3.556 \(\int e^{-3 \tanh ^{-1}(a x)} (c-\frac {c}{a x})^{5/2} \, dx\)

Optimal. Leaf size=181 \[ -\frac {11 a^{3/2} x^{5/2} \left (c-\frac {c}{a x}\right )^{5/2} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )}{(1-a x)^{5/2}}+\frac {a^2 x^3 (191-25 a x) \left (c-\frac {c}{a x}\right )^{5/2}}{3 (1-a x)^{5/2} \sqrt {a x+1}}+\frac {26 a x^2 \left (c-\frac {c}{a x}\right )^{5/2}}{3 \sqrt {1-a x} \sqrt {a x+1}}-\frac {2 x \sqrt {1-a x} \left (c-\frac {c}{a x}\right )^{5/2}}{3 \sqrt {a x+1}} \]

[Out]

-11*a^(3/2)*(c-c/a/x)^(5/2)*x^(5/2)*arcsinh(a^(1/2)*x^(1/2))/(-a*x+1)^(5/2)+1/3*a^2*(c-c/a/x)^(5/2)*x^3*(-25*a
*x+191)/(-a*x+1)^(5/2)/(a*x+1)^(1/2)+26/3*a*(c-c/a/x)^(5/2)*x^2/(-a*x+1)^(1/2)/(a*x+1)^(1/2)-2/3*(c-c/a/x)^(5/
2)*x*(-a*x+1)^(1/2)/(a*x+1)^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {6134, 6129, 98, 150, 143, 54, 215} \[ \frac {a^2 x^3 (191-25 a x) \left (c-\frac {c}{a x}\right )^{5/2}}{3 (1-a x)^{5/2} \sqrt {a x+1}}-\frac {11 a^{3/2} x^{5/2} \left (c-\frac {c}{a x}\right )^{5/2} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )}{(1-a x)^{5/2}}+\frac {26 a x^2 \left (c-\frac {c}{a x}\right )^{5/2}}{3 \sqrt {1-a x} \sqrt {a x+1}}-\frac {2 x \sqrt {1-a x} \left (c-\frac {c}{a x}\right )^{5/2}}{3 \sqrt {a x+1}} \]

Antiderivative was successfully verified.

[In]

Int[(c - c/(a*x))^(5/2)/E^(3*ArcTanh[a*x]),x]

[Out]

(a^2*(c - c/(a*x))^(5/2)*x^3*(191 - 25*a*x))/(3*(1 - a*x)^(5/2)*Sqrt[1 + a*x]) + (26*a*(c - c/(a*x))^(5/2)*x^2
)/(3*Sqrt[1 - a*x]*Sqrt[1 + a*x]) - (2*(c - c/(a*x))^(5/2)*x*Sqrt[1 - a*x])/(3*Sqrt[1 + a*x]) - (11*a^(3/2)*(c
- c/(a*x))^(5/2)*x^(5/2)*ArcSinh[Sqrt[a]*Sqrt[x]])/(1 - a*x)^(5/2)

Rule 54

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

Rule 98

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

Rule 143

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((b^2*d*e*g - a^2*d*f*h*m - a*b*(d*(f*g + e*h) - c*f*h*(m + 1)) + b*f*h*(b*c - a*d)*(m + 1)*x)*(a + b*x
)^(m + 1)*(c + d*x)^(n + 1))/(b^2*d*(b*c - a*d)*(m + 1)), x] + Dist[(a*d*f*h*m + b*(d*(f*g + e*h) - c*f*h*(m +
2)))/(b^2*d), Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && EqQ[m
+ n + 2, 0] && NeQ[m, -1] &&  !(SumSimplerQ[n, 1] &&  !SumSimplerQ[m, 1])

Rule 150

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegersQ[2*m, 2*n, 2*p]

