3.1088 \(\int e^{2 \tanh ^{-1}(a x)} x (c-a^2 c x^2)^{3/2} \, dx\)

Optimal. Leaf size=111 \[ \frac {c^{3/2} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{4 a^2}-\frac {1}{5} x^2 \left (c-a^2 c x^2\right )^{3/2}+\frac {c x \sqrt {c-a^2 c x^2}}{4 a}-\frac {(15 a x+14) \left (c-a^2 c x^2\right )^{3/2}}{30 a^2} \]

[Out]

-1/5*x^2*(-a^2*c*x^2+c)^(3/2)-1/30*(15*a*x+14)*(-a^2*c*x^2+c)^(3/2)/a^2+1/4*c^(3/2)*arctan(a*x*c^(1/2)/(-a^2*c
*x^2+c)^(1/2))/a^2+1/4*c*x*(-a^2*c*x^2+c)^(1/2)/a

________________________________________________________________________________________

Rubi [A]  time = 0.19, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {6151, 1809, 780, 195, 217, 203} \[ \frac {c^{3/2} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{4 a^2}-\frac {1}{5} x^2 \left (c-a^2 c x^2\right )^{3/2}+\frac {c x \sqrt {c-a^2 c x^2}}{4 a}-\frac {(15 a x+14) \left (c-a^2 c x^2\right )^{3/2}}{30 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*x*Sqrt[c - a^2*c*x^2])/(4*a) - (x^2*(c - a^2*c*x^2)^(3/2))/5 - ((14 + 15*a*x)*(c - a^2*c*x^2)^(3/2))/(30*a^
2) + (c^(3/2)*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2]])/(4*a^2)

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rule 6151

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

Rubi steps

\begin {align*} \int e^{2 \tanh ^{-1}(a x)} x \left (c-a^2 c x^2\right )^{3/2} \, dx &=c \int x (1+a x)^2 \sqrt {c-a^2 c x^2} \, dx\\ &=-\frac {1}{5} x^2 \left (c-a^2 c x^2\right )^{3/2}-\frac {\int x \left (-7 a^2 c-10 a^3 c x\right ) \sqrt {c-a^2 c x^2} \, dx}{5 a^2}\\ &=-\frac {1}{5} x^2 \left (c-a^2 c x^2\right )^{3/2}-\frac {(14+15 a x) \left (c-a^2 c x^2\right )^{3/2}}{30 a^2}+\frac {c \int \sqrt {c-a^2 c x^2} \, dx}{2 a}\\ &=\frac {c x \sqrt {c-a^2 c x^2}}{4 a}-\frac {1}{5} x^2 \left (c-a^2 c x^2\right )^{3/2}-\frac {(14+15 a x) \left (c-a^2 c x^2\right )^{3/2}}{30 a^2}+\frac {c^2 \int \frac {1}{\sqrt {c-a^2 c x^2}} \, dx}{4 a}\\ &=\frac {c x \sqrt {c-a^2 c x^2}}{4 a}-\frac {1}{5} x^2 \left (c-a^2 c x^2\right )^{3/2}-\frac {(14+15 a x) \left (c-a^2 c x^2\right )^{3/2}}{30 a^2}+\frac {c^2 \operatorname {Subst}\left (\int \frac {1}{1+a^2 c x^2} \, dx,x,\frac {x}{\sqrt {c-a^2 c x^2}}\right )}{4 a}\\ &=\frac {c x \sqrt {c-a^2 c x^2}}{4 a}-\frac {1}{5} x^2 \left (c-a^2 c x^2\right )^{3/2}-\frac {(14+15 a x) \left (c-a^2 c x^2\right )^{3/2}}{30 a^2}+\frac {c^{3/2} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{4 a^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.11, size = 97, normalized size = 0.87 \[ \frac {c \left (12 a^4 x^4+30 a^3 x^3+16 a^2 x^2-15 a x-28\right ) \sqrt {c-a^2 c x^2}-15 c^{3/2} \tan ^{-1}\left (\frac {a x \sqrt {c-a^2 c x^2}}{\sqrt {c} \left (a^2 x^2-1\right )}\right )}{60 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Sqrt[c - a^2*c*x^2]*(-28 - 15*a*x + 16*a^2*x^2 + 30*a^3*x^3 + 12*a^4*x^4) - 15*c^(3/2)*ArcTan[(a*x*Sqrt[c -
a^2*c*x^2])/(Sqrt[c]*(-1 + a^2*x^2))])/(60*a^2)

