3.1.90 \(\int \frac {(d-c^2 d x^2)^{5/2} (a+b \cosh ^{-1}(c x))}{x^2} \, dx\) [90]

Optimal. Leaf size=284 \[ \frac {9 b c^3 d^2 x^2 \sqrt {d-c^2 d x^2}}{16 \sqrt {-1+c x} \sqrt {1+c x}}-\frac {b c^5 d^2 x^4 \sqrt {d-c^2 d x^2}}{16 \sqrt {-1+c x} \sqrt {1+c x}}-\frac {15}{8} c^2 d^2 x \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )-\frac {5}{4} c^2 d x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \cosh ^{-1}(c x)\right )-\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \cosh ^{-1}(c x)\right )}{x}+\frac {15 c d^2 \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )^2}{16 b \sqrt {-1+c x} \sqrt {1+c x}}+\frac {b c d^2 \sqrt {d-c^2 d x^2} \log (x)}{\sqrt {-1+c x} \sqrt {1+c x}} \]

[Out]

-5/4*c^2*d*x*(-c^2*d*x^2+d)^(3/2)*(a+b*arccosh(c*x))-(-c^2*d*x^2+d)^(5/2)*(a+b*arccosh(c*x))/x-15/8*c^2*d^2*x*
(a+b*arccosh(c*x))*(-c^2*d*x^2+d)^(1/2)+9/16*b*c^3*d^2*x^2*(-c^2*d*x^2+d)^(1/2)/(c*x-1)^(1/2)/(c*x+1)^(1/2)-1/
16*b*c^5*d^2*x^4*(-c^2*d*x^2+d)^(1/2)/(c*x-1)^(1/2)/(c*x+1)^(1/2)+15/16*c*d^2*(a+b*arccosh(c*x))^2*(-c^2*d*x^2
+d)^(1/2)/b/(c*x-1)^(1/2)/(c*x+1)^(1/2)+b*c*d^2*ln(x)*(-c^2*d*x^2+d)^(1/2)/(c*x-1)^(1/2)/(c*x+1)^(1/2)

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Rubi [A]
time = 0.22, antiderivative size = 284, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5928, 5897, 5895, 5893, 30, 74, 14, 272, 45} \begin {gather*} -\frac {15}{8} c^2 d^2 x \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )+\frac {15 c d^2 \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )^2}{16 b \sqrt {c x-1} \sqrt {c x+1}}-\frac {5}{4} c^2 d x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \cosh ^{-1}(c x)\right )-\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \cosh ^{-1}(c x)\right )}{x}+\frac {b c d^2 \log (x) \sqrt {d-c^2 d x^2}}{\sqrt {c x-1} \sqrt {c x+1}}-\frac {b c^5 d^2 x^4 \sqrt {d-c^2 d x^2}}{16 \sqrt {c x-1} \sqrt {c x+1}}+\frac {9 b c^3 d^2 x^2 \sqrt {d-c^2 d x^2}}{16 \sqrt {c x-1} \sqrt {c x+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d - c^2*d*x^2)^(5/2)*(a + b*ArcCosh[c*x]))/x^2,x]

[Out]

(9*b*c^3*d^2*x^2*Sqrt[d - c^2*d*x^2])/(16*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) - (b*c^5*d^2*x^4*Sqrt[d - c^2*d*x^2])/
(16*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) - (15*c^2*d^2*x*Sqrt[d - c^2*d*x^2]*(a + b*ArcCosh[c*x]))/8 - (5*c^2*d*x*(d
- c^2*d*x^2)^(3/2)*(a + b*ArcCosh[c*x]))/4 - ((d - c^2*d*x^2)^(5/2)*(a + b*ArcCosh[c*x]))/x + (15*c*d^2*Sqrt[d
- c^2*d*x^2]*(a + b*ArcCosh[c*x])^2)/(16*b*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) + (b*c*d^2*Sqrt[d - c^2*d*x^2]*Log[x
])/(Sqrt[-1 + c*x]*Sqrt[1 + c*x])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 74

