3.345 \(\int \frac {e^{\tanh ^{-1}(a x)} x^4}{(c-a c x)^3} \, dx\)

Optimal. Leaf size=135 \[ -\frac {19 \sin ^{-1}(a x)}{2 a^5 c^3}+\frac {(a x+1)^4}{5 a^5 c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac {19 (a x+1)^3}{15 a^5 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {6 (a x+1)^2}{a^5 c^3 \sqrt {1-a^2 x^2}}+\frac {(a x+20) \sqrt {1-a^2 x^2}}{2 a^5 c^3} \]

[Out]

1/5*(a*x+1)^4/a^5/c^3/(-a^2*x^2+1)^(5/2)-19/15*(a*x+1)^3/a^5/c^3/(-a^2*x^2+1)^(3/2)-19/2*arcsin(a*x)/a^5/c^3+6
*(a*x+1)^2/a^5/c^3/(-a^2*x^2+1)^(1/2)+1/2*(a*x+20)*(-a^2*x^2+1)^(1/2)/a^5/c^3

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Rubi [A]  time = 0.42, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {6128, 852, 1635, 780, 216} \[ \frac {(a x+1)^4}{5 a^5 c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac {19 (a x+1)^3}{15 a^5 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {6 (a x+1)^2}{a^5 c^3 \sqrt {1-a^2 x^2}}+\frac {(a x+20) \sqrt {1-a^2 x^2}}{2 a^5 c^3}-\frac {19 \sin ^{-1}(a x)}{2 a^5 c^3} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*x^4)/(c - a*c*x)^3,x]

[Out]

(1 + a*x)^4/(5*a^5*c^3*(1 - a^2*x^2)^(5/2)) - (19*(1 + a*x)^3)/(15*a^5*c^3*(1 - a^2*x^2)^(3/2)) + (6*(1 + a*x)
^2)/(a^5*c^3*Sqrt[1 - a^2*x^2]) + ((20 + a*x)*Sqrt[1 - a^2*x^2])/(2*a^5*c^3) - (19*ArcSin[a*x])/(2*a^5*c^3)

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 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 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 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)^3} \, dx &=c \int \frac {x^4 \sqrt {1-a^2 x^2}}{(c-a c x)^4} \, dx\\ &=\frac {\int \frac {x^4 (c+a c x)^4}{\left (1-a^2 x^2\right )^{7/2}} \, dx}{c^7}\\ &=\frac {(1+a x)^4}{5 a^5 c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac {\int \frac {(c+a c x)^3 \left (\frac {4}{a^4}+\frac {5 x}{a^3}+\frac {5 x^2}{a^2}+\frac {5 x^3}{a}\right )}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{5 c^6}\\ &=\frac {(1+a x)^4}{5 a^5 c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac {19 (1+a x)^3}{15 a^5 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {\int \frac {(c+a c x)^2 \left (\frac {45}{a^4}+\frac {30 x}{a^3}+\frac {15 x^2}{a^2}\right )}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{15 c^5}\\ &=\frac {(1+a x)^4}{5 a^5 c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac {19 (1+a x)^3}{15 a^5 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {6 (1+a x)^2}{a^5 c^3 \sqrt {1-a^2 x^2}}-\frac {\int \frac {\left (\frac {135}{a^4}+\frac {15 x}{a^3}\right ) (c+a c x)}{\sqrt {1-a^2 x^2}} \, dx}{15 c^4}\\ &=\frac {(1+a x)^4}{5 a^5 c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac {19 (1+a x)^3}{15 a^5 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {6 (1+a x)^2}{a^5 c^3 \sqrt {1-a^2 x^2}}+\frac {(20+a x) \sqrt {1-a^2 x^2}}{2 a^5 c^3}-\frac {19 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{2 a^4 c^3}\\ &=\frac {(1+a x)^4}{5 a^5 c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac {19 (1+a x)^3}{15 a^5 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {6 (1+a x)^2}{a^5 c^3 \sqrt {1-a^2 x^2}}+\frac {(20+a x) \sqrt {1-a^2 x^2}}{2 a^5 c^3}-\frac {19 \sin ^{-1}(a x)}{2 a^5 c^3}\\ \end {align*}

