3.1.44 \(\int \frac {x^5 (d+e x)^2}{(d^2-e^2 x^2)^{7/2}} \, dx\) [44]

Optimal. Leaf size=143 \[ \frac {d^4 (d+e x)^2}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {22 d^3 (d+e x)}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 d (30 d+23 e x)}{15 e^6 \sqrt {d^2-e^2 x^2}}+\frac {\sqrt {d^2-e^2 x^2}}{e^6}-\frac {2 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^6} \]

[Out]

1/5*d^4*(e*x+d)^2/e^6/(-e^2*x^2+d^2)^(5/2)-22/15*d^3*(e*x+d)/e^6/(-e^2*x^2+d^2)^(3/2)-2*d*arctan(e*x/(-e^2*x^2
+d^2)^(1/2))/e^6+2/15*d*(23*e*x+30*d)/e^6/(-e^2*x^2+d^2)^(1/2)+(-e^2*x^2+d^2)^(1/2)/e^6

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Rubi [A]
time = 0.17, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1649, 1828, 655, 223, 209} \begin {gather*} -\frac {2 d \text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^6}+\frac {2 d (30 d+23 e x)}{15 e^6 \sqrt {d^2-e^2 x^2}}+\frac {\sqrt {d^2-e^2 x^2}}{e^6}+\frac {d^4 (d+e x)^2}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {22 d^3 (d+e x)}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^5*(d + e*x)^2)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(d^4*(d + e*x)^2)/(5*e^6*(d^2 - e^2*x^2)^(5/2)) - (22*d^3*(d + e*x))/(15*e^6*(d^2 - e^2*x^2)^(3/2)) + (2*d*(30
*d + 23*e*x))/(15*e^6*Sqrt[d^2 - e^2*x^2]) + Sqrt[d^2 - e^2*x^2]/e^6 - (2*d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])
/e^6

Rule 209

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

Rule 223

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 655

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 1649

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, Simp[(-d)*f*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2
*a*e*(p + 1))), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)
*Q + f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p
+ 1/2, 0] && GtQ[m, 0]

Rule 1828

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*
g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {x^5 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {d^4 (d+e x)^2}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {(d+e x) \left (\frac {2 d^5}{e^5}+\frac {5 d^4 x}{e^4}+\frac {5 d^3 x^2}{e^3}+\frac {5 d^2 x^3}{e^2}+\frac {5 d x^4}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d}\\ &=\frac {d^4 (d+e x)^2}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {22 d^3 (d+e x)}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {\frac {16 d^5}{e^5}+\frac {45 d^4 x}{e^4}+\frac {30 d^3 x^2}{e^3}+\frac {15 d^2 x^3}{e^2}}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2}\\ &=\frac {d^4 (d+e x)^2}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {22 d^3 (d+e x)}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 d (30 d+23 e x)}{15 e^6 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {\frac {30 d^5}{e^5}+\frac {15 d^4 x}{e^4}}{\sqrt {d^2-e^2 x^2}} \, dx}{15 d^4}\\ &=\frac {d^4 (d+e x)^2}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {22 d^3 (d+e x)}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 d (30 d+23 e x)}{15 e^6 \sqrt {d^2-e^2 x^2}}+\frac {\sqrt {d^2-e^2 x^2}}{e^6}-\frac {(2 d) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^5}\\ &=\frac {d^4 (d+e x)^2}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {22 d^3 (d+e x)}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 d (30 d+23 e x)}{15 e^6 \sqrt {d^2-e^2 x^2}}+\frac {\sqrt {d^2-e^2 x^2}}{e^6}-\frac {(2 d) \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5}\\ &=\frac {d^4 (d+e x)^2}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {22 d^3 (d+e x)}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 d (30 d+23 e x)}{15 e^6 \sqrt {d^2-e^2 x^2}}+\frac {\sqrt {d^2-e^2 x^2}}{e^6}-\frac {2 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^6}\\ \end {align*}

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Mathematica [A]
time = 0.41, size = 126, normalized size = 0.88 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-56 d^4+82 d^3 e x+32 d^2 e^2 x^2-76 d e^3 x^3+15 e^4 x^4\right )}{15 e^6 (-d+e x)^3 (d+e x)}+\frac {2 d \left (-e^2\right )^{3/2} \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{e^9} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(d + e*x)^2)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-56*d^4 + 82*d^3*e*x + 32*d^2*e^2*x^2 - 76*d*e^3*x^3 + 15*e^4*x^4))/(15*e^6*(-d + e*x)^3
*(d + e*x)) + (2*d*(-e^2)^(3/2)*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/e^9

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(302\) vs. \(2(127)=254\).
time = 0.08, size = 303, normalized size = 2.12

method	result	size
risch	\(\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{e^{6}}-\frac {2 d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{5} \sqrt {e^{2}}}+\frac {d \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{8 e^{7} \left (x +\frac {d}{e}\right )}-\frac {383 d \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d \left (x -\frac {d}{e}\right ) e}}{120 e^{7} \left (x -\frac {d}{e}\right )}-\frac {41 d^{2} \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d \left (x -\frac {d}{e}\right ) e}}{60 e^{8} \left (x -\frac {d}{e}\right )^{2}}-\frac {d^{3} \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d \left (x -\frac {d}{e}\right ) e}}{10 e^{9} \left (x -\frac {d}{e}\right )^{3}}\)	\(238\)
default	\(e^{2} \left (-\frac {x^{6}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {6 d^{2} \left (\frac {x^{4}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {4 d^{2} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )}{e^{2}}\right )}{e^{2}}\right )+2 e d \left (\frac {x^{5}}{5 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {\frac {x^{3}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}}{e^{2}}}{e^{2}}\right )+d^{2} \left (\frac {x^{4}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {4 d^{2} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )}{e^{2}}\right )\)	\(303\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x,method=_RETURNVERBOSE)

