3.849 \(\int \frac{(d+e x)^6}{(d^2-e^2 x^2)^{7/2}} \, dx\)

Optimal. Leaf size=112 \[ \frac{2 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac{2 (d+e x)^3}{3 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 (d+e x)}{e \sqrt{d^2-e^2 x^2}}-\frac{\tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e} \]

[Out]

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

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Rubi [A]  time = 0.0331972, antiderivative size = 112, 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, 653, 217, 203} \[ \frac{2 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac{2 (d+e x)^3}{3 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 (d+e x)}{e \sqrt{d^2-e^2 x^2}}-\frac{\tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^6/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(2*(d + e*x)^5)/(5*e*(d^2 - e^2*x^2)^(5/2)) - (2*(d + e*x)^3)/(3*e*(d^2 - e^2*x^2)^(3/2)) + (2*(d + e*x))/(e*S
qrt[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)^6}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac{2 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\int \frac{(d+e x)^4}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx\\ &=\frac{2 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\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)^5}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\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)^5}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\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)^5}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\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 = 0.193265, size = 113, normalized size = 1.01 \[ \frac{(d+e x) \left (2 d \left (13 d^2-24 d e x+23 e^2 x^2\right ) \sqrt{1-\frac{e^2 x^2}{d^2}}-15 (d-e x)^3 \sin ^{-1}\left (\frac{e x}{d}\right )\right )}{15 d e (d-e x)^2 \sqrt{d^2-e^2 x^2} \sqrt{1-\frac{e^2 x^2}{d^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^6/(d^2 - e^2*x^2)^(7/2),x]

[Out]

((d + e*x)*(2*d*(13*d^2 - 24*d*e*x + 23*e^2*x^2)*Sqrt[1 - (e^2*x^2)/d^2] - 15*(d - e*x)^3*ArcSin[(e*x)/d]))/(1
5*d*e*(d - e*x)^2*Sqrt[d^2 - e^2*x^2]*Sqrt[1 - (e^2*x^2)/d^2])

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Maple [B]  time = 0.084, size = 225, normalized size = 2. \begin{align*}{\frac{{e}^{4}{x}^{5}}{5} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}-{\frac{{e}^{2}{x}^{3}}{3} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{38\,x}{15}{\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}}}}}+6\,{\frac{{e}^{3}d{x}^{4}}{ \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{5/2}}}-{\frac{4\,{d}^{3}e{x}^{2}}{3} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{26\,{d}^{5}}{15\,e} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{15\,{d}^{2}{e}^{2}{x}^{3}}{2} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}-{\frac{13\,{d}^{4}x}{10} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{23\,{d}^{2}x}{30} \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)^(7/2),x)

[Out]

1/5*e^4*x^5/(-e^2*x^2+d^2)^(5/2)-1/3*e^2*x^3/(-e^2*x^2+d^2)^(3/2)+38/15*x/(-e^2*x^2+d^2)^(1/2)-1/(e^2)^(1/2)*a
rctan((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)^(5/2)-4/3*d^3*e*x^2/(-e^2*x^2+d^2)^(5/2)+
26/15*d^5/e/(-e^2*x^2+d^2)^(5/2)+15/2*e^2*d^2*x^3/(-e^2*x^2+d^2)^(5/2)-13/10*d^4*x/(-e^2*x^2+d^2)^(5/2)+23/30*
d^2*x/(-e^2*x^2+d^2)^(3/2)

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Maxima [B]  time = 1.98363, size = 405, normalized size = 3.62 \begin{align*} \frac{1}{15} \, e^{6} x{\left (\frac{15 \, x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e^{2}} - \frac{20 \, d^{2} x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e^{4}} + \frac{8 \, d^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e^{6}}\right )} - \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{6 \, d e^{3} x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}}} + \frac{15 \, d^{2} e^{2} x^{3}}{2 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}}} - \frac{4 \, d^{3} e x^{2}}{3 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}}} - \frac{13 \, d^{4} x}{10 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}}} + \frac{26 \, d^{5}}{15 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e} + \frac{31 \, d^{2} x}{30 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}}} + \frac{16 \, x}{15 \, \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)^6/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

1/15*e^6*x*(15*x^4/((-e^2*x^2 + d^2)^(5/2)*e^2) - 20*d^2*x^2/((-e^2*x^2 + d^2)^(5/2)*e^4) + 8*d^4/((-e^2*x^2 +
d^2)^(5/2)*e^6)) - 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)) + 6*d*
e^3*x^4/(-e^2*x^2 + d^2)^(5/2) + 15/2*d^2*e^2*x^3/(-e^2*x^2 + d^2)^(5/2) - 4/3*d^3*e*x^2/(-e^2*x^2 + d^2)^(5/2
) - 13/10*d^4*x/(-e^2*x^2 + d^2)^(5/2) + 26/15*d^5/((-e^2*x^2 + d^2)^(5/2)*e) + 31/30*d^2*x/(-e^2*x^2 + d^2)^(
3/2) + 16/15*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 = 2.20407, size = 333, normalized size = 2.97 \begin{align*} \frac{2 \,{\left (13 \, e^{3} x^{3} - 39 \, d e^{2} x^{2} + 39 \, d^{2} e x - 13 \, d^{3} + 15 \,{\left (e^{3} x^{3} - 3 \, d e^{2} x^{2} + 3 \, d^{2} e x - d^{3}\right )} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) -{\left (23 \, e^{2} x^{2} - 24 \, d e x + 13 \, d^{2}\right )} \sqrt{-e^{2} x^{2} + d^{2}}\right )}}{15 \,{\left (e^{4} x^{3} - 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x - d^{3} 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)^(7/2),x, algorithm="fricas")

[Out]

2/15*(13*e^3*x^3 - 39*d*e^2*x^2 + 39*d^2*e*x - 13*d^3 + 15*(e^3*x^3 - 3*d*e^2*x^2 + 3*d^2*e*x - d^3)*arctan(-(
d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (23*e^2*x^2 - 24*d*e*x + 13*d^2)*sqrt(-e^2*x^2 + d^2))/(e^4*x^3 - 3*d*e^3*x
^2 + 3*d^2*e^2*x - d^3*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{7}{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)**(7/2),x)

[Out]

Integral((d + e*x)**6/(-(-d + e*x)*(d + e*x))**(7/2), x)

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Giac [A]  time = 1.37681, size = 128, normalized size = 1.14 \begin{align*} -\arcsin \left (\frac{x e}{d}\right ) e^{\left (-1\right )} \mathrm{sgn}\left (d\right ) - \frac{2 \,{\left (13 \, d^{5} e^{\left (-1\right )} +{\left (15 \, d^{4} -{\left (10 \, d^{3} e -{\left (10 \, d^{2} e^{2} +{\left (23 \, x e^{4} + 45 \, d e^{3}\right )} x\right )} x\right )} x\right )} x\right )} \sqrt{-x^{2} e^{2} + d^{2}}}{15 \,{\left (x^{2} e^{2} - d^{2}\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^6/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-arcsin(x*e/d)*e^(-1)*sgn(d) - 2/15*(13*d^5*e^(-1) + (15*d^4 - (10*d^3*e - (10*d^2*e^2 + (23*x*e^4 + 45*d*e^3)
*x)*x)*x)*x)*sqrt(-x^2*e^2 + d^2)/(x^2*e^2 - d^2)^3