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

Optimal. Leaf size=182 \[ \frac {4 e^2 (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (90 d+107 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {3 e \sqrt {d^2-e^2 x^2}}{d^6 x}-\frac {13 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^6}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^5 x^2}+\frac {e^2 (25 d+31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.36, antiderivative size = 182, normalized size of antiderivative = 1.00, 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 = {1805, 1807, 807, 266, 63, 208} \begin {gather*} \frac {e^2 (90 d+107 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}+\frac {e^2 (25 d+31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {4 e^2 (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {3 e \sqrt {d^2-e^2 x^2}}{d^6 x}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^5 x^2}-\frac {13 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*e^2*(d + e*x))/(5*d^2*(d^2 - e^2*x^2)^(5/2)) + (e^2*(25*d + 31*e*x))/(15*d^4*(d^2 - e^2*x^2)^(3/2)) + (e^2*
(90*d + 107*e*x))/(15*d^6*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(2*d^5*x^2) - (3*e*Sqrt[d^2 - e^2*x^2])/(
d^6*x) - (13*e^2*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(2*d^6)

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]

Rule 1807

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

Rubi steps

\begin {align*} \int \frac {(d+e x)^3}{x^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {4 e^2 (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {-5 d^3-15 d^2 e x-20 d e^2 x^2-16 e^3 x^3}{x^3 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2}\\ &=\frac {4 e^2 (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (25 d+31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {15 d^3+45 d^2 e x+75 d e^2 x^2+62 e^3 x^3}{x^3 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4}\\ &=\frac {4 e^2 (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (25 d+31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e^2 (90 d+107 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {-15 d^3-45 d^2 e x-90 d e^2 x^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx}{15 d^6}\\ &=\frac {4 e^2 (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (25 d+31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e^2 (90 d+107 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^5 x^2}+\frac {\int \frac {90 d^4 e+195 d^3 e^2 x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{30 d^8}\\ &=\frac {4 e^2 (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (25 d+31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e^2 (90 d+107 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^5 x^2}-\frac {3 e \sqrt {d^2-e^2 x^2}}{d^6 x}+\frac {\left (13 e^2\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{2 d^5}\\ &=\frac {4 e^2 (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (25 d+31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e^2 (90 d+107 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^5 x^2}-\frac {3 e \sqrt {d^2-e^2 x^2}}{d^6 x}+\frac {\left (13 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{4 d^5}\\ &=\frac {4 e^2 (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (25 d+31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e^2 (90 d+107 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^5 x^2}-\frac {3 e \sqrt {d^2-e^2 x^2}}{d^6 x}-\frac {13 \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 )}{2 d^5}\\ &=\frac {4 e^2 (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (25 d+31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e^2 (90 d+107 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^5 x^2}-\frac {3 e \sqrt {d^2-e^2 x^2}}{d^6 x}-\frac {13 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^6}\\ \end {align*}

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Mathematica [C]  time = 0.07, size = 119, normalized size = 0.65 \begin {gather*} \frac {e \left (-45 d^6+285 d^4 e^2 x^2-380 d^2 e^4 x^4+9 d^5 e x \, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};1-\frac {e^2 x^2}{d^2}\right )+3 d^5 e x \, _2F_1\left (-\frac {5}{2},2;-\frac {3}{2};1-\frac {e^2 x^2}{d^2}\right )+152 e^6 x^6\right )}{15 d^6 x \left (d^2-e^2 x^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(-45*d^6 + 285*d^4*e^2*x^2 - 380*d^2*e^4*x^4 + 152*e^6*x^6 + 9*d^5*e*x*Hypergeometric2F1[-5/2, 1, -3/2, 1 -
(e^2*x^2)/d^2] + 3*d^5*e*x*Hypergeometric2F1[-5/2, 2, -3/2, 1 - (e^2*x^2)/d^2]))/(15*d^6*x*(d^2 - e^2*x^2)^(5
/2))

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IntegrateAlgebraic [A]  time = 0.72, size = 121, normalized size = 0.66 \begin {gather*} \frac {13 e^2 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^6}+\frac {\sqrt {d^2-e^2 x^2} \left (-15 d^4-45 d^3 e x+479 d^2 e^2 x^2-717 d e^3 x^3+304 e^4 x^4\right )}{30 d^6 x^2 (d-e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-15*d^4 - 45*d^3*e*x + 479*d^2*e^2*x^2 - 717*d*e^3*x^3 + 304*e^4*x^4))/(30*d^6*x^2*(d -
e*x)^3) + (13*e^2*ArcTanh[(Sqrt[-e^2]*x)/d - Sqrt[d^2 - e^2*x^2]/d])/d^6

