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

Optimal. Leaf size=150 \[ -\frac {2 \left (a+b \cosh ^{-1}(c+d x)\right )}{3 d e (e (c+d x))^{3/2}}-\frac {4 b \sqrt {-c-d x+1} \sqrt {e (c+d x)} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {c+d x+1}}{\sqrt {2}}\right )\right |2\right )}{3 d e^3 \sqrt {-c-d x} \sqrt {c+d x-1}}+\frac {4 b \sqrt {c+d x-1} \sqrt {c+d x+1}}{3 d e^2 \sqrt {e (c+d x)}} \]

[Out]

-2/3*(a+b*arccosh(d*x+c))/d/e/(e*(d*x+c))^(3/2)-4/3*b*EllipticE(1/2*(d*x+c+1)^(1/2)*2^(1/2),2^(1/2))*(-d*x-c+1
)^(1/2)*(e*(d*x+c))^(1/2)/d/e^3/(-d*x-c)^(1/2)/(d*x+c-1)^(1/2)+4/3*b*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)/d/e^2/(e*
(d*x+c))^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {5866, 5662, 104, 12, 16, 114, 113} \[ -\frac {2 \left (a+b \cosh ^{-1}(c+d x)\right )}{3 d e (e (c+d x))^{3/2}}+\frac {4 b \sqrt {c+d x-1} \sqrt {c+d x+1}}{3 d e^2 \sqrt {e (c+d x)}}-\frac {4 b \sqrt {-c-d x+1} \sqrt {e (c+d x)} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {c+d x+1}}{\sqrt {2}}\right )\right |2\right )}{3 d e^3 \sqrt {-c-d x} \sqrt {c+d x-1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b*Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x])/(3*d*e^2*Sqrt[e*(c + d*x)]) - (2*(a + b*ArcCosh[c + d*x]))/(3*d*e*(
e*(c + d*x))^(3/2)) - (4*b*Sqrt[1 - c - d*x]*Sqrt[e*(c + d*x)]*EllipticE[ArcSin[Sqrt[1 + c + d*x]/Sqrt[2]], 2]
)/(3*d*e^3*Sqrt[-c - d*x]*Sqrt[-1 + c + d*x])

Rule 12

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

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 104

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

Rule 113

Int[Sqrt[(e_.) + (f_.)*(x_)]/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[(2*Rt[-((b*e
- a*f)/d), 2]*EllipticE[ArcSin[Sqrt[a + b*x]/Rt[-((b*c - a*d)/d), 2]], (f*(b*c - a*d))/(d*(b*e - a*f))])/b, x
] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !LtQ[-((b*c - a*d)/d),
0] &&  !(SimplerQ[c + d*x, a + b*x] && GtQ[-(d/(b*c - a*d)), 0] && GtQ[d/(d*e - c*f), 0] &&  !LtQ[(b*c - a*d)
/b, 0])

Rule 114

Int[Sqrt[(e_.) + (f_.)*(x_)]/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Dist[(Sqrt[e + f*
x]*Sqrt[(b*(c + d*x))/(b*c - a*d)])/(Sqrt[c + d*x]*Sqrt[(b*(e + f*x))/(b*e - a*f)]), Int[Sqrt[(b*e)/(b*e - a*f
) + (b*f*x)/(b*e - a*f)]/(Sqrt[a + b*x]*Sqrt[(b*c)/(b*c - a*d) + (b*d*x)/(b*c - a*d)]), x], x] /; FreeQ[{a, b,
c, d, e, f}, x] &&  !(GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0]) &&  !LtQ[-((b*c - a*d)/d), 0]

Rule 5662

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcC
osh[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCosh[c*x])^(n - 1))/(Sqr
t[-1 + c*x]*Sqrt[1 + c*x]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 5866

