3.212 \(\int \frac {(a+b \cosh ^{-1}(c+d x))^2}{(c e+d e x)^{5/2}} \, dx\)

Optimal. Leaf size=153 \[ -\frac {16 b^2 \sqrt {e (c+d x)} \, _3F_2\left (\frac {1}{4},\frac {1}{4},1;\frac {3}{4},\frac {5}{4};(c+d x)^2\right )}{3 d e^3}-\frac {8 b \sqrt {-c-d x+1} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};(c+d x)^2\right ) \left (a+b \cosh ^{-1}(c+d x)\right )}{3 d e^2 \sqrt {c+d x-1} \sqrt {e (c+d x)}}-\frac {2 \left (a+b \cosh ^{-1}(c+d x)\right )^2}{3 d e (e (c+d x))^{3/2}} \]

[Out]

-2/3*(a+b*arccosh(d*x+c))^2/d/e/(e*(d*x+c))^(3/2)-8/3*b*(a+b*arccosh(d*x+c))*hypergeom([-1/4, 1/2],[3/4],(d*x+
c)^2)*(-d*x-c+1)^(1/2)/d/e^2/(d*x+c-1)^(1/2)/(e*(d*x+c))^(1/2)-16/3*b^2*HypergeometricPFQ([1/4, 1/4, 1],[3/4,
5/4],(d*x+c)^2)*(e*(d*x+c))^(1/2)/d/e^3

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Rubi [A]  time = 0.31, antiderivative size = 165, normalized size of antiderivative = 1.08, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {5866, 5662, 5763} \[ -\frac {16 b^2 \sqrt {e (c+d x)} \, _3F_2\left (\frac {1}{4},\frac {1}{4},1;\frac {3}{4},\frac {5}{4};(c+d x)^2\right )}{3 d e^3}-\frac {8 b \sqrt {1-(c+d x)^2} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};(c+d x)^2\right ) \left (a+b \cosh ^{-1}(c+d x)\right )}{3 d e^2 \sqrt {c+d x-1} \sqrt {c+d x+1} \sqrt {e (c+d x)}}-\frac {2 \left (a+b \cosh ^{-1}(c+d x)\right )^2}{3 d e (e (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCosh[c + d*x])^2/(c*e + d*e*x)^(5/2),x]

[Out]

(-2*(a + b*ArcCosh[c + d*x])^2)/(3*d*e*(e*(c + d*x))^(3/2)) - (8*b*Sqrt[1 - (c + d*x)^2]*(a + b*ArcCosh[c + d*
x])*Hypergeometric2F1[-1/4, 1/2, 3/4, (c + d*x)^2])/(3*d*e^2*Sqrt[-1 + c + d*x]*Sqrt[e*(c + d*x)]*Sqrt[1 + c +
d*x]) - (16*b^2*Sqrt[e*(c + d*x)]*HypergeometricPFQ[{1/4, 1/4, 1}, {3/4, 5/4}, (c + d*x)^2])/(3*d*e^3)

Rule 5662

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcC
osh[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCosh[c*x])^(n - 1))/(Sqr
t[-1 + c*x]*Sqrt[1 + c*x]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

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 5866

Int[((a_.) + ArcCosh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcCosh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
]

Rubi steps

\begin {align*} \int \frac {\left (a+b \cosh ^{-1}(c+d x)\right )^2}{(c e+d e x)^{5/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \cosh ^{-1}(x)\right )^2}{(e x)^{5/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {2 \left (a+b \cosh ^{-1}(c+d x)\right )^2}{3 d e (e (c+d x))^{3/2}}+\frac {(4 b) \operatorname {Subst}\left (\int \frac {a+b \cosh ^{-1}(x)}{\sqrt {-1+x} (e x)^{3/2} \sqrt {1+x}} \, dx,x,c+d x\right )}{3 d e}\\ &=-\frac {2 \left (a+b \cosh ^{-1}(c+d x)\right )^2}{3 d e (e (c+d x))^{3/2}}-\frac {8 b \sqrt {1-(c+d x)^2} \left (a+b \cosh ^{-1}(c+d x)\right ) \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};(c+d x)^2\right )}{3 d e^2 \sqrt {-1+c+d x} \sqrt {e (c+d x)} \sqrt {1+c+d x}}-\frac {16 b^2 \sqrt {e (c+d x)} \, _3F_2\left (\frac {1}{4},\frac {1}{4},1;\frac {3}{4},\frac {5}{4};(c+d x)^2\right )}{3 d e^3}\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 140, normalized size = 0.92 \[ \frac {2 \left (4 b (c+d x) \left (-2 b (c+d x) \, _3F_2\left (\frac {1}{4},\frac {1}{4},1;\frac {3}{4},\frac {5}{4};(c+d x)^2\right )-\frac {\sqrt {1-(c+d x)^2} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};(c+d x)^2\right ) \left (a+b \cosh ^{-1}(c+d x)\right )}{\sqrt {c+d x-1} \sqrt {c+d x+1}}\right )-\left (a+b \cosh ^{-1}(c+d x)\right )^2\right )}{3 d e (e (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCosh[c + d*x])^2/(c*e + d*e*x)^(5/2),x]

