3.413 \(\int e^{-\coth ^{-1}(a x)} (c-\frac {c}{a x})^4 \, dx\)

Optimal. Leaf size=135 \[ c^4 x \sqrt {1-\frac {1}{a^2 x^2}}-\frac {32 c^4 \sqrt {1-\frac {1}{a^2 x^2}}}{3 a}+\frac {5 c^4 \sqrt {1-\frac {1}{a^2 x^2}}}{2 a^2 x}-\frac {5 c^4 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a}-\frac {c^4 \sqrt {1-\frac {1}{a^2 x^2}}}{3 a^3 x^2}-\frac {25 c^4 \csc ^{-1}(a x)}{2 a} \]

[Out]

-25/2*c^4*arccsc(a*x)/a-5*c^4*arctanh((1-1/a^2/x^2)^(1/2))/a-32/3*c^4*(1-1/a^2/x^2)^(1/2)/a-1/3*c^4*(1-1/a^2/x
^2)^(1/2)/a^3/x^2+5/2*c^4*(1-1/a^2/x^2)^(1/2)/a^2/x+c^4*x*(1-1/a^2/x^2)^(1/2)

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Rubi [A]  time = 0.44, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {6177, 1807, 1809, 844, 216, 266, 63, 208} \[ c^4 x \sqrt {1-\frac {1}{a^2 x^2}}-\frac {32 c^4 \sqrt {1-\frac {1}{a^2 x^2}}}{3 a}+\frac {5 c^4 \sqrt {1-\frac {1}{a^2 x^2}}}{2 a^2 x}-\frac {c^4 \sqrt {1-\frac {1}{a^2 x^2}}}{3 a^3 x^2}-\frac {5 c^4 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a}-\frac {25 c^4 \csc ^{-1}(a x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[(c - c/(a*x))^4/E^ArcCoth[a*x],x]

[Out]

(-32*c^4*Sqrt[1 - 1/(a^2*x^2)])/(3*a) - (c^4*Sqrt[1 - 1/(a^2*x^2)])/(3*a^3*x^2) + (5*c^4*Sqrt[1 - 1/(a^2*x^2)]
)/(2*a^2*x) + c^4*Sqrt[1 - 1/(a^2*x^2)]*x - (25*c^4*ArcCsc[a*x])/(2*a) - (5*c^4*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]]
)/a

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 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 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 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[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])

Rule 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rule 6177

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.), x_Symbol] :> -Dist[c^n, Subst[Int[((c + d*x)^(p -
n)*(1 - x^2/a^2)^(n/2))/x^2, x], x, 1/x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] && IntegerQ[(n - 1)
/2] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p, n/2 + 1]) && IntegerQ[2*p]

