3.400 \(\int \frac {e^{2 \tanh ^{-1}(a x)} \sqrt {c-a c x}}{x^4} \, dx\)

Optimal. Leaf size=89 \[ -\frac {11}{8} a^3 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right )-\frac {11 a^2 \sqrt {c-a c x}}{8 x}-\frac {\sqrt {c-a c x}}{3 x^3}-\frac {11 a \sqrt {c-a c x}}{12 x^2} \]

[Out]

-11/8*a^3*arctanh((-a*c*x+c)^(1/2)/c^(1/2))*c^(1/2)-1/3*(-a*c*x+c)^(1/2)/x^3-11/12*a*(-a*c*x+c)^(1/2)/x^2-11/8
*a^2*(-a*c*x+c)^(1/2)/x

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Rubi [A]  time = 0.13, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ -\frac {11 a^2 \sqrt {c-a c x}}{8 x}-\frac {11}{8} a^3 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right )-\frac {11 a \sqrt {c-a c x}}{12 x^2}-\frac {\sqrt {c-a c x}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(E^(2*ArcTanh[a*x])*Sqrt[c - a*c*x])/x^4,x]

[Out]

-Sqrt[c - a*c*x]/(3*x^3) - (11*a*Sqrt[c - a*c*x])/(12*x^2) - (11*a^2*Sqrt[c - a*c*x])/(8*x) - (11*a^3*Sqrt[c]*
ArcTanh[Sqrt[c - a*c*x]/Sqrt[c]])/8

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^4} \, dx &=\int \frac {(1+a x) \sqrt {c-a c x}}{x^4 (1-a x)} \, dx\\ &=c \int \frac {1+a x}{x^4 \sqrt {c-a c x}} \, dx\\ &=-\frac {\sqrt {c-a c x}}{3 x^3}+\frac {1}{6} (11 a c) \int \frac {1}{x^3 \sqrt {c-a c x}} \, dx\\ &=-\frac {\sqrt {c-a c x}}{3 x^3}-\frac {11 a \sqrt {c-a c x}}{12 x^2}+\frac {1}{8} \left (11 a^2 c\right ) \int \frac {1}{x^2 \sqrt {c-a c x}} \, dx\\ &=-\frac {\sqrt {c-a c x}}{3 x^3}-\frac {11 a \sqrt {c-a c x}}{12 x^2}-\frac {11 a^2 \sqrt {c-a c x}}{8 x}+\frac {1}{16} \left (11 a^3 c\right ) \int \frac {1}{x \sqrt {c-a c x}} \, dx\\ &=-\frac {\sqrt {c-a c x}}{3 x^3}-\frac {11 a \sqrt {c-a c x}}{12 x^2}-\frac {11 a^2 \sqrt {c-a c x}}{8 x}-\frac {1}{8} \left (11 a^2\right ) \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}}{3 x^3}-\frac {11 a \sqrt {c-a c x}}{12 x^2}-\frac {11 a^2 \sqrt {c-a c x}}{8 x}-\frac {11}{8} a^3 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 63, normalized size = 0.71 \[ -\frac {11}{8} a^3 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right )-\frac {\left (33 a^2 x^2+22 a x+8\right ) \sqrt {c-a c x}}{24 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*ArcTanh[a*x])*Sqrt[c - a*c*x])/x^4,x]

[Out]

-1/24*(Sqrt[c - a*c*x]*(8 + 22*a*x + 33*a^2*x^2))/x^3 - (11*a^3*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/Sqrt[c]])/8

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fricas [A]  time = 0.80, size = 133, normalized size = 1.49 \[ \left [\frac {33 \, a^{3} \sqrt {c} x^{3} \log \left (\frac {a c x + 2 \, \sqrt {-a c x + c} \sqrt {c} - 2 \, c}{x}\right ) - 2 \, {\left (33 \, a^{2} x^{2} + 22 \, a x + 8\right )} \sqrt {-a c x + c}}{48 \, x^{3}}, \frac {33 \, a^{3} \sqrt {-c} x^{3} \arctan \left (\frac {\sqrt {-a c x + c} \sqrt {-c}}{c}\right ) - {\left (33 \, a^{2} x^{2} + 22 \, a x + 8\right )} \sqrt {-a c x + c}}{24 \, x^{3}}\right ] \]

