\(\int \frac {\sqrt {c+d x}}{1+x^2} \, dx\) [649]

Optimal result

Integrand size = 17, antiderivative size = 316 \[ \int \frac {\sqrt {c+d x}}{1+x^2} \, dx=\frac {d \text {arctanh}\left (\frac {\sqrt {c+\sqrt {c^2+d^2}}-\sqrt {2} \sqrt {c+d x}}{\sqrt {c-\sqrt {c^2+d^2}}}\right )}{\sqrt {2} \sqrt {c-\sqrt {c^2+d^2}}}-\frac {d \text {arctanh}\left (\frac {\sqrt {c+\sqrt {c^2+d^2}}+\sqrt {2} \sqrt {c+d x}}{\sqrt {c-\sqrt {c^2+d^2}}}\right )}{\sqrt {2} \sqrt {c-\sqrt {c^2+d^2}}}+\frac {d \log \left (c+\sqrt {c^2+d^2}+d x-\sqrt {2} \sqrt {c+\sqrt {c^2+d^2}} \sqrt {c+d x}\right )}{2 \sqrt {2} \sqrt {c+\sqrt {c^2+d^2}}}-\frac {d \log \left (c+\sqrt {c^2+d^2}+d x+\sqrt {2} \sqrt {c+\sqrt {c^2+d^2}} \sqrt {c+d x}\right )}{2 \sqrt {2} \sqrt {c+\sqrt {c^2+d^2}}} \]

[Out]

1/2*d*arctanh((-2^(1/2)*(d*x+c)^(1/2)+(c+(c^2+d^2)^(1/2))^(1/2))/(c-(c^2+d^2)^(1/2))^(1/2))*2^(1/2)/(c-(c^2+d^
2)^(1/2))^(1/2)-1/2*d*arctanh((2^(1/2)*(d*x+c)^(1/2)+(c+(c^2+d^2)^(1/2))^(1/2))/(c-(c^2+d^2)^(1/2))^(1/2))*2^(
1/2)/(c-(c^2+d^2)^(1/2))^(1/2)+1/4*d*ln(c+d*x+(c^2+d^2)^(1/2)-2^(1/2)*(d*x+c)^(1/2)*(c+(c^2+d^2)^(1/2))^(1/2))
*2^(1/2)/(c+(c^2+d^2)^(1/2))^(1/2)-1/4*d*ln(c+d*x+(c^2+d^2)^(1/2)+2^(1/2)*(d*x+c)^(1/2)*(c+(c^2+d^2)^(1/2))^(1
/2))*2^(1/2)/(c+(c^2+d^2)^(1/2))^(1/2)

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {714, 1143, 648, 632, 212, 642} \[ \int \frac {\sqrt {c+d x}}{1+x^2} \, dx=\frac {d \text {arctanh}\left (\frac {\sqrt {\sqrt {c^2+d^2}+c}-\sqrt {2} \sqrt {c+d x}}{\sqrt {c-\sqrt {c^2+d^2}}}\right )}{\sqrt {2} \sqrt {c-\sqrt {c^2+d^2}}}-\frac {d \text {arctanh}\left (\frac {\sqrt {\sqrt {c^2+d^2}+c}+\sqrt {2} \sqrt {c+d x}}{\sqrt {c-\sqrt {c^2+d^2}}}\right )}{\sqrt {2} \sqrt {c-\sqrt {c^2+d^2}}}+\frac {d \log \left (-\sqrt {2} \sqrt {\sqrt {c^2+d^2}+c} \sqrt {c+d x}+\sqrt {c^2+d^2}+c+d x\right )}{2 \sqrt {2} \sqrt {\sqrt {c^2+d^2}+c}}-\frac {d \log \left (\sqrt {2} \sqrt {\sqrt {c^2+d^2}+c} \sqrt {c+d x}+\sqrt {c^2+d^2}+c+d x\right )}{2 \sqrt {2} \sqrt {\sqrt {c^2+d^2}+c}} \]

[In]

Int[Sqrt[c + d*x]/(1 + x^2),x]

[Out]

