3.156 \(\int \frac{(f x)^m (a+b \cosh ^{-1}(c x))}{(d-c^2 d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=450 \[ -\frac{b c (2-m) m \sqrt{c x-1} \sqrt{c x+1} (f x)^{m+2} \text{HypergeometricPFQ}\left (\left \{1,\frac{m}{2}+1,\frac{m}{2}+1\right \},\left \{\frac{m}{2}+\frac{3}{2},\frac{m}{2}+2\right \},c^2 x^2\right )}{3 d^2 f^2 (m+1) (m+2) \sqrt{d-c^2 d x^2}}-\frac{(2-m) m \sqrt{1-c^2 x^2} (f x)^{m+1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )}{3 d^2 f (m+1) \sqrt{d-c^2 d x^2}}+\frac{b c (2-m) \sqrt{c x-1} \sqrt{c x+1} (f x)^{m+2} \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},c^2 x^2\right )}{3 d^2 f^2 (m+2) \sqrt{d-c^2 d x^2}}+\frac{b c \sqrt{c x-1} \sqrt{c x+1} (f x)^{m+2} \text{Hypergeometric2F1}\left (2,\frac{m+2}{2},\frac{m+4}{2},c^2 x^2\right )}{3 d^2 f^2 (m+2) \sqrt{d-c^2 d x^2}}+\frac{(2-m) (f x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{3 d^2 f \sqrt{d-c^2 d x^2}}+\frac{(f x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{3 d f \left (d-c^2 d x^2\right )^{3/2}} \]

[Out]

((f*x)^(1 + m)*(a + b*ArcCosh[c*x]))/(3*d*f*(d - c^2*d*x^2)^(3/2)) + ((2 - m)*(f*x)^(1 + m)*(a + b*ArcCosh[c*x
]))/(3*d^2*f*Sqrt[d - c^2*d*x^2]) - ((2 - m)*m*(f*x)^(1 + m)*Sqrt[1 - c^2*x^2]*(a + b*ArcCosh[c*x])*Hypergeome
tric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2])/(3*d^2*f*(1 + m)*Sqrt[d - c^2*d*x^2]) + (b*c*(2 - m)*(f*x)^(2 + m
)*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, c^2*x^2])/(3*d^2*f^2*(2 + m)*Sqrt[d
- c^2*d*x^2]) + (b*c*(f*x)^(2 + m)*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*Hypergeometric2F1[2, (2 + m)/2, (4 + m)/2, c^2
*x^2])/(3*d^2*f^2*(2 + m)*Sqrt[d - c^2*d*x^2]) - (b*c*(2 - m)*m*(f*x)^(2 + m)*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*Hyp
ergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2])/(3*d^2*f^2*(1 + m)*(2 + m)*Sqrt[d - c^2*
d*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.993008, antiderivative size = 465, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {5798, 5756, 5763, 364} \[ -\frac{b c (2-m) m \sqrt{c x-1} \sqrt{c x+1} (f x)^{m+2} \, _3F_2\left (1,\frac{m}{2}+1,\frac{m}{2}+1;\frac{m}{2}+\frac{3}{2},\frac{m}{2}+2;c^2 x^2\right )}{3 d^2 f^2 (m+1) (m+2) \sqrt{d-c^2 d x^2}}-\frac{(2-m) m \sqrt{1-c^2 x^2} (f x)^{m+1} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )}{3 d^2 f (m+1) \sqrt{d-c^2 d x^2}}+\frac{(2-m) (f x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{3 d^2 f \sqrt{d-c^2 d x^2}}+\frac{(f x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{3 d^2 f (1-c x) (c x+1) \sqrt{d-c^2 d x^2}}+\frac{b c (2-m) \sqrt{c x-1} \sqrt{c x+1} (f x)^{m+2} \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};c^2 x^2\right )}{3 d^2 f^2 (m+2) \sqrt{d-c^2 d x^2}}+\frac{b c \sqrt{c x-1} \sqrt{c x+1} (f x)^{m+2} \, _2F_1\left (2,\frac{m+2}{2};\frac{m+4}{2};c^2 x^2\right )}{3 d^2 f^2 (m+2) \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((f*x)^m*(a + b*ArcCosh[c*x]))/(d - c^2*d*x^2)^(5/2),x]

[Out]

((2 - m)*(f*x)^(1 + m)*(a + b*ArcCosh[c*x]))/(3*d^2*f*Sqrt[d - c^2*d*x^2]) + ((f*x)^(1 + m)*(a + b*ArcCosh[c*x
]))/(3*d^2*f*(1 - c*x)*(1 + c*x)*Sqrt[d - c^2*d*x^2]) - ((2 - m)*m*(f*x)^(1 + m)*Sqrt[1 - c^2*x^2]*(a + b*ArcC
osh[c*x])*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2])/(3*d^2*f*(1 + m)*Sqrt[d - c^2*d*x^2]) + (b*c*
(2 - m)*(f*x)^(2 + m)*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, c^2*x^2])/(3*d^2
*f^2*(2 + m)*Sqrt[d - c^2*d*x^2]) + (b*c*(f*x)^(2 + m)*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*Hypergeometric2F1[2, (2 +
m)/2, (4 + m)/2, c^2*x^2])/(3*d^2*f^2*(2 + m)*Sqrt[d - c^2*d*x^2]) - (b*c*(2 - m)*m*(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])/(3*d^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 5756

