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

Optimal. Leaf size=68 \[ -\frac {7}{4} a^2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right )-\frac {\sqrt {c-a c x}}{2 x^2}-\frac {7 a \sqrt {c-a c x}}{4 x} \]

[Out]

-7/4*a^2*arctanh((-a*c*x+c)^(1/2)/c^(1/2))*c^(1/2)-1/2*(-a*c*x+c)^(1/2)/x^2-7/4*a*(-a*c*x+c)^(1/2)/x

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Rubi [A]  time = 0.12, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {6130, 21, 78, 51, 63, 208} \[ -\frac {7}{4} a^2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right )-\frac {\sqrt {c-a c x}}{2 x^2}-\frac {7 a \sqrt {c-a c x}}{4 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[c - a*c*x]/(2*x^2) - (7*a*Sqrt[c - a*c*x])/(4*x) - (7*a^2*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/Sqrt[c]])/4

Rule 21

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

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 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^{2 \tanh ^{-1}(a x)} \sqrt {c-a c x}}{x^3} \, dx &=\int \frac {(1+a x) \sqrt {c-a c x}}{x^3 (1-a x)} \, dx\\ &=c \int \frac {1+a x}{x^3 \sqrt {c-a c x}} \, dx\\ &=-\frac {\sqrt {c-a c x}}{2 x^2}+\frac {1}{4} (7 a c) \int \frac {1}{x^2 \sqrt {c-a c x}} \, dx\\ &=-\frac {\sqrt {c-a c x}}{2 x^2}-\frac {7 a \sqrt {c-a c x}}{4 x}+\frac {1}{8} \left (7 a^2 c\right ) \int \frac {1}{x \sqrt {c-a c x}} \, dx\\ &=-\frac {\sqrt {c-a c x}}{2 x^2}-\frac {7 a \sqrt {c-a c x}}{4 x}-\frac {1}{4} (7 a) \operatorname {Subst}\left (\int \frac {1}{\frac {1}{a}-\frac {x^2}{a c}} \, dx,x,\sqrt {c-a c x}\right )\\ &=-\frac {\sqrt {c-a c x}}{2 x^2}-\frac {7 a \sqrt {c-a c x}}{4 x}-\frac {7}{4} a^2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 55, normalized size = 0.81 \[ -\frac {7}{4} a^2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right )-\frac {(7 a x+2) \sqrt {c-a c x}}{4 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*((2 + 7*a*x)*Sqrt[c - a*c*x])/x^2 - (7*a^2*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/Sqrt[c]])/4

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fricas [A]  time = 0.56, size = 117, normalized size = 1.72 \[ \left [\frac {7 \, a^{2} \sqrt {c} x^{2} \log \left (\frac {a c x + 2 \, \sqrt {-a c x + c} \sqrt {c} - 2 \, c}{x}\right ) - 2 \, \sqrt {-a c x + c} {\left (7 \, a x + 2\right )}}{8 \, x^{2}}, \frac {7 \, a^{2} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-a c x + c} \sqrt {-c}}{c}\right ) - \sqrt {-a c x + c} {\left (7 \, a x + 2\right )}}{4 \, x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(7*a^2*sqrt(c)*x^2*log((a*c*x + 2*sqrt(-a*c*x + c)*sqrt(c) - 2*c)/x) - 2*sqrt(-a*c*x + c)*(7*a*x + 2))/x^
2, 1/4*(7*a^2*sqrt(-c)*x^2*arctan(sqrt(-a*c*x + c)*sqrt(-c)/c) - sqrt(-a*c*x + c)*(7*a*x + 2))/x^2]

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giac [A]  time = 0.18, size = 76, normalized size = 1.12 \[ \frac {\frac {7 \, a^{3} c \arctan \left (\frac {\sqrt {-a c x + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} + \frac {7 \, {\left (-a c x + c\right )}^{\frac {3}{2}} a^{3} c - 9 \, \sqrt {-a c x + c} a^{3} c^{2}}{a^{2} c^{2} x^{2}}}{4 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(7*a^3*c*arctan(sqrt(-a*c*x + c)/sqrt(-c))/sqrt(-c) + (7*(-a*c*x + c)^(3/2)*a^3*c - 9*sqrt(-a*c*x + c)*a^3
*c^2)/(a^2*c^2*x^2))/a

