3.163 \(\int \frac {(f x)^m \cosh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx\)

Optimal. Leaf size=128 \[ \frac {a \sqrt {a x-1} (f x)^{m+2} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;a^2 x^2\right )}{f^2 (m+1) (m+2) \sqrt {1-a x}}+\frac {\cosh ^{-1}(a x) (f x)^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};a^2 x^2\right )}{f (m+1)} \]

[Out]

(f*x)^(1+m)*arccosh(a*x)*hypergeom([1/2, 1/2+1/2*m],[3/2+1/2*m],a^2*x^2)/f/(1+m)+a*(f*x)^(2+m)*HypergeometricP
FQ([1, 1+1/2*m, 1+1/2*m],[3/2+1/2*m, 2+1/2*m],a^2*x^2)*(a*x-1)^(1/2)/f^2/(1+m)/(2+m)/(-a*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 = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {5798, 5763} \[ \frac {a \sqrt {a x-1} \sqrt {a x+1} (f x)^{m+2} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;a^2 x^2\right )}{f^2 (m+1) (m+2) \sqrt {1-a^2 x^2}}+\frac {\cosh ^{-1}(a x) (f x)^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};a^2 x^2\right )}{f (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[((f*x)^m*ArcCosh[a*x])/Sqrt[1 - a^2*x^2],x]

[Out]

((f*x)^(1 + m)*ArcCosh[a*x]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, a^2*x^2])/(f*(1 + m)) + (a*(f*x)^(2 +
m)*Sqrt[-1 + a*x]*Sqrt[1 + a*x]*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, a^2*x^2])/(f^2
*(1 + m)*(2 + m)*Sqrt[1 - a^2*x^2])

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 5798

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Dist
[((-d)^IntPart[p]*(d + e*x^2)^FracPart[p])/((1 + c*x)^FracPart[p]*(-1 + c*x)^FracPart[p]), Int[(f*x)^m*(1 + c*
x)^p*(-1 + c*x)^p*(a + b*ArcCosh[c*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[c^2*d + e, 0]
&&  !IntegerQ[p]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 124, normalized size = 0.97 \[ \frac {x (f x)^m \left (\frac {a x \sqrt {a x-1} \sqrt {a x+1} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;a^2 x^2\right )}{(m+2) \sqrt {1-a^2 x^2}}+\cosh ^{-1}(a x) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};a^2 x^2\right )\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

Integrate[((f*x)^m*ArcCosh[a*x])/Sqrt[1 - a^2*x^2],x]

[Out]

(x*(f*x)^m*(ArcCosh[a*x]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, a^2*x^2] + (a*x*Sqrt[-1 + a*x]*Sqrt[1 +
a*x]*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, a^2*x^2])/((2 + m)*Sqrt[1 - a^2*x^2])))/(1
+ m)

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} \left (f x\right )^{m} \operatorname {arcosh}\left (a x\right )}{a^{2} x^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-a^2*x^2 + 1)*(f*x)^m*arccosh(a*x)/(a^2*x^2 - 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x\right )^{m} \operatorname {arcosh}\left (a x\right )}{\sqrt {-a^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((f*x)^m*arccosh(a*x)/sqrt(-a^2*x^2 + 1), x)

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maple [F]  time = 0.57, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x \right )^{m} \mathrm {arccosh}\left (a x \right )}{\sqrt {-a^{2} x^{2}+1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x)^m*arccosh(a*x)/(-a^2*x^2+1)^(1/2),x)

[Out]

int((f*x)^m*arccosh(a*x)/(-a^2*x^2+1)^(1/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((f*x)^m*arccosh(a*x)/sqrt(-a^2*x^2 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((acosh(a*x)*(f*x)^m)/(1 - a^2*x^2)^(1/2),x)

[Out]

int((acosh(a*x)*(f*x)^m)/(1 - a^2*x^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x\right )^{m} \operatorname {acosh}{\left (a x \right )}}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)**m*acosh(a*x)/(-a**2*x**2+1)**(1/2),x)

[Out]

Integral((f*x)**m*acosh(a*x)/sqrt(-(a*x - 1)*(a*x + 1)), x)

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