3.164 \(\int (d x)^m (a+b \cosh ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=181 \[ -\frac {2 b^2 c^2 (d x)^{m+3} \, _3F_2\left (1,\frac {m}{2}+\frac {3}{2},\frac {m}{2}+\frac {3}{2};\frac {m}{2}+2,\frac {m}{2}+\frac {5}{2};c^2 x^2\right )}{d^3 (m+1) (m+2) (m+3)}-\frac {2 b c \sqrt {1-c x} (d x)^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )}{d^2 (m+1) (m+2) \sqrt {c x-1}}+\frac {(d x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )^2}{d (m+1)} \]

[Out]

(d*x)^(1+m)*(a+b*arccosh(c*x))^2/d/(1+m)-2*b^2*c^2*(d*x)^(3+m)*HypergeometricPFQ([1, 3/2+1/2*m, 3/2+1/2*m],[2+
1/2*m, 5/2+1/2*m],c^2*x^2)/d^3/(3+m)/(m^2+3*m+2)-2*b*c*(d*x)^(2+m)*(a+b*arccosh(c*x))*hypergeom([1/2, 1+1/2*m]
,[2+1/2*m],c^2*x^2)*(-c*x+1)^(1/2)/d^2/(1+m)/(2+m)/(c*x-1)^(1/2)

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Rubi [A]  time = 0.31, antiderivative size = 194, normalized size of antiderivative = 1.07, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5662, 5763} \[ -\frac {2 b^2 c^2 (d x)^{m+3} \, _3F_2\left (1,\frac {m}{2}+\frac {3}{2},\frac {m}{2}+\frac {3}{2};\frac {m}{2}+2,\frac {m}{2}+\frac {5}{2};c^2 x^2\right )}{d^3 (m+1) (m+2) (m+3)}-\frac {2 b c \sqrt {1-c^2 x^2} (d x)^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )}{d^2 (m+1) (m+2) \sqrt {c x-1} \sqrt {c x+1}}+\frac {(d x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )^2}{d (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^m*(a + b*ArcCosh[c*x])^2,x]

[Out]

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

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 (d x)^m \left (a+b \cosh ^{-1}(c x)\right )^2 \, dx &=\frac {(d x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )^2}{d (1+m)}-\frac {(2 b c) \int \frac {(d x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{d (1+m)}\\ &=\frac {(d x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )^2}{d (1+m)}-\frac {2 b c (d x)^{2+m} \sqrt {1-c^2 x^2} \left (a+b \cosh ^{-1}(c x)\right ) \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};c^2 x^2\right )}{d^2 (1+m) (2+m) \sqrt {-1+c x} \sqrt {1+c x}}-\frac {2 b^2 c^2 (d x)^{3+m} \, _3F_2\left (1,\frac {3}{2}+\frac {m}{2},\frac {3}{2}+\frac {m}{2};2+\frac {m}{2},\frac {5}{2}+\frac {m}{2};c^2 x^2\right )}{d^3 (1+m) (2+m) (3+m)}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 164, normalized size = 0.91 \[ \frac {x (d x)^m \left (-\frac {2 b^2 c^2 x^2 \, _3F_2\left (1,\frac {m}{2}+\frac {3}{2},\frac {m}{2}+\frac {3}{2};\frac {m}{2}+2,\frac {m}{2}+\frac {5}{2};c^2 x^2\right )}{m^2+5 m+6}-\frac {2 b c x \sqrt {1-c^2 x^2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )}{(m+2) \sqrt {c x-1} \sqrt {c x+1}}+\left (a+b \cosh ^{-1}(c x)\right )^2\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(d*x)^m*((a + b*ArcCosh[c*x])^2 - (2*b*c*x*Sqrt[1 - c^2*x^2]*(a + b*ArcCosh[c*x])*Hypergeometric2F1[1/2, (2
+ m)/2, (4 + m)/2, c^2*x^2])/((2 + m)*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) - (2*b^2*c^2*x^2*HypergeometricPFQ[{1, 3/
2 + m/2, 3/2 + m/2}, {2 + m/2, 5/2 + m/2}, c^2*x^2])/(6 + 5*m + m^2)))/(1 + m)

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{2} \operatorname {arcosh}\left (c x\right )^{2} + 2 \, a b \operatorname {arcosh}\left (c x\right ) + a^{2}\right )} \left (d x\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*arccosh(c*x)^2 + 2*a*b*arccosh(c*x) + a^2)*(d*x)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arccosh(c*x) + a)^2*(d*x)^m, x)

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maple [F]  time = 3.19, size = 0, normalized size = 0.00 \[ \int \left (d x \right )^{m} \left (a +b \,\mathrm {arccosh}\left (c x \right )\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m*(a+b*arccosh(c*x))^2,x)

[Out]

int((d*x)^m*(a+b*arccosh(c*x))^2,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(c*x))^2*(d*x)^m,x)

[Out]

int((a + b*acosh(c*x))^2*(d*x)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**m*(a+b*acosh(c*x))**2,x)

[Out]

Integral((d*x)**m*(a + b*acosh(c*x))**2, x)

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