3.1106 \(\int \frac {e^{2 \tanh ^{-1}(a x)} (c-a^2 c x^2)^{5/2}}{x^5} \, dx\)

Optimal. Leaf size=155 \[ -\frac {\left (c-a^2 c x^2\right )^{5/2}}{4 x^4}-\frac {a c (9 a x+16) \left (c-a^2 c x^2\right )^{3/2}}{24 x^3}+2 a^4 c^{5/2} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )+\frac {9}{8} a^4 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )+\frac {a^3 c^2 (16-9 a x) \sqrt {c-a^2 c x^2}}{8 x} \]

[Out]

-1/24*a*c*(9*a*x+16)*(-a^2*c*x^2+c)^(3/2)/x^3-1/4*(-a^2*c*x^2+c)^(5/2)/x^4+2*a^4*c^(5/2)*arctan(a*x*c^(1/2)/(-
a^2*c*x^2+c)^(1/2))+9/8*a^4*c^(5/2)*arctanh((-a^2*c*x^2+c)^(1/2)/c^(1/2))+1/8*a^3*c^2*(-9*a*x+16)*(-a^2*c*x^2+
c)^(1/2)/x

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Rubi [A]  time = 0.33, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {6151, 1807, 811, 813, 844, 217, 203, 266, 63, 208} \[ \frac {a^3 c^2 (16-9 a x) \sqrt {c-a^2 c x^2}}{8 x}+2 a^4 c^{5/2} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )+\frac {9}{8} a^4 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )-\frac {a c (9 a x+16) \left (c-a^2 c x^2\right )^{3/2}}{24 x^3}-\frac {\left (c-a^2 c x^2\right )^{5/2}}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Int[(E^(2*ArcTanh[a*x])*(c - a^2*c*x^2)^(5/2))/x^5,x]

[Out]

(a^3*c^2*(16 - 9*a*x)*Sqrt[c - a^2*c*x^2])/(8*x) - (a*c*(16 + 9*a*x)*(c - a^2*c*x^2)^(3/2))/(24*x^3) - (c - a^
2*c*x^2)^(5/2)/(4*x^4) + 2*a^4*c^(5/2)*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2]] + (9*a^4*c^(5/2)*ArcTanh[Sqrt
[c - a^2*c*x^2]/Sqrt[c]])/8

