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

Optimal. Leaf size=145 \[ \frac {4 e (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^5 x}+\frac {e (15 d+19 e x)}{5 d^5 \sqrt {d^2-e^2 x^2}}-\frac {3 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^5}+\frac {e (5 d+7 e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.29, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1805, 807, 266, 63, 208} \begin {gather*} \frac {4 e (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^5 x}+\frac {e (15 d+19 e x)}{5 d^5 \sqrt {d^2-e^2 x^2}}+\frac {e (5 d+7 e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {3 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*e*(d + e*x))/(5*d*(d^2 - e^2*x^2)^(5/2)) + (e*(5*d + 7*e*x))/(5*d^3*(d^2 - e^2*x^2)^(3/2)) + (e*(15*d + 19*
e*x))/(5*d^5*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(d^5*x) - (3*e*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/d^5

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 807

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

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^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {4 e (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {-5 d^3-15 d^2 e x-16 d e^2 x^2}{x^2 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2}\\ &=\frac {4 e (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e (5 d+7 e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {15 d^3+45 d^2 e x+42 d e^2 x^2}{x^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4}\\ &=\frac {4 e (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e (5 d+7 e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e (15 d+19 e x)}{5 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {-15 d^3-45 d^2 e x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{15 d^6}\\ &=\frac {4 e (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e (5 d+7 e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e (15 d+19 e x)}{5 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^5 x}+\frac {(3 e) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{d^4}\\ &=\frac {4 e (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e (5 d+7 e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e (15 d+19 e x)}{5 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^5 x}+\frac {(3 e) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{2 d^4}\\ &=\frac {4 e (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e (5 d+7 e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e (15 d+19 e x)}{5 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^5 x}-\frac {3 \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^4 e}\\ &=\frac {4 e (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e (5 d+7 e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e (15 d+19 e x)}{5 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^5 x}-\frac {3 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^5}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 96, normalized size = 0.66 \begin {gather*} \frac {-5 d^6+d^5 e x+45 d^4 e^2 x^2-60 d^2 e^4 x^4+3 d^5 e x \, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};1-\frac {e^2 x^2}{d^2}\right )+24 e^6 x^6}{5 d^5 x \left (d^2-e^2 x^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*d^6 + d^5*e*x + 45*d^4*e^2*x^2 - 60*d^2*e^4*x^4 + 24*e^6*x^6 + 3*d^5*e*x*Hypergeometric2F1[-5/2, 1, -3/2,
1 - (e^2*x^2)/d^2])/(5*d^5*x*(d^2 - e^2*x^2)^(5/2))

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IntegrateAlgebraic [A]  time = 0.59, size = 108, normalized size = 0.74 \begin {gather*} \frac {6 e \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^5}+\frac {\sqrt {d^2-e^2 x^2} \left (-5 d^3+39 d^2 e x-57 d e^2 x^2+24 e^3 x^3\right )}{5 d^5 x (d-e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-5*d^3 + 39*d^2*e*x - 57*d*e^2*x^2 + 24*e^3*x^3))/(5*d^5*x*(d - e*x)^3) + (6*e*ArcTanh[(
Sqrt[-e^2]*x)/d - Sqrt[d^2 - e^2*x^2]/d])/d^5

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fricas [A]  time = 0.42, size = 184, normalized size = 1.27 \begin {gather*} \frac {24 \, e^{4} x^{4} - 72 \, d e^{3} x^{3} + 72 \, d^{2} e^{2} x^{2} - 24 \, d^{3} e x + 15 \, {\left (e^{4} x^{4} - 3 \, d e^{3} x^{3} + 3 \, d^{2} e^{2} x^{2} - d^{3} e x\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (24 \, e^{3} x^{3} - 57 \, d e^{2} x^{2} + 39 \, d^{2} e x - 5 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{5 \, {\left (d^{5} e^{3} x^{4} - 3 \, d^{6} e^{2} x^{3} + 3 \, d^{7} e x^{2} - d^{8} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(24*e^4*x^4 - 72*d*e^3*x^3 + 72*d^2*e^2*x^2 - 24*d^3*e*x + 15*(e^4*x^4 - 3*d*e^3*x^3 + 3*d^2*e^2*x^2 - d^3
*e*x)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - (24*e^3*x^3 - 57*d*e^2*x^2 + 39*d^2*e*x - 5*d^3)*sqrt(-e^2*x^2 + d^
2))/(d^5*e^3*x^4 - 3*d^6*e^2*x^3 + 3*d^7*e*x^2 - d^8*x)

