3.3.3 \(\int \frac {(d^2-e^2 x^2)^{5/2}}{(d+e x)^4} \, dx\)

Optimal. Leaf size=113 \[ -\frac {2 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^3}-\frac {5 \left (d^2-e^2 x^2\right )^{3/2}}{2 e (d+e x)}-\frac {15 d \sqrt {d^2-e^2 x^2}}{2 e}-\frac {15 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e} \]

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Rubi [A]  time = 0.05, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {663, 665, 217, 203} \begin {gather*} -\frac {2 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^3}-\frac {5 \left (d^2-e^2 x^2\right )^{3/2}}{2 e (d+e x)}-\frac {15 d \sqrt {d^2-e^2 x^2}}{2 e}-\frac {15 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d^2 - e^2*x^2)^(5/2)/(d + e*x)^4,x]

[Out]

(-15*d*Sqrt[d^2 - e^2*x^2])/(2*e) - (5*(d^2 - e^2*x^2)^(3/2))/(2*e*(d + e*x)) - (2*(d^2 - e^2*x^2)^(5/2))/(e*(
d + e*x)^3) - (15*d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 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 665

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

Rubi steps

\begin {align*} \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx &=-\frac {2 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^3}-5 \int \frac {\left (d^2-e^2 x^2\right )^{3/2}}{(d+e x)^2} \, dx\\ &=-\frac {5 \left (d^2-e^2 x^2\right )^{3/2}}{2 e (d+e x)}-\frac {2 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^3}-\frac {1}{2} (15 d) \int \frac {\sqrt {d^2-e^2 x^2}}{d+e x} \, dx\\ &=-\frac {15 d \sqrt {d^2-e^2 x^2}}{2 e}-\frac {5 \left (d^2-e^2 x^2\right )^{3/2}}{2 e (d+e x)}-\frac {2 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^3}-\frac {1}{2} \left (15 d^2\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=-\frac {15 d \sqrt {d^2-e^2 x^2}}{2 e}-\frac {5 \left (d^2-e^2 x^2\right )^{3/2}}{2 e (d+e x)}-\frac {2 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^3}-\frac {1}{2} \left (15 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )\\ &=-\frac {15 d \sqrt {d^2-e^2 x^2}}{2 e}-\frac {5 \left (d^2-e^2 x^2\right )^{3/2}}{2 e (d+e x)}-\frac {2 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^3}-\frac {15 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 75, normalized size = 0.66 \begin {gather*} \sqrt {d^2-e^2 x^2} \left (-\frac {8 d^2}{e (d+e x)}-\frac {4 d}{e}+\frac {x}{2}\right )-\frac {15 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d^2 - e^2*x^2)^(5/2)/(d + e*x)^4,x]

[Out]

Sqrt[d^2 - e^2*x^2]*((-4*d)/e + x/2 - (8*d^2)/(e*(d + e*x))) - (15*d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e
)

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IntegrateAlgebraic [A]  time = 0.45, size = 98, normalized size = 0.87 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-24 d^2-7 d e x+e^2 x^2\right )}{2 e (d+e x)}-\frac {15 d^2 \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{2 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d^2 - e^2*x^2)^(5/2)/(d + e*x)^4,x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-24*d^2 - 7*d*e*x + e^2*x^2))/(2*e*(d + e*x)) - (15*d^2*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) +
Sqrt[d^2 - e^2*x^2]])/(2*e^2)

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fricas [A]  time = 0.40, size = 99, normalized size = 0.88 \begin {gather*} -\frac {24 \, d^{2} e x + 24 \, d^{3} - 30 \, {\left (d^{2} e x + d^{3}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (e^{2} x^{2} - 7 \, d e x - 24 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{2 \, {\left (e^{2} x + d e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x, algorithm="fricas")

[Out]

