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

Optimal. Leaf size=132 \[ \frac {7}{24} d^2 x \left (d^2-e^2 x^2\right )^{3/2}+\frac {7 d \left (d^2-e^2 x^2\right )^{5/2}}{30 e}+\frac {(d-e x) \left (d^2-e^2 x^2\right )^{5/2}}{6 e}+\frac {7 d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e}+\frac {7}{16} d^4 x \sqrt {d^2-e^2 x^2} \]

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Rubi [A]  time = 0.05, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {655, 671, 641, 195, 217, 203} \begin {gather*} \frac {7}{16} d^4 x \sqrt {d^2-e^2 x^2}+\frac {7}{24} d^2 x \left (d^2-e^2 x^2\right )^{3/2}+\frac {7 d \left (d^2-e^2 x^2\right )^{5/2}}{30 e}+\frac {(d-e x) \left (d^2-e^2 x^2\right )^{5/2}}{6 e}+\frac {7 d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7*d^4*x*Sqrt[d^2 - e^2*x^2])/16 + (7*d^2*x*(d^2 - e^2*x^2)^(3/2))/24 + (7*d*(d^2 - e^2*x^2)^(5/2))/(30*e) + (
(d - e*x)*(d^2 - e^2*x^2)^(5/2))/(6*e) + (7*d^6*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(16*e)

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 641

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

Rule 655

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

Rule 671

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

Rubi steps

\begin {align*} \int \frac {\left (d^2-e^2 x^2\right )^{7/2}}{(d+e x)^2} \, dx &=\int (d-e x)^2 \left (d^2-e^2 x^2\right )^{3/2} \, dx\\ &=\frac {(d-e x) \left (d^2-e^2 x^2\right )^{5/2}}{6 e}+\frac {1}{6} (7 d) \int (d-e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx\\ &=\frac {7 d \left (d^2-e^2 x^2\right )^{5/2}}{30 e}+\frac {(d-e x) \left (d^2-e^2 x^2\right )^{5/2}}{6 e}+\frac {1}{6} \left (7 d^2\right ) \int \left (d^2-e^2 x^2\right )^{3/2} \, dx\\ &=\frac {7}{24} d^2 x \left (d^2-e^2 x^2\right )^{3/2}+\frac {7 d \left (d^2-e^2 x^2\right )^{5/2}}{30 e}+\frac {(d-e x) \left (d^2-e^2 x^2\right )^{5/2}}{6 e}+\frac {1}{8} \left (7 d^4\right ) \int \sqrt {d^2-e^2 x^2} \, dx\\ &=\frac {7}{16} d^4 x \sqrt {d^2-e^2 x^2}+\frac {7}{24} d^2 x \left (d^2-e^2 x^2\right )^{3/2}+\frac {7 d \left (d^2-e^2 x^2\right )^{5/2}}{30 e}+\frac {(d-e x) \left (d^2-e^2 x^2\right )^{5/2}}{6 e}+\frac {1}{16} \left (7 d^6\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=\frac {7}{16} d^4 x \sqrt {d^2-e^2 x^2}+\frac {7}{24} d^2 x \left (d^2-e^2 x^2\right )^{3/2}+\frac {7 d \left (d^2-e^2 x^2\right )^{5/2}}{30 e}+\frac {(d-e x) \left (d^2-e^2 x^2\right )^{5/2}}{6 e}+\frac {1}{16} \left (7 d^6\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 {7}{16} d^4 x \sqrt {d^2-e^2 x^2}+\frac {7}{24} d^2 x \left (d^2-e^2 x^2\right )^{3/2}+\frac {7 d \left (d^2-e^2 x^2\right )^{5/2}}{30 e}+\frac {(d-e x) \left (d^2-e^2 x^2\right )^{5/2}}{6 e}+\frac {7 d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 102, normalized size = 0.77 \begin {gather*} \frac {105 d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\sqrt {d^2-e^2 x^2} \left (96 d^5+135 d^4 e x-192 d^3 e^2 x^2+10 d^2 e^3 x^3+96 d e^4 x^4-40 e^5 x^5\right )}{240 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(96*d^5 + 135*d^4*e*x - 192*d^3*e^2*x^2 + 10*d^2*e^3*x^3 + 96*d*e^4*x^4 - 40*e^5*x^5) + 1
05*d^6*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(240*e)

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IntegrateAlgebraic [A]  time = 0.40, size = 125, normalized size = 0.95 \begin {gather*} \frac {7 d^6 \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{16 e^2}+\frac {\sqrt {d^2-e^2 x^2} \left (96 d^5+135 d^4 e x-192 d^3 e^2 x^2+10 d^2 e^3 x^3+96 d e^4 x^4-40 e^5 x^5\right )}{240 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(96*d^5 + 135*d^4*e*x - 192*d^3*e^2*x^2 + 10*d^2*e^3*x^3 + 96*d*e^4*x^4 - 40*e^5*x^5))/(2
40*e) + (7*d^6*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(16*e^2)

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fricas [A]  time = 0.42, size = 105, normalized size = 0.80 \begin {gather*} -\frac {210 \, d^{6} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (40 \, e^{5} x^{5} - 96 \, d e^{4} x^{4} - 10 \, d^{2} e^{3} x^{3} + 192 \, d^{3} e^{2} x^{2} - 135 \, d^{4} e x - 96 \, d^{5}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{240 \, e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(7/2)/(e*x+d)^2,x, algorithm="fricas")

[Out]

-1/240*(210*d^6*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (40*e^5*x^5 - 96*d*e^4*x^4 - 10*d^2*e^3*x^3 + 192*
d^3*e^2*x^2 - 135*d^4*e*x - 96*d^5)*sqrt(-e^2*x^2 + d^2))/e

