3.436 \(\int \frac {e^{-3 \tanh ^{-1}(a x)} \sqrt {c-a c x}}{x^5} \, dx\)

Optimal. Leaf size=249 \[ \frac {1115 a^4 c^2 (1-a x)^{3/2}}{64 \sqrt {a x+1} (c-a c x)^{3/2}}-\frac {1115 a^4 c^2 (1-a x)^{3/2} \tanh ^{-1}\left (\sqrt {a x+1}\right )}{64 (c-a c x)^{3/2}}+\frac {1115 a^3 c^2 (1-a x)^{3/2}}{192 x \sqrt {a x+1} (c-a c x)^{3/2}}-\frac {223 a^2 c^2 (1-a x)^{3/2}}{96 x^2 \sqrt {a x+1} (c-a c x)^{3/2}}-\frac {c^2 (1-a x)^{3/2}}{4 x^4 \sqrt {a x+1} (c-a c x)^{3/2}}+\frac {25 a c^2 (1-a x)^{3/2}}{24 x^3 \sqrt {a x+1} (c-a c x)^{3/2}} \]

[Out]

-1115/64*a^4*c^2*(-a*x+1)^(3/2)*arctanh((a*x+1)^(1/2))/(-a*c*x+c)^(3/2)+1115/64*a^4*c^2*(-a*x+1)^(3/2)/(-a*c*x
+c)^(3/2)/(a*x+1)^(1/2)-1/4*c^2*(-a*x+1)^(3/2)/x^4/(-a*c*x+c)^(3/2)/(a*x+1)^(1/2)+25/24*a*c^2*(-a*x+1)^(3/2)/x
^3/(-a*c*x+c)^(3/2)/(a*x+1)^(1/2)-223/96*a^2*c^2*(-a*x+1)^(3/2)/x^2/(-a*c*x+c)^(3/2)/(a*x+1)^(1/2)+1115/192*a^
3*c^2*(-a*x+1)^(3/2)/x/(-a*c*x+c)^(3/2)/(a*x+1)^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 252, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {6130, 23, 89, 78, 51, 63, 208} \[ -\frac {1115 a^2 c^2 (1-a x)^{3/2} \sqrt {a x+1}}{96 x^2 (c-a c x)^{3/2}}+\frac {223 a^2 c^2 (1-a x)^{3/2}}{24 x^2 \sqrt {a x+1} (c-a c x)^{3/2}}+\frac {1115 a^3 c^2 (1-a x)^{3/2} \sqrt {a x+1}}{64 x (c-a c x)^{3/2}}-\frac {1115 a^4 c^2 (1-a x)^{3/2} \tanh ^{-1}\left (\sqrt {a x+1}\right )}{64 (c-a c x)^{3/2}}+\frac {25 a c^2 (1-a x)^{3/2}}{24 x^3 \sqrt {a x+1} (c-a c x)^{3/2}}-\frac {c^2 (1-a x)^{3/2}}{4 x^4 \sqrt {a x+1} (c-a c x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c^2*(1 - a*x)^(3/2))/(4*x^4*Sqrt[1 + a*x]*(c - a*c*x)^(3/2)) + (25*a*c^2*(1 - a*x)^(3/2))/(24*x^3*Sqrt[1 + a
*x]*(c - a*c*x)^(3/2)) + (223*a^2*c^2*(1 - a*x)^(3/2))/(24*x^2*Sqrt[1 + a*x]*(c - a*c*x)^(3/2)) - (1115*a^2*c^
2*(1 - a*x)^(3/2)*Sqrt[1 + a*x])/(96*x^2*(c - a*c*x)^(3/2)) + (1115*a^3*c^2*(1 - a*x)^(3/2)*Sqrt[1 + a*x])/(64
*x*(c - a*c*x)^(3/2)) - (1115*a^4*c^2*(1 - a*x)^(3/2)*ArcTanh[Sqrt[1 + a*x]])/(64*(c - a*c*x)^(3/2))

Rule 23

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

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 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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 6130

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, n, p}, x] && EqQ[a^2*c^2 - d^2, 0] &&  !(IntegerQ[p] || GtQ[c, 0]
)

