3.3.47 \(\int e^{3 \tanh ^{-1}(a x)} \sqrt {c-a c x} \, dx\) [247]

Optimal. Leaf size=119 \[ -\frac {4 c \sqrt {1-a^2 x^2}}{a \sqrt {c-a c x}}-\frac {2 c^2 \left (1-a^2 x^2\right )^{3/2}}{3 a (c-a c x)^{3/2}}+\frac {4 \sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{a} \]

[Out]

-2/3*c^2*(-a^2*x^2+1)^(3/2)/a/(-a*c*x+c)^(3/2)+4*arctanh(1/2*c^(1/2)*(-a^2*x^2+1)^(1/2)*2^(1/2)/(-a*c*x+c)^(1/
2))*2^(1/2)*c^(1/2)/a-4*c*(-a^2*x^2+1)^(1/2)/a/(-a*c*x+c)^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6262, 679, 675, 214} \begin {gather*} -\frac {2 c^2 \left (1-a^2 x^2\right )^{3/2}}{3 a (c-a c x)^{3/2}}-\frac {4 c \sqrt {1-a^2 x^2}}{a \sqrt {c-a c x}}+\frac {4 \sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 675

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 679

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 + 2*p + 1))), x] - Dist[2*c*d*(p/(e^2*(m + 2*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] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[
m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 6262

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

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Mathematica [A]
time = 0.02, size = 67, normalized size = 0.56 \begin {gather*} -\frac {2 \sqrt {c-a c x} \left (\sqrt {1+a x} (7+a x)-6 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {1+a x}}{\sqrt {2}}\right )\right )}{3 a \sqrt {1-a x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c - a*c*x]*(Sqrt[1 + a*x]*(7 + a*x) - 6*Sqrt[2]*ArcTanh[Sqrt[1 + a*x]/Sqrt[2]]))/(3*a*Sqrt[1 - a*x])

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Maple [A]
time = 1.25, size = 95, normalized size = 0.80

method result size
default \(-\frac {2 \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}\, \left (6 \sqrt {c}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {\left (a x +1\right ) c}\, \sqrt {2}}{2 \sqrt {c}}\right )-\sqrt {\left (a x +1\right ) c}\, a x -7 \sqrt {\left (a x +1\right ) c}\right )}{3 \left (a x -1\right ) \sqrt {\left (a x +1\right ) c}\, a}\) \(95\)
risch \(\frac {2 \left (a x +7\right ) \left (a x +1\right ) \sqrt {-\frac {\left (-a^{2} x^{2}+1\right ) c}{a x -1}}\, \left (a x -1\right ) c}{3 a \sqrt {\left (a x +1\right ) c}\, \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}}-\frac {4 \sqrt {2}\, \sqrt {c}\, \arctanh \left (\frac {\sqrt {c x a +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {-\frac {\left (-a^{2} x^{2}+1\right ) c}{a x -1}}\, \left (a x -1\right )}{a \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}}\) \(153\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.37, size = 220, normalized size = 1.85 \begin {gather*} \left [\frac {2 \, {\left (3 \, \sqrt {2} {\left (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 ) + \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} {\left (a x + 7\right )}\right )}}{3 \, {\left (a^{2} x - a\right )}}, \frac {2 \, {\left (6 \, \sqrt {2} {\left (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 ) + \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} {\left (a x + 7\right )}\right )}}{3 \, {\left (a^{2} x - a\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2/3*(3*sqrt(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)*sq
rt(c) - 3*c)/(a^2*x^2 - 2*a*x + 1)) + sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*(a*x + 7))/(a^2*x - a), 2/3*(6*sqrt(
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)) + sqrt(-a^2
*x^2 + 1)*sqrt(-a*c*x + c)*(a*x + 7))/(a^2*x - a)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {c-a\,c\,x}\,{\left (a\,x+1\right )}^3}{{\left (1-a^2\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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