### 3.866 $$\int \frac{\sqrt{c d^2-c e^2 x^2}}{(d+e x)^{3/2}} \, dx$$

Optimal. Leaf size=99 $\frac{2 \sqrt{c d^2-c e^2 x^2}}{e \sqrt{d+e x}}-\frac{2 \sqrt{2} \sqrt{c} \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{c d^2-c e^2 x^2}}{\sqrt{2} \sqrt{c} \sqrt{d} \sqrt{d+e x}}\right )}{e}$

[Out]

(2*Sqrt[c*d^2 - c*e^2*x^2])/(e*Sqrt[d + e*x]) - (2*Sqrt[2]*Sqrt[c]*Sqrt[d]*ArcTanh[Sqrt[c*d^2 - c*e^2*x^2]/(Sq
rt[2]*Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])])/e

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Rubi [A]  time = 0.0553575, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 29, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.103, Rules used = {665, 661, 208} $\frac{2 \sqrt{c d^2-c e^2 x^2}}{e \sqrt{d+e x}}-\frac{2 \sqrt{2} \sqrt{c} \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{c d^2-c e^2 x^2}}{\sqrt{2} \sqrt{c} \sqrt{d} \sqrt{d+e x}}\right )}{e}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sqrt[c*d^2 - c*e^2*x^2]/(d + e*x)^(3/2),x]

[Out]

(2*Sqrt[c*d^2 - c*e^2*x^2])/(e*Sqrt[d + e*x]) - (2*Sqrt[2]*Sqrt[c]*Sqrt[d]*ArcTanh[Sqrt[c*d^2 - c*e^2*x^2]/(Sq
rt[2]*Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])])/e

Rule 665

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + c*x^2)^p)/(
e*(m + 2*p + 1)), x] - Dist[(2*c*d*p)/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1), x], x] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[
m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 661

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(2*c*d + e^2*x^2
), x], x, Sqrt[a + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0]

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

Mathematica [A]  time = 0.0925152, size = 98, normalized size = 0.99 $\frac{2 \sqrt{c \left (d^2-e^2 x^2\right )} \left (\frac{1}{\sqrt{d+e x}}-\frac{\sqrt{2} \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{\sqrt{2} \sqrt{d} \sqrt{d+e x}}\right )}{\sqrt{d^2-e^2 x^2}}\right )}{e}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sqrt[c*d^2 - c*e^2*x^2]/(d + e*x)^(3/2),x]

[Out]

(2*Sqrt[c*(d^2 - e^2*x^2)]*(1/Sqrt[d + e*x] - (Sqrt[2]*Sqrt[d]*ArcTanh[Sqrt[d^2 - e^2*x^2]/(Sqrt[2]*Sqrt[d]*Sq
rt[d + e*x])])/Sqrt[d^2 - e^2*x^2]))/e

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Maple [A]  time = 0.243, size = 97, normalized size = 1. \begin{align*} -2\,{\frac{\sqrt{-c \left ({e}^{2}{x}^{2}-{d}^{2} \right ) }}{\sqrt{ex+d}\sqrt{- \left ( ex-d \right ) c}e\sqrt{cd}} \left ( cd\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{- \left ( ex-d \right ) c}\sqrt{2}}{\sqrt{cd}}} \right ) -\sqrt{- \left ( ex-d \right ) c}\sqrt{cd} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((-c*e^2*x^2+c*d^2)^(1/2)/(e*x+d)^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-c e^{2} x^{2} + c d^{2}}}{{\left (e x + d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c*e^2*x^2+c*d^2)^(1/2)/(e*x+d)^(3/2),x, algorithm="maxima")

[Out]

integrate(sqrt(-c*e^2*x^2 + c*d^2)/(e*x + d)^(3/2), x)

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Fricas [A]  time = 2.30145, size = 533, normalized size = 5.38 \begin{align*} \left [\frac{\sqrt{2} \sqrt{c d}{\left (e x + d\right )} \log \left (-\frac{c e^{2} x^{2} - 2 \, c d e x - 3 \, c d^{2} + 2 \, \sqrt{2} \sqrt{-c e^{2} x^{2} + c d^{2}} \sqrt{c d} \sqrt{e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + 2 \, \sqrt{-c e^{2} x^{2} + c d^{2}} \sqrt{e x + d}}{e^{2} x + d e}, -\frac{2 \,{\left (\sqrt{2} \sqrt{-c d}{\left (e x + d\right )} \arctan \left (\frac{\sqrt{2} \sqrt{-c e^{2} x^{2} + c d^{2}} \sqrt{-c d} \sqrt{e x + d}}{c e^{2} x^{2} - c d^{2}}\right ) - \sqrt{-c e^{2} x^{2} + c d^{2}} \sqrt{e x + d}\right )}}{e^{2} x + d e}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c*e^2*x^2+c*d^2)^(1/2)/(e*x+d)^(3/2),x, algorithm="fricas")

[Out]

[(sqrt(2)*sqrt(c*d)*(e*x + d)*log(-(c*e^2*x^2 - 2*c*d*e*x - 3*c*d^2 + 2*sqrt(2)*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(
c*d)*sqrt(e*x + d))/(e^2*x^2 + 2*d*e*x + d^2)) + 2*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e*x + d))/(e^2*x + d*e), -2*(
sqrt(2)*sqrt(-c*d)*(e*x + d)*arctan(sqrt(2)*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(-c*d)*sqrt(e*x + d)/(c*e^2*x^2 - c*d
^2)) - sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e*x + d))/(e^2*x + d*e)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{- c \left (- d + e x\right ) \left (d + e x\right )}}{\left (d + e x\right )^{\frac{3}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c*e**2*x**2+c*d**2)**(1/2)/(e*x+d)**(3/2),x)

[Out]

Integral(sqrt(-c*(-d + e*x)*(d + e*x))/(d + e*x)**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-c e^{2} x^{2} + c d^{2}}}{{\left (e x + d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c*e^2*x^2+c*d^2)^(1/2)/(e*x+d)^(3/2),x, algorithm="giac")

[Out]

integrate(sqrt(-c*e^2*x^2 + c*d^2)/(e*x + d)^(3/2), x)