\(\int (f x)^m (d+e x^2)^2 (a+b \text {arccosh}(c x)) \, dx\) [520]

Optimal result

Integrand size = 23, antiderivative size = 353 \[ \int (f x)^m \left (d+e x^2\right )^2 (a+b \text {arccosh}(c x)) \, dx=\frac {b e \left (2 c^2 d (5+m)^2+e \left (12+7 m+m^2\right )\right ) (f x)^{2+m} \left (1-c^2 x^2\right )}{c^3 f^2 (3+m)^2 (5+m)^2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {b e^2 (f x)^{4+m} \left (1-c^2 x^2\right )}{c f^4 (5+m)^2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {d^2 (f x)^{1+m} (a+b \text {arccosh}(c x))}{f (1+m)}+\frac {2 d e (f x)^{3+m} (a+b \text {arccosh}(c x))}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} (a+b \text {arccosh}(c x))}{f^5 (5+m)}-\frac {b \left (\frac {c^4 d^2 (3+m) (5+m)}{1+m}+\frac {e (2+m) \left (2 c^2 d (5+m)^2+e \left (12+7 m+m^2\right )\right )}{(3+m) (5+m)}\right ) (f x)^{2+m} \sqrt {1-c^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{c^3 f^2 (2+m) (3+m) (5+m) \sqrt {-1+c x} \sqrt {1+c x}} \]

[Out]

d^2*(f*x)^(1+m)*(a+b*arccosh(c*x))/f/(1+m)+2*d*e*(f*x)^(3+m)*(a+b*arccosh(c*x))/f^3/(3+m)+e^2*(f*x)^(5+m)*(a+b
*arccosh(c*x))/f^5/(5+m)+b*e*(2*c^2*d*(5+m)^2+e*(m^2+7*m+12))*(f*x)^(2+m)*(-c^2*x^2+1)/c^3/f^2/(3+m)^2/(5+m)^2
/(c*x-1)^(1/2)/(c*x+1)^(1/2)+b*e^2*(f*x)^(4+m)*(-c^2*x^2+1)/c/f^4/(5+m)^2/(c*x-1)^(1/2)/(c*x+1)^(1/2)-b*(c^4*d
^2*(3+m)*(5+m)/(1+m)+e*(2+m)*(2*c^2*d*(5+m)^2+e*(m^2+7*m+12))/(3+m)/(5+m))*(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^3/f^2/(2+m)/(3+m)/(5+m)/(c*x-1)^(1/2)/(c*x+1)^(1/2)

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 332, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {276, 5958, 12, 534, 1281, 470, 372, 371} \[ \int (f x)^m \left (d+e x^2\right )^2 (a+b \text {arccosh}(c x)) \, dx=\frac {d^2 (f x)^{m+1} (a+b \text {arccosh}(c x))}{f (m+1)}+\frac {2 d e (f x)^{m+3} (a+b \text {arccosh}(c x))}{f^3 (m+3)}+\frac {e^2 (f x)^{m+5} (a+b \text {arccosh}(c x))}{f^5 (m+5)}+\frac {b e^2 \left (1-c^2 x^2\right ) (f x)^{m+4}}{c f^4 (m+5)^2 \sqrt {c x-1} \sqrt {c x+1}}-\frac {b c \sqrt {1-c^2 x^2} (f x)^{m+2} \left (\frac {e \left (2 c^2 d (m+5)^2+e \left (m^2+7 m+12\right )\right )}{c^4 (m+3)^2 (m+5)^2}+\frac {d^2}{m^2+3 m+2}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},c^2 x^2\right )}{f^2 \sqrt {c x-1} \sqrt {c x+1}}+\frac {b e \left (1-c^2 x^2\right ) (f x)^{m+2} \left (2 c^2 d (m+5)^2+e \left (m^2+7 m+12\right )\right )}{c^3 f^2 (m+3)^2 (m+5)^2 \sqrt {c x-1} \sqrt {c x+1}} \]

[In]

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

[Out]

