3.2.53 \(\int (f x)^m \sqrt {d-c^2 d x^2} (a+b \cosh ^{-1}(c x)) \, dx\) [153]

Optimal. Leaf size=278 \[ -\frac {b c (f x)^{2+m} \sqrt {d-c^2 d x^2}}{f^2 (2+m)^2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {(f x)^{1+m} \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{f (2+m)}+\frac {(f x)^{1+m} \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right ) \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};c^2 x^2\right )}{f \left (2+3 m+m^2\right ) \sqrt {1-c x} \sqrt {1+c x}}-\frac {b c (f x)^{2+m} \sqrt {d-c^2 d x^2} \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};c^2 x^2\right )}{f^2 (1+m) (2+m)^2 \sqrt {-1+c x} \sqrt {1+c x}} \]

[Out]

(f*x)^(1+m)*(a+b*arccosh(c*x))*(-c^2*d*x^2+d)^(1/2)/f/(2+m)+(f*x)^(1+m)*(a+b*arccosh(c*x))*hypergeom([1/2, 1/2
+1/2*m],[3/2+1/2*m],c^2*x^2)*(-c^2*d*x^2+d)^(1/2)/f/(m^2+3*m+2)/(-c*x+1)^(1/2)/(c*x+1)^(1/2)-b*c*(f*x)^(2+m)*(
-c^2*d*x^2+d)^(1/2)/f^2/(2+m)^2/(c*x-1)^(1/2)/(c*x+1)^(1/2)-b*c*(f*x)^(2+m)*hypergeom([1, 1+1/2*m, 1+1/2*m],[2
+1/2*m, 3/2+1/2*m],c^2*x^2)*(-c^2*d*x^2+d)^(1/2)/f^2/(1+m)/(2+m)^2/(c*x-1)^(1/2)/(c*x+1)^(1/2)

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Rubi [A]
time = 0.24, antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {5926, 5949, 32} \begin {gather*} -\frac {b c \sqrt {d-c^2 d x^2} (f x)^{m+2} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;c^2 x^2\right )}{f^2 (m+1) (m+2)^2 \sqrt {c x-1} \sqrt {c x+1}}+\frac {\sqrt {d-c^2 d x^2} (f x)^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )}{f \left (m^2+3 m+2\right ) \sqrt {1-c x} \sqrt {c x+1}}+\frac {\sqrt {d-c^2 d x^2} (f x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{f (m+2)}-\frac {b c \sqrt {d-c^2 d x^2} (f x)^{m+2}}{f^2 (m+2)^2 \sqrt {c x-1} \sqrt {c x+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*c*(f*x)^(2 + m)*Sqrt[d - c^2*d*x^2])/(f^2*(2 + m)^2*Sqrt[-1 + c*x]*Sqrt[1 + c*x])) + ((f*x)^(1 + m)*Sqrt[
d - c^2*d*x^2]*(a + b*ArcCosh[c*x]))/(f*(2 + m)) + ((f*x)^(1 + m)*Sqrt[d - c^2*d*x^2]*(a + b*ArcCosh[c*x])*Hyp
ergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2])/(f*(2 + 3*m + m^2)*Sqrt[1 - c*x]*Sqrt[1 + c*x]) - (b*c*(f*
x)^(2 + m)*Sqrt[d - c^2*d*x^2]*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2])/(f^2*(
1 + m)*(2 + m)^2*Sqrt[-1 + c*x]*Sqrt[1 + c*x])

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 5926

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(
f*x)^(m + 1)*Sqrt[d + e*x^2]*((a + b*ArcCosh[c*x])^n/(f*(m + 2))), x] + (-Dist[(1/(m + 2))*Simp[Sqrt[d + e*x^2
]/(Sqrt[1 + c*x]*Sqrt[-1 + c*x])], Int[(f*x)^m*((a + b*ArcCosh[c*x])^n/(Sqrt[1 + c*x]*Sqrt[-1 + c*x])), x], x]
 - Dist[b*c*(n/(f*(m + 2)))*Simp[Sqrt[d + e*x^2]/(Sqrt[1 + c*x]*Sqrt[-1 + c*x])], Int[(f*x)^(m + 1)*(a + b*Arc
Cosh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && IGtQ[n, 0] && (IGtQ[m,
-2] || EqQ[n, 1])