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d1_) + (e1_.)*(x_))^(p_)*((d2_) + (e2_.)*(x_
))^(p_), x_Symbol] :> -Simp[((f*x)^(m + 1)*(d1 + e1*x)^(p + 1)*(d2 + e2*x)^(p + 1)*(a + b*ArcCosh[c*x])^n)/(2*
d1*d2*f*(p + 1)), x] + (Dist[(m + 2*p + 3)/(2*d1*d2*(p + 1)), Int[(f*x)^m*(d1 + e1*x)^(p + 1)*(d2 + e2*x)^(p +
1)*(a + b*ArcCosh[c*x])^n, x], x] - Dist[(b*c*n*(-(d1*d2))^IntPart[p]*(d1 + e1*x)^FracPart[p]*(d2 + e2*x)^Fra
cPart[p])/(2*f*(p + 1)*(1 + c*x)^FracPart[p]*(-1 + c*x)^FracPart[p]), Int[(f*x)^(m + 1)*(-1 + c^2*x^2)^(p + 1/
2)*(a + b*ArcCosh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d1, e1, d2, e2, f, m}, x] && EqQ[e1 - c*d1, 0] &&
EqQ[e2 + c*d2, 0] && GtQ[n, 0] && LtQ[p, -1] &&  !GtQ[m, 1] && (IntegerQ[m] || EqQ[n, 1]) && IntegerQ[p + 1/2]

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 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

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

Mathematica [A]  time = 0.71891, size = 319, normalized size = 0.71 \[ \frac{x \sqrt{c x-1} \sqrt{c x+1} (f x)^m \left (\frac{(m-2) \left (b c m x \sqrt{c x-1} \sqrt{c x+1} \text{HypergeometricPFQ}\left (\left \{1,\frac{m}{2}+1,\frac{m}{2}+1\right \},\left \{\frac{m}{2}+\frac{3}{2},\frac{m}{2}+2\right \},c^2 x^2\right )+m (m+2) \sqrt{1-c^2 x^2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )-(m+1) \left (b c x \sqrt{c x-1} \sqrt{c x+1} \text{Hypergeometric2F1}\left (1,\frac{m}{2}+1,\frac{m}{2}+2,c^2 x^2\right )+(m+2) \left (a+b \cosh ^{-1}(c x)\right )\right )\right )}{(m+1) (m+2) \sqrt{c x-1} \sqrt{c x+1}}+\frac{b c x \text{Hypergeometric2F1}\left (2,\frac{m}{2}+1,\frac{m}{2}+2,c^2 x^2\right )}{m+2}-\frac{a+b \cosh ^{-1}(c x)}{(c x-1)^{3/2} (c x+1)^{3/2}}\right )}{3 d^2 \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((f*x)^m*(a + b*ArcCosh[c*x]))/(d - c^2*d*x^2)^(5/2),x]

[Out]

(x*(f*x)^m*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(-((a + b*ArcCosh[c*x])/((-1 + c*x)^(3/2)*(1 + c*x)^(3/2))) + (b*c*x*H
ypergeometric2F1[2, 1 + m/2, 2 + m/2, c^2*x^2])/(2 + m) + ((-2 + m)*(m*(2 + m)*Sqrt[1 - c^2*x^2]*(a + b*ArcCos
h[c*x])*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2] - (1 + m)*((2 + m)*(a + b*ArcCosh[c*x]) + b*c*x*
Sqrt[-1 + c*x]*Sqrt[1 + c*x]*Hypergeometric2F1[1, 1 + m/2, 2 + m/2, c^2*x^2]) + b*c*m*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[-1 + c*
x]*Sqrt[1 + c*x])))/(3*d^2*Sqrt[d - c^2*d*x^2])

________________________________________________________________________________________

Maple [F]  time = 0.582, size = 0, normalized size = 0. \begin{align*} \int{ \left ( fx \right ) ^{m} \left ( a+b{\rm arccosh} \left (cx\right ) \right ) \left ( -{c}^{2}d{x}^{2}+d \right ) ^{-{\frac{5}{2}}}}\, 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)^(5/2),x)

[Out]

int((f*x)^m*(a+b*arccosh(c*x))/(-c^2*d*x^2+d)^(5/2),x)

________________________________________________________________________________________

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}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{5}{2}}}\,{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)^(5/2),x, algorithm="maxima")

[Out]

integrate((b*arccosh(c*x) + a)*(f*x)^m/(-c^2*d*x^2 + d)^(5/2), x)

________________________________________________________________________________________

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^{6} d^{3} x^{6} - 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} - d^{3}}, 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)^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-c^2*d*x^2 + d)*(b*arccosh(c*x) + a)*(f*x)^m/(c^6*d^3*x^6 - 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 - d^3
), x)

________________________________________________________________________________________

Sympy [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((f*x)**m*(a+b*acosh(c*x))/(-c**2*d*x**2+d)**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

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}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{5}{2}}}\,{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)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arccosh(c*x) + a)*(f*x)^m/(-c^2*d*x^2 + d)^(5/2), x)