3.2.47 \(\int (f x)^m (d-c^2 d x^2) (a+b \cosh ^{-1}(c x)) \, dx\) [147]

Optimal. Leaf size=184 \[ \frac {b c d (f x)^{2+m} \sqrt {-1+c x} \sqrt {1+c x}}{f^2 (3+m)^2}+\frac {d (f x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )}{f (1+m)}-\frac {c^2 d (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}-\frac {b c d (7+3 m) (f x)^{2+m} \sqrt {1-c^2 x^2} \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};c^2 x^2\right )}{f^2 (1+m) (2+m) (3+m)^2 \sqrt {-1+c x} \sqrt {1+c x}} \]

[Out]

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

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Rubi [A]
time = 0.19, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {14, 5921, 12, 471, 127, 372, 371} \begin {gather*} -\frac {c^2 d (f x)^{m+3} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (m+3)}+\frac {d (f x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{f (m+1)}-\frac {b c d (3 m+7) \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 \sqrt {c x-1} \sqrt {c x+1}}+\frac {b c d \sqrt {c x-1} \sqrt {c x+1} (f x)^{m+2}}{f^2 (m+3)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c*d*(f*x)^(2 + m)*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(f^2*(3 + m)^2) + (d*(f*x)^(1 + m)*(a + b*ArcCosh[c*x]))/(f
*(1 + m)) - (c^2*d*(f*x)^(3 + m)*(a + b*ArcCosh[c*x]))/(f^3*(3 + m)) - (b*c*d*(7 + 3*m)*(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*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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 127

Int[((f_.)*(x_))^(p_.)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Dist[(a + b*x)^Frac
Part[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[n, m]

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 471

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 5921

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 ) \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 {c^2 d (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}-(b c) \int \frac {d (f x)^{1+m} \left (3+m-c^2 (1+m) x^2\right )}{f (1+m) (3+m) \sqrt {-1+c x} \sqrt {1+c x}} \, dx\\ &=\frac {d (f x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )}{f (1+m)}-\frac {c^2 d (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}-\frac {(b c d) \int \frac {(f x)^{1+m} \left (3+m-c^2 (1+m) x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{f \left (3+4 m+m^2\right )}\\ &=\frac {b c d (f x)^{2+m} \sqrt {-1+c x} \sqrt {1+c x}}{f^2 (3+m)^2}+\frac {d (f x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )}{f (1+m)}-\frac {c^2 d (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}-\frac {(b c d (7+3 m)) \int \frac {(f x)^{1+m}}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{f (1+m) (3+m)^2}\\ &=\frac {b c d (f x)^{2+m} \sqrt {-1+c x} \sqrt {1+c x}}{f^2 (3+m)^2}+\frac {d (f x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )}{f (1+m)}-\frac {c^2 d (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}-\frac {\left (b c d (7+3 m) \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 \sqrt {-1+c x} \sqrt {1+c x}}\\ &=\frac {b c d (f x)^{2+m} \sqrt {-1+c x} \sqrt {1+c x}}{f^2 (3+m)^2}+\frac {d (f x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )}{f (1+m)}-\frac {c^2 d (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}-\frac {\left (b c d (7+3 m) \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 \sqrt {-1+c x} \sqrt {1+c x}}\\ &=\frac {b c d (f x)^{2+m} \sqrt {-1+c x} \sqrt {1+c x}}{f^2 (3+m)^2}+\frac {d (f x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )}{f (1+m)}-\frac {c^2 d (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}-\frac {b c d (7+3 m) (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 \sqrt {-1+c x} \sqrt {1+c x}}\\ \end {align*}

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Mathematica [A]
time = 0.49, size = 193, normalized size = 1.05 \begin {gather*} d x (f x)^m \left (-\frac {b c x \sqrt {1-c^2 x^2} \, _2F_1\left (\frac {1}{2},1+\frac {m}{2};2+\frac {m}{2};c^2 x^2\right )}{\left (2+3 m+m^2\right ) \sqrt {-1+c x} \sqrt {1+c x}}+\frac {-\frac {\left (-3-m+c^2 x^2+c^2 m x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )}{1+m}+\frac {b c^3 x^3 \sqrt {1-c^2 x^2} \, _2F_1\left (\frac {1}{2},2+\frac {m}{2};3+\frac {m}{2};c^2 x^2\right )}{(4+m) \sqrt {-1+c x} \sqrt {1+c x}}}{3+m}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 9.05, size = 0, normalized size = 0.00 \[\int \left (f x \right )^{m} \left (-c^{2} d \,x^{2}+d \right ) \left (a +b \,\mathrm {arccosh}\left (c x \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*c^2*d*f^m*x^3*x^m/(m + 3) - (b*c^2*d*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^3*d*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^4*d*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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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-d*(Integral(-a*(f*x)**m, x) + Integral(-b*(f*x)**m*acosh(c*x), x) + Integral(a*c**2*x**2*(f*x)**m, x) + Integ
ral(b*c**2*x**2*(f*x)**m*acosh(c*x), x))

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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