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

Optimal. Leaf size=122 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{8 \sqrt {2} a c^{5/2}}-\frac {\sqrt {1-a^2 x^2}}{8 a c (c-a c x)^{3/2}}+\frac {\sqrt {1-a^2 x^2}}{2 a (c-a c x)^{5/2}} \]

[Out]

-1/16*arctanh(1/2*c^(1/2)*(-a^2*x^2+1)^(1/2)*2^(1/2)/(-a*c*x+c)^(1/2))/a/c^(5/2)*2^(1/2)+1/2*(-a^2*x^2+1)^(1/2
)/a/(-a*c*x+c)^(5/2)-1/8*(-a^2*x^2+1)^(1/2)/a/c/(-a*c*x+c)^(3/2)

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Rubi [A]  time = 0.11, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {6127, 663, 673, 661, 208} \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{8 \sqrt {2} a c^{5/2}}-\frac {\sqrt {1-a^2 x^2}}{8 a c (c-a c x)^{3/2}}+\frac {\sqrt {1-a^2 x^2}}{2 a (c-a c x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[1 - a^2*x^2]/(2*a*(c - a*c*x)^(5/2)) - Sqrt[1 - a^2*x^2]/(8*a*c*(c - a*c*x)^(3/2)) - ArcTanh[(Sqrt[c]*Sqr
t[1 - a^2*x^2])/(Sqrt[2]*Sqrt[c - a*c*x])]/(8*Sqrt[2]*a*c^(5/2))

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 661

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

Rule 663

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + c*x^2)^p)/(
e*(m + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1), x], x] /; FreeQ[
{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1
, 0] && IntegerQ[2*p]

Rule 673

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1)
)/(2*c*d*(m + p + 1)), x] + Dist[(m + 2*p + 2)/(2*d*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x]
/; FreeQ[{a, c, d, e, p}, x] && EqQ[c*d^2 + a*e^2, 0] && LtQ[m, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rule 6127

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

Rubi steps

\begin {align*} \int \frac {e^{\tanh ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx &=c \int \frac {\sqrt {1-a^2 x^2}}{(c-a c x)^{7/2}} \, dx\\ &=\frac {\sqrt {1-a^2 x^2}}{2 a (c-a c x)^{5/2}}-\frac {\int \frac {1}{(c-a c x)^{3/2} \sqrt {1-a^2 x^2}} \, dx}{4 c}\\ &=\frac {\sqrt {1-a^2 x^2}}{2 a (c-a c x)^{5/2}}-\frac {\sqrt {1-a^2 x^2}}{8 a c (c-a c x)^{3/2}}-\frac {\int \frac {1}{\sqrt {c-a c x} \sqrt {1-a^2 x^2}} \, dx}{16 c^2}\\ &=\frac {\sqrt {1-a^2 x^2}}{2 a (c-a c x)^{5/2}}-\frac {\sqrt {1-a^2 x^2}}{8 a c (c-a c x)^{3/2}}+\frac {a \operatorname {Subst}\left (\int \frac {1}{-2 a^2 c+a^2 c^2 x^2} \, dx,x,\frac {\sqrt {1-a^2 x^2}}{\sqrt {c-a c x}}\right )}{8 c}\\ &=\frac {\sqrt {1-a^2 x^2}}{2 a (c-a c x)^{5/2}}-\frac {\sqrt {1-a^2 x^2}}{8 a c (c-a c x)^{3/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{8 \sqrt {2} a c^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 57, normalized size = 0.47 \[ \frac {(a x+1)^{3/2} \sqrt {c-a c x} \, _2F_1\left (\frac {3}{2},3;\frac {5}{2};\frac {1}{2} (a x+1)\right )}{12 a c^3 \sqrt {1-a x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1 + a*x)^(3/2)*Sqrt[c - a*c*x]*Hypergeometric2F1[3/2, 3, 5/2, (1 + a*x)/2])/(12*a*c^3*Sqrt[1 - a*x])

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fricas [A]  time = 0.48, size = 308, normalized size = 2.52 \[ \left [\frac {\sqrt {2} {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} + 2 \, a c x + 2 \, \sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a^{2} x^{2} - 2 \, a x + 1}\right ) - 4 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} {\left (a x + 3\right )}}{32 \, {\left (a^{4} c^{3} x^{3} - 3 \, a^{3} c^{3} x^{2} + 3 \, a^{2} c^{3} x - a c^{3}\right )}}, -\frac {\sqrt {2} {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {-c}}{a^{2} c x^{2} - c}\right ) + 2 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} {\left (a x + 3\right )}}{16 \, {\left (a^{4} c^{3} x^{3} - 3 \, a^{3} c^{3} x^{2} + 3 \, a^{2} c^{3} x - a c^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/32*(sqrt(2)*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*sqrt(c)*log(-(a^2*c*x^2 + 2*a*c*x + 2*sqrt(2)*sqrt(-a^2*x^2 +
1)*sqrt(-a*c*x + c)*sqrt(c) - 3*c)/(a^2*x^2 - 2*a*x + 1)) - 4*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*(a*x + 3))/
(a^4*c^3*x^3 - 3*a^3*c^3*x^2 + 3*a^2*c^3*x - a*c^3), -1/16*(sqrt(2)*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*sqrt(-c)
*arctan(sqrt(2)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*sqrt(-c)/(a^2*c*x^2 - c)) + 2*sqrt(-a^2*x^2 + 1)*sqrt(-a*c
*x + c)*(a*x + 3))/(a^4*c^3*x^3 - 3*a^3*c^3*x^2 + 3*a^2*c^3*x - a*c^3)]

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giac [A]  time = 0.25, size = 76, normalized size = 0.62 \[ \frac {\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {a c x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c} c} + \frac {2 \, {\left ({\left (a c x + c\right )}^{\frac {3}{2}} + 2 \, \sqrt {a c x + c} c\right )}}{{\left (a c x - c\right )}^{2} c}}{16 \, a {\left | c \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(sqrt(2)*arctan(1/2*sqrt(2)*sqrt(a*c*x + c)/sqrt(-c))/(sqrt(-c)*c) + 2*((a*c*x + c)^(3/2) + 2*sqrt(a*c*x
+ c)*c)/((a*c*x - c)^2*c))/(a*abs(c))

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maple [A]  time = 0.05, size = 156, normalized size = 1.28 \[ \frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}\, \left (\sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) x^{2} a^{2} c -2 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) x a c -2 x a \sqrt {c \left (a x +1\right )}\, \sqrt {c}+\sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c -6 \sqrt {c \left (a x +1\right )}\, \sqrt {c}\right )}{16 c^{\frac {7}{2}} \left (a x -1\right )^{3} \sqrt {c \left (a x +1\right )}\, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(-a^2*x^2+1)^(1/2)*(-c*(a*x-1))^(1/2)/c^(7/2)*(2^(1/2)*arctanh(1/2*(c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2))*x^2
*a^2*c-2*2^(1/2)*arctanh(1/2*(c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2))*x*a*c-2*x*a*(c*(a*x+1))^(1/2)*c^(1/2)+2^(1/2)*
arctanh(1/2*(c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2))*c-6*(c*(a*x+1))^(1/2)*c^(1/2))/(a*x-1)^3/(c*(a*x+1))^(1/2)/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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