3.160 \(\int \frac {(f x)^m (a+b \cosh ^{-1}(c x))}{\sqrt {\text {d1}+c \text {d1} x} \sqrt {\text {d2}-c \text {d2} x}} \, dx\)

Optimal. Leaf size=188 \[ \frac {b c \sqrt {c x-1} \sqrt {c x+1} (f x)^{m+2} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;c^2 x^2\right )}{f^2 (m+1) (m+2) \sqrt {c \text {d1} x+\text {d1}} \sqrt {\text {d2}-c \text {d2} x}}+\frac {\sqrt {1-c^2 x^2} (f x)^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )}{f (m+1) \sqrt {c \text {d1} x+\text {d1}} \sqrt {\text {d2}-c \text {d2} x}} \]

[Out]

b*c*(f*x)^(2+m)*HypergeometricPFQ([1, 1+1/2*m, 1+1/2*m],[3/2+1/2*m, 2+1/2*m],c^2*x^2)*(c*x-1)^(1/2)*(c*x+1)^(1
/2)/f^2/(1+m)/(2+m)/(c*d1*x+d1)^(1/2)/(-c*d2*x+d2)^(1/2)+(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*x^2+1)^(1/2)/f/(1+m)/(c*d1*x+d1)^(1/2)/(-c*d2*x+d2)^(1/2)

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Rubi [A]  time = 0.56, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {5765, 5763} \[ \frac {b c \sqrt {c x-1} \sqrt {c x+1} (f x)^{m+2} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;c^2 x^2\right )}{f^2 (m+1) (m+2) \sqrt {c \text {d1} x+\text {d1}} \sqrt {\text {d2}-c \text {d2} x}}+\frac {\sqrt {1-c^2 x^2} (f x)^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )}{f (m+1) \sqrt {c \text {d1} x+\text {d1}} \sqrt {\text {d2}-c \text {d2} x}} \]

Antiderivative was successfully verified.

[In]

Int[((f*x)^m*(a + b*ArcCosh[c*x]))/(Sqrt[d1 + c*d1*x]*Sqrt[d2 - c*d2*x]),x]

[Out]

((f*x)^(1 + 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])/(
f*(1 + m)*Sqrt[d1 + c*d1*x]*Sqrt[d2 - c*d2*x]) + (b*c*(f*x)^(2 + m)*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*Hypergeometri
cPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2])/(f^2*(1 + m)*(2 + m)*Sqrt[d1 + c*d1*x]*Sqrt[d2 - c
*d2*x])

Rule 5763

Int[(((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_))/(Sqrt[(d1_) + (e1_.)*(x_)]*Sqrt[(d2_) + (e2_.)*(x_
)]), x_Symbol] :> Simp[((f*x)^(m + 1)*Sqrt[1 - c^2*x^2]*(a + b*ArcCosh[c*x])*Hypergeometric2F1[1/2, (1 + m)/2,
(3 + m)/2, c^2*x^2])/(f*(m + 1)*Sqrt[d1 + e1*x]*Sqrt[d2 + e2*x]), x] + Simp[(b*c*(f*x)^(m + 2)*Hypergeometric
PFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2])/(Sqrt[-(d1*d2)]*f^2*(m + 1)*(m + 2)), x] /; FreeQ[{
a, b, c, d1, e1, d2, e2, f, m}, x] && EqQ[e1 - c*d1, 0] && EqQ[e2 + c*d2, 0] && GtQ[d1, 0] && LtQ[d2, 0] &&  !
IntegerQ[m]

Rule 5765

Int[(((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/(Sqrt[(d1_) + (e1_.)*(x_)]*Sqrt[(d2_) + (e2_
.)*(x_)]), x_Symbol] :> Dist[(Sqrt[1 + c*x]*Sqrt[-1 + c*x])/(Sqrt[d1 + e1*x]*Sqrt[d2 + e2*x]), Int[((f*x)^m*(a
+ b*ArcCosh[c*x])^n)/(Sqrt[1 + c*x]*Sqrt[-1 + c*x]), x], x] /; FreeQ[{a, b, c, d1, e1, d2, e2, f, m}, x] && E
qQ[e1 - c*d1, 0] && EqQ[e2 + c*d2, 0] && GtQ[n, 0] &&  !(GtQ[d1, 0] && LtQ[d2, 0]) && (IntegerQ[m] || EqQ[n, 1
])

Rubi steps

\begin {align*} \int \frac {(f x)^m \left (a+b \cosh ^{-1}(c x)\right )}{\sqrt {\text {d1}+c \text {d1} x} \sqrt {\text {d2}-c \text {d2} x}} \, dx &=\frac {\left (\sqrt {-1+c x} \sqrt {1+c x}\right ) \int \frac {(f x)^m \left (a+b \cosh ^{-1}(c x)\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{\sqrt {\text {d1}+c \text {d1} x} \sqrt {\text {d2}-c \text {d2} x}}\\ &=\frac {(f x)^{1+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 )}{f (1+m) \sqrt {\text {d1}+c \text {d1} x} \sqrt {\text {d2}-c \text {d2} x}}+\frac {b c (f x)^{2+m} \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 )}{f^2 (1+m) (2+m) \sqrt {\text {d1}+c \text {d1} x} \sqrt {\text {d2}-c \text {d2} x}}\\ \end {align*}

