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

Optimal. Leaf size=307 \[ \frac {c^4 d^2 (f x)^{m+5} \left (a+b \cosh ^{-1}(c x)\right )}{f^5 (m+5)}-\frac {2 c^2 d^2 (f x)^{m+3} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (m+3)}+\frac {d^2 (f x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{f (m+1)}-\frac {b c d^2 \left (15 m^2+100 m+149\right ) \sqrt {1-c^2 x^2} (f x)^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{f^2 (m+1) (m+2) (m+3)^2 (m+5)^2 \sqrt {c x-1} \sqrt {c x+1}}-\frac {b c d^2 \left (m^2+13 m+38\right ) \left (1-c^2 x^2\right ) (f x)^{m+2}}{f^2 (m+3)^2 (m+5)^2 \sqrt {c x-1} \sqrt {c x+1}}+\frac {b c^3 d^2 \left (1-c^2 x^2\right ) (f x)^{m+4}}{f^4 (m+5)^2 \sqrt {c x-1} \sqrt {c x+1}} \]

[Out]

d^2*(f*x)^(1+m)*(a+b*arccosh(c*x))/f/(1+m)-2*c^2*d^2*(f*x)^(3+m)*(a+b*arccosh(c*x))/f^3/(3+m)+c^4*d^2*(f*x)^(5
+m)*(a+b*arccosh(c*x))/f^5/(5+m)-b*c*d^2*(m^2+13*m+38)*(f*x)^(2+m)*(-c^2*x^2+1)/f^2/(3+m)^2/(5+m)^2/(c*x-1)^(1
/2)/(c*x+1)^(1/2)+b*c^3*d^2*(f*x)^(4+m)*(-c^2*x^2+1)/f^4/(5+m)^2/(c*x-1)^(1/2)/(c*x+1)^(1/2)-b*c*d^2*(15*m^2+1
00*m+149)*(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)/f^2/(m^2+3*m+2)/(m^2+8*m+
15)^2/(c*x-1)^(1/2)/(c*x+1)^(1/2)

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Rubi [A]  time = 0.50, antiderivative size = 307, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {270, 5731, 12, 520, 1267, 459, 365, 364} \[ -\frac {2 c^2 d^2 (f x)^{m+3} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (m+3)}+\frac {c^4 d^2 (f x)^{m+5} \left (a+b \cosh ^{-1}(c x)\right )}{f^5 (m+5)}+\frac {d^2 (f x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{f (m+1)}-\frac {b c d^2 \left (15 m^2+100 m+149\right ) \sqrt {1-c^2 x^2} (f x)^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{f^2 (m+1) (m+2) (m+3)^2 (m+5)^2 \sqrt {c x-1} \sqrt {c x+1}}-\frac {b c d^2 \left (m^2+13 m+38\right ) \left (1-c^2 x^2\right ) (f x)^{m+2}}{f^2 (m+3)^2 (m+5)^2 \sqrt {c x-1} \sqrt {c x+1}}+\frac {b c^3 d^2 \left (1-c^2 x^2\right ) (f x)^{m+4}}{f^4 (m+5)^2 \sqrt {c x-1} \sqrt {c x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*c*d^2*(38 + 13*m + m^2)*(f*x)^(2 + m)*(1 - c^2*x^2))/(f^2*(3 + m)^2*(5 + m)^2*Sqrt[-1 + c*x]*Sqrt[1 + c*x
])) + (b*c^3*d^2*(f*x)^(4 + m)*(1 - c^2*x^2))/(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*c^2*d^2*(f*x)^(3 + m)*(a + b*ArcCosh[c*x]))/(f^3*(3 + m)) + (c^4*d^2*
(f*x)^(5 + m)*(a + b*ArcCosh[c*x]))/(f^5*(5 + m)) - (b*c*d^2*(149 + 100*m + 15*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*(1 + m)*(2 + m)*(3 + m)^2*(5 + m)^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 5731