________________________________________________________________________________________

fricas [A]  time = 0.88, size = 198, normalized size = 1.78 \[ \left [\frac {15 \, \sqrt {-c} c \log \left (2 \, a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right ) + 2 \, {\left (12 \, a^{4} c x^{4} + 30 \, a^{3} c x^{3} + 16 \, a^{2} c x^{2} - 15 \, a c x - 28 \, c\right )} \sqrt {-a^{2} c x^{2} + c}}{120 \, a^{2}}, -\frac {15 \, c^{\frac {3}{2}} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right ) - {\left (12 \, a^{4} c x^{4} + 30 \, a^{3} c x^{3} + 16 \, a^{2} c x^{2} - 15 \, a c x - 28 \, c\right )} \sqrt {-a^{2} c x^{2} + c}}{60 \, a^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/120*(15*sqrt(-c)*c*log(2*a^2*c*x^2 + 2*sqrt(-a^2*c*x^2 + c)*a*sqrt(-c)*x - c) + 2*(12*a^4*c*x^4 + 30*a^3*c*
x^3 + 16*a^2*c*x^2 - 15*a*c*x - 28*c)*sqrt(-a^2*c*x^2 + c))/a^2, -1/60*(15*c^(3/2)*arctan(sqrt(-a^2*c*x^2 + c)
*a*sqrt(c)*x/(a^2*c*x^2 - c)) - (12*a^4*c*x^4 + 30*a^3*c*x^3 + 16*a^2*c*x^2 - 15*a*c*x - 28*c)*sqrt(-a^2*c*x^2
+ c))/a^2]

________________________________________________________________________________________

giac [A]  time = 0.20, size = 98, normalized size = 0.88 \[ \frac {1}{60} \, \sqrt {-a^{2} c x^{2} + c} {\left ({\left (2 \, {\left (3 \, {\left (2 \, a^{2} c x + 5 \, a c\right )} x + 8 \, c\right )} x - \frac {15 \, c}{a}\right )} x - \frac {28 \, c}{a^{2}}\right )} - \frac {c^{2} \log \left ({\left | -\sqrt {-a^{2} c} x + \sqrt {-a^{2} c x^{2} + c} \right |}\right )}{4 \, a \sqrt {-c} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*sqrt(-a^2*c*x^2 + c)*((2*(3*(2*a^2*c*x + 5*a*c)*x + 8*c)*x - 15*c/a)*x - 28*c/a^2) - 1/4*c^2*log(abs(-sqr
t(-a^2*c)*x + sqrt(-a^2*c*x^2 + c)))/(a*sqrt(-c)*abs(a))

________________________________________________________________________________________

maple [B]  time = 0.04, size = 222, normalized size = 2.00 \[ \frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}{5 a^{2} c}-\frac {x \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{2 a}-\frac {3 c x \sqrt {-a^{2} c \,x^{2}+c}}{4 a}-\frac {3 c^{2} \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{4 a \sqrt {a^{2} c}}-\frac {2 \left (-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{3 a^{2}}+\frac {c \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )}\, x}{a}+\frac {c^{2} \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )}}\right )}{a \sqrt {a^{2} c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(-a^2*c*x^2+c)^(5/2)/a^2/c-1/2*x/a*(-a^2*c*x^2+c)^(3/2)-3/4*c*x*(-a^2*c*x^2+c)^(1/2)/a-3/4/a*c^2/(a^2*c)^(
1/2)*arctan((a^2*c)^(1/2)*x/(-a^2*c*x^2+c)^(1/2))-2/3/a^2*(-(x-1/a)^2*a^2*c-2*a*c*(x-1/a))^(3/2)+1/a*c*(-(x-1/
a)^2*a^2*c-2*a*c*(x-1/a))^(1/2)*x+1/a*c^2/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-(x-1/a)^2*a^2*c-2*a*c*(x-1/a)
)^(1/2))

