3.126 \(\int \frac {x^3 (a+b \cosh ^{-1}(c x))}{(d-c^2 d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=158 \[ -\frac {a+b \cosh ^{-1}(c x)}{c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {a+b \cosh ^{-1}(c x)}{3 c^4 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {5 b \sqrt {d-c^2 d x^2} \tanh ^{-1}(c x)}{6 c^4 d^3 \sqrt {c x-1} \sqrt {c x+1}}+\frac {b x \sqrt {d-c^2 d x^2}}{6 c^3 d^3 (c x-1)^{3/2} (c x+1)^{3/2}} \]

[Out]

1/3*(a+b*arccosh(c*x))/c^4/d/(-c^2*d*x^2+d)^(3/2)+(-a-b*arccosh(c*x))/c^4/d^2/(-c^2*d*x^2+d)^(1/2)+1/6*b*x*(-c
^2*d*x^2+d)^(1/2)/c^3/d^3/(c*x-1)^(3/2)/(c*x+1)^(3/2)+5/6*b*arctanh(c*x)*(-c^2*d*x^2+d)^(1/2)/c^4/d^3/(c*x-1)^
(1/2)/(c*x+1)^(1/2)

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Rubi [A]  time = 0.45, antiderivative size = 243, normalized size of antiderivative = 1.54, number of steps used = 5, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5798, 94, 89, 21, 37, 5733, 12, 385, 206} \[ \frac {x^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 c d^2 (1-c x) (c x+1) \sqrt {d-c^2 d x^2}}+\frac {(1-c x)^2 \left (a+b \cosh ^{-1}(c x)\right )}{3 c^4 d^2 (c x+1) \sqrt {d-c^2 d x^2}}-\frac {a+b \cosh ^{-1}(c x)}{c^4 d^2 (c x+1) \sqrt {d-c^2 d x^2}}+\frac {b x \sqrt {c x-1} \sqrt {c x+1}}{6 c^3 d^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}-\frac {5 b \sqrt {c x-1} \sqrt {c x+1} \tanh ^{-1}(c x)}{6 c^4 d^2 \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*ArcCosh[c*x]))/(d - c^2*d*x^2)^(5/2),x]

[Out]

(b*x*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(6*c^3*d^2*(1 - c^2*x^2)*Sqrt[d - c^2*d*x^2]) - (a + b*ArcCosh[c*x])/(c^4*d
^2*(1 + c*x)*Sqrt[d - c^2*d*x^2]) + (x^3*(a + b*ArcCosh[c*x]))/(3*c*d^2*(1 - c*x)*(1 + c*x)*Sqrt[d - c^2*d*x^2
]) + ((1 - c*x)^2*(a + b*ArcCosh[c*x]))/(3*c^4*d^2*(1 + c*x)*Sqrt[d - c^2*d*x^2]) - (5*b*Sqrt[-1 + c*x]*Sqrt[1
+ c*x]*ArcTanh[c*x])/(6*c^4*d^2*Sqrt[d - c^2*d*x^2])

Rule 12

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

Rule 21

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 94

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

Rule 206

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

Rule 385

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

Rule 5733

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d1_) + (e1_.)*(x_))^(p_)*((d2_) + (e2_.)*(x_))^(p_), x_Sym
bol] :> With[{u = IntHide[x^m*(1 + c*x)^p*(-1 + c*x)^p, x]}, Dist[(-(d1*d2))^p*(a + b*ArcCosh[c*x]), u, x] - D
ist[b*c*(-(d1*d2))^p, Int[SimplifyIntegrand[u/(Sqrt[1 + c*x]*Sqrt[-1 + c*x]), x], x], x]] /; FreeQ[{a, b, c, d
1, e1, d2, e2}, x] && EqQ[e1 - c*d1, 0] && EqQ[e2 + c*d2, 0] && IntegerQ[p - 1/2] && (IGtQ[(m + 1)/2, 0] || IL
tQ[(m + 2*p + 3)/2, 0]) && NeQ[p, -2^(-1)] && GtQ[d1, 0] && LtQ[d2, 0]

