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

Optimal. Leaf size=176 \[ \frac{b c \sqrt{c x-1} \sqrt{c x+1} (f x)^{m+2} \text{HypergeometricPFQ}\left (\left \{1,\frac{m}{2}+1,\frac{m}{2}+1\right \},\left \{\frac{m}{2}+\frac{3}{2},\frac{m}{2}+2\right \},c^2 x^2\right )}{f^2 (m+1) (m+2) \sqrt{d-c^2 d x^2}}+\frac{\sqrt{1-c^2 x^2} (f x)^{m+1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )}{f (m+1) \sqrt{d-c^2 d x^2}} \]

[Out]

((f*x)^(1 + m)*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*(1 + m)*Sqrt[d - c^2*d*x^2]) + (b*c*(f*x)^(2 + m)*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*HypergeometricPFQ[{1, 1 + m/2
, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2])/(f^2*(1 + m)*(2 + m)*Sqrt[d - c^2*d*x^2])

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Rubi [A]  time = 0.352383, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {5798, 5763} \[ \frac{b c \sqrt{c x-1} \sqrt{c x+1} (f x)^{m+2} \, _3F_2\left (1,\frac{m}{2}+1,\frac{m}{2}+1;\frac{m}{2}+\frac{3}{2},\frac{m}{2}+2;c^2 x^2\right )}{f^2 (m+1) (m+2) \sqrt{d-c^2 d x^2}}+\frac{\sqrt{1-c^2 x^2} (f x)^{m+1} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )}{f (m+1) \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0756048, size = 147, normalized size = 0.84 \[ \frac{x (f x)^m \left (b c x \sqrt{c x-1} \sqrt{c x+1} \text{HypergeometricPFQ}\left (\left \{1,\frac{m}{2}+1,\frac{m}{2}+1\right \},\left \{\frac{m}{2}+\frac{3}{2},\frac{m}{2}+2\right \},c^2 x^2\right )+(m+2) \sqrt{1-c^2 x^2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )\right )}{(m+1) (m+2) \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.45, size = 0, normalized size = 0. \begin{align*} \int{ \left ( fx \right ) ^{m} \left ( a+b{\rm arccosh} \left (cx\right ) \right ){\frac{1}{\sqrt{-{c}^{2}d{x}^{2}+d}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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