\(\int \frac {d+e x}{\sqrt {a+c x^2}} \, dx\) [564]

Optimal result

Integrand size = 17, antiderivative size = 43 \[ \int \frac {d+e x}{\sqrt {a+c x^2}} \, dx=\frac {e \sqrt {a+c x^2}}{c}+\frac {d \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{\sqrt {c}} \]

[Out]

d*arctanh(x*c^(1/2)/(c*x^2+a)^(1/2))/c^(1/2)+e*(c*x^2+a)^(1/2)/c

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {655, 223, 212} \[ \int \frac {d+e x}{\sqrt {a+c x^2}} \, dx=\frac {d \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{\sqrt {c}}+\frac {e \sqrt {a+c x^2}}{c} \]

[In]

Int[(d + e*x)/Sqrt[a + c*x^2],x]

[Out]

(e*Sqrt[a + c*x^2])/c + (d*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/Sqrt[c]

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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 655

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[e*((a + c*x^2)^(p + 1)/(2*c*(p + 1))),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.07 \[ \int \frac {d+e x}{\sqrt {a+c x^2}} \, dx=\frac {e \sqrt {a+c x^2}}{c}-\frac {d \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{\sqrt {c}} \]

[In]

Integrate[(d + e*x)/Sqrt[a + c*x^2],x]

[Out]

(e*Sqrt[a + c*x^2])/c - (d*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/Sqrt[c]

Maple [A] (verified)

Time = 2.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86

method	result	size
default	\(\frac {d \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{\sqrt {c}}+\frac {e \sqrt {c \,x^{2}+a}}{c}\)	\(37\)
risch	\(\frac {d \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{\sqrt {c}}+\frac {e \sqrt {c \,x^{2}+a}}{c}\)	\(37\)

[In]

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

[Out]

d*ln(c^(1/2)*x+(c*x^2+a)^(1/2))/c^(1/2)+e*(c*x^2+a)^(1/2)/c

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 2.14 \[ \int \frac {d+e x}{\sqrt {a+c x^2}} \, dx=\left [\frac {\sqrt {c} d \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, \sqrt {c x^{2} + a} e}{2 \, c}, -\frac {\sqrt {-c} d \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - \sqrt {c x^{2} + a} e}{c}\right ] \]

[In]

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

[Out]

[1/2*(sqrt(c)*d*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*sqrt(c*x^2 + a)*e)/c, -(sqrt(-c)*d*arctan(
sqrt(-c)*x/sqrt(c*x^2 + a)) - sqrt(c*x^2 + a)*e)/c]

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.65 \[ \int \frac {d+e x}{\sqrt {a+c x^2}} \, dx=\begin {cases} d \left (\begin {cases} \frac {\log {\left (2 \sqrt {c} \sqrt {a + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {c x^{2}}} & \text {otherwise} \end {cases}\right ) + \frac {e \sqrt {a + c x^{2}}}{c} & \text {for}\: c \neq 0 \\\frac {d x + \frac {e x^{2}}{2}}{\sqrt {a}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((d*Piecewise((log(2*sqrt(c)*sqrt(a + c*x**2) + 2*c*x)/sqrt(c), Ne(a, 0)), (x*log(x)/sqrt(c*x**2), Tr
ue)) + e*sqrt(a + c*x**2)/c, Ne(c, 0)), ((d*x + e*x**2/2)/sqrt(a), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.67 \[ \int \frac {d+e x}{\sqrt {a+c x^2}} \, dx=\frac {d \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {c}} + \frac {\sqrt {c x^{2} + a} e}{c} \]

[In]

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

[Out]

d*arcsinh(c*x/sqrt(a*c))/sqrt(c) + sqrt(c*x^2 + a)*e/c

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.91 \[ \int \frac {d+e x}{\sqrt {a+c x^2}} \, dx=-\frac {d \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{\sqrt {c}} + \frac {\sqrt {c x^{2} + a} e}{c} \]

[In]

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

[Out]

-d*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/sqrt(c) + sqrt(c*x^2 + a)*e/c

Mupad [B] (verification not implemented)

Time = 9.41 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84 \[ \int \frac {d+e x}{\sqrt {a+c x^2}} \, dx=\frac {e\,\sqrt {c\,x^2+a}}{c}+\frac {d\,\ln \left (\sqrt {c}\,x+\sqrt {c\,x^2+a}\right )}{\sqrt {c}} \]

[In]

int((d + e*x)/(a + c*x^2)^(1/2),x)

[Out]

(e*(a + c*x^2)^(1/2))/c + (d*log(c^(1/2)*x + (a + c*x^2)^(1/2)))/c^(1/2)