3.1108 \(\int \frac {e^{2 \tanh ^{-1}(a x)} x^3}{\sqrt {c-a^2 c x^2}} \, dx\)

Optimal. Leaf size=137 \[ \frac {x^2 \sqrt {c-a^2 c x^2}}{3 a^2 c}+\frac {11 \sqrt {c-a^2 c x^2}}{3 a^4 c}+\frac {(a x+1)^2}{a^4 \sqrt {c-a^2 c x^2}}-\frac {3 \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{a^4 \sqrt {c}}+\frac {x \sqrt {c-a^2 c x^2}}{a^3 c} \]

[Out]

-3*arctan(a*x*c^(1/2)/(-a^2*c*x^2+c)^(1/2))/a^4/c^(1/2)+(a*x+1)^2/a^4/(-a^2*c*x^2+c)^(1/2)+11/3*(-a^2*c*x^2+c)
^(1/2)/a^4/c+x*(-a^2*c*x^2+c)^(1/2)/a^3/c+1/3*x^2*(-a^2*c*x^2+c)^(1/2)/a^2/c

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Rubi [A]  time = 0.36, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6151, 1635, 1815, 641, 217, 203} \[ \frac {x^2 \sqrt {c-a^2 c x^2}}{3 a^2 c}+\frac {x \sqrt {c-a^2 c x^2}}{a^3 c}+\frac {11 \sqrt {c-a^2 c x^2}}{3 a^4 c}+\frac {(a x+1)^2}{a^4 \sqrt {c-a^2 c x^2}}-\frac {3 \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{a^4 \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + a*x)^2/(a^4*Sqrt[c - a^2*c*x^2]) + (11*Sqrt[c - a^2*c*x^2])/(3*a^4*c) + (x*Sqrt[c - a^2*c*x^2])/(a^3*c) +
(x^2*Sqrt[c - a^2*c*x^2])/(3*a^2*c) - (3*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2]])/(a^4*Sqrt[c])

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 641

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

Rule 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 0]

Rule 1815

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Si
mp[(e*x^(q - 1)*(a + b*x^2)^(p + 1))/(b*(q + 2*p + 1)), x] + Dist[1/(b*(q + 2*p + 1)), Int[(a + b*x^2)^p*Expan
dToSum[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x]
&& PolyQ[Pq, x] &&  !LeQ[p, -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 \frac {e^{2 \tanh ^{-1}(a x)} x^3}{\sqrt {c-a^2 c x^2}} \, dx &=c \int \frac {x^3 (1+a x)^2}{\left (c-a^2 c x^2\right )^{3/2}} \, dx\\ &=\frac {(1+a x)^2}{a^4 \sqrt {c-a^2 c x^2}}-\int \frac {(1+a x) \left (\frac {2}{a^3}+\frac {x}{a^2}+\frac {x^2}{a}\right )}{\sqrt {c-a^2 c x^2}} \, dx\\ &=\frac {(1+a x)^2}{a^4 \sqrt {c-a^2 c x^2}}+\frac {x^2 \sqrt {c-a^2 c x^2}}{3 a^2 c}+\frac {\int \frac {-\frac {6 c}{a}-11 c x-6 a c x^2}{\sqrt {c-a^2 c x^2}} \, dx}{3 a^2 c}\\ &=\frac {(1+a x)^2}{a^4 \sqrt {c-a^2 c x^2}}+\frac {x \sqrt {c-a^2 c x^2}}{a^3 c}+\frac {x^2 \sqrt {c-a^2 c x^2}}{3 a^2 c}-\frac {\int \frac {18 a c^2+22 a^2 c^2 x}{\sqrt {c-a^2 c x^2}} \, dx}{6 a^4 c^2}\\ &=\frac {(1+a x)^2}{a^4 \sqrt {c-a^2 c x^2}}+\frac {11 \sqrt {c-a^2 c x^2}}{3 a^4 c}+\frac {x \sqrt {c-a^2 c x^2}}{a^3 c}+\frac {x^2 \sqrt {c-a^2 c x^2}}{3 a^2 c}-\frac {3 \int \frac {1}{\sqrt {c-a^2 c x^2}} \, dx}{a^3}\\ &=\frac {(1+a x)^2}{a^4 \sqrt {c-a^2 c x^2}}+\frac {11 \sqrt {c-a^2 c x^2}}{3 a^4 c}+\frac {x \sqrt {c-a^2 c x^2}}{a^3 c}+\frac {x^2 \sqrt {c-a^2 c x^2}}{3 a^2 c}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{1+a^2 c x^2} \, dx,x,\frac {x}{\sqrt {c-a^2 c x^2}}\right )}{a^3}\\ &=\frac {(1+a x)^2}{a^4 \sqrt {c-a^2 c x^2}}+\frac {11 \sqrt {c-a^2 c x^2}}{3 a^4 c}+\frac {x \sqrt {c-a^2 c x^2}}{a^3 c}+\frac {x^2 \sqrt {c-a^2 c x^2}}{3 a^2 c}-\frac {3 \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{a^4 \sqrt {c}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 97, normalized size = 0.71 \[ \frac {9 \sqrt {c} \tan ^{-1}\left (\frac {a x \sqrt {c-a^2 c x^2}}{\sqrt {c} \left (a^2 x^2-1\right )}\right )+\frac {\left (a^3 x^3+2 a^2 x^2+5 a x-14\right ) \sqrt {c-a^2 c x^2}}{a x-1}}{3 a^4 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[c - a^2*c*x^2]*(-14 + 5*a*x + 2*a^2*x^2 + a^3*x^3))/(-1 + a*x) + 9*Sqrt[c]*ArcTan[(a*x*Sqrt[c - a^2*c*x
^2])/(Sqrt[c]*(-1 + a^2*x^2))])/(3*a^4*c)

