3.6.45 \(\int e^{-2 \tanh ^{-1}(a x)} (c-\frac {c}{a x})^{7/2} \, dx\) [545]

Optimal. Leaf size=164 \[ \frac {21 c^3 \sqrt {c-\frac {c}{a x}}}{a}+\frac {5 c^2 \left (c-\frac {c}{a x}\right )^{3/2}}{3 a}-\frac {3 c \left (c-\frac {c}{a x}\right )^{5/2}}{5 a}-\left (c-\frac {c}{a x}\right )^{7/2} x+\frac {11 c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a}-\frac {32 \sqrt {2} c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right )}{a} \]

[Out]

5/3*c^2*(c-c/a/x)^(3/2)/a-3/5*c*(c-c/a/x)^(5/2)/a-(c-c/a/x)^(7/2)*x+11*c^(7/2)*arctanh((c-c/a/x)^(1/2)/c^(1/2)
)/a-32*c^(7/2)*arctanh(1/2*(c-c/a/x)^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)/a+21*c^3*(c-c/a/x)^(1/2)/a

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Rubi [A]
time = 0.17, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6268, 25, 528, 382, 100, 159, 162, 65, 214} \begin {gather*} \frac {11 c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a}-\frac {32 \sqrt {2} c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right )}{a}+\frac {21 c^3 \sqrt {c-\frac {c}{a x}}}{a}+\frac {5 c^2 \left (c-\frac {c}{a x}\right )^{3/2}}{3 a}-\frac {3 c \left (c-\frac {c}{a x}\right )^{5/2}}{5 a}-x \left (c-\frac {c}{a x}\right )^{7/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c - c/(a*x))^(7/2)/E^(2*ArcTanh[a*x]),x]

[Out]

(21*c^3*Sqrt[c - c/(a*x)])/a + (5*c^2*(c - c/(a*x))^(3/2))/(3*a) - (3*c*(c - c/(a*x))^(5/2))/(5*a) - (c - c/(a
*x))^(7/2)*x + (11*c^(7/2)*ArcTanh[Sqrt[c - c/(a*x)]/Sqrt[c]])/a - (32*Sqrt[2]*c^(7/2)*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 65

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 100

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 159

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 162

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 214

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

Rule 382

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 528

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 6268

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 )^{7/2} \, dx &=\int \frac {\left (c-\frac {c}{a x}\right )^{7/2} (1-a x)}{1+a x} \, dx\\ &=-\frac {a \int \frac {\left (c-\frac {c}{a x}\right )^{9/2} x}{1+a x} \, dx}{c}\\ &=-\frac {a \int \frac {\left (c-\frac {c}{a x}\right )^{9/2}}{a+\frac {1}{x}} \, dx}{c}\\ &=\frac {a \text {Subst}\left (\int \frac {\left (c-\frac {c x}{a}\right )^{9/2}}{x^2 (a+x)} \, dx,x,\frac {1}{x}\right )}{c}\\ &=-\left (c-\frac {c}{a x}\right )^{7/2} x-\frac {\text {Subst}\left (\int \frac {\left (c-\frac {c x}{a}\right )^{5/2} \left (\frac {11 c^2}{2}+\frac {3 c^2 x}{2 a}\right )}{x (a+x)} \, dx,x,\frac {1}{x}\right )}{c}\\ &=-\frac {3 c \left (c-\frac {c}{a x}\right )^{5/2}}{5 a}-\left (c-\frac {c}{a x}\right )^{7/2} x-\frac {2 \text {Subst}\left (\int \frac {\left (c-\frac {c x}{a}\right )^{3/2} \left (\frac {55 c^3}{4}-\frac {25 c^3 x}{4 a}\right )}{x (a+x)} \, dx,x,\frac {1}{x}\right )}{5 c}\\ &=\frac {5 c^2 \left (c-\frac {c}{a x}\right )^{3/2}}{3 a}-\frac {3 c \left (c-\frac {c}{a x}\right )^{5/2}}{5 a}-\left (c-\frac {c}{a x}\right )^{7/2} x-\frac {4 \text {Subst}\left (\int \frac {\sqrt {c-\frac {c x}{a}} \left (\frac {165 c^4}{8}-\frac {315 c^4 x}{8 a}\right )}{x (a+x)} \, dx,x,\frac {1}{x}\right )}{15 c}\\ &=\frac {21 c^3 \sqrt {c-\frac {c}{a x}}}{a}+\frac {5 c^2 \left (c-\frac {c}{a x}\right )^{3/2}}{3 a}-\frac {3 c \left (c-\frac {c}{a x}\right )^{5/2}}{5 a}-\left (c-\frac {c}{a x}\right )^{7/2} x-\frac {8 \text {Subst}\left (\int \frac {\frac {165 c^5}{16}-\frac {795 c^5 x}{16 a}}{x (a+x) \sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )}{15 c}\\ &=\frac {21 c^3 \sqrt {c-\frac {c}{a x}}}{a}+\frac {5 c^2 \left (c-\frac {c}{a x}\right )^{3/2}}{3 a}-\frac {3 c \left (c-\frac {c}{a x}\right )^{5/2}}{5 a}-\left (c-\frac {c}{a x}\right )^{7/2} x-\frac {\left (11 c^4\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )}{2 a}+\frac {\left (32 c^4\right ) \text {Subst}\left (\int \frac {1}{(a+x) \sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )}{a}\\ &=\frac {21 c^3 \sqrt {c-\frac {c}{a x}}}{a}+\frac {5 c^2 \left (c-\frac {c}{a x}\right )^{3/2}}{3 a}-\frac {3 c \left (c-\frac {c}{a x}\right )^{5/2}}{5 a}-\left (c-\frac {c}{a x}\right )^{7/2} x+\left (11 c^3\right ) \text {Subst}\left (\int \frac {1}{a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-\frac {c}{a x}}\right )-\left (64 c^3\right ) \text {Subst}\left (\int \frac {1}{2 a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-\frac {c}{a x}}\right )\\ &=\frac {21 c^3 \sqrt {c-\frac {c}{a x}}}{a}+\frac {5 c^2 \left (c-\frac {c}{a x}\right )^{3/2}}{3 a}-\frac {3 c \left (c-\frac {c}{a x}\right )^{5/2}}{5 a}-\left (c-\frac {c}{a x}\right )^{7/2} x+\frac {11 c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a}-\frac {32 \sqrt {2} c^{7/2} \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 = 0.13, size = 125, normalized size = 0.76 \begin {gather*} \frac {c^3 \sqrt {c-\frac {c}{a x}} \left (6-52 a x+376 a^2 x^2-15 a^3 x^3\right )}{15 a^3 x^2}+\frac {11 c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a}-\frac {32 \sqrt {2} c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c - c/(a*x))^(7/2)/E^(2*ArcTanh[a*x]),x]

