(1/2)),x, algorithm="giac")
Output:
integrate((b*arctan(d*x + c) + a)/(f*sqrt(x) + e), x)
Timed out. \[ \int \frac {a+b \arctan (c+d x)}{e+f \sqrt {x}} \, dx=\int \frac {a+b\,\mathrm {atan}\left (c+d\,x\right )}{e+f\,\sqrt {x}} \,d x \] Input:
int((a + b*atan(c + d*x))/(e + f*x^(1/2)),x)
Output:
int((a + b*atan(c + d*x))/(e + f*x^(1/2)), x)
\[ \int \frac {a+b \arctan (c+d x)}{e+f \sqrt {x}} \, dx=\frac {2 \sqrt {x}\, a f +\left (\int \frac {\mathit {atan} \left (d x +c \right )}{-f^{2} x +e^{2}}d x \right ) b e \,f^{2}-\left (\int \frac {\sqrt {x}\, \mathit {atan} \left (d x +c \right )}{-f^{2} x +e^{2}}d x \right ) b \,f^{3}-2 \,\mathrm {log}\left (\sqrt {x}\, f +e \right ) a e}{f^{2}} \] Input:
int((a+b*atan(d*x+c))/(e+f*x^(1/2)),x)
Output:
(2*sqrt(x)*a*f + int(atan(c + d*x)/(e**2 - f**2*x),x)*b*e*f**2 - int((sqrt (x)*atan(c + d*x))/(e**2 - f**2*x),x)*b*f**3 - 2*log(sqrt(x)*f + e)*a*e)/f **2