3.544 \(\int e^{-2 \tanh ^{-1}(a x)} (c-\frac {c}{a x})^{9/2} \, dx\)

Optimal. Leaf size=189 \[ \frac {13 c^{9/2} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a}-\frac {64 \sqrt {2} c^{9/2} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right )}{a}+\frac {51 c^4 \sqrt {c-\frac {c}{a x}}}{a}+\frac {19 c^3 \left (c-\frac {c}{a x}\right )^{3/2}}{3 a}+\frac {3 c^2 \left (c-\frac {c}{a x}\right )^{5/2}}{5 a}-\frac {5 c \left (c-\frac {c}{a x}\right )^{7/2}}{7 a}-x \left (c-\frac {c}{a x}\right )^{9/2} \]

[Out]

19/3*c^3*(c-c/a/x)^(3/2)/a+3/5*c^2*(c-c/a/x)^(5/2)/a-5/7*c*(c-c/a/x)^(7/2)/a-(c-c/a/x)^(9/2)*x+13*c^(9/2)*arct
anh((c-c/a/x)^(1/2)/c^(1/2))/a-64*c^(9/2)*arctanh(1/2*(c-c/a/x)^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)/a+51*c^4*(c-c/a
/x)^(1/2)/a

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Rubi [A]  time = 0.26, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6133, 25, 514, 375, 98, 154, 156, 63, 208} \[ \frac {51 c^4 \sqrt {c-\frac {c}{a x}}}{a}+\frac {19 c^3 \left (c-\frac {c}{a x}\right )^{3/2}}{3 a}+\frac {3 c^2 \left (c-\frac {c}{a x}\right )^{5/2}}{5 a}+\frac {13 c^{9/2} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a}-\frac {64 \sqrt {2} c^{9/2} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right )}{a}-\frac {5 c \left (c-\frac {c}{a x}\right )^{7/2}}{7 a}-x \left (c-\frac {c}{a x}\right )^{9/2} \]

Antiderivative was successfully verified.

[In]

Int[(c - c/(a*x))^(9/2)/E^(2*ArcTanh[a*x]),x]

[Out]

(51*c^4*Sqrt[c - c/(a*x)])/a + (19*c^3*(c - c/(a*x))^(3/2))/(3*a) + (3*c^2*(c - c/(a*x))^(5/2))/(5*a) - (5*c*(
c - c/(a*x))^(7/2))/(7*a) - (c - c/(a*x))^(9/2)*x + (13*c^(9/2)*ArcTanh[Sqrt[c - c/(a*x)]/Sqrt[c]])/a - (64*Sq
rt[2]*c^(9/2)*ArcTanh[Sqrt[c - c/(a*x)]/(Sqrt[2]*Sqrt[c])])/a

Rule 25

Int[(u_.)*((a_) + (b_.)*(x_)^(n_.))^(m_.)*((c_) + (d_.)*(x_)^(q_.))^(p_.), x_Symbol] :> Dist[(d/a)^p, Int[(u*(
a + b*x^n)^(m + p))/x^(n*p), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[q, -n] && IntegerQ[p] && EqQ[a*c -
b*d, 0] &&  !(IntegerQ[m] && NegQ[n])

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 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 154

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2
*n, 2*p]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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 375

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Subst[Int[((a + b/x^n)^p*(c +
d/x^n)^q)/x^2, x], x, 1/x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]

Rule 514

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[x^(m - n*q)*
(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] |
|  !IntegerQ[p])