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Mathematica [C]  time = 0.15, size = 122, normalized size = 0.90 \[ \frac {\sqrt {a x+1} \left (-15 a^4 x^4-75 a^3 x^3+433 a^2 x^2-639 a x+308\right )+140 \sqrt {2} (a x-1) \, _2F_1\left (-\frac {3}{2},-\frac {3}{2};-\frac {1}{2};\frac {1}{2} (1-a x)\right )+360 (1-a x)^{5/2} \sin ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )}{30 a^5 c^3 (1-a x)^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^ArcTanh[a*x]*x^4)/(c - a*c*x)^3,x]

[Out]

(Sqrt[1 + a*x]*(308 - 639*a*x + 433*a^2*x^2 - 75*a^3*x^3 - 15*a^4*x^4) + 360*(1 - a*x)^(5/2)*ArcSin[Sqrt[1 - a
*x]/Sqrt[2]] + 140*Sqrt[2]*(-1 + a*x)*Hypergeometric2F1[-3/2, -3/2, -1/2, (1 - a*x)/2])/(30*a^5*c^3*(1 - a*x)^
(5/2))

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fricas [A]  time = 0.47, size = 153, normalized size = 1.13 \[ \frac {448 \, a^{3} x^{3} - 1344 \, a^{2} x^{2} + 1344 \, a x + 570 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (15 \, a^{4} x^{4} + 75 \, a^{3} x^{3} - 713 \, a^{2} x^{2} + 1059 \, a x - 448\right )} \sqrt {-a^{2} x^{2} + 1} - 448}{30 \, {\left (a^{8} c^{3} x^{3} - 3 \, a^{7} c^{3} x^{2} + 3 \, a^{6} c^{3} x - a^{5} c^{3}\right )}} \]

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)^3,x, algorithm="fricas")

[Out]

1/30*(448*a^3*x^3 - 1344*a^2*x^2 + 1344*a*x + 570*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*arctan((sqrt(-a^2*x^2 + 1)
- 1)/(a*x)) + (15*a^4*x^4 + 75*a^3*x^3 - 713*a^2*x^2 + 1059*a*x - 448)*sqrt(-a^2*x^2 + 1) - 448)/(a^8*c^3*x^3
- 3*a^7*c^3*x^2 + 3*a^6*c^3*x - a^5*c^3)

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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)^3,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 = 208, normalized size = 1.54 \[ \frac {x \sqrt {-a^{2} x^{2}+1}}{2 c^{3} a^{4}}-\frac {19 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 c^{3} a^{4} \sqrt {a^{2}}}+\frac {4 \sqrt {-a^{2} x^{2}+1}}{c^{3} a^{5}}-\frac {41 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{15 c^{3} a^{7} \left (x -\frac {1}{a}\right )^{2}}-\frac {199 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{15 c^{3} a^{6} \left (x -\frac {1}{a}\right )}-\frac {2 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{5 c^{3} a^{8} \left (x -\frac {1}{a}\right )^{3}} \]

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)^3,x)

[Out]

1/2/c^3/a^4*x*(-a^2*x^2+1)^(1/2)-19/2/c^3/a^4/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))+4/c^3/a^5*(
-a^2*x^2+1)^(1/2)-41/15/c^3/a^7/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)-199/15/c^3/a^6/(x-1/a)*(-a^2*(x-1
/a)^2-2*a*(x-1/a))^(1/2)-2/5/c^3/a^8/(x-1/a)^3*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)

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maxima [A]  time = 0.41, size = 185, normalized size = 1.37 \[ -\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{5 \, {\left (a^{8} c^{3} x^{3} - 3 \, a^{7} c^{3} x^{2} + 3 \, a^{6} c^{3} x - a^{5} c^{3}\right )}} - \frac {41 \, \sqrt {-a^{2} x^{2} + 1}}{15 \, {\left (a^{7} c^{3} x^{2} - 2 \, a^{6} c^{3} x + a^{5} c^{3}\right )}} - \frac {199 \, \sqrt {-a^{2} x^{2} + 1}}{15 \, {\left (a^{6} c^{3} x - a^{5} c^{3}\right )}} + \frac {\sqrt {-a^{2} x^{2} + 1} x}{2 \, a^{4} c^{3}} - \frac {19 \, \arcsin \left (a x\right )}{2 \, a^{5} c^{3}} + \frac {4 \, \sqrt {-a^{2} x^{2} + 1}}{a^{5} c^{3}} \]