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Mathematica [C]  time = 6.27, size = 264, normalized size = 1.40 \[ \frac {2^{-m-3} \sqrt {c \text {d1} x+\text {d1}} \left (\frac {c x}{c x+1}\right )^{1-m} (f x)^m \left (b m \left (\frac {c x}{c x+1}\right )^m \sinh \left (2 \cosh ^{-1}(c x)\right ) \left (\sqrt {\pi } c (m+1) x \sqrt {\frac {c x-1}{c x+1}} \Gamma (m+1) \, _3\tilde {F}_2\left (1,\frac {m+2}{2},\frac {m+2}{2};\frac {m+3}{2},\frac {m+4}{2};c^2 x^2\right )-2^{m+2} (c x-1) \cosh ^{-1}(c x) \, _2F_1\left (1,\frac {m+2}{2};\frac {m+3}{2};c^2 x^2\right )\right )+a 2^{m+3} (m+1) (c x-1) F_1\left (-m;-m,\frac {1}{2};1-m;\frac {1}{c x+1},\frac {2}{c x+1}\right )\right )}{c^2 \text {d1} m (m+1) x \sqrt {\frac {c x-1}{c x+1}} \sqrt {\text {d2}-c \text {d2} x}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((f*x)^m*(a + b*ArcCosh[c*x]))/(Sqrt[d1 + c*d1*x]*Sqrt[d2 - c*d2*x]),x]

[Out]

(2^(-3 - m)*(f*x)^m*((c*x)/(1 + c*x))^(1 - m)*Sqrt[d1 + c*d1*x]*(2^(3 + m)*a*(1 + m)*(-1 + c*x)*AppellF1[-m, -
m, 1/2, 1 - m, (1 + c*x)^(-1), 2/(1 + c*x)] + b*m*((c*x)/(1 + c*x))^m*(-(2^(2 + m)*(-1 + c*x)*ArcCosh[c*x]*Hyp
ergeometric2F1[1, (2 + m)/2, (3 + m)/2, c^2*x^2]) + c*(1 + m)*Sqrt[Pi]*x*Sqrt[(-1 + c*x)/(1 + c*x)]*Gamma[1 +
m]*HypergeometricPFQRegularized[{1, (2 + m)/2, (2 + m)/2}, {(3 + m)/2, (4 + m)/2}, c^2*x^2])*Sinh[2*ArcCosh[c*
x]]))/(c^2*d1*m*(1 + m)*x*Sqrt[(-1 + c*x)/(1 + c*x)]*Sqrt[d2 - c*d2*x])

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {c d_{1} x + d_{1}} \sqrt {-c d_{2} x + d_{2}} {\left (b \operatorname {arcosh}\left (c x\right ) + a\right )} \left (f x\right )^{m}}{c^{2} d_{1} d_{2} x^{2} - d_{1} d_{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*(a+b*arccosh(c*x))/(c*d1*x+d1)^(1/2)/(-c*d2*x+d2)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(c*d1*x + d1)*sqrt(-c*d2*x + d2)*(b*arccosh(c*x) + a)*(f*x)^m/(c^2*d1*d2*x^2 - d1*d2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arcosh}\left (c x\right ) + a\right )} \left (f x\right )^{m}}{\sqrt {c d_{1} x + d_{1}} \sqrt {-c d_{2} x + d_{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*(a+b*arccosh(c*x))/(c*d1*x+d1)^(1/2)/(-c*d2*x+d2)^(1/2),x, algorithm="giac")

[Out]

integrate((b*arccosh(c*x) + a)*(f*x)^m/(sqrt(c*d1*x + d1)*sqrt(-c*d2*x + d2)), x)

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maple [F]  time = 0.70, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x \right )^{m} \left (a +b \,\mathrm {arccosh}\left (c x \right )\right )}{\sqrt {c \mathit {d1} x +\mathit {d1}}\, \sqrt {-c \mathit {d2} x +\mathit {d2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x)^m*(a+b*arccosh(c*x))/(c*d1*x+d1)^(1/2)/(-c*d2*x+d2)^(1/2),x)

[Out]

int((f*x)^m*(a+b*arccosh(c*x))/(c*d1*x+d1)^(1/2)/(-c*d2*x+d2)^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arcosh}\left (c x\right ) + a\right )} \left (f x\right )^{m}}{\sqrt {c d_{1} x + d_{1}} \sqrt {-c d_{2} x + d_{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*(a+b*arccosh(c*x))/(c*d1*x+d1)^(1/2)/(-c*d2*x+d2)^(1/2),x, algorithm="maxima")

[Out]

integrate((b*arccosh(c*x) + a)*(f*x)^m/(sqrt(c*d1*x + d1)*sqrt(-c*d2*x + d2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )\,{\left (f\,x\right )}^m}{\sqrt {d_{1}+c\,d_{1}\,x}\,\sqrt {d_{2}-c\,d_{2}\,x}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*acosh(c*x))*(f*x)^m)/((d1 + c*d1*x)^(1/2)*(d2 - c*d2*x)^(1/2)),x)

[Out]

int(((a + b*acosh(c*x))*(f*x)^m)/((d1 + c*d1*x)^(1/2)*(d2 - c*d2*x)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x\right )^{m} \left (a + b \operatorname {acosh}{\left (c x \right )}\right )}{\sqrt {d_{1} \left (c x + 1\right )} \sqrt {- d_{2} \left (c x - 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)**m*(a+b*acosh(c*x))/(c*d1*x+d1)**(1/2)/(-c*d2*x+d2)**(1/2),x)

[Out]

Integral((f*x)**m*(a + b*acosh(c*x))/(sqrt(d1*(c*x + 1))*sqrt(-d2*(c*x - 1))), x)

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