3.1.4 \(\int x^2 (d+e x) (d^2-e^2 x^2)^{3/2} \, dx\)

Optimal. Leaf size=159 \[ -\frac {d x \left (d^2-e^2 x^2\right )^{5/2}}{6 e^2}-\frac {d^2 \left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}+\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac {d^7 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^3}+\frac {d^5 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e^2} \]

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Rubi [A]  time = 0.11, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {797, 641, 195, 217, 203} \begin {gather*} \frac {d^5 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e^2}-\frac {d^2 \left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}-\frac {d x \left (d^2-e^2 x^2\right )^{5/2}}{6 e^2}+\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac {d^7 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 797

Int[(x_)^2*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/c, Int[(f + g*x)*(a + c*x^2)^(p
+ 1), x], x] - Dist[a/c, Int[(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, f, g, p}, x] && EqQ[a*g^2 + f^2*
c, 0]

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 135, normalized size = 0.85 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (105 d^6 \sin ^{-1}\left (\frac {e x}{d}\right )-\sqrt {1-\frac {e^2 x^2}{d^2}} \left (96 d^6+105 d^5 e x+48 d^4 e^2 x^2-490 d^3 e^3 x^3-384 d^2 e^4 x^4+280 d e^5 x^5+240 e^6 x^6\right )\right )}{1680 e^3 \sqrt {1-\frac {e^2 x^2}{d^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-(Sqrt[1 - (e^2*x^2)/d^2]*(96*d^6 + 105*d^5*e*x + 48*d^4*e^2*x^2 - 490*d^3*e^3*x^3 - 384
*d^2*e^4*x^4 + 280*d*e^5*x^5 + 240*e^6*x^6)) + 105*d^6*ArcSin[(e*x)/d]))/(1680*e^3*Sqrt[1 - (e^2*x^2)/d^2])

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IntegrateAlgebraic [A]  time = 0.45, size = 136, normalized size = 0.86 \begin {gather*} \frac {d^7 \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{16 e^4}+\frac {\sqrt {d^2-e^2 x^2} \left (-96 d^6-105 d^5 e x-48 d^4 e^2 x^2+490 d^3 e^3 x^3+384 d^2 e^4 x^4-280 d e^5 x^5-240 e^6 x^6\right )}{1680 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^2*(d + e*x)*(d^2 - e^2*x^2)^(3/2),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-96*d^6 - 105*d^5*e*x - 48*d^4*e^2*x^2 + 490*d^3*e^3*x^3 + 384*d^2*e^4*x^4 - 280*d*e^5*x
^5 - 240*e^6*x^6))/(1680*e^3) + (d^7*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(16*e^4)

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fricas [A]  time = 0.40, size = 116, normalized size = 0.73 \begin {gather*} -\frac {210 \, d^{7} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (240 \, e^{6} x^{6} + 280 \, d e^{5} x^{5} - 384 \, d^{2} e^{4} x^{4} - 490 \, d^{3} e^{3} x^{3} + 48 \, d^{4} e^{2} x^{2} + 105 \, d^{5} e x + 96 \, d^{6}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{1680 \, e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1680*(210*d^7*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (240*e^6*x^6 + 280*d*e^5*x^5 - 384*d^2*e^4*x^4 -
490*d^3*e^3*x^3 + 48*d^4*e^2*x^2 + 105*d^5*e*x + 96*d^6)*sqrt(-e^2*x^2 + d^2))/e^3

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giac [A]  time = 0.23, size = 96, normalized size = 0.60 \begin {gather*} \frac {1}{16} \, d^{7} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-3\right )} \mathrm {sgn}\relax (d) - \frac {1}{1680} \, {\left (96 \, d^{6} e^{\left (-3\right )} + {\left (105 \, d^{5} e^{\left (-2\right )} + 2 \, {\left (24 \, d^{4} e^{\left (-1\right )} - {\left (245 \, d^{3} + 4 \, {\left (48 \, d^{2} e - 5 \, {\left (6 \, x e^{3} + 7 \, d e^{2}\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \sqrt {-x^{2} e^{2} + d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*d^7*arcsin(x*e/d)*e^(-3)*sgn(d) - 1/1680*(96*d^6*e^(-3) + (105*d^5*e^(-2) + 2*(24*d^4*e^(-1) - (245*d^3 +
4*(48*d^2*e - 5*(6*x*e^3 + 7*d*e^2)*x)*x)*x)*x)*x)*sqrt(-x^2*e^2 + d^2)

