3.124 \(\int \frac {a+b \sinh ^{-1}(c+d x)}{(c e+d e x)^5} \, dx\)

Optimal. Leaf size=90 \[ -\frac {a+b \sinh ^{-1}(c+d x)}{4 d e^5 (c+d x)^4}+\frac {b \sqrt {(c+d x)^2+1}}{6 d e^5 (c+d x)}-\frac {b \sqrt {(c+d x)^2+1}}{12 d e^5 (c+d x)^3} \]

[Out]

1/4*(-a-b*arcsinh(d*x+c))/d/e^5/(d*x+c)^4-1/12*b*(1+(d*x+c)^2)^(1/2)/d/e^5/(d*x+c)^3+1/6*b*(1+(d*x+c)^2)^(1/2)
/d/e^5/(d*x+c)

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Rubi [A]  time = 0.07, antiderivative size = 90, normalized size of antiderivative = 1.00, 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 = {5865, 12, 5661, 271, 264} \[ -\frac {a+b \sinh ^{-1}(c+d x)}{4 d e^5 (c+d x)^4}+\frac {b \sqrt {(c+d x)^2+1}}{6 d e^5 (c+d x)}-\frac {b \sqrt {(c+d x)^2+1}}{12 d e^5 (c+d x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*Sqrt[1 + (c + d*x)^2])/(12*d*e^5*(c + d*x)^3) + (b*Sqrt[1 + (c + d*x)^2])/(6*d*e^5*(c + d*x)) - (a + b*Arc
Sinh[c + d*x])/(4*d*e^5*(c + d*x)^4)

Rule 12

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

Rule 264

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

Rule 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 5661

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcS
inh[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt
[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 5865

Int[((a_.) + ArcSinh[(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*ArcSinh[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 \sinh ^{-1}(c+d x)}{(c e+d e x)^5} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {a+b \sinh ^{-1}(x)}{e^5 x^5} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {a+b \sinh ^{-1}(x)}{x^5} \, dx,x,c+d x\right )}{d e^5}\\ &=-\frac {a+b \sinh ^{-1}(c+d x)}{4 d e^5 (c+d x)^4}+\frac {b \operatorname {Subst}\left (\int \frac {1}{x^4 \sqrt {1+x^2}} \, dx,x,c+d x\right )}{4 d e^5}\\ &=-\frac {b \sqrt {1+(c+d x)^2}}{12 d e^5 (c+d x)^3}-\frac {a+b \sinh ^{-1}(c+d x)}{4 d e^5 (c+d x)^4}-\frac {b \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {1+x^2}} \, dx,x,c+d x\right )}{6 d e^5}\\ &=-\frac {b \sqrt {1+(c+d x)^2}}{12 d e^5 (c+d x)^3}+\frac {b \sqrt {1+(c+d x)^2}}{6 d e^5 (c+d x)}-\frac {a+b \sinh ^{-1}(c+d x)}{4 d e^5 (c+d x)^4}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 61, normalized size = 0.68 \[ -\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )+b (c+d x) \sqrt {(c+d x)^2+1} \left (1-2 (c+d x)^2\right )}{12 d e^5 (c+d x)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(b*(c + d*x)*(1 - 2*(c + d*x)^2)*Sqrt[1 + (c + d*x)^2] + 3*(a + b*ArcSinh[c + d*x]))/(d*e^5*(c + d*x)^4)

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fricas [B]  time = 0.57, size = 210, normalized size = 2.33 \[ \frac {3 \, a d^{4} x^{4} + 12 \, a c d^{3} x^{3} + 18 \, a c^{2} d^{2} x^{2} + 12 \, a c^{3} d x - 3 \, b c^{4} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) + {\left (2 \, b c^{4} d^{3} x^{3} + 6 \, b c^{5} d^{2} x^{2} + 2 \, b c^{7} - b c^{5} + {\left (6 \, b c^{6} - b c^{4}\right )} d x\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}{12 \, {\left (c^{4} d^{5} e^{5} x^{4} + 4 \, c^{5} d^{4} e^{5} x^{3} + 6 \, c^{6} d^{3} e^{5} x^{2} + 4 \, c^{7} d^{2} e^{5} x + c^{8} d e^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))/(d*e*x+c*e)^5,x, algorithm="fricas")

[Out]

1/12*(3*a*d^4*x^4 + 12*a*c*d^3*x^3 + 18*a*c^2*d^2*x^2 + 12*a*c^3*d*x - 3*b*c^4*log(d*x + c + sqrt(d^2*x^2 + 2*
c*d*x + c^2 + 1)) + (2*b*c^4*d^3*x^3 + 6*b*c^5*d^2*x^2 + 2*b*c^7 - b*c^5 + (6*b*c^6 - b*c^4)*d*x)*sqrt(d^2*x^2
+ 2*c*d*x + c^2 + 1))/(c^4*d^5*e^5*x^4 + 4*c^5*d^4*e^5*x^3 + 6*c^6*d^3*e^5*x^2 + 4*c^7*d^2*e^5*x + c^8*d*e^5)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arsinh}\left (d x + c\right ) + a}{{\left (d e x + c e\right )}^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))/(d*e*x+c*e)^5,x, algorithm="giac")

