3.1.84 \(\int \frac {x^4 (d+e x)^3}{(d^2-e^2 x^2)^{7/2}} \, dx\) [84]

Optimal. Leaf size=142 \[ \frac {d^3 (d+e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {6 d^2 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {24 d (d+e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}+\frac {\sqrt {d^2-e^2 x^2}}{e^5}-\frac {3 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5} \]

[Out]

1/5*d^3*(e*x+d)^3/e^5/(-e^2*x^2+d^2)^(5/2)-6/5*d^2*(e*x+d)^2/e^5/(-e^2*x^2+d^2)^(3/2)-3*d*arctan(e*x/(-e^2*x^2
+d^2)^(1/2))/e^5+24/5*d*(e*x+d)/e^5/(-e^2*x^2+d^2)^(1/2)+(-e^2*x^2+d^2)^(1/2)/e^5

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Rubi [A]
time = 0.19, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1649, 655, 223, 209} \begin {gather*} -\frac {3 d \text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5}-\frac {6 d^2 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {24 d (d+e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}+\frac {\sqrt {d^2-e^2 x^2}}{e^5}+\frac {d^3 (d+e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

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

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]

Rubi steps

\begin {align*} \int \frac {x^4 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {d^3 (d+e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {(d+e x)^2 \left (\frac {3 d^4}{e^4}+\frac {5 d^3 x}{e^3}+\frac {5 d^2 x^2}{e^2}+\frac {5 d x^3}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d}\\ &=\frac {d^3 (d+e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {6 d^2 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {(d+e x) \left (\frac {27 d^4}{e^4}+\frac {30 d^3 x}{e^3}+\frac {15 d^2 x^2}{e^2}\right )}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2}\\ &=\frac {d^3 (d+e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {6 d^2 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {24 d (d+e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {\frac {45 d^4}{e^4}+\frac {15 d^3 x}{e^3}}{\sqrt {d^2-e^2 x^2}} \, dx}{15 d^3}\\ &=\frac {d^3 (d+e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {6 d^2 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {24 d (d+e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}+\frac {\sqrt {d^2-e^2 x^2}}{e^5}-\frac {(3 d) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^4}\\ &=\frac {d^3 (d+e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {6 d^2 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {24 d (d+e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}+\frac {\sqrt {d^2-e^2 x^2}}{e^5}-\frac {(3 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^4}\\ &=\frac {d^3 (d+e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {6 d^2 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {24 d (d+e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}+\frac {\sqrt {d^2-e^2 x^2}}{e^5}-\frac {3 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5}\\ \end {align*}

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Mathematica [A]
time = 0.38, size = 108, normalized size = 0.76 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-24 d^3+57 d^2 e x-39 d e^2 x^2+5 e^3 x^3\right )}{5 e^5 (-d+e x)^3}+\frac {3 d \left (-e^2\right )^{3/2} \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{e^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-24*d^3 + 57*d^2*e*x - 39*d*e^2*x^2 + 5*e^3*x^3))/(5*e^5*(-d + e*x)^3) + (3*d*(-e^2)^(3/
2)*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/e^8

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(437\) vs. \(2(126)=252\).
time = 0.07, size = 438, normalized size = 3.08

method	result	size
risch	\(\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{e^{5}}-\frac {3 d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{4} \sqrt {e^{2}}}-\frac {24 d \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d \left (x -\frac {d}{e}\right ) e}}{5 e^{6} \left (x -\frac {d}{e}\right )}-\frac {d^{3} \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d \left (x -\frac {d}{e}\right ) e}}{5 e^{8} \left (x -\frac {d}{e}\right )^{3}}-\frac {6 d^{2} \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d \left (x -\frac {d}{e}\right ) e}}{5 e^{7} \left (x -\frac {d}{e}\right )^{2}}\)	\(195\)
default	\(e^{3} \left (-\frac {x^{6}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {6 d^{2} \left (\frac {x^{4}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {4 d^{2} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )}{e^{2}}\right )}{e^{2}}\right )+3 d \,e^{2} \left (\frac {x^{5}}{5 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {\frac {x^{3}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}}{e^{2}}}{e^{2}}\right )+3 d^{2} e \left (\frac {x^{4}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {4 d^{2} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )}{e^{2}}\right )+d^{3} \left (\frac {x^{3}}{2 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {3 d^{2} \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )}{2 e^{2}}\right )\)	\(438\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (120) = 240\).
time = 0.48, size = 299, normalized size = 2.11 \begin {gather*} -\frac {x^{6} e}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {9 \, d^{2} x^{4} e^{\left (-1\right )}}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {d^{3} x^{3} e^{\left (-2\right )}}{2 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} - \frac {12 \, d^{4} x^{2} e^{\left (-3\right )}}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} - \frac {3 \, d^{5} x e^{\left (-4\right )}}{10 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {24 \, d^{6} e^{\left (-5\right )}}{5 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {1}{5} \, {\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^{2} + \frac {9 \, d^{3} x e^{\left (-4\right )}}{10 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}}} - {\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 - 3 \, d \arcsin \left (\frac {x e}{d}\right ) e^{\left (-5\right )} - \frac {6 \, d x e^{\left (-4\right )}}{5 \, \sqrt {-x^{2} e^{2} + d^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

