3.3.9 \(\int \sqrt {c e+d e x} (a+b \cosh ^{-1}(c+d x))^2 \, dx\) [209]

Optimal. Leaf size=153 \[ \frac {2 (e (c+d x))^{3/2} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{3 d e}-\frac {8 b \sqrt {1-c-d x} (e (c+d x))^{5/2} \left (a+b \cosh ^{-1}(c+d x)\right ) \, _2F_1\left (\frac {1}{2},\frac {5}{4};\frac {9}{4};(c+d x)^2\right )}{15 d e^2 \sqrt {-1+c+d x}}-\frac {16 b^2 (e (c+d x))^{7/2} \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};(c+d x)^2\right )}{105 d e^3} \]

[Out]

2/3*(e*(d*x+c))^(3/2)*(a+b*arccosh(d*x+c))^2/d/e-16/105*b^2*(e*(d*x+c))^(7/2)*hypergeom([1, 7/4, 7/4],[9/4, 11
/4],(d*x+c)^2)/d/e^3-8/15*b*(e*(d*x+c))^(5/2)*(a+b*arccosh(d*x+c))*hypergeom([1/2, 5/4],[9/4],(d*x+c)^2)*(-d*x
-c+1)^(1/2)/d/e^2/(d*x+c-1)^(1/2)

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Rubi [A]
time = 0.21, antiderivative size = 153, normalized size of antiderivative = 1.00, 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 = {5996, 5883, 5949} \begin {gather*} -\frac {16 b^2 (e (c+d x))^{7/2} \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};(c+d x)^2\right )}{105 d e^3}-\frac {8 b \sqrt {-c-d x+1} (e (c+d x))^{5/2} \, _2F_1\left (\frac {1}{2},\frac {5}{4};\frac {9}{4};(c+d x)^2\right ) \left (a+b \cosh ^{-1}(c+d x)\right )}{15 d e^2 \sqrt {c+d x-1}}+\frac {2 (e (c+d x))^{3/2} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{3 d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(e*(c + d*x))^(3/2)*(a + b*ArcCosh[c + d*x])^2)/(3*d*e) - (8*b*Sqrt[1 - c - d*x]*(e*(c + d*x))^(5/2)*(a + b
*ArcCosh[c + d*x])*Hypergeometric2F1[1/2, 5/4, 9/4, (c + d*x)^2])/(15*d*e^2*Sqrt[-1 + c + d*x]) - (16*b^2*(e*(
c + d*x))^(7/2)*HypergeometricPFQ[{1, 7/4, 7/4}, {9/4, 11/4}, (c + d*x)^2])/(105*d*e^3)

Rule 5883

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)/(Sqrt
[1 + c*x]*Sqrt[-1 + c*x])), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

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]

Rule 5996

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 \sqrt {c e+d e x} \left (a+b \cosh ^{-1}(c+d x)\right )^2 \, dx &=\frac {\text {Subst}\left (\int \sqrt {e x} \left (a+b \cosh ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac {2 (e (c+d x))^{3/2} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{3 d e}-\frac {(4 b) \text {Subst}\left (\int \frac {(e x)^{3/2} \left (a+b \cosh ^{-1}(x)\right )}{\sqrt {-1+x} \sqrt {1+x}} \, dx,x,c+d x\right )}{3 d e}\\ &=\frac {2 (e (c+d x))^{3/2} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{3 d e}-\frac {8 b (e (c+d x))^{5/2} \sqrt {1-(c+d x)^2} \left (a+b \cosh ^{-1}(c+d x)\right ) \, _2F_1\left (\frac {1}{2},\frac {5}{4};\frac {9}{4};(c+d x)^2\right )}{15 d e^2 \sqrt {-1+c+d x} \sqrt {1+c+d x}}-\frac {16 b^2 (e (c+d x))^{7/2} \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};(c+d x)^2\right )}{105 d e^3}\\ \end {align*}

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Mathematica [A]
time = 0.32, size = 140, normalized size = 0.92 \begin {gather*} \frac {2 (e (c+d x))^{3/2} \left (35 \left (a+b \cosh ^{-1}(c+d x)\right )^2-4 b (c+d x) \left (\frac {7 \sqrt {1-(c+d x)^2} \left (a+b \cosh ^{-1}(c+d x)\right ) \, _2F_1\left (\frac {1}{2},\frac {5}{4};\frac {9}{4};(c+d x)^2\right )}{\sqrt {-1+c+d x} \sqrt {1+c+d x}}+2 b (c+d x) \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};(c+d x)^2\right )\right )\right )}{105 d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(e*(c + d*x))^(3/2)*(35*(a + b*ArcCosh[c + d*x])^2 - 4*b*(c + d*x)*((7*Sqrt[1 - (c + d*x)^2]*(a + b*ArcCosh
[c + d*x])*Hypergeometric2F1[1/2, 5/4, 9/4, (c + d*x)^2])/(Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x]) + 2*b*(c + d*
x)*HypergeometricPFQ[{1, 7/4, 7/4}, {9/4, 11/4}, (c + d*x)^2])))/(105*d*e)

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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \left (a +b \,\mathrm {arccosh}\left (d x +c \right )\right )^{2} \sqrt {d e x +c e}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a+b*arccosh(d*x+c))^2*(d*e*x+c*e)^(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((a+b*arccosh(d*x+c))^2*(d*e*x+c*e)^(1/2),x, algorithm="maxima")

[Out]

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

[Out]

integral((b^2*arccosh(d*x + c)^2 + 2*a*b*arccosh(d*x + c) + a^2)*sqrt(d*x + c)*e^(1/2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {e \left (c + d x\right )} \left (a + b \operatorname {acosh}{\left (c + d x \right )}\right )^{2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acosh(d*x+c))**2*(d*e*x+c*e)**(1/2),x)

[Out]

Integral(sqrt(e*(c + d*x))*(a + b*acosh(c + d*x))**2, 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((a+b*arccosh(d*x+c))^2*(d*e*x+c*e)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(d*e*x + c*e)*(b*arccosh(d*x + c) + a)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sqrt {c\,e+d\,e\,x}\,{\left (a+b\,\mathrm {acosh}\left (c+d\,x\right )\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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