### 3.836 $$\int \frac{(d+e x)^6}{(d^2-e^2 x^2)^{5/2}} \, dx$$

Optimal. Leaf size=143 $\frac{2 (d+e x)^5}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac{14 (d+e x)^3}{3 e \sqrt{d^2-e^2 x^2}}-\frac{35 \sqrt{d^2-e^2 x^2} (d+e x)}{6 e}-\frac{35 d \sqrt{d^2-e^2 x^2}}{2 e}+\frac{35 d^2 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e}$

[Out]

(2*(d + e*x)^5)/(3*e*(d^2 - e^2*x^2)^(3/2)) - (14*(d + e*x)^3)/(3*e*Sqrt[d^2 - e^2*x^2]) - (35*d*Sqrt[d^2 - e^
2*x^2])/(2*e) - (35*(d + e*x)*Sqrt[d^2 - e^2*x^2])/(6*e) + (35*d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e)

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Rubi [A]  time = 0.0584483, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.208, Rules used = {669, 671, 641, 217, 203} $\frac{2 (d+e x)^5}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac{14 (d+e x)^3}{3 e \sqrt{d^2-e^2 x^2}}-\frac{35 \sqrt{d^2-e^2 x^2} (d+e x)}{6 e}-\frac{35 d \sqrt{d^2-e^2 x^2}}{2 e}+\frac{35 d^2 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^6/(d^2 - e^2*x^2)^(5/2),x]

[Out]

(2*(d + e*x)^5)/(3*e*(d^2 - e^2*x^2)^(3/2)) - (14*(d + e*x)^3)/(3*e*Sqrt[d^2 - e^2*x^2]) - (35*d*Sqrt[d^2 - e^
2*x^2])/(2*e) - (35*(d + e*x)*Sqrt[d^2 - e^2*x^2])/(6*e) + (35*d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(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 671

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*(m + 2*p + 1)), x] + Dist[(2*c*d*(m + p))/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p, x]
, x] /; FreeQ[{a, c, d, e, p}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && 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)^6}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx &=\frac{2 (d+e x)^5}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac{7}{3} \int \frac{(d+e x)^4}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx\\ &=\frac{2 (d+e x)^5}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac{14 (d+e x)^3}{3 e \sqrt{d^2-e^2 x^2}}+\frac{35}{3} \int \frac{(d+e x)^2}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=\frac{2 (d+e x)^5}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac{14 (d+e x)^3}{3 e \sqrt{d^2-e^2 x^2}}-\frac{35 (d+e x) \sqrt{d^2-e^2 x^2}}{6 e}+\frac{1}{2} (35 d) \int \frac{d+e x}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=\frac{2 (d+e x)^5}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac{14 (d+e x)^3}{3 e \sqrt{d^2-e^2 x^2}}-\frac{35 d \sqrt{d^2-e^2 x^2}}{2 e}-\frac{35 (d+e x) \sqrt{d^2-e^2 x^2}}{6 e}+\frac{1}{2} \left (35 d^2\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=\frac{2 (d+e x)^5}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac{14 (d+e x)^3}{3 e \sqrt{d^2-e^2 x^2}}-\frac{35 d \sqrt{d^2-e^2 x^2}}{2 e}-\frac{35 (d+e x) \sqrt{d^2-e^2 x^2}}{6 e}+\frac{1}{2} \left (35 d^2\right ) \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)^5}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac{14 (d+e x)^3}{3 e \sqrt{d^2-e^2 x^2}}-\frac{35 d \sqrt{d^2-e^2 x^2}}{2 e}-\frac{35 (d+e x) \sqrt{d^2-e^2 x^2}}{6 e}+\frac{35 d^2 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e}\\ \end{align*}

Mathematica [A]  time = 0.21352, size = 121, normalized size = 0.85 $\frac{(d+e x) \left (\sqrt{1-\frac{e^2 x^2}{d^2}} \left (-229 d^2 e x+164 d^3+30 d e^2 x^2+3 e^3 x^3\right )-105 d (d-e x)^2 \sin ^{-1}\left (\frac{e x}{d}\right )\right )}{6 e (e x-d) \sqrt{d^2-e^2 x^2} \sqrt{1-\frac{e^2 x^2}{d^2}}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^6/(d^2 - e^2*x^2)^(5/2),x]

[Out]

((d + e*x)*(Sqrt[1 - (e^2*x^2)/d^2]*(164*d^3 - 229*d^2*e*x + 30*d*e^2*x^2 + 3*e^3*x^3) - 105*d*(d - e*x)^2*Arc
Sin[(e*x)/d]))/(6*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 = 0.08, size = 189, normalized size = 1.3 \begin{align*} -{\frac{{e}^{4}{x}^{5}}{2} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{35\,{d}^{2}{e}^{2}{x}^{3}}{6} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}-{\frac{131\,{d}^{2}x}{6}{\frac{1}{\sqrt{-{e}^{2}{x}^{2}+{d}^{2}}}}}+{\frac{35\,{d}^{2}}{2}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{e}^{2}{x}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}-6\,{\frac{{e}^{3}d{x}^{4}}{ \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{3/2}}}+44\,{\frac{{d}^{3}e{x}^{2}}{ \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{3/2}}}-{\frac{82\,{d}^{5}}{3\,e} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{16\,{d}^{4}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)^6/(-e^2*x^2+d^2)^(5/2),x)