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] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int (f x)^m \left (d-c^2 d 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 c^2 d^2 (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {c^4 d^2 (f x)^{5+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^5 (5+m)}-(b c) \int \frac {d^2 (f x)^{1+m} \left (\frac {1}{1+m}-\frac {2 c^2 x^2}{3+m}+\frac {c^4 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 c^2 d^2 (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {c^4 d^2 (f x)^{5+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^5 (5+m)}-\frac {\left (b c d^2\right ) \int \frac {(f x)^{1+m} \left (\frac {1}{1+m}-\frac {2 c^2 x^2}{3+m}+\frac {c^4 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 c^2 d^2 (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {c^4 d^2 (f x)^{5+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^5 (5+m)}-\frac {\left (b c d^2 \sqrt {-1+c^2 x^2}\right ) \int \frac {(f x)^{1+m} \left (\frac {1}{1+m}-\frac {2 c^2 x^2}{3+m}+\frac {c^4 x^4}{5+m}\right )}{\sqrt {-1+c^2 x^2}} \, dx}{f \sqrt {-1+c x} \sqrt {1+c x}}\\ &=\frac {b c^3 d^2 (f x)^{4+m} \left (1-c^2 x^2\right )}{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 c^2 d^2 (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {c^4 d^2 (f x)^{5+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^5 (5+m)}-\frac {\left (b d^2 \sqrt {-1+c^2 x^2}\right ) \int \frac {(f x)^{1+m} \left (\frac {c^2 (5+m)}{1+m}-\frac {c^4 \left (38+13 m+m^2\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 c d^2 \left (38+13 m+m^2\right ) (f x)^{2+m} \left (1-c^2 x^2\right )}{f^2 (3+m)^2 (5+m)^2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {b c^3 d^2 (f x)^{4+m} \left (1-c^2 x^2\right )}{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 c^2 d^2 (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {c^4 d^2 (f x)^{5+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^5 (5+m)}-\frac {\left (b c d^2 \left (149+100 m+15 m^2\right ) \sqrt {-1+c^2 x^2}\right ) \int \frac {(f x)^{1+m}}{\sqrt {-1+c^2 x^2}} \, dx}{f (1+m) (3+m)^2 (5+m)^2 \sqrt {-1+c x} \sqrt {1+c x}}\\ &=-\frac {b c d^2 \left (38+13 m+m^2\right ) (f x)^{2+m} \left (1-c^2 x^2\right )}{f^2 (3+m)^2 (5+m)^2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {b c^3 d^2 (f x)^{4+m} \left (1-c^2 x^2\right )}{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 c^2 d^2 (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {c^4 d^2 (f x)^{5+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^5 (5+m)}-\frac {\left (b c d^2 \left (149+100 m+15 m^2\right ) \sqrt {1-c^2 x^2}\right ) \int \frac {(f x)^{1+m}}{\sqrt {1-c^2 x^2}} \, dx}{f (1+m) (3+m)^2 (5+m)^2 \sqrt {-1+c x} \sqrt {1+c x}}\\ &=-\frac {b c d^2 \left (38+13 m+m^2\right ) (f x)^{2+m} \left (1-c^2 x^2\right )}{f^2 (3+m)^2 (5+m)^2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {b c^3 d^2 (f x)^{4+m} \left (1-c^2 x^2\right )}{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 c^2 d^2 (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {c^4 d^2 (f x)^{5+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^5 (5+m)}-\frac {b c d^2 \left (149+100 m+15 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 )}{f^2 (1+m) (2+m) (3+m)^2 (5+m)^2 \sqrt {-1+c x} \sqrt {1+c x}}\\ \end {align*}

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Mathematica [A]  time = 0.45, size = 290, normalized size = 0.94 \[ d^2 x (f x)^m \left (\frac {c^4 x^4 \left (a+b \cosh ^{-1}(c x)\right )}{m+5}-\frac {2 c^2 x^2 \left (a+b \cosh ^{-1}(c x)\right )}{m+3}+\frac {a+b \cosh ^{-1}(c x)}{m+1}-\frac {b c x \sqrt {1-c^2 x^2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{\left (m^2+3 m+2\right ) \sqrt {c x-1} \sqrt {c x+1}}-\frac {b c^5 x^5 \sqrt {1-c^2 x^2} \, _2F_1\left (\frac {1}{2},\frac {m+6}{2};\frac {m+8}{2};c^2 x^2\right )}{(m+5) (m+6) \sqrt {c x-1} \sqrt {c x+1}}+\frac {2 b c^3 x^3 \sqrt {1-c^2 x^2} \, _2F_1\left (\frac {1}{2},\frac {m+4}{2};\frac {m+6}{2};c^2 x^2\right )}{\left (m^2+7 m+12\right ) \sqrt {c x-1} \sqrt {c x+1}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

