3.277 \(\int \frac {e^{-3 \tanh ^{-1}(a x)}}{(c-a c x)^{9/2}} \, dx\)

Optimal. Leaf size=160 \[ \frac {15 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{32 \sqrt {2} a c^{9/2}}-\frac {15 \sqrt {c-a c x}}{32 a c^5 \sqrt {1-a^2 x^2}}+\frac {5}{16 a c^4 \sqrt {1-a^2 x^2} \sqrt {c-a c x}}+\frac {1}{4 a c^3 \sqrt {1-a^2 x^2} (c-a c x)^{3/2}} \]

[Out]

15/64*arctanh(1/2*c^(1/2)*(-a^2*x^2+1)^(1/2)*2^(1/2)/(-a*c*x+c)^(1/2))/a/c^(9/2)*2^(1/2)+1/4/a/c^3/(-a*c*x+c)^
(3/2)/(-a^2*x^2+1)^(1/2)+5/16/a/c^4/(-a*c*x+c)^(1/2)/(-a^2*x^2+1)^(1/2)-15/32*(-a*c*x+c)^(1/2)/a/c^5/(-a^2*x^2
+1)^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 160, 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 = {6127, 673, 667, 661, 208} \[ -\frac {15 \sqrt {c-a c x}}{32 a c^5 \sqrt {1-a^2 x^2}}+\frac {5}{16 a c^4 \sqrt {1-a^2 x^2} \sqrt {c-a c x}}+\frac {1}{4 a c^3 \sqrt {1-a^2 x^2} (c-a c x)^{3/2}}+\frac {15 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{32 \sqrt {2} a c^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcTanh[a*x])*(c - a*c*x)^(9/2)),x]

[Out]

1/(4*a*c^3*(c - a*c*x)^(3/2)*Sqrt[1 - a^2*x^2]) + 5/(16*a*c^4*Sqrt[c - a*c*x]*Sqrt[1 - a^2*x^2]) - (15*Sqrt[c
- a*c*x])/(32*a*c^5*Sqrt[1 - a^2*x^2]) + (15*ArcTanh[(Sqrt[c]*Sqrt[1 - a^2*x^2])/(Sqrt[2]*Sqrt[c - a*c*x])])/(
32*Sqrt[2]*a*c^(9/2))

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 667

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(d*(d + e*x)^m*(a + c*x^2)^(p + 1)
)/(2*a*e*(p + 1)), x] + Dist[(d*(m + 2*p + 2))/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x], x
] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && LtQ[0, m, 1] && 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^{-3 \tanh ^{-1}(a x)}}{(c-a c x)^{9/2}} \, dx &=\frac {\int \frac {1}{(c-a c x)^{3/2} \left (1-a^2 x^2\right )^{3/2}} \, dx}{c^3}\\ &=\frac {1}{4 a c^3 (c-a c x)^{3/2} \sqrt {1-a^2 x^2}}+\frac {5 \int \frac {1}{\sqrt {c-a c x} \left (1-a^2 x^2\right )^{3/2}} \, dx}{8 c^4}\\ &=\frac {1}{4 a c^3 (c-a c x)^{3/2} \sqrt {1-a^2 x^2}}+\frac {5}{16 a c^4 \sqrt {c-a c x} \sqrt {1-a^2 x^2}}+\frac {15 \int \frac {\sqrt {c-a c x}}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{32 c^5}\\ &=\frac {1}{4 a c^3 (c-a c x)^{3/2} \sqrt {1-a^2 x^2}}+\frac {5}{16 a c^4 \sqrt {c-a c x} \sqrt {1-a^2 x^2}}-\frac {15 \sqrt {c-a c x}}{32 a c^5 \sqrt {1-a^2 x^2}}+\frac {15 \int \frac {1}{\sqrt {c-a c x} \sqrt {1-a^2 x^2}} \, dx}{64 c^4}\\ &=\frac {1}{4 a c^3 (c-a c x)^{3/2} \sqrt {1-a^2 x^2}}+\frac {5}{16 a c^4 \sqrt {c-a c x} \sqrt {1-a^2 x^2}}-\frac {15 \sqrt {c-a c x}}{32 a c^5 \sqrt {1-a^2 x^2}}-\frac {(15 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 )}{32 c^3}\\ &=\frac {1}{4 a c^3 (c-a c x)^{3/2} \sqrt {1-a^2 x^2}}+\frac {5}{16 a c^4 \sqrt {c-a c x} \sqrt {1-a^2 x^2}}-\frac {15 \sqrt {c-a c x}}{32 a c^5 \sqrt {1-a^2 x^2}}+\frac {15 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{32 \sqrt {2} a c^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 57, normalized size = 0.36 \[ -\frac {(1-a x)^{3/2} \, _2F_1\left (-\frac {1}{2},3;\frac {1}{2};\frac {1}{2} (a x+1)\right )}{4 a c^3 \sqrt {a x+1} (c-a c x)^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^(3*ArcTanh[a*x])*(c - a*c*x)^(9/2)),x]