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 203

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

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 217

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

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 811

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

Rule 813

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

Rule 844

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

Rule 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rule 6151

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

Rubi steps

\begin {align*} \int \frac {e^{2 \tanh ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2}}{x^5} \, dx &=c \int \frac {(1+a x)^2 \left (c-a^2 c x^2\right )^{3/2}}{x^5} \, dx\\ &=-\frac {\left (c-a^2 c x^2\right )^{5/2}}{4 x^4}-\frac {1}{4} \int \frac {\left (-8 a c-3 a^2 c x\right ) \left (c-a^2 c x^2\right )^{3/2}}{x^4} \, dx\\ &=-\frac {a c (16+9 a x) \left (c-a^2 c x^2\right )^{3/2}}{24 x^3}-\frac {\left (c-a^2 c x^2\right )^{5/2}}{4 x^4}+\frac {\int \frac {\left (-32 a^3 c^3-18 a^4 c^3 x\right ) \sqrt {c-a^2 c x^2}}{x^2} \, dx}{16 c}\\ &=\frac {a^3 c^2 (16-9 a x) \sqrt {c-a^2 c x^2}}{8 x}-\frac {a c (16+9 a x) \left (c-a^2 c x^2\right )^{3/2}}{24 x^3}-\frac {\left (c-a^2 c x^2\right )^{5/2}}{4 x^4}-\frac {\int \frac {36 a^4 c^4-64 a^5 c^4 x}{x \sqrt {c-a^2 c x^2}} \, dx}{32 c}\\ &=\frac {a^3 c^2 (16-9 a x) \sqrt {c-a^2 c x^2}}{8 x}-\frac {a c (16+9 a x) \left (c-a^2 c x^2\right )^{3/2}}{24 x^3}-\frac {\left (c-a^2 c x^2\right )^{5/2}}{4 x^4}-\frac {1}{8} \left (9 a^4 c^3\right ) \int \frac {1}{x \sqrt {c-a^2 c x^2}} \, dx+\left (2 a^5 c^3\right ) \int \frac {1}{\sqrt {c-a^2 c x^2}} \, dx\\ &=\frac {a^3 c^2 (16-9 a x) \sqrt {c-a^2 c x^2}}{8 x}-\frac {a c (16+9 a x) \left (c-a^2 c x^2\right )^{3/2}}{24 x^3}-\frac {\left (c-a^2 c x^2\right )^{5/2}}{4 x^4}-\frac {1}{16} \left (9 a^4 c^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c-a^2 c x}} \, dx,x,x^2\right )+\left (2 a^5 c^3\right ) \operatorname {Subst}\left (\int \frac {1}{1+a^2 c x^2} \, dx,x,\frac {x}{\sqrt {c-a^2 c x^2}}\right )\\ &=\frac {a^3 c^2 (16-9 a x) \sqrt {c-a^2 c x^2}}{8 x}-\frac {a c (16+9 a x) \left (c-a^2 c x^2\right )^{3/2}}{24 x^3}-\frac {\left (c-a^2 c x^2\right )^{5/2}}{4 x^4}+2 a^4 c^{5/2} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )+\frac {1}{8} \left (9 a^2 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {1}{a^2}-\frac {x^2}{a^2 c}} \, dx,x,\sqrt {c-a^2 c x^2}\right )\\ &=\frac {a^3 c^2 (16-9 a x) \sqrt {c-a^2 c x^2}}{8 x}-\frac {a c (16+9 a x) \left (c-a^2 c x^2\right )^{3/2}}{24 x^3}-\frac {\left (c-a^2 c x^2\right )^{5/2}}{4 x^4}+2 a^4 c^{5/2} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )+\frac {9}{8} a^4 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 151, normalized size = 0.97 \[ -\frac {9}{8} a^4 c^{5/2} \log (x)+\frac {9}{8} a^4 c^{5/2} \log \left (\sqrt {c} \sqrt {c-a^2 c x^2}+c\right )-2 a^4 c^{5/2} \tan ^{-1}\left (\frac {a x \sqrt {c-a^2 c x^2}}{\sqrt {c} \left (a^2 x^2-1\right )}\right )-\frac {c^2 \left (24 a^4 x^4-64 a^3 x^3-3 a^2 x^2+16 a x+6\right ) \sqrt {c-a^2 c x^2}}{24 x^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^(2*ArcTanh[a*x])*(c - a^2*c*x^2)^(5/2))/x^5,x]

[Out]

-1/24*(c^2*Sqrt[c - a^2*c*x^2]*(6 + 16*a*x - 3*a^2*x^2 - 64*a^3*x^3 + 24*a^4*x^4))/x^4 - 2*a^4*c^(5/2)*ArcTan[
(a*x*Sqrt[c - a^2*c*x^2])/(Sqrt[c]*(-1 + a^2*x^2))] - (9*a^4*c^(5/2)*Log[x])/8 + (9*a^4*c^(5/2)*Log[c + Sqrt[c
]*Sqrt[c - a^2*c*x^2]])/8

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fricas [A]  time = 0.83, size = 329, normalized size = 2.12 \[ \left [-\frac {96 \, a^{4} c^{\frac {5}{2}} x^{4} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right ) - 27 \, a^{4} c^{\frac {5}{2}} x^{4} \log \left (-\frac {a^{2} c x^{2} - 2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {c} - 2 \, c}{x^{2}}\right ) + 2 \, {\left (24 \, a^{4} c^{2} x^{4} - 64 \, a^{3} c^{2} x^{3} - 3 \, a^{2} c^{2} x^{2} + 16 \, a c^{2} x + 6 \, c^{2}\right )} \sqrt {-a^{2} c x^{2} + c}}{48 \, x^{4}}, \frac {27 \, a^{4} \sqrt {-c} c^{2} x^{4} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-c}}{a^{2} c x^{2} - c}\right ) + 24 \, a^{4} \sqrt {-c} c^{2} x^{4} \log \left (2 \, a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right ) - {\left (24 \, a^{4} c^{2} x^{4} - 64 \, a^{3} c^{2} x^{3} - 3 \, a^{2} c^{2} x^{2} + 16 \, a c^{2} x + 6 \, c^{2}\right )} \sqrt {-a^{2} c x^{2} + c}}{24 \, x^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^(5/2)/x^5,x, algorithm="fricas")