Rubi steps

\begin {align*} \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^4 \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {\left (c-\frac {c x}{a}\right )^5}{x^2 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{c}\\ &=c^4 \sqrt {1-\frac {1}{a^2 x^2}} x+\frac {\operatorname {Subst}\left (\int \frac {\frac {5 c^5}{a}-\frac {10 c^5 x}{a^2}+\frac {10 c^5 x^2}{a^3}-\frac {5 c^5 x^3}{a^4}+\frac {c^5 x^4}{a^5}}{x \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{c}\\ &=-\frac {c^4 \sqrt {1-\frac {1}{a^2 x^2}}}{3 a^3 x^2}+c^4 \sqrt {1-\frac {1}{a^2 x^2}} x-\frac {a^2 \operatorname {Subst}\left (\int \frac {-\frac {15 c^5}{a^3}+\frac {30 c^5 x}{a^4}-\frac {32 c^5 x^2}{a^5}+\frac {15 c^5 x^3}{a^6}}{x \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{3 c}\\ &=-\frac {c^4 \sqrt {1-\frac {1}{a^2 x^2}}}{3 a^3 x^2}+\frac {5 c^4 \sqrt {1-\frac {1}{a^2 x^2}}}{2 a^2 x}+c^4 \sqrt {1-\frac {1}{a^2 x^2}} x+\frac {a^4 \operatorname {Subst}\left (\int \frac {\frac {30 c^5}{a^5}-\frac {75 c^5 x}{a^6}+\frac {64 c^5 x^2}{a^7}}{x \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{6 c}\\ &=-\frac {32 c^4 \sqrt {1-\frac {1}{a^2 x^2}}}{3 a}-\frac {c^4 \sqrt {1-\frac {1}{a^2 x^2}}}{3 a^3 x^2}+\frac {5 c^4 \sqrt {1-\frac {1}{a^2 x^2}}}{2 a^2 x}+c^4 \sqrt {1-\frac {1}{a^2 x^2}} x-\frac {a^6 \operatorname {Subst}\left (\int \frac {-\frac {30 c^5}{a^7}+\frac {75 c^5 x}{a^8}}{x \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{6 c}\\ &=-\frac {32 c^4 \sqrt {1-\frac {1}{a^2 x^2}}}{3 a}-\frac {c^4 \sqrt {1-\frac {1}{a^2 x^2}}}{3 a^3 x^2}+\frac {5 c^4 \sqrt {1-\frac {1}{a^2 x^2}}}{2 a^2 x}+c^4 \sqrt {1-\frac {1}{a^2 x^2}} x-\frac {\left (25 c^4\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{2 a^2}+\frac {\left (5 c^4\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{a}\\ &=-\frac {32 c^4 \sqrt {1-\frac {1}{a^2 x^2}}}{3 a}-\frac {c^4 \sqrt {1-\frac {1}{a^2 x^2}}}{3 a^3 x^2}+\frac {5 c^4 \sqrt {1-\frac {1}{a^2 x^2}}}{2 a^2 x}+c^4 \sqrt {1-\frac {1}{a^2 x^2}} x-\frac {25 c^4 \csc ^{-1}(a x)}{2 a}+\frac {\left (5 c^4\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x}{a^2}}} \, dx,x,\frac {1}{x^2}\right )}{2 a}\\ &=-\frac {32 c^4 \sqrt {1-\frac {1}{a^2 x^2}}}{3 a}-\frac {c^4 \sqrt {1-\frac {1}{a^2 x^2}}}{3 a^3 x^2}+\frac {5 c^4 \sqrt {1-\frac {1}{a^2 x^2}}}{2 a^2 x}+c^4 \sqrt {1-\frac {1}{a^2 x^2}} x-\frac {25 c^4 \csc ^{-1}(a x)}{2 a}-\left (5 a c^4\right ) \operatorname {Subst}\left (\int \frac {1}{a^2-a^2 x^2} \, dx,x,\sqrt {1-\frac {1}{a^2 x^2}}\right )\\ &=-\frac {32 c^4 \sqrt {1-\frac {1}{a^2 x^2}}}{3 a}-\frac {c^4 \sqrt {1-\frac {1}{a^2 x^2}}}{3 a^3 x^2}+\frac {5 c^4 \sqrt {1-\frac {1}{a^2 x^2}}}{2 a^2 x}+c^4 \sqrt {1-\frac {1}{a^2 x^2}} x-\frac {25 c^4 \csc ^{-1}(a x)}{2 a}-\frac {5 c^4 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 175, normalized size = 1.30 \[ \frac {c^4 \left (6 a^5 x^5-64 a^4 x^4+9 a^3 x^3+62 a^2 x^2+90 a^4 x^4 \sqrt {1-\frac {1}{a^2 x^2}} \sin ^{-1}\left (\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {2}}\right )-30 a^4 x^4 \sqrt {1-\frac {1}{a^2 x^2}} \sin ^{-1}\left (\frac {1}{a x}\right )-30 a^4 x^4 \sqrt {1-\frac {1}{a^2 x^2}} \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )-15 a x+2\right )}{6 a^5 x^4 \sqrt {1-\frac {1}{a^2 x^2}}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c - c/(a*x))^4/E^ArcCoth[a*x],x]

[Out]

(c^4*(2 - 15*a*x + 62*a^2*x^2 + 9*a^3*x^3 - 64*a^4*x^4 + 6*a^5*x^5 + 90*a^4*Sqrt[1 - 1/(a^2*x^2)]*x^4*ArcSin[S
qrt[1 - 1/(a*x)]/Sqrt[2]] - 30*a^4*Sqrt[1 - 1/(a^2*x^2)]*x^4*ArcSin[1/(a*x)] - 30*a^4*Sqrt[1 - 1/(a^2*x^2)]*x^
4*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]]))/(6*a^5*Sqrt[1 - 1/(a^2*x^2)]*x^4)

