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

Optimal. Leaf size=198 \[ \frac {d (f x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{f (m+1)}+\frac {e (f x)^{m+3} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (m+3)}-\frac {b \sqrt {1-c^2 x^2} (f x)^{m+2} \left (c^2 d (m+3)^2+e (m+1) (m+2)\right ) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{c f^2 (m+1) (m+2) (m+3)^2 \sqrt {c x-1} \sqrt {c x+1}}-\frac {b e \sqrt {c x-1} \sqrt {c x+1} (f x)^{m+2}}{c f^2 (m+3)^2} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 187, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {5786, 460, 126, 365, 364} \[ \frac {d (f x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{f (m+1)}+\frac {e (f x)^{m+3} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (m+3)}-\frac {b \sqrt {1-c^2 x^2} (f x)^{m+2} \left (\frac {c^2 d}{m^2+3 m+2}+\frac {e}{(m+3)^2}\right ) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{c f^2 \sqrt {c x-1} \sqrt {c x+1}}-\frac {b e \sqrt {c x-1} \sqrt {c x+1} (f x)^{m+2}}{c f^2 (m+3)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 126

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

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 365

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

Rule 460

Int[((e_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b2_.)*(x_)^(non2_.))^(p_.)*((c_) + (d_.)
*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m + 1)*(a1 + b1*x^(n/2))^(p + 1)*(a2 + b2*x^(n/2))^(p + 1))/(b1*b2*e*
(m + n*(p + 1) + 1)), x] - Dist[(a1*a2*d*(m + 1) - b1*b2*c*(m + n*(p + 1) + 1))/(b1*b2*(m + n*(p + 1) + 1)), I
nt[(e*x)^m*(a1 + b1*x^(n/2))^p*(a2 + b2*x^(n/2))^p, x], x] /; FreeQ[{a1, b1, a2, b2, c, d, e, m, n, p}, x] &&
EqQ[non2, n/2] && EqQ[a2*b1 + a1*b2, 0] && NeQ[m + n*(p + 1) + 1, 0]

Rule 5786

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(d*(f*x)^(
m + 1)*(a + b*ArcCosh[c*x]))/(f*(m + 1)), x] + (-Dist[(b*c)/(f*(m + 1)*(m + 3)), Int[((f*x)^(m + 1)*(d*(m + 3)
+ e*(m + 1)*x^2))/(Sqrt[1 + c*x]*Sqrt[-1 + c*x]), x], x] + Simp[(e*(f*x)^(m + 3)*(a + b*ArcCosh[c*x]))/(f^3*(
m + 3)), x]) /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[c^2*d + e, 0] && NeQ[m, -1] && NeQ[m, -3]

Rubi steps

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

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Mathematica [A]  time = 0.61, size = 186, normalized size = 0.94 \[ x (f x)^m \left (\frac {\frac {\left (d (m+3)+e (m+1) x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )}{m+1}-\frac {b c e x^3 \sqrt {1-c^2 x^2} \, _2F_1\left (\frac {1}{2},\frac {m+4}{2};\frac {m+6}{2};c^2 x^2\right )}{(m+4) \sqrt {c x-1} \sqrt {c x+1}}}{m+3}-\frac {b c d x \sqrt {1-c^2 x^2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{\left (m^2+3 m+2\right ) \sqrt {c x-1} \sqrt {c x+1}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x*(f*x)^m*(-((b*c*d*x*Sqrt[1 - c^2*x^2]*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, c^2*x^2])/((2 + 3*m + m^2
)*Sqrt[-1 + c*x]*Sqrt[1 + c*x])) + (((d*(3 + m) + e*(1 + m)*x^2)*(a + b*ArcCosh[c*x]))/(1 + m) - (b*c*e*x^3*Sq
rt[1 - c^2*x^2]*Hypergeometric2F1[1/2, (4 + m)/2, (6 + m)/2, c^2*x^2])/((4 + m)*Sqrt[-1 + c*x]*Sqrt[1 + c*x]))
/(3 + m))

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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a e x^{2} + a d + {\left (b e x^{2} + b d\right )} \operatorname {arcosh}\left (c x\right )\right )} \left (f x\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \left (f x \right )^{m} \left (e \,x^{2}+d \right ) \left (a +b \,\mathrm {arccosh}\left (c x \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {a e f^{m} x^{3} x^{m}}{m + 3} + \frac {{\left (b e f^{m} {\left (m + 1\right )} x^{3} + b d f^{m} {\left (m + 3\right )} x\right )} x^{m} \log \left (c x + \sqrt {c x + 1} \sqrt {c x - 1}\right )}{m^{2} + 4 \, m + 3} + \frac {\left (f x\right )^{m + 1} a d}{f {\left (m + 1\right )}} + \int \frac {{\left (b c e f^{m} {\left (m + 1\right )} x^{3} + b c d f^{m} {\left (m + 3\right )} x\right )} x^{m}}{{\left (m^{2} + 4 \, m + 3\right )} c^{3} x^{3} - {\left (m^{2} + 4 \, m + 3\right )} c x + {\left ({\left (m^{2} + 4 \, m + 3\right )} c^{2} x^{2} - m^{2} - 4 \, m - 3\right )} \sqrt {c x + 1} \sqrt {c x - 1}}\,{d x} - \int \frac {{\left (b c^{2} e f^{m} {\left (m + 1\right )} x^{4} + b c^{2} d f^{m} {\left (m + 3\right )} x^{2}\right )} x^{m}}{{\left (m^{2} + 4 \, m + 3\right )} c^{2} x^{2} - m^{2} - 4 \, m - 3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*e*f^m*x^3*x^m/(m + 3) + (b*e*f^m*(m + 1)*x^3 + b*d*f^m*(m + 3)*x)*x^m*log(c*x + sqrt(c*x + 1)*sqrt(c*x - 1))
/(m^2 + 4*m + 3) + (f*x)^(m + 1)*a*d/(f*(m + 1)) + integrate((b*c*e*f^m*(m + 1)*x^3 + b*c*d*f^m*(m + 3)*x)*x^m
/((m^2 + 4*m + 3)*c^3*x^3 - (m^2 + 4*m + 3)*c*x + ((m^2 + 4*m + 3)*c^2*x^2 - m^2 - 4*m - 3)*sqrt(c*x + 1)*sqrt
(c*x - 1)), x) - integrate((b*c^2*e*f^m*(m + 1)*x^4 + b*c^2*d*f^m*(m + 3)*x^2)*x^m/((m^2 + 4*m + 3)*c^2*x^2 -
m^2 - 4*m - 3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (f x\right )^{m} \left (a + b \operatorname {acosh}{\left (c x \right )}\right ) \left (d + e x^{2}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)**m*(e*x**2+d)*(a+b*acosh(c*x)),x)

[Out]

Integral((f*x)**m*(a + b*acosh(c*x))*(d + e*x**2), x)

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