Rule 215

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

Rule 6129

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

Rule 6134

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

Rubi steps

\begin {align*} \int e^{-3 \tanh ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^{5/2} \, dx &=\frac {\left (\left (c-\frac {c}{a x}\right )^{5/2} x^{5/2}\right ) \int \frac {e^{-3 \tanh ^{-1}(a x)} (1-a x)^{5/2}}{x^{5/2}} \, dx}{(1-a x)^{5/2}}\\ &=\frac {\left (\left (c-\frac {c}{a x}\right )^{5/2} x^{5/2}\right ) \int \frac {(1-a x)^4}{x^{5/2} (1+a x)^{3/2}} \, dx}{(1-a x)^{5/2}}\\ &=-\frac {2 \left (c-\frac {c}{a x}\right )^{5/2} x \sqrt {1-a x}}{3 \sqrt {1+a x}}-\frac {\left (2 \left (c-\frac {c}{a x}\right )^{5/2} x^{5/2}\right ) \int \frac {(1-a x)^2 \left (\frac {13 a}{2}-\frac {a^2 x}{2}\right )}{x^{3/2} (1+a x)^{3/2}} \, dx}{3 (1-a x)^{5/2}}\\ &=\frac {26 a \left (c-\frac {c}{a x}\right )^{5/2} x^2}{3 \sqrt {1-a x} \sqrt {1+a x}}-\frac {2 \left (c-\frac {c}{a x}\right )^{5/2} x \sqrt {1-a x}}{3 \sqrt {1+a x}}-\frac {\left (4 \left (c-\frac {c}{a x}\right )^{5/2} x^{5/2}\right ) \int \frac {(1-a x) \left (-\frac {79 a^2}{4}-\frac {25 a^3 x}{4}\right )}{\sqrt {x} (1+a x)^{3/2}} \, dx}{3 (1-a x)^{5/2}}\\ &=\frac {a^2 \left (c-\frac {c}{a x}\right )^{5/2} x^3 (191-25 a x)}{3 (1-a x)^{5/2} \sqrt {1+a x}}+\frac {26 a \left (c-\frac {c}{a x}\right )^{5/2} x^2}{3 \sqrt {1-a x} \sqrt {1+a x}}-\frac {2 \left (c-\frac {c}{a x}\right )^{5/2} x \sqrt {1-a x}}{3 \sqrt {1+a x}}-\frac {\left (11 a^2 \left (c-\frac {c}{a x}\right )^{5/2} x^{5/2}\right ) \int \frac {1}{\sqrt {x} \sqrt {1+a x}} \, dx}{2 (1-a x)^{5/2}}\\ &=\frac {a^2 \left (c-\frac {c}{a x}\right )^{5/2} x^3 (191-25 a x)}{3 (1-a x)^{5/2} \sqrt {1+a x}}+\frac {26 a \left (c-\frac {c}{a x}\right )^{5/2} x^2}{3 \sqrt {1-a x} \sqrt {1+a x}}-\frac {2 \left (c-\frac {c}{a x}\right )^{5/2} x \sqrt {1-a x}}{3 \sqrt {1+a x}}-\frac {\left (11 a^2 \left (c-\frac {c}{a x}\right )^{5/2} x^{5/2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+a x^2}} \, dx,x,\sqrt {x}\right )}{(1-a x)^{5/2}}\\ &=\frac {a^2 \left (c-\frac {c}{a x}\right )^{5/2} x^3 (191-25 a x)}{3 (1-a x)^{5/2} \sqrt {1+a x}}+\frac {26 a \left (c-\frac {c}{a x}\right )^{5/2} x^2}{3 \sqrt {1-a x} \sqrt {1+a x}}-\frac {2 \left (c-\frac {c}{a x}\right )^{5/2} x \sqrt {1-a x}}{3 \sqrt {1+a x}}-\frac {11 a^{3/2} \left (c-\frac {c}{a x}\right )^{5/2} x^{5/2} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )}{(1-a x)^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 97, normalized size = 0.54 \[ \frac {c^2 \sqrt {c-\frac {c}{a x}} \left (-33 a^{3/2} x^{3/2} \sqrt {a x+1} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )+3 a^3 x^3+133 a^2 x^2+32 a x-2\right )}{3 a^2 x \sqrt {1-a^2 x^2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c - c/(a*x))^(5/2)/E^(3*ArcTanh[a*x]),x]

[Out]

(c^2*Sqrt[c - c/(a*x)]*(-2 + 32*a*x + 133*a^2*x^2 + 3*a^3*x^3 - 33*a^(3/2)*x^(3/2)*Sqrt[1 + a*x]*ArcSinh[Sqrt[
a]*Sqrt[x]]))/(3*a^2*x*Sqrt[1 - a^2*x^2])

