3.270 \(\int e^{-3 \tanh ^{-1}(a x)} (c-a c x)^{5/2} \, dx\)

Optimal. Leaf size=171 \[ \frac {2 (c-a c x)^{9/2}}{7 a c^2 \sqrt {1-a^2 x^2}}-\frac {4096 c^2 \sqrt {c-a c x}}{35 a \sqrt {1-a^2 x^2}}+\frac {32 (c-a c x)^{7/2}}{35 a c \sqrt {1-a^2 x^2}}+\frac {128 (c-a c x)^{5/2}}{35 a \sqrt {1-a^2 x^2}}+\frac {1024 c (c-a c x)^{3/2}}{35 a \sqrt {1-a^2 x^2}} \]

[Out]

1024/35*c*(-a*c*x+c)^(3/2)/a/(-a^2*x^2+1)^(1/2)+128/35*(-a*c*x+c)^(5/2)/a/(-a^2*x^2+1)^(1/2)+32/35*(-a*c*x+c)^
(7/2)/a/c/(-a^2*x^2+1)^(1/2)+2/7*(-a*c*x+c)^(9/2)/a/c^2/(-a^2*x^2+1)^(1/2)-4096/35*c^2*(-a*c*x+c)^(1/2)/a/(-a^
2*x^2+1)^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6127, 657, 649} \[ \frac {2 (c-a c x)^{9/2}}{7 a c^2 \sqrt {1-a^2 x^2}}-\frac {4096 c^2 \sqrt {c-a c x}}{35 a \sqrt {1-a^2 x^2}}+\frac {32 (c-a c x)^{7/2}}{35 a c \sqrt {1-a^2 x^2}}+\frac {128 (c-a c x)^{5/2}}{35 a \sqrt {1-a^2 x^2}}+\frac {1024 c (c-a c x)^{3/2}}{35 a \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(c - a*c*x)^(5/2)/E^(3*ArcTanh[a*x]),x]

[Out]

(-4096*c^2*Sqrt[c - a*c*x])/(35*a*Sqrt[1 - a^2*x^2]) + (1024*c*(c - a*c*x)^(3/2))/(35*a*Sqrt[1 - a^2*x^2]) + (
128*(c - a*c*x)^(5/2))/(35*a*Sqrt[1 - a^2*x^2]) + (32*(c - a*c*x)^(7/2))/(35*a*c*Sqrt[1 - a^2*x^2]) + (2*(c -
a*c*x)^(9/2))/(7*a*c^2*Sqrt[1 - a^2*x^2])

Rule 649

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

Rule 657

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

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 e^{-3 \tanh ^{-1}(a x)} (c-a c x)^{5/2} \, dx &=\frac {\int \frac {(c-a c x)^{11/2}}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{c^3}\\ &=\frac {2 (c-a c x)^{9/2}}{7 a c^2 \sqrt {1-a^2 x^2}}+\frac {16 \int \frac {(c-a c x)^{9/2}}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{7 c^2}\\ &=\frac {32 (c-a c x)^{7/2}}{35 a c \sqrt {1-a^2 x^2}}+\frac {2 (c-a c x)^{9/2}}{7 a c^2 \sqrt {1-a^2 x^2}}+\frac {192 \int \frac {(c-a c x)^{7/2}}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{35 c}\\ &=\frac {128 (c-a c x)^{5/2}}{35 a \sqrt {1-a^2 x^2}}+\frac {32 (c-a c x)^{7/2}}{35 a c \sqrt {1-a^2 x^2}}+\frac {2 (c-a c x)^{9/2}}{7 a c^2 \sqrt {1-a^2 x^2}}+\frac {512}{35} \int \frac {(c-a c x)^{5/2}}{\left (1-a^2 x^2\right )^{3/2}} \, dx\\ &=\frac {1024 c (c-a c x)^{3/2}}{35 a \sqrt {1-a^2 x^2}}+\frac {128 (c-a c x)^{5/2}}{35 a \sqrt {1-a^2 x^2}}+\frac {32 (c-a c x)^{7/2}}{35 a c \sqrt {1-a^2 x^2}}+\frac {2 (c-a c x)^{9/2}}{7 a c^2 \sqrt {1-a^2 x^2}}+\frac {1}{35} (2048 c) \int \frac {(c-a c x)^{3/2}}{\left (1-a^2 x^2\right )^{3/2}} \, dx\\ &=-\frac {4096 c^2 \sqrt {c-a c x}}{35 a \sqrt {1-a^2 x^2}}+\frac {1024 c (c-a c x)^{3/2}}{35 a \sqrt {1-a^2 x^2}}+\frac {128 (c-a c x)^{5/2}}{35 a \sqrt {1-a^2 x^2}}+\frac {32 (c-a c x)^{7/2}}{35 a c \sqrt {1-a^2 x^2}}+\frac {2 (c-a c x)^{9/2}}{7 a c^2 \sqrt {1-a^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 70, normalized size = 0.41 \[ \frac {2 c^3 \sqrt {1-a x} \left (5 a^4 x^4-36 a^3 x^3+142 a^2 x^2-708 a x-1451\right )}{35 a \sqrt {a x+1} \sqrt {c-a c x}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c - a*c*x)^(5/2)/E^(3*ArcTanh[a*x]),x]