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maple [A]  time = 0.04, size = 148, normalized size = 0.93 \begin {gather*} \frac {d^{7} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{16 \sqrt {e^{2}}\, e^{2}}+\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, d^{5} x}{16 e^{2}}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{3} x}{24 e^{2}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} x^{2}}{7 e}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d x}{6 e^{2}}-\frac {2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d^{2}}{35 e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7*x^2*(-e^2*x^2+d^2)^(5/2)/e-2/35*d^2*(-e^2*x^2+d^2)^(5/2)/e^3-1/6*d*x*(-e^2*x^2+d^2)^(5/2)/e^2+1/24*d^3*x*
(-e^2*x^2+d^2)^(3/2)/e^2+1/16*d^5*x*(-e^2*x^2+d^2)^(1/2)/e^2+1/16*d^7/e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2
*x^2+d^2)^(1/2)*x)

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maxima [A]  time = 0.95, size = 127, normalized size = 0.80 \begin {gather*} \frac {d^{7} \arcsin \left (\frac {e x}{d}\right )}{16 \, e^{3}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{5} x}{16 \, e^{2}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3} x}{24 \, e^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} x^{2}}{7 \, e} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d x}{6 \, e^{2}} - \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2}}{35 \, e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*d^7*arcsin(e*x/d)/e^3 + 1/16*sqrt(-e^2*x^2 + d^2)*d^5*x/e^2 + 1/24*(-e^2*x^2 + d^2)^(3/2)*d^3*x/e^2 - 1/7
*(-e^2*x^2 + d^2)^(5/2)*x^2/e - 1/6*(-e^2*x^2 + d^2)^(5/2)*d*x/e^2 - 2/35*(-e^2*x^2 + d^2)^(5/2)*d^2/e^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 12.27, size = 653, normalized size = 4.11 \begin {gather*} d^{3} \left (\begin {cases} - \frac {i d^{4} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{8 e^{3}} + \frac {i d^{3} x}{8 e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {3 i d x^{3}}{8 \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {i e^{2} x^{5}}{4 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{4} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{8 e^{3}} - \frac {d^{3} x}{8 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {3 d x^{3}}{8 \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - \frac {e^{2} x^{5}}{4 d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) + d^{2} e \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 ) - d e^{2} \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 ) - e^{3} \left (\begin {cases} - \frac {8 d^{6} \sqrt {d^{2} - e^{2} x^{2}}}{105 e^{6}} - \frac {4 d^{4} x^{2} \sqrt {d^{2} - e^{2} x^{2}}}{105 e^{4}} - \frac {d^{2} x^{4} \sqrt {d^{2} - e^{2} x^{2}}}{35 e^{2}} + \frac {x^{6} \sqrt {d^{2} - e^{2} x^{2}}}{7} & \text {for}\: e \neq 0 \\\frac {x^{6} \sqrt {d^{2}}}{6} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(e*x+d)*(-e**2*x**2+d**2)**(3/2),x)

[Out]

d**3*Piecewise((-I*d**4*acosh(e*x/d)/(8*e**3) + I*d**3*x/(8*e**2*sqrt(-1 + e**2*x**2/d**2)) - 3*I*d*x**3/(8*sq
rt(-1 + e**2*x**2/d**2)) + I*e**2*x**5/(4*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (d**4*asin(e
*x/d)/(8*e**3) - d**3*x/(8*e**2*sqrt(1 - e**2*x**2/d**2)) + 3*d*x**3/(8*sqrt(1 - e**2*x**2/d**2)) - e**2*x**5/
(4*d*sqrt(1 - e**2*x**2/d**2)), True)) + d**2*e*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)) - d*
e**2*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/(4
8*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*asin(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*sq
rt(1 - e**2*x**2/d**2)), True)) - e**3*Piecewise((-8*d**6*sqrt(d**2 - e**2*x**2)/(105*e**6) - 4*d**4*x**2*sqrt
(d**2 - e**2*x**2)/(105*e**4) - d**2*x**4*sqrt(d**2 - e**2*x**2)/(35*e**2) + x**6*sqrt(d**2 - e**2*x**2)/7, Ne
(e, 0)), (x**6*sqrt(d**2)/6, True))

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