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

Optimal. Leaf size=353 \[ -\frac{b \sqrt{1-c^2 x^2} (f x)^{m+2} \left (\frac{c^4 d^2 (m+3) (m+5)}{m+1}+\frac{e (m+2) \left (2 c^2 d (m+5)^2+e \left (m^2+7 m+12\right )\right )}{(m+3) (m+5)}\right ) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},c^2 x^2\right )}{c^3 f^2 (m+2) (m+3) (m+5) \sqrt{c x-1} \sqrt{c x+1}}+\frac{d^2 (f x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{f (m+1)}+\frac{2 d e (f x)^{m+3} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (m+3)}+\frac{e^2 (f x)^{m+5} \left (a+b \cosh ^{-1}(c x)\right )}{f^5 (m+5)}+\frac{b e \left (1-c^2 x^2\right ) (f x)^{m+2} \left (2 c^2 d (m+5)^2+e \left (m^2+7 m+12\right )\right )}{c^3 f^2 (m+3)^2 (m+5)^2 \sqrt{c x-1} \sqrt{c x+1}}+\frac{b e^2 \left (1-c^2 x^2\right ) (f x)^{m+4}}{c f^4 (m+5)^2 \sqrt{c x-1} \sqrt{c x+1}} \]

[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^4*d^2*(3 + m)*(5 + m))/(1 + m) + (e*(2 +
m)*(2*c^2*d*(5 + m)^2 + e*(12 + 7*m + m^2)))/((3 + m)*(5 + m)))*(f*x)^(2 + m)*Sqrt[1 - c^2*x^2]*Hypergeometric
2F1[1/2, (2 + m)/2, (4 + m)/2, c^2*x^2])/(c^3*f^2*(2 + m)*(3 + m)*(5 + m)*Sqrt[-1 + c*x]*Sqrt[1 + c*x])

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Rubi [A]  time = 0.560145, 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 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 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]))

Rule 12

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

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 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 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 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 (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*}

Mathematica [A]  time = 0.49686, size = 293, normalized size = 0.83 \[ x (f x)^m \left (-\frac{b c d^2 x \sqrt{1-c^2 x^2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},c^2 x^2\right )}{\left (m^2+3 m+2\right ) \sqrt{c x-1} \sqrt{c x+1}}-\frac{2 b c d e x^3 \sqrt{1-c^2 x^2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+4}{2},\frac{m+6}{2},c^2 x^2\right )}{\left (m^2+7 m+12\right ) \sqrt{c x-1} \sqrt{c x+1}}-\frac{b c e^2 x^5 \sqrt{1-c^2 x^2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+6}{2},\frac{m+8}{2},c^2 x^2\right )}{(m+5) (m+6) \sqrt{c x-1} \sqrt{c x+1}}+\frac{d^2 \left (a+b \cosh ^{-1}(c x)\right )}{m+1}+\frac{2 d e x^2 \left (a+b \cosh ^{-1}(c x)\right )}{m+3}+\frac{e^2 x^4 \left (a+b \cosh ^{-1}(c x)\right )}{m+5}\right ) \]

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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Maple [F]  time = 4.069, size = 0, normalized size = 0. \begin{align*} \int \left ( fx \right ) ^{m} \left ( e{x}^{2}+d \right ) ^{2} \left ( a+b{\rm arccosh} \left (cx\right ) \right ) \, dx \end{align*}

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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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

[Out]

Timed out

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Giac [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*(e*x^2+d)^2*(a+b*arccosh(c*x)),x, algorithm="giac")

[Out]

Timed out