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maple [A]  time = 0.04, size = 65, normalized size = 0.96 \[ -2 a^{2} c^{2} \left (\frac {-\frac {7 \left (-a c x +c \right )^{\frac {3}{2}}}{8 c}+\frac {9 \sqrt {-a c x +c}}{8}}{x^{2} a^{2} c^{2}}+\frac {7 \arctanh \left (\frac {\sqrt {-a c x +c}}{\sqrt {c}}\right )}{8 c^{\frac {3}{2}}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a^2*c^2*((-7/8/c*(-a*c*x+c)^(3/2)+9/8*(-a*c*x+c)^(1/2))/x^2/a^2/c^2+7/8/c^(3/2)*arctanh((-a*c*x+c)^(1/2)/c^
(1/2)))

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maxima [A]  time = 0.40, size = 103, normalized size = 1.51 \[ \frac {1}{8} \, a^{2} c^{2} {\left (\frac {2 \, {\left (7 \, {\left (-a c x + c\right )}^{\frac {3}{2}} - 9 \, \sqrt {-a c x + c} c\right )}}{{\left (a c x - c\right )}^{2} c + 2 \, {\left (a c x - c\right )} c^{2} + c^{3}} + \frac {7 \, \log \left (\frac {\sqrt {-a c x + c} - \sqrt {c}}{\sqrt {-a c x + c} + \sqrt {c}}\right )}{c^{\frac {3}{2}}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*a^2*c^2*(2*(7*(-a*c*x + c)^(3/2) - 9*sqrt(-a*c*x + c)*c)/((a*c*x - c)^2*c + 2*(a*c*x - c)*c^2 + c^3) + 7*l
og((sqrt(-a*c*x + c) - sqrt(c))/(sqrt(-a*c*x + c) + sqrt(c)))/c^(3/2))

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mupad [B]  time = 0.08, size = 54, normalized size = 0.79 \[ \frac {7\,{\left (c-a\,c\,x\right )}^{3/2}}{4\,c\,x^2}-\frac {7\,a^2\,\sqrt {c}\,\mathrm {atanh}\left (\frac {\sqrt {c-a\,c\,x}}{\sqrt {c}}\right )}{4}-\frac {9\,\sqrt {c-a\,c\,x}}{4\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(7*(c - a*c*x)^(3/2))/(4*c*x^2) - (7*a^2*c^(1/2)*atanh((c - a*c*x)^(1/2)/c^(1/2)))/4 - (9*(c - a*c*x)^(1/2))/(
4*x^2)

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sympy [B]  time = 37.00, size = 270, normalized size = 3.97 \[ - \frac {10 a^{2} c^{4} \sqrt {- a c x + c}}{16 a c^{4} x - 8 c^{4} + 8 c^{2} \left (- a c x + c\right )^{2}} + \frac {6 a^{2} c^{3} \left (- a c x + c\right )^{\frac {3}{2}}}{16 a c^{4} x - 8 c^{4} + 8 c^{2} \left (- a c x + c\right )^{2}} + \frac {3 a^{2} c^{3} \sqrt {\frac {1}{c^{5}}} \log {\left (- c^{3} \sqrt {\frac {1}{c^{5}}} + \sqrt {- a c x + c} \right )}}{8} - \frac {3 a^{2} c^{3} \sqrt {\frac {1}{c^{5}}} \log {\left (c^{3} \sqrt {\frac {1}{c^{5}}} + \sqrt {- a c x + c} \right )}}{8} + \frac {a^{2} c^{2} \sqrt {\frac {1}{c^{3}}} \log {\left (- c^{2} \sqrt {\frac {1}{c^{3}}} + \sqrt {- a c x + c} \right )}}{2} - \frac {a^{2} c^{2} \sqrt {\frac {1}{c^{3}}} \log {\left (c^{2} \sqrt {\frac {1}{c^{3}}} + \sqrt {- a c x + c} \right )}}{2} - \frac {a \sqrt {- a c x + c}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-10*a**2*c**4*sqrt(-a*c*x + c)/(16*a*c**4*x - 8*c**4 + 8*c**2*(-a*c*x + c)**2) + 6*a**2*c**3*(-a*c*x + c)**(3/
2)/(16*a*c**4*x - 8*c**4 + 8*c**2*(-a*c*x + c)**2) + 3*a**2*c**3*sqrt(c**(-5))*log(-c**3*sqrt(c**(-5)) + sqrt(
-a*c*x + c))/8 - 3*a**2*c**3*sqrt(c**(-5))*log(c**3*sqrt(c**(-5)) + sqrt(-a*c*x + c))/8 + a**2*c**2*sqrt(c**(-
3))*log(-c**2*sqrt(c**(-3)) + sqrt(-a*c*x + c))/2 - a**2*c**2*sqrt(c**(-3))*log(c**2*sqrt(c**(-3)) + sqrt(-a*c
*x + c))/2 - a*sqrt(-a*c*x + c)/x

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