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fricas [A]  time = 0.58, size = 156, normalized size = 1.16 \[ \frac {150 \, a^{3} c^{4} x^{3} \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right ) - 30 \, a^{3} c^{4} x^{3} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) + 30 \, a^{3} c^{4} x^{3} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) + {\left (6 \, a^{4} c^{4} x^{4} - 58 \, a^{3} c^{4} x^{3} - 49 \, a^{2} c^{4} x^{2} + 13 \, a c^{4} x - 2 \, c^{4}\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{6 \, a^{4} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^4*((a*x-1)/(a*x+1))^(1/2),x, algorithm="fricas")

[Out]

1/6*(150*a^3*c^4*x^3*arctan(sqrt((a*x - 1)/(a*x + 1))) - 30*a^3*c^4*x^3*log(sqrt((a*x - 1)/(a*x + 1)) + 1) + 3
0*a^3*c^4*x^3*log(sqrt((a*x - 1)/(a*x + 1)) - 1) + (6*a^4*c^4*x^4 - 58*a^3*c^4*x^3 - 49*a^2*c^4*x^2 + 13*a*c^4
*x - 2*c^4)*sqrt((a*x - 1)/(a*x + 1)))/(a^4*x^3)

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giac [B]  time = 0.17, size = 265, normalized size = 1.96 \[ \frac {25 \, c^{4} \arctan \left (-x {\left | a \right |} + \sqrt {a^{2} x^{2} - 1}\right ) \mathrm {sgn}\left (a x + 1\right )}{a} + \frac {5 \, c^{4} \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} - 1} \right |}\right ) \mathrm {sgn}\left (a x + 1\right )}{{\left | a \right |}} + \frac {\sqrt {a^{2} x^{2} - 1} c^{4} \mathrm {sgn}\left (a x + 1\right )}{a} - \frac {15 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} - 1}\right )}^{5} c^{4} {\left | a \right |} \mathrm {sgn}\left (a x + 1\right ) + 60 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} - 1}\right )}^{4} a c^{4} \mathrm {sgn}\left (a x + 1\right ) + 132 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} - 1}\right )}^{2} a c^{4} \mathrm {sgn}\left (a x + 1\right ) - 15 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} - 1}\right )} c^{4} {\left | a \right |} \mathrm {sgn}\left (a x + 1\right ) + 64 \, a c^{4} \mathrm {sgn}\left (a x + 1\right )}{3 \, {\left ({\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} - 1}\right )}^{2} + 1\right )}^{3} a {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^4*((a*x-1)/(a*x+1))^(1/2),x, algorithm="giac")

[Out]

25*c^4*arctan(-x*abs(a) + sqrt(a^2*x^2 - 1))*sgn(a*x + 1)/a + 5*c^4*log(abs(-x*abs(a) + sqrt(a^2*x^2 - 1)))*sg
n(a*x + 1)/abs(a) + sqrt(a^2*x^2 - 1)*c^4*sgn(a*x + 1)/a - 1/3*(15*(x*abs(a) - sqrt(a^2*x^2 - 1))^5*c^4*abs(a)
*sgn(a*x + 1) + 60*(x*abs(a) - sqrt(a^2*x^2 - 1))^4*a*c^4*sgn(a*x + 1) + 132*(x*abs(a) - sqrt(a^2*x^2 - 1))^2*
a*c^4*sgn(a*x + 1) - 15*(x*abs(a) - sqrt(a^2*x^2 - 1))*c^4*abs(a)*sgn(a*x + 1) + 64*a*c^4*sgn(a*x + 1))/(((x*a
bs(a) - sqrt(a^2*x^2 - 1))^2 + 1)^3*a*abs(a))

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maple [B]  time = 0.06, size = 290, normalized size = 2.15 \[ \frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right ) c^{4} \left (-66 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, x^{4} a^{4}+96 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, x^{3} a^{3}+66 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\, x^{2} a^{2}-75 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, x^{3} a^{3}-75 a^{3} x^{3} \sqrt {a^{2}}\, \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )+66 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{3} a^{4}-96 \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{3} a^{4}-15 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\, x a +2 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\right )}{6 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a^{4} x^{3} \sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c/a/x)^4*((a*x-1)/(a*x+1))^(1/2),x)

[Out]