(d*ArcTanh[(Sqrt[c + Sqrt[c^2 + d^2]] - Sqrt[2]*Sqrt[c + d*x])/Sqrt[c - Sqrt[c^2 + d^2]]])/(Sqrt[2]*Sqrt[c - S
qrt[c^2 + d^2]]) - (d*ArcTanh[(Sqrt[c + Sqrt[c^2 + d^2]] + Sqrt[2]*Sqrt[c + d*x])/Sqrt[c - Sqrt[c^2 + d^2]]])/
(Sqrt[2]*Sqrt[c - Sqrt[c^2 + d^2]]) + (d*Log[c + Sqrt[c^2 + d^2] + d*x - Sqrt[2]*Sqrt[c + Sqrt[c^2 + d^2]]*Sqr
t[c + d*x]])/(2*Sqrt[2]*Sqrt[c + Sqrt[c^2 + d^2]]) - (d*Log[c + Sqrt[c^2 + d^2] + d*x + Sqrt[2]*Sqrt[c + Sqrt[
c^2 + d^2]]*Sqrt[c + d*x]])/(2*Sqrt[2]*Sqrt[c + Sqrt[c^2 + d^2]])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 714

Int[Sqrt[(d_) + (e_.)*(x_)]/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[2*e, Subst[Int[x^2/(c*d^2 + a*e^2 - 2*c*d
*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0]

Rule 1143

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r = Rt[2*q - b/
c, 2]}, Dist[1/(2*c*r), Int[x^(m - 1)/(q - r*x + x^2), x], x] - Dist[1/(2*c*r), Int[x^(m - 1)/(q + r*x + x^2),
x], x]]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && GeQ[m, 1] && LtQ[m, 3] && NegQ[b^2 - 4*a*c]

Rubi steps \begin{align*} \text {integral}& = (2 d) \text {Subst}\left (\int \frac {x^2}{c^2+d^2-2 c x^2+x^4} \, dx,x,\sqrt {c+d x}\right ) \\ & = \frac {d \text {Subst}\left (\int \frac {x}{\sqrt {c^2+d^2}-\sqrt {2} \sqrt {c+\sqrt {c^2+d^2}} x+x^2} \, dx,x,\sqrt {c+d x}\right )}{\sqrt {2} \sqrt {c+\sqrt {c^2+d^2}}}-\frac {d \text {Subst}\left (\int \frac {x}{\sqrt {c^2+d^2}+\sqrt {2} \sqrt {c+\sqrt {c^2+d^2}} x+x^2} \, dx,x,\sqrt {c+d x}\right )}{\sqrt {2} \sqrt {c+\sqrt {c^2+d^2}}} \\ & = \frac {1}{2} d \text {Subst}\left (\int \frac {1}{\sqrt {c^2+d^2}-\sqrt {2} \sqrt {c+\sqrt {c^2+d^2}} x+x^2} \, dx,x,\sqrt {c+d x}\right )+\frac {1}{2} d \text {Subst}\left (\int \frac {1}{\sqrt {c^2+d^2}+\sqrt {2} \sqrt {c+\sqrt {c^2+d^2}} x+x^2} \, dx,x,\sqrt {c+d x}\right )+\frac {d \text {Subst}\left (\int \frac {-\sqrt {2} \sqrt {c+\sqrt {c^2+d^2}}+2 x}{\sqrt {c^2+d^2}-\sqrt {2} \sqrt {c+\sqrt {c^2+d^2}} x+x^2} \, dx,x,\sqrt {c+d x}\right )}{2 \sqrt {2} \sqrt {c+\sqrt {c^2+d^2}}}-\frac {d \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {c+\sqrt {c^2+d^2}}+2 x}{\sqrt {c^2+d^2}+\sqrt {2} \sqrt {c+\sqrt {c^2+d^2}} x+x^2} \, dx,x,\sqrt {c+d x}\right )}{2 \sqrt {2} \sqrt {c+\sqrt {c^2+d^2}}} \\ & = \frac {d \log \left (c+\sqrt {c^2+d^2}+d x-\sqrt {2} \sqrt {c+\sqrt {c^2+d^2}} \sqrt {c+d x}\right )}{2 \sqrt {2} \sqrt {c+\sqrt {c^2+d^2}}}-\frac {d \log \left (c+\sqrt {c^2+d^2}+d x+\sqrt {2} \sqrt {c+\sqrt {c^2+d^2}} \sqrt {c+d x}\right )}{2 \sqrt {2} \sqrt {c+\sqrt {c^2+d^2}}}-d \text {Subst}\left (\int \frac {1}{2 \left (c-\sqrt {c^2+d^2}\right )-x^2} \, dx,x,-\sqrt {2} \sqrt {c+\sqrt {c^2+d^2}}+2 \sqrt {c+d x}\right )-d \text {Subst}\left (\int \frac {1}{2 \left (c-\sqrt {c^2+d^2}\right )-x^2} \, dx,x,\sqrt {2} \sqrt {c+\sqrt {c^2+d^2}}+2 \sqrt {c+d x}\right ) \\ & = \frac {d \tanh ^{-1}\left (\frac {\sqrt {c+\sqrt {c^2+d^2}}-\sqrt {2} \sqrt {c+d x}}{\sqrt {c-\sqrt {c^2+d^2}}}\right )}{\sqrt {2} \sqrt {c-\sqrt {c^2+d^2}}}-\frac {d \tanh ^{-1}\left (\frac {\sqrt {c+\sqrt {c^2+d^2}}+\sqrt {2} \sqrt {c+d x}}{\sqrt {c-\sqrt {c^2+d^2}}}\right )}{\sqrt {2} \sqrt {c-\sqrt {c^2+d^2}}}+\frac {d \log \left (c+\sqrt {c^2+d^2}+d x-\sqrt {2} \sqrt {c+\sqrt {c^2+d^2}} \sqrt {c+d x}\right )}{2 \sqrt {2} \sqrt {c+\sqrt {c^2+d^2}}}-\frac {d \log \left (c+\sqrt {c^2+d^2}+d x+\sqrt {2} \sqrt {c+\sqrt {c^2+d^2}} \sqrt {c+d x}\right )}{2 \sqrt {2} \sqrt {c+\sqrt {c^2+d^2}}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.19 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.26 \[ \int \frac {\sqrt {c+d x}}{1+x^2} \, dx=i \left (\sqrt {-c-i d} \arctan \left (\frac {\sqrt {c+d x}}{\sqrt {-c-i d}}\right )-\sqrt {-c+i d} \arctan \left (\frac {\sqrt {c+d x}}{\sqrt {-c+i d}}\right )\right ) \]

