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

Optimal. Leaf size=128 \[ \frac{a \sqrt{a x-1} (f x)^{m+2} \text{HypergeometricPFQ}\left (\left \{1,\frac{m}{2}+1,\frac{m}{2}+1\right \},\left \{\frac{m}{2}+\frac{3}{2},\frac{m}{2}+2\right \},a^2 x^2\right )}{f^2 (m+1) (m+2) \sqrt{1-a x}}+\frac{\cosh ^{-1}(a x) (f x)^{m+1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},a^2 x^2\right )}{f (m+1)} \]

[Out]

((f*x)^(1 + m)*ArcCosh[a*x]*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]*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*x])

________________________________________________________________________________________

Rubi [A]  time = 0.286776, antiderivative size = 141, normalized size of antiderivative = 1.1, 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 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]

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 \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*}

Mathematica [A]  time = 0.0765463, size = 124, normalized size = 0.97 \[ \frac{x (f x)^m \left (\frac{a x \sqrt{a x-1} \sqrt{a x+1} \text{HypergeometricPFQ}\left (\left \{1,\frac{m}{2}+1,\frac{m}{2}+1\right \},\left \{\frac{m}{2}+\frac{3}{2},\frac{m}{2}+2\right \},a^2 x^2\right )}{(m+2) \sqrt{1-a^2 x^2}}+\cosh ^{-1}(a x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},a^2 x^2\right )\right )}{m+1} \]

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)

________________________________________________________________________________________

Maple [F]  time = 0.368, size = 0, normalized size = 0. \begin{align*} \int{ \left ( fx \right ) ^{m}{\rm arccosh} \left (ax\right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}\, dx \end{align*}

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)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (f x\right )^{m} \operatorname{arcosh}\left (a x\right )}{\sqrt{-a^{2} x^{2} + 1}}\,{d x} \end{align*}

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (f x\right )^{m} \operatorname{arcosh}\left (a x\right )}{\sqrt{-a^{2} x^{2} + 1}}\,{d x} \end{align*}

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)