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fricas [A]  time = 0.42, size = 205, normalized size = 1.13 \begin {gather*} \frac {254 \, e^{5} x^{5} - 762 \, d e^{4} x^{4} + 762 \, d^{2} e^{3} x^{3} - 254 \, d^{3} e^{2} x^{2} + 195 \, {\left (e^{5} x^{5} - 3 \, d e^{4} x^{4} + 3 \, d^{2} e^{3} x^{3} - d^{3} e^{2} x^{2}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (304 \, e^{4} x^{4} - 717 \, d e^{3} x^{3} + 479 \, d^{2} e^{2} x^{2} - 45 \, d^{3} e x - 15 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{30 \, {\left (d^{6} e^{3} x^{5} - 3 \, d^{7} e^{2} x^{4} + 3 \, d^{8} e x^{3} - d^{9} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(254*e^5*x^5 - 762*d*e^4*x^4 + 762*d^2*e^3*x^3 - 254*d^3*e^2*x^2 + 195*(e^5*x^5 - 3*d*e^4*x^4 + 3*d^2*e^3
*x^3 - d^3*e^2*x^2)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - (304*e^4*x^4 - 717*d*e^3*x^3 + 479*d^2*e^2*x^2 - 45*d
^3*e*x - 15*d^4)*sqrt(-e^2*x^2 + d^2))/(d^6*e^3*x^5 - 3*d^7*e^2*x^4 + 3*d^8*e*x^3 - d^9*x^2)

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giac [A]  time = 0.33, size = 259, normalized size = 1.42 \begin {gather*} -\frac {\sqrt {-x^{2} e^{2} + d^{2}} {\left ({\left ({\left ({\left (x {\left (\frac {107 \, x e^{7}}{d^{6}} + \frac {90 \, e^{6}}{d^{5}}\right )} - \frac {245 \, e^{5}}{d^{4}}\right )} x - \frac {205 \, e^{4}}{d^{3}}\right )} x + \frac {150 \, e^{3}}{d^{2}}\right )} x + \frac {127 \, e^{2}}{d}\right )}}{15 \, {\left (x^{2} e^{2} - d^{2}\right )}^{3}} - \frac {13 \, e^{2} \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 )}{2 \, d^{6}} + \frac {x^{2} {\left (\frac {12 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{4}}{x} + e^{6}\right )}}{8 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{6}} - \frac {{\left (\frac {12 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{6} e^{8}}{x} + \frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{6} e^{6}}{x^{2}}\right )} e^{\left (-8\right )}}{8 \, d^{12}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*sqrt(-x^2*e^2 + d^2)*((((x*(107*x*e^7/d^6 + 90*e^6/d^5) - 245*e^5/d^4)*x - 205*e^4/d^3)*x + 150*e^3/d^2)
*x + 127*e^2/d)/(x^2*e^2 - d^2)^3 - 13/2*e^2*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^6
+ 1/8*x^2*(12*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^4/x + e^6)/((d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^6) - 1/8*(12*(d
*e + sqrt(-x^2*e^2 + d^2)*e)*d^6*e^8/x + (d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^6*e^6/x^2)*e^(-8)/d^12

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maple [A]  time = 0.01, size = 222, normalized size = 1.22 \begin {gather*} \frac {19 e^{3} x}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d^{2}}+\frac {13 e^{2}}{10 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d}+\frac {76 e^{3} x}{15 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{4}}-\frac {3 e}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} x}-\frac {d}{2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} x^{2}}+\frac {13 e^{2}}{6 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{3}}-\frac {13 e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 \sqrt {d^{2}}\, d^{5}}+\frac {152 e^{3} x}{15 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{6}}+\frac {13 e^{2}}{2 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

19/5*e^3*x/d^2/(-e^2*x^2+d^2)^(5/2)+76/15*e^3/d^4*x/(-e^2*x^2+d^2)^(3/2)+152/15*e^3/d^6*x/(-e^2*x^2+d^2)^(1/2)
-3*e/x/(-e^2*x^2+d^2)^(5/2)+13/10*e^2/d/(-e^2*x^2+d^2)^(5/2)+13/6*e^2/d^3/(-e^2*x^2+d^2)^(3/2)+13/2*e^2/d^5/(-
e^2*x^2+d^2)^(1/2)-13/2*e^2/d^5/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-1/2*d/x^2/(-e^2*x
^2+d^2)^(5/2)

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maxima [A]  time = 0.48, size = 216, normalized size = 1.19 \begin {gather*} \frac {19 \, e^{3} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2}} + \frac {13 \, e^{2}}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d} + \frac {76 \, e^{3} x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{4}} + \frac {13 \, e^{2}}{6 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3}} - \frac {3 \, e}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} x} + \frac {152 \, e^{3} x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{6}} - \frac {13 \, e^{2} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{2 \, d^{6}} + \frac {13 \, e^{2}}{2 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{5}} - \frac {d}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

19/5*e^3*x/((-e^2*x^2 + d^2)^(5/2)*d^2) + 13/10*e^2/((-e^2*x^2 + d^2)^(5/2)*d) + 76/15*e^3*x/((-e^2*x^2 + d^2)
^(3/2)*d^4) + 13/6*e^2/((-e^2*x^2 + d^2)^(3/2)*d^3) - 3*e/((-e^2*x^2 + d^2)^(5/2)*x) + 152/15*e^3*x/(sqrt(-e^2
*x^2 + d^2)*d^6) - 13/2*e^2*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d^6 + 13/2*e^2/(sqrt(-e^2*x^2
+ d^2)*d^5) - 1/2*d/((-e^2*x^2 + d^2)^(5/2)*x^2)

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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^3\,{\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^3*(d^2 - e^2*x^2)^(7/2)),x)

[Out]

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

[Out]

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

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