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)^3,x, algorithm="maxima")

[Out]

-2/5*sqrt(-a^2*x^2 + 1)/(a^8*c^3*x^3 - 3*a^7*c^3*x^2 + 3*a^6*c^3*x - a^5*c^3) - 41/15*sqrt(-a^2*x^2 + 1)/(a^7*
c^3*x^2 - 2*a^6*c^3*x + a^5*c^3) - 199/15*sqrt(-a^2*x^2 + 1)/(a^6*c^3*x - a^5*c^3) + 1/2*sqrt(-a^2*x^2 + 1)*x/
(a^4*c^3) - 19/2*arcsin(a*x)/(a^5*c^3) + 4*sqrt(-a^2*x^2 + 1)/(a^5*c^3)

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mupad [B]  time = 0.83, size = 302, normalized size = 2.24 \[ \frac {4\,a^4\,\sqrt {1-a^2\,x^2}}{15\,\left (a^{11}\,c^3\,x^2-2\,a^{10}\,c^3\,x+a^9\,c^3\right )}-\frac {2\,\sqrt {1-a^2\,x^2}}{5\,\sqrt {-a^2}\,\left (a^3\,c^3\,\sqrt {-a^2}+3\,a^5\,c^3\,x^2\,\sqrt {-a^2}-a^6\,c^3\,x^3\,\sqrt {-a^2}-3\,a^4\,c^3\,x\,\sqrt {-a^2}\right )}-\frac {3\,\sqrt {1-a^2\,x^2}}{a^7\,c^3\,x^2-2\,a^6\,c^3\,x+a^5\,c^3}-\frac {199\,\sqrt {1-a^2\,x^2}}{15\,\left (a^3\,c^3\,\sqrt {-a^2}-a^4\,c^3\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}+\frac {4\,\sqrt {1-a^2\,x^2}}{a^5\,c^3}+\frac {x\,\sqrt {1-a^2\,x^2}}{2\,a^4\,c^3}-\frac {19\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{2\,a^4\,c^3\,\sqrt {-a^2}} \]

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)^3),x)

[Out]

(4*a^4*(1 - a^2*x^2)^(1/2))/(15*(a^9*c^3 - 2*a^10*c^3*x + a^11*c^3*x^2)) - (2*(1 - a^2*x^2)^(1/2))/(5*(-a^2)^(
1/2)*(a^3*c^3*(-a^2)^(1/2) + 3*a^5*c^3*x^2*(-a^2)^(1/2) - a^6*c^3*x^3*(-a^2)^(1/2) - 3*a^4*c^3*x*(-a^2)^(1/2))
) - (3*(1 - a^2*x^2)^(1/2))/(a^5*c^3 - 2*a^6*c^3*x + a^7*c^3*x^2) - (199*(1 - a^2*x^2)^(1/2))/(15*(a^3*c^3*(-a
^2)^(1/2) - a^4*c^3*x*(-a^2)^(1/2))*(-a^2)^(1/2)) + (4*(1 - a^2*x^2)^(1/2))/(a^5*c^3) + (x*(1 - a^2*x^2)^(1/2)
)/(2*a^4*c^3) - (19*asinh(x*(-a^2)^(1/2)))/(2*a^4*c^3*(-a^2)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {x^{4}}{a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} - 3 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + 3 a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {a x^{5}}{a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} - 3 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + 3 a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx}{c^{3}} \]

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)**3,x)

[Out]

-(Integral(x**4/(a**3*x**3*sqrt(-a**2*x**2 + 1) - 3*a**2*x**2*sqrt(-a**2*x**2 + 1) + 3*a*x*sqrt(-a**2*x**2 + 1
) - sqrt(-a**2*x**2 + 1)), x) + Integral(a*x**5/(a**3*x**3*sqrt(-a**2*x**2 + 1) - 3*a**2*x**2*sqrt(-a**2*x**2
+ 1) + 3*a*x*sqrt(-a**2*x**2 + 1) - sqrt(-a**2*x**2 + 1)), x))/c**3

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