3.327 \(\int \frac {e^{\tanh ^{-1}(a x)} x^4}{c-a c x} \, dx\)

Optimal. Leaf size=146 \[ -\frac {27 \sin ^{-1}(a x)}{8 a^5 c}+\frac {x^3 \sqrt {1-a^2 x^2}}{4 a^2 c}+\frac {13 \sqrt {1-a^2 x^2}}{3 a^5 c}+\frac {(a x+1)^2}{a^5 c \sqrt {1-a^2 x^2}}+\frac {11 x \sqrt {1-a^2 x^2}}{8 a^4 c}+\frac {2 x^2 \sqrt {1-a^2 x^2}}{3 a^3 c} \]

[Out]

-27/8*arcsin(a*x)/a^5/c+(a*x+1)^2/a^5/c/(-a^2*x^2+1)^(1/2)+13/3*(-a^2*x^2+1)^(1/2)/a^5/c+11/8*x*(-a^2*x^2+1)^(
1/2)/a^4/c+2/3*x^2*(-a^2*x^2+1)^(1/2)/a^3/c+1/4*x^3*(-a^2*x^2+1)^(1/2)/a^2/c

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Rubi [A]  time = 0.34, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {6128, 852, 1635, 1815, 641, 216} \[ \frac {x^3 \sqrt {1-a^2 x^2}}{4 a^2 c}+\frac {2 x^2 \sqrt {1-a^2 x^2}}{3 a^3 c}+\frac {11 x \sqrt {1-a^2 x^2}}{8 a^4 c}+\frac {13 \sqrt {1-a^2 x^2}}{3 a^5 c}+\frac {(a x+1)^2}{a^5 c \sqrt {1-a^2 x^2}}-\frac {27 \sin ^{-1}(a x)}{8 a^5 c} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*x^4)/(c - a*c*x),x]

[Out]

(1 + a*x)^2/(a^5*c*Sqrt[1 - a^2*x^2]) + (13*Sqrt[1 - a^2*x^2])/(3*a^5*c) + (11*x*Sqrt[1 - a^2*x^2])/(8*a^4*c)
+ (2*x^2*Sqrt[1 - a^2*x^2])/(3*a^3*c) + (x^3*Sqrt[1 - a^2*x^2])/(4*a^2*c) - (27*ArcSin[a*x])/(8*a^5*c)

Rule 216

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

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 852

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^(2*m)/a
^m, Int[((f + g*x)^n*(a + c*x^2)^(m + p))/(d - e*x)^m, x], x] /; FreeQ[{a, c, d, e, f, g, n, p}, x] && NeQ[e*f
- d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[f, 0] && ILtQ[m, -1] &&  !(IGtQ[n, 0] && ILtQ[m +
n, 0] &&  !GtQ[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 6128

