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

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

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Rubi [A]  time = 0.16, antiderivative size = 114, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 12

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

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]

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 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 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]

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*}

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Mathematica [C]  time = 0.06, size = 81, normalized size = 0.71 \begin {gather*} \frac {9 d^5+45 d^4 e x-55 d^2 e^3 x^3+3 d^5 \, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};1-\frac {e^2 x^2}{d^2}\right )+22 e^5 x^5}{15 d^4 \left (d^2-e^2 x^2\right )^{5/2}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.67, size = 93, normalized size = 0.82 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (32 d^2-51 d e x+22 e^2 x^2\right )}{15 d^4 (d-e x)^3}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(32*d^2 - 51*d*e*x + 22*e^2*x^2))/(15*d^4*(d - e*x)^3) + (2*ArcTanh[(Sqrt[-e^2]*x)/d - Sq
rt[d^2 - e^2*x^2]/d])/d^4

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fricas [A]  time = 0.41, size = 158, normalized size = 1.39 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.29, size = 117, normalized size = 1.03 \begin {gather*} -\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 {gather*}

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

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maple [A]  time = 0.01, size = 158, normalized size = 1.39 \begin {gather*} \frac {4 e x}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {4 d}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {11 e x}{15 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{2}}+\frac {1}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}\, d^{3}}+\frac {22 e x}{15 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{4}}+\frac {1}{\sqrt {-e^{2} x^{2}+d^{2}}\, d^{3}} \end {gather*}

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 [A]  time = 0.46, size = 152, normalized size = 1.33 \begin {gather*} \frac {4 \, e x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {4 \, d}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {11 \, e x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}} + \frac {1}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d} + \frac {22 \, e x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{4}} - \frac {\log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{d^{4}} + \frac {1}{\sqrt {-e^{2} x^{2} + d^{2}} d^{3}} \end {gather*}

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^3}{x\,{\left (d^2-e^2\,x^2\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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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