(b*e*(2*c^2*d*(5 + m)^2 + e*(12 + 7*m + m^2))*(f*x)^(2 + m)*(1 - c^2*x^2))/(c^3*f^2*(3 + m)^2*(5 + m)^2*Sqrt[-
1 + c*x]*Sqrt[1 + c*x]) + (b*e^2*(f*x)^(4 + m)*(1 - c^2*x^2))/(c*f^4*(5 + m)^2*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) +
(d^2*(f*x)^(1 + m)*(a + b*ArcCosh[c*x]))/(f*(1 + m)) + (2*d*e*(f*x)^(3 + m)*(a + b*ArcCosh[c*x]))/(f^3*(3 + m
)) + (e^2*(f*x)^(5 + m)*(a + b*ArcCosh[c*x]))/(f^5*(5 + m)) - (b*c*(d^2/(2 + 3*m + m^2) + (e*(2*c^2*d*(5 + m)^
2 + e*(12 + 7*m + m^2)))/(c^4*(3 + m)^2*(5 + 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])/(f^2*Sqrt[-1 + c*x]*Sqrt[1 + c*x])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 534

Int[(u_.)*((c_) + (d_.)*(x_)^(n_.) + (e_.)*(x_)^(n2_.))^(q_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b
2_.)*(x_)^(non2_.))^(p_.), x_Symbol] :> Dist[(a1 + b1*x^(n/2))^FracPart[p]*((a2 + b2*x^(n/2))^FracPart[p]/(a1*
a2 + b1*b2*x^n)^FracPart[p]), Int[u*(a1*a2 + b1*b2*x^n)^p*(c + d*x^n + e*x^(2*n))^q, x], x] /; FreeQ[{a1, b1,
a2, b2, c, d, e, n, p, q}, x] && EqQ[non2, n/2] && EqQ[n2, 2*n] && EqQ[a2*b1 + a1*b2, 0]

Rule 1281

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Si
mp[c^p*(f*x)^(m + 4*p - 1)*((d + e*x^2)^(q + 1)/(e*f^(4*p - 1)*(m + 4*p + 2*q + 1))), x] + Dist[1/(e*(m + 4*p
+ 2*q + 1)), Int[(f*x)^m*(d + e*x^2)^q*ExpandToSum[e*(m + 4*p + 2*q + 1)*((a + b*x^2 + c*x^4)^p - c^p*x^(4*p))
- d*c^p*(m + 4*p - 1)*x^(4*p - 2), x], x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[b^2 - 4*a*c, 0] &&
IGtQ[p, 0] &&  !IntegerQ[q] && NeQ[m + 4*p + 2*q + 1, 0]

Rule 5958

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcCosh[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/(Sqrt[
1 + c*x]*Sqrt[-1 + c*x]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[c^2*d + e, 0] && IntegerQ[p] &
& (GtQ[p, 0] || (IGtQ[(m - 1)/2, 0] && LeQ[m + p, 0]))