[In]

Integrate[Sqrt[c + d*x]/(1 + x^2),x]

[Out]

I*(Sqrt[-c - I*d]*ArcTan[Sqrt[c + d*x]/Sqrt[-c - I*d]] - Sqrt[-c + I*d]*ArcTan[Sqrt[c + d*x]/Sqrt[-c + I*d]])

Maple [A] (verified)

Time = 4.62 (sec) , antiderivative size = 255, normalized size of antiderivative = 0.81

method	result	size
pseudoelliptic	\(\frac {\frac {\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \left (\ln \left (d x +c +\sqrt {d x +c}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )-\ln \left (c +d x -\sqrt {d x +c}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )\right ) \left (c -\sqrt {c^{2}+d^{2}}\right )}{4}+d^{2} \left (\arctan \left (\frac {2 \sqrt {d x +c}-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )+\arctan \left (\frac {2 \sqrt {d x +c}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}\, d}\)	\(255\)
derivativedivides	\(2 d \left (\frac {\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \left (\sqrt {c^{2}+d^{2}}-c \right ) \left (\frac {\ln \left (\sqrt {d x +c}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-d x -c -\sqrt {c^{2}+d^{2}}\right )}{2}-\frac {\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \arctan \left (\frac {\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-2 \sqrt {d x +c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{4 d^{2}}-\frac {\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \left (\sqrt {c^{2}+d^{2}}-c \right ) \left (\frac {\ln \left (d x +c +\sqrt {d x +c}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}-\frac {\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \arctan \left (\frac {2 \sqrt {d x +c}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{4 d^{2}}\right )\)	\(330\)
default	\(2 d \left (\frac {\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \left (\sqrt {c^{2}+d^{2}}-c \right ) \left (\frac {\ln \left (\sqrt {d x +c}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-d x -c -\sqrt {c^{2}+d^{2}}\right )}{2}-\frac {\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \arctan \left (\frac {\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-2 \sqrt {d x +c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{4 d^{2}}-\frac {\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \left (\sqrt {c^{2}+d^{2}}-c \right ) \left (\frac {\ln \left (d x +c +\sqrt {d x +c}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}-\frac {\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \arctan \left (\frac {2 \sqrt {d x +c}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{4 d^{2}}\right )\)	\(330\)

[In]

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

[Out]

(1/4*(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(ln(d*x+c+(d*x+c)^(1/2)*(2*(c^2+d^2)^(1/2)+2*
c)^(1/2)+(c^2+d^2)^(1/2))-ln(c+d*x-(d*x+c)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2)))*(c-(c^2+d^2)^
(1/2))+d^2*(arctan((2*(d*x+c)^(1/2)-(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))+arctan((2*(d
*x+c)^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)/d

