3.1.9 \(\int \frac {(d+c^2 d x^2) (a+b \sinh ^{-1}(c x))}{x^4} \, dx\) [9]

Optimal. Leaf size=80 \[ -\frac {b c d \sqrt {1+c^2 x^2}}{6 x^2}-\frac {d \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac {c^2 d \left (a+b \sinh ^{-1}(c x)\right )}{x}-\frac {5}{6} b c^3 d \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right ) \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {14, 5803, 12, 457, 79, 65, 214} \begin {gather*} -\frac {c^2 d \left (a+b \sinh ^{-1}(c x)\right )}{x}-\frac {d \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac {b c d \sqrt {c^2 x^2+1}}{6 x^2}-\frac {5}{6} b c^3 d \tanh ^{-1}\left (\sqrt {c^2 x^2+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + c^2*d*x^2)*(a + b*ArcSinh[c*x]))/x^4,x]

[Out]

-1/6*(b*c*d*Sqrt[1 + c^2*x^2])/x^2 - (d*(a + b*ArcSinh[c*x]))/(3*x^3) - (c^2*d*(a + b*ArcSinh[c*x]))/x - (5*b*
c^3*d*ArcTanh[Sqrt[1 + c^2*x^2]])/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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 65

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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 214

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

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 5803

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1
+ c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {\left (d+c^2 d x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{x^4} \, dx &=-\frac {d \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac {c^2 d \left (a+b \sinh ^{-1}(c x)\right )}{x}-(b c) \int \frac {d \left (-1-3 c^2 x^2\right )}{3 x^3 \sqrt {1+c^2 x^2}} \, dx\\ &=-\frac {d \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac {c^2 d \left (a+b \sinh ^{-1}(c x)\right )}{x}-\frac {1}{3} (b c d) \int \frac {-1-3 c^2 x^2}{x^3 \sqrt {1+c^2 x^2}} \, dx\\ &=-\frac {d \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac {c^2 d \left (a+b \sinh ^{-1}(c x)\right )}{x}-\frac {1}{6} (b c d) \text {Subst}\left (\int \frac {-1-3 c^2 x}{x^2 \sqrt {1+c^2 x}} \, dx,x,x^2\right )\\ &=-\frac {b c d \sqrt {1+c^2 x^2}}{6 x^2}-\frac {d \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac {c^2 d \left (a+b \sinh ^{-1}(c x)\right )}{x}+\frac {1}{12} \left (5 b c^3 d\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {1+c^2 x}} \, dx,x,x^2\right )\\ &=-\frac {b c d \sqrt {1+c^2 x^2}}{6 x^2}-\frac {d \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac {c^2 d \left (a+b \sinh ^{-1}(c x)\right )}{x}+\frac {1}{6} (5 b c d) \text {Subst}\left (\int \frac {1}{-\frac {1}{c^2}+\frac {x^2}{c^2}} \, dx,x,\sqrt {1+c^2 x^2}\right )\\ &=-\frac {b c d \sqrt {1+c^2 x^2}}{6 x^2}-\frac {d \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac {c^2 d \left (a+b \sinh ^{-1}(c x)\right )}{x}-\frac {5}{6} b c^3 d \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 93, normalized size = 1.16 \begin {gather*} -\frac {a d}{3 x^3}-\frac {a c^2 d}{x}-\frac {b c d \sqrt {1+c^2 x^2}}{6 x^2}-\frac {b d \sinh ^{-1}(c x)}{3 x^3}-\frac {b c^2 d \sinh ^{-1}(c x)}{x}-\frac {5}{6} b c^3 d \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + c^2*d*x^2)*(a + b*ArcSinh[c*x]))/x^4,x]

[Out]

-1/3*(a*d)/x^3 - (a*c^2*d)/x - (b*c*d*Sqrt[1 + c^2*x^2])/(6*x^2) - (b*d*ArcSinh[c*x])/(3*x^3) - (b*c^2*d*ArcSi
nh[c*x])/x - (5*b*c^3*d*ArcTanh[Sqrt[1 + c^2*x^2]])/6

