3.125 \(\int \frac {a+b \sinh ^{-1}(c+d x)}{(c e+d e x)^6} \, dx\)

Optimal. Leaf size=115 \[ -\frac {a+b \sinh ^{-1}(c+d x)}{5 d e^6 (c+d x)^5}+\frac {3 b \sqrt {(c+d x)^2+1}}{40 d e^6 (c+d x)^2}-\frac {b \sqrt {(c+d x)^2+1}}{20 d e^6 (c+d x)^4}-\frac {3 b \tanh ^{-1}\left (\sqrt {(c+d x)^2+1}\right )}{40 d e^6} \]

[Out]

1/5*(-a-b*arcsinh(d*x+c))/d/e^6/(d*x+c)^5-3/40*b*arctanh((1+(d*x+c)^2)^(1/2))/d/e^6-1/20*b*(1+(d*x+c)^2)^(1/2)
/d/e^6/(d*x+c)^4+3/40*b*(1+(d*x+c)^2)^(1/2)/d/e^6/(d*x+c)^2

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ -\frac {a+b \sinh ^{-1}(c+d x)}{5 d e^6 (c+d x)^5}+\frac {3 b \sqrt {(c+d x)^2+1}}{40 d e^6 (c+d x)^2}-\frac {b \sqrt {(c+d x)^2+1}}{20 d e^6 (c+d x)^4}-\frac {3 b \tanh ^{-1}\left (\sqrt {(c+d x)^2+1}\right )}{40 d e^6} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c + d*x])/(c*e + d*e*x)^6,x]

[Out]

-(b*Sqrt[1 + (c + d*x)^2])/(20*d*e^6*(c + d*x)^4) + (3*b*Sqrt[1 + (c + d*x)^2])/(40*d*e^6*(c + d*x)^2) - (a +
b*ArcSinh[c + d*x])/(5*d*e^6*(c + d*x)^5) - (3*b*ArcTanh[Sqrt[1 + (c + d*x)^2]])/(40*d*e^6)

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)^6} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {a+b \sinh ^{-1}(x)}{e^6 x^6} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {a+b \sinh ^{-1}(x)}{x^6} \, dx,x,c+d x\right )}{d e^6}\\ &=-\frac {a+b \sinh ^{-1}(c+d x)}{5 d e^6 (c+d x)^5}+\frac {b \operatorname {Subst}\left (\int \frac {1}{x^5 \sqrt {1+x^2}} \, dx,x,c+d x\right )}{5 d e^6}\\ &=-\frac {a+b \sinh ^{-1}(c+d x)}{5 d e^6 (c+d x)^5}+\frac {b \operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt {1+x}} \, dx,x,(c+d x)^2\right )}{10 d e^6}\\ &=-\frac {b \sqrt {1+(c+d x)^2}}{20 d e^6 (c+d x)^4}-\frac {a+b \sinh ^{-1}(c+d x)}{5 d e^6 (c+d x)^5}-\frac {(3 b) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {1+x}} \, dx,x,(c+d x)^2\right )}{40 d e^6}\\ &=-\frac {b \sqrt {1+(c+d x)^2}}{20 d e^6 (c+d x)^4}+\frac {3 b \sqrt {1+(c+d x)^2}}{40 d e^6 (c+d x)^2}-\frac {a+b \sinh ^{-1}(c+d x)}{5 d e^6 (c+d x)^5}+\frac {(3 b) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1+x}} \, dx,x,(c+d x)^2\right )}{80 d e^6}\\ &=-\frac {b \sqrt {1+(c+d x)^2}}{20 d e^6 (c+d x)^4}+\frac {3 b \sqrt {1+(c+d x)^2}}{40 d e^6 (c+d x)^2}-\frac {a+b \sinh ^{-1}(c+d x)}{5 d e^6 (c+d x)^5}+\frac {(3 b) \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sqrt {1+(c+d x)^2}\right )}{40 d e^6}\\ &=-\frac {b \sqrt {1+(c+d x)^2}}{20 d e^6 (c+d x)^4}+\frac {3 b \sqrt {1+(c+d x)^2}}{40 d e^6 (c+d x)^2}-\frac {a+b \sinh ^{-1}(c+d x)}{5 d e^6 (c+d x)^5}-\frac {3 b \tanh ^{-1}\left (\sqrt {1+(c+d x)^2}\right )}{40 d e^6}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.08, size = 61, normalized size = 0.53 \[ -\frac {\frac {a+b \sinh ^{-1}(c+d x)}{(c+d x)^5}+b \sqrt {(c+d x)^2+1} \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};(c+d x)^2+1\right )}{5 d e^6} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[c + d*x])/(c*e + d*e*x)^6,x]