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(7/2)/(e*x+d)^2,x, algorithm="giac")

[Out]

sage0*x

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maple [B]  time = 0.05, size = 228, normalized size = 1.73 \begin {gather*} \frac {7 d^{6} \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 )}{16 \sqrt {e^{2}}}+\frac {7 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{4} x}{16}+\frac {7 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} d^{2} x}{24}+\frac {7 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}} x}{30}+\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}}}{5 d e}+\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {9}{2}}}{5 \left (x +\frac {d}{e}\right )^{2} d \,e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-e^2*x^2+d^2)^(7/2)/(e*x+d)^2,x)

[Out]

1/5/e^3/d/(x+d/e)^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(9/2)+1/5/e/d*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)+7/30*(2*(x
+d/e)*d*e-(x+d/e)^2*e^2)^(5/2)*x+7/24*d^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)*x+7/16*d^4*(2*(x+d/e)*d*e-(x+d/e
)^2*e^2)^(1/2)*x+7/16*d^6/(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 [C]  time = 3.10, size = 139, normalized size = 1.05 \begin {gather*} -\frac {7 i \, d^{6} \arcsin \left (\frac {e x}{d} + 2\right )}{16 \, e} + \frac {7}{16} \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{4} x + \frac {7 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{5}}{8 \, e} + \frac {7}{24} \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2} x + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}}}{6 \, {\left (e^{2} x + d e\right )}} + \frac {7 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d}{30 \, e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(7/2)/(e*x+d)^2,x, algorithm="maxima")

[Out]

-7/16*I*d^6*arcsin(e*x/d + 2)/e + 7/16*sqrt(e^2*x^2 + 4*d*e*x + 3*d^2)*d^4*x + 7/8*sqrt(e^2*x^2 + 4*d*e*x + 3*
d^2)*d^5/e + 7/24*(-e^2*x^2 + d^2)^(3/2)*d^2*x + 1/6*(-e^2*x^2 + d^2)^(7/2)/(e^2*x + d*e) + 7/30*(-e^2*x^2 + d
^2)^(5/2)*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 )}^{7/2}}{{\left (d+e\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d^2 - e^2*x^2)^(7/2)/(d + e*x)^2,x)

[Out]

int((d^2 - e^2*x^2)^(7/2)/(d + e*x)^2, x)

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sympy [C]  time = 17.75, size = 495, normalized size = 3.75 \begin {gather*} d^{4} \left (\begin {cases} - \frac {i d^{2} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{2 e} - \frac {i d x}{2 \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {i e^{2} x^{3}}{2 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{2} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{2 e} + \frac {d x \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{2} & \text {otherwise} \end {cases}\right ) - 2 d^{3} e \left (\begin {cases} \frac {x^{2} \sqrt {d^{2}}}{2} & \text {for}\: e^{2} = 0 \\- \frac {\left (d^{2} - e^{2} x^{2}\right )^{\frac {3}{2}}}{3 e^{2}} & \text {otherwise} \end {cases}\right ) + 2 d e^{3} \left (\begin {cases} - \frac {2 d^{4} \sqrt {d^{2} - e^{2} x^{2}}}{15 e^{4}} - \frac {d^{2} x^{2} \sqrt {d^{2} - e^{2} x^{2}}}{15 e^{2}} + \frac {x^{4} \sqrt {d^{2} - e^{2} x^{2}}}{5} & \text {for}\: e \neq 0 \\\frac {x^{4} \sqrt {d^{2}}}{4} & \text {otherwise} \end {cases}\right ) - e^{4} \left (\begin {cases} - \frac {i d^{6} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{16 e^{5}} + \frac {i d^{5} x}{16 e^{4} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {i d^{3} x^{3}}{48 e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {5 i d x^{5}}{24 \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {i e^{2} x^{7}}{6 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{6} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{16 e^{5}} - \frac {d^{5} x}{16 e^{4} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {d^{3} x^{3}}{48 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {5 d x^{5}}{24 \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - \frac {e^{2} x^{7}}{6 d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**4*Piecewise((-I*d**2*acosh(e*x/d)/(2*e) - I*d*x/(2*sqrt(-1 + e**2*x**2/d**2)) + I*e**2*x**3/(2*d*sqrt(-1 +
e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (d**2*asin(e*x/d)/(2*e) + d*x*sqrt(1 - e**2*x**2/d**2)/2, True)) -
2*d**3*e*Piecewise((x**2*sqrt(d**2)/2, Eq(e**2, 0)), (-(d**2 - e**2*x**2)**(3/2)/(3*e**2), True)) + 2*d*e**3*
Piecewise((-2*d**4*sqrt(d**2 - e**2*x**2)/(15*e**4) - d**2*x**2*sqrt(d**2 - e**2*x**2)/(15*e**2) + x**4*sqrt(d
**2 - e**2*x**2)/5, Ne(e, 0)), (x**4*sqrt(d**2)/4, True)) - e**4*Piecewise((-I*d**6*acosh(e*x/d)/(16*e**5) + I
*d**5*x/(16*e**4*sqrt(-1 + e**2*x**2/d**2)) - I*d**3*x**3/(48*e**2*sqrt(-1 + e**2*x**2/d**2)) - 5*I*d*x**5/(24
*sqrt(-1 + e**2*x**2/d**2)) + I*e**2*x**7/(6*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (d**6*asi
n(e*x/d)/(16*e**5) - d**5*x/(16*e**4*sqrt(1 - e**2*x**2/d**2)) + d**3*x**3/(48*e**2*sqrt(1 - e**2*x**2/d**2))
+ 5*d*x**5/(24*sqrt(1 - e**2*x**2/d**2)) - e**2*x**7/(6*d*sqrt(1 - e**2*x**2/d**2)), True))

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