3.58 \(\int \frac {e^{-3 \coth ^{-1}(a x)}}{x^5} \, dx\)

Optimal. Leaf size=133 \[ -\frac {51}{8} a^4 \csc ^{-1}(a x)-\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{x^2}-\frac {a \left (a-\frac {1}{x}\right )^3}{\sqrt {1-\frac {1}{a^2 x^2}}}+\frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{4 x^3}-\frac {27}{4} a^4 \sqrt {1-\frac {1}{a^2 x^2}}-\frac {9}{8} a^3 \sqrt {1-\frac {1}{a^2 x^2}} \left (2 a-\frac {3}{x}\right ) \]

[Out]

-51/8*a^4*arccsc(a*x)-a*(a-1/x)^3/(1-1/a^2/x^2)^(1/2)-27/4*a^4*(1-1/a^2/x^2)^(1/2)-9/8*a^3*(2*a-3/x)*(1-1/a^2/
x^2)^(1/2)+1/4*a*(1-1/a^2/x^2)^(1/2)/x^3-a^2*(1-1/a^2/x^2)^(1/2)/x^2

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Rubi [A]  time = 0.84, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.917, Rules used = {6169, 1633, 1593, 12, 852, 1635, 1815, 27, 743, 641, 216} \[ -\frac {27}{4} a^4 \sqrt {1-\frac {1}{a^2 x^2}}-\frac {9}{8} a^3 \sqrt {1-\frac {1}{a^2 x^2}} \left (2 a-\frac {3}{x}\right )-\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{x^2}-\frac {a \left (a-\frac {1}{x}\right )^3}{\sqrt {1-\frac {1}{a^2 x^2}}}+\frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{4 x^3}-\frac {51}{8} a^4 \csc ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcCoth[a*x])*x^5),x]

[Out]

(-27*a^4*Sqrt[1 - 1/(a^2*x^2)])/4 - (9*a^3*Sqrt[1 - 1/(a^2*x^2)]*(2*a - 3/x))/8 - (a*(a - x^(-1))^3)/Sqrt[1 -
1/(a^2*x^2)] + (a*Sqrt[1 - 1/(a^2*x^2)])/(4*x^3) - (a^2*Sqrt[1 - 1/(a^2*x^2)])/x^2 - (51*a^4*ArcCsc[a*x])/8

Rule 12

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 743

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - a*e^
2*(m - 1) + 2*c*d*e*(m + p)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2,
0] && If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m
, p, x]

Rule 852

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

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1633

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d*e, Int[(d + e*x)^(m - 1)*
PolynomialQuotient[Pq, a*e + c*d*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq
, x] && EqQ[c*d^2 + a*e^2, 0] && EqQ[PolynomialRemainder[Pq, a*e + c*d*x, x], 0]

Rule 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 0]

Rule 1815

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Si
mp[(e*x^(q - 1)*(a + b*x^2)^(p + 1))/(b*(q + 2*p + 1)), x] + Dist[1/(b*(q + 2*p + 1)), Int[(a + b*x^2)^p*Expan
dToSum[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x]
&& PolyQ[Pq, x] &&  !LeQ[p, -1]