[Out]

-1/5*((a + b*ArcSinh[c + d*x])/(c + d*x)^5 + b*Sqrt[1 + (c + d*x)^2]*Hypergeometric2F1[1/2, 3, 3/2, 1 + (c + d
*x)^2])/(d*e^6)

________________________________________________________________________________________

fricas [B]  time = 0.72, size = 509, normalized size = 4.43 \[ -\frac {8 \, a c^{5} - 8 \, {\left (b d^{5} x^{5} + 5 \, b c d^{4} x^{4} + 10 \, b c^{2} d^{3} x^{3} + 10 \, b c^{3} d^{2} x^{2} + 5 \, b c^{4} d x\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) + 3 \, {\left (b c^{5} d^{5} x^{5} + 5 \, b c^{6} d^{4} x^{4} + 10 \, b c^{7} d^{3} x^{3} + 10 \, b c^{8} d^{2} x^{2} + 5 \, b c^{9} d x + b c^{10}\right )} \log \left (-d x - c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} + 1\right ) - 8 \, {\left (b d^{5} x^{5} + 5 \, b c d^{4} x^{4} + 10 \, b c^{2} d^{3} x^{3} + 10 \, b c^{3} d^{2} x^{2} + 5 \, b c^{4} d x + b c^{5}\right )} \log \left (-d x - c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) - 3 \, {\left (b c^{5} d^{5} x^{5} + 5 \, b c^{6} d^{4} x^{4} + 10 \, b c^{7} d^{3} x^{3} + 10 \, b c^{8} d^{2} x^{2} + 5 \, b c^{9} d x + b c^{10}\right )} \log \left (-d x - c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} - 1\right ) - {\left (3 \, b c^{5} d^{3} x^{3} + 9 \, b c^{6} d^{2} x^{2} + 3 \, b c^{8} - 2 \, b c^{6} + {\left (9 \, b c^{7} - 2 \, b c^{5}\right )} d x\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}{40 \, {\left (c^{5} d^{6} e^{6} x^{5} + 5 \, c^{6} d^{5} e^{6} x^{4} + 10 \, c^{7} d^{4} e^{6} x^{3} + 10 \, c^{8} d^{3} e^{6} x^{2} + 5 \, c^{9} d^{2} e^{6} x + c^{10} d e^{6}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))/(d*e*x+c*e)^6,x, algorithm="fricas")

[Out]

-1/40*(8*a*c^5 - 8*(b*d^5*x^5 + 5*b*c*d^4*x^4 + 10*b*c^2*d^3*x^3 + 10*b*c^3*d^2*x^2 + 5*b*c^4*d*x)*log(d*x + c
+ sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) + 3*(b*c^5*d^5*x^5 + 5*b*c^6*d^4*x^4 + 10*b*c^7*d^3*x^3 + 10*b*c^8*d^2*x
^2 + 5*b*c^9*d*x + b*c^10)*log(-d*x - c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1) + 1) - 8*(b*d^5*x^5 + 5*b*c*d^4*x^
4 + 10*b*c^2*d^3*x^3 + 10*b*c^3*d^2*x^2 + 5*b*c^4*d*x + b*c^5)*log(-d*x - c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1
)) - 3*(b*c^5*d^5*x^5 + 5*b*c^6*d^4*x^4 + 10*b*c^7*d^3*x^3 + 10*b*c^8*d^2*x^2 + 5*b*c^9*d*x + b*c^10)*log(-d*x
- c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1) - 1) - (3*b*c^5*d^3*x^3 + 9*b*c^6*d^2*x^2 + 3*b*c^8 - 2*b*c^6 + (9*b*
c^7 - 2*b*c^5)*d*x)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))/(c^5*d^6*e^6*x^5 + 5*c^6*d^5*e^6*x^4 + 10*c^7*d^4*e^6*x
^3 + 10*c^8*d^3*e^6*x^2 + 5*c^9*d^2*e^6*x + c^10*d*e^6)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arsinh}\left (d x + c\right ) + a}{{\left (d e x + c e\right )}^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))/(d*e*x+c*e)^6,x, algorithm="giac")