[Out]

e^2*(-x^6/e^2/(-e^2*x^2+d^2)^(5/2)+6*d^2/e^2*(x^4/e^2/(-e^2*x^2+d^2)^(5/2)-4*d^2/e^2*(1/3*x^2/e^2/(-e^2*x^2+d^
2)^(5/2)-2/15*d^2/e^4/(-e^2*x^2+d^2)^(5/2))))+2*e*d*(1/5*x^5/e^2/(-e^2*x^2+d^2)^(5/2)-1/e^2*(1/3*x^3/e^2/(-e^2
*x^2+d^2)^(3/2)-1/e^2*(x/e^2/(-e^2*x^2+d^2)^(1/2)-1/e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))
)))+d^2*(x^4/e^2/(-e^2*x^2+d^2)^(5/2)-4*d^2/e^2*(1/3*x^2/e^2/(-e^2*x^2+d^2)^(5/2)-2/15*d^2/e^4/(-e^2*x^2+d^2)^
(5/2)))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 255 vs. \(2 (121) = 242\).
time = 0.50, size = 255, normalized size = 1.78 \begin {gather*} \frac {7 \, d^{2} x^{4} e^{\left (-2\right )}}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} - \frac {28 \, d^{4} x^{2} e^{\left (-4\right )}}{3 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {56 \, d^{6} e^{\left (-6\right )}}{15 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {2}{15} \, {\left (\frac {15 \, x^{4} e^{\left (-2\right )}}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} - \frac {20 \, d^{2} x^{2} e^{\left (-4\right )}}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {8 \, d^{4} e^{\left (-6\right )}}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}}\right )} d x e - \frac {2}{3} \, {\left (\frac {3 \, x^{2} e^{\left (-2\right )}}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, d^{2} e^{\left (-4\right )}}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}}}\right )} d x e^{\left (-1\right )} - \frac {x^{6}}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {8 \, d^{3} x e^{\left (-5\right )}}{15 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}}} - 2 \, d \arcsin \left (\frac {x e}{d}\right ) e^{\left (-6\right )} - \frac {14 \, d x e^{\left (-5\right )}}{15 \, \sqrt {-x^{2} e^{2} + d^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

7*d^2*x^4*e^(-2)/(-x^2*e^2 + d^2)^(5/2) - 28/3*d^4*x^2*e^(-4)/(-x^2*e^2 + d^2)^(5/2) + 56/15*d^6*e^(-6)/(-x^2*
e^2 + d^2)^(5/2) + 2/15*(15*x^4*e^(-2)/(-x^2*e^2 + d^2)^(5/2) - 20*d^2*x^2*e^(-4)/(-x^2*e^2 + d^2)^(5/2) + 8*d
^4*e^(-6)/(-x^2*e^2 + d^2)^(5/2))*d*x*e - 2/3*(3*x^2*e^(-2)/(-x^2*e^2 + d^2)^(3/2) - 2*d^2*e^(-4)/(-x^2*e^2 +
d^2)^(3/2))*d*x*e^(-1) - x^6/(-x^2*e^2 + d^2)^(5/2) + 8/15*d^3*x*e^(-5)/(-x^2*e^2 + d^2)^(3/2) - 2*d*arcsin(x*
e/d)*e^(-6) - 14/15*d*x*e^(-5)/sqrt(-x^2*e^2 + d^2)

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Fricas [A]
time = 2.05, size = 177, normalized size = 1.24 \begin {gather*} \frac {56 \, d x^{4} e^{4} - 112 \, d^{2} x^{3} e^{3} + 112 \, d^{4} x e - 56 \, d^{5} + 60 \, {\left (d x^{4} e^{4} - 2 \, d^{2} x^{3} e^{3} + 2 \, d^{4} x e - d^{5}\right )} \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) + {\left (15 \, x^{4} e^{4} - 76 \, d x^{3} e^{3} + 32 \, d^{2} x^{2} e^{2} + 82 \, d^{3} x e - 56 \, d^{4}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{15 \, {\left (x^{4} e^{10} - 2 \, d x^{3} e^{9} + 2 \, d^{3} x e^{7} - d^{4} e^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

1/15*(56*d*x^4*e^4 - 112*d^2*x^3*e^3 + 112*d^4*x*e - 56*d^5 + 60*(d*x^4*e^4 - 2*d^2*x^3*e^3 + 2*d^4*x*e - d^5)
*arctan(-(d - sqrt(-x^2*e^2 + d^2))*e^(-1)/x) + (15*x^4*e^4 - 76*d*x^3*e^3 + 32*d^2*x^2*e^2 + 82*d^3*x*e - 56*
d^4)*sqrt(-x^2*e^2 + d^2))/(x^4*e^10 - 2*d*x^3*e^9 + 2*d^3*x*e^7 - d^4*e^6)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{5} \left (d + e x\right )^{2}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(e*x+d)**2/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral(x**5*(d + e*x)**2/(-(-d + e*x)*(d + e*x))**(7/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

integrate((x*e + d)^2*x^5/(-x^2*e^2 + d^2)^(7/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^5\,{\left (d+e\,x\right )}^2}{{\left (d^2-e^2\,x^2\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*(d + e*x)^2)/(d^2 - e^2*x^2)^(7/2),x)

[Out]

int((x^5*(d + e*x)^2)/(d^2 - e^2*x^2)^(7/2), x)

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