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

Optimal. Leaf size=145 \[ \frac {2 (e (c+d x))^{5/2} \left (a+b \cosh ^{-1}(c+d x)\right )}{5 d e}-\frac {4 b \sqrt {c+d x-1} \sqrt {c+d x+1} (e (c+d x))^{3/2}}{25 d}-\frac {12 b e \sqrt {-c-d x+1} \sqrt {e (c+d x)} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {c+d x+1}}{\sqrt {2}}\right )\right |2\right )}{25 d \sqrt {-c-d x} \sqrt {c+d x-1}} \]

[Out]

2/5*(e*(d*x+c))^(5/2)*(a+b*arccosh(d*x+c))/d/e-12/25*b*e*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/(-d*x-c)^(1/2)/(d*x+c-1)^(1/2)-4/25*b*(e*(d*x+c))^(3/2)*(d*x+c-1)^(1/2)*(d*x+c+
1)^(1/2)/d

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Rubi [A]  time = 0.11, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {5866, 5662, 102, 12, 114, 113} \[ \frac {2 (e (c+d x))^{5/2} \left (a+b \cosh ^{-1}(c+d x)\right )}{5 d e}-\frac {4 b \sqrt {c+d x-1} \sqrt {c+d x+1} (e (c+d x))^{3/2}}{25 d}-\frac {12 b e \sqrt {-c-d x+1} \sqrt {e (c+d x)} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {c+d x+1}}{\sqrt {2}}\right )\right |2\right )}{25 d \sqrt {-c-d x} \sqrt {c+d x-1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*b*Sqrt[-1 + c + d*x]*(e*(c + d*x))^(3/2)*Sqrt[1 + c + d*x])/(25*d) + (2*(e*(c + d*x))^(5/2)*(a + b*ArcCosh
[c + d*x]))/(5*d*e) - (12*b*e*Sqrt[1 - c - d*x]*Sqrt[e*(c + d*x)]*EllipticE[ArcSin[Sqrt[1 + c + d*x]/Sqrt[2]],
2])/(25*d*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 102

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))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && 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 (c e+d e x)^{3/2} \left (a+b \cosh ^{-1}(c+d x)\right ) \, dx &=\frac {\operatorname {Subst}\left (\int (e x)^{3/2} \left (a+b \cosh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {2 (e (c+d x))^{5/2} \left (a+b \cosh ^{-1}(c+d x)\right )}{5 d e}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {(e x)^{5/2}}{\sqrt {-1+x} \sqrt {1+x}} \, dx,x,c+d x\right )}{5 d e}\\ &=-\frac {4 b \sqrt {-1+c+d x} (e (c+d x))^{3/2} \sqrt {1+c+d x}}{25 d}+\frac {2 (e (c+d x))^{5/2} \left (a+b \cosh ^{-1}(c+d x)\right )}{5 d e}-\frac {(4 b) \operatorname {Subst}\left (\int \frac {3 e^2 \sqrt {e x}}{2 \sqrt {-1+x} \sqrt {1+x}} \, dx,x,c+d x\right )}{25 d e}\\ &=-\frac {4 b \sqrt {-1+c+d x} (e (c+d x))^{3/2} \sqrt {1+c+d x}}{25 d}+\frac {2 (e (c+d x))^{5/2} \left (a+b \cosh ^{-1}(c+d x)\right )}{5 d e}-\frac {(6 b e) \operatorname {Subst}\left (\int \frac {\sqrt {e x}}{\sqrt {-1+x} \sqrt {1+x}} \, dx,x,c+d x\right )}{25 d}\\ &=-\frac {4 b \sqrt {-1+c+d x} (e (c+d x))^{3/2} \sqrt {1+c+d x}}{25 d}+\frac {2 (e (c+d x))^{5/2} \left (a+b \cosh ^{-1}(c+d x)\right )}{5 d e}-\frac {\left (3 \sqrt {2} b e \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 )}{25 d \sqrt {-c-d x} \sqrt {-1+c+d x}}\\ &=-\frac {4 b \sqrt {-1+c+d x} (e (c+d x))^{3/2} \sqrt {1+c+d x}}{25 d}+\frac {2 (e (c+d x))^{5/2} \left (a+b \cosh ^{-1}(c+d x)\right )}{5 d e}-\frac {12 b e \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 )}{25 d \sqrt {-c-d x} \sqrt {-1+c+d x}}\\ \end {align*}

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Mathematica [C]  time = 0.49, size = 109, normalized size = 0.75 \[ \frac {2 (e (c+d x))^{3/2} \left (5 (c+d x) \left (a+b \cosh ^{-1}(c+d x)\right )-\frac {2 b \left (c^2+\sqrt {1-(c+d x)^2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};(c+d x)^2\right )+2 c d x+d^2 x^2-1\right )}{\sqrt {c+d x-1} \sqrt {c+d x+1}}\right )}{25 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(e*(c + d*x))^(3/2)*(5*(c + d*x)*(a + b*ArcCosh[c + d*x]) - (2*b*(-1 + c^2 + 2*c*d*x + d^2*x^2 + Sqrt[1 - (
c + d*x)^2]*Hypergeometric2F1[1/2, 3/4, 7/4, (c + d*x)^2]))/(Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x])))/(25*d)

