3.205 \(\int \frac {a+b \cosh ^{-1}(c+d x)}{(c e+d e x)^{7/2}} \, dx\)

Optimal. Leaf size=130 \[ -\frac {2 \left (a+b \cosh ^{-1}(c+d x)\right )}{5 d e (e (c+d x))^{5/2}}+\frac {4 b \sqrt {-c-d x+1} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )\right |-1\right )}{15 d e^{7/2} \sqrt {c+d x-1}}+\frac {4 b \sqrt {c+d x-1} \sqrt {c+d x+1}}{15 d e^2 (e (c+d x))^{3/2}} \]

[Out]

-2/5*(a+b*arccosh(d*x+c))/d/e/(e*(d*x+c))^(5/2)+4/15*b*EllipticF((e*(d*x+c))^(1/2)/e^(1/2),I)*(-d*x-c+1)^(1/2)
/d/e^(7/2)/(d*x+c-1)^(1/2)+4/15*b*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)/d/e^2/(e*(d*x+c))^(3/2)

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Rubi [A]  time = 0.11, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {5866, 5662, 104, 12, 16, 117, 116} \[ -\frac {2 \left (a+b \cosh ^{-1}(c+d x)\right )}{5 d e (e (c+d x))^{5/2}}+\frac {4 b \sqrt {c+d x-1} \sqrt {c+d x+1}}{15 d e^2 (e (c+d x))^{3/2}}+\frac {4 b \sqrt {-c-d x+1} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )\right |-1\right )}{15 d e^{7/2} \sqrt {c+d x-1}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCosh[c + d*x])/(c*e + d*e*x)^(7/2),x]

[Out]

(4*b*Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x])/(15*d*e^2*(e*(c + d*x))^(3/2)) - (2*(a + b*ArcCosh[c + d*x]))/(5*d*
e*(e*(c + d*x))^(5/2)) + (4*b*Sqrt[1 - c - d*x]*EllipticF[ArcSin[Sqrt[e*(c + d*x)]/Sqrt[e]], -1])/(15*d*e^(7/2
)*Sqrt[-1 + c + d*x])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 104

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegersQ[2*m, 2*n, 2*p]

Rule 116

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(e_) + (f_.)*(x_)]), x_Symbol] :> Simp[(2*Rt[-(b/d), 2]*E
llipticF[ArcSin[Sqrt[b*x]/(Sqrt[c]*Rt[-(b/d), 2])], (c*f)/(d*e)])/(b*Sqrt[e]), x] /; FreeQ[{b, c, d, e, f}, x]
&& GtQ[c, 0] && GtQ[e, 0] && (PosQ[-(b/d)] || NegQ[-(b/f)])

Rule 117

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(e_) + (f_.)*(x_)]), x_Symbol] :> Dist[(Sqrt[1 + (d*x)/c]
*Sqrt[1 + (f*x)/e])/(Sqrt[c + d*x]*Sqrt[e + f*x]), Int[1/(Sqrt[b*x]*Sqrt[1 + (d*x)/c]*Sqrt[1 + (f*x)/e]), x],
x] /; FreeQ[{b, c, d, e, f}, x] &&  !(GtQ[c, 0] && GtQ[e, 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 \frac {a+b \cosh ^{-1}(c+d x)}{(c e+d e x)^{7/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {a+b \cosh ^{-1}(x)}{(e x)^{7/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {2 \left (a+b \cosh ^{-1}(c+d x)\right )}{5 d e (e (c+d x))^{5/2}}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x} (e x)^{5/2} \sqrt {1+x}} \, dx,x,c+d x\right )}{5 d e}\\ &=\frac {4 b \sqrt {-1+c+d x} \sqrt {1+c+d x}}{15 d e^2 (e (c+d x))^{3/2}}-\frac {2 \left (a+b \cosh ^{-1}(c+d x)\right )}{5 d e (e (c+d x))^{5/2}}+\frac {(4 b) \operatorname {Subst}\left (\int \frac {e x}{2 \sqrt {-1+x} (e x)^{3/2} \sqrt {1+x}} \, dx,x,c+d x\right )}{15 d e^3}\\ &=\frac {4 b \sqrt {-1+c+d x} \sqrt {1+c+d x}}{15 d e^2 (e (c+d x))^{3/2}}-\frac {2 \left (a+b \cosh ^{-1}(c+d x)\right )}{5 d e (e (c+d x))^{5/2}}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {x}{\sqrt {-1+x} (e x)^{3/2} \sqrt {1+x}} \, dx,x,c+d x\right )}{15 d e^2}\\ &=\frac {4 b \sqrt {-1+c+d x} \sqrt {1+c+d x}}{15 d e^2 (e (c+d x))^{3/2}}-\frac {2 \left (a+b \cosh ^{-1}(c+d x)\right )}{5 d e (e (c+d x))^{5/2}}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x} \sqrt {e x} \sqrt {1+x}} \, dx,x,c+d x\right )}{15 d e^3}\\ &=\frac {4 b \sqrt {-1+c+d x} \sqrt {1+c+d x}}{15 d e^2 (e (c+d x))^{3/2}}-\frac {2 \left (a+b \cosh ^{-1}(c+d x)\right )}{5 d e (e (c+d x))^{5/2}}+\frac {\left (2 b \sqrt {1-c-d x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {e x} \sqrt {1+x}} \, dx,x,c+d x\right )}{15 d e^3 \sqrt {-1+c+d x}}\\ &=\frac {4 b \sqrt {-1+c+d x} \sqrt {1+c+d x}}{15 d e^2 (e (c+d x))^{3/2}}-\frac {2 \left (a+b \cosh ^{-1}(c+d x)\right )}{5 d e (e (c+d x))^{5/2}}+\frac {4 b \sqrt {1-c-d x} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )\right |-1\right )}{15 d e^{7/2} \sqrt {-1+c+d x}}\\ \end {align*}

