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

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

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Rubi [A]  time = 0.27, 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 = {1635, 1814, 641, 217, 203} \begin {gather*} \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 {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 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 1635

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 1814

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) \operatorname {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 = 0.21, size = 111, normalized size = 0.78 \begin {gather*} \frac {56 d^4-82 d^3 e x-32 d^2 e^2 x^2-\frac {30 (d-e x)^3 (d+e x) \sin ^{-1}\left (\frac {e x}{d}\right )}{\sqrt {1-\frac {e^2 x^2}{d^2}}}+76 d e^3 x^3-15 e^4 x^4}{15 e^6 (d-e x)^2 \sqrt {d^2-e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(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 - (30*(d - e*x)^3*(d + e*x)*ArcSin[(e*x)/d])
/Sqrt[1 - (e^2*x^2)/d^2])/(15*e^6*(d - e*x)^2*Sqrt[d^2 - e^2*x^2])

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IntegrateAlgebraic [A]  time = 0.65, size = 126, normalized size = 0.88 \begin {gather*} -\frac {2 d \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{e^7}-\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 (e x-d)^3 (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*(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))/(e^6*(-d + e*x)
^3*(d + e*x)) - (2*d*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/e^7

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fricas [A]  time = 0.40, size = 188, normalized size = 1.31 \begin {gather*} \frac {56 \, d e^{4} x^{4} - 112 \, d^{2} e^{3} x^{3} + 112 \, d^{4} e x - 56 \, d^{5} + 60 \, {\left (d e^{4} x^{4} - 2 \, d^{2} e^{3} x^{3} + 2 \, d^{4} e x - d^{5}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (15 \, e^{4} x^{4} - 76 \, d e^{3} x^{3} + 32 \, d^{2} e^{2} x^{2} + 82 \, d^{3} e x - 56 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{10} x^{4} - 2 \, d e^{9} x^{3} + 2 \, d^{3} e^{7} x - 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*e^4*x^4 - 112*d^2*e^3*x^3 + 112*d^4*e*x - 56*d^5 + 60*(d*e^4*x^4 - 2*d^2*e^3*x^3 + 2*d^4*e*x - d^5)
*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (15*e^4*x^4 - 76*d*e^3*x^3 + 32*d^2*e^2*x^2 + 82*d^3*e*x - 56*d^4
)*sqrt(-e^2*x^2 + d^2))/(e^10*x^4 - 2*d*e^9*x^3 + 2*d^3*e^7*x - d^4*e^6)

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giac [A]  time = 0.30, size = 106, normalized size = 0.74 \begin {gather*} -2 \, d \arcsin \left (\frac {x e}{d}\right ) e^{\left (-6\right )} \mathrm {sgn}\relax (d) - \frac {{\left (56 \, d^{6} e^{\left (-6\right )} + {\left (30 \, d^{5} e^{\left (-5\right )} - {\left (140 \, d^{4} e^{\left (-4\right )} + {\left (70 \, d^{3} e^{\left (-3\right )} - {\left (105 \, d^{2} e^{\left (-2\right )} + {\left (46 \, d e^{\left (-1\right )} - 15 \, x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{15 \, {\left (x^{2} e^{2} - d^{2}\right )}^{3}} \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]

-2*d*arcsin(x*e/d)*e^(-6)*sgn(d) - 1/15*(56*d^6*e^(-6) + (30*d^5*e^(-5) - (140*d^4*e^(-4) + (70*d^3*e^(-3) - (
105*d^2*e^(-2) + (46*d*e^(-1) - 15*x)*x)*x)*x)*x)*x)*sqrt(-x^2*e^2 + d^2)/(x^2*e^2 - d^2)^3

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maple [A]  time = 0.01, size = 193, normalized size = 1.35 \begin {gather*} -\frac {x^{6}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {2 d \,x^{5}}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e}+\frac {7 d^{2} x^{4}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{2}}-\frac {28 d^{4} x^{2}}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{4}}-\frac {2 d \,x^{3}}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{3}}+\frac {56 d^{6}}{15 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{6}}+\frac {2 d x}{\sqrt {-e^{2} x^{2}+d^{2}}\, e^{5}}-\frac {2 d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}\, e^{5}} \end {gather*}

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)

[Out]

-x^6/(-e^2*x^2+d^2)^(5/2)+7/e^2*d^2*x^4/(-e^2*x^2+d^2)^(5/2)-28/3/e^4*d^4*x^2/(-e^2*x^2+d^2)^(5/2)+56/15/e^6*d
^6/(-e^2*x^2+d^2)^(5/2)+2/5/e*d*x^5/(-e^2*x^2+d^2)^(5/2)-2/3/e^3*d*x^3/(-e^2*x^2+d^2)^(3/2)+2/e^5*d*x/(-e^2*x^
2+d^2)^(1/2)-2/e^5*d/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)

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maxima [B]  time = 1.02, size = 276, normalized size = 1.93 \begin {gather*} \frac {2}{15} \, d e x {\left (\frac {15 \, x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {20 \, d^{2} x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {8 \, d^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{6}}\right )} - \frac {x^{6}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} - \frac {2 \, d x {\left (\frac {3 \, x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}} - \frac {2 \, d^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}}\right )}}{3 \, e} + \frac {7 \, d^{2} x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {28 \, d^{4} x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {56 \, d^{6}}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{6}} + \frac {8 \, d^{3} x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{5}} - \frac {14 \, d x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{5}} - \frac {2 \, d \arcsin \left (\frac {e x}{d}\right )}{e^{6}} \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]

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

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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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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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