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

Optimal. Leaf size=128 \[ -\frac {4 (7-3 a x)}{35 c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac {166 a x+105}{105 c^4 \sqrt {1-a^2 x^2}}+\frac {83 a x+35}{105 c^4 \left (1-a^2 x^2\right )^{3/2}}+\frac {16 (a x+1)}{7 c^4 \left (1-a^2 x^2\right )^{7/2}}-\frac {\tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{c^4} \]

[Out]

16/7*(a*x+1)/c^4/(-a^2*x^2+1)^(7/2)-4/35*(-3*a*x+7)/c^4/(-a^2*x^2+1)^(5/2)+1/105*(83*a*x+35)/c^4/(-a^2*x^2+1)^
(3/2)-arctanh((-a^2*x^2+1)^(1/2))/c^4+1/105*(166*a*x+105)/c^4/(-a^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.31, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {6128, 852, 1805, 823, 12, 266, 63, 208} \[ -\frac {4 (7-3 a x)}{35 c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac {166 a x+105}{105 c^4 \sqrt {1-a^2 x^2}}+\frac {83 a x+35}{105 c^4 \left (1-a^2 x^2\right )^{3/2}}+\frac {16 (a x+1)}{7 c^4 \left (1-a^2 x^2\right )^{7/2}}-\frac {\tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*(1 + a*x))/(7*c^4*(1 - a^2*x^2)^(7/2)) - (4*(7 - 3*a*x))/(35*c^4*(1 - a^2*x^2)^(5/2)) + (35 + 83*a*x)/(105
*c^4*(1 - a^2*x^2)^(3/2)) + (105 + 166*a*x)/(105*c^4*Sqrt[1 - a^2*x^2]) - ArcTanh[Sqrt[1 - a^2*x^2]]/c^4

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[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 (c-a c x)^4} \, dx &=c \int \frac {\sqrt {1-a^2 x^2}}{x (c-a c x)^5} \, dx\\ &=\frac {\int \frac {(c+a c x)^5}{x \left (1-a^2 x^2\right )^{9/2}} \, dx}{c^9}\\ &=\frac {16 (1+a x)}{7 c^4 \left (1-a^2 x^2\right )^{7/2}}-\frac {\int \frac {-7 c^5-19 a c^5 x+35 a^2 c^5 x^2+7 a^3 c^5 x^3}{x \left (1-a^2 x^2\right )^{7/2}} \, dx}{7 c^9}\\ &=\frac {16 (1+a x)}{7 c^4 \left (1-a^2 x^2\right )^{7/2}}-\frac {4 (7-3 a x)}{35 c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac {\int \frac {35 c^5+83 a c^5 x}{x \left (1-a^2 x^2\right )^{5/2}} \, dx}{35 c^9}\\ &=\frac {16 (1+a x)}{7 c^4 \left (1-a^2 x^2\right )^{7/2}}-\frac {4 (7-3 a x)}{35 c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac {35+83 a x}{105 c^4 \left (1-a^2 x^2\right )^{3/2}}+\frac {\int \frac {105 a^2 c^5+166 a^3 c^5 x}{x \left (1-a^2 x^2\right )^{3/2}} \, dx}{105 a^2 c^9}\\ &=\frac {16 (1+a x)}{7 c^4 \left (1-a^2 x^2\right )^{7/2}}-\frac {4 (7-3 a x)}{35 c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac {35+83 a x}{105 c^4 \left (1-a^2 x^2\right )^{3/2}}+\frac {105+166 a x}{105 c^4 \sqrt {1-a^2 x^2}}+\frac {\int \frac {105 a^4 c^5}{x \sqrt {1-a^2 x^2}} \, dx}{105 a^4 c^9}\\ &=\frac {16 (1+a x)}{7 c^4 \left (1-a^2 x^2\right )^{7/2}}-\frac {4 (7-3 a x)}{35 c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac {35+83 a x}{105 c^4 \left (1-a^2 x^2\right )^{3/2}}+\frac {105+166 a x}{105 c^4 \sqrt {1-a^2 x^2}}+\frac {\int \frac {1}{x \sqrt {1-a^2 x^2}} \, dx}{c^4}\\ &=\frac {16 (1+a x)}{7 c^4 \left (1-a^2 x^2\right )^{7/2}}-\frac {4 (7-3 a x)}{35 c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac {35+83 a x}{105 c^4 \left (1-a^2 x^2\right )^{3/2}}+\frac {105+166 a x}{105 c^4 \sqrt {1-a^2 x^2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-a^2 x}} \, dx,x,x^2\right )}{2 c^4}\\ &=\frac {16 (1+a x)}{7 c^4 \left (1-a^2 x^2\right )^{7/2}}-\frac {4 (7-3 a x)}{35 c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac {35+83 a x}{105 c^4 \left (1-a^2 x^2\right )^{3/2}}+\frac {105+166 a x}{105 c^4 \sqrt {1-a^2 x^2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{\frac {1}{a^2}-\frac {x^2}{a^2}} \, dx,x,\sqrt {1-a^2 x^2}\right )}{a^2 c^4}\\ &=\frac {16 (1+a x)}{7 c^4 \left (1-a^2 x^2\right )^{7/2}}-\frac {4 (7-3 a x)}{35 c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac {35+83 a x}{105 c^4 \left (1-a^2 x^2\right )^{3/2}}+\frac {105+166 a x}{105 c^4 \sqrt {1-a^2 x^2}}-\frac {\tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{c^4}\\ \end {align*}