Rule 6133

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

Rubi steps

\begin {align*} \int e^{-2 \tanh ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^{9/2} \, dx &=\int \frac {\left (c-\frac {c}{a x}\right )^{9/2} (1-a x)}{1+a x} \, dx\\ &=-\frac {a \int \frac {\left (c-\frac {c}{a x}\right )^{11/2} x}{1+a x} \, dx}{c}\\ &=-\frac {a \int \frac {\left (c-\frac {c}{a x}\right )^{11/2}}{a+\frac {1}{x}} \, dx}{c}\\ &=\frac {a \operatorname {Subst}\left (\int \frac {\left (c-\frac {c x}{a}\right )^{11/2}}{x^2 (a+x)} \, dx,x,\frac {1}{x}\right )}{c}\\ &=-\left (c-\frac {c}{a x}\right )^{9/2} x-\frac {\operatorname {Subst}\left (\int \frac {\left (c-\frac {c x}{a}\right )^{7/2} \left (\frac {13 c^2}{2}+\frac {5 c^2 x}{2 a}\right )}{x (a+x)} \, dx,x,\frac {1}{x}\right )}{c}\\ &=-\frac {5 c \left (c-\frac {c}{a x}\right )^{7/2}}{7 a}-\left (c-\frac {c}{a x}\right )^{9/2} x-\frac {2 \operatorname {Subst}\left (\int \frac {\left (c-\frac {c x}{a}\right )^{5/2} \left (\frac {91 c^3}{4}-\frac {21 c^3 x}{4 a}\right )}{x (a+x)} \, dx,x,\frac {1}{x}\right )}{7 c}\\ &=\frac {3 c^2 \left (c-\frac {c}{a x}\right )^{5/2}}{5 a}-\frac {5 c \left (c-\frac {c}{a x}\right )^{7/2}}{7 a}-\left (c-\frac {c}{a x}\right )^{9/2} x-\frac {4 \operatorname {Subst}\left (\int \frac {\left (c-\frac {c x}{a}\right )^{3/2} \left (\frac {455 c^4}{8}-\frac {665 c^4 x}{8 a}\right )}{x (a+x)} \, dx,x,\frac {1}{x}\right )}{35 c}\\ &=\frac {19 c^3 \left (c-\frac {c}{a x}\right )^{3/2}}{3 a}+\frac {3 c^2 \left (c-\frac {c}{a x}\right )^{5/2}}{5 a}-\frac {5 c \left (c-\frac {c}{a x}\right )^{7/2}}{7 a}-\left (c-\frac {c}{a x}\right )^{9/2} x-\frac {8 \operatorname {Subst}\left (\int \frac {\sqrt {c-\frac {c x}{a}} \left (\frac {1365 c^5}{16}-\frac {5355 c^5 x}{16 a}\right )}{x (a+x)} \, dx,x,\frac {1}{x}\right )}{105 c}\\ &=\frac {51 c^4 \sqrt {c-\frac {c}{a x}}}{a}+\frac {19 c^3 \left (c-\frac {c}{a x}\right )^{3/2}}{3 a}+\frac {3 c^2 \left (c-\frac {c}{a x}\right )^{5/2}}{5 a}-\frac {5 c \left (c-\frac {c}{a x}\right )^{7/2}}{7 a}-\left (c-\frac {c}{a x}\right )^{9/2} x-\frac {16 \operatorname {Subst}\left (\int \frac {\frac {1365 c^6}{32}-\frac {12075 c^6 x}{32 a}}{x (a+x) \sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )}{105 c}\\ &=\frac {51 c^4 \sqrt {c-\frac {c}{a x}}}{a}+\frac {19 c^3 \left (c-\frac {c}{a x}\right )^{3/2}}{3 a}+\frac {3 c^2 \left (c-\frac {c}{a x}\right )^{5/2}}{5 a}-\frac {5 c \left (c-\frac {c}{a x}\right )^{7/2}}{7 a}-\left (c-\frac {c}{a x}\right )^{9/2} x-\frac {\left (13 c^5\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )}{2 a}+\frac {\left (64 c^5\right ) \operatorname {Subst}\left (\int \frac {1}{(a+x) \sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )}{a}\\ &=\frac {51 c^4 \sqrt {c-\frac {c}{a x}}}{a}+\frac {19 c^3 \left (c-\frac {c}{a x}\right )^{3/2}}{3 a}+\frac {3 c^2 \left (c-\frac {c}{a x}\right )^{5/2}}{5 a}-\frac {5 c \left (c-\frac {c}{a x}\right )^{7/2}}{7 a}-\left (c-\frac {c}{a x}\right )^{9/2} x+\left (13 c^4\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-\frac {c}{a x}}\right )-\left (128 c^4\right ) \operatorname {Subst}\left (\int \frac {1}{2 a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-\frac {c}{a x}}\right )\\ &=\frac {51 c^4 \sqrt {c-\frac {c}{a x}}}{a}+\frac {19 c^3 \left (c-\frac {c}{a x}\right )^{3/2}}{3 a}+\frac {3 c^2 \left (c-\frac {c}{a x}\right )^{5/2}}{5 a}-\frac {5 c \left (c-\frac {c}{a x}\right )^{7/2}}{7 a}-\left (c-\frac {c}{a x}\right )^{9/2} x+\frac {13 c^{9/2} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a}-\frac {64 \sqrt {2} c^{9/2} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right )}{a}\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 133, normalized size = 0.70 \[ \frac {c^4 \left (-105 a^4 x^4+6428 a^3 x^3-1196 a^2 x^2+258 a x-30\right ) \sqrt {c-\frac {c}{a x}}}{105 a^4 x^3}+\frac {13 c^{9/2} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a}-\frac {64 \sqrt {2} c^{9/2} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right )}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - c/(a*x))^(9/2)/E^(2*ArcTanh[a*x]),x]

