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

Optimal. Leaf size=128 \[ -\frac{16 b^2 c^2 (f x)^{7/2} \text{HypergeometricPFQ}\left (\left \{1,\frac{7}{4},\frac{7}{4}\right \},\left \{\frac{9}{4},\frac{11}{4}\right \},c^2 x^2\right )}{105 f^3}-\frac{8 b c \sqrt{1-c x} (f x)^{5/2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{5}{4},\frac{9}{4},c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )}{15 f^2 \sqrt{c x-1}}+\frac{2 (f x)^{3/2} \left (a+b \cosh ^{-1}(c x)\right )^2}{3 f} \]

[Out]

(2*(f*x)^(3/2)*(a + b*ArcCosh[c*x])^2)/(3*f) - (8*b*c*(f*x)^(5/2)*Sqrt[1 - c*x]*(a + b*ArcCosh[c*x])*Hypergeom
etric2F1[1/2, 5/4, 9/4, c^2*x^2])/(15*f^2*Sqrt[-1 + c*x]) - (16*b^2*c^2*(f*x)^(7/2)*HypergeometricPFQ[{1, 7/4,
7/4}, {9/4, 11/4}, c^2*x^2])/(105*f^3)

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Rubi [A]  time = 0.285291, antiderivative size = 141, normalized size of antiderivative = 1.1, 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*}

Mathematica [A]  time = 0.397293, size = 118, normalized size = 0.92 \[ \frac{2}{105} x \sqrt{f x} \left (35 \left (a+b \cosh ^{-1}(c x)\right )^2-4 b c x \left (2 b c x \text{HypergeometricPFQ}\left (\left \{1,\frac{7}{4},\frac{7}{4}\right \},\left \{\frac{9}{4},\frac{11}{4}\right \},c^2 x^2\right )+\frac{7 \sqrt{1-c^2 x^2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{5}{4},\frac{9}{4},c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )}{\sqrt{c x-1} \sqrt{c x+1}}\right )\right ) \]

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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Maple [F]  time = 0.29, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b{\rm arccosh} \left (cx\right ) \right ) ^{2}\sqrt{fx}\, dx \end{align*}

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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{f x} \left (a + b \operatorname{acosh}{\left (c x \right )}\right )^{2}\, dx \end{align*}

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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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out