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a d e x + a c e + {\left (b d e x + b c e\right )} \operatorname {arcosh}\left (d x + c\right )\right )} \sqrt {d e x + c e}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*d*e*x + a*c*e + (b*d*e*x + b*c*e)*arccosh(d*x + c))*sqrt(d*e*x + c*e), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^(3/2)*(a+b*arccosh(d*x+c)),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.02, size = 254, normalized size = 1.75 \[ \frac {\frac {2 \left (d e x +c e \right )^{\frac {5}{2}} a}{5}+2 b \left (\frac {\left (d e x +c e \right )^{\frac {5}{2}} \mathrm {arccosh}\left (\frac {d e x +c e}{e}\right )}{5}-\frac {2 \left (\sqrt {-\frac {1}{e}}\, \left (d e x +c e \right )^{\frac {7}{2}}+3 \sqrt {\frac {d e x +c e +e}{e}}\, \sqrt {-\frac {d e x +c e -e}{e}}\, e^{3} \EllipticF \left (\sqrt {d e x +c e}\, \sqrt {-\frac {1}{e}}, i\right )-3 e^{3} \sqrt {\frac {d e x +c e +e}{e}}\, \sqrt {-\frac {d e x +c e -e}{e}}\, \EllipticE \left (\sqrt {d e x +c e}\, \sqrt {-\frac {1}{e}}, i\right )-\sqrt {-\frac {1}{e}}\, \left (d e x +c e \right )^{\frac {3}{2}} e^{2}\right )}{25 e \sqrt {-\frac {1}{e}}\, \sqrt {\frac {d e x +c e +e}{e}}\, \sqrt {\frac {d e x +c e -e}{e}}}\right )}{d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/e*(1/5*(d*e*x+c*e)^(5/2)*a+b*(1/5*(d*e*x+c*e)^(5/2)*arccosh((d*e*x+c*e)/e)-2/25/e*((-1/e)^(1/2)*(d*e*x+c*e
)^(7/2)+3*((d*e*x+c*e+e)/e)^(1/2)*(-(d*e*x+c*e-e)/e)^(1/2)*e^3*EllipticF((d*e*x+c*e)^(1/2)*(-1/e)^(1/2),I)-3*e
^3*((d*e*x+c*e+e)/e)^(1/2)*(-(d*e*x+c*e-e)/e)^(1/2)*EllipticE((d*e*x+c*e)^(1/2)*(-1/e)^(1/2),I)-(-1/e)^(1/2)*(
d*e*x+c*e)^(3/2)*e^2)/(-1/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 \[ \frac {1}{25} \, {\left (\frac {10 \, {\left (d^{2} e^{\frac {3}{2}} x^{2} + 2 \, c d e^{\frac {3}{2}} x + c^{2} e^{\frac {3}{2}}\right )} \sqrt {d x + c} \log \left (d x + \sqrt {d x + c + 1} \sqrt {d x + c - 1} + c\right )}{d} - \frac {4 \, {\left (d x + c\right )}^{\frac {5}{2}} e^{\frac {3}{2}} + 5 i \, e^{\frac {3}{2}} {\left (\log \left (i \, \sqrt {d x + c} + 1\right ) - \log \left (-i \, \sqrt {d x + c} + 1\right )\right )} - 5 \, e^{\frac {3}{2}} \log \left (\sqrt {d x + c} + 1\right ) + 5 \, e^{\frac {3}{2}} \log \left (\sqrt {d x + c} - 1\right ) + 20 \, \sqrt {d x + c} e^{\frac {3}{2}}}{d} + 25 \, \int \frac {2 \, {\left (d^{2} e^{\frac {3}{2}} x^{2} + 2 \, c d e^{\frac {3}{2}} x + c^{2} e^{\frac {3}{2}}\right )} \sqrt {d x + c}}{5 \, {\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + c^{3} + {\left (d^{2} x^{2} + 2 \, c d x + c^{2} - 1\right )} \sqrt {d x + c + 1} \sqrt {d x + c - 1} + {\left (3 \, c^{2} d - d\right )} x - c\right )}}\,{d x}\right )} b + \frac {2 \, {\left (d e x + c e\right )}^{\frac {5}{2}} a}{5 \, d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/25*(10*(d^2*e^(3/2)*x^2 + 2*c*d*e^(3/2)*x + c^2*e^(3/2))*sqrt(d*x + c)*log(d*x + sqrt(d*x + c + 1)*sqrt(d*x
+ c - 1) + c)/d - (4*(d*x + c)^(5/2)*e^(3/2) + 5*I*e^(3/2)*(log(I*sqrt(d*x + c) + 1) - log(-I*sqrt(d*x + c) +
1)) - 5*e^(3/2)*log(sqrt(d*x + c) + 1) + 5*e^(3/2)*log(sqrt(d*x + c) - 1) + 20*sqrt(d*x + c)*e^(3/2))/d + 25*i
ntegrate(2/5*(d^2*e^(3/2)*x^2 + 2*c*d*e^(3/2)*x + c^2*e^(3/2))*sqrt(d*x + c)/(d^3*x^3 + 3*c*d^2*x^2 + c^3 + (d
^2*x^2 + 2*c*d*x + c^2 - 1)*sqrt(d*x + c + 1)*sqrt(d*x + c - 1) + (3*c^2*d - d)*x - c), x))*b + 2/5*(d*e*x + c
*e)^(5/2)*a/(d*e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((e*(c + d*x))**(3/2)*(a + b*acosh(c + d*x)), x)

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