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Mathematica [C]  time = 0.18, size = 79, normalized size = 0.62 \[ \frac {-166 a^7 x^7+581 a^5 x^5-700 a^3 x^3+15 \, _2F_1\left (-\frac {7}{2},1;-\frac {5}{2};1-a^2 x^2\right )+105 a^2 x^2+525 a x+120}{105 c^4 \left (1-a^2 x^2\right )^{7/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(120 + 525*a*x + 105*a^2*x^2 - 700*a^3*x^3 + 581*a^5*x^5 - 166*a^7*x^7 + 15*Hypergeometric2F1[-7/2, 1, -5/2, 1
- a^2*x^2])/(105*c^4*(1 - a^2*x^2)^(7/2))

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fricas [A]  time = 0.46, size = 163, normalized size = 1.27 \[ \frac {296 \, a^{4} x^{4} - 1184 \, a^{3} x^{3} + 1776 \, a^{2} x^{2} - 1184 \, a x + 105 \, {\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 4 \, a x + 1\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - {\left (166 \, a^{3} x^{3} - 559 \, a^{2} x^{2} + 659 \, a x - 296\right )} \sqrt {-a^{2} x^{2} + 1} + 296}{105 \, {\left (a^{4} c^{4} x^{4} - 4 \, a^{3} c^{4} x^{3} + 6 \, a^{2} c^{4} x^{2} - 4 \, a c^{4} x + c^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(296*a^4*x^4 - 1184*a^3*x^3 + 1776*a^2*x^2 - 1184*a*x + 105*(a^4*x^4 - 4*a^3*x^3 + 6*a^2*x^2 - 4*a*x + 1
)*log((sqrt(-a^2*x^2 + 1) - 1)/x) - (166*a^3*x^3 - 559*a^2*x^2 + 659*a*x - 296)*sqrt(-a^2*x^2 + 1) + 296)/(a^4
*c^4*x^4 - 4*a^3*c^4*x^3 + 6*a^2*c^4*x^2 - 4*a*c^4*x + c^4)

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giac [B]  time = 0.26, size = 243, normalized size = 1.90 \[ -\frac {a \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{c^{4} {\left | a \right |}} + \frac {2 \, {\left (296 \, a - \frac {1547 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}}{a x} + \frac {4011 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{a^{3} x^{2}} - \frac {5600 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3}}{a^{5} x^{3}} + \frac {4760 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{4}}{a^{7} x^{4}} - \frac {2205 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{5}}{a^{9} x^{5}} + \frac {525 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{6}}{a^{11} x^{6}}\right )}}{105 \, c^{4} {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )}^{7} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/(c^4*abs(a)) + 2/105*(296*a - 1547*(sqrt(-a^2
*x^2 + 1)*abs(a) + a)/(a*x) + 4011*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2/(a^3*x^2) - 5600*(sqrt(-a^2*x^2 + 1)*abs(
a) + a)^3/(a^5*x^3) + 4760*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4/(a^7*x^4) - 2205*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^
5/(a^9*x^5) + 525*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^6/(a^11*x^6))/(c^4*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x)
- 1)^7*abs(a))