[Out]

integrate((b*arcsinh(d*x + c) + a)/(d*e*x + c*e)^6, x)

________________________________________________________________________________________

maple [A]  time = 0.01, size = 94, normalized size = 0.82 \[ \frac {-\frac {a}{5 e^{6} \left (d x +c \right )^{5}}+\frac {b \left (-\frac {\arcsinh \left (d x +c \right )}{5 \left (d x +c \right )^{5}}-\frac {\sqrt {1+\left (d x +c \right )^{2}}}{20 \left (d x +c \right )^{4}}+\frac {3 \sqrt {1+\left (d x +c \right )^{2}}}{40 \left (d x +c \right )^{2}}-\frac {3 \arctanh \left (\frac {1}{\sqrt {1+\left (d x +c \right )^{2}}}\right )}{40}\right )}{e^{6}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(d*x+c))/(d*e*x+c*e)^6,x)

[Out]

1/d*(-1/5*a/e^6/(d*x+c)^5+b/e^6*(-1/5/(d*x+c)^5*arcsinh(d*x+c)-1/20/(d*x+c)^4*(1+(d*x+c)^2)^(1/2)+3/40/(d*x+c)
^2*(1+(d*x+c)^2)^(1/2)-3/40*arctanh(1/(1+(d*x+c)^2)^(1/2))))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{30} \, b {\left (\frac {2 \, {\left (3 \, d^{4} x^{4} + 12 \, c d^{3} x^{3} + 3 \, c^{4} + {\left (18 \, c^{2} d^{2} - d^{2}\right )} x^{2} - c^{2} + 2 \, {\left (6 \, c^{3} d - c d\right )} x - 3 \, \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )\right )}}{d^{6} e^{6} x^{5} + 5 \, c d^{5} e^{6} x^{4} + 10 \, c^{2} d^{4} e^{6} x^{3} + 10 \, c^{3} d^{3} e^{6} x^{2} + 5 \, c^{4} d^{2} e^{6} x + c^{5} d e^{6}} - \frac {3 i \, {\left (\log \left (\frac {i \, {\left (d^{2} x + c d\right )}}{d} + 1\right ) - \log \left (-\frac {i \, {\left (d^{2} x + c d\right )}}{d} + 1\right )\right )}}{d e^{6}} + 30 \, \int \frac {1}{5 \, {\left (d^{8} e^{6} x^{8} + 8 \, c d^{7} e^{6} x^{7} + c^{8} e^{6} + c^{6} e^{6} + {\left (28 \, c^{2} d^{6} e^{6} + d^{6} e^{6}\right )} x^{6} + 2 \, {\left (28 \, c^{3} d^{5} e^{6} + 3 \, c d^{5} e^{6}\right )} x^{5} + 5 \, {\left (14 \, c^{4} d^{4} e^{6} + 3 \, c^{2} d^{4} e^{6}\right )} x^{4} + 4 \, {\left (14 \, c^{5} d^{3} e^{6} + 5 \, c^{3} d^{3} e^{6}\right )} x^{3} + {\left (28 \, c^{6} d^{2} e^{6} + 15 \, c^{4} d^{2} e^{6}\right )} x^{2} + 2 \, {\left (4 \, c^{7} d e^{6} + 3 \, c^{5} d e^{6}\right )} x + {\left (d^{7} e^{6} x^{7} + 7 \, c d^{6} e^{6} x^{6} + c^{7} e^{6} + c^{5} e^{6} + {\left (21 \, c^{2} d^{5} e^{6} + d^{5} e^{6}\right )} x^{5} + 5 \, {\left (7 \, c^{3} d^{4} e^{6} + c d^{4} e^{6}\right )} x^{4} + 5 \, {\left (7 \, c^{4} d^{3} e^{6} + 2 \, c^{2} d^{3} e^{6}\right )} x^{3} + {\left (21 \, c^{5} d^{2} e^{6} + 10 \, c^{3} d^{2} e^{6}\right )} x^{2} + {\left (7 \, c^{6} d e^{6} + 5 \, c^{4} d e^{6}\right )} x\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )}}\,{d x}\right )} - \frac {a}{5 \, {\left (d^{6} e^{6} x^{5} + 5 \, c d^{5} e^{6} x^{4} + 10 \, c^{2} d^{4} e^{6} x^{3} + 10 \, c^{3} d^{3} e^{6} x^{2} + 5 \, c^{4} d^{2} e^{6} x + c^{5} d e^{6}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))/(d*e*x+c*e)^6,x, algorithm="maxima")