Rule 5949

Int[(((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_))/(Sqrt[(d1_) + (e1_.)*(x_)]*Sqrt[(d2_) + (e2_.)*(x_
)]), x_Symbol] :> Simp[((f*x)^(m + 1)/(f*(m + 1)))*Simp[Sqrt[1 - c^2*x^2]/(Sqrt[d1 + e1*x]*Sqrt[d2 + e2*x])]*(
a + b*ArcCosh[c*x])*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2], x] + Simp[b*c*((f*x)^(m + 2)/(f^2*(
m + 1)*(m + 2)))*Simp[Sqrt[1 + c*x]/Sqrt[d1 + e1*x]]*Simp[Sqrt[-1 + c*x]/Sqrt[d2 + e2*x]]*HypergeometricPFQ[{1
, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2], x] /; FreeQ[{a, b, c, d1, e1, d2, e2, f, m}, x] && EqQ[e1
, c*d1] && EqQ[e2, (-c)*d2] &&  !IntegerQ[m]

Rubi steps

\begin {align*} \int (f x)^m \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right ) \, dx &=\frac {\sqrt {d-c^2 d x^2} \int (f x)^m \sqrt {-1+c x} \sqrt {1+c x} \left (a+b \cosh ^{-1}(c x)\right ) \, dx}{\sqrt {-1+c x} \sqrt {1+c x}}\\ &=\frac {(f x)^{1+m} \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{f (2+m)}-\frac {\sqrt {d-c^2 d x^2} \int \frac {(f x)^m \left (a+b \cosh ^{-1}(c x)\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{(2+m) \sqrt {-1+c x} \sqrt {1+c x}}-\frac {\left (b c \sqrt {d-c^2 d x^2}\right ) \int (f x)^{1+m} \, dx}{f (2+m) \sqrt {-1+c x} \sqrt {1+c x}}\\ &=-\frac {b c (f x)^{2+m} \sqrt {d-c^2 d x^2}}{f^2 (2+m)^2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {(f x)^{1+m} \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{f (2+m)}+\frac {(f x)^{1+m} \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right ) \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};c^2 x^2\right )}{f \left (2+3 m+m^2\right ) (1-c x) (1+c x)}-\frac {b c (f x)^{2+m} \sqrt {d-c^2 d x^2} \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};c^2 x^2\right )}{f^2 (1+m) (2+m)^2 \sqrt {-1+c x} \sqrt {1+c x}}\\ \end {align*}

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Mathematica [A]
time = 0.21, size = 223, normalized size = 0.80 \begin {gather*} \frac {x (f x)^m \sqrt {d-c^2 d x^2} \left ((1+m) \left (-b c x \sqrt {-1+c x} \sqrt {1+c x}+a (2+m) \left (-1+c^2 x^2\right )+b (2+m) \left (-1+c^2 x^2\right ) \cosh ^{-1}(c x)\right )-(2+m) \sqrt {1-c^2 x^2} \left (a+b \cosh ^{-1}(c x)\right ) \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};c^2 x^2\right )-b c x \sqrt {-1+c x} \sqrt {1+c x} \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};c^2 x^2\right )\right )}{(1+m) (2+m)^2 (-1+c x) (1+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(f*x)^m*Sqrt[d - c^2*d*x^2]*((1 + m)*(-(b*c*x*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) + a*(2 + m)*(-1 + c^2*x^2) + b*
(2 + m)*(-1 + c^2*x^2)*ArcCosh[c*x]) - (2 + m)*Sqrt[1 - c^2*x^2]*(a + b*ArcCosh[c*x])*Hypergeometric2F1[1/2, (
1 + m)/2, (3 + m)/2, c^2*x^2] - b*c*x*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3
/2 + m/2, 2 + m/2}, c^2*x^2]))/((1 + m)*(2 + m)^2*(-1 + c*x)*(1 + c*x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((f*x)^m*(-c^2*d*x^2+d)^(1/2)*(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)^(1/2)*(a+b*arccosh(c*x)),x, algorithm="maxima")

[Out]

integrate(sqrt(-c^2*d*x^2 + d)*(b*arccosh(c*x) + a)*(f*x)^m, 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)^(1/2)*(a+b*arccosh(c*x)),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (f x\right )^{m} \sqrt {- d \left (c x - 1\right ) \left (c x + 1\right )} \left (a + b \operatorname {acosh}{\left (c x \right )}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)**m*(-c**2*d*x**2+d)**(1/2)*(a+b*acosh(c*x)),x)

[Out]

Integral((f*x)**m*sqrt(-d*(c*x - 1)*(c*x + 1))*(a + b*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)^(1/2)*(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 = -0.00 \begin {gather*} \int \left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )\,\sqrt {d-c^2\,d\,x^2}\,{\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)^(1/2)*(f*x)^m,x)

[Out]

int((a + b*acosh(c*x))*(d - c^2*d*x^2)^(1/2)*(f*x)^m, x)

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