Rule 5798

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

Rubi steps

\begin {align*} \int \frac {x^3 \left (a+b \cosh ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^{5/2}} \, dx &=\frac {\left (\sqrt {-1+c x} \sqrt {1+c x}\right ) \int \frac {x^3 \left (a+b \cosh ^{-1}(c x)\right )}{(-1+c x)^{5/2} (1+c x)^{5/2}} \, dx}{d^2 \sqrt {d-c^2 d x^2}}\\ &=-\frac {a+b \cosh ^{-1}(c x)}{c^4 d^2 (1+c x) \sqrt {d-c^2 d x^2}}+\frac {x^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 c d^2 (1-c x) (1+c x) \sqrt {d-c^2 d x^2}}+\frac {(1-c x)^2 \left (a+b \cosh ^{-1}(c x)\right )}{3 c^4 d^2 (1+c x) \sqrt {d-c^2 d x^2}}-\frac {\left (b c \sqrt {-1+c x} \sqrt {1+c x}\right ) \int \frac {2-3 c^2 x^2}{3 c^4 \left (1-c^2 x^2\right )^2} \, dx}{d^2 \sqrt {d-c^2 d x^2}}\\ &=-\frac {a+b \cosh ^{-1}(c x)}{c^4 d^2 (1+c x) \sqrt {d-c^2 d x^2}}+\frac {x^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 c d^2 (1-c x) (1+c x) \sqrt {d-c^2 d x^2}}+\frac {(1-c x)^2 \left (a+b \cosh ^{-1}(c x)\right )}{3 c^4 d^2 (1+c x) \sqrt {d-c^2 d x^2}}-\frac {\left (b \sqrt {-1+c x} \sqrt {1+c x}\right ) \int \frac {2-3 c^2 x^2}{\left (1-c^2 x^2\right )^2} \, dx}{3 c^3 d^2 \sqrt {d-c^2 d x^2}}\\ &=\frac {b x \sqrt {-1+c x} \sqrt {1+c x}}{6 c^3 d^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}-\frac {a+b \cosh ^{-1}(c x)}{c^4 d^2 (1+c x) \sqrt {d-c^2 d x^2}}+\frac {x^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 c d^2 (1-c x) (1+c x) \sqrt {d-c^2 d x^2}}+\frac {(1-c x)^2 \left (a+b \cosh ^{-1}(c x)\right )}{3 c^4 d^2 (1+c x) \sqrt {d-c^2 d x^2}}-\frac {\left (5 b \sqrt {-1+c x} \sqrt {1+c x}\right ) \int \frac {1}{1-c^2 x^2} \, dx}{6 c^3 d^2 \sqrt {d-c^2 d x^2}}\\ &=\frac {b x \sqrt {-1+c x} \sqrt {1+c x}}{6 c^3 d^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}-\frac {a+b \cosh ^{-1}(c x)}{c^4 d^2 (1+c x) \sqrt {d-c^2 d x^2}}+\frac {x^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 c d^2 (1-c x) (1+c x) \sqrt {d-c^2 d x^2}}+\frac {(1-c x)^2 \left (a+b \cosh ^{-1}(c x)\right )}{3 c^4 d^2 (1+c x) \sqrt {d-c^2 d x^2}}-\frac {5 b \sqrt {-1+c x} \sqrt {1+c x} \tanh ^{-1}(c x)}{6 c^4 d^2 \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 122, normalized size = 0.77 \[ \frac {-6 a c^2 x^2+4 a+b \left (4-6 c^2 x^2\right ) \cosh ^{-1}(c x)-5 b \sqrt {c x-1} \sqrt {c x+1} \left (c^2 x^2-1\right ) \tanh ^{-1}(c x)-b c x \sqrt {c x-1} \sqrt {c x+1}}{6 c^4 d^2 \left (c^2 x^2-1\right ) \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(a + b*ArcCosh[c*x]))/(d - c^2*d*x^2)^(5/2),x]

[Out]

(4*a - 6*a*c^2*x^2 - b*c*x*Sqrt[-1 + c*x]*Sqrt[1 + c*x] + b*(4 - 6*c^2*x^2)*ArcCosh[c*x] - 5*b*Sqrt[-1 + c*x]*
Sqrt[1 + c*x]*(-1 + c^2*x^2)*ArcTanh[c*x])/(6*c^4*d^2*(-1 + c^2*x^2)*Sqrt[d - c^2*d*x^2])

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fricas [A]  time = 0.69, size = 469, normalized size = 2.97 \[ \left [\frac {4 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} - 1} b c x + 8 \, {\left (3 \, b c^{2} x^{2} - 2 \, b\right )} \sqrt {-c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) - 5 \, {\left (b c^{4} x^{4} - 2 \, b c^{2} x^{2} + b\right )} \sqrt {-d} \log \left (-\frac {c^{6} d x^{6} + 5 \, c^{4} d x^{4} - 5 \, c^{2} d x^{2} + 4 \, {\left (c^{3} x^{3} + c x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} - 1} \sqrt {-d} - d}{c^{6} x^{6} - 3 \, c^{4} x^{4} + 3 \, c^{2} x^{2} - 1}\right ) + 8 \, {\left (3 \, a c^{2} x^{2} - 2 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{24 \, {\left (c^{8} d^{3} x^{4} - 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )}}, \frac {2 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} - 1} b c x + 5 \, {\left (b c^{4} x^{4} - 2 \, b c^{2} x^{2} + b\right )} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} - 1} c \sqrt {d} x}{c^{4} d x^{4} - d}\right ) + 4 \, {\left (3 \, b c^{2} x^{2} - 2 \, b\right )} \sqrt {-c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) + 4 \, {\left (3 \, a c^{2} x^{2} - 2 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{12 \, {\left (c^{8} d^{3} x^{4} - 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arccosh(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