[Out]

(c^3*Sqrt[c - c/(a*x)]*(6 - 52*a*x + 376*a^2*x^2 - 15*a^3*x^3))/(15*a^3*x^2) + (11*c^(7/2)*ArcTanh[Sqrt[c - c/
(a*x)]/Sqrt[c]])/a - (32*Sqrt[2]*c^(7/2)*ArcTanh[Sqrt[c - c/(a*x)]/(Sqrt[2]*Sqrt[c])])/a

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(280\) vs. \(2(137)=274\).
time = 0.40, size = 281, normalized size = 1.71

method	result	size
risch	\(-\frac {\left (15 a^{4} x^{4}-391 a^{3} x^{3}+428 a^{2} x^{2}-58 a x +6\right ) c^{3} \sqrt {\frac {c \left (a x -1\right )}{a x}}}{15 x^{2} a^{3} \left (a x -1\right )}-\frac {\left (-\frac {11 a^{3} \ln \left (\frac {-\frac {1}{2} a c +a^{2} c x}{\sqrt {a^{2} c}}+\sqrt {a^{2} c \,x^{2}-c x a}\right )}{2 \sqrt {a^{2} c}}-\frac {16 a^{2} \sqrt {2}\, \ln \left (\frac {4 c -3 \left (x +\frac {1}{a}\right ) a c +2 \sqrt {2}\, \sqrt {c}\, \sqrt {a^{2} c \left (x +\frac {1}{a}\right )^{2}-3 \left (x +\frac {1}{a}\right ) a c +2 c}}{x +\frac {1}{a}}\right )}{\sqrt {c}}\right ) c^{3} \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {a c x \left (a x -1\right )}}{a^{3} \left (a x -1\right )}\)	\(224\)
default	\(-\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, c^{3} \left (480 a^{\frac {7}{2}} \sqrt {\left (a x -1\right ) x}\, \sqrt {\frac {1}{a}}\, x^{4}-1110 a^{\frac {7}{2}} \sqrt {a \,x^{2}-x}\, \sqrt {\frac {1}{a}}\, x^{4}-480 a^{\frac {5}{2}} \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {\frac {1}{a}}\, \sqrt {\left (a x -1\right ) x}\, a -3 a x +1}{a x +1}\right ) x^{4}+660 a^{\frac {5}{2}} \left (a \,x^{2}-x \right )^{\frac {3}{2}} x^{2} \sqrt {\frac {1}{a}}+555 \ln \left (\frac {2 \sqrt {a \,x^{2}-x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) \sqrt {\frac {1}{a}}\, a^{3} x^{4}-720 \ln \left (\frac {2 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) \sqrt {\frac {1}{a}}\, a^{3} x^{4}-92 a^{\frac {3}{2}} \left (a \,x^{2}-x \right )^{\frac {3}{2}} x \sqrt {\frac {1}{a}}+12 \left (a \,x^{2}-x \right )^{\frac {3}{2}} \sqrt {a}\, \sqrt {\frac {1}{a}}\right )}{30 x^{3} a^{\frac {7}{2}} \sqrt {\left (a x -1\right ) x}\, \sqrt {\frac {1}{a}}}\)	\(281\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c/a/x)^(7/2)/(a*x+1)^2*(-a^2*x^2+1),x,method=_RETURNVERBOSE)