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Mathematica [C]  time = 0.16, size = 94, normalized size = 0.72 \[ \frac {2 \left (-3 \left (a+b \cosh ^{-1}(c+d x)\right )-\frac {2 b (c+d x) \sqrt {1-(c+d x)^2} \, _2F_1\left (-\frac {3}{4},\frac {1}{2};\frac {1}{4};(c+d x)^2\right )}{\sqrt {c+d x-1} \sqrt {c+d x+1}}\right )}{15 d e (e (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCosh[c + d*x])/(c*e + d*e*x)^(7/2),x]

[Out]

(2*(-3*(a + b*ArcCosh[c + d*x]) - (2*b*(c + d*x)*Sqrt[1 - (c + d*x)^2]*Hypergeometric2F1[-3/4, 1/2, 1/4, (c +
d*x)^2])/(Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x])))/(15*d*e*(e*(c + d*x))^(5/2))

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d e x + c e} {\left (b \operatorname {arcosh}\left (d x + c\right ) + a\right )}}{d^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(d*x+c))/(d*e*x+c*e)^(7/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*e*x + c*e)*(b*arccosh(d*x + c) + a)/(d^4*e^4*x^4 + 4*c*d^3*e^4*x^3 + 6*c^2*d^2*e^4*x^2 + 4*c^3
*d*e^4*x + c^4*e^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arcosh}\left (d x + c\right ) + a}{{\left (d e x + c e\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(d*x+c))/(d*e*x+c*e)^(7/2),x, algorithm="giac")

[Out]

integrate((b*arccosh(d*x + c) + a)/(d*e*x + c*e)^(7/2), x)

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maple [A]  time = 0.03, size = 201, normalized size = 1.55 \[ \frac {-\frac {2 a}{5 \left (d e x +c e \right )^{\frac {5}{2}}}+2 b \left (-\frac {\mathrm {arccosh}\left (\frac {d e x +c e}{e}\right )}{5 \left (d e x +c e \right )^{\frac {5}{2}}}+\frac {\frac {2 \sqrt {\frac {d e x +c e +e}{e}}\, \sqrt {-\frac {d e x +c e -e}{e}}\, \EllipticF \left (\sqrt {d e x +c e}\, \sqrt {-\frac {1}{e}}, i\right ) \left (d e x +c e \right )^{\frac {3}{2}}}{15}+\frac {2 \sqrt {-\frac {1}{e}}\, \left (d e x +c e \right )^{2}}{15}-\frac {2 \sqrt {-\frac {1}{e}}\, e^{2}}{15}}{e^{3} \sqrt {-\frac {1}{e}}\, \left (d e x +c e \right )^{\frac {3}{2}} \sqrt {\frac {d e x +c e +e}{e}}\, \sqrt {\frac {d e x +c e -e}{e}}}\right )}{d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccosh(d*x+c))/(d*e*x+c*e)^(7/2),x)

[Out]

2/d/e*(-1/5*a/(d*e*x+c*e)^(5/2)+b*(-1/5/(d*e*x+c*e)^(5/2)*arccosh((d*e*x+c*e)/e)+2/15/e^3*(((d*e*x+c*e+e)/e)^(
1/2)*(-(d*e*x+c*e-e)/e)^(1/2)*EllipticF((d*e*x+c*e)^(1/2)*(-1/e)^(1/2),I)*(d*e*x+c*e)^(3/2)+(-1/e)^(1/2)*(d*e*
x+c*e)^2-(-1/e)^(1/2)*e^2)/(-1/e)^(1/2)/(d*e*x+c*e)^(3/2)/((d*e*x+c*e+e)/e)^(1/2)/((d*e*x+c*e-e)/e)^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(d*x+c))/(d*e*x+c*e)^(7/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {acosh}\left (c+d\,x\right )}{{\left (c\,e+d\,e\,x\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(c + d*x))/(c*e + d*e*x)^(7/2),x)

[Out]

int((a + b*acosh(c + d*x))/(c*e + d*e*x)^(7/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acosh(d*x+c))/(d*e*x+c*e)**(7/2),x)

[Out]

Timed out

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