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[(a*c + b*
d*x^2)^m*(e + f*x)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m] && Integer
Q[m] && (NeQ[m, -1] || (EqQ[e, 0] && (EqQ[p, 1] ||  !IntegerQ[p])))

Rule 272

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 5893

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)/(Sqrt[(d1_) + (e1_.)*(x_)]*Sqrt[(d2_) + (e2_.)*(x_)]), x_Symbol]
:> Simp[(1/(b*c*(n + 1)))*Simp[Sqrt[1 + c*x]/Sqrt[d1 + e1*x]]*Simp[Sqrt[-1 + c*x]/Sqrt[d2 + e2*x]]*(a + b*Arc
Cosh[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d1, e1, d2, e2, n}, x] && EqQ[e1, c*d1] && EqQ[e2, (-c)*d2] && NeQ[n
, -1]

Rule 5895

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[x*Sqrt[d + e*x^2]*(
(a + b*ArcCosh[c*x])^n/2), x] + (-Dist[(1/2)*Simp[Sqrt[d + e*x^2]/(Sqrt[1 + c*x]*Sqrt[-1 + c*x])], Int[(a + b*
ArcCosh[c*x])^n/(Sqrt[1 + c*x]*Sqrt[-1 + c*x]), x], x] - Dist[b*c*(n/2)*Simp[Sqrt[d + e*x^2]/(Sqrt[1 + c*x]*Sq
rt[-1 + c*x])], Int[x*(a + b*ArcCosh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0]
&& GtQ[n, 0]

Rule 5897

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[x*(d + e*x^2)^p*(
(a + b*ArcCosh[c*x])^n/(2*p + 1)), x] + (Dist[2*d*(p/(2*p + 1)), Int[(d + e*x^2)^(p - 1)*(a + b*ArcCosh[c*x])^
n, x], x] - Dist[b*c*(n/(2*p + 1))*Simp[(d + e*x^2)^p/((1 + c*x)^p*(-1 + c*x)^p)], Int[x*(1 + c*x)^(p - 1/2)*(
-1 + c*x)^(p - 1/2)*(a + b*ArcCosh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] &&
GtQ[n, 0] && GtQ[p, 0]