________________________________________________________________________________________

maxima [A]  time = 0.53, size = 167, normalized size = 1.50 \[ -\frac {1}{60} \, a {\left (\frac {30 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} x}{a^{2}} - \frac {60 \, \sqrt {a^{2} c x^{2} - 4 \, a c x + 3 \, c} c x}{a^{2}} + \frac {45 \, \sqrt {-a^{2} c x^{2} + c} c x}{a^{2}} + \frac {45 \, c^{\frac {3}{2}} \arcsin \left (a x\right )}{a^{3}} + \frac {40 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}}}{a^{3}} - \frac {12 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}}}{a^{3} c} + \frac {120 \, \sqrt {a^{2} c x^{2} - 4 \, a c x + 3 \, c} c}{a^{3}} - \frac {60 \, c^{3} \arcsin \left (a x - 2\right )}{a^{6} \left (-\frac {c}{a^{2}}\right )^{\frac {3}{2}}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*a*(30*(-a^2*c*x^2 + c)^(3/2)*x/a^2 - 60*sqrt(a^2*c*x^2 - 4*a*c*x + 3*c)*c*x/a^2 + 45*sqrt(-a^2*c*x^2 + c
)*c*x/a^2 + 45*c^(3/2)*arcsin(a*x)/a^3 + 40*(-a^2*c*x^2 + c)^(3/2)/a^3 - 12*(-a^2*c*x^2 + c)^(5/2)/(a^3*c) + 1
20*sqrt(a^2*c*x^2 - 4*a*c*x + 3*c)*c/a^3 - 60*c^3*arcsin(a*x - 2)/(a^6*(-c/a^2)^(3/2)))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 16.12, size = 306, normalized size = 2.76 \[ a^{2} c \left (\begin {cases} \frac {x^{4} \sqrt {- a^{2} c x^{2} + c}}{5} - \frac {x^{2} \sqrt {- a^{2} c x^{2} + c}}{15 a^{2}} - \frac {2 \sqrt {- a^{2} c x^{2} + c}}{15 a^{4}} & \text {for}\: a \neq 0 \\\frac {\sqrt {c} x^{4}}{4} & \text {otherwise} \end {cases}\right ) + 2 a c \left (\begin {cases} \frac {i a^{2} \sqrt {c} x^{5}}{4 \sqrt {a^{2} x^{2} - 1}} - \frac {3 i \sqrt {c} x^{3}}{8 \sqrt {a^{2} x^{2} - 1}} + \frac {i \sqrt {c} x}{8 a^{2} \sqrt {a^{2} x^{2} - 1}} - \frac {i \sqrt {c} \operatorname {acosh}{\left (a x \right )}}{8 a^{3}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {a^{2} \sqrt {c} x^{5}}{4 \sqrt {- a^{2} x^{2} + 1}} + \frac {3 \sqrt {c} x^{3}}{8 \sqrt {- a^{2} x^{2} + 1}} - \frac {\sqrt {c} x}{8 a^{2} \sqrt {- a^{2} x^{2} + 1}} + \frac {\sqrt {c} \operatorname {asin}{\left (a x \right )}}{8 a^{3}} & \text {otherwise} \end {cases}\right ) + c \left (\begin {cases} 0 & \text {for}\: c = 0 \\\frac {\sqrt {c} x^{2}}{2} & \text {for}\: a^{2} = 0 \\- \frac {\left (- a^{2} c x^{2} + c\right )^{\frac {3}{2}}}{3 a^{2} c} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*c*Piecewise((x**4*sqrt(-a**2*c*x**2 + c)/5 - x**2*sqrt(-a**2*c*x**2 + c)/(15*a**2) - 2*sqrt(-a**2*c*x**2
+ c)/(15*a**4), Ne(a, 0)), (sqrt(c)*x**4/4, True)) + 2*a*c*Piecewise((I*a**2*sqrt(c)*x**5/(4*sqrt(a**2*x**2 -
1)) - 3*I*sqrt(c)*x**3/(8*sqrt(a**2*x**2 - 1)) + I*sqrt(c)*x/(8*a**2*sqrt(a**2*x**2 - 1)) - I*sqrt(c)*acosh(a*
x)/(8*a**3), Abs(a**2*x**2) > 1), (-a**2*sqrt(c)*x**5/(4*sqrt(-a**2*x**2 + 1)) + 3*sqrt(c)*x**3/(8*sqrt(-a**2*
x**2 + 1)) - sqrt(c)*x/(8*a**2*sqrt(-a**2*x**2 + 1)) + sqrt(c)*asin(a*x)/(8*a**3), True)) + c*Piecewise((0, Eq
(c, 0)), (sqrt(c)*x**2/2, Eq(a**2, 0)), (-(-a**2*c*x**2 + c)**(3/2)/(3*a**2*c), True))

________________________________________________________________________________________