[Out]

[-1/48*(96*a^4*c^(5/2)*x^4*arctan(sqrt(-a^2*c*x^2 + c)*a*sqrt(c)*x/(a^2*c*x^2 - c)) - 27*a^4*c^(5/2)*x^4*log(-
(a^2*c*x^2 - 2*sqrt(-a^2*c*x^2 + c)*sqrt(c) - 2*c)/x^2) + 2*(24*a^4*c^2*x^4 - 64*a^3*c^2*x^3 - 3*a^2*c^2*x^2 +
16*a*c^2*x + 6*c^2)*sqrt(-a^2*c*x^2 + c))/x^4, 1/24*(27*a^4*sqrt(-c)*c^2*x^4*arctan(sqrt(-a^2*c*x^2 + c)*sqrt
(-c)/(a^2*c*x^2 - c)) + 24*a^4*sqrt(-c)*c^2*x^4*log(2*a^2*c*x^2 + 2*sqrt(-a^2*c*x^2 + c)*a*sqrt(-c)*x - c) - (
24*a^4*c^2*x^4 - 64*a^3*c^2*x^3 - 3*a^2*c^2*x^2 + 16*a*c^2*x + 6*c^2)*sqrt(-a^2*c*x^2 + c))/x^4]

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giac [B]  time = 0.37, size = 440, normalized size = 2.84 \[ -\frac {9 \, a^{4} c^{3} \arctan \left (-\frac {\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}}{\sqrt {-c}}\right )}{4 \, \sqrt {-c}} + \frac {2 \, a^{5} \sqrt {-c} c^{2} \log \left ({\left | -\sqrt {-a^{2} c} x + \sqrt {-a^{2} c x^{2} + c} \right |}\right )}{{\left | a \right |}} - \sqrt {-a^{2} c x^{2} + c} a^{4} c^{2} + \frac {3 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{7} a^{4} c^{3} {\left | a \right |} - 96 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{6} a^{5} \sqrt {-c} c^{3} + 21 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{5} a^{4} c^{4} {\left | a \right |} + 192 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{4} a^{5} \sqrt {-c} c^{4} + 21 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{3} a^{4} c^{5} {\left | a \right |} - 160 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{2} a^{5} \sqrt {-c} c^{5} + 3 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )} a^{4} c^{6} {\left | a \right |} + 64 \, a^{5} \sqrt {-c} c^{6}}{12 \, {\left ({\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{2} - c\right )}^{4} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^(5/2)/x^5,x, algorithm="giac")

[Out]

-9/4*a^4*c^3*arctan(-(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))/sqrt(-c))/sqrt(-c) + 2*a^5*sqrt(-c)*c^2*log(abs(-
sqrt(-a^2*c)*x + sqrt(-a^2*c*x^2 + c)))/abs(a) - sqrt(-a^2*c*x^2 + c)*a^4*c^2 + 1/12*(3*(sqrt(-a^2*c)*x - sqrt
(-a^2*c*x^2 + c))^7*a^4*c^3*abs(a) - 96*(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^6*a^5*sqrt(-c)*c^3 + 21*(sqrt(
-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^5*a^4*c^4*abs(a) + 192*(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^4*a^5*sqrt(-c
)*c^4 + 21*(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^3*a^4*c^5*abs(a) - 160*(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 +
c))^2*a^5*sqrt(-c)*c^5 + 3*(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))*a^4*c^6*abs(a) + 64*a^5*sqrt(-c)*c^6)/(((sq
rt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^2 - c)^4*abs(a))