[Out]

1/30*b*(2*(3*d^4*x^4 + 12*c*d^3*x^3 + 3*c^4 + (18*c^2*d^2 - d^2)*x^2 - c^2 + 2*(6*c^3*d - c*d)*x - 3*log(d*x +
c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)))/(d^6*e^6*x^5 + 5*c*d^5*e^6*x^4 + 10*c^2*d^4*e^6*x^3 + 10*c^3*d^3*e^6*
x^2 + 5*c^4*d^2*e^6*x + c^5*d*e^6) - 3*I*(log(I*(d^2*x + c*d)/d + 1) - log(-I*(d^2*x + c*d)/d + 1))/(d*e^6) +
30*integrate(1/5/(d^8*e^6*x^8 + 8*c*d^7*e^6*x^7 + c^8*e^6 + c^6*e^6 + (28*c^2*d^6*e^6 + d^6*e^6)*x^6 + 2*(28*c
^3*d^5*e^6 + 3*c*d^5*e^6)*x^5 + 5*(14*c^4*d^4*e^6 + 3*c^2*d^4*e^6)*x^4 + 4*(14*c^5*d^3*e^6 + 5*c^3*d^3*e^6)*x^
3 + (28*c^6*d^2*e^6 + 15*c^4*d^2*e^6)*x^2 + 2*(4*c^7*d*e^6 + 3*c^5*d*e^6)*x + (d^7*e^6*x^7 + 7*c*d^6*e^6*x^6 +
c^7*e^6 + c^5*e^6 + (21*c^2*d^5*e^6 + d^5*e^6)*x^5 + 5*(7*c^3*d^4*e^6 + c*d^4*e^6)*x^4 + 5*(7*c^4*d^3*e^6 + 2
*c^2*d^3*e^6)*x^3 + (21*c^5*d^2*e^6 + 10*c^3*d^2*e^6)*x^2 + (7*c^6*d*e^6 + 5*c^4*d*e^6)*x)*sqrt(d^2*x^2 + 2*c*
d*x + c^2 + 1)), x)) - 1/5*a/(d^6*e^6*x^5 + 5*c*d^5*e^6*x^4 + 10*c^2*d^4*e^6*x^3 + 10*c^3*d^3*e^6*x^2 + 5*c^4*
d^2*e^6*x + c^5*d*e^6)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {asinh}\left (c+d\,x\right )}{{\left (c\,e+d\,e\,x\right )}^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c + d*x))/(c*e + d*e*x)^6,x)

[Out]

int((a + b*asinh(c + d*x))/(c*e + d*e*x)^6, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a}{c^{6} + 6 c^{5} d x + 15 c^{4} d^{2} x^{2} + 20 c^{3} d^{3} x^{3} + 15 c^{2} d^{4} x^{4} + 6 c d^{5} x^{5} + d^{6} x^{6}}\, dx + \int \frac {b \operatorname {asinh}{\left (c + d x \right )}}{c^{6} + 6 c^{5} d x + 15 c^{4} d^{2} x^{2} + 20 c^{3} d^{3} x^{3} + 15 c^{2} d^{4} x^{4} + 6 c d^{5} x^{5} + d^{6} x^{6}}\, dx}{e^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(d*x+c))/(d*e*x+c*e)**6,x)

[Out]

(Integral(a/(c**6 + 6*c**5*d*x + 15*c**4*d**2*x**2 + 20*c**3*d**3*x**3 + 15*c**2*d**4*x**4 + 6*c*d**5*x**5 + d
**6*x**6), x) + Integral(b*asinh(c + d*x)/(c**6 + 6*c**5*d*x + 15*c**4*d**2*x**2 + 20*c**3*d**3*x**3 + 15*c**2
*d**4*x**4 + 6*c*d**5*x**5 + d**6*x**6), x))/e**6

________________________________________________________________________________________