Rule 5928

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp
[(f*x)^(m + 1)*(d + e*x^2)^p*((a + b*ArcCosh[c*x])^n/(f*(m + 1))), x] + (-Dist[2*e*(p/(f^2*(m + 1))), Int[(f*x
)^(m + 2)*(d + e*x^2)^(p - 1)*(a + b*ArcCosh[c*x])^n, x], x] - Dist[b*c*(n/(f*(m + 1)))*Simp[(d + e*x^2)^p/((1
+ c*x)^p*(-1 + c*x)^p)], Int[(f*x)^(m + 1)*(1 + c*x)^(p - 1/2)*(-1 + c*x)^(p - 1/2)*(a + b*ArcCosh[c*x])^(n -
1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \cosh ^{-1}(c x)\right )}{x^2} \, dx &=\frac {\left (d^2 \sqrt {d-c^2 d x^2}\right ) \int \frac {(-1+c x)^{5/2} (1+c x)^{5/2} \left (a+b \cosh ^{-1}(c x)\right )}{x^2} \, dx}{\sqrt {-1+c x} \sqrt {1+c x}}\\ &=-\frac {d^2 (1-c x)^2 (1+c x)^2 \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{x}+\frac {\left (b c d^2 \sqrt {d-c^2 d x^2}\right ) \int \frac {\left (-1+c^2 x^2\right )^2}{x} \, dx}{\sqrt {-1+c x} \sqrt {1+c x}}+\frac {\left (5 c^2 d^2 \sqrt {d-c^2 d x^2}\right ) \int (-1+c x)^{3/2} (1+c x)^{3/2} \left (a+b \cosh ^{-1}(c x)\right ) \, dx}{\sqrt {-1+c x} \sqrt {1+c x}}\\ &=-\frac {5}{4} c^2 d^2 x (1-c x) (1+c x) \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )-\frac {d^2 (1-c x)^2 (1+c x)^2 \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{x}+\frac {\left (b c d^2 \sqrt {d-c^2 d x^2}\right ) \text {Subst}\left (\int \frac {\left (-1+c^2 x\right )^2}{x} \, dx,x,x^2\right )}{2 \sqrt {-1+c x} \sqrt {1+c x}}-\frac {\left (15 c^2 d^2 \sqrt {d-c^2 d x^2}\right ) \int \sqrt {-1+c x} \sqrt {1+c x} \left (a+b \cosh ^{-1}(c x)\right ) \, dx}{4 \sqrt {-1+c x} \sqrt {1+c x}}-\frac {\left (5 b c^3 d^2 \sqrt {d-c^2 d x^2}\right ) \int x \left (-1+c^2 x^2\right ) \, dx}{4 \sqrt {-1+c x} \sqrt {1+c x}}\\ &=-\frac {15}{8} c^2 d^2 x \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )-\frac {5}{4} c^2 d^2 x (1-c x) (1+c x) \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )-\frac {d^2 (1-c x)^2 (1+c x)^2 \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{x}+\frac {\left (b c d^2 \sqrt {d-c^2 d x^2}\right ) \text {Subst}\left (\int \left (-2 c^2+\frac {1}{x}+c^4 x\right ) \, dx,x,x^2\right )}{2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {\left (15 c^2 d^2 \sqrt {d-c^2 d x^2}\right ) \int \frac {a+b \cosh ^{-1}(c x)}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{8 \sqrt {-1+c x} \sqrt {1+c x}}-\frac {\left (5 b c^3 d^2 \sqrt {d-c^2 d x^2}\right ) \int \left (-x+c^2 x^3\right ) \, dx}{4 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {\left (15 b c^3 d^2 \sqrt {d-c^2 d x^2}\right ) \int x \, dx}{8 \sqrt {-1+c x} \sqrt {1+c x}}\\ &=\frac {9 b c^3 d^2 x^2 \sqrt {d-c^2 d x^2}}{16 \sqrt {-1+c x} \sqrt {1+c x}}-\frac {b c^5 d^2 x^4 \sqrt {d-c^2 d x^2}}{16 \sqrt {-1+c x} \sqrt {1+c x}}-\frac {15}{8} c^2 d^2 x \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )-\frac {5}{4} c^2 d^2 x (1-c x) (1+c x) \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )-\frac {d^2 (1-c x)^2 (1+c x)^2 \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{x}+\frac {15 c d^2 \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )^2}{16 b \sqrt {-1+c x} \sqrt {1+c x}}+\frac {b c d^2 \sqrt {d-c^2 d x^2} \log (x)}{\sqrt {-1+c x} \sqrt {1+c x}}\\ \end {align*}

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Mathematica [A]
time = 1.15, size = 305, normalized size = 1.07 \begin {gather*} \frac {1}{128} d^2 \left (\frac {16 a \sqrt {d-c^2 d x^2} \left (-8-9 c^2 x^2+2 c^4 x^4\right )}{x}+240 a c \sqrt {d} \text {ArcTan}\left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (-1+c^2 x^2\right )}\right )+64 b c \sqrt {d-c^2 d x^2} \left (-\frac {2 \cosh ^{-1}(c x)}{c x}+\frac {\cosh ^{-1}(c x)^2+2 \log (c x)}{\sqrt {\frac {-1+c x}{1+c x}} (1+c x)}\right )+\frac {32 b c \sqrt {d-c^2 d x^2} \left (2 \cosh ^{-1}(c x)^2+\cosh \left (2 \cosh ^{-1}(c x)\right )-2 \cosh ^{-1}(c x) \sinh \left (2 \cosh ^{-1}(c x)\right )\right )}{\sqrt {\frac {-1+c x}{1+c x}} (1+c x)}-\frac {b c \sqrt {d-c^2 d x^2} \left (8 \cosh ^{-1}(c x)^2+\cosh \left (4 \cosh ^{-1}(c x)\right )-4 \cosh ^{-1}(c x) \sinh \left (4 \cosh ^{-1}(c x)\right )\right )}{\sqrt {\frac {-1+c x}{1+c x}} (1+c x)}\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[((d - c^2*d*x^2)^(5/2)*(a + b*ArcCosh[c*x]))/x^2,x]

