3.2.60 \(\int \frac {(f x)^m (a+b \cosh ^{-1}(c x))}{\sqrt {\text {d1}+c \text {d1} x} \sqrt {\text {d2}-c \text {d2} x}} \, dx\) [160]

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

[Out]

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*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 = 0.22, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {5949} \begin {gather*} \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}} \end {gather*}

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 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 \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] Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.
time = 4.22, size = 264, normalized size = 1.40 \begin {gather*} \frac {2^{-3-m} (f x)^m \left (\frac {c x}{1+c x}\right )^{1-m} \sqrt {\text {d1}+c \text {d1} x} \left (2^{3+m} a (1+m) (-1+c x) F_1\left (-m;-m,\frac {1}{2};1-m;\frac {1}{1+c x},\frac {2}{1+c x}\right )+b m \left (\frac {c x}{1+c x}\right )^m \left (-2^{2+m} (-1+c x) \cosh ^{-1}(c x) \, _2F_1\left (1,\frac {2+m}{2};\frac {3+m}{2};c^2 x^2\right )+c (1+m) \sqrt {\pi } x \sqrt {\frac {-1+c x}{1+c x}} \Gamma (1+m) \, _3\tilde {F}_2\left (1,\frac {2+m}{2},\frac {2+m}{2};\frac {3+m}{2},\frac {4+m}{2};c^2 x^2\right )\right ) \sinh \left (2 \cosh ^{-1}(c x)\right )\right )}{c^2 \text {d1} m (1+m) x \sqrt {\frac {-1+c x}{1+c x}} \sqrt {\text {d2}-c \text {d2} x}} \end {gather*}

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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Maple [F]
time = 180.00, 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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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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*(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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

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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Giac [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*(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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

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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