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maple [B]  time = 0.05, size = 451, normalized size = 3.52 \[ \frac {-\arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )+\frac {\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}}{a}-\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{a \left (x -\frac {1}{a}\right )}+\frac {\frac {2 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{7 a \left (x -\frac {1}{a}\right )^{4}}-\frac {6 a \left (\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{5 a \left (x -\frac {1}{a}\right )^{3}}-\frac {2 a \left (\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}\right )}{5}\right )}{7}}{a^{3}}-\frac {\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{5 a \left (x -\frac {1}{a}\right )^{3}}-\frac {2 a \left (\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}\right )}{5}}{a^{2}}}{c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^4*(-arctanh(1/(-a^2*x^2+1)^(1/2))+1/a*(1/3/a/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)-1/3/(x-1/a)*(-a^
2*(x-1/a)^2-2*a*(x-1/a))^(1/2))-1/a/(x-1/a)*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)+2/a^3*(1/7/a/(x-1/a)^4*(-a^2*(x
-1/a)^2-2*a*(x-1/a))^(1/2)-3/7*a*(1/5/a/(x-1/a)^3*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)-2/5*a*(1/3/a/(x-1/a)^2*(-
a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)-1/3/(x-1/a)*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2))))-1/a^2*(1/5/a/(x-1/a)^3*(-a^
2*(x-1/a)^2-2*a*(x-1/a))^(1/2)-2/5*a*(1/3/a/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)-1/3/(x-1/a)*(-a^2*(x-
1/a)^2-2*a*(x-1/a))^(1/2))))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1} {\left (a c x - c\right )}^{4} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.83, size = 327, normalized size = 2.55 \[ \frac {7\,a^2\,\sqrt {1-a^2\,x^2}}{15\,\left (a^4\,c^4\,x^2-2\,a^3\,c^4\,x+a^2\,c^4\right )}+\frac {4\,a^4\,\sqrt {1-a^2\,x^2}}{35\,\left (a^6\,c^4\,x^2-2\,a^5\,c^4\,x+a^4\,c^4\right )}+\frac {2\,a^2\,\sqrt {1-a^2\,x^2}}{7\,\left (a^6\,c^4\,x^4-4\,a^5\,c^4\,x^3+6\,a^4\,c^4\,x^2-4\,a^3\,c^4\,x+a^2\,c^4\right )}+\frac {166\,a\,\sqrt {1-a^2\,x^2}}{105\,\sqrt {-a^2}\,\left (c^4\,x\,\sqrt {-a^2}-\frac {c^4\,\sqrt {-a^2}}{a}\right )}+\frac {13\,a\,\sqrt {1-a^2\,x^2}}{35\,\sqrt {-a^2}\,\left (3\,c^4\,x\,\sqrt {-a^2}-\frac {c^4\,\sqrt {-a^2}}{a}+a^2\,c^4\,x^3\,\sqrt {-a^2}-3\,a\,c^4\,x^2\,\sqrt {-a^2}\right )}+\frac {\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{c^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((1 - a^2*x^2)^(1/2)*1i)*1i)/c^4 + (7*a^2*(1 - a^2*x^2)^(1/2))/(15*(a^2*c^4 - 2*a^3*c^4*x + a^4*c^4*x^2))
+ (4*a^4*(1 - a^2*x^2)^(1/2))/(35*(a^4*c^4 - 2*a^5*c^4*x + a^6*c^4*x^2)) + (2*a^2*(1 - a^2*x^2)^(1/2))/(7*(a^
2*c^4 - 4*a^3*c^4*x + 6*a^4*c^4*x^2 - 4*a^5*c^4*x^3 + a^6*c^4*x^4)) + (166*a*(1 - a^2*x^2)^(1/2))/(105*(-a^2)^
(1/2)*(c^4*x*(-a^2)^(1/2) - (c^4*(-a^2)^(1/2))/a)) + (13*a*(1 - a^2*x^2)^(1/2))/(35*(-a^2)^(1/2)*(3*c^4*x*(-a^
2)^(1/2) - (c^4*(-a^2)^(1/2))/a + a^2*c^4*x^3*(-a^2)^(1/2) - 3*a*c^4*x^2*(-a^2)^(1/2)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a x}{a^{4} x^{5} \sqrt {- a^{2} x^{2} + 1} - 4 a^{3} x^{4} \sqrt {- a^{2} x^{2} + 1} + 6 a^{2} x^{3} \sqrt {- a^{2} x^{2} + 1} - 4 a x^{2} \sqrt {- a^{2} x^{2} + 1} + x \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {1}{a^{4} x^{5} \sqrt {- a^{2} x^{2} + 1} - 4 a^{3} x^{4} \sqrt {- a^{2} x^{2} + 1} + 6 a^{2} x^{3} \sqrt {- a^{2} x^{2} + 1} - 4 a x^{2} \sqrt {- a^{2} x^{2} + 1} + x \sqrt {- a^{2} x^{2} + 1}}\, dx}{c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a*x/(a**4*x**5*sqrt(-a**2*x**2 + 1) - 4*a**3*x**4*sqrt(-a**2*x**2 + 1) + 6*a**2*x**3*sqrt(-a**2*x**2
+ 1) - 4*a*x**2*sqrt(-a**2*x**2 + 1) + x*sqrt(-a**2*x**2 + 1)), x) + Integral(1/(a**4*x**5*sqrt(-a**2*x**2 +
1) - 4*a**3*x**4*sqrt(-a**2*x**2 + 1) + 6*a**2*x**3*sqrt(-a**2*x**2 + 1) - 4*a*x**2*sqrt(-a**2*x**2 + 1) + x*s
qrt(-a**2*x**2 + 1)), x))/c**4

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