### 3.837 $$\int \frac{(d+e x)^5}{(d^2-e^2 x^2)^{5/2}} \, dx$$

Optimal. Leaf size=108 $\frac{2 (d+e x)^4}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac{10 (d+e x)^2}{3 e \sqrt{d^2-e^2 x^2}}-\frac{5 \sqrt{d^2-e^2 x^2}}{e}+\frac{5 d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e}$

[Out]

(2*(d + e*x)^4)/(3*e*(d^2 - e^2*x^2)^(3/2)) - (10*(d + e*x)^2)/(3*e*Sqrt[d^2 - e^2*x^2]) - (5*Sqrt[d^2 - e^2*x
^2])/e + (5*d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/e

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Rubi [A]  time = 0.0398871, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {669, 641, 217, 203} $\frac{2 (d+e x)^4}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac{10 (d+e x)^2}{3 e \sqrt{d^2-e^2 x^2}}-\frac{5 \sqrt{d^2-e^2 x^2}}{e}+\frac{5 d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*(d + e*x)^4)/(3*e*(d^2 - e^2*x^2)^(3/2)) - (10*(d + e*x)^2)/(3*e*Sqrt[d^2 - e^2*x^2]) - (5*Sqrt[d^2 - e^2*x
^2])/e + (5*d*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 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 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)^5}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx &=\frac{2 (d+e x)^4}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac{5}{3} \int \frac{(d+e x)^3}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx\\ &=\frac{2 (d+e x)^4}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac{10 (d+e x)^2}{3 e \sqrt{d^2-e^2 x^2}}+5 \int \frac{d+e x}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=\frac{2 (d+e x)^4}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac{10 (d+e x)^2}{3 e \sqrt{d^2-e^2 x^2}}-\frac{5 \sqrt{d^2-e^2 x^2}}{e}+(5 d) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=\frac{2 (d+e x)^4}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac{10 (d+e x)^2}{3 e \sqrt{d^2-e^2 x^2}}-\frac{5 \sqrt{d^2-e^2 x^2}}{e}+(5 d) \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)^4}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac{10 (d+e x)^2}{3 e \sqrt{d^2-e^2 x^2}}-\frac{5 \sqrt{d^2-e^2 x^2}}{e}+\frac{5 d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e}\\ \end{align*}

Mathematica [A]  time = 0.157079, size = 109, normalized size = 1.01 $\frac{(d+e x) \left (\left (23 d^2-34 d e x+3 e^2 x^2\right ) \sqrt{1-\frac{e^2 x^2}{d^2}}-15 (d-e x)^2 \sin ^{-1}\left (\frac{e x}{d}\right )\right )}{3 e (e x-d) \sqrt{d^2-e^2 x^2} \sqrt{1-\frac{e^2 x^2}{d^2}}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((d + e*x)*((23*d^2 - 34*d*e*x + 3*e^2*x^2)*Sqrt[1 - (e^2*x^2)/d^2] - 15*(d - e*x)^2*ArcSin[(e*x)/d]))/(3*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 = 0.062, size = 160, normalized size = 1.5 \begin{align*} -{{e}^{3}{x}^{4} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}+14\,{\frac{e{d}^{2}{x}^{2}}{ \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{3/2}}}-{\frac{23\,{d}^{4}}{3\,e} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{5\,d{e}^{2}{x}^{3}}{3} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}-{\frac{23\,dx}{3}{\frac{1}{\sqrt{-{e}^{2}{x}^{2}+{d}^{2}}}}}+5\,{\frac{d}{\sqrt{{e}^{2}}}\arctan \left ({\frac{\sqrt{{e}^{2}}x}{\sqrt{-{e}^{2}{x}^{2}+{d}^{2}}}} \right ) }+{\frac{11\,{d}^{3}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)^5/(-e^2*x^2+d^2)^(5/2),x)

[Out]

-e^3*x^4/(-e^2*x^2+d^2)^(3/2)+14*e*d^2*x^2/(-e^2*x^2+d^2)^(3/2)-23/3*d^4/e/(-e^2*x^2+d^2)^(3/2)+5/3*d*e^2*x^3/
(-e^2*x^2+d^2)^(3/2)-23/3*d*x/(-e^2*x^2+d^2)^(1/2)+5*d/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))+
11/3*d^3*x/(-e^2*x^2+d^2)^(3/2)

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Maxima [A]  time = 1.87467, size = 244, normalized size = 2.26 \begin{align*} \frac{5}{3} \, d 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{e^{3} x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}}} + \frac{14 \, d^{2} e x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}}} + \frac{11 \, d^{3} x}{3 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}}} - \frac{23 \, d^{4}}{3 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} e} - \frac{13 \, d x}{3 \, \sqrt{-e^{2} x^{2} + d^{2}}} + \frac{5 \, d \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)^5/(-e^2*x^2+d^2)^(5/2),x, algorithm="maxima")

[Out]

5/3*d*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)) - e^3*x^4/(-e^2*x^2 + d^
2)^(3/2) + 14*d^2*e*x^2/(-e^2*x^2 + d^2)^(3/2) + 11/3*d^3*x/(-e^2*x^2 + d^2)^(3/2) - 23/3*d^4/((-e^2*x^2 + d^2
)^(3/2)*e) - 13/3*d*x/sqrt(-e^2*x^2 + d^2) + 5*d*arcsin(e^2*x/sqrt(d^2*e^2))/sqrt(e^2)

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Fricas [A]  time = 2.1472, size = 277, normalized size = 2.56 \begin{align*} -\frac{23 \, d e^{2} x^{2} - 46 \, d^{2} e x + 23 \, d^{3} + 30 \,{\left (d e^{2} x^{2} - 2 \, d^{2} e x + d^{3}\right )} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) +{\left (3 \, e^{2} x^{2} - 34 \, d e x + 23 \, d^{2}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{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)^5/(-e^2*x^2+d^2)^(5/2),x, algorithm="fricas")

[Out]

-1/3*(23*d*e^2*x^2 - 46*d^2*e*x + 23*d^3 + 30*(d*e^2*x^2 - 2*d^2*e*x + d^3)*arctan(-(d - sqrt(-e^2*x^2 + d^2))
/(e*x)) + (3*e^2*x^2 - 34*d*e*x + 23*d^2)*sqrt(-e^2*x^2 + d^2))/(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 )^{5}}{\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)**5/(-e**2*x**2+d**2)**(5/2),x)

[Out]

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

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Giac [A]  time = 1.34395, size = 116, normalized size = 1.07 \begin{align*} 5 \, d \arcsin \left (\frac{x e}{d}\right ) e^{\left (-1\right )} \mathrm{sgn}\left (d\right ) - \frac{{\left (23 \, d^{4} e^{\left (-1\right )} +{\left (12 \, d^{3} -{\left (42 \, d^{2} e -{\left (3 \, x e^{3} - 28 \, d e^{2}\right )} x\right )} x\right )} x\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)^5/(-e^2*x^2+d^2)^(5/2),x, algorithm="giac")

[Out]

5*d*arcsin(x*e/d)*e^(-1)*sgn(d) - 1/3*(23*d^4*e^(-1) + (12*d^3 - (42*d^2*e - (3*x*e^3 - 28*d*e^2)*x)*x)*x)*sqr
t(-x^2*e^2 + d^2)/(x^2*e^2 - d^2)^2