-1/2*(24*d^2*e*x + 24*d^3 - 30*(d^2*e*x + d^3)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (e^2*x^2 - 7*d*e*x
- 24*d^2)*sqrt(-e^2*x^2 + d^2))/(e^2*x + d*e)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: (12*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^
2*exp(2))*exp(1))/x/exp(2))^4*exp(1)^12*exp(2)^2+6*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp
(2))^5*exp(1)^10*exp(2)^3+36*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(1)^10*exp(2
)^3+18*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(1)^8*exp(2)^4+12*d^2*(-1/2*(-2*d*
exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^12*exp(2)^2+280*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2
*exp(2))*exp(1))/x/exp(2))^3*exp(1)^10*exp(2)^3+288*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/ex
p(2))^4*exp(1)^8*exp(2)^4+72*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(1)^6*exp(2)
^5-36*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^10*exp(2)^3+360*d^2*(-1/2*(-2*d
*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^8*exp(2)^4+324*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2
*exp(2))*exp(1))/x/exp(2))^4*exp(1)^6*exp(2)^5+72*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(
2))^5*exp(1)^4*exp(2)^6+522*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^8*exp(2)^
4+546*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^6*exp(2)^5+189*d^2*(-1/2*(-2*d*
exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(1)^4*exp(2)^6+27*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*e
xp(2))*exp(1))/x/exp(2))^5*exp(2)^8+756*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(
1)^6*exp(2)^5+540*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^4*exp(2)^6+2*d^2*ex
p(1)^8*exp(2)^4+126*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(2)^8+444*d^2*(-1/2*(
-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^4*exp(2)^6+216*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2
-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(2)^8+67*d^2*exp(1)^4*exp(2)^6+252*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*e
xp(2))*exp(1))/x/exp(2))^2*exp(2)^8+126*d^2*exp(2)^8+8*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x
/exp(2))^3*exp(1)^14*exp(2)-189/2*d^2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(2)^8/x/exp(2)-234*d^2*(-
2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^4*exp(2)^6/x/exp(2)-165*d^2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2
))*exp(1))*exp(1)^6*exp(2)^5/x/exp(2)+9*d^2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^8*exp(2)^4/x/ex
p(2)-3*d^2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^10*exp(2)^3/x/exp(2))/((-1/2*(-2*d*exp(1)-2*sqrt
(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(2)-(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x+exp(2))^3/(3*exp(1)^
11+9*exp(1)^7*exp(2)^2+3*exp(1)^5*exp(2)^3+9*exp(1)^9*exp(2))+1/2*(48*d^2*exp(1)^8*exp(2)^3+40*d^2*exp(1)^6*ex
p(2)^4-66*d^2*exp(1)^4*exp(2)^5-148*d^2*exp(2)^7)*atan((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x+exp
(2))/sqrt(-exp(1)^4+exp(2)^2))/sqrt(-exp(1)^4+exp(2)^2)/(exp(1)^13+3*exp(1)^9*exp(2)^2+exp(1)^7*exp(2)^3+3*exp
(1)^11*exp(2))-15/2*d^2*sign(d)*asin(x*exp(2)/d/exp(1))/exp(1)+2*(2*exp(1)*1/8/exp(1)*x-16*d*1/8/exp(1))*sqrt(
d^2-x^2*exp(2))

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maple [B]  time = 0.01, size = 284, normalized size = 2.51 \begin {gather*} -\frac {15 d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{2 \sqrt {e^{2}}}-\frac {15 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, x}{2}-\frac {5 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} x}{d^{2}}-\frac {4 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}}}{d^{3} e}-\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}}}{\left (x +\frac {d}{e}\right )^{4} d \,e^{5}}-\frac {3 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}}}{\left (x +\frac {d}{e}\right )^{3} d^{2} e^{4}}-\frac {4 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}}}{\left (x +\frac {d}{e}\right )^{2} d^{3} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x)

[Out]

-1/e^5/d/(x+d/e)^4*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)-3/e^4/d^2/(x+d/e)^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)
-4/e^3/d^3/(x+d/e)^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)-4/e/d^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(5/2)-5/d^2*(2*
(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)*x-15/2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x-15/2*d^2/(e^2)^(1/2)*arctan((e^2
)^(1/2)/(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x)

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maxima [A]  time = 0.98, size = 134, normalized size = 1.19 \begin {gather*} -\frac {15 \, d^{2} \arcsin \left (\frac {e x}{d}\right )}{2 \, e} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{2 \, {\left (e^{4} x^{3} + 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x + d^{3} e\right )}} + \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d}{2 \, {\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}} - \frac {15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{e^{2} x + d e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x, algorithm="maxima")

[Out]

-15/2*d^2*arcsin(e*x/d)/e + 1/2*(-e^2*x^2 + d^2)^(5/2)/(e^4*x^3 + 3*d*e^3*x^2 + 3*d^2*e^2*x + d^3*e) + 5/2*(-e
^2*x^2 + d^2)^(3/2)*d/(e^3*x^2 + 2*d*e^2*x + d^2*e) - 15*sqrt(-e^2*x^2 + d^2)*d^2/(e^2*x + d*e)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d^2-e^2\,x^2\right )}^{5/2}}{{\left (d+e\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d^2 - e^2*x^2)^(5/2)/(d + e*x)^4,x)

[Out]

int((d^2 - e^2*x^2)^(5/2)/(d + e*x)^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}}}{\left (d + e x\right )^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e**2*x**2+d**2)**(5/2)/(e*x+d)**4,x)

[Out]

Integral((-(-d + e*x)*(d + e*x))**(5/2)/(d + e*x)**4, x)

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