3.891 $$\int \frac{1}{\sqrt{d+e x} (c d^2-c e^2 x^2)^{3/2}} \, dx$$

Optimal. Leaf size=150 $-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{c d^2-c e^2 x^2}}{\sqrt{2} \sqrt{c} \sqrt{d} \sqrt{d+e x}}\right )}{4 \sqrt{2} c^{3/2} d^{5/2} e}+\frac{3 \sqrt{d+e x}}{4 c d^2 e \sqrt{c d^2-c e^2 x^2}}-\frac{1}{2 c d e \sqrt{d+e x} \sqrt{c d^2-c e^2 x^2}}$

[Out]

-1/(2*c*d*e*Sqrt[d + e*x]*Sqrt[c*d^2 - c*e^2*x^2]) + (3*Sqrt[d + e*x])/(4*c*d^2*e*Sqrt[c*d^2 - c*e^2*x^2]) - (
3*ArcTanh[Sqrt[c*d^2 - c*e^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])])/(4*Sqrt[2]*c^(3/2)*d^(5/2)*e)

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Rubi [A]  time = 0.0727255, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.138, Rules used = {673, 667, 661, 208} $-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{c d^2-c e^2 x^2}}{\sqrt{2} \sqrt{c} \sqrt{d} \sqrt{d+e x}}\right )}{4 \sqrt{2} c^{3/2} d^{5/2} e}+\frac{3 \sqrt{d+e x}}{4 c d^2 e \sqrt{c d^2-c e^2 x^2}}-\frac{1}{2 c d e \sqrt{d+e x} \sqrt{c d^2-c e^2 x^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-1/(2*c*d*e*Sqrt[d + e*x]*Sqrt[c*d^2 - c*e^2*x^2]) + (3*Sqrt[d + e*x])/(4*c*d^2*e*Sqrt[c*d^2 - c*e^2*x^2]) - (
3*ArcTanh[Sqrt[c*d^2 - c*e^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])])/(4*Sqrt[2]*c^(3/2)*d^(5/2)*e)

Rule 673

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

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{1}{\sqrt{d+e x} \left (c d^2-c e^2 x^2\right )^{3/2}} \, dx &=-\frac{1}{2 c d e \sqrt{d+e x} \sqrt{c d^2-c e^2 x^2}}+\frac{3 \int \frac{\sqrt{d+e x}}{\left (c d^2-c e^2 x^2\right )^{3/2}} \, dx}{4 d}\\ &=-\frac{1}{2 c d e \sqrt{d+e x} \sqrt{c d^2-c e^2 x^2}}+\frac{3 \sqrt{d+e x}}{4 c d^2 e \sqrt{c d^2-c e^2 x^2}}+\frac{3 \int \frac{1}{\sqrt{d+e x} \sqrt{c d^2-c e^2 x^2}} \, dx}{8 c d^2}\\ &=-\frac{1}{2 c d e \sqrt{d+e x} \sqrt{c d^2-c e^2 x^2}}+\frac{3 \sqrt{d+e x}}{4 c d^2 e \sqrt{c d^2-c e^2 x^2}}+\frac{(3 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 )}{4 c d^2}\\ &=-\frac{1}{2 c d e \sqrt{d+e x} \sqrt{c d^2-c e^2 x^2}}+\frac{3 \sqrt{d+e x}}{4 c d^2 e \sqrt{c d^2-c e^2 x^2}}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{c d^2-c e^2 x^2}}{\sqrt{2} \sqrt{c} \sqrt{d} \sqrt{d+e x}}\right )}{4 \sqrt{2} c^{3/2} d^{5/2} e}\\ \end{align*}

