3.8.8 \(\int \frac {\sqrt {c d^2-c e^2 x^2}}{(d+e x)^{5/2}} \, dx\)

Optimal. Leaf size=98 \[ \frac {\sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{\sqrt {2} \sqrt {d} e}-\frac {\sqrt {c d^2-c e^2 x^2}}{e (d+e x)^{3/2}} \]

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Rubi [A]  time = 0.05, antiderivative size = 98, 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 = {663, 661, 208} \begin {gather*} \frac {\sqrt {c} \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} \sqrt {d} e}-\frac {\sqrt {c d^2-c e^2 x^2}}{e (d+e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[c*d^2 - c*e^2*x^2]/(d + e*x)^(5/2),x]

[Out]

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

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]

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 663

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 + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(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] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1
, 0] && IntegerQ[2*p]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 101, normalized size = 1.03 \begin {gather*} \frac {\sqrt {c \left (d^2-e^2 x^2\right )} \left (\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{\sqrt {2} \sqrt {d} \sqrt {d+e x}}\right )}{\sqrt {d} \sqrt {d^2-e^2 x^2}}-\frac {2}{(d+e x)^{3/2}}\right )}{2 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c*d^2 - c*e^2*x^2]/(d + e*x)^(5/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.34, size = 127, normalized size = 1.30 \begin {gather*} -\frac {\sqrt {2 c d (d+e x)-c (d+e x)^2}}{e (d+e x)^{3/2}}-\frac {\sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c} (d+e x)-\sqrt {2 c d (d+e x)-c (d+e x)^2}}\right )}{\sqrt {d} e} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[c*d^2 - c*e^2*x^2]/(d + e*x)^(5/2),x]

[Out]

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

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fricas [A]  time = 0.42, size = 292, normalized size = 2.98 \begin {gather*} \left [\frac {\sqrt {\frac {1}{2}} {\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )} \sqrt {\frac {c}{d}} \log \left (-\frac {c e^{2} x^{2} - 2 \, c d e x - 3 \, c d^{2} - 4 \, \sqrt {\frac {1}{2}} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d} d \sqrt {\frac {c}{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}}{2 \, {\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}}, \frac {\sqrt {\frac {1}{2}} {\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )} \sqrt {-\frac {c}{d}} \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d} d \sqrt {-\frac {c}{d}}}{c e^{2} x^{2} - c d^{2}}\right ) - \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d}}{e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e}\right ] \end {gather*}

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)^(5/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {-c e^{2} x^{2} + c d^{2}}}{{\left (e x + d\right )}^{\frac {5}{2}}}\,{d x} \end {gather*}

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)^(5/2),x, algorithm="giac")

[Out]

integrate(sqrt(-c*e^2*x^2 + c*d^2)/(e*x + d)^(5/2), x)

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maple [A]  time = 0.07, size = 127, normalized size = 1.30 \begin {gather*} \frac {\sqrt {-\left (e^{2} x^{2}-d^{2}\right ) c}\, \left (\sqrt {2}\, c e x \arctanh \left (\frac {\sqrt {-\left (e x -d \right ) c}\, \sqrt {2}}{2 \sqrt {c d}}\right )+\sqrt {2}\, c d \arctanh \left (\frac {\sqrt {-\left (e x -d \right ) c}\, \sqrt {2}}{2 \sqrt {c d}}\right )-2 \sqrt {-\left (e x -d \right ) c}\, \sqrt {c d}\right )}{2 \left (e x +d \right )^{\frac {3}{2}} \sqrt {-\left (e x -d \right ) c}\, \sqrt {c d}\, e} \end {gather*}

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)^(5/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {-c e^{2} x^{2} + c d^{2}}}{{\left (e x + d\right )}^{\frac {5}{2}}}\,{d x} \end {gather*}

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)^(5/2),x, algorithm="maxima")

[Out]

integrate(sqrt(-c*e^2*x^2 + c*d^2)/(e*x + d)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {c\,d^2-c\,e^2\,x^2}}{{\left (d+e\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*d^2 - c*e^2*x^2)^(1/2)/(d + e*x)^(5/2),x)

[Out]

int((c*d^2 - c*e^2*x^2)^(1/2)/(d + e*x)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {- c \left (- d + e x\right ) \left (d + e x\right )}}{\left (d + e x\right )^{\frac {5}{2}}}\, dx \end {gather*}

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)**(5/2),x)

[Out]

Integral(sqrt(-c*(-d + e*x)*(d + e*x))/(d + e*x)**(5/2), x)

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