\(\int \frac {(d+e x)^3}{x (d^2-e^2 x^2)^{7/2}} \, dx\) [89]

Optimal result

Integrand size = 27, antiderivative size = 114 \[ \int \frac {(d+e x)^3}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {4 (d+e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d+22 e x}{15 d^4 \sqrt {d^2-e^2 x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1819, 837, 12, 272, 65, 214} \[ \int \frac {(d+e x)^3}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^4}+\frac {4 (d+e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d+22 e x}{15 d^4 \sqrt {d^2-e^2 x^2}} \]

[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 65

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 214

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

Rule 272

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 837

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] +
Dist[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 1819

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*} \text {integral}& = \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 {\text {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 {\text {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 [A] (verified)

Time = 0.41 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.88 \[ \int \frac {(d+e x)^3}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\frac {d \sqrt {d^2-e^2 x^2} \left (32 d^2-51 d e x+22 e^2 x^2\right )}{(d-e x)^3}-15 \sqrt {d^2} \log (x)+15 \sqrt {d^2} \log \left (\sqrt {d^2}-\sqrt {d^2-e^2 x^2}\right )}{15 d^5} \]

[In]

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

[Out]

((d*Sqrt[d^2 - e^2*x^2]*(32*d^2 - 51*d*e*x + 22*e^2*x^2))/(d - e*x)^3 - 15*Sqrt[d^2]*Log[x] + 15*Sqrt[d^2]*Log
[Sqrt[d^2] - Sqrt[d^2 - e^2*x^2]])/(15*d^5)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(304\) vs. \(2(100)=200\).

Time = 0.38 (sec) , antiderivative size = 305, normalized size of antiderivative = 2.68

method	result	size
default	\(e^{3} \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )+d^{3} \left (\frac {1}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {1}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {\frac {1}{d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{2} \sqrt {d^{2}}}}{d^{2}}}{d^{2}}\right )+3 d^{2} e \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )+\frac {3 d}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\)	\(305\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.39 \[ \int \frac {(d+e x)^3}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {32 \, e^{3} x^{3} - 96 \, d e^{2} x^{2} + 96 \, d^{2} e x - 32 \, d^{3} + 15 \, {\left (e^{3} x^{3} - 3 \, d e^{2} x^{2} + 3 \, d^{2} e x - d^{3}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (22 \, e^{2} x^{2} - 51 \, d e x + 32 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{4} e^{3} x^{3} - 3 \, d^{5} e^{2} x^{2} + 3 \, d^{6} e x - d^{7}\right )}} \]

[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)

Sympy [F]

\[ \int \frac {(d+e x)^3}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {\left (d + e x\right )^{3}}{x \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.33 \[ \int \frac {(d+e x)^3}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {4 \, e x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {4 \, d}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {11 \, e x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}} + \frac {1}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d} + \frac {22 \, e x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{4}} - \frac {\log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{d^{4}} + \frac {1}{\sqrt {-e^{2} x^{2} + d^{2}} d^{3}} \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 212 vs. \(2 (100) = 200\).

Time = 0.31 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.86 \[ \int \frac {(d+e x)^3}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {e \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |} \right |}}{2 \, e^{2} {\left | x \right |}}\right )}{d^{4} {\left | e \right |}} + \frac {2 \, {\left (32 \, e - \frac {115 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}}{e x} + \frac {185 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2}}{e^{3} x^{2}} - \frac {135 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3}}{e^{5} x^{3}} + \frac {45 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4}}{e^{7} x^{4}}\right )}}{15 \, d^{4} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} - 1\right )}^{5} {\left | e \right |}} \]

[In]

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

[Out]

-e*log(1/2*abs(-2*d*e - 2*sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*abs(x)))/(d^4*abs(e)) + 2/15*(32*e - 115*(d*e + sq
rt(-e^2*x^2 + d^2)*abs(e))/(e*x) + 185*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^2/(e^3*x^2) - 135*(d*e + sqrt(-e^2*
x^2 + d^2)*abs(e))^3/(e^5*x^3) + 45*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^4/(e^7*x^4))/(d^4*((d*e + sqrt(-e^2*x^
2 + d^2)*abs(e))/(e^2*x) - 1)^5*abs(e))

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^3}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^3}{x\,{\left (d^2-e^2\,x^2\right )}^{7/2}} \,d x \]

[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)