3.163 \(\int \sqrt {f x} (a+b \cosh ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=128 \[ -\frac {16 b^2 c^2 (f x)^{7/2} \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};c^2 x^2\right )}{105 f^3}-\frac {8 b c \sqrt {1-c x} (f x)^{5/2} \, _2F_1\left (\frac {1}{2},\frac {5}{4};\frac {9}{4};c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )}{15 f^2 \sqrt {c x-1}}+\frac {2 (f x)^{3/2} \left (a+b \cosh ^{-1}(c x)\right )^2}{3 f} \]

[Out]

2/3*(f*x)^(3/2)*(a+b*arccosh(c*x))^2/f-16/105*b^2*c^2*(f*x)^(7/2)*HypergeometricPFQ([1, 7/4, 7/4],[9/4, 11/4],
c^2*x^2)/f^3-8/15*b*c*(f*x)^(5/2)*(a+b*arccosh(c*x))*hypergeom([1/2, 5/4],[9/4],c^2*x^2)*(-c*x+1)^(1/2)/f^2/(c
*x-1)^(1/2)

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Rubi [A]  time = 0.29, antiderivative size = 141, normalized size of antiderivative = 1.10, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {5662, 5763} \[ -\frac {16 b^2 c^2 (f x)^{7/2} \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};c^2 x^2\right )}{105 f^3}-\frac {8 b c \sqrt {1-c^2 x^2} (f x)^{5/2} \, _2F_1\left (\frac {1}{2},\frac {5}{4};\frac {9}{4};c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )}{15 f^2 \sqrt {c x-1} \sqrt {c x+1}}+\frac {2 (f x)^{3/2} \left (a+b \cosh ^{-1}(c x)\right )^2}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(f*x)^(3/2)*(a + b*ArcCosh[c*x])^2)/(3*f) - (8*b*c*(f*x)^(5/2)*Sqrt[1 - c^2*x^2]*(a + b*ArcCosh[c*x])*Hyper
geometric2F1[1/2, 5/4, 9/4, c^2*x^2])/(15*f^2*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) - (16*b^2*c^2*(f*x)^(7/2)*Hypergeo
metricPFQ[{1, 7/4, 7/4}, {9/4, 11/4}, c^2*x^2])/(105*f^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]

Rubi steps

\begin {align*} \int \sqrt {f x} \left (a+b \cosh ^{-1}(c x)\right )^2 \, dx &=\frac {2 (f x)^{3/2} \left (a+b \cosh ^{-1}(c x)\right )^2}{3 f}-\frac {(4 b c) \int \frac {(f x)^{3/2} \left (a+b \cosh ^{-1}(c x)\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{3 f}\\ &=\frac {2 (f x)^{3/2} \left (a+b \cosh ^{-1}(c x)\right )^2}{3 f}-\frac {8 b c (f x)^{5/2} \sqrt {1-c^2 x^2} \left (a+b \cosh ^{-1}(c x)\right ) \, _2F_1\left (\frac {1}{2},\frac {5}{4};\frac {9}{4};c^2 x^2\right )}{15 f^2 \sqrt {-1+c x} \sqrt {1+c x}}-\frac {16 b^2 c^2 (f x)^{7/2} \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};c^2 x^2\right )}{105 f^3}\\ \end {align*}

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Mathematica [A]  time = 0.43, size = 118, normalized size = 0.92 \[ \frac {2}{105} x \sqrt {f x} \left (35 \left (a+b \cosh ^{-1}(c x)\right )^2-4 b c x \left (2 b c x \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};c^2 x^2\right )+\frac {7 \sqrt {1-c^2 x^2} \, _2F_1\left (\frac {1}{2},\frac {5}{4};\frac {9}{4};c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )}{\sqrt {c x-1} \sqrt {c x+1}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*Sqrt[f*x]*(35*(a + b*ArcCosh[c*x])^2 - 4*b*c*x*((7*Sqrt[1 - c^2*x^2]*(a + b*ArcCosh[c*x])*Hypergeometric2
F1[1/2, 5/4, 9/4, c^2*x^2])/(Sqrt[-1 + c*x]*Sqrt[1 + c*x]) + 2*b*c*x*HypergeometricPFQ[{1, 7/4, 7/4}, {9/4, 11
/4}, c^2*x^2])))/105

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{2} \operatorname {arcosh}\left (c x\right )^{2} + 2 \, a b \operatorname {arcosh}\left (c x\right ) + a^{2}\right )} \sqrt {f x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*arccosh(c*x)^2 + 2*a*b*arccosh(c*x) + a^2)*sqrt(f*x), x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:sym2poly/r2sym(co
nst gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*b^2*sqrt(f)*x^(3/2)*log(c*x + sqrt(c*x + 1)*sqrt(c*x - 1))^2 + 2/3*(f*x)^(3/2)*a^2/f + integrate(2/3*(((3*
a*b*c^2*sqrt(f) - 2*b^2*c^2*sqrt(f))*x^2 - 3*a*b*sqrt(f))*sqrt(c*x + 1)*sqrt(c*x - 1)*sqrt(x) + ((3*a*b*c^3*sq
rt(f) - 2*b^2*c^3*sqrt(f))*x^3 - (3*a*b*c*sqrt(f) - 2*b^2*c*sqrt(f))*x)*sqrt(x))*log(c*x + sqrt(c*x + 1)*sqrt(
c*x - 1))/(c^3*x^3 + (c^2*x^2 - 1)*sqrt(c*x + 1)*sqrt(c*x - 1) - c*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(f*x)*(a + b*acosh(c*x))**2, x)

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