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^4,x, algorithm="fricas")

[Out]

[1/48*(33*a^3*sqrt(c)*x^3*log((a*c*x + 2*sqrt(-a*c*x + c)*sqrt(c) - 2*c)/x) - 2*(33*a^2*x^2 + 22*a*x + 8)*sqrt
(-a*c*x + c))/x^3, 1/24*(33*a^3*sqrt(-c)*x^3*arctan(sqrt(-a*c*x + c)*sqrt(-c)/c) - (33*a^2*x^2 + 22*a*x + 8)*s
qrt(-a*c*x + c))/x^3]

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giac [A]  time = 0.17, size = 104, normalized size = 1.17 \[ \frac {\frac {33 \, a^{4} c \arctan \left (\frac {\sqrt {-a c x + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} - \frac {33 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} a^{4} c - 88 \, {\left (-a c x + c\right )}^{\frac {3}{2}} a^{4} c^{2} + 63 \, \sqrt {-a c x + c} a^{4} c^{3}}{a^{3} c^{3} x^{3}}}{24 \, a} \]

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^4,x, algorithm="giac")

[Out]

1/24*(33*a^4*c*arctan(sqrt(-a*c*x + c)/sqrt(-c))/sqrt(-c) - (33*(a*c*x - c)^2*sqrt(-a*c*x + c)*a^4*c - 88*(-a*
c*x + c)^(3/2)*a^4*c^2 + 63*sqrt(-a*c*x + c)*a^4*c^3)/(a^3*c^3*x^3))/a

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maple [A]  time = 0.04, size = 80, normalized size = 0.90 \[ 2 a^{3} c^{3} \left (-\frac {\frac {11 \left (-a c x +c \right )^{\frac {5}{2}}}{16 c^{2}}-\frac {11 \left (-a c x +c \right )^{\frac {3}{2}}}{6 c}+\frac {21 \sqrt {-a c x +c}}{16}}{x^{3} a^{3} c^{3}}-\frac {11 \arctanh \left (\frac {\sqrt {-a c x +c}}{\sqrt {c}}\right )}{16 c^{\frac {5}{2}}}\right ) \]

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^4,x)

[Out]

2*a^3*c^3*(-(11/16/c^2*(-a*c*x+c)^(5/2)-11/6/c*(-a*c*x+c)^(3/2)+21/16*(-a*c*x+c)^(1/2))/x^3/a^3/c^3-11/16/c^(5
/2)*arctanh((-a*c*x+c)^(1/2)/c^(1/2)))

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maxima [A]  time = 0.47, size = 134, normalized size = 1.51 \[ -\frac {1}{48} \, a^{3} c^{3} {\left (\frac {2 \, {\left (33 \, {\left (-a c x + c\right )}^{\frac {5}{2}} - 88 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c + 63 \, \sqrt {-a c x + c} c^{2}\right )}}{{\left (a c x - c\right )}^{3} c^{2} + 3 \, {\left (a c x - c\right )}^{2} c^{3} + 3 \, {\left (a c x - c\right )} c^{4} + c^{5}} - \frac {33 \, \log \left (\frac {\sqrt {-a c x + c} - \sqrt {c}}{\sqrt {-a c x + c} + \sqrt {c}}\right )}{c^{\frac {5}{2}}}\right )} \]

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^4,x, algorithm="maxima")

[Out]

-1/48*a^3*c^3*(2*(33*(-a*c*x + c)^(5/2) - 88*(-a*c*x + c)^(3/2)*c + 63*sqrt(-a*c*x + c)*c^2)/((a*c*x - c)^3*c^
2 + 3*(a*c*x - c)^2*c^3 + 3*(a*c*x - c)*c^4 + c^5) - 33*log((sqrt(-a*c*x + c) - sqrt(c))/(sqrt(-a*c*x + c) + s
qrt(c)))/c^(5/2))