[Out]

-1/2*e^4*x^5/(-e^2*x^2+d^2)^(3/2)+35/6*e^2*d^2*x^3/(-e^2*x^2+d^2)^(3/2)-131/6*d^2*x/(-e^2*x^2+d^2)^(1/2)+35/2*
d^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))-6*d*e^3*x^4/(-e^2*x^2+d^2)^(3/2)+44*d^3*e*x^2/(-e^2
*x^2+d^2)^(3/2)-82/3*d^5/e/(-e^2*x^2+d^2)^(3/2)+16/3*d^4*x/(-e^2*x^2+d^2)^(3/2)

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Maxima [A]  time = 1.65743, size = 284, normalized size = 1.99 \begin{align*} \frac{35}{6} \, d^{2} 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^{4} x^{5}}{2 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}}} - \frac{6 \, d e^{3} x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}}} + \frac{44 \, d^{3} e x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}}} + \frac{16 \, d^{4} x}{3 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}}} - \frac{82 \, d^{5}}{3 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} e} - \frac{61 \, d^{2} x}{6 \, \sqrt{-e^{2} x^{2} + d^{2}}} + \frac{35 \, d^{2} \arcsin \left (\frac{e^{2} x}{\sqrt{d^{2} e^{2}}}\right )}{2 \, \sqrt{e^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^6/(-e^2*x^2+d^2)^(5/2),x, algorithm="maxima")

[Out]

35/6*d^2*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)) - 1/2*e^4*x^5/(-e^2*x
^2 + d^2)^(3/2) - 6*d*e^3*x^4/(-e^2*x^2 + d^2)^(3/2) + 44*d^3*e*x^2/(-e^2*x^2 + d^2)^(3/2) + 16/3*d^4*x/(-e^2*
x^2 + d^2)^(3/2) - 82/3*d^5/((-e^2*x^2 + d^2)^(3/2)*e) - 61/6*d^2*x/sqrt(-e^2*x^2 + d^2) + 35/2*d^2*arcsin(e^2
*x/sqrt(d^2*e^2))/sqrt(e^2)

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Fricas [A]  time = 2.22356, size = 313, normalized size = 2.19 \begin{align*} -\frac{164 \, d^{2} e^{2} x^{2} - 328 \, d^{3} e x + 164 \, d^{4} + 210 \,{\left (d^{2} e^{2} x^{2} - 2 \, d^{3} e x + d^{4}\right )} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) +{\left (3 \, e^{3} x^{3} + 30 \, d e^{2} x^{2} - 229 \, d^{2} e x + 164 \, d^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{6 \,{\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)^6/(-e^2*x^2+d^2)^(5/2),x, algorithm="fricas")

[Out]

-1/6*(164*d^2*e^2*x^2 - 328*d^3*e*x + 164*d^4 + 210*(d^2*e^2*x^2 - 2*d^3*e*x + d^4)*arctan(-(d - sqrt(-e^2*x^2
+ d^2))/(e*x)) + (3*e^3*x^3 + 30*d*e^2*x^2 - 229*d^2*e*x + 164*d^3)*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 )^{6}}{\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)**6/(-e**2*x**2+d**2)**(5/2),x)

[Out]

Integral((d + e*x)**6/(-(-d + e*x)*(d + e*x))**(5/2), x)

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Giac [A]  time = 1.2638, size = 131, normalized size = 0.92 \begin{align*} \frac{35}{2} \, d^{2} \arcsin \left (\frac{x e}{d}\right ) e^{\left (-1\right )} \mathrm{sgn}\left (d\right ) - \frac{{\left (164 \, d^{5} e^{\left (-1\right )} +{\left (99 \, d^{4} -{\left (264 \, d^{3} e +{\left (166 \, d^{2} e^{2} - 3 \,{\left (x e^{4} + 12 \, d e^{3}\right )} x\right )} x\right )} x\right )} x\right )} \sqrt{-x^{2} e^{2} + d^{2}}}{6 \,{\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)^6/(-e^2*x^2+d^2)^(5/2),x, algorithm="giac")

[Out]

35/2*d^2*arcsin(x*e/d)*e^(-1)*sgn(d) - 1/6*(164*d^5*e^(-1) + (99*d^4 - (264*d^3*e + (166*d^2*e^2 - 3*(x*e^4 +
12*d*e^3)*x)*x)*x)*x)*sqrt(-x^2*e^2 + d^2)/(x^2*e^2 - d^2)^2