[Out]

(2*(-(a + b*ArcCosh[c + d*x])^2 + 4*b*(c + d*x)*(-((Sqrt[1 - (c + d*x)^2]*(a + b*ArcCosh[c + d*x])*Hypergeomet
ric2F1[-1/4, 1/2, 3/4, (c + d*x)^2])/(Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x])) - 2*b*(c + d*x)*HypergeometricPFQ
[{1/4, 1/4, 1}, {3/4, 5/4}, (c + d*x)^2])))/(3*d*e*(e*(c + d*x))^(3/2))

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} \operatorname {arcosh}\left (d x + c\right )^{2} + 2 \, a b \operatorname {arcosh}\left (d x + c\right ) + a^{2}\right )} \sqrt {d e x + c e}}{d^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + 3 \, c^{2} d e^{3} x + c^{3} e^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(d*x+c))^2/(d*e*x+c*e)^(5/2),x, algorithm="fricas")

[Out]

integral((b^2*arccosh(d*x + c)^2 + 2*a*b*arccosh(d*x + c) + a^2)*sqrt(d*e*x + c*e)/(d^3*e^3*x^3 + 3*c*d^2*e^3*
x^2 + 3*c^2*d*e^3*x + c^3*e^3), x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(d*x+c))^2/(d*e*x+c*e)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is
real):Check [abs(t_nostep)]Evaluation time: 0.82sym2poly/r2sym(const gen & e,const index_m & i,const vecteur
& l) Error: Bad Argument Value

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \,\mathrm {arccosh}\left (d x +c \right )\right )^{2}}{\left (d e x +c e \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccosh(d*x+c))^2/(d*e*x+c*e)^(5/2),x)

[Out]

int((a+b*arccosh(d*x+c))^2/(d*e*x+c*e)^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {2 \, \sqrt {d x + c} b^{2} \sqrt {e} \log \left (d x + \sqrt {d x + c + 1} \sqrt {d x + c - 1} + c\right )^{2}}{3 \, {\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} - \frac {2 \, a^{2}}{3 \, {\left (d e x + c e\right )}^{\frac {3}{2}} d e} + \int \frac {2 \, {\left ({\left (2 \, b^{2} c^{2} + 3 \, {\left (c^{2} - 1\right )} a b + {\left (3 \, a b d^{2} + 2 \, b^{2} d^{2}\right )} x^{2} + 2 \, {\left (3 \, a b c d + 2 \, b^{2} c d\right )} x\right )} \sqrt {d x + c + 1} \sqrt {d x + c} \sqrt {d x + c - 1} + {\left ({\left (3 \, a b d^{3} + 2 \, b^{2} d^{3}\right )} x^{3} + 3 \, {\left (c^{3} - c\right )} a b + 2 \, {\left (c^{3} - c\right )} b^{2} + 3 \, {\left (3 \, a b c d^{2} + 2 \, b^{2} c d^{2}\right )} x^{2} + {\left (3 \, {\left (3 \, c^{2} d - d\right )} a b + 2 \, {\left (3 \, c^{2} d - d\right )} b^{2}\right )} x\right )} \sqrt {d x + c}\right )} \log \left (d x + \sqrt {d x + c + 1} \sqrt {d x + c - 1} + c\right )}{3 \, {\left (d^{6} e^{\frac {5}{2}} x^{6} + 6 \, c d^{5} e^{\frac {5}{2}} x^{5} + c^{6} e^{\frac {5}{2}} - c^{4} e^{\frac {5}{2}} + {\left (15 \, c^{2} d^{4} e^{\frac {5}{2}} - d^{4} e^{\frac {5}{2}}\right )} x^{4} + 4 \, {\left (5 \, c^{3} d^{3} e^{\frac {5}{2}} - c d^{3} e^{\frac {5}{2}}\right )} x^{3} + 3 \, {\left (5 \, c^{4} d^{2} e^{\frac {5}{2}} - 2 \, c^{2} d^{2} e^{\frac {5}{2}}\right )} x^{2} + {\left (d^{5} e^{\frac {5}{2}} x^{5} + 5 \, c d^{4} e^{\frac {5}{2}} x^{4} + c^{5} e^{\frac {5}{2}} - c^{3} e^{\frac {5}{2}} + {\left (10 \, c^{2} d^{3} e^{\frac {5}{2}} - d^{3} e^{\frac {5}{2}}\right )} x^{3} + {\left (10 \, c^{3} d^{2} e^{\frac {5}{2}} - 3 \, c d^{2} e^{\frac {5}{2}}\right )} x^{2} + {\left (5 \, c^{4} d e^{\frac {5}{2}} - 3 \, c^{2} d e^{\frac {5}{2}}\right )} x\right )} \sqrt {d x + c + 1} \sqrt {d x + c - 1} + 2 \, {\left (3 \, c^{5} d e^{\frac {5}{2}} - 2 \, c^{3} d e^{\frac {5}{2}}\right )} x\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(d*x+c))^2/(d*e*x+c*e)^(5/2),x, algorithm="maxima")

