3.498 \(\int e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}} x^3 \, dx\)

Optimal. Leaf size=130 \[ \frac {75 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{64 a^4}+\frac {75 x \sqrt {c-\frac {c}{a x}}}{64 a^3}+\frac {25 x^2 \sqrt {c-\frac {c}{a x}}}{32 a^2}+\frac {1}{4} x^4 \sqrt {c-\frac {c}{a x}}+\frac {5 x^3 \sqrt {c-\frac {c}{a x}}}{8 a} \]

[Out]

75/64*arctanh((c-c/a/x)^(1/2)/c^(1/2))*c^(1/2)/a^4+75/64*x*(c-c/a/x)^(1/2)/a^3+25/32*x^2*(c-c/a/x)^(1/2)/a^2+5
/8*x^3*(c-c/a/x)^(1/2)/a+1/4*x^4*(c-c/a/x)^(1/2)

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Rubi [A]  time = 0.38, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6167, 6133, 25, 514, 446, 78, 51, 63, 208} \[ \frac {25 x^2 \sqrt {c-\frac {c}{a x}}}{32 a^2}+\frac {75 x \sqrt {c-\frac {c}{a x}}}{64 a^3}+\frac {75 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{64 a^4}+\frac {1}{4} x^4 \sqrt {c-\frac {c}{a x}}+\frac {5 x^3 \sqrt {c-\frac {c}{a x}}}{8 a} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcCoth[a*x])*Sqrt[c - c/(a*x)]*x^3,x]

[Out]

(75*Sqrt[c - c/(a*x)]*x)/(64*a^3) + (25*Sqrt[c - c/(a*x)]*x^2)/(32*a^2) + (5*Sqrt[c - c/(a*x)]*x^3)/(8*a) + (S
qrt[c - c/(a*x)]*x^4)/4 + (75*Sqrt[c]*ArcTanh[Sqrt[c - c/(a*x)]/Sqrt[c]])/(64*a^4)

Rule 25

Int[(u_.)*((a_) + (b_.)*(x_)^(n_.))^(m_.)*((c_) + (d_.)*(x_)^(q_.))^(p_.), x_Symbol] :> Dist[(d/a)^p, Int[(u*(
a + b*x^n)^(m + p))/x^(n*p), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[q, -n] && IntegerQ[p] && EqQ[a*c -
b*d, 0] &&  !(IntegerQ[m] && NegQ[n])

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 514

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[x^(m - n*q)*
(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] |
|  !IntegerQ[p])

Rule 6133

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

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rubi steps

\begin {align*} \int e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}} x^3 \, dx &=-\int e^{2 \tanh ^{-1}(a x)} \sqrt {c-\frac {c}{a x}} x^3 \, dx\\ &=-\int \frac {\sqrt {c-\frac {c}{a x}} x^3 (1+a x)}{1-a x} \, dx\\ &=\frac {c \int \frac {x^2 (1+a x)}{\sqrt {c-\frac {c}{a x}}} \, dx}{a}\\ &=\frac {c \int \frac {\left (a+\frac {1}{x}\right ) x^3}{\sqrt {c-\frac {c}{a x}}} \, dx}{a}\\ &=-\frac {c \operatorname {Subst}\left (\int \frac {a+x}{x^5 \sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )}{a}\\ &=\frac {1}{4} \sqrt {c-\frac {c}{a x}} x^4-\frac {(15 c) \operatorname {Subst}\left (\int \frac {1}{x^4 \sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )}{8 a}\\ &=\frac {5 \sqrt {c-\frac {c}{a x}} x^3}{8 a}+\frac {1}{4} \sqrt {c-\frac {c}{a x}} x^4-\frac {(25 c) \operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )}{16 a^2}\\ &=\frac {25 \sqrt {c-\frac {c}{a x}} x^2}{32 a^2}+\frac {5 \sqrt {c-\frac {c}{a x}} x^3}{8 a}+\frac {1}{4} \sqrt {c-\frac {c}{a x}} x^4-\frac {(75 c) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )}{64 a^3}\\ &=\frac {75 \sqrt {c-\frac {c}{a x}} x}{64 a^3}+\frac {25 \sqrt {c-\frac {c}{a x}} x^2}{32 a^2}+\frac {5 \sqrt {c-\frac {c}{a x}} x^3}{8 a}+\frac {1}{4} \sqrt {c-\frac {c}{a x}} x^4-\frac {(75 c) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )}{128 a^4}\\ &=\frac {75 \sqrt {c-\frac {c}{a x}} x}{64 a^3}+\frac {25 \sqrt {c-\frac {c}{a x}} x^2}{32 a^2}+\frac {5 \sqrt {c-\frac {c}{a x}} x^3}{8 a}+\frac {1}{4} \sqrt {c-\frac {c}{a x}} x^4+\frac {75 \operatorname {Subst}\left (\int \frac {1}{a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-\frac {c}{a x}}\right )}{64 a^3}\\ &=\frac {75 \sqrt {c-\frac {c}{a x}} x}{64 a^3}+\frac {25 \sqrt {c-\frac {c}{a x}} x^2}{32 a^2}+\frac {5 \sqrt {c-\frac {c}{a x}} x^3}{8 a}+\frac {1}{4} \sqrt {c-\frac {c}{a x}} x^4+\frac {75 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{64 a^4}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 50, normalized size = 0.38 \[ \frac {\sqrt {c-\frac {c}{a x}} \left (a^4 x^4+15 \, _2F_1\left (\frac {1}{2},4;\frac {3}{2};1-\frac {1}{a x}\right )\right )}{4 a^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(2*ArcCoth[a*x])*Sqrt[c - c/(a*x)]*x^3,x]

