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

Optimal. Leaf size=151 \[ -\frac {16 b^2 (e (c+d x))^{5/2} \, _3F_2\left (1,\frac {5}{4},\frac {5}{4};\frac {7}{4},\frac {9}{4};(c+d x)^2\right )}{15 d e^3}-\frac {8 b \sqrt {-c-d x+1} (e (c+d x))^{3/2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};(c+d x)^2\right ) \left (a+b \cosh ^{-1}(c+d x)\right )}{3 d e^2 \sqrt {c+d x-1}}+\frac {2 \sqrt {e (c+d x)} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{d e} \]

[Out]

-16/15*b^2*(e*(d*x+c))^(5/2)*HypergeometricPFQ([1, 5/4, 5/4],[7/4, 9/4],(d*x+c)^2)/d/e^3-8/3*b*(e*(d*x+c))^(3/
2)*(a+b*arccosh(d*x+c))*hypergeom([1/2, 3/4],[7/4],(d*x+c)^2)*(-d*x-c+1)^(1/2)/d/e^2/(d*x+c-1)^(1/2)+2*(a+b*ar
ccosh(d*x+c))^2*(e*(d*x+c))^(1/2)/d/e

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Rubi [A]  time = 0.29, antiderivative size = 163, normalized size of antiderivative = 1.08, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {5866, 5662, 5763} \[ -\frac {16 b^2 (e (c+d x))^{5/2} \, _3F_2\left (1,\frac {5}{4},\frac {5}{4};\frac {7}{4},\frac {9}{4};(c+d x)^2\right )}{15 d e^3}-\frac {8 b \sqrt {1-(c+d x)^2} (e (c+d x))^{3/2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};(c+d x)^2\right ) \left (a+b \cosh ^{-1}(c+d x)\right )}{3 d e^2 \sqrt {c+d x-1} \sqrt {c+d x+1}}+\frac {2 \sqrt {e (c+d x)} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[e*(c + d*x)]*(a + b*ArcCosh[c + d*x])^2)/(d*e) - (8*b*(e*(c + d*x))^(3/2)*Sqrt[1 - (c + d*x)^2]*(a + b
*ArcCosh[c + d*x])*Hypergeometric2F1[1/2, 3/4, 7/4, (c + d*x)^2])/(3*d*e^2*Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x
]) - (16*b^2*(e*(c + d*x))^(5/2)*HypergeometricPFQ[{1, 5/4, 5/4}, {7/4, 9/4}, (c + d*x)^2])/(15*d*e^3)

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 5763

Int[(((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_))/(Sqrt[(d1_) + (e1_.)*(x_)]*Sqrt[(d2_) + (e2_.)*(x_
)]), x_Symbol] :> Simp[((f*x)^(m + 1)*Sqrt[1 - c^2*x^2]*(a + b*ArcCosh[c*x])*Hypergeometric2F1[1/2, (1 + m)/2,
(3 + m)/2, c^2*x^2])/(f*(m + 1)*Sqrt[d1 + e1*x]*Sqrt[d2 + e2*x]), x] + Simp[(b*c*(f*x)^(m + 2)*Hypergeometric
PFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2])/(Sqrt[-(d1*d2)]*f^2*(m + 1)*(m + 2)), x] /; FreeQ[{
a, b, c, d1, e1, d2, e2, f, m}, x] && EqQ[e1 - c*d1, 0] && EqQ[e2 + c*d2, 0] && GtQ[d1, 0] && LtQ[d2, 0] &&  !
IntegerQ[m]

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 {\left (a+b \cosh ^{-1}(c+d x)\right )^2}{\sqrt {c e+d e x}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \cosh ^{-1}(x)\right )^2}{\sqrt {e x}} \, dx,x,c+d x\right )}{d}\\ &=\frac {2 \sqrt {e (c+d x)} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{d e}-\frac {(4 b) \operatorname {Subst}\left (\int \frac {\sqrt {e x} \left (a+b \cosh ^{-1}(x)\right )}{\sqrt {-1+x} \sqrt {1+x}} \, dx,x,c+d x\right )}{d e}\\ &=\frac {2 \sqrt {e (c+d x)} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{d e}-\frac {8 b (e (c+d x))^{3/2} \sqrt {1-(c+d x)^2} \left (a+b \cosh ^{-1}(c+d x)\right ) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};(c+d x)^2\right )}{3 d e^2 \sqrt {-1+c+d x} \sqrt {1+c+d x}}-\frac {16 b^2 (e (c+d x))^{5/2} \, _3F_2\left (1,\frac {5}{4},\frac {5}{4};\frac {7}{4},\frac {9}{4};(c+d x)^2\right )}{15 d e^3}\\ \end {align*}

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Mathematica [A]  time = 0.30, size = 140, normalized size = 0.93 \[ \frac {2 \sqrt {e (c+d x)} \left (15 \left (a+b \cosh ^{-1}(c+d x)\right )^2-4 b (c+d x) \left (2 b (c+d x) \, _3F_2\left (1,\frac {5}{4},\frac {5}{4};\frac {7}{4},\frac {9}{4};(c+d x)^2\right )+\frac {5 \sqrt {1-(c+d x)^2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};(c+d x)^2\right ) \left (a+b \cosh ^{-1}(c+d x)\right )}{\sqrt {c+d x-1} \sqrt {c+d x+1}}\right )\right )}{15 d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[e*(c + d*x)]*(15*(a + b*ArcCosh[c + d*x])^2 - 4*b*(c + d*x)*((5*Sqrt[1 - (c + d*x)^2]*(a + b*ArcCosh[c
+ d*x])*Hypergeometric2F1[1/2, 3/4, 7/4, (c + d*x)^2])/(Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x]) + 2*b*(c + d*x)
*HypergeometricPFQ[{1, 5/4, 5/4}, {7/4, 9/4}, (c + d*x)^2])))/(15*d*e)

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \operatorname {arcosh}\left (d x + c\right )^{2} + 2 \, a b \operatorname {arcosh}\left (d x + c\right ) + a^{2}}{\sqrt {d e x + c e}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(d*x+c))^2/(d*e*x+c*e)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(d*x+c))^2/(d*e*x+c*e)^(1/2),x, algorithm="giac")

[Out]

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

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \,\mathrm {arccosh}\left (d x +c \right )\right )^{2}}{\sqrt {d e x +c e}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccosh(d*x+c))^2/(d*e*x+c*e)^(1/2),x)

[Out]

int((a+b*arccosh(d*x+c))^2/(d*e*x+c*e)^(1/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(d*x+c))^2/(d*e*x+c*e)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(c + d*x))^2/(c*e + d*e*x)^(1/2),x)

[Out]

int((a + b*acosh(c + d*x))^2/(c*e + d*e*x)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acosh(d*x+c))**2/(d*e*x+c*e)**(1/2),x)

[Out]

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

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