Rubi steps \begin{align*} \text {integral}& = \frac {d^2 (f x)^{1+m} (a+b \text {arccosh}(c x))}{f (1+m)}+\frac {2 d e (f x)^{3+m} (a+b \text {arccosh}(c x))}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} (a+b \text {arccosh}(c x))}{f^5 (5+m)}-(b c) \int \frac {(f x)^{1+m} \left (\frac {d^2}{1+m}+\frac {2 d e x^2}{3+m}+\frac {e^2 x^4}{5+m}\right )}{f \sqrt {-1+c x} \sqrt {1+c x}} \, dx \\ & = \frac {d^2 (f x)^{1+m} (a+b \text {arccosh}(c x))}{f (1+m)}+\frac {2 d e (f x)^{3+m} (a+b \text {arccosh}(c x))}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} (a+b \text {arccosh}(c x))}{f^5 (5+m)}-\frac {(b c) \int \frac {(f x)^{1+m} \left (\frac {d^2}{1+m}+\frac {2 d e x^2}{3+m}+\frac {e^2 x^4}{5+m}\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{f} \\ & = \frac {d^2 (f x)^{1+m} (a+b \text {arccosh}(c x))}{f (1+m)}+\frac {2 d e (f x)^{3+m} (a+b \text {arccosh}(c x))}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} (a+b \text {arccosh}(c x))}{f^5 (5+m)}-\frac {\left (b c \sqrt {-1+c^2 x^2}\right ) \int \frac {(f x)^{1+m} \left (\frac {d^2}{1+m}+\frac {2 d e x^2}{3+m}+\frac {e^2 x^4}{5+m}\right )}{\sqrt {-1+c^2 x^2}} \, dx}{f \sqrt {-1+c x} \sqrt {1+c x}} \\ & = \frac {b e^2 (f x)^{4+m} \left (1-c^2 x^2\right )}{c f^4 (5+m)^2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {d^2 (f x)^{1+m} (a+b \text {arccosh}(c x))}{f (1+m)}+\frac {2 d e (f x)^{3+m} (a+b \text {arccosh}(c x))}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} (a+b \text {arccosh}(c x))}{f^5 (5+m)}-\frac {\left (b \sqrt {-1+c^2 x^2}\right ) \int \frac {(f x)^{1+m} \left (\frac {c^2 d^2 (5+m)}{1+m}+\frac {e \left (2 c^2 d (5+m)^2+e \left (12+7 m+m^2\right )\right ) x^2}{(3+m) (5+m)}\right )}{\sqrt {-1+c^2 x^2}} \, dx}{c f (5+m) \sqrt {-1+c x} \sqrt {1+c x}} \\ & = \frac {b e \left (2 c^2 d (5+m)^2+e \left (12+7 m+m^2\right )\right ) (f x)^{2+m} \left (1-c^2 x^2\right )}{c^3 f^2 (3+m)^2 (5+m)^2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {b e^2 (f x)^{4+m} \left (1-c^2 x^2\right )}{c f^4 (5+m)^2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {d^2 (f x)^{1+m} (a+b \text {arccosh}(c x))}{f (1+m)}+\frac {2 d e (f x)^{3+m} (a+b \text {arccosh}(c x))}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} (a+b \text {arccosh}(c x))}{f^5 (5+m)}-\frac {\left (b \left (\frac {c^2 d^2 (5+m)}{1+m}+\frac {e (2+m) \left (2 c^2 d (5+m)^2+e \left (12+7 m+m^2\right )\right )}{c^2 (3+m)^2 (5+m)}\right ) \sqrt {-1+c^2 x^2}\right ) \int \frac {(f x)^{1+m}}{\sqrt {-1+c^2 x^2}} \, dx}{c f (5+m) \sqrt {-1+c x} \sqrt {1+c x}} \\ & = \frac {b e \left (2 c^2 d (5+m)^2+e \left (12+7 m+m^2\right )\right ) (f x)^{2+m} \left (1-c^2 x^2\right )}{c^3 f^2 (3+m)^2 (5+m)^2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {b e^2 (f x)^{4+m} \left (1-c^2 x^2\right )}{c f^4 (5+m)^2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {d^2 (f x)^{1+m} (a+b \text {arccosh}(c x))}{f (1+m)}+\frac {2 d e (f x)^{3+m} (a+b \text {arccosh}(c x))}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} (a+b \text {arccosh}(c x))}{f^5 (5+m)}-\frac {\left (b \left (\frac {c^2 d^2 (5+m)}{1+m}+\frac {e (2+m) \left (2 c^2 d (5+m)^2+e \left (12+7 m+m^2\right )\right )}{c^2 (3+m)^2 (5+m)}\right ) \sqrt {1-c^2 x^2}\right ) \int \frac {(f x)^{1+m}}{\sqrt {1-c^2 x^2}} \, dx}{c f (5+m) \sqrt {-1+c x} \sqrt {1+c x}} \\ & = \frac {b e \left (2 c^2 d (5+m)^2+e \left (12+7 m+m^2\right )\right ) (f x)^{2+m} \left (1-c^2 x^2\right )}{c^3 f^2 (3+m)^2 (5+m)^2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {b e^2 (f x)^{4+m} \left (1-c^2 x^2\right )}{c f^4 (5+m)^2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {d^2 (f x)^{1+m} (a+b \text {arccosh}(c x))}{f (1+m)}+\frac {2 d e (f x)^{3+m} (a+b \text {arccosh}(c x))}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} (a+b \text {arccosh}(c x))}{f^5 (5+m)}-\frac {b \left (\frac {c^4 d^2}{2+3 m+m^2}+\frac {e \left (2 c^2 d (5+m)^2+e \left (12+7 m+m^2\right )\right )}{(3+m)^2 (5+m)^2}\right ) (f x)^{2+m} \sqrt {1-c^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{c^3 f^2 \sqrt {-1+c x} \sqrt {1+c x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 293, normalized size of antiderivative = 0.83 \[ \int (f x)^m \left (d+e x^2\right )^2 (a+b \text {arccosh}(c x)) \, dx=x (f x)^m \left (\frac {d^2 (a+b \text {arccosh}(c x))}{1+m}+\frac {2 d e x^2 (a+b \text {arccosh}(c x))}{3+m}+\frac {e^2 x^4 (a+b \text {arccosh}(c x))}{5+m}-\frac {b c d^2 x \sqrt {1-c^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{\left (2+3 m+m^2\right ) \sqrt {-1+c x} \sqrt {1+c x}}-\frac {2 b c d e x^3 \sqrt {1-c^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4+m}{2},\frac {6+m}{2},c^2 x^2\right )}{\left (12+7 m+m^2\right ) \sqrt {-1+c x} \sqrt {1+c x}}-\frac {b c e^2 x^5 \sqrt {1-c^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {6+m}{2},\frac {8+m}{2},c^2 x^2\right )}{(5+m) (6+m) \sqrt {-1+c x} \sqrt {1+c x}}\right ) \]