[Out]

(Sqrt[c - c/(a*x)]*(a^4*x^4 + 15*Hypergeometric2F1[1/2, 4, 3/2, 1 - 1/(a*x)]))/(4*a^4)

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fricas [A]  time = 0.51, size = 179, normalized size = 1.38 \[ \left [\frac {2 \, {\left (16 \, a^{4} x^{4} + 40 \, a^{3} x^{3} + 50 \, a^{2} x^{2} + 75 \, a x\right )} \sqrt {\frac {a c x - c}{a x}} + 75 \, \sqrt {c} \log \left (-2 \, a c x - 2 \, a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}} + c\right )}{128 \, a^{4}}, \frac {{\left (16 \, a^{4} x^{4} + 40 \, a^{3} x^{3} + 50 \, a^{2} x^{2} + 75 \, a x\right )} \sqrt {\frac {a c x - c}{a x}} - 75 \, \sqrt {-c} \arctan \left (\frac {\sqrt {-c} \sqrt {\frac {a c x - c}{a x}}}{c}\right )}{64 \, a^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/128*(2*(16*a^4*x^4 + 40*a^3*x^3 + 50*a^2*x^2 + 75*a*x)*sqrt((a*c*x - c)/(a*x)) + 75*sqrt(c)*log(-2*a*c*x -
2*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x)) + c))/a^4, 1/64*((16*a^4*x^4 + 40*a^3*x^3 + 50*a^2*x^2 + 75*a*x)*sqrt((a
*c*x - c)/(a*x)) - 75*sqrt(-c)*arctan(sqrt(-c)*sqrt((a*c*x - c)/(a*x))/c))/a^4]

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giac [A]  time = 0.17, size = 142, normalized size = 1.09 \[ \frac {1}{64} \, \sqrt {a^{2} c x^{2} - a c x} {\left (2 \, {\left (4 \, x {\left (\frac {2 \, x {\left | a \right |}}{a^{2} \mathrm {sgn}\relax (x)} + \frac {5 \, {\left | a \right |}}{a^{3} \mathrm {sgn}\relax (x)}\right )} + \frac {25 \, {\left | a \right |}}{a^{4} \mathrm {sgn}\relax (x)}\right )} x + \frac {75 \, {\left | a \right |}}{a^{5} \mathrm {sgn}\relax (x)}\right )} + \frac {75 \, \sqrt {c} \log \left ({\left | a \right |} {\left | c \right |}\right ) \mathrm {sgn}\relax (x)}{128 \, a^{4}} - \frac {75 \, \sqrt {c} \log \left ({\left | -2 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )} \sqrt {c} {\left | a \right |} + a c \right |}\right )}{128 \, a^{4} \mathrm {sgn}\relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*sqrt(a^2*c*x^2 - a*c*x)*(2*(4*x*(2*x*abs(a)/(a^2*sgn(x)) + 5*abs(a)/(a^3*sgn(x))) + 25*abs(a)/(a^4*sgn(x)
))*x + 75*abs(a)/(a^5*sgn(x))) + 75/128*sqrt(c)*log(abs(a)*abs(c))*sgn(x)/a^4 - 75/128*sqrt(c)*log(abs(-2*(sqr
t(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))*sqrt(c)*abs(a) + a*c))/(a^4*sgn(x))

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maple [A]  time = 0.05, size = 172, normalized size = 1.32 \[ \frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, x \left (32 x \left (a \,x^{2}-x \right )^{\frac {3}{2}} a^{\frac {7}{2}}+112 \left (a \,x^{2}-x \right )^{\frac {3}{2}} a^{\frac {5}{2}}+212 \sqrt {a \,x^{2}-x}\, a^{\frac {5}{2}} x -106 \sqrt {a \,x^{2}-x}\, a^{\frac {3}{2}}+256 a^{\frac {3}{2}} \sqrt {\left (a x -1\right ) x}+128 a \ln \left (\frac {2 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right )-53 \ln \left (\frac {2 \sqrt {a \,x^{2}-x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a \right )}{128 \sqrt {\left (a x -1\right ) x}\, a^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*(c*(a*x-1)/a/x)^(1/2)*x*(32*x*(a*x^2-x)^(3/2)*a^(7/2)+112*(a*x^2-x)^(3/2)*a^(5/2)+212*(a*x^2-x)^(1/2)*a^
(5/2)*x-106*(a*x^2-x)^(1/2)*a^(3/2)+256*a^(3/2)*((a*x-1)*x)^(1/2)+128*a*ln(1/2*(2*((a*x-1)*x)^(1/2)*a^(1/2)+2*
a*x-1)/a^(1/2))-53*ln(1/2*(2*(a*x^2-x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*a)/((a*x-1)*x)^(1/2)/a^(9/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*x + 1)*sqrt(c - c/(a*x))*x^3/(a*x - 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3\,\sqrt {c-\frac {c}{a\,x}}\,\left (a\,x+1\right )}{a\,x-1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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