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

Rubi steps

\begin {align*} \int \frac {e^{\tanh ^{-1}(a x)} x^4}{c-a c x} \, dx &=c \int \frac {x^4 \sqrt {1-a^2 x^2}}{(c-a c x)^2} \, dx\\ &=\frac {\int \frac {x^4 (c+a c x)^2}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{c^3}\\ &=\frac {(1+a x)^2}{a^5 c \sqrt {1-a^2 x^2}}-\frac {\int \frac {(c+a c x) \left (\frac {2}{a^4}+\frac {x}{a^3}+\frac {x^2}{a^2}+\frac {x^3}{a}\right )}{\sqrt {1-a^2 x^2}} \, dx}{c^2}\\ &=\frac {(1+a x)^2}{a^5 c \sqrt {1-a^2 x^2}}+\frac {x^3 \sqrt {1-a^2 x^2}}{4 a^2 c}+\frac {\int \frac {-\frac {8 c}{a^2}-\frac {12 c x}{a}-11 c x^2-8 a c x^3}{\sqrt {1-a^2 x^2}} \, dx}{4 a^2 c^2}\\ &=\frac {(1+a x)^2}{a^5 c \sqrt {1-a^2 x^2}}+\frac {2 x^2 \sqrt {1-a^2 x^2}}{3 a^3 c}+\frac {x^3 \sqrt {1-a^2 x^2}}{4 a^2 c}-\frac {\int \frac {24 c+52 a c x+33 a^2 c x^2}{\sqrt {1-a^2 x^2}} \, dx}{12 a^4 c^2}\\ &=\frac {(1+a x)^2}{a^5 c \sqrt {1-a^2 x^2}}+\frac {11 x \sqrt {1-a^2 x^2}}{8 a^4 c}+\frac {2 x^2 \sqrt {1-a^2 x^2}}{3 a^3 c}+\frac {x^3 \sqrt {1-a^2 x^2}}{4 a^2 c}+\frac {\int \frac {-81 a^2 c-104 a^3 c x}{\sqrt {1-a^2 x^2}} \, dx}{24 a^6 c^2}\\ &=\frac {(1+a x)^2}{a^5 c \sqrt {1-a^2 x^2}}+\frac {13 \sqrt {1-a^2 x^2}}{3 a^5 c}+\frac {11 x \sqrt {1-a^2 x^2}}{8 a^4 c}+\frac {2 x^2 \sqrt {1-a^2 x^2}}{3 a^3 c}+\frac {x^3 \sqrt {1-a^2 x^2}}{4 a^2 c}-\frac {27 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{8 a^4 c}\\ &=\frac {(1+a x)^2}{a^5 c \sqrt {1-a^2 x^2}}+\frac {13 \sqrt {1-a^2 x^2}}{3 a^5 c}+\frac {11 x \sqrt {1-a^2 x^2}}{8 a^4 c}+\frac {2 x^2 \sqrt {1-a^2 x^2}}{3 a^3 c}+\frac {x^3 \sqrt {1-a^2 x^2}}{4 a^2 c}-\frac {27 \sin ^{-1}(a x)}{8 a^5 c}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 81, normalized size = 0.55 \[ \frac {162 \sin ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )-\frac {\sqrt {a x+1} \left (6 a^4 x^4+10 a^3 x^3+17 a^2 x^2+47 a x-128\right )}{\sqrt {1-a x}}}{24 a^5 c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^ArcTanh[a*x]*x^4)/(c - a*c*x),x]

[Out]

(-((Sqrt[1 + a*x]*(-128 + 47*a*x + 17*a^2*x^2 + 10*a^3*x^3 + 6*a^4*x^4))/Sqrt[1 - a*x]) + 162*ArcSin[Sqrt[1 -
a*x]/Sqrt[2]])/(24*a^5*c)

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fricas [A]  time = 0.51, size = 95, normalized size = 0.65 \[ \frac {128 \, a x + 162 \, {\left (a x - 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (6 \, a^{4} x^{4} + 10 \, a^{3} x^{3} + 17 \, a^{2} x^{2} + 47 \, a x - 128\right )} \sqrt {-a^{2} x^{2} + 1} - 128}{24 \, {\left (a^{6} c x - a^{5} c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4/(-a*c*x+c),x, algorithm="fricas")

[Out]

1/24*(128*a*x + 162*(a*x - 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (6*a^4*x^4 + 10*a^3*x^3 + 17*a^2*x^2 +
47*a*x - 128)*sqrt(-a^2*x^2 + 1) - 128)/(a^6*c*x - a^5*c)

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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*x+1)/(-a^2*x^2+1)^(1/2)*x^4/(-a*c*x+c),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 = 0.05, size = 166, normalized size = 1.14 \[ \frac {x^{3} \sqrt {-a^{2} x^{2}+1}}{4 a^{2} c}+\frac {11 x \sqrt {-a^{2} x^{2}+1}}{8 a^{4} c}-\frac {27 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 c \,a^{4} \sqrt {a^{2}}}+\frac {2 x^{2} \sqrt {-a^{2} x^{2}+1}}{3 a^{3} c}+\frac {10 \sqrt {-a^{2} x^{2}+1}}{3 a^{5} c}-\frac {2 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{c \,a^{6} \left (x -\frac {1}{a}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4/(-a*c*x+c),x)