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maple [B]  time = 0.06, size = 447, normalized size = 2.88 \[ \frac {2 a^{3} \left (-a^{2} c \,x^{2}+c \right )^{\frac {7}{2}}}{3 c x}+\frac {5 a^{5} c x \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{6}+\frac {5 a^{5} c^{2} x \sqrt {-a^{2} c \,x^{2}+c}}{4}+\frac {5 a^{5} c^{3} \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{4 \sqrt {a^{2} c}}-\frac {2 a \left (-a^{2} c \,x^{2}+c \right )^{\frac {7}{2}}}{3 c \,x^{3}}-\frac {2 a^{4} \left (-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )\right )^{\frac {5}{2}}}{5}+\frac {2 a^{5} x \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}{3}-\frac {5 a^{2} \left (-a^{2} c \,x^{2}+c \right )^{\frac {7}{2}}}{8 c \,x^{2}}-\frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {7}{2}}}{4 c \,x^{4}}-\frac {3 a^{4} c \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{8}+\frac {9 a^{4} c^{\frac {5}{2}} \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )}{8}-\frac {9 a^{4} \sqrt {-a^{2} c \,x^{2}+c}\, c^{2}}{8}+\frac {a^{5} c \left (-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}} x}{2}+\frac {3 a^{5} c^{2} \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )}\, x}{4}+\frac {3 a^{5} c^{3} \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )}}\right )}{4 \sqrt {a^{2} c}}-\frac {9 a^{4} \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}{40} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^(5/2)/x^5,x)

[Out]

2/3*a^3/c/x*(-a^2*c*x^2+c)^(7/2)+5/6*a^5*c*x*(-a^2*c*x^2+c)^(3/2)+5/4*a^5*c^2*x*(-a^2*c*x^2+c)^(1/2)+5/4*a^5*c
^3/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-a^2*c*x^2+c)^(1/2))-2/3*a/c/x^3*(-a^2*c*x^2+c)^(7/2)-2/5*a^4*(-(x-1/
a)^2*a^2*c-2*a*c*(x-1/a))^(5/2)+2/3*a^5*x*(-a^2*c*x^2+c)^(5/2)-5/8*a^2/c/x^2*(-a^2*c*x^2+c)^(7/2)-1/4/c/x^4*(-
a^2*c*x^2+c)^(7/2)-3/8*a^4*c*(-a^2*c*x^2+c)^(3/2)+9/8*a^4*c^(5/2)*ln((2*c+2*c^(1/2)*(-a^2*c*x^2+c)^(1/2))/x)-9
/8*a^4*(-a^2*c*x^2+c)^(1/2)*c^2+1/2*a^5*c*(-(x-1/a)^2*a^2*c-2*a*c*(x-1/a))^(3/2)*x+3/4*a^5*c^2*(-(x-1/a)^2*a^2
*c-2*a*c*(x-1/a))^(1/2)*x+3/4*a^5*c^3/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-(x-1/a)^2*a^2*c-2*a*c*(x-1/a))^(1
/2))-9/40*a^4*(-a^2*c*x^2+c)^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} {\left (a x + 1\right )}^{2}}{{\left (a^{2} x^{2} - 1\right )} x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^(5/2)/x^5,x, algorithm="maxima")

[Out]

-integrate((-a^2*c*x^2 + c)^(5/2)*(a*x + 1)^2/((a^2*x^2 - 1)*x^5), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {{\left (c-a^2\,c\,x^2\right )}^{5/2}\,{\left (a\,x+1\right )}^2}{x^5\,\left (a^2\,x^2-1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((c - a^2*c*x^2)^(5/2)*(a*x + 1)^2)/(x^5*(a^2*x^2 - 1)),x)

[Out]

-int(((c - a^2*c*x^2)^(5/2)*(a*x + 1)^2)/(x^5*(a^2*x^2 - 1)), x)