[In]

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

[Out]

x*(f*x)^m*((d^2*(a + b*ArcCosh[c*x]))/(1 + m) + (2*d*e*x^2*(a + b*ArcCosh[c*x]))/(3 + m) + (e^2*x^4*(a + b*Arc
Cosh[c*x]))/(5 + m) - (b*c*d^2*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]) - (2*b*c*d*e*x^3*Sqrt[1 - c^2*x^2]*Hypergeometric2F1[1/2, (4 + m)/2
, (6 + m)/2, c^2*x^2])/((12 + 7*m + m^2)*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) - (b*c*e^2*x^5*Sqrt[1 - c^2*x^2]*Hyperg
eometric2F1[1/2, (6 + m)/2, (8 + m)/2, c^2*x^2])/((5 + m)*(6 + m)*Sqrt[-1 + c*x]*Sqrt[1 + c*x]))

Maple [F]

\[\int \left (f x \right )^{m} \left (e \,x^{2}+d \right )^{2} \left (a +b \,\operatorname {arccosh}\left (c x \right )\right )d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int (f x)^m \left (d+e x^2\right )^2 (a+b \text {arccosh}(c x)) \, dx=\int { {\left (e x^{2} + d\right )}^{2} {\left (b \operatorname {arcosh}\left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]

[In]

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

[Out]

integral((a*e^2*x^4 + 2*a*d*e*x^2 + a*d^2 + (b*e^2*x^4 + 2*b*d*e*x^2 + b*d^2)*arccosh(c*x))*(f*x)^m, x)

Sympy [F]

\[ \int (f x)^m \left (d+e x^2\right )^2 (a+b \text {arccosh}(c x)) \, dx=\int \left (f x\right )^{m} \left (a + b \operatorname {acosh}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{2}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int (f x)^m \left (d+e x^2\right )^2 (a+b \text {arccosh}(c x)) \, dx=\int { {\left (e x^{2} + d\right )}^{2} {\left (b \operatorname {arcosh}\left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]

[In]

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

[Out]

a*e^2*f^m*x^5*x^m/(m + 5) + 2*a*d*e*f^m*x^3*x^m/(m + 3) + (f*x)^(m + 1)*a*d^2/(f*(m + 1)) + ((m^2 + 4*m + 3)*b
*e^2*f^m*x^5 + 2*(m^2 + 6*m + 5)*b*d*e*f^m*x^3 + (m^2 + 8*m + 15)*b*d^2*f^m*x)*x^m*log(c*x + sqrt(c*x + 1)*sqr
t(c*x - 1))/(m^3 + 9*m^2 + 23*m + 15) + integrate(((m^2 + 4*m + 3)*b*c*e^2*f^m*x^5 + 2*(m^2 + 6*m + 5)*b*c*d*e
*f^m*x^3 + (m^2 + 8*m + 15)*b*c*d^2*f^m*x)*x^m/((m^3 + 9*m^2 + 23*m + 15)*c^3*x^3 - (m^3 + 9*m^2 + 23*m + 15)*
c*x + ((m^3 + 9*m^2 + 23*m + 15)*c^2*x^2 - m^3 - 9*m^2 - 23*m - 15)*sqrt(c*x + 1)*sqrt(c*x - 1)), x) - integra
te(((m^2 + 4*m + 3)*b*c^2*e^2*f^m*x^6 + 2*(m^2 + 6*m + 5)*b*c^2*d*e*f^m*x^4 + (m^2 + 8*m + 15)*b*c^2*d^2*f^m*x
^2)*x^m/((m^3 + 9*m^2 + 23*m + 15)*c^2*x^2 - m^3 - 9*m^2 - 23*m - 15), x)

Giac [F]

\[ \int (f x)^m \left (d+e x^2\right )^2 (a+b \text {arccosh}(c x)) \, dx=\int { {\left (e x^{2} + d\right )}^{2} {\left (b \operatorname {arcosh}\left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (f x)^m \left (d+e x^2\right )^2 (a+b \text {arccosh}(c x)) \, dx=\int \left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )\,{\left (f\,x\right )}^m\,{\left (e\,x^2+d\right )}^2 \,d x \]

[In]

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

[Out]

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