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

Optimal. Leaf size=316 \[ \frac{d \log \left (-\sqrt{2} \sqrt{\sqrt{c^2+d^2}+c} \sqrt{c+d x}+\sqrt{c^2+d^2}+c+d x\right )}{2 \sqrt{2} \sqrt{\sqrt{c^2+d^2}+c}}-\frac{d \log \left (\sqrt{2} \sqrt{\sqrt{c^2+d^2}+c} \sqrt{c+d x}+\sqrt{c^2+d^2}+c+d x\right )}{2 \sqrt{2} \sqrt{\sqrt{c^2+d^2}+c}}+\frac{d \tanh ^{-1}\left (\frac{\sqrt{\sqrt{c^2+d^2}+c}-\sqrt{2} \sqrt{c+d x}}{\sqrt{c-\sqrt{c^2+d^2}}}\right )}{\sqrt{2} \sqrt{c-\sqrt{c^2+d^2}}}-\frac{d \tanh ^{-1}\left (\frac{\sqrt{\sqrt{c^2+d^2}+c}+\sqrt{2} \sqrt{c+d x}}{\sqrt{c-\sqrt{c^2+d^2}}}\right )}{\sqrt{2} \sqrt{c-\sqrt{c^2+d^2}}} \]

[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]])

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Rubi [A]  time = 0.346685, antiderivative size = 316, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {700, 1129, 634, 618, 206, 628} \[ \frac{d \log \left (-\sqrt{2} \sqrt{\sqrt{c^2+d^2}+c} \sqrt{c+d x}+\sqrt{c^2+d^2}+c+d x\right )}{2 \sqrt{2} \sqrt{\sqrt{c^2+d^2}+c}}-\frac{d \log \left (\sqrt{2} \sqrt{\sqrt{c^2+d^2}+c} \sqrt{c+d x}+\sqrt{c^2+d^2}+c+d x\right )}{2 \sqrt{2} \sqrt{\sqrt{c^2+d^2}+c}}+\frac{d \tanh ^{-1}\left (\frac{\sqrt{\sqrt{c^2+d^2}+c}-\sqrt{2} \sqrt{c+d x}}{\sqrt{c-\sqrt{c^2+d^2}}}\right )}{\sqrt{2} \sqrt{c-\sqrt{c^2+d^2}}}-\frac{d \tanh ^{-1}\left (\frac{\sqrt{\sqrt{c^2+d^2}+c}+\sqrt{2} \sqrt{c+d x}}{\sqrt{c-\sqrt{c^2+d^2}}}\right )}{\sqrt{2} \sqrt{c-\sqrt{c^2+d^2}}} \]

Antiderivative was successfully verified.

[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 700

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 1129

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]

Rule 634

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 618

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 206

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

Rule 628

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]

Rubi steps

\begin{align*} \int \frac{\sqrt{c+d x}}{1+x^2} \, dx &=(2 d) \operatorname{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 \operatorname{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 \operatorname{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 \operatorname{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 \operatorname{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 \operatorname{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 \operatorname{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 \operatorname{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 \operatorname{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]  time = 0.0424746, size = 75, normalized size = 0.24 \[ i \sqrt{c+i d} \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c+i d}}\right )-i \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c-i d}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c + d*x]/(1 + x^2),x]

[Out]

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

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Maple [B]  time = 0.209, size = 570, normalized size = 1.8 \begin{align*}{\frac{c}{4\,d}\sqrt{2\,\sqrt{{c}^{2}+{d}^{2}}+2\,c}\ln \left ( dx+c+\sqrt{dx+c}\sqrt{2\,\sqrt{{c}^{2}+{d}^{2}}+2\,c}+\sqrt{{c}^{2}+{d}^{2}} \right ) }-{\frac{{c}^{2}}{d}\arctan \left ({ \left ( 2\,\sqrt{dx+c}+\sqrt{2\,\sqrt{{c}^{2}+{d}^{2}}+2\,c} \right ){\frac{1}{\sqrt{2\,\sqrt{{c}^{2}+{d}^{2}}-2\,c}}}} \right ){\frac{1}{\sqrt{2\,\sqrt{{c}^{2}+{d}^{2}}-2\,c}}}}-{\frac{1}{4\,d}\sqrt{2\,\sqrt{{c}^{2}+{d}^{2}}+2\,c}\sqrt{{c}^{2}+{d}^{2}}\ln \left ( dx+c+\sqrt{dx+c}\sqrt{2\,\sqrt{{c}^{2}+{d}^{2}}+2\,c}+\sqrt{{c}^{2}+{d}^{2}} \right ) }+{\frac{{c}^{2}+{d}^{2}}{d}\arctan \left ({ \left ( 2\,\sqrt{dx+c}+\sqrt{2\,\sqrt{{c}^{2}+{d}^{2}}+2\,c} \right ){\frac{1}{\sqrt{2\,\sqrt{{c}^{2}+{d}^{2}}-2\,c}}}} \right ){\frac{1}{\sqrt{2\,\sqrt{{c}^{2}+{d}^{2}}-2\,c}}}}-{\frac{c}{4\,d}\sqrt{2\,\sqrt{{c}^{2}+{d}^{2}}+2\,c}\ln \left ( \sqrt{dx+c}\sqrt{2\,\sqrt{{c}^{2}+{d}^{2}}+2\,c}-dx-c-\sqrt{{c}^{2}+{d}^{2}} \right ) }+{\frac{{c}^{2}}{d}\arctan \left ({ \left ( \sqrt{2\,\sqrt{{c}^{2}+{d}^{2}}+2\,c}-2\,\sqrt{dx+c} \right ){\frac{1}{\sqrt{2\,\sqrt{{c}^{2}+{d}^{2}}-2\,c}}}} \right ){\frac{1}{\sqrt{2\,\sqrt{{c}^{2}+{d}^{2}}-2\,c}}}}+{\frac{1}{4\,d}\sqrt{2\,\sqrt{{c}^{2}+{d}^{2}}+2\,c}\sqrt{{c}^{2}+{d}^{2}}\ln \left ( \sqrt{dx+c}\sqrt{2\,\sqrt{{c}^{2}+{d}^{2}}+2\,c}-dx-c-\sqrt{{c}^{2}+{d}^{2}} \right ) }-{\frac{{c}^{2}+{d}^{2}}{d}\arctan \left ({ \left ( \sqrt{2\,\sqrt{{c}^{2}+{d}^{2}}+2\,c}-2\,\sqrt{dx+c} \right ){\frac{1}{\sqrt{2\,\sqrt{{c}^{2}+{d}^{2}}-2\,c}}}} \right ){\frac{1}{\sqrt{2\,\sqrt{{c}^{2}+{d}^{2}}-2\,c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d x + c}}{x^{2} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(d*x + c)/(x^2 + 1), x)

