3.3.7 \(\int (c e+d e x)^{5/2} (a+b \cosh ^{-1}(c+d x))^2 \, dx\) [207]

Optimal. Leaf size=153 \[ \frac {2 (e (c+d x))^{7/2} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{7 d e}-\frac {8 b \sqrt {1-c-d x} (e (c+d x))^{9/2} \left (a+b \cosh ^{-1}(c+d x)\right ) \, _2F_1\left (\frac {1}{2},\frac {9}{4};\frac {13}{4};(c+d x)^2\right )}{63 d e^2 \sqrt {-1+c+d x}}-\frac {16 b^2 (e (c+d x))^{11/2} \, _3F_2\left (1,\frac {11}{4},\frac {11}{4};\frac {13}{4},\frac {15}{4};(c+d x)^2\right )}{693 d e^3} \]

[Out]

2/7*(e*(d*x+c))^(7/2)*(a+b*arccosh(d*x+c))^2/d/e-16/693*b^2*(e*(d*x+c))^(11/2)*hypergeom([1, 11/4, 11/4],[13/4
, 15/4],(d*x+c)^2)/d/e^3-8/63*b*(e*(d*x+c))^(9/2)*(a+b*arccosh(d*x+c))*hypergeom([1/2, 9/4],[13/4],(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 = 0.22, 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))^{11/2} \, _3F_2\left (1,\frac {11}{4},\frac {11}{4};\frac {13}{4},\frac {15}{4};(c+d x)^2\right )}{693 d e^3}-\frac {8 b \sqrt {-c-d x+1} (e (c+d x))^{9/2} \, _2F_1\left (\frac {1}{2},\frac {9}{4};\frac {13}{4};(c+d x)^2\right ) \left (a+b \cosh ^{-1}(c+d x)\right )}{63 d e^2 \sqrt {c+d x-1}}+\frac {2 (e (c+d x))^{7/2} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{7 d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(e*(c + d*x))^(7/2)*(a + b*ArcCosh[c + d*x])^2)/(7*d*e) - (8*b*Sqrt[1 - c - d*x]*(e*(c + d*x))^(9/2)*(a + b
*ArcCosh[c + d*x])*Hypergeometric2F1[1/2, 9/4, 13/4, (c + d*x)^2])/(63*d*e^2*Sqrt[-1 + c + d*x]) - (16*b^2*(e*
(c + d*x))^(11/2)*HypergeometricPFQ[{1, 11/4, 11/4}, {13/4, 15/4}, (c + d*x)^2])/(693*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 (c e+d e x)^{5/2} \left (a+b \cosh ^{-1}(c+d x)\right )^2 \, dx &=\frac {\text {Subst}\left (\int (e x)^{5/2} \left (a+b \cosh ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac {2 (e (c+d x))^{7/2} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{7 d e}-\frac {(4 b) \text {Subst}\left (\int \frac {(e x)^{7/2} \left (a+b \cosh ^{-1}(x)\right )}{\sqrt {-1+x} \sqrt {1+x}} \, dx,x,c+d x\right )}{7 d e}\\ &=\frac {2 (e (c+d x))^{7/2} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{7 d e}-\frac {8 b (e (c+d x))^{9/2} \sqrt {1-(c+d x)^2} \left (a+b \cosh ^{-1}(c+d x)\right ) \, _2F_1\left (\frac {1}{2},\frac {9}{4};\frac {13}{4};(c+d x)^2\right )}{63 d e^2 \sqrt {-1+c+d x} \sqrt {1+c+d x}}-\frac {16 b^2 (e (c+d x))^{11/2} \, _3F_2\left (1,\frac {11}{4},\frac {11}{4};\frac {13}{4},\frac {15}{4};(c+d x)^2\right )}{693 d e^3}\\ \end {align*}