[Out]

(2*c^3*Sqrt[1 - a*x]*(-1451 - 708*a*x + 142*a^2*x^2 - 36*a^3*x^3 + 5*a^4*x^4))/(35*a*Sqrt[1 + a*x]*Sqrt[c - a*
c*x])

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fricas [A]  time = 0.45, size = 82, normalized size = 0.48 \[ -\frac {2 \, {\left (5 \, a^{4} c^{2} x^{4} - 36 \, a^{3} c^{2} x^{3} + 142 \, a^{2} c^{2} x^{2} - 708 \, a c^{2} x - 1451 \, c^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{35 \, {\left (a^{3} x^{2} - a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^(5/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="fricas")

[Out]

-2/35*(5*a^4*c^2*x^4 - 36*a^3*c^2*x^3 + 142*a^2*c^2*x^2 - 708*a*c^2*x - 1451*c^2)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c
*x + c)/(a^3*x^2 - a)

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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((-a*c*x+c)^(5/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),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 = 0.03, size = 71, normalized size = 0.42 \[ \frac {2 \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}} \left (-a c x +c \right )^{\frac {5}{2}} \left (5 x^{4} a^{4}-36 x^{3} a^{3}+142 a^{2} x^{2}-708 a x -1451\right )}{35 \left (a x +1\right )^{2} \left (a x -1\right )^{4} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-a*c*x+c)^(5/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x)

[Out]

2/35*(-a^2*x^2+1)^(3/2)*(-a*c*x+c)^(5/2)*(5*a^4*x^4-36*a^3*x^3+142*a^2*x^2-708*a*x-1451)/(a*x+1)^2/(a*x-1)^4/a

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maxima [A]  time = 0.34, size = 73, normalized size = 0.43 \[ \frac {2 \, {\left (5 \, a^{4} c^{\frac {5}{2}} x^{4} - 36 \, a^{3} c^{\frac {5}{2}} x^{3} + 142 \, a^{2} c^{\frac {5}{2}} x^{2} - 708 \, a c^{\frac {5}{2}} x - 1451 \, c^{\frac {5}{2}}\right )} \sqrt {a x + 1} {\left (a x - 1\right )}}{35 \, {\left (a^{3} x^{2} - a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^(5/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="maxima")

[Out]

2/35*(5*a^4*c^(5/2)*x^4 - 36*a^3*c^(5/2)*x^3 + 142*a^2*c^(5/2)*x^2 - 708*a*c^(5/2)*x - 1451*c^(5/2))*sqrt(a*x
+ 1)*(a*x - 1)/(a^3*x^2 - a)

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mupad [B]  time = 1.10, size = 116, normalized size = 0.68 \[ \frac {2048\,c^2\,\sqrt {1-a^2\,x^2}\,\sqrt {c-a\,c\,x}}{35\,a\,\left (a\,x-1\right )}-\frac {16\,c^2\,\sqrt {1-a^2\,x^2}\,\sqrt {c-a\,c\,x}}{a\,\left (a\,x+1\right )}-\frac {2\,c^2\,\sqrt {1-a^2\,x^2}\,\sqrt {c-a\,c\,x}\,\left (5\,a^2\,x^2-36\,a\,x+147\right )}{35\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1 - a^2*x^2)^(3/2)*(c - a*c*x)^(5/2))/(a*x + 1)^3,x)

[Out]

(2048*c^2*(1 - a^2*x^2)^(1/2)*(c - a*c*x)^(1/2))/(35*a*(a*x - 1)) - (16*c^2*(1 - a^2*x^2)^(1/2)*(c - a*c*x)^(1
/2))/(a*(a*x + 1)) - (2*c^2*(1 - a^2*x^2)^(1/2)*(c - a*c*x)^(1/2)*(5*a^2*x^2 - 36*a*x + 147))/(35*a)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- c \left (a x - 1\right )\right )^{\frac {5}{2}} \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}{\left (a x + 1\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)**(5/2)/(a*x+1)**3*(-a**2*x**2+1)**(3/2),x)

[Out]

Integral((-c*(a*x - 1))**(5/2)*(-(a*x - 1)*(a*x + 1))**(3/2)/(a*x + 1)**3, x)

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