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

Optimal. Leaf size=353 \[ \frac {d^2 (f x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{f (m+1)}+\frac {2 d e (f x)^{m+3} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (m+3)}+\frac {e^2 (f x)^{m+5} \left (a+b \cosh ^{-1}(c x)\right )}{f^5 (m+5)}+\frac {b e^2 \left (1-c^2 x^2\right ) (f x)^{m+4}}{c f^4 (m+5)^2 \sqrt {c x-1} \sqrt {c x+1}}+\frac {b e \left (1-c^2 x^2\right ) (f x)^{m+2} \left (2 c^2 d (m+5)^2+e \left (m^2+7 m+12\right )\right )}{c^3 f^2 (m+3)^2 (m+5)^2 \sqrt {c x-1} \sqrt {c x+1}}-\frac {b \sqrt {1-c^2 x^2} (f x)^{m+2} \left (\frac {c^4 d^2 (m+3) (m+5)}{m+1}+\frac {e (m+2) \left (2 c^2 d (m+5)^2+e \left (m^2+7 m+12\right )\right )}{(m+3) (m+5)}\right ) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{c^3 f^2 (m+2) (m+3) (m+5) \sqrt {c x-1} \sqrt {c x+1}} \]

[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)

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Rubi [A]  time = 0.56, antiderivative size = 332, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {270, 5790, 12, 520, 1267, 459, 365, 364} \[ \frac {d^2 (f x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{f (m+1)}+\frac {2 d e (f x)^{m+3} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (m+3)}+\frac {e^2 (f x)^{m+5} \left (a+b \cosh ^{-1}(c x)\right )}{f^5 (m+5)}-\frac {b c \sqrt {1-c^2 x^2} (f x)^{m+2} \left (\frac {e \left (2 c^2 d (m+5)^2+e \left (m^2+7 m+12\right )\right )}{c^4 (m+3)^2 (m+5)^2}+\frac {d^2}{m^2+3 m+2}\right ) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{f^2 \sqrt {c x-1} \sqrt {c x+1}}+\frac {b e \left (1-c^2 x^2\right ) (f x)^{m+2} \left (2 c^2 d (m+5)^2+e \left (m^2+7 m+12\right )\right )}{c^3 f^2 (m+3)^2 (m+5)^2 \sqrt {c x-1} \sqrt {c x+1}}+\frac {b e^2 \left (1-c^2 x^2\right ) (f x)^{m+4}}{c f^4 (m+5)^2 \sqrt {c x-1} \sqrt {c x+1}} \]

Antiderivative was successfully verified.

[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 270

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 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 459

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 520

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 1267

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 5790

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*} \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \cosh ^{-1}(c x)\right ) \, dx &=\frac {d^2 (f x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \cosh ^{-1}(c x)\right )}{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} \left (a+b \cosh ^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \cosh ^{-1}(c x)\right )}{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} \left (a+b \cosh ^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \cosh ^{-1}(c x)\right )}{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} \left (a+b \cosh ^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \cosh ^{-1}(c x)\right )}{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} \left (a+b \cosh ^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \cosh ^{-1}(c x)\right )}{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} \left (a+b \cosh ^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \cosh ^{-1}(c x)\right )}{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} \left (a+b \cosh ^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \cosh ^{-1}(c x)\right )}{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} \, _2F_1\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*}

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Mathematica [A]  time = 0.49, size = 293, normalized size = 0.83 \[ x (f x)^m \left (\frac {d^2 \left (a+b \cosh ^{-1}(c x)\right )}{m+1}+\frac {2 d e x^2 \left (a+b \cosh ^{-1}(c x)\right )}{m+3}+\frac {e^2 x^4 \left (a+b \cosh ^{-1}(c x)\right )}{m+5}-\frac {b c d^2 x \sqrt {1-c^2 x^2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{\left (m^2+3 m+2\right ) \sqrt {c x-1} \sqrt {c x+1}}-\frac {2 b c d e x^3 \sqrt {1-c^2 x^2} \, _2F_1\left (\frac {1}{2},\frac {m+4}{2};\frac {m+6}{2};c^2 x^2\right )}{\left (m^2+7 m+12\right ) \sqrt {c x-1} \sqrt {c x+1}}-\frac {b c e^2 x^5 \sqrt {1-c^2 x^2} \, _2F_1\left (\frac {1}{2},\frac {m+6}{2};\frac {m+8}{2};c^2 x^2\right )}{(m+5) (m+6) \sqrt {c x-1} \sqrt {c x+1}}\right ) \]

Antiderivative was successfully verified.

[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]))

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a e^{2} x^{4} + 2 \, a d e x^{2} + a d^{2} + {\left (b e^{2} x^{4} + 2 \, b d e x^{2} + b d^{2}\right )} \operatorname {arcosh}\left (c x\right )\right )} \left (f x\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*(e*x^2+d)^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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {a e^{2} f^{m} x^{5} x^{m}}{m + 5} + \frac {2 \, a d e f^{m} x^{3} x^{m}}{m + 3} + \frac {\left (f x\right )^{m + 1} a d^{2}}{f {\left (m + 1\right )}} + \frac {{\left ({\left (m^{2} + 4 \, m + 3\right )} b e^{2} f^{m} x^{5} + 2 \, {\left (m^{2} + 6 \, m + 5\right )} b d e f^{m} x^{3} + {\left (m^{2} + 8 \, m + 15\right )} b d^{2} f^{m} x\right )} x^{m} \log \left (c x + \sqrt {c x + 1} \sqrt {c x - 1}\right )}{m^{3} + 9 \, m^{2} + 23 \, m + 15} + \int \frac {{\left ({\left (m^{2} + 4 \, m + 3\right )} b c e^{2} f^{m} x^{5} + 2 \, {\left (m^{2} + 6 \, m + 5\right )} b c d e f^{m} x^{3} + {\left (m^{2} + 8 \, m + 15\right )} b c d^{2} f^{m} x\right )} x^{m}}{{\left (m^{3} + 9 \, m^{2} + 23 \, m + 15\right )} c^{3} x^{3} - {\left (m^{3} + 9 \, m^{2} + 23 \, m + 15\right )} c x + {\left ({\left (m^{3} + 9 \, m^{2} + 23 \, m + 15\right )} c^{2} x^{2} - m^{3} - 9 \, m^{2} - 23 \, m - 15\right )} \sqrt {c x + 1} \sqrt {c x - 1}}\,{d x} - \int \frac {{\left ({\left (m^{2} + 4 \, m + 3\right )} b c^{2} e^{2} f^{m} x^{6} + 2 \, {\left (m^{2} + 6 \, m + 5\right )} b c^{2} d e f^{m} x^{4} + {\left (m^{2} + 8 \, m + 15\right )} b c^{2} d^{2} f^{m} x^{2}\right )} x^{m}}{{\left (m^{3} + 9 \, m^{2} + 23 \, m + 15\right )} c^{2} x^{2} - m^{3} - 9 \, m^{2} - 23 \, m - 15}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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