1/6*((a*x-1)/(a*x+1))^(1/2)*(a*x+1)*c^4*(-66*(a^2*x^2-1)^(1/2)*(a^2)^(1/2)*x^4*a^4+96*(a^2)^(1/2)*((a*x-1)*(a*
x+1))^(1/2)*x^3*a^3+66*(a^2*x^2-1)^(3/2)*(a^2)^(1/2)*x^2*a^2-75*(a^2*x^2-1)^(1/2)*(a^2)^(1/2)*x^3*a^3-75*a^3*x
^3*(a^2)^(1/2)*arctan(1/(a^2*x^2-1)^(1/2))+66*ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x^3*a^4-96
*ln((a^2*x+((a*x-1)*(a*x+1))^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x^3*a^4-15*(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))^(1/2)/a^4/x^3/(a^2)^(1/2)

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maxima [A]  time = 0.41, size = 223, normalized size = 1.65 \[ \frac {1}{3} \, {\left (\frac {75 \, c^{4} \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right )}{a^{2}} - \frac {15 \, c^{4} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2}} + \frac {15 \, c^{4} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2}} + \frac {87 \, c^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{2}} + 61 \, c^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} - 55 \, c^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} - 45 \, c^{4} \sqrt {\frac {a x - 1}{a x + 1}}}{\frac {2 \, {\left (a x - 1\right )} a^{2}}{a x + 1} - \frac {2 \, {\left (a x - 1\right )}^{3} a^{2}}{{\left (a x + 1\right )}^{3}} - \frac {{\left (a x - 1\right )}^{4} a^{2}}{{\left (a x + 1\right )}^{4}} + a^{2}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^4*((a*x-1)/(a*x+1))^(1/2),x, algorithm="maxima")

[Out]

1/3*(75*c^4*arctan(sqrt((a*x - 1)/(a*x + 1)))/a^2 - 15*c^4*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^2 + 15*c^4*log
(sqrt((a*x - 1)/(a*x + 1)) - 1)/a^2 + (87*c^4*((a*x - 1)/(a*x + 1))^(7/2) + 61*c^4*((a*x - 1)/(a*x + 1))^(5/2)
- 55*c^4*((a*x - 1)/(a*x + 1))^(3/2) - 45*c^4*sqrt((a*x - 1)/(a*x + 1)))/(2*(a*x - 1)*a^2/(a*x + 1) - 2*(a*x
- 1)^3*a^2/(a*x + 1)^3 - (a*x - 1)^4*a^2/(a*x + 1)^4 + a^2))*a

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mupad [B]  time = 0.12, size = 185, normalized size = 1.37 \[ \frac {25\,c^4\,\mathrm {atan}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a}-\frac {15\,c^4\,\sqrt {\frac {a\,x-1}{a\,x+1}}+\frac {55\,c^4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{3}-\frac {61\,c^4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}}{3}-29\,c^4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/2}}{a+\frac {2\,a\,\left (a\,x-1\right )}{a\,x+1}-\frac {2\,a\,{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}-\frac {a\,{\left (a\,x-1\right )}^4}{{\left (a\,x+1\right )}^4}}-\frac {10\,c^4\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c/(a*x))^4*((a*x - 1)/(a*x + 1))^(1/2),x)

[Out]

(25*c^4*atan(((a*x - 1)/(a*x + 1))^(1/2)))/a - (15*c^4*((a*x - 1)/(a*x + 1))^(1/2) + (55*c^4*((a*x - 1)/(a*x +
1))^(3/2))/3 - (61*c^4*((a*x - 1)/(a*x + 1))^(5/2))/3 - 29*c^4*((a*x - 1)/(a*x + 1))^(7/2))/(a + (2*a*(a*x -
1))/(a*x + 1) - (2*a*(a*x - 1)^3)/(a*x + 1)^3 - (a*(a*x - 1)^4)/(a*x + 1)^4) - (10*c^4*atanh(((a*x - 1)/(a*x +
1))^(1/2)))/a

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {c^{4} \left (\int a^{4} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}\, dx + \int \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{x^{4}}\, dx + \int \left (- \frac {4 a \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{x^{3}}\right )\, dx + \int \frac {6 a^{2} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{x^{2}}\, dx + \int \left (- \frac {4 a^{3} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{x}\right )\, dx\right )}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)**4*((a*x-1)/(a*x+1))**(1/2),x)

[Out]

c**4*(Integral(a**4*sqrt(a*x/(a*x + 1) - 1/(a*x + 1)), x) + Integral(sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/x**4, x
) + Integral(-4*a*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/x**3, x) + Integral(6*a**2*sqrt(a*x/(a*x + 1) - 1/(a*x + 1
))/x**2, x) + Integral(-4*a**3*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/x, x))/a**4

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