3.161 \(\int \frac {(f x)^m (a+b \cosh ^{-1}(c x))}{(\text {d1}+c \text {d1} x)^{3/2} (\text {d2}-c \text {d2} x)^{3/2}} \, dx\)

Optimal. Leaf size=336 \[ -\frac {b c m \sqrt {c x-1} \sqrt {c x+1} (f x)^{m+2} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;c^2 x^2\right )}{\text {d1} \text {d2} f^2 (m+1) (m+2) \sqrt {c \text {d1} x+\text {d1}} \sqrt {\text {d2}-c \text {d2} x}}-\frac {m \sqrt {1-c^2 x^2} (f x)^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )}{\text {d1} \text {d2} f (m+1) \sqrt {c \text {d1} x+\text {d1}} \sqrt {\text {d2}-c \text {d2} x}}+\frac {(f x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{\text {d1} \text {d2} f \sqrt {c \text {d1} x+\text {d1}} \sqrt {\text {d2}-c \text {d2} x}}+\frac {b c \sqrt {c x-1} \sqrt {c x+1} (f x)^{m+2} \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{\text {d1} \text {d2} f^2 (m+2) \sqrt {c \text {d1} x+\text {d1}} \sqrt {\text {d2}-c \text {d2} x}} \]

[Out]

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

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Rubi [A]  time = 0.94, antiderivative size = 336, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {5756, 5765, 5763, 364} \[ -\frac {b c m \sqrt {c x-1} \sqrt {c x+1} (f x)^{m+2} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;c^2 x^2\right )}{\text {d1} \text {d2} f^2 (m+1) (m+2) \sqrt {c \text {d1} x+\text {d1}} \sqrt {\text {d2}-c \text {d2} x}}-\frac {m \sqrt {1-c^2 x^2} (f x)^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )}{\text {d1} \text {d2} f (m+1) \sqrt {c \text {d1} x+\text {d1}} \sqrt {\text {d2}-c \text {d2} x}}+\frac {(f x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{\text {d1} \text {d2} f \sqrt {c \text {d1} x+\text {d1}} \sqrt {\text {d2}-c \text {d2} x}}+\frac {b c \sqrt {c x-1} \sqrt {c x+1} (f x)^{m+2} \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{\text {d1} \text {d2} f^2 (m+2) \sqrt {c \text {d1} x+\text {d1}} \sqrt {\text {d2}-c \text {d2} x}} \]

Antiderivative was successfully verified.

[In]

Int[((f*x)^m*(a + b*ArcCosh[c*x]))/((d1 + c*d1*x)^(3/2)*(d2 - c*d2*x)^(3/2)),x]

[Out]

((f*x)^(1 + m)*(a + b*ArcCosh[c*x]))/(d1*d2*f*Sqrt[d1 + c*d1*x]*Sqrt[d2 - c*d2*x]) - (m*(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])/(d1*d2*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]*Hypergeometric2F1[1, (2 + m)/2
, (4 + m)/2, c^2*x^2])/(d1*d2*f^2*(2 + m)*Sqrt[d1 + c*d1*x]*Sqrt[d2 - c*d2*x]) - (b*c*m*(f*x)^(2 + m)*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])/(d1*d2*f^2*(1 +
m)*(2 + m)*Sqrt[d1 + c*d1*x]*Sqrt[d2 - c*d2*x])

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 5756

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d1_) + (e1_.)*(x_))^(p_)*((d2_) + (e2_.)*(x_
))^(p_), x_Symbol] :> -Simp[((f*x)^(m + 1)*(d1 + e1*x)^(p + 1)*(d2 + e2*x)^(p + 1)*(a + b*ArcCosh[c*x])^n)/(2*
d1*d2*f*(p + 1)), x] + (Dist[(m + 2*p + 3)/(2*d1*d2*(p + 1)), Int[(f*x)^m*(d1 + e1*x)^(p + 1)*(d2 + e2*x)^(p +
1)*(a + b*ArcCosh[c*x])^n, x], x] - Dist[(b*c*n*(-(d1*d2))^IntPart[p]*(d1 + e1*x)^FracPart[p]*(d2 + e2*x)^Fra
cPart[p])/(2*f*(p + 1)*(1 + c*x)^FracPart[p]*(-1 + c*x)^FracPart[p]), Int[(f*x)^(m + 1)*(-1 + c^2*x^2)^(p + 1/
2)*(a + b*ArcCosh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d1, e1, d2, e2, f, m}, x] && EqQ[e1 - c*d1, 0] &&
EqQ[e2 + c*d2, 0] && GtQ[n, 0] && LtQ[p, -1] &&  !GtQ[m, 1] && (IntegerQ[m] || EqQ[n, 1]) && IntegerQ[p + 1/2]

