3.144 \(\int \frac {(f x)^{3/2} (a+b \cosh ^{-1}(c x))}{\sqrt {d-c^2 d x^2}} \, dx\)

Optimal. Leaf size=141 \[ \frac {4 b c \sqrt {c x-1} \sqrt {c x+1} (f x)^{7/2} \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};c^2 x^2\right )}{35 f^2 \sqrt {d-c^2 d x^2}}+\frac {2 \sqrt {1-c^2 x^2} (f x)^{5/2} \, _2F_1\left (\frac {1}{2},\frac {5}{4};\frac {9}{4};c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )}{5 f \sqrt {d-c^2 d x^2}} \]

[Out]

4/35*b*c*(f*x)^(7/2)*HypergeometricPFQ([1, 7/4, 7/4],[9/4, 11/4],c^2*x^2)*(c*x-1)^(1/2)*(c*x+1)^(1/2)/f^2/(-c^
2*d*x^2+d)^(1/2)+2/5*(f*x)^(5/2)*(a+b*arccosh(c*x))*hypergeom([1/2, 5/4],[9/4],c^2*x^2)*(-c^2*x^2+1)^(1/2)/f/(
-c^2*d*x^2+d)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.38, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {5798, 5763} \[ \frac {4 b c \sqrt {c x-1} \sqrt {c x+1} (f x)^{7/2} \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};c^2 x^2\right )}{35 f^2 \sqrt {d-c^2 d x^2}}+\frac {2 \sqrt {1-c^2 x^2} (f x)^{5/2} \, _2F_1\left (\frac {1}{2},\frac {5}{4};\frac {9}{4};c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )}{5 f \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((f*x)^(3/2)*(a + b*ArcCosh[c*x]))/Sqrt[d - c^2*d*x^2],x]

[Out]

(2*(f*x)^(5/2)*Sqrt[1 - c^2*x^2]*(a + b*ArcCosh[c*x])*Hypergeometric2F1[1/2, 5/4, 9/4, c^2*x^2])/(5*f*Sqrt[d -
c^2*d*x^2]) + (4*b*c*(f*x)^(7/2)*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*HypergeometricPFQ[{1, 7/4, 7/4}, {9/4, 11/4}, c
^2*x^2])/(35*f^2*Sqrt[d - c^2*d*x^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 5798

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Dist
[((-d)^IntPart[p]*(d + e*x^2)^FracPart[p])/((1 + c*x)^FracPart[p]*(-1 + c*x)^FracPart[p]), Int[(f*x)^m*(1 + c*
x)^p*(-1 + c*x)^p*(a + b*ArcCosh[c*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[c^2*d + e, 0]
&&  !IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {(f x)^{3/2} \left (a+b \cosh ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2}} \, dx &=\frac {\left (\sqrt {-1+c x} \sqrt {1+c x}\right ) \int \frac {(f x)^{3/2} \left (a+b \cosh ^{-1}(c x)\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{\sqrt {d-c^2 d x^2}}\\ &=\frac {2 (f x)^{5/2} \sqrt {1-c^2 x^2} \left (a+b \cosh ^{-1}(c x)\right ) \, _2F_1\left (\frac {1}{2},\frac {5}{4};\frac {9}{4};c^2 x^2\right )}{5 f \sqrt {d-c^2 d x^2}}+\frac {4 b c (f x)^{7/2} \sqrt {-1+c x} \sqrt {1+c x} \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};c^2 x^2\right )}{35 f^2 \sqrt {d-c^2 d x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 115, normalized size = 0.82 \[ \frac {2 x (f x)^{3/2} \left (2 b c x \sqrt {c x-1} \sqrt {c x+1} \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};c^2 x^2\right )+7 \sqrt {1-c^2 x^2} \, _2F_1\left (\frac {1}{2},\frac {5}{4};\frac {9}{4};c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )\right )}{35 \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((f*x)^(3/2)*(a + b*ArcCosh[c*x]))/Sqrt[d - c^2*d*x^2],x]

[Out]

(2*x*(f*x)^(3/2)*(7*Sqrt[1 - c^2*x^2]*(a + b*ArcCosh[c*x])*Hypergeometric2F1[1/2, 5/4, 9/4, c^2*x^2] + 2*b*c*x
*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*HypergeometricPFQ[{1, 7/4, 7/4}, {9/4, 11/4}, c^2*x^2]))/(35*Sqrt[d - c^2*d*x^2]
)

________________________________________________________________________________________

fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b f x \operatorname {arcosh}\left (c x\right ) + a f x\right )} \sqrt {f x}}{c^{2} d x^{2} - d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^(3/2)*(a+b*arccosh(c*x))/(-c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^(3/2)*(a+b*arccosh(c*x))/(-c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

integrate((f*x)^(3/2)*(b*arccosh(c*x) + a)/sqrt(-c^2*d*x^2 + d), x)

________________________________________________________________________________________

maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x \right )^{\frac {3}{2}} \left (a +b \,\mathrm {arccosh}\left (c x \right )\right )}{\sqrt {-c^{2} d \,x^{2}+d}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^(3/2)*(a+b*arccosh(c*x))/(-c^2*d*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

integrate((f*x)^(3/2)*(b*arccosh(c*x) + a)/sqrt(-c^2*d*x^2 + d), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )\,{\left (f\,x\right )}^{3/2}}{\sqrt {d-c^2\,d\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*acosh(c*x))*(f*x)^(3/2))/(d - c^2*d*x^2)^(1/2),x)

[Out]

int(((a + b*acosh(c*x))*(f*x)^(3/2))/(d - c^2*d*x^2)^(1/2), x)

________________________________________________________________________________________

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)**(3/2)*(a+b*acosh(c*x))/(-c**2*d*x**2+d)**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________