[Out]

(c^4*Sqrt[c - c/(a*x)]*(-30 + 258*a*x - 1196*a^2*x^2 + 6428*a^3*x^3 - 105*a^4*x^4))/(105*a^4*x^3) + (13*c^(9/2
)*ArcTanh[Sqrt[c - c/(a*x)]/Sqrt[c]])/a - (64*Sqrt[2]*c^(9/2)*ArcTanh[Sqrt[c - c/(a*x)]/(Sqrt[2]*Sqrt[c])])/a

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fricas [A]  time = 0.59, size = 342, normalized size = 1.81 \[ \left [\frac {6720 \, \sqrt {2} a^{3} c^{\frac {9}{2}} x^{3} \log \left (\frac {2 \, \sqrt {2} a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}} - 3 \, a c x + c}{a x + 1}\right ) + 1365 \, a^{3} c^{\frac {9}{2}} x^{3} \log \left (-2 \, a c x - 2 \, a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}} + c\right ) - 2 \, {\left (105 \, a^{4} c^{4} x^{4} - 6428 \, a^{3} c^{4} x^{3} + 1196 \, a^{2} c^{4} x^{2} - 258 \, a c^{4} x + 30 \, c^{4}\right )} \sqrt {\frac {a c x - c}{a x}}}{210 \, a^{4} x^{3}}, \frac {6720 \, \sqrt {2} a^{3} \sqrt {-c} c^{4} x^{3} \arctan \left (\frac {\sqrt {2} \sqrt {-c} \sqrt {\frac {a c x - c}{a x}}}{2 \, c}\right ) - 1365 \, a^{3} \sqrt {-c} c^{4} x^{3} \arctan \left (\frac {\sqrt {-c} \sqrt {\frac {a c x - c}{a x}}}{c}\right ) - {\left (105 \, a^{4} c^{4} x^{4} - 6428 \, a^{3} c^{4} x^{3} + 1196 \, a^{2} c^{4} x^{2} - 258 \, a c^{4} x + 30 \, c^{4}\right )} \sqrt {\frac {a c x - c}{a x}}}{105 \, a^{4} x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^(9/2)/(a*x+1)^2*(-a^2*x^2+1),x, algorithm="fricas")

[Out]

[1/210*(6720*sqrt(2)*a^3*c^(9/2)*x^3*log((2*sqrt(2)*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x)) - 3*a*c*x + c)/(a*x +
1)) + 1365*a^3*c^(9/2)*x^3*log(-2*a*c*x - 2*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x)) + c) - 2*(105*a^4*c^4*x^4 - 64
28*a^3*c^4*x^3 + 1196*a^2*c^4*x^2 - 258*a*c^4*x + 30*c^4)*sqrt((a*c*x - c)/(a*x)))/(a^4*x^3), 1/105*(6720*sqrt
(2)*a^3*sqrt(-c)*c^4*x^3*arctan(1/2*sqrt(2)*sqrt(-c)*sqrt((a*c*x - c)/(a*x))/c) - 1365*a^3*sqrt(-c)*c^4*x^3*ar
ctan(sqrt(-c)*sqrt((a*c*x - c)/(a*x))/c) - (105*a^4*c^4*x^4 - 6428*a^3*c^4*x^3 + 1196*a^2*c^4*x^2 - 258*a*c^4*
x + 30*c^4)*sqrt((a*c*x - c)/(a*x)))/(a^4*x^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((c-c/a/x)^(9/2)/(a*x+1)^2*(-a^2*x^2+1),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x)]sym2poly/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 = 305, normalized size = 1.61 \[ -\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, c^{4} \left (-17430 \sqrt {a \,x^{2}-x}\, \sqrt {\frac {1}{a}}\, a^{\frac {9}{2}} x^{5}+6720 \sqrt {\frac {1}{a}}\, a^{\frac {9}{2}} \sqrt {\left (a x -1\right ) x}\, x^{5}+10920 \left (a \,x^{2}-x \right )^{\frac {3}{2}} \sqrt {\frac {1}{a}}\, a^{\frac {7}{2}} x^{3}+8715 \ln \left (\frac {2 \sqrt {a \,x^{2}-x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) \sqrt {\frac {1}{a}}\, x^{5} a^{4}-10080 \ln \left (\frac {2 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) \sqrt {\frac {1}{a}}\, x^{5} a^{4}-6720 \ln \left (\frac {2 \sqrt {2}\, \sqrt {\frac {1}{a}}\, \sqrt {\left (a x -1\right ) x}\, a -3 a x +1}{a x +1}\right ) \sqrt {2}\, a^{\frac {7}{2}} x^{5}-1936 a^{\frac {5}{2}} \left (a \,x^{2}-x \right )^{\frac {3}{2}} x^{2} \sqrt {\frac {1}{a}}+456 a^{\frac {3}{2}} \left (a \,x^{2}-x \right )^{\frac {3}{2}} x \sqrt {\frac {1}{a}}-60 \left (a \,x^{2}-x \right )^{\frac {3}{2}} \sqrt {a}\, \sqrt {\frac {1}{a}}\right )}{210 x^{4} a^{\frac {9}{2}} \sqrt {\left (a x -1\right ) x}\, \sqrt {\frac {1}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c/a/x)^(9/2)/(a*x+1)^2*(-a^2*x^2+1),x)

