3.143 \(\int \frac {(d+c^2 d x^2)^{5/2} (a+b \sinh ^{-1}(c x))}{x^4} \, dx\)

Optimal. Leaf size=266 \[ -\frac {5 c^2 d \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x}-\frac {\left (c^2 d x^2+d\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}+\frac {5}{2} c^4 d^2 x \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5 c^3 d^2 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b \sqrt {c^2 x^2+1}}-\frac {b c d^2 \sqrt {c^2 d x^2+d}}{6 x^2 \sqrt {c^2 x^2+1}}-\frac {b c^5 d^2 x^2 \sqrt {c^2 d x^2+d}}{4 \sqrt {c^2 x^2+1}}+\frac {7 b c^3 d^2 \log (x) \sqrt {c^2 d x^2+d}}{3 \sqrt {c^2 x^2+1}} \]

[Out]

-5/3*c^2*d*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))/x-1/3*(c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))/x^3+5/2*c^4*d^2
*x*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)-1/6*b*c*d^2*(c^2*d*x^2+d)^(1/2)/x^2/(c^2*x^2+1)^(1/2)-1/4*b*c^5*d^2*
x^2*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)+5/4*c^3*d^2*(a+b*arcsinh(c*x))^2*(c^2*d*x^2+d)^(1/2)/b/(c^2*x^2+1)^(
1/2)+7/3*b*c^3*d^2*ln(x)*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.30, antiderivative size = 266, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {5739, 5682, 5675, 30, 14, 266, 43} \[ \frac {5}{2} c^4 d^2 x \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5 c^3 d^2 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b \sqrt {c^2 x^2+1}}-\frac {5 c^2 d \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x}-\frac {\left (c^2 d x^2+d\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac {b c^5 d^2 x^2 \sqrt {c^2 d x^2+d}}{4 \sqrt {c^2 x^2+1}}-\frac {b c d^2 \sqrt {c^2 d x^2+d}}{6 x^2 \sqrt {c^2 x^2+1}}+\frac {7 b c^3 d^2 \log (x) \sqrt {c^2 d x^2+d}}{3 \sqrt {c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Int[((d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/x^4,x]

[Out]

-(b*c*d^2*Sqrt[d + c^2*d*x^2])/(6*x^2*Sqrt[1 + c^2*x^2]) - (b*c^5*d^2*x^2*Sqrt[d + c^2*d*x^2])/(4*Sqrt[1 + c^2
*x^2]) + (5*c^4*d^2*x*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/2 - (5*c^2*d*(d + c^2*d*x^2)^(3/2)*(a + b*ArcS
inh[c*x]))/(3*x) - ((d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/(3*x^3) + (5*c^3*d^2*Sqrt[d + c^2*d*x^2]*(a +
b*ArcSinh[c*x])^2)/(4*b*Sqrt[1 + c^2*x^2]) + (7*b*c^3*d^2*Sqrt[d + c^2*d*x^2]*Log[x])/(3*Sqrt[1 + c^2*x^2])

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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 5675

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSinh[c*x]
)^(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && GtQ[d, 0] && NeQ[n, -1
]

Rule 5682

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(x*Sqrt[d + e*x^2]*
(a + b*ArcSinh[c*x])^n)/2, x] + (Dist[Sqrt[d + e*x^2]/(2*Sqrt[1 + c^2*x^2]), Int[(a + b*ArcSinh[c*x])^n/Sqrt[1
+ c^2*x^2], x], x] - Dist[(b*c*n*Sqrt[d + e*x^2])/(2*Sqrt[1 + c^2*x^2]), Int[x*(a + b*ArcSinh[c*x])^(n - 1),
x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0]