d^2*x*(f*x)^m*((a + b*ArcCosh[c*x])/(1 + m) - (2*c^2*x^2*(a + b*ArcCosh[c*x]))/(3 + m) + (c^4*x^4*(a + b*ArcCo
sh[c*x]))/(5 + m) - (b*c*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^3*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^5*x^5*Sqrt[1 - c^2*x^2]*Hypergeometric2F
1[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 = 1.00, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a c^{4} d^{2} x^{4} - 2 \, a c^{2} d^{2} x^{2} + a d^{2} + {\left (b c^{4} d^{2} x^{4} - 2 \, b c^{2} d^{2} x^{2} + b d^{2}\right )} \operatorname {arcosh}\left (c x\right )\right )} \left (f x\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((f*x)^m*(-c^2*d*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 \[ \frac {a c^{4} d^{2} f^{m} x^{5} x^{m}}{m + 5} - \frac {2 \, a c^{2} d^{2} f^{m} x^{3} x^{m}}{m + 3} + \frac {\left (f x\right )^{m + 1} a d^{2}}{f {\left (m + 1\right )}} + \frac {{\left ({\left (m^{2} + 4 \, m + 3\right )} b c^{4} d^{2} f^{m} x^{5} - 2 \, {\left (m^{2} + 6 \, m + 5\right )} b c^{2} d^{2} f^{m} x^{3} + {\left (m^{2} + 8 \, m + 15\right )} b d^{2} f^{m} x\right )} x^{m} \log \left (c x + \sqrt {c x + 1} \sqrt {c x - 1}\right )}{m^{3} + 9 \, m^{2} + 23 \, m + 15} + \int \frac {{\left ({\left (m^{2} + 4 \, m + 3\right )} b c^{5} d^{2} f^{m} x^{5} - 2 \, {\left (m^{2} + 6 \, m + 5\right )} b c^{3} d^{2} f^{m} x^{3} + {\left (m^{2} + 8 \, m + 15\right )} b c d^{2} f^{m} x\right )} x^{m}}{{\left (m^{3} + 9 \, m^{2} + 23 \, m + 15\right )} c^{3} x^{3} - {\left (m^{3} + 9 \, m^{2} + 23 \, m + 15\right )} c x + {\left ({\left (m^{3} + 9 \, m^{2} + 23 \, m + 15\right )} c^{2} x^{2} - m^{3} - 9 \, m^{2} - 23 \, m - 15\right )} \sqrt {c x + 1} \sqrt {c x - 1}}\,{d x} - \int \frac {{\left ({\left (m^{2} + 4 \, m + 3\right )} b c^{6} d^{2} f^{m} x^{6} - 2 \, {\left (m^{2} + 6 \, m + 5\right )} b c^{4} d^{2} f^{m} x^{4} + {\left (m^{2} + 8 \, m + 15\right )} b c^{2} d^{2} f^{m} x^{2}\right )} x^{m}}{{\left (m^{3} + 9 \, m^{2} + 23 \, m + 15\right )} c^{2} x^{2} - m^{3} - 9 \, m^{2} - 23 \, m - 15}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*c^4*d^2*f^m*x^5*x^m/(m + 5) - 2*a*c^2*d^2*f^m*x^3*x^m/(m + 3) + (f*x)^(m + 1)*a*d^2/(f*(m + 1)) + ((m^2 + 4*
m + 3)*b*c^4*d^2*f^m*x^5 - 2*(m^2 + 6*m + 5)*b*c^2*d^2*f^m*x^3 + (m^2 + 8*m + 15)*b*d^2*f^m*x)*x^m*log(c*x + s
qrt(c*x + 1)*sqrt(c*x - 1))/(m^3 + 9*m^2 + 23*m + 15) + integrate(((m^2 + 4*m + 3)*b*c^5*d^2*f^m*x^5 - 2*(m^2
+ 6*m + 5)*b*c^3*d^2*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) - integrate(((m^2 + 4*m + 3)*b*c^6*d^2*f^m*x^6 - 2*(m^2 + 6*m + 5)*b*c^4*d^2*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 \[ \int \left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )\,{\left (d-c^2\,d\,x^2\right )}^2\,{\left (f\,x\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ d^{2} \left (\int a \left (f x\right )^{m}\, dx + \int b \left (f x\right )^{m} \operatorname {acosh}{\left (c x \right )}\, dx + \int \left (- 2 a c^{2} x^{2} \left (f x\right )^{m}\right )\, dx + \int a c^{4} x^{4} \left (f x\right )^{m}\, dx + \int \left (- 2 b c^{2} x^{2} \left (f x\right )^{m} \operatorname {acosh}{\left (c x \right )}\right )\, dx + \int b c^{4} x^{4} \left (f x\right )^{m} \operatorname {acosh}{\left (c x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**2*(Integral(a*(f*x)**m, x) + Integral(b*(f*x)**m*acosh(c*x), x) + Integral(-2*a*c**2*x**2*(f*x)**m, x) + In
tegral(a*c**4*x**4*(f*x)**m, x) + Integral(-2*b*c**2*x**2*(f*x)**m*acosh(c*x), x) + Integral(b*c**4*x**4*(f*x)
**m*acosh(c*x), x))

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