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

Optimal. Leaf size=136 \[ \frac {256 c^3 \sqrt {1-a^2 x^2}}{35 a \sqrt {c-a c x}}+\frac {64 c^2 \sqrt {1-a^2 x^2} \sqrt {c-a c x}}{35 a}+\frac {24 c \sqrt {1-a^2 x^2} (c-a c x)^{3/2}}{35 a}+\frac {2 \sqrt {1-a^2 x^2} (c-a c x)^{5/2}}{7 a} \]

[Out]

24/35*c*(-a*c*x+c)^(3/2)*(-a^2*x^2+1)^(1/2)/a+2/7*(-a*c*x+c)^(5/2)*(-a^2*x^2+1)^(1/2)/a+256/35*c^3*(-a^2*x^2+1
)^(1/2)/a/(-a*c*x+c)^(1/2)+64/35*c^2*(-a*c*x+c)^(1/2)*(-a^2*x^2+1)^(1/2)/a

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Rubi [A]  time = 0.11, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6127, 657, 649} \[ \frac {256 c^3 \sqrt {1-a^2 x^2}}{35 a \sqrt {c-a c x}}+\frac {64 c^2 \sqrt {1-a^2 x^2} \sqrt {c-a c x}}{35 a}+\frac {24 c \sqrt {1-a^2 x^2} (c-a c x)^{3/2}}{35 a}+\frac {2 \sqrt {1-a^2 x^2} (c-a c x)^{5/2}}{7 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(256*c^3*Sqrt[1 - a^2*x^2])/(35*a*Sqrt[c - a*c*x]) + (64*c^2*Sqrt[c - a*c*x]*Sqrt[1 - a^2*x^2])/(35*a) + (24*c
*(c - a*c*x)^(3/2)*Sqrt[1 - a^2*x^2])/(35*a) + (2*(c - a*c*x)^(5/2)*Sqrt[1 - a^2*x^2])/(7*a)

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^{-\tanh ^{-1}(a x)} (c-a c x)^{5/2} \, dx &=\frac {\int \frac {(c-a c x)^{7/2}}{\sqrt {1-a^2 x^2}} \, dx}{c}\\ &=\frac {2 (c-a c x)^{5/2} \sqrt {1-a^2 x^2}}{7 a}+\frac {12}{7} \int \frac {(c-a c x)^{5/2}}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {24 c (c-a c x)^{3/2} \sqrt {1-a^2 x^2}}{35 a}+\frac {2 (c-a c x)^{5/2} \sqrt {1-a^2 x^2}}{7 a}+\frac {1}{35} (96 c) \int \frac {(c-a c x)^{3/2}}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {64 c^2 \sqrt {c-a c x} \sqrt {1-a^2 x^2}}{35 a}+\frac {24 c (c-a c x)^{3/2} \sqrt {1-a^2 x^2}}{35 a}+\frac {2 (c-a c x)^{5/2} \sqrt {1-a^2 x^2}}{7 a}+\frac {1}{35} \left (128 c^2\right ) \int \frac {\sqrt {c-a c x}}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {256 c^3 \sqrt {1-a^2 x^2}}{35 a \sqrt {c-a c x}}+\frac {64 c^2 \sqrt {c-a c x} \sqrt {1-a^2 x^2}}{35 a}+\frac {24 c (c-a c x)^{3/2} \sqrt {1-a^2 x^2}}{35 a}+\frac {2 (c-a c x)^{5/2} \sqrt {1-a^2 x^2}}{7 a}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 57, normalized size = 0.42 \[ -\frac {2 c^3 \sqrt {1-a^2 x^2} \left (5 a^3 x^3-27 a^2 x^2+71 a x-177\right )}{35 a \sqrt {c-a c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c^3*Sqrt[1 - a^2*x^2]*(-177 + 71*a*x - 27*a^2*x^2 + 5*a^3*x^3))/(35*a*Sqrt[c - a*c*x])

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fricas [A]  time = 0.41, size = 69, normalized size = 0.51 \[ \frac {2 \, {\left (5 \, a^{3} c^{2} x^{3} - 27 \, a^{2} c^{2} x^{2} + 71 \, a c^{2} x - 177 \, c^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{35 \, {\left (a^{2} x - a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(5*a^3*c^2*x^3 - 27*a^2*c^2*x^2 + 71*a*c^2*x - 177*c^2)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)/(a^2*x - 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)*(-a^2*x^2+1)^(1/2),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 = 56, normalized size = 0.41 \[ \frac {2 \sqrt {-a^{2} x^{2}+1}\, \left (-a c x +c \right )^{\frac {5}{2}} \left (5 x^{3} a^{3}-27 a^{2} x^{2}+71 a x -177\right )}{35 \left (a x -1\right )^{3} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^(5/2)*(5*a^3*x^3-27*a^2*x^2+71*a*x-177)/(a*x-1)^3/a

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maxima [A]  time = 0.39, size = 60, normalized size = 0.44 \[ -\frac {2 \, {\left (5 \, a^{3} c^{\frac {5}{2}} x^{3} - 27 \, a^{2} c^{\frac {5}{2}} x^{2} + 71 \, a c^{\frac {5}{2}} x - 177 \, c^{\frac {5}{2}}\right )} \sqrt {a x + 1} {\left (a x - 1\right )}}{35 \, {\left (a^{2} x - a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/35*(5*a^3*c^(5/2)*x^3 - 27*a^2*c^(5/2)*x^2 + 71*a*c^(5/2)*x - 177*c^(5/2))*sqrt(a*x + 1)*(a*x - 1)/(a^2*x -
a)

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mupad [B]  time = 0.99, size = 80, normalized size = 0.59 \[ \frac {2\,c^2\,\sqrt {1-a^2\,x^2}\,\sqrt {c-a\,c\,x}\,\left (5\,a^2\,x^2-22\,a\,x+49\right )}{35\,a}-\frac {256\,c^2\,\sqrt {1-a^2\,x^2}\,\sqrt {c-a\,c\,x}}{35\,a\,\left (a\,x-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*c^2*(1 - a^2*x^2)^(1/2)*(c - a*c*x)^(1/2)*(5*a^2*x^2 - 22*a*x + 49))/(35*a) - (256*c^2*(1 - a^2*x^2)^(1/2)*
(c - a*c*x)^(1/2))/(35*a*(a*x - 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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