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

Rubi steps

\begin {align*} \int \frac {a+b \cosh ^{-1}(c+d x)}{(c e+d e x)^{5/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {a+b \cosh ^{-1}(x)}{(e x)^{5/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {2 \left (a+b \cosh ^{-1}(c+d x)\right )}{3 d e (e (c+d x))^{3/2}}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x} (e x)^{3/2} \sqrt {1+x}} \, dx,x,c+d x\right )}{3 d e}\\ &=\frac {4 b \sqrt {-1+c+d x} \sqrt {1+c+d x}}{3 d e^2 \sqrt {e (c+d x)}}-\frac {2 \left (a+b \cosh ^{-1}(c+d x)\right )}{3 d e (e (c+d x))^{3/2}}+\frac {(4 b) \operatorname {Subst}\left (\int -\frac {e x}{2 \sqrt {-1+x} \sqrt {e x} \sqrt {1+x}} \, dx,x,c+d x\right )}{3 d e^3}\\ &=\frac {4 b \sqrt {-1+c+d x} \sqrt {1+c+d x}}{3 d e^2 \sqrt {e (c+d x)}}-\frac {2 \left (a+b \cosh ^{-1}(c+d x)\right )}{3 d e (e (c+d x))^{3/2}}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {x}{\sqrt {-1+x} \sqrt {e x} \sqrt {1+x}} \, dx,x,c+d x\right )}{3 d e^2}\\ &=\frac {4 b \sqrt {-1+c+d x} \sqrt {1+c+d x}}{3 d e^2 \sqrt {e (c+d x)}}-\frac {2 \left (a+b \cosh ^{-1}(c+d x)\right )}{3 d e (e (c+d x))^{3/2}}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {\sqrt {e x}}{\sqrt {-1+x} \sqrt {1+x}} \, dx,x,c+d x\right )}{3 d e^3}\\ &=\frac {4 b \sqrt {-1+c+d x} \sqrt {1+c+d x}}{3 d e^2 \sqrt {e (c+d x)}}-\frac {2 \left (a+b \cosh ^{-1}(c+d x)\right )}{3 d e (e (c+d x))^{3/2}}-\frac {\left (\sqrt {2} b \sqrt {1-c-d x} \sqrt {e (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {-x}}{\sqrt {\frac {1}{2}-\frac {x}{2}} \sqrt {1+x}} \, dx,x,c+d x\right )}{3 d e^3 \sqrt {-c-d x} \sqrt {-1+c+d x}}\\ &=\frac {4 b \sqrt {-1+c+d x} \sqrt {1+c+d x}}{3 d e^2 \sqrt {e (c+d x)}}-\frac {2 \left (a+b \cosh ^{-1}(c+d x)\right )}{3 d e (e (c+d x))^{3/2}}-\frac {4 b \sqrt {1-c-d x} \sqrt {e (c+d x)} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {1+c+d x}}{\sqrt {2}}\right )\right |2\right )}{3 d e^3 \sqrt {-c-d x} \sqrt {-1+c+d x}}\\ \end {align*}

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Mathematica [C]  time = 0.15, size = 94, normalized size = 0.63 \[ \frac {2 \left (-a-\frac {2 b (c+d x) \sqrt {1-(c+d x)^2} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};(c+d x)^2\right )}{\sqrt {c+d x-1} \sqrt {c+d x+1}}-b \cosh ^{-1}(c+d x)\right )}{3 d e (e (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-a - b*ArcCosh[c + d*x] - (2*b*(c + d*x)*Sqrt[1 - (c + d*x)^2]*Hypergeometric2F1[-1/4, 1/2, 3/4, (c + d*x)
^2])/(Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x])))/(3*d*e*(e*(c + d*x))^(3/2))

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fricas [F]  time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d e x + c e} {\left (b \operatorname {arcosh}\left (d x + c\right ) + a\right )}}{d^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + 3 \, c^{2} d e^{3} x + c^{3} e^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(d*e*x + c*e)*(b*arccosh(d*x + c) + a)/(d^3*e^3*x^3 + 3*c*d^2*e^3*x^2 + 3*c^2*d*e^3*x + c^3*e^3),
x)

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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((a+b*arccosh(d*x+c))/(d*e*x+c*e)^(5/2),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(t_nostep)]Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is
real):Check [abs(t_nostep)]sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argum
ent Value

