3.268 \(\int \frac {e^{-2 \tanh ^{-1}(a x)}}{(c-a c x)^{7/2}} \, dx\)

Optimal. Leaf size=83 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{2 \sqrt {2} a c^{7/2}}+\frac {1}{2 a c^3 \sqrt {c-a c x}}+\frac {1}{3 a c^2 (c-a c x)^{3/2}} \]

[Out]

1/3/a/c^2/(-a*c*x+c)^(3/2)-1/4*arctanh(1/2*(-a*c*x+c)^(1/2)*2^(1/2)/c^(1/2))/a/c^(7/2)*2^(1/2)+1/2/a/c^3/(-a*c
*x+c)^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6130, 21, 51, 63, 206} \[ \frac {1}{2 a c^3 \sqrt {c-a c x}}+\frac {1}{3 a c^2 (c-a c x)^{3/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{2 \sqrt {2} a c^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(2*ArcTanh[a*x])*(c - a*c*x)^(7/2)),x]

[Out]

1/(3*a*c^2*(c - a*c*x)^(3/2)) + 1/(2*a*c^3*Sqrt[c - a*c*x]) - ArcTanh[Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c])]/(2*Sq
rt[2]*a*c^(7/2))

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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)}}{(c-a c x)^{7/2}} \, dx &=\int \frac {1-a x}{(1+a x) (c-a c x)^{7/2}} \, dx\\ &=\frac {\int \frac {1}{(1+a x) (c-a c x)^{5/2}} \, dx}{c}\\ &=\frac {1}{3 a c^2 (c-a c x)^{3/2}}+\frac {\int \frac {1}{(1+a x) (c-a c x)^{3/2}} \, dx}{2 c^2}\\ &=\frac {1}{3 a c^2 (c-a c x)^{3/2}}+\frac {1}{2 a c^3 \sqrt {c-a c x}}+\frac {\int \frac {1}{(1+a x) \sqrt {c-a c x}} \, dx}{4 c^3}\\ &=\frac {1}{3 a c^2 (c-a c x)^{3/2}}+\frac {1}{2 a c^3 \sqrt {c-a c x}}-\frac {\operatorname {Subst}\left (\int \frac {1}{2-\frac {x^2}{c}} \, dx,x,\sqrt {c-a c x}\right )}{2 a c^4}\\ &=\frac {1}{3 a c^2 (c-a c x)^{3/2}}+\frac {1}{2 a c^3 \sqrt {c-a c x}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{2 \sqrt {2} a c^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 39, normalized size = 0.47 \[ \frac {\, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {1}{2} (1-a x)\right )}{3 a c^2 (c-a c x)^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^(2*ArcTanh[a*x])*(c - a*c*x)^(7/2)),x]

[Out]

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

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fricas [A]  time = 0.62, size = 196, normalized size = 2.36 \[ \left [\frac {3 \, \sqrt {2} {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \sqrt {c} \log \left (\frac {a c x + 2 \, \sqrt {2} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a x + 1}\right ) - 4 \, \sqrt {-a c x + c} {\left (3 \, a x - 5\right )}}{24 \, {\left (a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\right )}}, \frac {3 \, \sqrt {2} {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c} \sqrt {-c}}{2 \, c}\right ) - 2 \, \sqrt {-a c x + c} {\left (3 \, a x - 5\right )}}{12 \, {\left (a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^2*(-a^2*x^2+1)/(-a*c*x+c)^(7/2),x, algorithm="fricas")

[Out]

[1/24*(3*sqrt(2)*(a^2*x^2 - 2*a*x + 1)*sqrt(c)*log((a*c*x + 2*sqrt(2)*sqrt(-a*c*x + c)*sqrt(c) - 3*c)/(a*x + 1
)) - 4*sqrt(-a*c*x + c)*(3*a*x - 5))/(a^3*c^4*x^2 - 2*a^2*c^4*x + a*c^4), 1/12*(3*sqrt(2)*(a^2*x^2 - 2*a*x + 1
)*sqrt(-c)*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)*sqrt(-c)/c) - 2*sqrt(-a*c*x + c)*(3*a*x - 5))/(a^3*c^4*x^2 - 2*
a^2*c^4*x + a*c^4)]