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giac [A]  time = 0.29, size = 185, normalized size = 1.28 \begin {gather*} -\frac {\sqrt {-x^{2} e^{2} + d^{2}} {\left ({\left ({\left ({\left (x {\left (\frac {19 \, x e^{6}}{d^{5}} + \frac {15 \, e^{5}}{d^{4}}\right )} - \frac {45 \, e^{4}}{d^{3}}\right )} x - \frac {35 \, e^{3}}{d^{2}}\right )} x + \frac {30 \, e^{2}}{d}\right )} x + 24 \, e\right )}}{5 \, {\left (x^{2} e^{2} - d^{2}\right )}^{3}} - \frac {3 \, e \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^{5}} + \frac {x e^{3}}{2 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{5}} - \frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-1\right )}}{2 \, d^{5} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*sqrt(-x^2*e^2 + d^2)*((((x*(19*x*e^6/d^5 + 15*e^5/d^4) - 45*e^4/d^3)*x - 35*e^3/d^2)*x + 30*e^2/d)*x + 24
*e)/(x^2*e^2 - d^2)^3 - 3*e*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^5 + 1/2*x*e^3/((d*
e + sqrt(-x^2*e^2 + d^2)*e)*d^5) - 1/2*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^(-1)/(d^5*x)

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maple [A]  time = 0.01, size = 190, normalized size = 1.31 \begin {gather*} \frac {9 e^{2} x}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d}+\frac {4 e}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} x}+\frac {12 e^{2} x}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{3}}+\frac {e}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{2}}-\frac {3 e \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}\, d^{4}}+\frac {24 e^{2} x}{5 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{5}}+\frac {3 e}{\sqrt {-e^{2} x^{2}+d^{2}}\, d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/5*e/(-e^2*x^2+d^2)^(5/2)+9/5*e^2/d*x/(-e^2*x^2+d^2)^(5/2)+12/5*e^2/d^3*x/(-e^2*x^2+d^2)^(3/2)+24/5*e^2/d^5*x
/(-e^2*x^2+d^2)^(1/2)-d/x/(-e^2*x^2+d^2)^(5/2)+e/d^2/(-e^2*x^2+d^2)^(3/2)+3*e/d^4/(-e^2*x^2+d^2)^(1/2)-3*e/d^4
/(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 = 0.47, size = 184, normalized size = 1.27 \begin {gather*} \frac {9 \, e^{2} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d} + \frac {4 \, e}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {12 \, e^{2} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3}} + \frac {e}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}} - \frac {d}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} x} + \frac {24 \, e^{2} x}{5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{5}} - \frac {3 \, e \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{d^{5}} + \frac {3 \, e}{\sqrt {-e^{2} x^{2} + d^{2}} d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9/5*e^2*x/((-e^2*x^2 + d^2)^(5/2)*d) + 4/5*e/(-e^2*x^2 + d^2)^(5/2) + 12/5*e^2*x/((-e^2*x^2 + d^2)^(3/2)*d^3)
+ e/((-e^2*x^2 + d^2)^(3/2)*d^2) - d/((-e^2*x^2 + d^2)^(5/2)*x) + 24/5*e^2*x/(sqrt(-e^2*x^2 + d^2)*d^5) - 3*e*
log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d^5 + 3*e/(sqrt(-e^2*x^2 + d^2)*d^4)

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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^2\,{\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^2*(d^2 - e^2*x^2)^(7/2)),x)

[Out]

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

[Out]

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

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