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fricas [A]  time = 0.62, size = 200, normalized size = 1.46 \[ \left [-\frac {9 \, {\left (a x - 1\right )} \sqrt {-c} \log \left (2 \, a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right ) - 2 \, {\left (a^{3} x^{3} + 2 \, a^{2} x^{2} + 5 \, a x - 14\right )} \sqrt {-a^{2} c x^{2} + c}}{6 \, {\left (a^{5} c x - a^{4} c\right )}}, \frac {9 \, {\left (a x - 1\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right ) + {\left (a^{3} x^{3} + 2 \, a^{2} x^{2} + 5 \, a x - 14\right )} \sqrt {-a^{2} c x^{2} + c}}{3 \, {\left (a^{5} c x - a^{4} c\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(9*(a*x - 1)*sqrt(-c)*log(2*a^2*c*x^2 + 2*sqrt(-a^2*c*x^2 + c)*a*sqrt(-c)*x - c) - 2*(a^3*x^3 + 2*a^2*x^
2 + 5*a*x - 14)*sqrt(-a^2*c*x^2 + c))/(a^5*c*x - a^4*c), 1/3*(9*(a*x - 1)*sqrt(c)*arctan(sqrt(-a^2*c*x^2 + c)*
a*sqrt(c)*x/(a^2*c*x^2 - c)) + (a^3*x^3 + 2*a^2*x^2 + 5*a*x - 14)*sqrt(-a^2*c*x^2 + c))/(a^5*c*x - a^4*c)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {undef} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef

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maple [A]  time = 0.04, size = 149, normalized size = 1.09 \[ \frac {x^{2} \sqrt {-a^{2} c \,x^{2}+c}}{3 a^{2} c}+\frac {8 \sqrt {-a^{2} c \,x^{2}+c}}{3 a^{4} c}+\frac {x \sqrt {-a^{2} c \,x^{2}+c}}{a^{3} c}-\frac {3 \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{a^{3} \sqrt {a^{2} c}}-\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )}}{a^{5} c \left (x -\frac {1}{a}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^2*(-a^2*c*x^2+c)^(1/2)/a^2/c+8/3*(-a^2*c*x^2+c)^(1/2)/a^4/c+x*(-a^2*c*x^2+c)^(1/2)/a^3/c-3/a^3/(a^2*c)^(
1/2)*arctan((a^2*c)^(1/2)*x/(-a^2*c*x^2+c)^(1/2))-2/a^5/c/(x-1/a)*(-(x-1/a)^2*a^2*c-2*a*c*(x-1/a))^(1/2)

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maxima [A]  time = 0.55, size = 113, normalized size = 0.82 \[ -\frac {1}{3} \, a {\left (\frac {6 \, \sqrt {-a^{2} c x^{2} + c}}{a^{6} c x - a^{5} c} - \frac {\sqrt {-a^{2} c x^{2} + c} x^{2}}{a^{3} c} - \frac {3 \, \sqrt {-a^{2} c x^{2} + c} x}{a^{4} c} + \frac {9 \, \arcsin \left (a x\right )}{a^{5} \sqrt {c}} - \frac {8 \, \sqrt {-a^{2} c x^{2} + c}}{a^{5} c}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a*(6*sqrt(-a^2*c*x^2 + c)/(a^6*c*x - a^5*c) - sqrt(-a^2*c*x^2 + c)*x^2/(a^3*c) - 3*sqrt(-a^2*c*x^2 + c)*x
/(a^4*c) + 9*arcsin(a*x)/(a^5*sqrt(c)) - 8*sqrt(-a^2*c*x^2 + c)/(a^5*c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {x^{3}}{a x \sqrt {- a^{2} c x^{2} + c} - \sqrt {- a^{2} c x^{2} + c}}\, dx - \int \frac {a x^{4}}{a x \sqrt {- a^{2} c x^{2} + c} - \sqrt {- a^{2} c x^{2} + c}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(x**3/(a*x*sqrt(-a**2*c*x**2 + c) - sqrt(-a**2*c*x**2 + c)), x) - Integral(a*x**4/(a*x*sqrt(-a**2*c*x
**2 + c) - sqrt(-a**2*c*x**2 + c)), x)

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