3.179 \(\int \frac{x^5}{(d+e x)^3 \sqrt{d^2-e^2 x^2}} \, dx\)

Optimal. Leaf size=177 \[ \frac{d^4 (d-e x)^3}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{23 d^3 (d-e x)^2}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{127 d^2 (d-e x)}{15 e^6 \sqrt{d^2-e^2 x^2}}+\frac{3 d \sqrt{d^2-e^2 x^2}}{e^6}-\frac{x \sqrt{d^2-e^2 x^2}}{2 e^5}+\frac{13 d^2 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e^6} \]

[Out]

(d^4*(d - e*x)^3)/(5*e^6*(d^2 - e^2*x^2)^(5/2)) - (23*d^3*(d - e*x)^2)/(15*e^6*(d^2 - e^2*x^2)^(3/2)) + (127*d
^2*(d - e*x))/(15*e^6*Sqrt[d^2 - e^2*x^2]) + (3*d*Sqrt[d^2 - e^2*x^2])/e^6 - (x*Sqrt[d^2 - e^2*x^2])/(2*e^5) +
(13*d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e^6)

________________________________________________________________________________________

Rubi [A]  time = 0.439337, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {852, 1635, 1815, 641, 217, 203} \[ \frac{d^4 (d-e x)^3}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{23 d^3 (d-e x)^2}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{127 d^2 (d-e x)}{15 e^6 \sqrt{d^2-e^2 x^2}}+\frac{3 d \sqrt{d^2-e^2 x^2}}{e^6}-\frac{x \sqrt{d^2-e^2 x^2}}{2 e^5}+\frac{13 d^2 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e^6} \]

Antiderivative was successfully verified.

[In]

Int[x^5/((d + e*x)^3*Sqrt[d^2 - e^2*x^2]),x]

[Out]

(d^4*(d - e*x)^3)/(5*e^6*(d^2 - e^2*x^2)^(5/2)) - (23*d^3*(d - e*x)^2)/(15*e^6*(d^2 - e^2*x^2)^(3/2)) + (127*d
^2*(d - e*x))/(15*e^6*Sqrt[d^2 - e^2*x^2]) + (3*d*Sqrt[d^2 - e^2*x^2])/e^6 - (x*Sqrt[d^2 - e^2*x^2])/(2*e^5) +
(13*d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e^6)

Rule 852

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^(2*m)/a
^m, Int[((f + g*x)^n*(a + c*x^2)^(m + p))/(d - e*x)^m, x], x] /; FreeQ[{a, c, d, e, f, g, n, p}, x] && NeQ[e*f
- d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[f, 0] && ILtQ[m, -1] &&  !(IGtQ[n, 0] && ILtQ[m +
n, 0] &&  !GtQ[p, 1])

Rule 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 0]

Rule 1815

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Si
mp[(e*x^(q - 1)*(a + b*x^2)^(p + 1))/(b*(q + 2*p + 1)), x] + Dist[1/(b*(q + 2*p + 1)), Int[(a + b*x^2)^p*Expan
dToSum[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x]
&& PolyQ[Pq, x] &&  !LeQ[p, -1]

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{x^5}{(d+e x)^3 \sqrt{d^2-e^2 x^2}} \, dx &=\int \frac{x^5 (d-e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=\frac{d^4 (d-e x)^3}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{\int \frac{(d-e x)^2 \left (-\frac{3 d^5}{e^5}+\frac{5 d^4 x}{e^4}-\frac{5 d^3 x^2}{e^3}+\frac{5 d^2 x^3}{e^2}-\frac{5 d x^4}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d}\\ &=\frac{d^4 (d-e x)^3}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{23 d^3 (d-e x)^2}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{\int \frac{(d-e x) \left (-\frac{37 d^5}{e^5}+\frac{45 d^4 x}{e^4}-\frac{30 d^3 x^2}{e^3}+\frac{15 d^2 x^3}{e^2}\right )}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2}\\ &=\frac{d^4 (d-e x)^3}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{23 d^3 (d-e x)^2}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{127 d^2 (d-e x)}{15 e^6 \sqrt{d^2-e^2 x^2}}-\frac{\int \frac{-\frac{90 d^5}{e^5}+\frac{45 d^4 x}{e^4}-\frac{15 d^3 x^2}{e^3}}{\sqrt{d^2-e^2 x^2}} \, dx}{15 d^3}\\ &=\frac{d^4 (d-e x)^3}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{23 d^3 (d-e x)^2}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{127 d^2 (d-e x)}{15 e^6 \sqrt{d^2-e^2 x^2}}-\frac{x \sqrt{d^2-e^2 x^2}}{2 e^5}+\frac{\int \frac{\frac{195 d^5}{e^3}-\frac{90 d^4 x}{e^2}}{\sqrt{d^2-e^2 x^2}} \, dx}{30 d^3 e^2}\\ &=\frac{d^4 (d-e x)^3}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{23 d^3 (d-e x)^2}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{127 d^2 (d-e x)}{15 e^6 \sqrt{d^2-e^2 x^2}}+\frac{3 d \sqrt{d^2-e^2 x^2}}{e^6}-\frac{x \sqrt{d^2-e^2 x^2}}{2 e^5}+\frac{\left (13 d^2\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx}{2 e^5}\\ &=\frac{d^4 (d-e x)^3}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{23 d^3 (d-e x)^2}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{127 d^2 (d-e x)}{15 e^6 \sqrt{d^2-e^2 x^2}}+\frac{3 d \sqrt{d^2-e^2 x^2}}{e^6}-\frac{x \sqrt{d^2-e^2 x^2}}{2 e^5}+\frac{\left (13 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 )}{2 e^5}\\ &=\frac{d^4 (d-e x)^3}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{23 d^3 (d-e x)^2}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{127 d^2 (d-e x)}{15 e^6 \sqrt{d^2-e^2 x^2}}+\frac{3 d \sqrt{d^2-e^2 x^2}}{e^6}-\frac{x \sqrt{d^2-e^2 x^2}}{2 e^5}+\frac{13 d^2 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e^6}\\ \end{align*}