Mathematica [A]  time = 0.0873345, size = 128, normalized size = 0.85 $\frac{2 \sqrt{d} \sqrt{d+e x} (d+3 e x)-3 \sqrt{2} (d+e x) \sqrt{d^2-e^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{\sqrt{2} \sqrt{d} \sqrt{d+e x}}\right )}{8 c d^{5/2} e (d+e x) \sqrt{c \left (d^2-e^2 x^2\right )}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*Sqrt[d]*Sqrt[d + e*x]*(d + 3*e*x) - 3*Sqrt[2]*(d + e*x)*Sqrt[d^2 - e^2*x^2]*ArcTanh[Sqrt[d^2 - e^2*x^2]/(Sq
rt[2]*Sqrt[d]*Sqrt[d + e*x])])/(8*c*d^(5/2)*e*(d + e*x)*Sqrt[c*(d^2 - e^2*x^2)])

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Maple [A]  time = 0.194, size = 152, normalized size = 1. \begin{align*}{\frac{1}{8\,{c}^{2} \left ( ex-d \right ) e{d}^{2}}\sqrt{-c \left ({e}^{2}{x}^{2}-{d}^{2} \right ) } \left ( 3\,\sqrt{- \left ( ex-d \right ) c}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{- \left ( ex-d \right ) c}\sqrt{2}}{\sqrt{cd}}} \right ) xe+3\,\sqrt{- \left ( ex-d \right ) c}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{- \left ( ex-d \right ) c}\sqrt{2}}{\sqrt{cd}}} \right ) d-6\,\sqrt{cd}xe-2\,\sqrt{cd}d \right ) \left ( ex+d \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{cd}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/(e*x+d)^(3/2)*(-c*(e^2*x^2-d^2))^(1/2)/c^2*(3*(-(e*x-d)*c)^(1/2)*2^(1/2)*arctanh(1/2*(-(e*x-d)*c)^(1/2)*2^
(1/2)/(c*d)^(1/2))*x*e+3*(-(e*x-d)*c)^(1/2)*2^(1/2)*arctanh(1/2*(-(e*x-d)*c)^(1/2)*2^(1/2)/(c*d)^(1/2))*d-6*(c
*d)^(1/2)*x*e-2*(c*d)^(1/2)*d)/(e*x-d)/e/d^2/(c*d)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-c e^{2} x^{2} + c d^{2}\right )}^{\frac{3}{2}} \sqrt{e x + d}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.2437, size = 805, normalized size = 5.37 \begin{align*} \left [\frac{3 \, \sqrt{2}{\left (e^{3} x^{3} + d e^{2} x^{2} - d^{2} e x - d^{3}\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}}{\left (3 \, d e x + d^{2}\right )} \sqrt{e x + d}}{16 \,{\left (c^{2} d^{3} e^{4} x^{3} + c^{2} d^{4} e^{3} x^{2} - c^{2} d^{5} e^{2} x - c^{2} d^{6} e\right )}}, -\frac{3 \, \sqrt{2}{\left (e^{3} x^{3} + d e^{2} x^{2} - d^{2} e x - d^{3}\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}}{\left (3 \, d e x + d^{2}\right )} \sqrt{e x + d}}{8 \,{\left (c^{2} d^{3} e^{4} x^{3} + c^{2} d^{4} e^{3} x^{2} - c^{2} d^{5} e^{2} x - c^{2} d^{6} e\right )}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(3*sqrt(2)*(e^3*x^3 + d*e^2*x^2 - d^2*e*x - d^3)*sqrt(c*d)*log(-(c*e^2*x^2 - 2*c*d*e*x - 3*c*d^2 + 2*sqr
t(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)
*(3*d*e*x + d^2)*sqrt(e*x + d))/(c^2*d^3*e^4*x^3 + c^2*d^4*e^3*x^2 - c^2*d^5*e^2*x - c^2*d^6*e), -1/8*(3*sqrt(
2)*(e^3*x^3 + d*e^2*x^2 - d^2*e*x - d^3)*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)*(3*d*e*x + d^2)*sqrt(e*x + d))/(c^2*d^3*e^4*x^3 + c^2
*d^4*e^3*x^2 - c^2*d^5*e^2*x - c^2*d^6*e)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (- c \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac{3}{2}} \sqrt{d + e x}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

sage0*x