[Out]

(d^2*((16*a*Sqrt[d - c^2*d*x^2]*(-8 - 9*c^2*x^2 + 2*c^4*x^4))/x + 240*a*c*Sqrt[d]*ArcTan[(c*x*Sqrt[d - c^2*d*x
^2])/(Sqrt[d]*(-1 + c^2*x^2))] + 64*b*c*Sqrt[d - c^2*d*x^2]*((-2*ArcCosh[c*x])/(c*x) + (ArcCosh[c*x]^2 + 2*Log
[c*x])/(Sqrt[(-1 + c*x)/(1 + c*x)]*(1 + c*x))) + (32*b*c*Sqrt[d - c^2*d*x^2]*(2*ArcCosh[c*x]^2 + Cosh[2*ArcCos
h[c*x]] - 2*ArcCosh[c*x]*Sinh[2*ArcCosh[c*x]]))/(Sqrt[(-1 + c*x)/(1 + c*x)]*(1 + c*x)) - (b*c*Sqrt[d - c^2*d*x
^2]*(8*ArcCosh[c*x]^2 + Cosh[4*ArcCosh[c*x]] - 4*ArcCosh[c*x]*Sinh[4*ArcCosh[c*x]]))/(Sqrt[(-1 + c*x)/(1 + c*x
)]*(1 + c*x))))/128

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(549\) vs. \(2(244)=488\).
time = 4.01, size = 550, normalized size = 1.94

method	result	size
default	\(-\frac {a \left (-c^{2} d \,x^{2}+d \right )^{\frac {7}{2}}}{d x}-a \,c^{2} x \left (-c^{2} d \,x^{2}+d \right )^{\frac {5}{2}}-\frac {5 a \,c^{2} d x \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{4}-\frac {15 a \,c^{2} d^{2} x \sqrt {-c^{2} d \,x^{2}+d}}{8}-\frac {15 a \,c^{2} d^{3} \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{8 \sqrt {c^{2} d}}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{5} d^{2} x^{4}}{16 \sqrt {c x +1}\, \sqrt {c x -1}}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \mathrm {arccosh}\left (c x \right ) c \,d^{2}}{\sqrt {c x -1}\, \sqrt {c x +1}}+\frac {9 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{3} d^{2} x^{2}}{16 \sqrt {c x +1}\, \sqrt {c x -1}}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \ln \left (1+\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}\right ) c \,d^{2}}{\sqrt {c x -1}\, \sqrt {c x +1}}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{6} d^{2} \mathrm {arccosh}\left (c x \right ) x^{5}}{4 \left (c x +1\right ) \left (c x -1\right )}-\frac {11 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{4} d^{2} \mathrm {arccosh}\left (c x \right ) x^{3}}{8 \left (c x +1\right ) \left (c x -1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{2} d^{2} \mathrm {arccosh}\left (c x \right ) x}{8 \left (c x +1\right ) \left (c x -1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \mathrm {arccosh}\left (c x \right ) d^{2}}{\left (c x +1\right ) \left (c x -1\right ) x}-\frac {33 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c \,d^{2}}{128 \sqrt {c x +1}\, \sqrt {c x -1}}+\frac {15 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \mathrm {arccosh}\left (c x \right )^{2} c \,d^{2}}{16 \sqrt {c x -1}\, \sqrt {c x +1}}\)	\(550\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-c^2*d*x^2+d)^(5/2)*(a+b*arccosh(c*x))/x^2,x,method=_RETURNVERBOSE)