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sympy [C]  time = 65.52, size = 575, normalized size = 3.71 \[ - a^{4} c^{2} \left (\begin {cases} i \sqrt {c} \sqrt {a^{2} x^{2} - 1} - \sqrt {c} \log {\left (a x \right )} + \frac {\sqrt {c} \log {\left (a^{2} x^{2} \right )}}{2} + i \sqrt {c} \operatorname {asin}{\left (\frac {1}{a x} \right )} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\\sqrt {c} \sqrt {- a^{2} x^{2} + 1} + \frac {\sqrt {c} \log {\left (a^{2} x^{2} \right )}}{2} - \sqrt {c} \log {\left (\sqrt {- a^{2} x^{2} + 1} + 1 \right )} & \text {otherwise} \end {cases}\right ) - 2 a^{3} c^{2} \left (\begin {cases} - \frac {i a^{2} \sqrt {c} x}{\sqrt {a^{2} x^{2} - 1}} + i a \sqrt {c} \operatorname {acosh}{\left (a x \right )} + \frac {i \sqrt {c}}{x \sqrt {a^{2} x^{2} - 1}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\\frac {a^{2} \sqrt {c} x}{\sqrt {- a^{2} x^{2} + 1}} - a \sqrt {c} \operatorname {asin}{\left (a x \right )} - \frac {\sqrt {c}}{x \sqrt {- a^{2} x^{2} + 1}} & \text {otherwise} \end {cases}\right ) + 2 a c^{2} \left (\begin {cases} \frac {a^{3} \sqrt {c} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}}{3} - \frac {a \sqrt {c} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}}{3 x^{2}} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\\frac {i a^{3} \sqrt {c} \sqrt {1 - \frac {1}{a^{2} x^{2}}}}{3} - \frac {i a \sqrt {c} \sqrt {1 - \frac {1}{a^{2} x^{2}}}}{3 x^{2}} & \text {otherwise} \end {cases}\right ) + c^{2} \left (\begin {cases} \frac {a^{4} \sqrt {c} \operatorname {acosh}{\left (\frac {1}{a x} \right )}}{8} - \frac {a^{3} \sqrt {c}}{8 x \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} + \frac {3 a \sqrt {c}}{8 x^{3} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} - \frac {\sqrt {c}}{4 a x^{5} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\- \frac {i a^{4} \sqrt {c} \operatorname {asin}{\left (\frac {1}{a x} \right )}}{8} + \frac {i a^{3} \sqrt {c}}{8 x \sqrt {1 - \frac {1}{a^{2} x^{2}}}} - \frac {3 i a \sqrt {c}}{8 x^{3} \sqrt {1 - \frac {1}{a^{2} x^{2}}}} + \frac {i \sqrt {c}}{4 a x^{5} \sqrt {1 - \frac {1}{a^{2} x^{2}}}} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**2/(-a**2*x**2+1)*(-a**2*c*x**2+c)**(5/2)/x**5,x)

[Out]

-a**4*c**2*Piecewise((I*sqrt(c)*sqrt(a**2*x**2 - 1) - sqrt(c)*log(a*x) + sqrt(c)*log(a**2*x**2)/2 + I*sqrt(c)*
asin(1/(a*x)), Abs(a**2*x**2) > 1), (sqrt(c)*sqrt(-a**2*x**2 + 1) + sqrt(c)*log(a**2*x**2)/2 - sqrt(c)*log(sqr
t(-a**2*x**2 + 1) + 1), True)) - 2*a**3*c**2*Piecewise((-I*a**2*sqrt(c)*x/sqrt(a**2*x**2 - 1) + I*a*sqrt(c)*ac
osh(a*x) + I*sqrt(c)/(x*sqrt(a**2*x**2 - 1)), Abs(a**2*x**2) > 1), (a**2*sqrt(c)*x/sqrt(-a**2*x**2 + 1) - a*sq
rt(c)*asin(a*x) - sqrt(c)/(x*sqrt(-a**2*x**2 + 1)), True)) + 2*a*c**2*Piecewise((a**3*sqrt(c)*sqrt(-1 + 1/(a**
2*x**2))/3 - a*sqrt(c)*sqrt(-1 + 1/(a**2*x**2))/(3*x**2), 1/Abs(a**2*x**2) > 1), (I*a**3*sqrt(c)*sqrt(1 - 1/(a
**2*x**2))/3 - I*a*sqrt(c)*sqrt(1 - 1/(a**2*x**2))/(3*x**2), True)) + c**2*Piecewise((a**4*sqrt(c)*acosh(1/(a*
x))/8 - a**3*sqrt(c)/(8*x*sqrt(-1 + 1/(a**2*x**2))) + 3*a*sqrt(c)/(8*x**3*sqrt(-1 + 1/(a**2*x**2))) - sqrt(c)/
(4*a*x**5*sqrt(-1 + 1/(a**2*x**2))), 1/Abs(a**2*x**2) > 1), (-I*a**4*sqrt(c)*asin(1/(a*x))/8 + I*a**3*sqrt(c)/
(8*x*sqrt(1 - 1/(a**2*x**2))) - 3*I*a*sqrt(c)/(8*x**3*sqrt(1 - 1/(a**2*x**2))) + I*sqrt(c)/(4*a*x**5*sqrt(1 -
1/(a**2*x**2))), True))

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