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.63 \[ \int \frac {\sqrt {c+d x}}{1+x^2} \, dx=\frac {1}{2} \, \sqrt {-c + \sqrt {-d^{2}}} \log \left (\sqrt {d x + c} d + \sqrt {-d^{2}} \sqrt {-c + \sqrt {-d^{2}}}\right ) - \frac {1}{2} \, \sqrt {-c + \sqrt {-d^{2}}} \log \left (\sqrt {d x + c} d - \sqrt {-d^{2}} \sqrt {-c + \sqrt {-d^{2}}}\right ) - \frac {1}{2} \, \sqrt {-c - \sqrt {-d^{2}}} \log \left (\sqrt {d x + c} d + \sqrt {-d^{2}} \sqrt {-c - \sqrt {-d^{2}}}\right ) + \frac {1}{2} \, \sqrt {-c - \sqrt {-d^{2}}} \log \left (\sqrt {d x + c} d - \sqrt {-d^{2}} \sqrt {-c - \sqrt {-d^{2}}}\right ) \]

[In]

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

[Out]

1/2*sqrt(-c + sqrt(-d^2))*log(sqrt(d*x + c)*d + sqrt(-d^2)*sqrt(-c + sqrt(-d^2))) - 1/2*sqrt(-c + sqrt(-d^2))*
log(sqrt(d*x + c)*d - sqrt(-d^2)*sqrt(-c + sqrt(-d^2))) - 1/2*sqrt(-c - sqrt(-d^2))*log(sqrt(d*x + c)*d + sqrt
(-d^2)*sqrt(-c - sqrt(-d^2))) + 1/2*sqrt(-c - sqrt(-d^2))*log(sqrt(d*x + c)*d - sqrt(-d^2)*sqrt(-c - sqrt(-d^2
)))

Sympy [F]

\[ \int \frac {\sqrt {c+d x}}{1+x^2} \, dx=\int \frac {\sqrt {c + d x}}{x^{2} + 1}\, dx \]

[In]

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

[Out]

Integral(sqrt(c + d*x)/(x**2 + 1), x)

Maxima [F]

\[ \int \frac {\sqrt {c+d x}}{1+x^2} \, dx=\int { \frac {\sqrt {d x + c}}{x^{2} + 1} \,d x } \]

[In]

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

[Out]

integrate(sqrt(d*x + c)/(x^2 + 1), x)

Giac [F(-1)]

Timed out. \[ \int \frac {\sqrt {c+d x}}{1+x^2} \, dx=\text {Timed out} \]

[In]

integrate((d*x+c)^(1/2)/(x^2+1),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.42 \[ \int \frac {\sqrt {c+d x}}{1+x^2} \, dx=-\mathrm {atan}\left (\frac {2\,c\,\sqrt {-\frac {c}{4}-\frac {d\,1{}\mathrm {i}}{4}}\,\sqrt {c+d\,x}-d\,\sqrt {-\frac {c}{4}-\frac {d\,1{}\mathrm {i}}{4}}\,\sqrt {c+d\,x}\,2{}\mathrm {i}}{c^2+d^2}\right )\,\sqrt {-\frac {c}{4}-\frac {d\,1{}\mathrm {i}}{4}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {2\,c\,\sqrt {-\frac {c}{4}+\frac {d\,1{}\mathrm {i}}{4}}\,\sqrt {c+d\,x}+d\,\sqrt {-\frac {c}{4}+\frac {d\,1{}\mathrm {i}}{4}}\,\sqrt {c+d\,x}\,2{}\mathrm {i}}{c^2+d^2}\right )\,\sqrt {-\frac {c}{4}+\frac {d\,1{}\mathrm {i}}{4}}\,2{}\mathrm {i} \]

[In]

int((c + d*x)^(1/2)/(x^2 + 1),x)

[Out]

atan((2*c*((d*1i)/4 - c/4)^(1/2)*(c + d*x)^(1/2) + d*((d*1i)/4 - c/4)^(1/2)*(c + d*x)^(1/2)*2i)/(c^2 + d^2))*(
(d*1i)/4 - c/4)^(1/2)*2i - atan((2*c*(- c/4 - (d*1i)/4)^(1/2)*(c + d*x)^(1/2) - d*(- c/4 - (d*1i)/4)^(1/2)*(c
+ d*x)^(1/2)*2i)/(c^2 + d^2))*(- c/4 - (d*1i)/4)^(1/2)*2i