3.89 \(\int \frac{(d+e x)^3}{x (d^2-e^2 x^2)^{7/2}} \, dx\)

Optimal. Leaf size=114 \[ \frac{4 (d+e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{15 d+22 e x}{15 d^4 \sqrt{d^2-e^2 x^2}}+\frac{5 d+11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^4} \]

[Out]

(4*(d + e*x))/(5*(d^2 - e^2*x^2)^(5/2)) + (5*d + 11*e*x)/(15*d^2*(d^2 - e^2*x^2)^(3/2)) + (15*d + 22*e*x)/(15*
d^4*Sqrt[d^2 - e^2*x^2]) - ArcTanh[Sqrt[d^2 - e^2*x^2]/d]/d^4

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Rubi [A]  time = 0.158929, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {1805, 823, 12, 266, 63, 208} \[ \frac{4 (d+e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{15 d+22 e x}{15 d^4 \sqrt{d^2-e^2 x^2}}+\frac{5 d+11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^4} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3/(x*(d^2 - e^2*x^2)^(7/2)),x]

[Out]

(4*(d + e*x))/(5*(d^2 - e^2*x^2)^(5/2)) + (5*d + 11*e*x)/(15*d^2*(d^2 - e^2*x^2)^(3/2)) + (15*d + 22*e*x)/(15*
d^4*Sqrt[d^2 - e^2*x^2]) - ArcTanh[Sqrt[d^2 - e^2*x^2]/d]/d^4

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{(d+e x)^3}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac{4 (d+e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{\int \frac{-5 d^3-11 d^2 e x}{x \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2}\\ &=\frac{4 (d+e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{5 d+11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{\int \frac{-15 d^5 e^2-22 d^4 e^3 x}{x \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^6 e^2}\\ &=\frac{4 (d+e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{5 d+11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{15 d+22 e x}{15 d^4 \sqrt{d^2-e^2 x^2}}-\frac{\int -\frac{15 d^7 e^4}{x \sqrt{d^2-e^2 x^2}} \, dx}{15 d^{10} e^4}\\ &=\frac{4 (d+e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{5 d+11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{15 d+22 e x}{15 d^4 \sqrt{d^2-e^2 x^2}}+\frac{\int \frac{1}{x \sqrt{d^2-e^2 x^2}} \, dx}{d^3}\\ &=\frac{4 (d+e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{5 d+11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{15 d+22 e x}{15 d^4 \sqrt{d^2-e^2 x^2}}+\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )}{2 d^3}\\ &=\frac{4 (d+e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{5 d+11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{15 d+22 e x}{15 d^4 \sqrt{d^2-e^2 x^2}}-\frac{\operatorname{Subst}\left (\int \frac{1}{\frac{d^2}{e^2}-\frac{x^2}{e^2}} \, dx,x,\sqrt{d^2-e^2 x^2}\right )}{d^3 e^2}\\ &=\frac{4 (d+e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{5 d+11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{15 d+22 e x}{15 d^4 \sqrt{d^2-e^2 x^2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^4}\\ \end{align*}

Mathematica [C]  time = 0.0619555, size = 81, normalized size = 0.71 \[ \frac{3 d^5 \, _2F_1\left (-\frac{5}{2},1;-\frac{3}{2};1-\frac{e^2 x^2}{d^2}\right )-55 d^2 e^3 x^3+45 d^4 e x+9 d^5+22 e^5 x^5}{15 d^4 \left (d^2-e^2 x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^3/(x*(d^2 - e^2*x^2)^(7/2)),x]

[Out]

(9*d^5 + 45*d^4*e*x - 55*d^2*e^3*x^3 + 22*e^5*x^5 + 3*d^5*Hypergeometric2F1[-5/2, 1, -3/2, 1 - (e^2*x^2)/d^2])
/(15*d^4*(d^2 - e^2*x^2)^(5/2))

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Maple [A]  time = 0.066, size = 158, normalized size = 1.4 \begin{align*}{\frac{4\,ex}{5} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{11\,ex}{15\,{d}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{22\,ex}{15\,{d}^{4}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}}+{\frac{4\,d}{5} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{1}{3\,d} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{1}{{d}^{3}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}}-{\frac{1}{{d}^{3}}\ln \left ({\frac{1}{x} \left ( 2\,{d}^{2}+2\,\sqrt{{d}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}} \right ) } \right ){\frac{1}{\sqrt{{d}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3/x/(-e^2*x^2+d^2)^(7/2),x)

[Out]

4/5*e*x/(-e^2*x^2+d^2)^(5/2)+11/15*e/d^2*x/(-e^2*x^2+d^2)^(3/2)+22/15*e/d^4*x/(-e^2*x^2+d^2)^(1/2)+4/5*d/(-e^2
*x^2+d^2)^(5/2)+1/3/d/(-e^2*x^2+d^2)^(3/2)+1/d^3/(-e^2*x^2+d^2)^(1/2)-1/d^3/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2
)*(-e^2*x^2+d^2)^(1/2))/x)

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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((e*x+d)^3/x/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.60921, size = 327, normalized size = 2.87 \begin{align*} \frac{32 \, e^{3} x^{3} - 96 \, d e^{2} x^{2} + 96 \, d^{2} e x - 32 \, d^{3} + 15 \,{\left (e^{3} x^{3} - 3 \, d e^{2} x^{2} + 3 \, d^{2} e x - d^{3}\right )} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) -{\left (22 \, e^{2} x^{2} - 51 \, d e x + 32 \, d^{2}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{15 \,{\left (d^{4} e^{3} x^{3} - 3 \, d^{5} e^{2} x^{2} + 3 \, d^{6} e x - d^{7}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/x/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

1/15*(32*e^3*x^3 - 96*d*e^2*x^2 + 96*d^2*e*x - 32*d^3 + 15*(e^3*x^3 - 3*d*e^2*x^2 + 3*d^2*e*x - d^3)*log(-(d -
sqrt(-e^2*x^2 + d^2))/x) - (22*e^2*x^2 - 51*d*e*x + 32*d^2)*sqrt(-e^2*x^2 + d^2))/(d^4*e^3*x^3 - 3*d^5*e^2*x^
2 + 3*d^6*e*x - d^7)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{3}}{x \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)**3/x/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral((d + e*x)**3/(x*(-(-d + e*x)*(d + e*x))**(7/2)), x)

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Giac [A]  time = 1.2307, size = 158, normalized size = 1.39 \begin{align*} -\frac{\sqrt{-x^{2} e^{2} + d^{2}}{\left ({\left ({\left ({\left (x{\left (\frac{22 \, x e^{5}}{d^{4}} + \frac{15 \, e^{4}}{d^{3}}\right )} - \frac{55 \, e^{3}}{d^{2}}\right )} x - \frac{35 \, e^{2}}{d}\right )} x + 45 \, e\right )} x + 32 \, d\right )}}{15 \,{\left (x^{2} e^{2} - d^{2}\right )}^{3}} - \frac{\log \left (\frac{{\left | -2 \, d e - 2 \, \sqrt{-x^{2} e^{2} + d^{2}} e \right |} e^{\left (-2\right )}}{2 \,{\left | x \right |}}\right )}{d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/x/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-1/15*sqrt(-x^2*e^2 + d^2)*((((x*(22*x*e^5/d^4 + 15*e^4/d^3) - 55*e^3/d^2)*x - 35*e^2/d)*x + 45*e)*x + 32*d)/(
x^2*e^2 - d^2)^3 - log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^4