### 3.838 $$\int \frac{(d+e x)^4}{(d^2-e^2 x^2)^{5/2}} \, dx$$

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

[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

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Rubi [A]  time = 0.0231172, antiderivative size = 81, normalized size of antiderivative = 1., 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} $\frac{2 (d+e x)^3}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac{2 (d+e x)}{e \sqrt{d^2-e^2 x^2}}+\frac{\tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e}$

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 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]

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 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 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])

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*}

Mathematica [A]  time = 0.120354, size = 100, normalized size = 1.23 $-\frac{(d+e x) \left (4 d (d-2 e x) \sqrt{1-\frac{e^2 x^2}{d^2}}-3 (d-e x)^2 \sin ^{-1}\left (\frac{e x}{d}\right )\right )}{3 d e (d-e x) \sqrt{d^2-e^2 x^2} \sqrt{1-\frac{e^2 x^2}{d^2}}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.059, size = 132, normalized size = 1.6 \begin{align*}{\frac{{e}^{2}{x}^{3}}{3} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}-{\frac{7\,x}{3}{\frac{1}{\sqrt{-{e}^{2}{x}^{2}+{d}^{2}}}}}+{\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{e}^{2}{x}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}+4\,{\frac{de{x}^{2}}{ \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{3/2}}}-{\frac{4\,{d}^{3}}{3\,e} \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}}}} \end{align*}

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)*x/(-e^2*x^2+d^2)^
(1/2))+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 [B]  time = 1.68638, size = 207, normalized size = 2.56 \begin{align*} \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^{2} x}{\sqrt{d^{2} e^{2}}}\right )}{\sqrt{e^{2}}} \end{align*}

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^2*x/sqrt(d^2*e^2))/sqrt(e^2)

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Fricas [A]  time = 2.03055, size = 236, normalized size = 2.91 \begin{align*} -\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{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{align*}

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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Giac [A]  time = 1.41552, size = 89, normalized size = 1.1 \begin{align*} \arcsin \left (\frac{x e}{d}\right ) e^{\left (-1\right )} \mathrm{sgn}\left (d\right ) - \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{align*}

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