[Out]

-1/4*((1 - a*x)^(3/2)*Hypergeometric2F1[-1/2, 3, 1/2, (1 + a*x)/2])/(a*c^3*Sqrt[1 + a*x]*(c - a*c*x)^(3/2))

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fricas [A]  time = 0.43, size = 328, normalized size = 2.05 \[ \left [\frac {15 \, \sqrt {2} {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} + 2 \, a c x - 2 \, \sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a^{2} x^{2} - 2 \, a x + 1}\right ) + 4 \, {\left (15 \, a^{2} x^{2} - 20 \, a x - 3\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{128 \, {\left (a^{5} c^{5} x^{4} - 2 \, a^{4} c^{5} x^{3} + 2 \, a^{2} c^{5} x - a c^{5}\right )}}, \frac {15 \, \sqrt {2} {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {-c}}{a^{2} c x^{2} - c}\right ) + 2 \, {\left (15 \, a^{2} x^{2} - 20 \, a x - 3\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{64 \, {\left (a^{5} c^{5} x^{4} - 2 \, a^{4} c^{5} x^{3} + 2 \, a^{2} c^{5} x - a c^{5}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(9/2),x, algorithm="fricas")

[Out]

[1/128*(15*sqrt(2)*(a^4*x^4 - 2*a^3*x^3 + 2*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*(15*a^2*x^2 - 20*a*x - 3)*sqrt(-a^2*x^2 + 1
)*sqrt(-a*c*x + c))/(a^5*c^5*x^4 - 2*a^4*c^5*x^3 + 2*a^2*c^5*x - a*c^5), 1/64*(15*sqrt(2)*(a^4*x^4 - 2*a^3*x^3
+ 2*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*(15*a^
2*x^2 - 20*a*x - 3)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c))/(a^5*c^5*x^4 - 2*a^4*c^5*x^3 + 2*a^2*c^5*x - a*c^5)]

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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(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(9/2),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 = 173, normalized size = 1.08 \[ -\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}\, \left (15 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) x^{2} a^{2} \sqrt {c \left (a x +1\right )}-30 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) x a \sqrt {c \left (a x +1\right )}-30 x^{2} a^{2} \sqrt {c}+15 \arctanh \left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, \sqrt {c \left (a x +1\right )}+40 x a \sqrt {c}+6 \sqrt {c}\right )}{64 c^{\frac {11}{2}} \left (a x -1\right )^{3} \left (a x +1\right ) a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(9/2),x)

[Out]

-1/64*(-a^2*x^2+1)^(1/2)*(-c*(a*x-1))^(1/2)/c^(11/2)*(15*2^(1/2)*arctanh(1/2*(c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2)
)*x^2*a^2*(c*(a*x+1))^(1/2)-30*2^(1/2)*arctanh(1/2*(c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2))*x*a*(c*(a*x+1))^(1/2)-30
*x^2*a^2*c^(1/2)+15*arctanh(1/2*(c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*(c*(a*x+1))^(1/2)+40*x*a*c^(1/2)+6*
c^(1/2))/(a*x-1)^3/(a*x+1)/a

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (-a c x + c\right )}^{\frac {9}{2}} {\left (a x + 1\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(9/2),x, algorithm="maxima")

[Out]

integrate((-a^2*x^2 + 1)^(3/2)/((-a*c*x + c)^(9/2)*(a*x + 1)^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (1-a^2\,x^2\right )}^{3/2}}{{\left (c-a\,c\,x\right )}^{9/2}\,{\left (a\,x+1\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - a^2*x^2)^(3/2)/((c - a*c*x)^(9/2)*(a*x + 1)^3),x)

[Out]

int((1 - a^2*x^2)^(3/2)/((c - a*c*x)^(9/2)*(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(1/(a*x+1)**3*(-a**2*x**2+1)**(3/2)/(-a*c*x+c)**(9/2),x)

[Out]

Timed out

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