3.8.9 \(\int \frac {\sqrt {c d^2-c e^2 x^2}}{(d+e x)^{7/2}} \, dx\)

Optimal. Leaf size=141 \[ \frac {\sqrt {c d^2-c e^2 x^2}}{8 d e (d+e x)^{3/2}}-\frac {\sqrt {c d^2-c e^2 x^2}}{2 e (d+e x)^{5/2}}+\frac {\sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{8 \sqrt {2} d^{3/2} e} \]

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Rubi [A]  time = 0.08, antiderivative size = 141, normalized size of antiderivative = 1.00, 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 = {663, 673, 661, 208} \begin {gather*} \frac {\sqrt {c d^2-c e^2 x^2}}{8 d e (d+e x)^{3/2}}-\frac {\sqrt {c d^2-c e^2 x^2}}{2 e (d+e x)^{5/2}}+\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 )}{8 \sqrt {2} d^{3/2} e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[c*d^2 - c*e^2*x^2]/(d + e*x)^(7/2),x]

[Out]

-Sqrt[c*d^2 - c*e^2*x^2]/(2*e*(d + e*x)^(5/2)) + Sqrt[c*d^2 - c*e^2*x^2]/(8*d*e*(d + e*x)^(3/2)) + (Sqrt[c]*Ar
cTanh[Sqrt[c*d^2 - c*e^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])])/(8*Sqrt[2]*d^(3/2)*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]

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]

Rubi steps

\begin {align*} \int \frac {\sqrt {c d^2-c e^2 x^2}}{(d+e x)^{7/2}} \, dx &=-\frac {\sqrt {c d^2-c e^2 x^2}}{2 e (d+e x)^{5/2}}-\frac {1}{4} c \int \frac {1}{(d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}} \, dx\\ &=-\frac {\sqrt {c d^2-c e^2 x^2}}{2 e (d+e x)^{5/2}}+\frac {\sqrt {c d^2-c e^2 x^2}}{8 d e (d+e x)^{3/2}}-\frac {c \int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}} \, dx}{16 d}\\ &=-\frac {\sqrt {c d^2-c e^2 x^2}}{2 e (d+e x)^{5/2}}+\frac {\sqrt {c d^2-c e^2 x^2}}{8 d e (d+e x)^{3/2}}-\frac {(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 )}{8 d}\\ &=-\frac {\sqrt {c d^2-c e^2 x^2}}{2 e (d+e x)^{5/2}}+\frac {\sqrt {c d^2-c e^2 x^2}}{8 d 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 )}{8 \sqrt {2} d^{3/2} e}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 111, normalized size = 0.79 \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 )}{d^{3/2} \sqrt {d^2-e^2 x^2}}+\frac {2 e x-6 d}{d (d+e x)^{5/2}}\right )}{16 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c*d^2 - c*e^2*x^2]/(d + e*x)^(7/2),x]

[Out]

(Sqrt[c*(d^2 - e^2*x^2)]*((-6*d + 2*e*x)/(d*(d + e*x)^(5/2)) + (Sqrt[2]*ArcTanh[Sqrt[d^2 - e^2*x^2]/(Sqrt[2]*S
qrt[d]*Sqrt[d + e*x])])/(d^(3/2)*Sqrt[d^2 - e^2*x^2])))/(16*e)

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IntegrateAlgebraic [A]  time = 0.49, size = 141, normalized size = 1.00 \begin {gather*} \frac {(e x-3 d) \sqrt {2 c d (d+e x)-c (d+e x)^2}}{8 d e (d+e x)^{5/2}}-\frac {\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 )}{4 \sqrt {2} d^{3/2} e} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[c*d^2 - c*e^2*x^2]/(d + e*x)^(7/2),x]

[Out]

((-3*d + e*x)*Sqrt[2*c*d*(d + e*x) - c*(d + e*x)^2])/(8*d*e*(d + e*x)^(5/2)) - (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])])/(4*Sqrt[2]*d^(3/2)*e)

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fricas [A]  time = 0.41, size = 356, normalized size = 2.52 \begin {gather*} \left [\frac {\sqrt {\frac {1}{2}} {\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\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} {\left (e x - 3 \, d\right )}}{16 \, {\left (d e^{4} x^{3} + 3 \, d^{2} e^{3} x^{2} + 3 \, d^{3} e^{2} x + d^{4} e\right )}}, \frac {\sqrt {\frac {1}{2}} {\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\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} {\left (e x - 3 \, d\right )}}{8 \, {\left (d e^{4} x^{3} + 3 \, d^{2} e^{3} x^{2} + 3 \, d^{3} e^{2} x + d^{4} e\right )}}\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)^(7/2),x, algorithm="fricas")

[Out]

[1/16*(sqrt(1/2)*(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3)*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*x - 3*d))/(d*e^4*x^3 + 3*d^2*e^3*x^2 + 3*d^3*e^2*x + d^4*e), 1/8*(sqrt(1/2)*(e^3*x^3
+ 3*d*e^2*x^2 + 3*d^2*e*x + d^3)*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*x - 3*d))/(d*e^4*x^3 + 3*d^2*e^3*x^2 +
3*d^3*e^2*x + d^4*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 {7}{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)^(7/2),x, algorithm="giac")

[Out]

integrate(sqrt(-c*e^2*x^2 + c*d^2)/(e*x + d)^(7/2), x)

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maple [A]  time = 0.07, size = 190, normalized size = 1.35 \begin {gather*} \frac {\sqrt {-\left (e^{2} x^{2}-d^{2}\right ) c}\, \left (\sqrt {2}\, c \,e^{2} x^{2} \arctanh \left (\frac {\sqrt {-\left (e x -d \right ) c}\, \sqrt {2}}{2 \sqrt {c d}}\right )+2 \sqrt {2}\, c d e x \arctanh \left (\frac {\sqrt {-\left (e x -d \right ) c}\, \sqrt {2}}{2 \sqrt {c d}}\right )+\sqrt {2}\, c \,d^{2} \arctanh \left (\frac {\sqrt {-\left (e x -d \right ) c}\, \sqrt {2}}{2 \sqrt {c d}}\right )+2 \sqrt {c d}\, \sqrt {-\left (e x -d \right ) c}\, e x -6 \sqrt {-\left (e x -d \right ) c}\, \sqrt {c d}\, d \right )}{16 \left (e x +d \right )^{\frac {5}{2}} \sqrt {-\left (e x -d \right ) c}\, \sqrt {c d}\, 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)^(7/2),x)

[Out]

1/16*(-(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^2*c*e^2+2*2^(1/2)
*arctanh(1/2*(-(e*x-d)*c)^(1/2)*2^(1/2)/(c*d)^(1/2))*x*c*d*e+2^(1/2)*arctanh(1/2*(-(e*x-d)*c)^(1/2)*2^(1/2)/(c
*d)^(1/2))*c*d^2+2*x*e*(c*d)^(1/2)*(-(e*x-d)*c)^(1/2)-6*(-(e*x-d)*c)^(1/2)*(c*d)^(1/2)*d)/(e*x+d)^(5/2)/(-(e*x
-d)*c)^(1/2)/e/d/(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 {7}{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)^(7/2),x, algorithm="maxima")

[Out]

integrate(sqrt(-c*e^2*x^2 + c*d^2)/(e*x + d)^(7/2), 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 )}^{7/2}} \,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)^(7/2),x)

[Out]

int((c*d^2 - c*e^2*x^2)^(1/2)/(d + e*x)^(7/2), 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 {7}{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)**(7/2),x)

[Out]

Integral(sqrt(-c*(-d + e*x)*(d + e*x))/(d + e*x)**(7/2), x)

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