### 3.890 $$\int \frac{\sqrt{d+e x}}{(c d^2-c e^2 x^2)^{3/2}} \, dx$$

Optimal. Leaf size=104 $\frac{\sqrt{d+e x}}{c d e \sqrt{c d^2-c e^2 x^2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c d^2-c e^2 x^2}}{\sqrt{2} \sqrt{c} \sqrt{d} \sqrt{d+e x}}\right )}{\sqrt{2} c^{3/2} d^{3/2} e}$

[Out]

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

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Rubi [A]  time = 0.0506586, antiderivative size = 104, 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 = {667, 661, 208} $\frac{\sqrt{d+e x}}{c d e \sqrt{c d^2-c e^2 x^2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c d^2-c e^2 x^2}}{\sqrt{2} \sqrt{c} \sqrt{d} \sqrt{d+e x}}\right )}{\sqrt{2} c^{3/2} d^{3/2} e}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 667

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(d*(d + e*x)^m*(a + c*x^2)^(p + 1)
)/(2*a*e*(p + 1)), x] + Dist[(d*(m + 2*p + 2))/(2*a*(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] && LtQ[p, -1] && LtQ[0, m, 1] && 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{d+e x}}{\left (c d^2-c e^2 x^2\right )^{3/2}} \, dx &=\frac{\sqrt{d+e x}}{c d e \sqrt{c d^2-c e^2 x^2}}+\frac{\int \frac{1}{\sqrt{d+e x} \sqrt{c d^2-c e^2 x^2}} \, dx}{2 c d}\\ &=\frac{\sqrt{d+e x}}{c d e \sqrt{c d^2-c e^2 x^2}}+\frac{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 )}{c d}\\ &=\frac{\sqrt{d+e x}}{c d e \sqrt{c d^2-c e^2 x^2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c d^2-c e^2 x^2}}{\sqrt{2} \sqrt{c} \sqrt{d} \sqrt{d+e x}}\right )}{\sqrt{2} c^{3/2} d^{3/2} e}\\ \end{align*}

Mathematica [A]  time = 0.0535869, size = 110, normalized size = 1.06 $\frac{2 \sqrt{d} \sqrt{d+e x}-\sqrt{2} \sqrt{d^2-e^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{\sqrt{2} \sqrt{d} \sqrt{d+e x}}\right )}{2 c d^{3/2} e \sqrt{c \left (d^2-e^2 x^2\right )}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.169, size = 98, normalized size = 0.9 \begin{align*}{\frac{1}{2\,{c}^{2} \left ( ex-d \right ) ed}\sqrt{-c \left ({e}^{2}{x}^{2}-{d}^{2} \right ) } \left ( \sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{- \left ( ex-d \right ) c}{\frac{1}{\sqrt{cd}}}} \right ) \sqrt{- \left ( ex-d \right ) c}-2\,\sqrt{cd} \right ){\frac{1}{\sqrt{ex+d}}}{\frac{1}{\sqrt{cd}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.2346, size = 609, normalized size = 5.86 \begin{align*} \left [\frac{\sqrt{2}{\left (e^{2} x^{2} - d^{2}\right )} \sqrt{c d} \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 ) - 4 \, \sqrt{-c e^{2} x^{2} + c d^{2}} \sqrt{e x + d} d}{4 \,{\left (c^{2} d^{2} e^{3} x^{2} - c^{2} d^{4} e\right )}}, -\frac{\sqrt{2}{\left (e^{2} x^{2} - d^{2}\right )} \sqrt{-c d} \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 ) + 2 \, \sqrt{-c e^{2} x^{2} + c d^{2}} \sqrt{e x + d} d}{2 \,{\left (c^{2} d^{2} e^{3} x^{2} - c^{2} d^{4} e\right )}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(sqrt(2)*(e^2*x^2 - d^2)*sqrt(c*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)) - 4*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e*x + d)*d)/(c^2*d^
2*e^3*x^2 - c^2*d^4*e), -1/2*(sqrt(2)*(e^2*x^2 - d^2)*sqrt(-c*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)) + 2*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e*x + d)*d)/(c^2*d^2*e^3*x^2 - c^2*
d^4*e)]

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x