[1/24*(4*sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1)*b*c*x + 8*(3*b*c^2*x^2 - 2*b)*sqrt(-c^2*d*x^2 + d)*log(c*x + s
qrt(c^2*x^2 - 1)) - 5*(b*c^4*x^4 - 2*b*c^2*x^2 + b)*sqrt(-d)*log(-(c^6*d*x^6 + 5*c^4*d*x^4 - 5*c^2*d*x^2 + 4*(
c^3*x^3 + c*x)*sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1)*sqrt(-d) - d)/(c^6*x^6 - 3*c^4*x^4 + 3*c^2*x^2 - 1)) + 8
*(3*a*c^2*x^2 - 2*a)*sqrt(-c^2*d*x^2 + d))/(c^8*d^3*x^4 - 2*c^6*d^3*x^2 + c^4*d^3), 1/12*(2*sqrt(-c^2*d*x^2 +
d)*sqrt(c^2*x^2 - 1)*b*c*x + 5*(b*c^4*x^4 - 2*b*c^2*x^2 + b)*sqrt(d)*arctan(2*sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^
2 - 1)*c*sqrt(d)*x/(c^4*d*x^4 - d)) + 4*(3*b*c^2*x^2 - 2*b)*sqrt(-c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 - 1))
+ 4*(3*a*c^2*x^2 - 2*a)*sqrt(-c^2*d*x^2 + d))/(c^8*d^3*x^4 - 2*c^6*d^3*x^2 + c^4*d^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(x^3*(a+b*arccosh(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [B]  time = 0.54, size = 313, normalized size = 1.98 \[ \frac {a \,x^{2}}{c^{2} d \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-\frac {2 a}{3 d \,c^{4} \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \mathrm {arccosh}\left (c x \right ) x^{2}}{d^{3} \left (c^{2} x^{2}-1\right )^{2} c^{2}}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {c x +1}\, \sqrt {c x -1}\, x}{6 d^{3} \left (c^{2} x^{2}-1\right )^{2} c^{3}}-\frac {2 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \mathrm {arccosh}\left (c x \right )}{3 d^{3} \left (c^{2} x^{2}-1\right )^{2} c^{4}}-\frac {5 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {c x -1}\, \sqrt {c x +1}\, \ln \left (c x +\sqrt {c x -1}\, \sqrt {c x +1}-1\right )}{6 d^{3} c^{4} \left (c^{2} x^{2}-1\right )}+\frac {5 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {c x -1}\, \sqrt {c x +1}\, \ln \left (1+c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )}{6 d^{3} c^{4} \left (c^{2} x^{2}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arccosh(c*x))/(-c^2*d*x^2+d)^(5/2),x)

[Out]

a*x^2/c^2/d/(-c^2*d*x^2+d)^(3/2)-2/3*a/d/c^4/(-c^2*d*x^2+d)^(3/2)+b*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^2*x^2-1)^2/c
^2*arccosh(c*x)*x^2+1/6*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^2*x^2-1)^2/c^3*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x-2/3*b*(-d
*(c^2*x^2-1))^(1/2)/d^3/(c^2*x^2-1)^2/c^4*arccosh(c*x)-5/6*b*(-d*(c^2*x^2-1))^(1/2)*(c*x-1)^(1/2)*(c*x+1)^(1/2
)/d^3/c^4/(c^2*x^2-1)*ln(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2)-1)+5/6*b*(-d*(c^2*x^2-1))^(1/2)*(c*x-1)^(1/2)*(c*x+1)
^(1/2)/d^3/c^4/(c^2*x^2-1)*ln(1+c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))

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maxima [A]  time = 0.70, size = 175, normalized size = 1.11 \[ \frac {1}{12} \, b c {\left (\frac {2 \, \sqrt {-d} x}{c^{6} d^{3} x^{2} - c^{4} d^{3}} + \frac {5 \, \sqrt {-d} \log \left (c x + 1\right )}{c^{5} d^{3}} - \frac {5 \, \sqrt {-d} \log \left (c x - 1\right )}{c^{5} d^{3}}\right )} + \frac {1}{3} \, b {\left (\frac {3 \, x^{2}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2} d} - \frac {2}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{4} d}\right )} \operatorname {arcosh}\left (c x\right ) + \frac {1}{3} \, a {\left (\frac {3 \, x^{2}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2} d} - \frac {2}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{4} d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arccosh(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

1/12*b*c*(2*sqrt(-d)*x/(c^6*d^3*x^2 - c^4*d^3) + 5*sqrt(-d)*log(c*x + 1)/(c^5*d^3) - 5*sqrt(-d)*log(c*x - 1)/(
c^5*d^3)) + 1/3*b*(3*x^2/((-c^2*d*x^2 + d)^(3/2)*c^2*d) - 2/((-c^2*d*x^2 + d)^(3/2)*c^4*d))*arccosh(c*x) + 1/3
*a*(3*x^2/((-c^2*d*x^2 + d)^(3/2)*c^2*d) - 2/((-c^2*d*x^2 + d)^(3/2)*c^4*d))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3\,\left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a + b*acosh(c*x)))/(d - c^2*d*x^2)^(5/2),x)

[Out]

int((x^3*(a + b*acosh(c*x)))/(d - c^2*d*x^2)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*acosh(c*x))/(-c**2*d*x**2+d)**(5/2),x)

[Out]

Integral(x**3*(a + b*acosh(c*x))/(-d*(c*x - 1)*(c*x + 1))**(5/2), x)

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