[Out]

1/4*x^3*(-a^2*x^2+1)^(1/2)/a^2/c+11/8*x*(-a^2*x^2+1)^(1/2)/a^4/c-27/8/c/a^4/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(
-a^2*x^2+1)^(1/2))+2/3*x^2*(-a^2*x^2+1)^(1/2)/a^3/c+10/3*(-a^2*x^2+1)^(1/2)/a^5/c-2/c/a^6/(x-1/a)*(-a^2*(x-1/a
)^2-2*a*(x-1/a))^(1/2)

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maxima [A]  time = 0.45, size = 129, normalized size = 0.88 \[ \frac {\sqrt {-a^{2} x^{2} + 1} x^{3}}{4 \, a^{2} c} - \frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{a^{6} c x - a^{5} c} + \frac {2 \, \sqrt {-a^{2} x^{2} + 1} x^{2}}{3 \, a^{3} c} + \frac {11 \, \sqrt {-a^{2} x^{2} + 1} x}{8 \, a^{4} c} - \frac {27 \, \arcsin \left (a x\right )}{8 \, a^{5} c} + \frac {10 \, \sqrt {-a^{2} x^{2} + 1}}{3 \, a^{5} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4/(-a*c*x+c),x, algorithm="maxima")

[Out]

1/4*sqrt(-a^2*x^2 + 1)*x^3/(a^2*c) - 2*sqrt(-a^2*x^2 + 1)/(a^6*c*x - a^5*c) + 2/3*sqrt(-a^2*x^2 + 1)*x^2/(a^3*
c) + 11/8*sqrt(-a^2*x^2 + 1)*x/(a^4*c) - 27/8*arcsin(a*x)/(a^5*c) + 10/3*sqrt(-a^2*x^2 + 1)/(a^5*c)

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mupad [B]  time = 0.82, size = 163, normalized size = 1.12 \[ \frac {10\,\sqrt {1-a^2\,x^2}}{3\,a^5\,c}-\frac {2\,\sqrt {1-a^2\,x^2}}{\sqrt {-a^2}\,\left (a^3\,c\,\sqrt {-a^2}-a^4\,c\,x\,\sqrt {-a^2}\right )}+\frac {11\,x\,\sqrt {1-a^2\,x^2}}{8\,a^4\,c}-\frac {27\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{8\,a^4\,c\,\sqrt {-a^2}}+\frac {x^3\,\sqrt {1-a^2\,x^2}}{4\,a^2\,c}+\frac {2\,x^2\,\sqrt {1-a^2\,x^2}}{3\,a^3\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(a*x + 1))/((1 - a^2*x^2)^(1/2)*(c - a*c*x)),x)

[Out]

(10*(1 - a^2*x^2)^(1/2))/(3*a^5*c) - (2*(1 - a^2*x^2)^(1/2))/((-a^2)^(1/2)*(a^3*c*(-a^2)^(1/2) - a^4*c*x*(-a^2
)^(1/2))) + (11*x*(1 - a^2*x^2)^(1/2))/(8*a^4*c) - (27*asinh(x*(-a^2)^(1/2)))/(8*a^4*c*(-a^2)^(1/2)) + (x^3*(1
- a^2*x^2)^(1/2))/(4*a^2*c) + (2*x^2*(1 - a^2*x^2)^(1/2))/(3*a^3*c)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {x^{4}}{a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {a x^{5}}{a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*x**4/(-a*c*x+c),x)

[Out]

-(Integral(x**4/(a*x*sqrt(-a**2*x**2 + 1) - sqrt(-a**2*x**2 + 1)), x) + Integral(a*x**5/(a*x*sqrt(-a**2*x**2 +
1) - sqrt(-a**2*x**2 + 1)), x))/c

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