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Fricas [B]  time = 2.98742, size = 2176, normalized size = 6.89 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*sqrt(2)*(c^2 + d^2)^(3/4)*sqrt(d^2)*sqrt((c^2 + d^2 + sqrt(c^2 + d^2)*c)/d^2)*arctan(-(sqrt(2)*(c^2 +
d^2)^(3/4)*sqrt(d^2)*sqrt(d*x + c)*d*sqrt((c^2 + d^2 + sqrt(c^2 + d^2)*c)/d^2) - sqrt(2)*(c^2 + d^2)^(3/4)*sqr
t(d^2)*sqrt((sqrt(2)*(c^2 + d^2)^(3/4)*sqrt(d*x + c)*d^3*sqrt((c^2 + d^2 + sqrt(c^2 + d^2)*c)/d^2) + c^3*d^2 +
c*d^4 + (c^2*d^3 + d^5)*x + (c^2*d^2 + d^4)*sqrt(c^2 + d^2))/(c^2 + d^2))*sqrt((c^2 + d^2 + sqrt(c^2 + d^2)*c
)/d^2) + (c^2 + d^2)^(3/2)*sqrt(d^2) + (c^3 + c*d^2)*sqrt(d^2))/(c^2*d^2 + d^4)) + 4*sqrt(2)*(c^2 + d^2)^(3/4)
*sqrt(d^2)*sqrt((c^2 + d^2 + sqrt(c^2 + d^2)*c)/d^2)*arctan(-(sqrt(2)*(c^2 + d^2)^(3/4)*sqrt(d^2)*sqrt(d*x + c
)*d*sqrt((c^2 + d^2 + sqrt(c^2 + d^2)*c)/d^2) - sqrt(2)*(c^2 + d^2)^(3/4)*sqrt(d^2)*sqrt(-(sqrt(2)*(c^2 + d^2)
^(3/4)*sqrt(d*x + c)*d^3*sqrt((c^2 + d^2 + sqrt(c^2 + d^2)*c)/d^2) - c^3*d^2 - c*d^4 - (c^2*d^3 + d^5)*x - (c^
2*d^2 + d^4)*sqrt(c^2 + d^2))/(c^2 + d^2))*sqrt((c^2 + d^2 + sqrt(c^2 + d^2)*c)/d^2) - (c^2 + d^2)^(3/2)*sqrt(
d^2) - (c^3 + c*d^2)*sqrt(d^2))/(c^2*d^2 + d^4)) + sqrt(2)*(c^2 + d^2 - sqrt(c^2 + d^2)*c)*(c^2 + d^2)^(1/4)*s
qrt((c^2 + d^2 + sqrt(c^2 + d^2)*c)/d^2)*log((sqrt(2)*(c^2 + d^2)^(3/4)*sqrt(d*x + c)*d^3*sqrt((c^2 + d^2 + sq
rt(c^2 + d^2)*c)/d^2) + c^3*d^2 + c*d^4 + (c^2*d^3 + d^5)*x + (c^2*d^2 + d^4)*sqrt(c^2 + d^2))/(c^2 + d^2)) -
sqrt(2)*(c^2 + d^2 - sqrt(c^2 + d^2)*c)*(c^2 + d^2)^(1/4)*sqrt((c^2 + d^2 + sqrt(c^2 + d^2)*c)/d^2)*log(-(sqrt
(2)*(c^2 + d^2)^(3/4)*sqrt(d*x + c)*d^3*sqrt((c^2 + d^2 + sqrt(c^2 + d^2)*c)/d^2) - c^3*d^2 - c*d^4 - (c^2*d^3
+ d^5)*x - (c^2*d^2 + d^4)*sqrt(c^2 + d^2))/(c^2 + d^2)))/(c^2 + d^2)

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Sympy [A]  time = 3.88021, size = 53, normalized size = 0.17 \begin{align*} 2 d \operatorname{RootSum}{\left (256 t^{4} d^{4} + 32 t^{2} c d^{2} + c^{2} + d^{2}, \left ( t \mapsto t \log{\left (64 t^{3} d^{2} + 4 t c + \sqrt{c + d x} \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*d*RootSum(256*_t**4*d**4 + 32*_t**2*c*d**2 + c**2 + d**2, Lambda(_t, _t*log(64*_t**3*d**2 + 4*_t*c + sqrt(c
+ d*x))))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d x + c}}{x^{2} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(d*x + c)/(x^2 + 1), x)