3.3.4 \(\int \frac {a+b \cosh ^{-1}(c+d x)}{(c e+d e x)^{5/2}} \, dx\) [204]

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

[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 = 0.09, 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 = {5996, 5883, 106, 12, 16, 115, 114} \begin {gather*} -\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 .\text {ArcSin}\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)}} \end {gather*}

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 106

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 114

Int[Sqrt[(e_.) + (f_.)*(x_)]/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[(2/b)*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)))], 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 115

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 5883

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)/(Sqrt
[1 + c*x]*Sqrt[-1 + c*x])), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 5996

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 {\text {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) \text {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) \text {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) \text {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) \text {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 ) \text {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] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.10, size = 94, normalized size = 0.63 \begin {gather*} \frac {2 \left (-a-b \cosh ^{-1}(c+d x)-\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 {-1+c+d x} \sqrt {1+c+d x}}\right )}{3 d e (e (c+d x))^{3/2}} \end {gather*}

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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Maple [C] Result contains complex when optimal does not.
time = 0.06, size = 268, normalized size = 1.79

method	result	size
derivativedivides	\(\frac {-\frac {2 a}{3 \left (d e x +c e \right )^{\frac {3}{2}}}+2 b \left (-\frac {\mathrm {arccosh}\left (\frac {d e x +c e}{e}\right )}{3 \left (d e x +c e \right )^{\frac {3}{2}}}+\frac {-\frac {2 \sqrt {\frac {d e x +c e +e}{e}}\, \sqrt {\frac {-d e x -c e +e}{e}}\, \sqrt {d e x +c e}\, \EllipticF \left (\sqrt {d e x +c e}\, \sqrt {-\frac {1}{e}}, i\right ) e}{3}+\frac {2 \sqrt {\frac {d e x +c e +e}{e}}\, \sqrt {\frac {-d e x -c e +e}{e}}\, \sqrt {d e x +c e}\, \EllipticE \left (\sqrt {d e x +c e}\, \sqrt {-\frac {1}{e}}, i\right ) e}{3}+\frac {2 \sqrt {-\frac {1}{e}}\, \left (d e x +c e \right )^{2}}{3}-\frac {2 \sqrt {-\frac {1}{e}}\, e^{2}}{3}}{e^{3} \sqrt {-\frac {1}{e}}\, \sqrt {d e x +c e}\, \sqrt {\frac {d e x +c e +e}{e}}\, \sqrt {-\frac {-d e x -c e +e}{e}}}\right )}{d e}\)	\(268\)
default	\(\frac {-\frac {2 a}{3 \left (d e x +c e \right )^{\frac {3}{2}}}+2 b \left (-\frac {\mathrm {arccosh}\left (\frac {d e x +c e}{e}\right )}{3 \left (d e x +c e \right )^{\frac {3}{2}}}+\frac {-\frac {2 \sqrt {\frac {d e x +c e +e}{e}}\, \sqrt {\frac {-d e x -c e +e}{e}}\, \sqrt {d e x +c e}\, \EllipticF \left (\sqrt {d e x +c e}\, \sqrt {-\frac {1}{e}}, i\right ) e}{3}+\frac {2 \sqrt {\frac {d e x +c e +e}{e}}\, \sqrt {\frac {-d e x -c e +e}{e}}\, \sqrt {d e x +c e}\, \EllipticE \left (\sqrt {d e x +c e}\, \sqrt {-\frac {1}{e}}, i\right ) e}{3}+\frac {2 \sqrt {-\frac {1}{e}}\, \left (d e x +c e \right )^{2}}{3}-\frac {2 \sqrt {-\frac {1}{e}}\, e^{2}}{3}}{e^{3} \sqrt {-\frac {1}{e}}\, \sqrt {d e x +c e}\, \sqrt {\frac {d e x +c e +e}{e}}\, \sqrt {-\frac {-d e x -c e +e}{e}}}\right )}{d e}\)	\(268\)

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,method=_RETURNVERBOSE)

[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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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*e^(1/2)*integrate(1/3/((d^4*x^4*e^3 + 4*c*d^3*x^3*e^3 + (6*c^2*d^2 - d^2)*x^2*e^3 + 2*(2*c^3*d - c*d)*
x*e^3 + (d^3*x^3*e^3 + 3*c*d^2*x^2*e^3 + (3*c^2*d - d)*x*e^3 + (c^3 - c)*e^3)*sqrt(d*x + c + 1)*sqrt(d*x + c -
1) + (c^4 - c^2)*e^3)*sqrt(d*x + c)), x) + 2*e^(1/2)*log(d*x + sqrt(d*x + c + 1)*sqrt(d*x + c - 1) + c)/((d^2
*x*e^3 + c*d*e^3)*sqrt(d*x + c)) + (-I*(log(I*sqrt(d*x + c) + 1) - log(-I*sqrt(d*x + c) + 1))*e^(-5/2) - e^(-5
/2)*log(-(e^(1/2) - e^(1/2*log(d*x + c) + 1/2))/(e^(1/2) + e^(1/2*log(d*x + c) + 1/2))))/d)*b - 2/3*a*e^(-1)/(
(d*x*e + c*e)^(3/2)*d)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 283, normalized size = 1.89 \begin {gather*} -\frac {2 \, {\left (\sqrt {{\left (d x + c\right )} \cosh \left (1\right ) + {\left (d x + c\right )} \sinh \left (1\right )} b d \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\right ) - 2 \, {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \sqrt {d^{3} \cosh \left (1\right ) + d^{3} \sinh \left (1\right )} {\rm weierstrassZeta}\left (\frac {4}{d^{2}}, 0, {\rm weierstrassPInverse}\left (\frac {4}{d^{2}}, 0, \frac {d x + c}{d}\right )\right ) + {\left (a d - 2 \, {\left (b d^{2} x + b c d\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\right )} \sqrt {{\left (d x + c\right )} \cosh \left (1\right ) + {\left (d x + c\right )} \sinh \left (1\right )}\right )}}{3 \, {\left ({\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )} \cosh \left (1\right )^{3} + 3 \, {\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )} \cosh \left (1\right )^{2} \sinh \left (1\right ) + 3 \, {\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )} \cosh \left (1\right ) \sinh \left (1\right )^{2} + {\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )} \sinh \left (1\right )^{3}\right )}} \end {gather*}

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]

-2/3*(sqrt((d*x + c)*cosh(1) + (d*x + c)*sinh(1))*b*d*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 - 1)) - 2*(b*
d^2*x^2 + 2*b*c*d*x + b*c^2)*sqrt(d^3*cosh(1) + d^3*sinh(1))*weierstrassZeta(4/d^2, 0, weierstrassPInverse(4/d
^2, 0, (d*x + c)/d)) + (a*d - 2*(b*d^2*x + b*c*d)*sqrt(d^2*x^2 + 2*c*d*x + c^2 - 1))*sqrt((d*x + c)*cosh(1) +
(d*x + c)*sinh(1)))/((d^4*x^2 + 2*c*d^3*x + c^2*d^2)*cosh(1)^3 + 3*(d^4*x^2 + 2*c*d^3*x + c^2*d^2)*cosh(1)^2*s
inh(1) + 3*(d^4*x^2 + 2*c*d^3*x + c^2*d^2)*cosh(1)*sinh(1)^2 + (d^4*x^2 + 2*c*d^3*x + c^2*d^2)*sinh(1)^3)

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

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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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((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,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(t_

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

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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