3.9.90 \(\int \frac {\sqrt {d+e x}}{(c d^2-c e^2 x^2)^{3/2}} \, dx\) [890]

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]

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

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Rubi [A]
time = 0.03, antiderivative size = 104, normalized size of antiderivative = 1.00, 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 = {681, 675, 214} \begin {gather*} \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 {gather*}

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 214

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

Rule 675

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 681

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]

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 \text {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*}

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Mathematica [A]
time = 0.35, size = 124, normalized size = 1.19 \begin {gather*} \frac {2 \sqrt {d} (d+e x)-\sqrt {2} \sqrt {d+e x} \sqrt {d^2-e^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {d} \sqrt {d+e x}}{\sqrt {d^2-e^2 x^2}}\right )}{2 c d^{3/2} e \sqrt {d+e x} \sqrt {c \left (d^2-e^2 x^2\right )}} \end {gather*}

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

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Maple [A]
time = 0.51, size = 91, normalized size = 0.88

method	result	size
default	\(-\frac {\sqrt {c \left (-e^{2} x^{2}+d^{2}\right )}\, \left (\sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) \sqrt {c \left (-e x +d \right )}-2 \sqrt {c d}\right )}{2 \sqrt {e x +d}\, c^{2} \left (-e x +d \right ) e d \sqrt {c d}}\)	\(91\)

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,method=_RETURNVERBOSE)

[Out]

-1/2/(e*x+d)^(1/2)*(c*(-e^2*x^2+d^2))^(1/2)*(2^(1/2)*arctanh(1/2*(c*(-e*x+d))^(1/2)*2^(1/2)/(c*d)^(1/2))*(c*(-
e*x+d))^(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [A]
time = 2.39, size = 281, normalized size = 2.70 \begin {gather*} \left [\frac {\sqrt {2} {\left (x^{2} e^{2} - d^{2}\right )} \sqrt {c d} \log \left (-\frac {c x^{2} e^{2} - 2 \, c d x e - 3 \, c d^{2} + 2 \, \sqrt {2} \sqrt {-c x^{2} e^{2} + c d^{2}} \sqrt {c d} \sqrt {x e + d}}{x^{2} e^{2} + 2 \, d x e + d^{2}}\right ) - 4 \, \sqrt {-c x^{2} e^{2} + c d^{2}} \sqrt {x e + d} d}{4 \, {\left (c^{2} d^{2} x^{2} e^{3} - c^{2} d^{4} e\right )}}, -\frac {\sqrt {2} {\left (x^{2} e^{2} - d^{2}\right )} \sqrt {-c d} \arctan \left (\frac {\sqrt {2} \sqrt {-c x^{2} e^{2} + c d^{2}} \sqrt {-c d} \sqrt {x e + d}}{c x^{2} e^{2} - c d^{2}}\right ) + 2 \, \sqrt {-c x^{2} e^{2} + c d^{2}} \sqrt {x e + d} d}{2 \, {\left (c^{2} d^{2} x^{2} e^{3} - c^{2} d^{4} e\right )}}\right ] \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {d + e x}}{\left (- c \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

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 [A]
time = 2.48, size = 123, normalized size = 1.18 \begin {gather*} \frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-{\left (x e + d\right )} c + 2 \, c d}}{2 \, \sqrt {-c d}}\right ) e^{\left (-1\right )}}{2 \, \sqrt {-c d} c d} - \frac {\sqrt {2} {\left (\sqrt {c d} \arctan \left (\frac {\sqrt {c d}}{\sqrt {-c d}}\right ) + \sqrt {-c d}\right )} e^{\left (-1\right )}}{2 \, \sqrt {c d} \sqrt {-c d} c d} + \frac {e^{\left (-1\right )}}{\sqrt {-{\left (x e + d\right )} c + 2 \, c d} c d} \end {gather*}

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]

1/2*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-(x*e + d)*c + 2*c*d)/sqrt(-c*d))*e^(-1)/(sqrt(-c*d)*c*d) - 1/2*sqrt(2)*(s
qrt(c*d)*arctan(sqrt(c*d)/sqrt(-c*d)) + sqrt(-c*d))*e^(-1)/(sqrt(c*d)*sqrt(-c*d)*c*d) + e^(-1)/(sqrt(-(x*e + d
)*c + 2*c*d)*c*d)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {d+e\,x}}{{\left (c\,d^2-c\,e^2\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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