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 )}{(\text {d1}+c \text {d1} x)^{3/2} (\text {d2}-c \text {d2} x)^{3/2}} \, dx &=\frac {(f x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )}{\text {d1} \text {d2} f \sqrt {\text {d1}+c \text {d1} x} \sqrt {\text {d2}-c \text {d2} x}}-\frac {m \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}{\text {d1} \text {d2}}-\frac {\left (b c \sqrt {-1+c x} \sqrt {1+c x}\right ) \int \frac {(f x)^{1+m}}{-1+c^2 x^2} \, dx}{\text {d1} \text {d2} f \sqrt {\text {d1}+c \text {d1} x} \sqrt {\text {d2}-c \text {d2} x}}\\ &=\frac {(f x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )}{\text {d1} \text {d2} f \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} \, _2F_1\left (1,\frac {2+m}{2};\frac {4+m}{2};c^2 x^2\right )}{\text {d1} \text {d2} f^2 (2+m) \sqrt {\text {d1}+c \text {d1} x} \sqrt {\text {d2}-c \text {d2} x}}-\frac {\left (m \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}{\text {d1} \text {d2} \sqrt {\text {d1}+c \text {d1} x} \sqrt {\text {d2}-c \text {d2} x}}\\ &=\frac {(f x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )}{\text {d1} \text {d2} f \sqrt {\text {d1}+c \text {d1} x} \sqrt {\text {d2}-c \text {d2} x}}-\frac {m (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 )}{\text {d1} \text {d2} 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} \, _2F_1\left (1,\frac {2+m}{2};\frac {4+m}{2};c^2 x^2\right )}{\text {d1} \text {d2} f^2 (2+m) \sqrt {\text {d1}+c \text {d1} x} \sqrt {\text {d2}-c \text {d2} x}}-\frac {b c m (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 )}{\text {d1} \text {d2} 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 [F]  time = 2.56, size = 0, normalized size = 0.00 \[ \int \frac {(f x)^m \left (a+b \cosh ^{-1}(c x)\right )}{(\text {d1}+c \text {d1} x)^{3/2} (\text {d2}-c \text {d2} x)^{3/2}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[((f*x)^m*(a + b*ArcCosh[c*x]))/((d1 + c*d1*x)^(3/2)*(d2 - c*d2*x)^(3/2)),x]

[Out]

Integrate[((f*x)^m*(a + b*ArcCosh[c*x]))/((d1 + c*d1*x)^(3/2)*(d2 - c*d2*x)^(3/2)), x]

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c d_{1} x + d_{1}} \sqrt {-c d_{2} x + d_{2}} {\left (b \operatorname {arcosh}\left (c x\right ) + a\right )} \left (f x\right )^{m}}{c^{4} d_{1}^{2} d_{2}^{2} x^{4} - 2 \, c^{2} d_{1}^{2} d_{2}^{2} x^{2} + d_{1}^{2} d_{2}^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*(a+b*arccosh(c*x))/(c*d1*x+d1)^(3/2)/(-c*d2*x+d2)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*d1*x + d1)*sqrt(-c*d2*x + d2)*(b*arccosh(c*x) + a)*(f*x)^m/(c^4*d1^2*d2^2*x^4 - 2*c^2*d1^2*d2^
2*x^2 + d1^2*d2^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arcosh}\left (c x\right ) + a\right )} \left (f x\right )^{m}}{{\left (c d_{1} x + d_{1}\right )}^{\frac {3}{2}} {\left (-c d_{2} x + d_{2}\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*(a+b*arccosh(c*x))/(c*d1*x+d1)^(3/2)/(-c*d2*x+d2)^(3/2),x, algorithm="giac")

[Out]

integrate((b*arccosh(c*x) + a)*(f*x)^m/((c*d1*x + d1)^(3/2)*(-c*d2*x + d2)^(3/2)), x)

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maple [F]  time = 0.96, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x \right )^{m} \left (a +b \,\mathrm {arccosh}\left (c x \right )\right )}{\left (c \mathit {d1} x +\mathit {d1} \right )^{\frac {3}{2}} \left (-c \mathit {d2} x +\mathit {d2} \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x)^m*(a+b*arccosh(c*x))/(c*d1*x+d1)^(3/2)/(-c*d2*x+d2)^(3/2),x)

[Out]

int((f*x)^m*(a+b*arccosh(c*x))/(c*d1*x+d1)^(3/2)/(-c*d2*x+d2)^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arcosh}\left (c x\right ) + a\right )} \left (f x\right )^{m}}{{\left (c d_{1} x + d_{1}\right )}^{\frac {3}{2}} {\left (-c d_{2} x + d_{2}\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*(a+b*arccosh(c*x))/(c*d1*x+d1)^(3/2)/(-c*d2*x+d2)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*arccosh(c*x) + a)*(f*x)^m/((c*d1*x + d1)^(3/2)*(-c*d2*x + d2)^(3/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )\,{\left (f\,x\right )}^m}{{\left (d_{1}+c\,d_{1}\,x\right )}^{3/2}\,{\left (d_{2}-c\,d_{2}\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*acosh(c*x))*(f*x)^m)/((d1 + c*d1*x)^(3/2)*(d2 - c*d2*x)^(3/2)),x)

[Out]

int(((a + b*acosh(c*x))*(f*x)^m)/((d1 + c*d1*x)^(3/2)*(d2 - c*d2*x)^(3/2)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)**m*(a+b*acosh(c*x))/(c*d1*x+d1)**(3/2)/(-c*d2*x+d2)**(3/2),x)

[Out]

Timed out

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