Mathematica [A]  time = 0.198671, size = 98, normalized size = 0.55 \[ \frac{\frac{\sqrt{d^2-e^2 x^2} \left (479 d^2 e^2 x^2+717 d^3 e x+304 d^4+45 d e^3 x^3-15 e^4 x^4\right )}{(d+e x)^3}+195 d^2 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{30 e^6} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/((d + e*x)^3*Sqrt[d^2 - e^2*x^2]),x]

[Out]

((Sqrt[d^2 - e^2*x^2]*(304*d^4 + 717*d^3*e*x + 479*d^2*e^2*x^2 + 45*d*e^3*x^3 - 15*e^4*x^4))/(d + e*x)^3 + 195
*d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(30*e^6)

________________________________________________________________________________________

Maple [A]  time = 0.069, size = 212, normalized size = 1.2 \begin{align*} -{\frac{x}{2\,{e}^{5}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{13\,{d}^{2}}{2\,{e}^{5}}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}+3\,{\frac{d\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}{{e}^{6}}}+{\frac{127\,{d}^{2}}{15\,{e}^{7}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) } \left ({\frac{d}{e}}+x \right ) ^{-1}}-{\frac{23\,{d}^{3}}{15\,{e}^{8}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) } \left ({\frac{d}{e}}+x \right ) ^{-2}}+{\frac{{d}^{4}}{5\,{e}^{9}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) } \left ({\frac{d}{e}}+x \right ) ^{-3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x)

[Out]

-1/2*x*(-e^2*x^2+d^2)^(1/2)/e^5+13/2/e^5*d^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))+3*d*(-e^2*
x^2+d^2)^(1/2)/e^6+127/15/e^7*d^2/(d/e+x)*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2)-23/15/e^8*d^3/(d/e+x)^2*(-(d/e+
x)^2*e^2+2*d*e*(d/e+x))^(1/2)+1/5*d^4/e^9/(d/e+x)^3*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.75441, size = 409, normalized size = 2.31 \begin{align*} \frac{304 \, d^{2} e^{3} x^{3} + 912 \, d^{3} e^{2} x^{2} + 912 \, d^{4} e x + 304 \, d^{5} - 390 \,{\left (d^{2} e^{3} x^{3} + 3 \, d^{3} e^{2} x^{2} + 3 \, d^{4} e x + d^{5}\right )} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) -{\left (15 \, e^{4} x^{4} - 45 \, d e^{3} x^{3} - 479 \, d^{2} e^{2} x^{2} - 717 \, d^{3} e x - 304 \, d^{4}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{30 \,{\left (e^{9} x^{3} + 3 \, d e^{8} x^{2} + 3 \, d^{2} e^{7} x + d^{3} e^{6}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

1/30*(304*d^2*e^3*x^3 + 912*d^3*e^2*x^2 + 912*d^4*e*x + 304*d^5 - 390*(d^2*e^3*x^3 + 3*d^3*e^2*x^2 + 3*d^4*e*x
+ d^5)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (15*e^4*x^4 - 45*d*e^3*x^3 - 479*d^2*e^2*x^2 - 717*d^3*e*x
- 304*d^4)*sqrt(-e^2*x^2 + d^2))/(e^9*x^3 + 3*d*e^8*x^2 + 3*d^2*e^7*x + d^3*e^6)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5}}{\sqrt{- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(e*x+d)**3/(-e**2*x**2+d**2)**(1/2),x)

[Out]

Integral(x**5/(sqrt(-(-d + e*x)*(d + e*x))*(d + e*x)**3), x)

________________________________________________________________________________________

Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError