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

Optimal. Leaf size=118 \[ \frac {(e (c+d x))^{m+1} \left (a+b \cosh ^{-1}(c+d x)\right )}{d e (m+1)}-\frac {b \left (1-(c+d x)^2\right ) (e (c+d x))^{m+2} \, _2F_1\left (1,\frac {m+3}{2};\frac {m+4}{2};(c+d x)^2\right )}{d e^2 (m+1) (m+2) \sqrt {c+d x-1} \sqrt {c+d x+1}} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 124, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {5866, 5662, 126, 365, 364} \[ \frac {(e (c+d x))^{m+1} \left (a+b \cosh ^{-1}(c+d x)\right )}{d e (m+1)}-\frac {b \sqrt {1-(c+d x)^2} (e (c+d x))^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};(c+d x)^2\right )}{d e^2 (m+1) (m+2) \sqrt {c+d x-1} \sqrt {c+d x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*(c + d*x))^(1 + m)*(a + b*ArcCosh[c + d*x]))/(d*e*(1 + m)) - (b*(e*(c + d*x))^(2 + m)*Sqrt[1 - (c + d*x)^2
]*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])

Rule 126

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

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

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

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Mathematica [A]  time = 0.20, size = 106, normalized size = 0.90 \[ \frac {(c+d x) (e (c+d x))^m \left (a-\frac {b (c+d x) \sqrt {1-(c+d x)^2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};(c+d x)^2\right )}{(m+2) \sqrt {c+d x-1} \sqrt {c+d x+1}}+b \cosh ^{-1}(c+d x)\right )}{d (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \operatorname {arcosh}\left (d x + c\right ) + a\right )} {\left (d e x + c e\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 2.69, size = 0, normalized size = 0.00 \[ \int \left (d e x +c e \right )^{m} \left (a +b \,\mathrm {arccosh}\left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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