Rule 6169

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> -Subst[Int[(1 + x/a)^((n + 1)/2)/(x^(m + 2)*(1 - x/
a)^((n - 1)/2)*Sqrt[1 - x^2/a^2]), x], x, 1/x] /; FreeQ[a, x] && IntegerQ[(n - 1)/2] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {e^{-3 \coth ^{-1}(a x)}}{x^5} \, dx &=-\operatorname {Subst}\left (\int \frac {x^3 \left (1-\frac {x}{a}\right )^2}{\left (1+\frac {x}{a}\right ) \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {1-\frac {x^2}{a^2}} \left (a x^3-x^4\right )}{\left (1+\frac {x}{a}\right )^2} \, dx,x,\frac {1}{x}\right )}{a}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {(a-x) x^3 \sqrt {1-\frac {x^2}{a^2}}}{\left (1+\frac {x}{a}\right )^2} \, dx,x,\frac {1}{x}\right )}{a}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {a^2 x^3 \left (1-\frac {x^2}{a^2}\right )^{3/2}}{\left (1+\frac {x}{a}\right )^3} \, dx,x,\frac {1}{x}\right )}{a^2}\\ &=-\operatorname {Subst}\left (\int \frac {x^3 \left (1-\frac {x^2}{a^2}\right )^{3/2}}{\left (1+\frac {x}{a}\right )^3} \, dx,x,\frac {1}{x}\right )\\ &=-\operatorname {Subst}\left (\int \frac {x^3 \left (1-\frac {x}{a}\right )^3}{\left (1-\frac {x^2}{a^2}\right )^{3/2}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {a \left (a-\frac {1}{x}\right )^3}{\sqrt {1-\frac {1}{a^2 x^2}}}+\operatorname {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right )^2 \left (-3 a^3+a^2 x-a x^2\right )}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {a \left (a-\frac {1}{x}\right )^3}{\sqrt {1-\frac {1}{a^2 x^2}}}+\frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{4 x^3}-\frac {1}{4} a^2 \operatorname {Subst}\left (\int \frac {12 a-28 x+\frac {27 x^2}{a}-\frac {12 x^3}{a^2}}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {a \left (a-\frac {1}{x}\right )^3}{\sqrt {1-\frac {1}{a^2 x^2}}}+\frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{4 x^3}-\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{x^2}+\frac {1}{12} a^4 \operatorname {Subst}\left (\int \frac {-\frac {36}{a}+\frac {108 x}{a^2}-\frac {81 x^2}{a^3}}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {a \left (a-\frac {1}{x}\right )^3}{\sqrt {1-\frac {1}{a^2 x^2}}}+\frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{4 x^3}-\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{x^2}+\frac {1}{12} a^4 \operatorname {Subst}\left (\int -\frac {9 (2 a-3 x)^2}{a^3 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {a \left (a-\frac {1}{x}\right )^3}{\sqrt {1-\frac {1}{a^2 x^2}}}+\frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{4 x^3}-\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{x^2}-\frac {1}{4} (3 a) \operatorname {Subst}\left (\int \frac {(2 a-3 x)^2}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {9}{8} a^3 \sqrt {1-\frac {1}{a^2 x^2}} \left (2 a-\frac {3}{x}\right )-\frac {a \left (a-\frac {1}{x}\right )^3}{\sqrt {1-\frac {1}{a^2 x^2}}}+\frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{4 x^3}-\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{x^2}+\frac {1}{8} \left (3 a^3\right ) \operatorname {Subst}\left (\int \frac {-17+\frac {18 x}{a}}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {27}{4} a^4 \sqrt {1-\frac {1}{a^2 x^2}}-\frac {9}{8} a^3 \sqrt {1-\frac {1}{a^2 x^2}} \left (2 a-\frac {3}{x}\right )-\frac {a \left (a-\frac {1}{x}\right )^3}{\sqrt {1-\frac {1}{a^2 x^2}}}+\frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{4 x^3}-\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{x^2}-\frac {1}{8} \left (51 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {27}{4} a^4 \sqrt {1-\frac {1}{a^2 x^2}}-\frac {9}{8} a^3 \sqrt {1-\frac {1}{a^2 x^2}} \left (2 a-\frac {3}{x}\right )-\frac {a \left (a-\frac {1}{x}\right )^3}{\sqrt {1-\frac {1}{a^2 x^2}}}+\frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{4 x^3}-\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{x^2}-\frac {51}{8} a^4 \csc ^{-1}(a x)\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 75, normalized size = 0.56 \[ -\frac {51}{8} a^4 \sin ^{-1}\left (\frac {1}{a x}\right )-\frac {a \sqrt {1-\frac {1}{a^2 x^2}} \left (80 a^4 x^4+29 a^3 x^3-11 a^2 x^2+6 a x-2\right )}{8 x^3 (a x+1)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^(3*ArcCoth[a*x])*x^5),x]