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Mathematica [A]
time = 0.35, size = 140, normalized size = 0.92 \begin {gather*} \frac {2 (e (c+d x))^{7/2} \left (99 \left (a+b \cosh ^{-1}(c+d x)\right )^2-4 b (c+d x) \left (\frac {11 \sqrt {1-(c+d x)^2} \left (a+b \cosh ^{-1}(c+d x)\right ) \, _2F_1\left (\frac {1}{2},\frac {9}{4};\frac {13}{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 {11}{4},\frac {11}{4};\frac {13}{4},\frac {15}{4};(c+d x)^2\right )\right )\right )}{693 d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(e*(c + d*x))^(7/2)*(99*(a + b*ArcCosh[c + d*x])^2 - 4*b*(c + d*x)*((11*Sqrt[1 - (c + d*x)^2]*(a + b*ArcCos
h[c + d*x])*Hypergeometric2F1[1/2, 9/4, 13/4, (c + d*x)^2])/(Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x]) + 2*b*(c +
d*x)*HypergeometricPFQ[{1, 11/4, 11/4}, {13/4, 15/4}, (c + d*x)^2])))/(693*d*e)

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \left (d e x +c e \right )^{\frac {5}{2}} \left (a +b \,\mathrm {arccosh}\left (d x +c \right )\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

2/7*(d*x*e + c*e)^(7/2)*a^2*e^(-1)/d + 2/7*(b^2*d^3*x^3*e^(5/2) + 3*b^2*c*d^2*x^2*e^(5/2) + 3*b^2*c^2*d*x*e^(5
/2) + b^2*c^3*e^(5/2))*sqrt(d*x + c)*log(d*x + sqrt(d*x + c + 1)*sqrt(d*x + c - 1) + c)^2/d + integrate(2/7*((
(7*a*b*d^4 - 2*b^2*d^4)*x^4*e^(5/2) + 4*(7*a*b*c*d^3 - 2*b^2*c*d^3)*x^3*e^(5/2) - (12*b^2*c^2*d^2 - 7*(6*c^2*d
^2 - d^2)*a*b)*x^2*e^(5/2) - 2*(4*b^2*c^3*d - 7*(2*c^3*d - c*d)*a*b)*x*e^(5/2) - (2*b^2*c^4 - 7*(c^4 - c^2)*a*
b)*e^(5/2))*sqrt(d*x + c + 1)*sqrt(d*x + c)*sqrt(d*x + c - 1) + ((7*a*b*d^5 - 2*b^2*d^5)*x^5*e^(5/2) + 5*(7*a*
b*c*d^4 - 2*b^2*c*d^4)*x^4*e^(5/2) + (7*(10*c^2*d^3 - d^3)*a*b - 2*(10*c^2*d^3 - d^3)*b^2)*x^3*e^(5/2) + (7*(1
0*c^3*d^2 - 3*c*d^2)*a*b - 2*(10*c^3*d^2 - 3*c*d^2)*b^2)*x^2*e^(5/2) + (7*(5*c^4*d - 3*c^2*d)*a*b - 2*(5*c^4*d
- 3*c^2*d)*b^2)*x*e^(5/2) + (7*(c^5 - c^3)*a*b - 2*(c^5 - c^3)*b^2)*e^(5/2))*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((d*e*x+c*e)^(5/2)*(a+b*arccosh(d*x+c))^2,x, algorithm="fricas")

[Out]

integral(((b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*arccosh(d*x + c)^2*e^2 + 2*(a*b*d^2*x^2 + 2*a*b*c*d*x + a*b*c^
2)*arccosh(d*x + c)*e^2 + (a^2*d^2*x^2 + 2*a^2*c*d*x + a^2*c^2)*e^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 \left (e \left (c + d x\right )\right )^{\frac {5}{2}} \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((d*e*x+c*e)**(5/2)*(a+b*acosh(d*x+c))**2,x)

[Out]

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

[Out]

integrate((d*e*x + c*e)^(5/2)*(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 {\left (c\,e+d\,e\,x\right )}^{5/2}\,{\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)^(5/2)*(a + b*acosh(c + d*x))^2,x)

[Out]

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

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