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Maple [A]
time = 0.49, size = 87, normalized size = 1.09

method	result	size
derivativedivides	\(c^{3} \left (a d \left (-\frac {1}{c x}-\frac {1}{3 c^{3} x^{3}}\right )+b d \left (-\frac {\arcsinh \left (c x \right )}{c x}-\frac {\arcsinh \left (c x \right )}{3 c^{3} x^{3}}-\frac {5 \arctanh \left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )}{6}-\frac {\sqrt {c^{2} x^{2}+1}}{6 c^{2} x^{2}}\right )\right )\)	\(87\)
default	\(c^{3} \left (a d \left (-\frac {1}{c x}-\frac {1}{3 c^{3} x^{3}}\right )+b d \left (-\frac {\arcsinh \left (c x \right )}{c x}-\frac {\arcsinh \left (c x \right )}{3 c^{3} x^{3}}-\frac {5 \arctanh \left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )}{6}-\frac {\sqrt {c^{2} x^{2}+1}}{6 c^{2} x^{2}}\right )\right )\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c^2*d*x^2+d)*(a+b*arcsinh(c*x))/x^4,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 91, normalized size = 1.14 \begin {gather*} -{\left (c \operatorname {arsinh}\left (\frac {1}{c {\left | x \right |}}\right ) + \frac {\operatorname {arsinh}\left (c x\right )}{x}\right )} b c^{2} d + \frac {1}{6} \, {\left ({\left (c^{2} \operatorname {arsinh}\left (\frac {1}{c {\left | x \right |}}\right ) - \frac {\sqrt {c^{2} x^{2} + 1}}{x^{2}}\right )} c - \frac {2 \, \operatorname {arsinh}\left (c x\right )}{x^{3}}\right )} b d - \frac {a c^{2} d}{x} - \frac {a d}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)*(a+b*arcsinh(c*x))/x^4,x, algorithm="maxima")

[Out]

-(c*arcsinh(1/(c*abs(x))) + arcsinh(c*x)/x)*b*c^2*d + 1/6*((c^2*arcsinh(1/(c*abs(x))) - sqrt(c^2*x^2 + 1)/x^2)
*c - 2*arcsinh(c*x)/x^3)*b*d - a*c^2*d/x - 1/3*a*d/x^3

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 169 vs. \(2 (70) = 140\).
time = 0.38, size = 169, normalized size = 2.11 \begin {gather*} -\frac {5 \, b c^{3} d x^{3} \log \left (-c x + \sqrt {c^{2} x^{2} + 1} + 1\right ) - 5 \, b c^{3} d x^{3} \log \left (-c x + \sqrt {c^{2} x^{2} + 1} - 1\right ) + 6 \, a c^{2} d x^{2} - 2 \, {\left (3 \, b c^{2} + b\right )} d x^{3} \log \left (-c x + \sqrt {c^{2} x^{2} + 1}\right ) + \sqrt {c^{2} x^{2} + 1} b c d x + 2 \, a d + 2 \, {\left (3 \, b c^{2} d x^{2} - {\left (3 \, b c^{2} + b\right )} d x^{3} + b d\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{6 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)*(a+b*arcsinh(c*x))/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} d \left (\int \frac {a}{x^{4}}\, dx + \int \frac {a c^{2}}{x^{2}}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{x^{4}}\, dx + \int \frac {b c^{2} \operatorname {asinh}{\left (c x \right )}}{x^{2}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c**2*d*x**2+d)*(a+b*asinh(c*x))/x**4,x)

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)*(a+b*arcsinh(c*x))/x^4,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,\left (d\,c^2\,x^2+d\right )}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*asinh(c*x))*(d + c^2*d*x^2))/x^4,x)

[Out]

int(((a + b*asinh(c*x))*(d + c^2*d*x^2))/x^4, x)

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