[Out]

-1/8*(a*Sqrt[1 - 1/(a^2*x^2)]*(-2 + 6*a*x - 11*a^2*x^2 + 29*a^3*x^3 + 80*a^4*x^4))/(x^3*(1 + a*x)) - (51*a^4*A
rcSin[1/(a*x)])/8

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fricas [A]  time = 0.67, size = 77, normalized size = 0.58 \[ \frac {102 \, a^{4} x^{4} \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right ) - {\left (80 \, a^{4} x^{4} + 29 \, a^{3} x^{3} - 11 \, a^{2} x^{2} + 6 \, a x - 2\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{8 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(3/2)/x^5,x, algorithm="fricas")

[Out]

1/8*(102*a^4*x^4*arctan(sqrt((a*x - 1)/(a*x + 1))) - (80*a^4*x^4 + 29*a^3*x^3 - 11*a^2*x^2 + 6*a*x - 2)*sqrt((
a*x - 1)/(a*x + 1)))/x^4

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {undef} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(3/2)/x^5,x, algorithm="giac")

[Out]

undef

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maple [B]  time = 0.06, size = 690, normalized size = 5.19 \[ \frac {\left (-56 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, x^{7} a^{7}+56 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\, x^{5} a^{5}-163 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, x^{6} a^{6}-51 \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right ) \sqrt {a^{2}}\, x^{6} a^{6}+56 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{6} a^{7}+56 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, x^{6} a^{6}-56 \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{6} a^{7}+91 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\, x^{4} a^{4}-158 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, x^{5} a^{5}-102 \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right ) \sqrt {a^{2}}\, x^{5} a^{5}+112 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{5} a^{6}+16 \sqrt {a^{2}}\, \left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} x^{4} a^{4}+112 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, x^{5} a^{5}-112 \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{5} a^{6}+22 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\, x^{3} a^{3}-51 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, x^{4} a^{4}-51 a^{4} x^{4} \sqrt {a^{2}}\, \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )+56 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{4} a^{5}+56 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, x^{4} a^{4}-56 \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{4} a^{5}-7 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\, x^{2} a^{2}+4 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\, x a -2 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{8 \sqrt {a^{2}}\, x^{4} \left (a x -1\right ) \sqrt {\left (a x -1\right ) \left (a x +1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x-1)/(a*x+1))^(3/2)/x^5,x)

[Out]

1/8*(-56*(a^2*x^2-1)^(1/2)*(a^2)^(1/2)*x^7*a^7+56*(a^2*x^2-1)^(3/2)*(a^2)^(1/2)*x^5*a^5-163*(a^2*x^2-1)^(1/2)*
(a^2)^(1/2)*x^6*a^6-51*arctan(1/(a^2*x^2-1)^(1/2))*(a^2)^(1/2)*x^6*a^6+56*ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1
/2))/(a^2)^(1/2))*x^6*a^7+56*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*x^6*a^6-56*ln((a^2*x+((a*x-1)*(a*x+1))^(1/2)*
(a^2)^(1/2))/(a^2)^(1/2))*x^6*a^7+91*(a^2*x^2-1)^(3/2)*(a^2)^(1/2)*x^4*a^4-158*(a^2*x^2-1)^(1/2)*(a^2)^(1/2)*x
^5*a^5-102*arctan(1/(a^2*x^2-1)^(1/2))*(a^2)^(1/2)*x^5*a^5+112*ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))/(a^2)^
(1/2))*x^5*a^6+16*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(3/2)*x^4*a^4+112*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*x^5*a^5-
112*ln((a^2*x+((a*x-1)*(a*x+1))^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x^5*a^6+22*(a^2*x^2-1)^(3/2)*(a^2)^(1/2)*x^3*a
^3-51*(a^2*x^2-1)^(1/2)*(a^2)^(1/2)*x^4*a^4-51*a^4*x^4*(a^2)^(1/2)*arctan(1/(a^2*x^2-1)^(1/2))+56*ln((a^2*x+(a
^2*x^2-1)^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x^4*a^5+56*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*x^4*a^4-56*ln((a^2*x+
((a*x-1)*(a*x+1))^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x^4*a^5-7*(a^2*x^2-1)^(3/2)*(a^2)^(1/2)*x^2*a^2+4*(a^2*x^2-1
)^(3/2)*(a^2)^(1/2)*x*a-2*(a^2*x^2-1)^(3/2)*(a^2)^(1/2))*((a*x-1)/(a*x+1))^(3/2)/(a^2)^(1/2)/x^4/(a*x-1)/((a*x
-1)*(a*x+1))^(1/2)