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fricas [A]  time = 0.55, size = 352, normalized size = 1.94 \[ \left [\frac {33 \, {\left (a^{3} c^{2} x^{3} - a c^{2} x\right )} \sqrt {-c} \log \left (-\frac {8 \, a^{3} c x^{3} - 7 \, a c x - 4 \, {\left (2 \, a^{2} x^{2} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a c x - c}{a x}} - c}{a x - 1}\right ) - 4 \, {\left (3 \, a^{3} c^{2} x^{3} + 133 \, a^{2} c^{2} x^{2} + 32 \, a c^{2} x - 2 \, c^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{12 \, {\left (a^{4} x^{3} - a^{2} x\right )}}, \frac {33 \, {\left (a^{3} c^{2} x^{3} - a c^{2} x\right )} \sqrt {c} \arctan \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1} a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}}}{2 \, a^{2} c x^{2} - a c x - c}\right ) - 2 \, {\left (3 \, a^{3} c^{2} x^{3} + 133 \, a^{2} c^{2} x^{2} + 32 \, a c^{2} x - 2 \, c^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{6 \, {\left (a^{4} x^{3} - a^{2} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(33*(a^3*c^2*x^3 - a*c^2*x)*sqrt(-c)*log(-(8*a^3*c*x^3 - 7*a*c*x - 4*(2*a^2*x^2 + a*x)*sqrt(-a^2*x^2 + 1
)*sqrt(-c)*sqrt((a*c*x - c)/(a*x)) - c)/(a*x - 1)) - 4*(3*a^3*c^2*x^3 + 133*a^2*c^2*x^2 + 32*a*c^2*x - 2*c^2)*
sqrt(-a^2*x^2 + 1)*sqrt((a*c*x - c)/(a*x)))/(a^4*x^3 - a^2*x), 1/6*(33*(a^3*c^2*x^3 - a*c^2*x)*sqrt(c)*arctan(
2*sqrt(-a^2*x^2 + 1)*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x))/(2*a^2*c*x^2 - a*c*x - c)) - 2*(3*a^3*c^2*x^3 + 133*a
^2*c^2*x^2 + 32*a*c^2*x - 2*c^2)*sqrt(-a^2*x^2 + 1)*sqrt((a*c*x - c)/(a*x)))/(a^4*x^3 - a^2*x)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x)]sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.07, size = 191, normalized size = 1.06 \[ -\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, c^{2} \left (6 a^{\frac {7}{2}} x^{3} \sqrt {-\left (a x +1\right ) x}+33 \arctan \left (\frac {2 a x +1}{2 \sqrt {a}\, \sqrt {-\left (a x +1\right ) x}}\right ) x^{3} a^{3}+266 a^{\frac {5}{2}} x^{2} \sqrt {-\left (a x +1\right ) x}+33 \arctan \left (\frac {2 a x +1}{2 \sqrt {a}\, \sqrt {-\left (a x +1\right ) x}}\right ) x^{2} a^{2}+64 a^{\frac {3}{2}} x \sqrt {-\left (a x +1\right ) x}-4 \sqrt {a}\, \sqrt {-\left (a x +1\right ) x}\right ) \sqrt {-a^{2} x^{2}+1}}{6 x \,a^{\frac {5}{2}} \left (a x +1\right ) \sqrt {-\left (a x +1\right ) x}\, \left (a x -1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(c*(a*x-1)/a/x)^(1/2)/x*c^2/a^(5/2)*(6*a^(7/2)*x^3*(-(a*x+1)*x)^(1/2)+33*arctan(1/2/a^(1/2)*(2*a*x+1)/(-(
a*x+1)*x)^(1/2))*x^3*a^3+266*a^(5/2)*x^2*(-(a*x+1)*x)^(1/2)+33*arctan(1/2/a^(1/2)*(2*a*x+1)/(-(a*x+1)*x)^(1/2)
)*x^2*a^2+64*a^(3/2)*x*(-(a*x+1)*x)^(1/2)-4*a^(1/2)*(-(a*x+1)*x)^(1/2))*(-a^2*x^2+1)^(1/2)/(a*x+1)/(-(a*x+1)*x
)^(1/2)/(a*x-1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c-\frac {c}{a\,x}\right )}^{5/2}\,{\left (1-a^2\,x^2\right )}^{3/2}}{{\left (a\,x+1\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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