-x^6*e/(-x^2*e^2 + d^2)^(5/2) + 9*d^2*x^4*e^(-1)/(-x^2*e^2 + d^2)^(5/2) + 1/2*d^3*x^3*e^(-2)/(-x^2*e^2 + d^2)^
(5/2) - 12*d^4*x^2*e^(-3)/(-x^2*e^2 + d^2)^(5/2) - 3/10*d^5*x*e^(-4)/(-x^2*e^2 + d^2)^(5/2) + 24/5*d^6*e^(-5)/
(-x^2*e^2 + d^2)^(5/2) + 1/5*(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 + 9/10*d^3*x*e^(-4)/(-x^2*e^2 + d^2)^(3/2) - (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 - 3*d*arcsin(x*e/d)*e^(-5) - 6/5*d*x*e^(-4)/sqrt
(-x^2*e^2 + d^2)

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Fricas [A]
time = 3.04, size = 167, normalized size = 1.18 \begin {gather*} \frac {24 \, d x^{3} e^{3} - 72 \, d^{2} x^{2} e^{2} + 72 \, d^{3} x e - 24 \, d^{4} + 30 \, {\left (d x^{3} e^{3} - 3 \, d^{2} x^{2} e^{2} + 3 \, d^{3} x e - d^{4}\right )} \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) + {\left (5 \, x^{3} e^{3} - 39 \, d x^{2} e^{2} + 57 \, d^{2} x e - 24 \, d^{3}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{5 \, {\left (x^{3} e^{8} - 3 \, d x^{2} e^{7} + 3 \, d^{2} x e^{6} - d^{3} e^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

1/5*(24*d*x^3*e^3 - 72*d^2*x^2*e^2 + 72*d^3*x*e - 24*d^4 + 30*(d*x^3*e^3 - 3*d^2*x^2*e^2 + 3*d^3*x*e - d^4)*ar
ctan(-(d - sqrt(-x^2*e^2 + d^2))*e^(-1)/x) + (5*x^3*e^3 - 39*d*x^2*e^2 + 57*d^2*x*e - 24*d^3)*sqrt(-x^2*e^2 +
d^2))/(x^3*e^8 - 3*d*x^2*e^7 + 3*d^2*x*e^6 - d^3*e^5)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4} \left (d + e x\right )^{3}}{\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**4*(e*x+d)**3/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral(x**4*(d + e*x)**3/(-(-d + e*x)*(d + e*x))**(7/2), x)

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Giac [A]
time = 0.69, size = 193, normalized size = 1.36 \begin {gather*} -3 \, d \arcsin \left (\frac {x e}{d}\right ) e^{\left (-5\right )} \mathrm {sgn}\left (d\right ) + \sqrt {-x^{2} e^{2} + d^{2}} e^{\left (-5\right )} - \frac {2 \, {\left (\frac {80 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d e^{\left (-2\right )}}{x} - \frac {120 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d e^{\left (-4\right )}}{x^{2}} + \frac {70 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} d e^{\left (-6\right )}}{x^{3}} - \frac {15 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} d e^{\left (-8\right )}}{x^{4}} - 19 \, d\right )} e^{\left (-5\right )}}{5 \, {\left (\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} - 1\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-3*d*arcsin(x*e/d)*e^(-5)*sgn(d) + sqrt(-x^2*e^2 + d^2)*e^(-5) - 2/5*(80*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d*e^(-
2)/x - 120*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d*e^(-4)/x^2 + 70*(d*e + sqrt(-x^2*e^2 + d^2)*e)^3*d*e^(-6)/x^3 -
15*(d*e + sqrt(-x^2*e^2 + d^2)*e)^4*d*e^(-8)/x^4 - 19*d)*e^(-5)/((d*e + sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/x - 1)^
5

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^4\,{\left (d+e\,x\right )}^3}{{\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^4*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2),x)

[Out]

int((x^4*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2), x)

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