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maxima [A]  time = 0.42, size = 193, normalized size = 1.45 \[ \frac {1}{4} \, {\left (51 \, a^{3} \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right ) - 16 \, a^{3} \sqrt {\frac {a x - 1}{a x + 1}} - \frac {77 \, a^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{2}} + 149 \, a^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} + 123 \, a^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} + 35 \, a^{3} \sqrt {\frac {a x - 1}{a x + 1}}}{\frac {4 \, {\left (a x - 1\right )}}{a x + 1} + \frac {6 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} + \frac {4 \, {\left (a x - 1\right )}^{3}}{{\left (a x + 1\right )}^{3}} + \frac {{\left (a x - 1\right )}^{4}}{{\left (a x + 1\right )}^{4}} + 1}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(3/2)/x^5,x, algorithm="maxima")

[Out]

1/4*(51*a^3*arctan(sqrt((a*x - 1)/(a*x + 1))) - 16*a^3*sqrt((a*x - 1)/(a*x + 1)) - (77*a^3*((a*x - 1)/(a*x + 1
))^(7/2) + 149*a^3*((a*x - 1)/(a*x + 1))^(5/2) + 123*a^3*((a*x - 1)/(a*x + 1))^(3/2) + 35*a^3*sqrt((a*x - 1)/(
a*x + 1)))/(4*(a*x - 1)/(a*x + 1) + 6*(a*x - 1)^2/(a*x + 1)^2 + 4*(a*x - 1)^3/(a*x + 1)^3 + (a*x - 1)^4/(a*x +
1)^4 + 1))*a

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mupad [B]  time = 1.24, size = 190, normalized size = 1.43 \[ \frac {51\,a^4\,\mathrm {atan}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{4}-4\,a^4\,\sqrt {\frac {a\,x-1}{a\,x+1}}-\frac {\frac {35\,a^4\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{4}+\frac {123\,a^4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{4}+\frac {149\,a^4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}}{4}+\frac {77\,a^4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/2}}{4}}{\frac {6\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}+\frac {4\,{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}+\frac {{\left (a\,x-1\right )}^4}{{\left (a\,x+1\right )}^4}+\frac {4\,\left (a\,x-1\right )}{a\,x+1}+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x - 1)/(a*x + 1))^(3/2)/x^5,x)

[Out]

(51*a^4*atan(((a*x - 1)/(a*x + 1))^(1/2)))/4 - 4*a^4*((a*x - 1)/(a*x + 1))^(1/2) - ((35*a^4*((a*x - 1)/(a*x +
1))^(1/2))/4 + (123*a^4*((a*x - 1)/(a*x + 1))^(3/2))/4 + (149*a^4*((a*x - 1)/(a*x + 1))^(5/2))/4 + (77*a^4*((a
*x - 1)/(a*x + 1))^(7/2))/4)/((6*(a*x - 1)^2)/(a*x + 1)^2 + (4*(a*x - 1)^3)/(a*x + 1)^3 + (a*x - 1)^4/(a*x + 1
)^4 + (4*(a*x - 1))/(a*x + 1) + 1)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))**(3/2)/x**5,x)

[Out]

Integral(((a*x - 1)/(a*x + 1))**(3/2)/x**5, x)

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