Rubi steps

\begin {align*} \int \frac {e^{-3 \tanh ^{-1}(a x)} \sqrt {c-a c x}}{x^5} \, dx &=\int \frac {(1-a x)^{3/2} \sqrt {c-a c x}}{x^5 (1+a x)^{3/2}} \, dx\\ &=\frac {(1-a x)^{3/2} \int \frac {(c-a c x)^2}{x^5 (1+a x)^{3/2}} \, dx}{(c-a c x)^{3/2}}\\ &=-\frac {c^2 (1-a x)^{3/2}}{4 x^4 \sqrt {1+a x} (c-a c x)^{3/2}}+\frac {(1-a x)^{3/2} \int \frac {-\frac {25 a c^2}{2}+4 a^2 c^2 x}{x^4 (1+a x)^{3/2}} \, dx}{4 (c-a c x)^{3/2}}\\ &=-\frac {c^2 (1-a x)^{3/2}}{4 x^4 \sqrt {1+a x} (c-a c x)^{3/2}}+\frac {25 a c^2 (1-a x)^{3/2}}{24 x^3 \sqrt {1+a x} (c-a c x)^{3/2}}+\frac {\left (223 a^2 c^2 (1-a x)^{3/2}\right ) \int \frac {1}{x^3 (1+a x)^{3/2}} \, dx}{48 (c-a c x)^{3/2}}\\ &=-\frac {c^2 (1-a x)^{3/2}}{4 x^4 \sqrt {1+a x} (c-a c x)^{3/2}}+\frac {25 a c^2 (1-a x)^{3/2}}{24 x^3 \sqrt {1+a x} (c-a c x)^{3/2}}+\frac {223 a^2 c^2 (1-a x)^{3/2}}{24 x^2 \sqrt {1+a x} (c-a c x)^{3/2}}+\frac {\left (1115 a^2 c^2 (1-a x)^{3/2}\right ) \int \frac {1}{x^3 \sqrt {1+a x}} \, dx}{48 (c-a c x)^{3/2}}\\ &=-\frac {c^2 (1-a x)^{3/2}}{4 x^4 \sqrt {1+a x} (c-a c x)^{3/2}}+\frac {25 a c^2 (1-a x)^{3/2}}{24 x^3 \sqrt {1+a x} (c-a c x)^{3/2}}+\frac {223 a^2 c^2 (1-a x)^{3/2}}{24 x^2 \sqrt {1+a x} (c-a c x)^{3/2}}-\frac {1115 a^2 c^2 (1-a x)^{3/2} \sqrt {1+a x}}{96 x^2 (c-a c x)^{3/2}}-\frac {\left (1115 a^3 c^2 (1-a x)^{3/2}\right ) \int \frac {1}{x^2 \sqrt {1+a x}} \, dx}{64 (c-a c x)^{3/2}}\\ &=-\frac {c^2 (1-a x)^{3/2}}{4 x^4 \sqrt {1+a x} (c-a c x)^{3/2}}+\frac {25 a c^2 (1-a x)^{3/2}}{24 x^3 \sqrt {1+a x} (c-a c x)^{3/2}}+\frac {223 a^2 c^2 (1-a x)^{3/2}}{24 x^2 \sqrt {1+a x} (c-a c x)^{3/2}}-\frac {1115 a^2 c^2 (1-a x)^{3/2} \sqrt {1+a x}}{96 x^2 (c-a c x)^{3/2}}+\frac {1115 a^3 c^2 (1-a x)^{3/2} \sqrt {1+a x}}{64 x (c-a c x)^{3/2}}+\frac {\left (1115 a^4 c^2 (1-a x)^{3/2}\right ) \int \frac {1}{x \sqrt {1+a x}} \, dx}{128 (c-a c x)^{3/2}}\\ &=-\frac {c^2 (1-a x)^{3/2}}{4 x^4 \sqrt {1+a x} (c-a c x)^{3/2}}+\frac {25 a c^2 (1-a x)^{3/2}}{24 x^3 \sqrt {1+a x} (c-a c x)^{3/2}}+\frac {223 a^2 c^2 (1-a x)^{3/2}}{24 x^2 \sqrt {1+a x} (c-a c x)^{3/2}}-\frac {1115 a^2 c^2 (1-a x)^{3/2} \sqrt {1+a x}}{96 x^2 (c-a c x)^{3/2}}+\frac {1115 a^3 c^2 (1-a x)^{3/2} \sqrt {1+a x}}{64 x (c-a c x)^{3/2}}+\frac {\left (1115 a^3 c^2 (1-a x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {1}{a}+\frac {x^2}{a}} \, dx,x,\sqrt {1+a x}\right )}{64 (c-a c x)^{3/2}}\\ &=-\frac {c^2 (1-a x)^{3/2}}{4 x^4 \sqrt {1+a x} (c-a c x)^{3/2}}+\frac {25 a c^2 (1-a x)^{3/2}}{24 x^3 \sqrt {1+a x} (c-a c x)^{3/2}}+\frac {223 a^2 c^2 (1-a x)^{3/2}}{24 x^2 \sqrt {1+a x} (c-a c x)^{3/2}}-\frac {1115 a^2 c^2 (1-a x)^{3/2} \sqrt {1+a x}}{96 x^2 (c-a c x)^{3/2}}+\frac {1115 a^3 c^2 (1-a x)^{3/2} \sqrt {1+a x}}{64 x (c-a c x)^{3/2}}-\frac {1115 a^4 c^2 (1-a x)^{3/2} \tanh ^{-1}\left (\sqrt {1+a x}\right )}{64 (c-a c x)^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 65, normalized size = 0.26 \[ \frac {c \sqrt {1-a x} \left (223 a^4 x^4 \, _2F_1\left (-\frac {1}{2},3;\frac {1}{2};a x+1\right )+25 a x-6\right )}{24 x^4 \sqrt {a x+1} \sqrt {c-a c x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c*Sqrt[1 - a*x]*(-6 + 25*a*x + 223*a^4*x^4*Hypergeometric2F1[-1/2, 3, 1/2, 1 + a*x]))/(24*x^4*Sqrt[1 + a*x]*S
qrt[c - a*c*x])