[Out]

-a/d/x*(-c^2*d*x^2+d)^(7/2)-a*c^2*x*(-c^2*d*x^2+d)^(5/2)-5/4*a*c^2*d*x*(-c^2*d*x^2+d)^(3/2)-15/8*a*c^2*d^2*x*(
-c^2*d*x^2+d)^(1/2)-15/8*a*c^2*d^3/(c^2*d)^(1/2)*arctan((c^2*d)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))-1/16*b*(-d*(c^2*
x^2-1))^(1/2)*c^5*d^2/(c*x+1)^(1/2)/(c*x-1)^(1/2)*x^4-b*(-d*(c^2*x^2-1))^(1/2)/(c*x-1)^(1/2)/(c*x+1)^(1/2)*arc
cosh(c*x)*c*d^2+9/16*b*(-d*(c^2*x^2-1))^(1/2)*c^3*d^2/(c*x+1)^(1/2)/(c*x-1)^(1/2)*x^2+b*(-d*(c^2*x^2-1))^(1/2)
/(c*x-1)^(1/2)/(c*x+1)^(1/2)*ln(1+(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))^2)*c*d^2+1/4*b*(-d*(c^2*x^2-1))^(1/2)*c^6*
d^2/(c*x+1)/(c*x-1)*arccosh(c*x)*x^5-11/8*b*(-d*(c^2*x^2-1))^(1/2)*c^4*d^2/(c*x+1)/(c*x-1)*arccosh(c*x)*x^3+1/
8*b*(-d*(c^2*x^2-1))^(1/2)*c^2*d^2/(c*x+1)/(c*x-1)*arccosh(c*x)*x+b*(-d*(c^2*x^2-1))^(1/2)*arccosh(c*x)*d^2/(c
*x+1)/(c*x-1)/x-33/128*b*(-d*(c^2*x^2-1))^(1/2)*c*d^2/(c*x+1)^(1/2)/(c*x-1)^(1/2)+15/16*b*(-d*(c^2*x^2-1))^(1/
2)/(c*x-1)^(1/2)/(c*x+1)^(1/2)*arccosh(c*x)^2*c*d^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^2*d*x^2+d)^(5/2)*(a+b*arccosh(c*x))/x^2,x, algorithm="maxima")

[Out]

-1/8*(10*(-c^2*d*x^2 + d)^(3/2)*c^2*d*x + 15*sqrt(-c^2*d*x^2 + d)*c^2*d^2*x + 15*c*d^(5/2)*arcsin(c*x) + 8*(-c
^2*d*x^2 + d)^(5/2)/x)*a + b*integrate((-c^2*d*x^2 + d)^(5/2)*log(c*x + sqrt(c*x + 1)*sqrt(c*x - 1))/x^2, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*c^4*d^2*x^4 - 2*a*c^2*d^2*x^2 + a*d^2 + (b*c^4*d^2*x^4 - 2*b*c^2*d^2*x^2 + b*d^2)*arccosh(c*x))*sq
rt(-c^2*d*x^2 + d)/x^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}} \left (a + b \operatorname {acosh}{\left (c x \right )}\right )}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c**2*d*x**2+d)**(5/2)*(a+b*acosh(c*x))/x**2,x)

[Out]

Integral((-d*(c*x - 1)*(c*x + 1))**(5/2)*(a + b*acosh(c*x))/x**2, 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^2*d*x^2+d)^(5/2)*(a+b*arccosh(c*x))/x^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )\,{\left (d-c^2\,d\,x^2\right )}^{5/2}}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*acosh(c*x))*(d - c^2*d*x^2)^(5/2))/x^2,x)

[Out]

int(((a + b*acosh(c*x))*(d - c^2*d*x^2)^(5/2))/x^2, x)

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