3.7.79 \(\int \frac {(d+e x)^4}{(d^2-e^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=81 \[ \frac {2 (d+e x)^3}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 (d+e x)}{e \sqrt {d^2-e^2 x^2}}+\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e} \]

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Rubi [A]  time = 0.02, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {669, 653, 217, 203} \begin {gather*} \frac {2 (d+e x)^3}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 (d+e x)}{e \sqrt {d^2-e^2 x^2}}+\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^4/(d^2 - e^2*x^2)^(5/2),x]

[Out]

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

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 653

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

Rule 669

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*(p + 1)), x] - Dist[(e^2*(m + p))/(c*(p + 1)), Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 100, normalized size = 1.23 \begin {gather*} -\frac {(d+e x) \left (4 d (d-2 e x) \sqrt {1-\frac {e^2 x^2}{d^2}}-3 (d-e x)^2 \sin ^{-1}\left (\frac {e x}{d}\right )\right )}{3 d e (d-e x) \sqrt {d^2-e^2 x^2} \sqrt {1-\frac {e^2 x^2}{d^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^4/(d^2 - e^2*x^2)^(5/2),x]

[Out]

-1/3*((d + e*x)*(4*d*(d - 2*e*x)*Sqrt[1 - (e^2*x^2)/d^2] - 3*(d - e*x)^2*ArcSin[(e*x)/d]))/(d*e*(d - e*x)*Sqrt
[d^2 - e^2*x^2]*Sqrt[1 - (e^2*x^2)/d^2])

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IntegrateAlgebraic [A]  time = 0.43, size = 82, normalized size = 1.01 \begin {gather*} \frac {\sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{e^2}-\frac {4 (d-2 e x) \sqrt {d^2-e^2 x^2}}{3 e (e x-d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d + e*x)^4/(d^2 - e^2*x^2)^(5/2),x]

[Out]

(-4*(d - 2*e*x)*Sqrt[d^2 - e^2*x^2])/(3*e*(-d + e*x)^2) + (Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2
]])/e^2

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fricas [A]  time = 0.42, size = 112, normalized size = 1.38 \begin {gather*} -\frac {2 \, {\left (2 \, e^{2} x^{2} - 4 \, d e x + 2 \, d^{2} + 3 \, {\left (e^{2} x^{2} - 2 \, d e x + d^{2}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - 2 \, \sqrt {-e^{2} x^{2} + d^{2}} {\left (2 \, e x - d\right )}\right )}}{3 \, {\left (e^{3} x^{2} - 2 \, d e^{2} x + d^{2} e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(-e^2*x^2+d^2)^(5/2),x, algorithm="fricas")

[Out]

-2/3*(2*e^2*x^2 - 4*d*e*x + 2*d^2 + 3*(e^2*x^2 - 2*d*e*x + d^2)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - 2*
sqrt(-e^2*x^2 + d^2)*(2*e*x - d))/(e^3*x^2 - 2*d*e^2*x + d^2*e)

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giac [A]  time = 0.34, size = 66, normalized size = 0.81 \begin {gather*} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} \mathrm {sgn}\relax (d) - \frac {4 \, {\left (d^{3} e^{\left (-1\right )} - {\left (2 \, x e^{2} + 3 \, d e\right )} x^{2}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{3 \, {\left (x^{2} e^{2} - d^{2}\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(-e^2*x^2+d^2)^(5/2),x, algorithm="giac")

[Out]

arcsin(x*e/d)*e^(-1)*sgn(d) - 4/3*(d^3*e^(-1) - (2*x*e^2 + 3*d*e)*x^2)*sqrt(-x^2*e^2 + d^2)/(x^2*e^2 - d^2)^2

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maple [A]  time = 0.07, size = 132, normalized size = 1.63 \begin {gather*} \frac {e^{2} x^{3}}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {4 d e \,x^{2}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {7 d^{2} x}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {4 d^{3}}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e}-\frac {7 x}{3 \sqrt {-e^{2} x^{2}+d^{2}}}+\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^4/(-e^2*x^2+d^2)^(5/2),x)

[Out]

1/3*e^2*x^3/(-e^2*x^2+d^2)^(3/2)-7/3*x/(-e^2*x^2+d^2)^(1/2)+1/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1
/2)*x)+4*e*d*x^2/(-e^2*x^2+d^2)^(3/2)-4/3*d^3/e/(-e^2*x^2+d^2)^(3/2)+7/3*d^2*x/(-e^2*x^2+d^2)^(3/2)

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maxima [A]  time = 3.08, size = 143, normalized size = 1.77 \begin {gather*} \frac {1}{3} \, e^{4} 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 )} + \frac {4 \, d e x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}} + \frac {7 \, d^{2} x}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}} - \frac {4 \, d^{3}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e} - \frac {5 \, x}{3 \, \sqrt {-e^{2} x^{2} + d^{2}}} + \frac {\arcsin \left (\frac {e x}{d}\right )}{e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(-e^2*x^2+d^2)^(5/2),x, algorithm="maxima")

[Out]

1/3*e^4*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)) + 4*d*e*x^2/(-e^2*x^2 + d^
2)^(3/2) + 7/3*d^2*x/(-e^2*x^2 + d^2)^(3/2) - 4/3*d^3/((-e^2*x^2 + d^2)^(3/2)*e) - 5/3*x/sqrt(-e^2*x^2 + d^2)
+ arcsin(e*x/d)/e

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^4}{{\left (d^2-e^2\,x^2\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^4/(d^2 - e^2*x^2)^(5/2),x)

[Out]

int((d + e*x)^4/(d^2 - e^2*x^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{4}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**4/(-e**2*x**2+d**2)**(5/2),x)

[Out]

Integral((d + e*x)**4/(-(-d + e*x)*(d + e*x))**(5/2), x)

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