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fricas [A]  time = 0.47, size = 284, normalized size = 1.14 \[ \left [\frac {3345 \, {\left (a^{6} x^{6} - a^{4} x^{4}\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} + a c x + 2 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {c} - 2 \, c}{a x^{2} - x}\right ) - 2 \, {\left (3345 \, a^{4} x^{4} + 1115 \, a^{3} x^{3} - 446 \, a^{2} x^{2} + 200 \, a x - 48\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{384 \, {\left (a^{2} x^{6} - x^{4}\right )}}, -\frac {3345 \, {\left (a^{6} x^{6} - a^{4} x^{4}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {-c}}{a^{2} c x^{2} - c}\right ) + {\left (3345 \, a^{4} x^{4} + 1115 \, a^{3} x^{3} - 446 \, a^{2} x^{2} + 200 \, a x - 48\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{192 \, {\left (a^{2} x^{6} - x^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/384*(3345*(a^6*x^6 - a^4*x^4)*sqrt(c)*log(-(a^2*c*x^2 + a*c*x + 2*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*sqrt(
c) - 2*c)/(a*x^2 - x)) - 2*(3345*a^4*x^4 + 1115*a^3*x^3 - 446*a^2*x^2 + 200*a*x - 48)*sqrt(-a^2*x^2 + 1)*sqrt(
-a*c*x + c))/(a^2*x^6 - x^4), -1/192*(3345*(a^6*x^6 - a^4*x^4)*sqrt(-c)*arctan(sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x
+ c)*sqrt(-c)/(a^2*c*x^2 - c)) + (3345*a^4*x^4 + 1115*a^3*x^3 - 446*a^2*x^2 + 200*a*x - 48)*sqrt(-a^2*x^2 + 1)
*sqrt(-a*c*x + c))/(a^2*x^6 - x^4)]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.06, size = 122, normalized size = 0.49 \[ \frac {\sqrt {-c \left (a x -1\right )}\, \sqrt {-a^{2} x^{2}+1}\, \left (3345 \arctanh \left (\frac {\sqrt {c \left (a x +1\right )}}{\sqrt {c}}\right ) x^{4} a^{4} \sqrt {c \left (a x +1\right )}-3345 x^{4} a^{4} \sqrt {c}-1115 x^{3} a^{3} \sqrt {c}+446 x^{2} a^{2} \sqrt {c}-200 x a \sqrt {c}+48 \sqrt {c}\right )}{192 \sqrt {c}\, \left (a x -1\right ) \left (a x +1\right ) x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(-c*(a*x-1))^(1/2)*(-a^2*x^2+1)^(1/2)*(3345*arctanh((c*(a*x+1))^(1/2)/c^(1/2))*x^4*a^4*(c*(a*x+1))^(1/2)
-3345*x^4*a^4*c^(1/2)-1115*x^3*a^3*c^(1/2)+446*x^2*a^2*c^(1/2)-200*x*a*c^(1/2)+48*c^(1/2))/c^(1/2)/(a*x-1)/(a*
x+1)/x^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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