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 \[ -\frac {b c (2-m) m \sqrt {c x-1} \sqrt {c x+1} (f x)^{m+2} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;c^2 x^2\right )}{3 d^2 f^2 (m+1) (m+2) \sqrt {d-c^2 d x^2}}-\frac {(2-m) m \sqrt {1-c^2 x^2} (f x)^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )}{3 d^2 f (m+1) \sqrt {d-c^2 d x^2}}+\frac {(2-m) (f x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{3 d^2 f \sqrt {d-c^2 d x^2}}+\frac {(f x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{3 d f \left (d-c^2 d x^2\right )^{3/2}}+\frac {b c (2-m) \sqrt {c x-1} \sqrt {c x+1} (f x)^{m+2} \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{3 d^2 f^2 (m+2) \sqrt {d-c^2 d x^2}}+\frac {b c \sqrt {c x-1} \sqrt {c x+1} (f x)^{m+2} \, _2F_1\left (2,\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{3 d^2 f^2 (m+2) \sqrt {d-c^2 d x^2}} \]

[Out]

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

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Rubi [A]  time = 0.99, 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 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])

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 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 \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*}

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Mathematica [A]  time = 0.71, size = 319, normalized size = 0.71 \[ \frac {x \sqrt {c x-1} \sqrt {c x+1} (f x)^m \left (\frac {(m-2) \left (b c m x \sqrt {c x-1} \sqrt {c x+1} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;c^2 x^2\right )+m (m+2) \sqrt {1-c^2 x^2} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )-(m+1) \left ((m+2) \left (a+b \cosh ^{-1}(c x)\right )+b c x \sqrt {c x-1} \sqrt {c x+1} \, _2F_1\left (1,\frac {m}{2}+1;\frac {m}{2}+2;c^2 x^2\right )\right )\right )}{(m+1) (m+2) \sqrt {c x-1} \sqrt {c x+1}}-\frac {a+b \cosh ^{-1}(c x)}{(c x-1)^{3/2} (c x+1)^{3/2}}+\frac {b c x \, _2F_1\left (2,\frac {m}{2}+1;\frac {m}{2}+2;c^2 x^2\right )}{m+2}\right )}{3 d^2 \sqrt {d-c^2 d x^2}} \]

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])

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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b \operatorname {arcosh}\left (c x\right ) + a\right )} \left (f x\right )^{m}}{c^{6} d^{3} x^{6} - 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} - d^{3}}, x\right ) \]

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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arcosh}\left (c x\right ) + a\right )} \left (f x\right )^{m}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}}}\,{d x} \]

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)

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maple [F]  time = 0.91, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x \right )^{m} \left (a +b \,\mathrm {arccosh}\left (c x \right )\right )}{\left (-c^{2} d \,x^{2}+d \right )^{\frac {5}{2}}}\, dx \]

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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arcosh}\left (c x\right ) + a\right )} \left (f x\right )^{m}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}}}\,{d x} \]

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )\,{\left (f\,x\right )}^m}{{\left (d-c^2\,d\,x^2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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