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giac [A]  time = 0.14, size = 73, normalized size = 0.88 \[ \frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c}}{2 \, \sqrt {-c}}\right )}{4 \, a \sqrt {-c} c^{3}} + \frac {3 \, a c x - 5 \, c}{6 \, {\left (a c x - c\right )} \sqrt {-a c x + c} a c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^2*(-a^2*x^2+1)/(-a*c*x+c)^(7/2),x, algorithm="giac")

[Out]

1/4*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)/sqrt(-c))/(a*sqrt(-c)*c^3) + 1/6*(3*a*c*x - 5*c)/((a*c*x - c)*
sqrt(-a*c*x + c)*a*c^3)

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maple [A]  time = 0.04, size = 64, normalized size = 0.77 \[ \frac {\frac {1}{2 c^{2} \sqrt {-a c x +c}}+\frac {1}{3 c \left (-a c x +c \right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \arctanh \left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 c^{\frac {5}{2}}}}{a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x+1)^2*(-a^2*x^2+1)/(-a*c*x+c)^(7/2),x)

[Out]

2/c/a*(1/4/c^2/(-a*c*x+c)^(1/2)+1/6/c/(-a*c*x+c)^(3/2)-1/8/c^(5/2)*2^(1/2)*arctanh(1/2*(-a*c*x+c)^(1/2)*2^(1/2
)/c^(1/2)))

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maxima [A]  time = 0.50, size = 81, normalized size = 0.98 \[ \frac {\frac {3 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-a c x + c}}{\sqrt {2} \sqrt {c} + \sqrt {-a c x + c}}\right )}{c^{\frac {5}{2}}} - \frac {4 \, {\left (3 \, a c x - 5 \, c\right )}}{{\left (-a c x + c\right )}^{\frac {3}{2}} c^{2}}}{24 \, a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^2*(-a^2*x^2+1)/(-a*c*x+c)^(7/2),x, algorithm="maxima")

[Out]

1/24*(3*sqrt(2)*log(-(sqrt(2)*sqrt(c) - sqrt(-a*c*x + c))/(sqrt(2)*sqrt(c) + sqrt(-a*c*x + c)))/c^(5/2) - 4*(3
*a*c*x - 5*c)/((-a*c*x + c)^(3/2)*c^2))/(a*c)

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mupad [B]  time = 0.84, size = 64, normalized size = 0.77 \[ \frac {\frac {c-a\,c\,x}{2\,c^2}+\frac {1}{3\,c}}{a\,c\,{\left (c-a\,c\,x\right )}^{3/2}}-\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-a\,c\,x}}{2\,\sqrt {c}}\right )}{4\,a\,c^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(a^2*x^2 - 1)/((c - a*c*x)^(7/2)*(a*x + 1)^2),x)

[Out]

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

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sympy [A]  time = 66.36, size = 80, normalized size = 0.96 \[ \frac {1}{3 a c^{2} \left (- a c x + c\right )^{\frac {3}{2}}} + \frac {1}{2 a c^{3} \sqrt {- a c x + c}} + \frac {\sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} \sqrt {- a c x + c}}{2 \sqrt {- c}} \right )}}{4 a c^{3} \sqrt {- c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)**2*(-a**2*x**2+1)/(-a*c*x+c)**(7/2),x)

[Out]

1/(3*a*c**2*(-a*c*x + c)**(3/2)) + 1/(2*a*c**3*sqrt(-a*c*x + c)) + sqrt(2)*atan(sqrt(2)*sqrt(-a*c*x + c)/(2*sq
rt(-c)))/(4*a*c**3*sqrt(-c))

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