3.614 \(\int \frac{\sqrt{d+e x}}{a-c x^2} \, dx\)

Optimal. Leaf size=134 \[ \frac{\sqrt{\sqrt{a} e+\sqrt{c} d} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{\sqrt{a} c^{3/4}}-\frac{\sqrt{\sqrt{c} d-\sqrt{a} e} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{\sqrt{a} c^{3/4}} \]

[Out]

-((Sqrt[Sqrt[c]*d - Sqrt[a]*e]*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]])/(Sqrt[a]*c^(3/4))
) + (Sqrt[Sqrt[c]*d + Sqrt[a]*e]*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]])/(Sqrt[a]*c^(3/4
))

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Rubi [A]  time = 0.108612, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {700, 1130, 208} \[ \frac{\sqrt{\sqrt{a} e+\sqrt{c} d} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{\sqrt{a} c^{3/4}}-\frac{\sqrt{\sqrt{c} d-\sqrt{a} e} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{\sqrt{a} c^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[d + e*x]/(a - c*x^2),x]

[Out]

-((Sqrt[Sqrt[c]*d - Sqrt[a]*e]*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]])/(Sqrt[a]*c^(3/4))
) + (Sqrt[Sqrt[c]*d + Sqrt[a]*e]*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]])/(Sqrt[a]*c^(3/4
))

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 1130

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

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\sqrt{d+e x}}{a-c x^2} \, dx &=(2 e) \operatorname{Subst}\left (\int \frac{x^2}{-c d^2+a e^2+2 c d x^2-c x^4} \, dx,x,\sqrt{d+e x}\right )\\ &=-\left (\left (\frac{\sqrt{c} d}{\sqrt{a}}-e\right ) \operatorname{Subst}\left (\int \frac{1}{c d-\sqrt{a} \sqrt{c} e-c x^2} \, dx,x,\sqrt{d+e x}\right )\right )+\left (\frac{\sqrt{c} d}{\sqrt{a}}+e\right ) \operatorname{Subst}\left (\int \frac{1}{c d+\sqrt{a} \sqrt{c} e-c x^2} \, dx,x,\sqrt{d+e x}\right )\\ &=-\frac{\sqrt{\sqrt{c} d-\sqrt{a} e} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{\sqrt{a} c^{3/4}}+\frac{\sqrt{\sqrt{c} d+\sqrt{a} e} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d+\sqrt{a} e}}\right )}{\sqrt{a} c^{3/4}}\\ \end{align*}

Mathematica [A]  time = 0.0605512, size = 125, normalized size = 0.93 \[ \frac{\sqrt{\sqrt{a} e+\sqrt{c} d} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )-\sqrt{\sqrt{c} d-\sqrt{a} e} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{\sqrt{a} c^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[Sqrt[c]*d - Sqrt[a]*e]*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]]) + Sqrt[Sqrt[c]*d
+ Sqrt[a]*e]*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]])/(Sqrt[a]*c^(3/4))

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Maple [B]  time = 0.207, size = 203, normalized size = 1.5 \begin{align*}{ced\arctan \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( -cd+\sqrt{ac{e}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{ac{e}^{2}}}}{\frac{1}{\sqrt{ \left ( -cd+\sqrt{ac{e}^{2}} \right ) c}}}}-{e\arctan \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( -cd+\sqrt{ac{e}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{ \left ( -cd+\sqrt{ac{e}^{2}} \right ) c}}}}+{ced{\it Artanh} \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( cd+\sqrt{ac{e}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{ac{e}^{2}}}}{\frac{1}{\sqrt{ \left ( cd+\sqrt{ac{e}^{2}} \right ) c}}}}+{e{\it Artanh} \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( cd+\sqrt{ac{e}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{ \left ( cd+\sqrt{ac{e}^{2}} \right ) c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-integrate(sqrt(e*x + d)/(c*x^2 - a), x)

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Fricas [B]  time = 1.9826, size = 725, normalized size = 5.41 \begin{align*} \frac{1}{2} \, \sqrt{\frac{a c \sqrt{\frac{e^{2}}{a c^{3}}} + d}{a c}} \log \left (a c^{2} \sqrt{\frac{a c \sqrt{\frac{e^{2}}{a c^{3}}} + d}{a c}} \sqrt{\frac{e^{2}}{a c^{3}}} + \sqrt{e x + d} e\right ) - \frac{1}{2} \, \sqrt{\frac{a c \sqrt{\frac{e^{2}}{a c^{3}}} + d}{a c}} \log \left (-a c^{2} \sqrt{\frac{a c \sqrt{\frac{e^{2}}{a c^{3}}} + d}{a c}} \sqrt{\frac{e^{2}}{a c^{3}}} + \sqrt{e x + d} e\right ) - \frac{1}{2} \, \sqrt{-\frac{a c \sqrt{\frac{e^{2}}{a c^{3}}} - d}{a c}} \log \left (a c^{2} \sqrt{-\frac{a c \sqrt{\frac{e^{2}}{a c^{3}}} - d}{a c}} \sqrt{\frac{e^{2}}{a c^{3}}} + \sqrt{e x + d} e\right ) + \frac{1}{2} \, \sqrt{-\frac{a c \sqrt{\frac{e^{2}}{a c^{3}}} - d}{a c}} \log \left (-a c^{2} \sqrt{-\frac{a c \sqrt{\frac{e^{2}}{a c^{3}}} - d}{a c}} \sqrt{\frac{e^{2}}{a c^{3}}} + \sqrt{e x + d} e\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt((a*c*sqrt(e^2/(a*c^3)) + d)/(a*c))*log(a*c^2*sqrt((a*c*sqrt(e^2/(a*c^3)) + d)/(a*c))*sqrt(e^2/(a*c^3)
) + sqrt(e*x + d)*e) - 1/2*sqrt((a*c*sqrt(e^2/(a*c^3)) + d)/(a*c))*log(-a*c^2*sqrt((a*c*sqrt(e^2/(a*c^3)) + d)
/(a*c))*sqrt(e^2/(a*c^3)) + sqrt(e*x + d)*e) - 1/2*sqrt(-(a*c*sqrt(e^2/(a*c^3)) - d)/(a*c))*log(a*c^2*sqrt(-(a
*c*sqrt(e^2/(a*c^3)) - d)/(a*c))*sqrt(e^2/(a*c^3)) + sqrt(e*x + d)*e) + 1/2*sqrt(-(a*c*sqrt(e^2/(a*c^3)) - d)/
(a*c))*log(-a*c^2*sqrt(-(a*c*sqrt(e^2/(a*c^3)) - d)/(a*c))*sqrt(e^2/(a*c^3)) + sqrt(e*x + d)*e)

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Sympy [A]  time = 5.73174, size = 76, normalized size = 0.57 \begin{align*} - 2 e \operatorname{RootSum}{\left (256 t^{4} a^{2} c^{3} e^{4} - 32 t^{2} a c^{2} d e^{2} - a e^{2} + c d^{2}, \left ( t \mapsto t \log{\left (- 64 t^{3} a c^{2} e^{2} + 4 t c d + \sqrt{d + e x} \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out