[Out]

-1/210*(c*(a*x-1)/a/x)^(1/2)/x^4*c^4/a^(9/2)*(-17430*(a*x^2-x)^(1/2)*(1/a)^(1/2)*a^(9/2)*x^5+6720*(1/a)^(1/2)*
a^(9/2)*((a*x-1)*x)^(1/2)*x^5+10920*(a*x^2-x)^(3/2)*(1/a)^(1/2)*a^(7/2)*x^3+8715*ln(1/2*(2*(a*x^2-x)^(1/2)*a^(
1/2)+2*a*x-1)/a^(1/2))*(1/a)^(1/2)*x^5*a^4-10080*ln(1/2*(2*((a*x-1)*x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*(1/a)^(
1/2)*x^5*a^4-6720*ln((2*2^(1/2)*(1/a)^(1/2)*((a*x-1)*x)^(1/2)*a-3*a*x+1)/(a*x+1))*2^(1/2)*a^(7/2)*x^5-1936*a^(
5/2)*(a*x^2-x)^(3/2)*x^2*(1/a)^(1/2)+456*a^(3/2)*(a*x^2-x)^(3/2)*x*(1/a)^(1/2)-60*(a*x^2-x)^(3/2)*a^(1/2)*(1/a
)^(1/2))/((a*x-1)*x)^(1/2)/(1/a)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {{\left (a^{2} x^{2} - 1\right )} {\left (c - \frac {c}{a x}\right )}^{\frac {9}{2}}}{{\left (a x + 1\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^(9/2)/(a*x+1)^2*(-a^2*x^2+1),x, algorithm="maxima")

[Out]

-integrate((a^2*x^2 - 1)*(c - c/(a*x))^(9/2)/(a*x + 1)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {{\left (c-\frac {c}{a\,x}\right )}^{9/2}\,\left (a^2\,x^2-1\right )}{{\left (a\,x+1\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((c - c/(a*x))^(9/2)*(a^2*x^2 - 1))/(a*x + 1)^2,x)

[Out]

-int(((c - c/(a*x))^(9/2)*(a^2*x^2 - 1))/(a*x + 1)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \left (- \frac {5 c^{4} \sqrt {c - \frac {c}{a x}}}{a x + 1}\right )\, dx - \int \frac {10 c^{4} \sqrt {c - \frac {c}{a x}}}{a^{2} x^{2} + a x}\, dx - \int \left (- \frac {10 c^{4} \sqrt {c - \frac {c}{a x}}}{a^{3} x^{3} + a^{2} x^{2}}\right )\, dx - \int \frac {5 c^{4} \sqrt {c - \frac {c}{a x}}}{a^{4} x^{4} + a^{3} x^{3}}\, dx - \int \left (- \frac {c^{4} \sqrt {c - \frac {c}{a x}}}{a^{5} x^{5} + a^{4} x^{4}}\right )\, dx - \int \frac {a c^{4} x \sqrt {c - \frac {c}{a x}}}{a x + 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)**(9/2)/(a*x+1)**2*(-a**2*x**2+1),x)

[Out]

-Integral(-5*c**4*sqrt(c - c/(a*x))/(a*x + 1), x) - Integral(10*c**4*sqrt(c - c/(a*x))/(a**2*x**2 + a*x), x) -
Integral(-10*c**4*sqrt(c - c/(a*x))/(a**3*x**3 + a**2*x**2), x) - Integral(5*c**4*sqrt(c - c/(a*x))/(a**4*x**
4 + a**3*x**3), x) - Integral(-c**4*sqrt(c - c/(a*x))/(a**5*x**5 + a**4*x**4), x) - Integral(a*c**4*x*sqrt(c -
c/(a*x))/(a*x + 1), x)

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