3.123 \(\int \frac {a+b \sinh ^{-1}(c+d x)}{(c e+d e x)^4} \, dx\)

Optimal. Leaf size=84 \[ -\frac {a+b \sinh ^{-1}(c+d x)}{3 d e^4 (c+d x)^3}-\frac {b \sqrt {(c+d x)^2+1}}{6 d e^4 (c+d x)^2}+\frac {b \tanh ^{-1}\left (\sqrt {(c+d x)^2+1}\right )}{6 d e^4} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5865, 12, 5661, 266, 51, 63, 207} \[ -\frac {a+b \sinh ^{-1}(c+d x)}{3 d e^4 (c+d x)^3}-\frac {b \sqrt {(c+d x)^2+1}}{6 d e^4 (c+d x)^2}+\frac {b \tanh ^{-1}\left (\sqrt {(c+d x)^2+1}\right )}{6 d e^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c + d*x])/(c*e + d*e*x)^4,x]

[Out]

-(b*Sqrt[1 + (c + d*x)^2])/(6*d*e^4*(c + d*x)^2) - (a + b*ArcSinh[c + d*x])/(3*d*e^4*(c + d*x)^3) + (b*ArcTanh
[Sqrt[1 + (c + d*x)^2]])/(6*d*e^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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 266

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

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

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Mathematica [A]  time = 0.09, size = 82, normalized size = 0.98 \[ -\frac {2 a+b \sqrt {c^2+2 c d x+d^2 x^2+1} (c+d x)+2 b \sinh ^{-1}(c+d x)-b (c+d x)^3 \tanh ^{-1}\left (\sqrt {(c+d x)^2+1}\right )}{6 d e^4 (c+d x)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[c + d*x])/(c*e + d*e*x)^4,x]

[Out]

-1/6*(2*a + b*(c + d*x)*Sqrt[1 + c^2 + 2*c*d*x + d^2*x^2] + 2*b*ArcSinh[c + d*x] - b*(c + d*x)^3*ArcTanh[Sqrt[
1 + (c + d*x)^2]])/(d*e^4*(c + d*x)^3)

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fricas [B]  time = 0.56, size = 343, normalized size = 4.08 \[ -\frac {2 \, a c^{3} - 2 \, {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) - {\left (b c^{3} d^{3} x^{3} + 3 \, b c^{4} d^{2} x^{2} + 3 \, b c^{5} d x + b c^{6}\right )} \log \left (-d x - c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} + 1\right ) - 2 \, {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (-d x - c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) + {\left (b c^{3} d^{3} x^{3} + 3 \, b c^{4} d^{2} x^{2} + 3 \, b c^{5} d x + b c^{6}\right )} \log \left (-d x - c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} - 1\right ) + {\left (b c^{3} d x + b c^{4}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}{6 \, {\left (c^{3} d^{4} e^{4} x^{3} + 3 \, c^{4} d^{3} e^{4} x^{2} + 3 \, c^{5} d^{2} e^{4} x + c^{6} d e^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))/(d*e*x+c*e)^4,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arsinh}\left (d x + c\right ) + a}{{\left (d e x + c e\right )}^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))/(d*e*x+c*e)^4,x, algorithm="giac")

[Out]

integrate((b*arcsinh(d*x + c) + a)/(d*e*x + c*e)^4, x)

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maple [A]  time = 0.01, size = 74, normalized size = 0.88 \[ \frac {-\frac {a}{3 e^{4} \left (d x +c \right )^{3}}+\frac {b \left (-\frac {\arcsinh \left (d x +c \right )}{3 \left (d x +c \right )^{3}}-\frac {\sqrt {1+\left (d x +c \right )^{2}}}{6 \left (d x +c \right )^{2}}+\frac {\arctanh \left (\frac {1}{\sqrt {1+\left (d x +c \right )^{2}}}\right )}{6}\right )}{e^{4}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(d*x+c))/(d*e*x+c*e)^4,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{6} \, b {\left (\frac {2 \, {\left (d^{2} x^{2} + 2 \, c d x + c^{2} + \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )\right )}}{d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}} - \frac {i \, {\left (\log \left (\frac {i \, {\left (d^{2} x + c d\right )}}{d} + 1\right ) - \log \left (-\frac {i \, {\left (d^{2} x + c d\right )}}{d} + 1\right )\right )}}{d e^{4}} - 6 \, \int \frac {1}{3 \, {\left (d^{6} e^{4} x^{6} + 6 \, c d^{5} e^{4} x^{5} + c^{6} e^{4} + c^{4} e^{4} + {\left (15 \, c^{2} d^{4} e^{4} + d^{4} e^{4}\right )} x^{4} + 4 \, {\left (5 \, c^{3} d^{3} e^{4} + c d^{3} e^{4}\right )} x^{3} + 3 \, {\left (5 \, c^{4} d^{2} e^{4} + 2 \, c^{2} d^{2} e^{4}\right )} x^{2} + 2 \, {\left (3 \, c^{5} d e^{4} + 2 \, c^{3} d e^{4}\right )} x + {\left (d^{5} e^{4} x^{5} + 5 \, c d^{4} e^{4} x^{4} + c^{5} e^{4} + c^{3} e^{4} + {\left (10 \, c^{2} d^{3} e^{4} + d^{3} e^{4}\right )} x^{3} + {\left (10 \, c^{3} d^{2} e^{4} + 3 \, c d^{2} e^{4}\right )} x^{2} + {\left (5 \, c^{4} d e^{4} + 3 \, c^{2} d e^{4}\right )} x\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )}}\,{d x}\right )} - \frac {a}{3 \, {\left (d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))/(d*e*x+c*e)^4,x, algorithm="maxima")

[Out]

-1/6*b*(2*(d^2*x^2 + 2*c*d*x + c^2 + log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)))/(d^4*e^4*x^3 + 3*c*d^3*
e^4*x^2 + 3*c^2*d^2*e^4*x + c^3*d*e^4) - I*(log(I*(d^2*x + c*d)/d + 1) - log(-I*(d^2*x + c*d)/d + 1))/(d*e^4)
- 6*integrate(1/3/(d^6*e^4*x^6 + 6*c*d^5*e^4*x^5 + c^6*e^4 + c^4*e^4 + (15*c^2*d^4*e^4 + d^4*e^4)*x^4 + 4*(5*c
^3*d^3*e^4 + c*d^3*e^4)*x^3 + 3*(5*c^4*d^2*e^4 + 2*c^2*d^2*e^4)*x^2 + 2*(3*c^5*d*e^4 + 2*c^3*d*e^4)*x + (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 + (5*c^4*d*e^4 + 3*c^2*d*e^4)*x)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)), x)) - 1/3*a/(d^4*e^4*x^3 + 3*c*d^3*e
^4*x^2 + 3*c^2*d^2*e^4*x + c^3*d*e^4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {asinh}\left (c+d\,x\right )}{{\left (c\,e+d\,e\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c + d*x))/(c*e + d*e*x)^4,x)

[Out]

int((a + b*asinh(c + d*x))/(c*e + d*e*x)^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {b \operatorname {asinh}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx}{e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(d*x+c))/(d*e*x+c*e)**4,x)

[Out]

(Integral(a/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(b*asinh(c + d*x)
/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x))/e**4

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