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maple [C]  time = 0.03, size = 269, normalized size = 1.79 \[ \frac {-\frac {2 a}{3 \left (d e x +c e \right )^{\frac {3}{2}}}+2 b \left (-\frac {\mathrm {arccosh}\left (\frac {d e x +c e}{e}\right )}{3 \left (d e x +c e \right )^{\frac {3}{2}}}+\frac {-\frac {2 \sqrt {\frac {d e x +c e +e}{e}}\, \sqrt {-\frac {d e x +c e -e}{e}}\, \sqrt {d e x +c e}\, \EllipticF \left (\sqrt {d e x +c e}\, \sqrt {-\frac {1}{e}}, i\right ) e}{3}+\frac {2 \sqrt {\frac {d e x +c e +e}{e}}\, \sqrt {-\frac {d e x +c e -e}{e}}\, \sqrt {d e x +c e}\, \EllipticE \left (\sqrt {d e x +c e}\, \sqrt {-\frac {1}{e}}, i\right ) e}{3}+\frac {2 \sqrt {-\frac {1}{e}}\, \left (d e x +c e \right )^{2}}{3}-\frac {2 \sqrt {-\frac {1}{e}}\, e^{2}}{3}}{e^{3} \sqrt {-\frac {1}{e}}\, \sqrt {d e x +c e}\, \sqrt {\frac {d e x +c e +e}{e}}\, \sqrt {\frac {d e x +c e -e}{e}}}\right )}{d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/e*(-1/3*a/(d*e*x+c*e)^(3/2)+b*(-1/3/(d*e*x+c*e)^(3/2)*arccosh((d*e*x+c*e)/e)+2/3/e^3*(-((d*e*x+c*e+e)/e)^(
1/2)*(-(d*e*x+c*e-e)/e)^(1/2)*(d*e*x+c*e)^(1/2)*EllipticF((d*e*x+c*e)^(1/2)*(-1/e)^(1/2),I)*e+((d*e*x+c*e+e)/e
)^(1/2)*(-(d*e*x+c*e-e)/e)^(1/2)*(d*e*x+c*e)^(1/2)*EllipticE((d*e*x+c*e)^(1/2)*(-1/e)^(1/2),I)*e+(-1/e)^(1/2)*
(d*e*x+c*e)^2-(-1/e)^(1/2)*e^2)/(-1/e)^(1/2)/(d*e*x+c*e)^(1/2)/((d*e*x+c*e+e)/e)^(1/2)/((d*e*x+c*e-e)/e)^(1/2)
))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{3} \, {\left (6 \, \sqrt {e} \int \frac {1}{3 \, {\left (d^{4} e^{3} x^{4} + 4 \, c d^{3} e^{3} x^{3} + c^{4} e^{3} - c^{2} e^{3} + {\left (6 \, c^{2} d^{2} e^{3} - d^{2} e^{3}\right )} x^{2} + {\left (d^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + c^{3} e^{3} - c e^{3} + {\left (3 \, c^{2} d e^{3} - d e^{3}\right )} x\right )} \sqrt {d x + c + 1} \sqrt {d x + c - 1} + 2 \, {\left (2 \, c^{3} d e^{3} - c d e^{3}\right )} x\right )} \sqrt {d x + c}}\,{d x} + \frac {\sqrt {e} {\left (-\frac {i \, {\left (\log \left (i \, \sqrt {d x + c} + 1\right ) - \log \left (-i \, \sqrt {d x + c} + 1\right )\right )}}{e^{3}} + \frac {\log \left (\sqrt {d x + c} + 1\right )}{e^{3}} - \frac {\log \left (\sqrt {d x + c} - 1\right )}{e^{3}}\right )}}{d} + \frac {2 \, \sqrt {e} \log \left (d x + \sqrt {d x + c + 1} \sqrt {d x + c - 1} + c\right )}{{\left (d^{2} e^{3} x + c d e^{3}\right )} \sqrt {d x + c}}\right )} b - \frac {2 \, a}{3 \, {\left (d e x + c e\right )}^{\frac {3}{2}} d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(6*sqrt(e)*integrate(1/3/((d^4*e^3*x^4 + 4*c*d^3*e^3*x^3 + c^4*e^3 - c^2*e^3 + (6*c^2*d^2*e^3 - d^2*e^3)*
x^2 + (d^3*e^3*x^3 + 3*c*d^2*e^3*x^2 + c^3*e^3 - c*e^3 + (3*c^2*d*e^3 - d*e^3)*x)*sqrt(d*x + c + 1)*sqrt(d*x +
c - 1) + 2*(2*c^3*d*e^3 - c*d*e^3)*x)*sqrt(d*x + c)), x) + sqrt(e)*(-I*(log(I*sqrt(d*x + c) + 1) - log(-I*sqr
t(d*x + c) + 1))/e^3 + log(sqrt(d*x + c) + 1)/e^3 - log(sqrt(d*x + c) - 1)/e^3)/d + 2*sqrt(e)*log(d*x + sqrt(d
*x + c + 1)*sqrt(d*x + c - 1) + c)/((d^2*e^3*x + c*d*e^3)*sqrt(d*x + c)))*b - 2/3*a/((d*e*x + c*e)^(3/2)*d*e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*acosh(c + d*x))/(e*(c + d*x))**(5/2), x)

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