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

Optimal. Leaf size=101 \[ \frac {64 c^2 \sqrt {1-a^2 x^2}}{15 a \sqrt {c-a c x}}+\frac {16 c \sqrt {1-a^2 x^2} \sqrt {c-a c x}}{15 a}+\frac {2 \sqrt {1-a^2 x^2} (c-a c x)^{3/2}}{5 a} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6127, 657, 649} \[ \frac {64 c^2 \sqrt {1-a^2 x^2}}{15 a \sqrt {c-a c x}}+\frac {16 c \sqrt {1-a^2 x^2} \sqrt {c-a c x}}{15 a}+\frac {2 \sqrt {1-a^2 x^2} (c-a c x)^{3/2}}{5 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*c^2*Sqrt[1 - a^2*x^2])/(15*a*Sqrt[c - a*c*x]) + (16*c*Sqrt[c - a*c*x]*Sqrt[1 - a^2*x^2])/(15*a) + (2*(c -
a*c*x)^(3/2)*Sqrt[1 - a^2*x^2])/(5*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)^{3/2} \, dx &=\frac {\int \frac {(c-a c x)^{5/2}}{\sqrt {1-a^2 x^2}} \, dx}{c}\\ &=\frac {2 (c-a c x)^{3/2} \sqrt {1-a^2 x^2}}{5 a}+\frac {8}{5} \int \frac {(c-a c x)^{3/2}}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {16 c \sqrt {c-a c x} \sqrt {1-a^2 x^2}}{15 a}+\frac {2 (c-a c x)^{3/2} \sqrt {1-a^2 x^2}}{5 a}+\frac {1}{15} (32 c) \int \frac {\sqrt {c-a c x}}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {64 c^2 \sqrt {1-a^2 x^2}}{15 a \sqrt {c-a c x}}+\frac {16 c \sqrt {c-a c x} \sqrt {1-a^2 x^2}}{15 a}+\frac {2 (c-a c x)^{3/2} \sqrt {1-a^2 x^2}}{5 a}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 49, normalized size = 0.49 \[ \frac {2 c^2 \sqrt {1-a^2 x^2} \left (3 a^2 x^2-14 a x+43\right )}{15 a \sqrt {c-a c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c^2*Sqrt[1 - a^2*x^2]*(43 - 14*a*x + 3*a^2*x^2))/(15*a*Sqrt[c - a*c*x])

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fricas [A]  time = 0.51, size = 52, normalized size = 0.51 \[ -\frac {2 \, {\left (3 \, a^{2} c x^{2} - 14 \, a c x + 43 \, c\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{15 \, {\left (a^{2} x - a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(3*a^2*c*x^2 - 14*a*c*x + 43*c)*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)^(3/2)/(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 = 48, normalized size = 0.48 \[ \frac {2 \sqrt {-a^{2} x^{2}+1}\, \left (-a c x +c \right )^{\frac {3}{2}} \left (3 a^{2} x^{2}-14 a x +43\right )}{15 \left (a x -1\right )^{2} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^(3/2)*(3*a^2*x^2-14*a*x+43)/(a*x-1)^2/a

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maxima [A]  time = 0.35, size = 49, normalized size = 0.49 \[ \frac {2 \, {\left (3 \, a^{2} c^{\frac {3}{2}} x^{2} - 14 \, a c^{\frac {3}{2}} x + 43 \, c^{\frac {3}{2}}\right )} \sqrt {a x + 1} {\left (a x - 1\right )}}{15 \, {\left (a^{2} x - a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*a^2*c^(3/2)*x^2 - 14*a*c^(3/2)*x + 43*c^(3/2))*sqrt(a*x + 1)*(a*x - 1)/(a^2*x - a)

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mupad [B]  time = 0.94, size = 76, normalized size = 0.75 \[ -\frac {\sqrt {c-a\,c\,x}\,\left (\frac {86\,c\,\sqrt {1-a^2\,x^2}}{15\,a^2}+\frac {2\,c\,x^2\,\sqrt {1-a^2\,x^2}}{5}-\frac {28\,c\,x\,\sqrt {1-a^2\,x^2}}{15\,a}\right )}{x-\frac {1}{a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((c - a*c*x)^(1/2)*((86*c*(1 - a^2*x^2)^(1/2))/(15*a^2) + (2*c*x^2*(1 - a^2*x^2)^(1/2))/5 - (28*c*x*(1 - a^2*
x^2)^(1/2))/(15*a)))/(x - 1/a)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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