[Out]

-1/30*(c*(a*x-1)/a/x)^(1/2)*c^3*(480*a^(7/2)*((a*x-1)*x)^(1/2)*(1/a)^(1/2)*x^4-1110*a^(7/2)*(a*x^2-x)^(1/2)*(1
/a)^(1/2)*x^4-480*a^(5/2)*2^(1/2)*ln((2*2^(1/2)*(1/a)^(1/2)*((a*x-1)*x)^(1/2)*a-3*a*x+1)/(a*x+1))*x^4+660*a^(5
/2)*(a*x^2-x)^(3/2)*x^2*(1/a)^(1/2)+555*ln(1/2*(2*(a*x^2-x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*(1/a)^(1/2)*a^3*x^
4-720*ln(1/2*(2*((a*x-1)*x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*(1/a)^(1/2)*a^3*x^4-92*a^(3/2)*(a*x^2-x)^(3/2)*x*(
1/a)^(1/2)+12*(a*x^2-x)^(3/2)*a^(1/2)*(1/a)^(1/2))/x^3/a^(7/2)/((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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^(7/2)/(a*x+1)^2*(-a^2*x^2+1),x, algorithm="maxima")

[Out]

-integrate((a^2*x^2 - 1)*(c - c/(a*x))^(7/2)/(a*x + 1)^2, x)

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Fricas [A]
time = 0.39, size = 320, normalized size = 1.95 \begin {gather*} \left [\frac {480 \, \sqrt {2} a^{2} c^{\frac {7}{2}} x^{2} \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 ) + 165 \, a^{2} c^{\frac {7}{2}} x^{2} \log \left (-2 \, a c x - 2 \, a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}} + c\right ) - 2 \, {\left (15 \, a^{3} c^{3} x^{3} - 376 \, a^{2} c^{3} x^{2} + 52 \, a c^{3} x - 6 \, c^{3}\right )} \sqrt {\frac {a c x - c}{a x}}}{30 \, a^{3} x^{2}}, \frac {480 \, \sqrt {2} a^{2} \sqrt {-c} c^{3} x^{2} \arctan \left (\frac {\sqrt {2} \sqrt {-c} \sqrt {\frac {a c x - c}{a x}}}{2 \, c}\right ) - 165 \, a^{2} \sqrt {-c} c^{3} x^{2} \arctan \left (\frac {\sqrt {-c} \sqrt {\frac {a c x - c}{a x}}}{c}\right ) - {\left (15 \, a^{3} c^{3} x^{3} - 376 \, a^{2} c^{3} x^{2} + 52 \, a c^{3} x - 6 \, c^{3}\right )} \sqrt {\frac {a c x - c}{a x}}}{15 \, a^{3} x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^(7/2)/(a*x+1)^2*(-a^2*x^2+1),x, algorithm="fricas")

[Out]

[1/30*(480*sqrt(2)*a^2*c^(7/2)*x^2*log((2*sqrt(2)*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x)) - 3*a*c*x + c)/(a*x + 1)
) + 165*a^2*c^(7/2)*x^2*log(-2*a*c*x - 2*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x)) + c) - 2*(15*a^3*c^3*x^3 - 376*a^
2*c^3*x^2 + 52*a*c^3*x - 6*c^3)*sqrt((a*c*x - c)/(a*x)))/(a^3*x^2), 1/15*(480*sqrt(2)*a^2*sqrt(-c)*c^3*x^2*arc
tan(1/2*sqrt(2)*sqrt(-c)*sqrt((a*c*x - c)/(a*x))/c) - 165*a^2*sqrt(-c)*c^3*x^2*arctan(sqrt(-c)*sqrt((a*c*x - c
)/(a*x))/c) - (15*a^3*c^3*x^3 - 376*a^2*c^3*x^2 + 52*a*c^3*x - 6*c^3)*sqrt((a*c*x - c)/(a*x)))/(a^3*x^2)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \left (- \frac {4 c^{3} \sqrt {c - \frac {c}{a x}}}{a x + 1}\right )\, dx - \int \frac {6 c^{3} \sqrt {c - \frac {c}{a x}}}{a^{2} x^{2} + a x}\, dx - \int \left (- \frac {4 c^{3} \sqrt {c - \frac {c}{a x}}}{a^{3} x^{3} + a^{2} x^{2}}\right )\, dx - \int \frac {c^{3} \sqrt {c - \frac {c}{a x}}}{a^{4} x^{4} + a^{3} x^{3}}\, dx - \int \frac {a c^{3} x \sqrt {c - \frac {c}{a x}}}{a x + 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)**(7/2)/(a*x+1)**2*(-a**2*x**2+1),x)

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^(7/2)/(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,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {{\left (c-\frac {c}{a\,x}\right )}^{7/2}\,\left (a^2\,x^2-1\right )}{{\left (a\,x+1\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((c - c/(a*x))^(7/2)*(a^2*x^2 - 1))/(a*x + 1)^2,x)

[Out]

-int(((c - c/(a*x))^(7/2)*(a^2*x^2 - 1))/(a*x + 1)^2, x)

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