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mupad [B]  time = 0.07, size = 74, normalized size = 0.83 \[ \frac {11\,{\left (c-a\,c\,x\right )}^{3/2}}{3\,c\,x^3}-\frac {21\,\sqrt {c-a\,c\,x}}{8\,x^3}-\frac {11\,{\left (c-a\,c\,x\right )}^{5/2}}{8\,c^2\,x^3}+\frac {a^3\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c-a\,c\,x}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,11{}\mathrm {i}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*c^(1/2)*atan(((c - a*c*x)^(1/2)*1i)/c^(1/2))*11i)/8 - (21*(c - a*c*x)^(1/2))/(8*x^3) + (11*(c - a*c*x)^(3
/2))/(3*c*x^3) - (11*(c - a*c*x)^(5/2))/(8*c^2*x^3)

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sympy [B]  time = 55.92, size = 439, normalized size = 4.93 \[ \frac {66 a^{3} c^{6} \sqrt {- a c x + c}}{- 144 a c^{6} x + 96 c^{6} - 144 c^{4} \left (- a c x + c\right )^{2} + 48 c^{3} \left (- a c x + c\right )^{3}} - \frac {80 a^{3} c^{5} \left (- a c x + c\right )^{\frac {3}{2}}}{- 144 a c^{6} x + 96 c^{6} - 144 c^{4} \left (- a c x + c\right )^{2} + 48 c^{3} \left (- a c x + c\right )^{3}} + \frac {30 a^{3} c^{4} \left (- a c x + c\right )^{\frac {5}{2}}}{- 144 a c^{6} x + 96 c^{6} - 144 c^{4} \left (- a c x + c\right )^{2} + 48 c^{3} \left (- a c x + c\right )^{3}} - \frac {10 a^{3} c^{4} \sqrt {- a c x + c}}{16 a c^{4} x - 8 c^{4} + 8 c^{2} \left (- a c x + c\right )^{2}} + \frac {5 a^{3} c^{4} \sqrt {\frac {1}{c^{7}}} \log {\left (- c^{4} \sqrt {\frac {1}{c^{7}}} + \sqrt {- a c x + c} \right )}}{16} - \frac {5 a^{3} c^{4} \sqrt {\frac {1}{c^{7}}} \log {\left (c^{4} \sqrt {\frac {1}{c^{7}}} + \sqrt {- a c x + c} \right )}}{16} + \frac {6 a^{3} c^{3} \left (- a c x + c\right )^{\frac {3}{2}}}{16 a c^{4} x - 8 c^{4} + 8 c^{2} \left (- a c x + c\right )^{2}} + \frac {3 a^{3} c^{3} \sqrt {\frac {1}{c^{5}}} \log {\left (- c^{3} \sqrt {\frac {1}{c^{5}}} + \sqrt {- a c x + c} \right )}}{8} - \frac {3 a^{3} c^{3} \sqrt {\frac {1}{c^{5}}} \log {\left (c^{3} \sqrt {\frac {1}{c^{5}}} + \sqrt {- a c x + c} \right )}}{8} \]

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**4,x)

[Out]

66*a**3*c**6*sqrt(-a*c*x + c)/(-144*a*c**6*x + 96*c**6 - 144*c**4*(-a*c*x + c)**2 + 48*c**3*(-a*c*x + c)**3) -
80*a**3*c**5*(-a*c*x + c)**(3/2)/(-144*a*c**6*x + 96*c**6 - 144*c**4*(-a*c*x + c)**2 + 48*c**3*(-a*c*x + c)**
3) + 30*a**3*c**4*(-a*c*x + c)**(5/2)/(-144*a*c**6*x + 96*c**6 - 144*c**4*(-a*c*x + c)**2 + 48*c**3*(-a*c*x +
c)**3) - 10*a**3*c**4*sqrt(-a*c*x + c)/(16*a*c**4*x - 8*c**4 + 8*c**2*(-a*c*x + c)**2) + 5*a**3*c**4*sqrt(c**(
-7))*log(-c**4*sqrt(c**(-7)) + sqrt(-a*c*x + c))/16 - 5*a**3*c**4*sqrt(c**(-7))*log(c**4*sqrt(c**(-7)) + sqrt(
-a*c*x + c))/16 + 6*a**3*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**3*c**
3*sqrt(c**(-5))*log(-c**3*sqrt(c**(-5)) + sqrt(-a*c*x + c))/8 - 3*a**3*c**3*sqrt(c**(-5))*log(c**3*sqrt(c**(-5
)) + sqrt(-a*c*x + c))/8

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