[Out]

integrate((b*arcsinh(d*x + c) + a)/(d*e*x + c*e)^5, x)

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maple [A]  time = 0.01, size = 80, normalized size = 0.89 \[ \frac {-\frac {a}{4 e^{5} \left (d x +c \right )^{4}}+\frac {b \left (-\frac {\arcsinh \left (d x +c \right )}{4 \left (d x +c \right )^{4}}-\frac {\sqrt {1+\left (d x +c \right )^{2}}}{12 \left (d x +c \right )^{3}}+\frac {\sqrt {1+\left (d x +c \right )^{2}}}{6 d x +6 c}\right )}{e^{5}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-1/4*a/e^5/(d*x+c)^4+b/e^5*(-1/4/(d*x+c)^4*arcsinh(d*x+c)-1/12/(d*x+c)^3*(1+(d*x+c)^2)^(1/2)+1/6/(d*x+c)*
(1+(d*x+c)^2)^(1/2)))

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maxima [B]  time = 0.36, size = 258, normalized size = 2.87 \[ \frac {1}{12} \, b {\left (\frac {{\left (2 \, d^{4} x^{4} + 8 \, c d^{3} x^{3} + 2 \, c^{4} + {\left (12 \, c^{2} d^{2} + d^{2}\right )} x^{2} + c^{2} + 2 \, {\left (4 \, c^{3} d + c d\right )} x - 1\right )} d}{{\left (d^{5} e^{5} x^{3} + 3 \, c d^{4} e^{5} x^{2} + 3 \, c^{2} d^{3} e^{5} x + c^{3} d^{2} e^{5}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}} - \frac {3 \, \operatorname {arsinh}\left (d x + c\right )}{d^{5} e^{5} x^{4} + 4 \, c d^{4} e^{5} x^{3} + 6 \, c^{2} d^{3} e^{5} x^{2} + 4 \, c^{3} d^{2} e^{5} x + c^{4} d e^{5}}\right )} - \frac {a}{4 \, {\left (d^{5} e^{5} x^{4} + 4 \, c d^{4} e^{5} x^{3} + 6 \, c^{2} d^{3} e^{5} x^{2} + 4 \, c^{3} d^{2} e^{5} x + c^{4} d e^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))/(d*e*x+c*e)^5,x, algorithm="maxima")

[Out]

1/12*b*((2*d^4*x^4 + 8*c*d^3*x^3 + 2*c^4 + (12*c^2*d^2 + d^2)*x^2 + c^2 + 2*(4*c^3*d + c*d)*x - 1)*d/((d^5*e^5
*x^3 + 3*c*d^4*e^5*x^2 + 3*c^2*d^3*e^5*x + c^3*d^2*e^5)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) - 3*arcsinh(d*x + c
)/(d^5*e^5*x^4 + 4*c*d^4*e^5*x^3 + 6*c^2*d^3*e^5*x^2 + 4*c^3*d^2*e^5*x + c^4*d*e^5)) - 1/4*a/(d^5*e^5*x^4 + 4*
c*d^4*e^5*x^3 + 6*c^2*d^3*e^5*x^2 + 4*c^3*d^2*e^5*x + c^4*d*e^5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {asinh}\left (c+d\,x\right )}{{\left (c\,e+d\,e\,x\right )}^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a}{c^{5} + 5 c^{4} d x + 10 c^{3} d^{2} x^{2} + 10 c^{2} d^{3} x^{3} + 5 c d^{4} x^{4} + d^{5} x^{5}}\, dx + \int \frac {b \operatorname {asinh}{\left (c + d x \right )}}{c^{5} + 5 c^{4} d x + 10 c^{3} d^{2} x^{2} + 10 c^{2} d^{3} x^{3} + 5 c d^{4} x^{4} + d^{5} x^{5}}\, dx}{e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(d*x+c))/(d*e*x+c*e)**5,x)

[Out]

(Integral(a/(c**5 + 5*c**4*d*x + 10*c**3*d**2*x**2 + 10*c**2*d**3*x**3 + 5*c*d**4*x**4 + d**5*x**5), x) + Inte
gral(b*asinh(c + d*x)/(c**5 + 5*c**4*d*x + 10*c**3*d**2*x**2 + 10*c**2*d**3*x**3 + 5*c*d**4*x**4 + d**5*x**5),
x))/e**5

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