[Out]

-2/3*sqrt(d*x + c)*b^2*sqrt(e)*log(d*x + sqrt(d*x + c + 1)*sqrt(d*x + c - 1) + c)^2/(d^3*e^3*x^2 + 2*c*d^2*e^3
*x + c^2*d*e^3) - 2/3*a^2/((d*e*x + c*e)^(3/2)*d*e) + integrate(2/3*((2*b^2*c^2 + 3*(c^2 - 1)*a*b + (3*a*b*d^2
+ 2*b^2*d^2)*x^2 + 2*(3*a*b*c*d + 2*b^2*c*d)*x)*sqrt(d*x + c + 1)*sqrt(d*x + c)*sqrt(d*x + c - 1) + ((3*a*b*d
^3 + 2*b^2*d^3)*x^3 + 3*(c^3 - c)*a*b + 2*(c^3 - c)*b^2 + 3*(3*a*b*c*d^2 + 2*b^2*c*d^2)*x^2 + (3*(3*c^2*d - d)
*a*b + 2*(3*c^2*d - d)*b^2)*x)*sqrt(d*x + c))*log(d*x + sqrt(d*x + c + 1)*sqrt(d*x + c - 1) + c)/(d^6*e^(5/2)*
x^6 + 6*c*d^5*e^(5/2)*x^5 + c^6*e^(5/2) - c^4*e^(5/2) + (15*c^2*d^4*e^(5/2) - d^4*e^(5/2))*x^4 + 4*(5*c^3*d^3*
e^(5/2) - c*d^3*e^(5/2))*x^3 + 3*(5*c^4*d^2*e^(5/2) - 2*c^2*d^2*e^(5/2))*x^2 + (d^5*e^(5/2)*x^5 + 5*c*d^4*e^(5
/2)*x^4 + c^5*e^(5/2) - c^3*e^(5/2) + (10*c^2*d^3*e^(5/2) - d^3*e^(5/2))*x^3 + (10*c^3*d^2*e^(5/2) - 3*c*d^2*e
^(5/2))*x^2 + (5*c^4*d*e^(5/2) - 3*c^2*d*e^(5/2))*x)*sqrt(d*x + c + 1)*sqrt(d*x + c - 1) + 2*(3*c^5*d*e^(5/2)
- 2*c^3*d*e^(5/2))*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {acosh}\left (c+d\,x\right )\right )}^2}{{\left (c\,e+d\,e\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(c + d*x))^2/(c*e + d*e*x)^(5/2),x)

[Out]

int((a + b*acosh(c + d*x))^2/(c*e + d*e*x)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {acosh}{\left (c + d x \right )}\right )^{2}}{\left (e \left (c + d x\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acosh(d*x+c))**2/(d*e*x+c*e)**(5/2),x)

[Out]

Integral((a + b*acosh(c + d*x))**2/(e*(c + d*x))**(5/2), x)

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