3.228 \(\int (c e+d e x)^m (a+b \cosh ^{-1}(c+d x))^2 \, dx\)

Optimal. Leaf size=206 \[ -\frac {2 b^2 (e (c+d x))^{m+3} \, _3F_2\left (1,\frac {m}{2}+\frac {3}{2},\frac {m}{2}+\frac {3}{2};\frac {m}{2}+2,\frac {m}{2}+\frac {5}{2};(c+d x)^2\right )}{d e^3 (m+1) (m+2) (m+3)}-\frac {2 b \sqrt {-c-d x+1} (e (c+d x))^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};(c+d x)^2\right ) \left (a+b \cosh ^{-1}(c+d x)\right )}{d e^2 (m+1) (m+2) \sqrt {c+d x-1}}+\frac {(e (c+d x))^{m+1} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{d e (m+1)} \]

[Out]

(e*(d*x+c))^(1+m)*(a+b*arccosh(d*x+c))^2/d/e/(1+m)-2*b^2*(e*(d*x+c))^(3+m)*HypergeometricPFQ([1, 3/2+1/2*m, 3/
2+1/2*m],[2+1/2*m, 5/2+1/2*m],(d*x+c)^2)/d/e^3/(3+m)/(m^2+3*m+2)-2*b*(e*(d*x+c))^(2+m)*(a+b*arccosh(d*x+c))*hy
pergeom([1/2, 1+1/2*m],[2+1/2*m],(d*x+c)^2)*(-d*x-c+1)^(1/2)/d/e^2/(1+m)/(2+m)/(d*x+c-1)^(1/2)

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Rubi [A]  time = 0.32, antiderivative size = 218, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {5866, 5662, 5763} \[ -\frac {2 b^2 (e (c+d x))^{m+3} \, _3F_2\left (1,\frac {m}{2}+\frac {3}{2},\frac {m}{2}+\frac {3}{2};\frac {m}{2}+2,\frac {m}{2}+\frac {5}{2};(c+d x)^2\right )}{d e^3 (m+1) (m+2) (m+3)}-\frac {2 b \sqrt {1-(c+d x)^2} (e (c+d x))^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};(c+d x)^2\right ) \left (a+b \cosh ^{-1}(c+d x)\right )}{d e^2 (m+1) (m+2) \sqrt {c+d x-1} \sqrt {c+d x+1}}+\frac {(e (c+d x))^{m+1} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{d e (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.43, size = 178, normalized size = 0.86 \[ \frac {(c+d x) (e (c+d x))^m \left (\left (a+b \cosh ^{-1}(c+d x)\right )^2-\frac {2 b (c+d x) \left (\frac {b (c+d x) \, _3F_2\left (1,\frac {m}{2}+\frac {3}{2},\frac {m}{2}+\frac {3}{2};\frac {m}{2}+2,\frac {m}{2}+\frac {5}{2};(c+d x)^2\right )}{m+3}+\frac {\sqrt {1-(c+d x)^2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};(c+d x)^2\right ) \left (a+b \cosh ^{-1}(c+d x)\right )}{\sqrt {c+d x-1} \sqrt {c+d x+1}}\right )}{m+2}\right )}{d (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{2} \operatorname {arcosh}\left (d x + c\right )^{2} + 2 \, a b \operatorname {arcosh}\left (d x + c\right ) + a^{2}\right )} {\left (d e x + c e\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*arccosh(d*x + c)^2 + 2*a*b*arccosh(d*x + c) + a^2)*(d*e*x + c*e)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 3.00, size = 0, normalized size = 0.00 \[ \int \left (d e x +c e \right )^{m} \left (a +b \,\mathrm {arccosh}\left (d x +c \right )\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^2*d*e^m*x + b^2*c*e^m)*(d*x + c)^m*log(d*x + sqrt(d*x + c + 1)*sqrt(d*x + c - 1) + c)^2/(d*(m + 1)) + (d*e*
x + c*e)^(m + 1)*a^2/(d*e*(m + 1)) + integrate(-2*((b^2*c^2*e^m - (c^2*e^m*(m + 1) - e^m*(m + 1))*a*b - (a*b*d
^2*e^m*(m + 1) - b^2*d^2*e^m)*x^2 - 2*(a*b*c*d*e^m*(m + 1) - b^2*c*d*e^m)*x)*sqrt(d*x + c + 1)*sqrt(d*x + c -
1)*(d*x + c)^m - ((a*b*d^3*e^m*(m + 1) - b^2*d^3*e^m)*x^3 + (c^3*e^m*(m + 1) - c*e^m*(m + 1))*a*b - (c^3*e^m -
c*e^m)*b^2 + 3*(a*b*c*d^2*e^m*(m + 1) - b^2*c*d^2*e^m)*x^2 + ((3*c^2*d*e^m*(m + 1) - d*e^m*(m + 1))*a*b - (3*
c^2*d*e^m - d*e^m)*b^2)*x)*(d*x + c)^m)*log(d*x + sqrt(d*x + c + 1)*sqrt(d*x + c - 1) + c)/(d^3*(m + 1)*x^3 +
3*c*d^2*(m + 1)*x^2 + c^3*(m + 1) + (d^2*(m + 1)*x^2 + 2*c*d*(m + 1)*x + c^2*(m + 1) - m - 1)*sqrt(d*x + c + 1
)*sqrt(d*x + c - 1) - c*(m + 1) + (3*c^2*d*(m + 1) - d*(m + 1))*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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