Rule 5739

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

Rubi steps

\begin {align*} \int \frac {\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{x^4} \, dx &=-\frac {\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}+\frac {1}{3} \left (5 c^2 d\right ) \int \frac {\left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{x^2} \, dx+\frac {\left (b c d^2 \sqrt {d+c^2 d x^2}\right ) \int \frac {\left (1+c^2 x^2\right )^2}{x^3} \, dx}{3 \sqrt {1+c^2 x^2}}\\ &=-\frac {5 c^2 d \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x}-\frac {\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}+\left (5 c^4 d^2\right ) \int \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx+\frac {\left (b c d^2 \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int \frac {\left (1+c^2 x\right )^2}{x^2} \, dx,x,x^2\right )}{6 \sqrt {1+c^2 x^2}}+\frac {\left (5 b c^3 d^2 \sqrt {d+c^2 d x^2}\right ) \int \frac {1+c^2 x^2}{x} \, dx}{3 \sqrt {1+c^2 x^2}}\\ &=\frac {5}{2} c^4 d^2 x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {5 c^2 d \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x}-\frac {\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}+\frac {\left (b c d^2 \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int \left (c^4+\frac {1}{x^2}+\frac {2 c^2}{x}\right ) \, dx,x,x^2\right )}{6 \sqrt {1+c^2 x^2}}+\frac {\left (5 b c^3 d^2 \sqrt {d+c^2 d x^2}\right ) \int \left (\frac {1}{x}+c^2 x\right ) \, dx}{3 \sqrt {1+c^2 x^2}}+\frac {\left (5 c^4 d^2 \sqrt {d+c^2 d x^2}\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}} \, dx}{2 \sqrt {1+c^2 x^2}}-\frac {\left (5 b c^5 d^2 \sqrt {d+c^2 d x^2}\right ) \int x \, dx}{2 \sqrt {1+c^2 x^2}}\\ &=-\frac {b c d^2 \sqrt {d+c^2 d x^2}}{6 x^2 \sqrt {1+c^2 x^2}}-\frac {b c^5 d^2 x^2 \sqrt {d+c^2 d x^2}}{4 \sqrt {1+c^2 x^2}}+\frac {5}{2} c^4 d^2 x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {5 c^2 d \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x}-\frac {\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}+\frac {5 c^3 d^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b \sqrt {1+c^2 x^2}}+\frac {7 b c^3 d^2 \sqrt {d+c^2 d x^2} \log (x)}{3 \sqrt {1+c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.97, size = 287, normalized size = 1.08 \[ \frac {d^2 \left (4 a \sqrt {c^2 x^2+1} \left (3 c^4 x^4-14 c^2 x^2-2\right ) \sqrt {c^2 d x^2+d}+60 a c^3 \sqrt {d} x^3 \sqrt {c^2 x^2+1} \log \left (\sqrt {d} \sqrt {c^2 d x^2+d}+c d x\right )+24 b c^2 x^2 \sqrt {c^2 d x^2+d} \left (-2 \sqrt {c^2 x^2+1} \sinh ^{-1}(c x)+2 c x \log (c x)+c x \sinh ^{-1}(c x)^2\right )+4 b \sqrt {c^2 d x^2+d} \left (2 c^3 x^3 \log (c x)-2 \left (c^2 x^2+1\right )^{3/2} \sinh ^{-1}(c x)-c x\right )-3 b c^3 x^3 \sqrt {c^2 d x^2+d} \left (\cosh \left (2 \sinh ^{-1}(c x)\right )-2 \sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)+\sinh \left (2 \sinh ^{-1}(c x)\right )\right )\right )\right )}{24 x^3 \sqrt {c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/x^4,x]

[Out]

(d^2*(4*a*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2]*(-2 - 14*c^2*x^2 + 3*c^4*x^4) + 24*b*c^2*x^2*Sqrt[d + c^2*d*x^
2]*(-2*Sqrt[1 + c^2*x^2]*ArcSinh[c*x] + c*x*ArcSinh[c*x]^2 + 2*c*x*Log[c*x]) + 4*b*Sqrt[d + c^2*d*x^2]*(-(c*x)
- 2*(1 + c^2*x^2)^(3/2)*ArcSinh[c*x] + 2*c^3*x^3*Log[c*x]) + 60*a*c^3*Sqrt[d]*x^3*Sqrt[1 + c^2*x^2]*Log[c*d*x
+ Sqrt[d]*Sqrt[d + c^2*d*x^2]] - 3*b*c^3*x^3*Sqrt[d + c^2*d*x^2]*(Cosh[2*ArcSinh[c*x]] - 2*ArcSinh[c*x]*(ArcS
inh[c*x] + Sinh[2*ArcSinh[c*x]]))))/(24*x^3*Sqrt[1 + c^2*x^2])

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))/x^4,x, algorithm="fricas")

[Out]

integral((a*c^4*d^2*x^4 + 2*a*c^2*d^2*x^2 + a*d^2 + (b*c^4*d^2*x^4 + 2*b*c^2*d^2*x^2 + b*d^2)*arcsinh(c*x))*sq
rt(c^2*d*x^2 + d)/x^4, x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)^(5/2)*(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,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [B]  time = 0.30, size = 1316, normalized size = 4.95 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))/x^4,x)

[Out]

-1/3*a/d/x^3*(c^2*d*x^2+d)^(7/2)+4/3*a*c^4*x*(c^2*d*x^2+d)^(5/2)-4/3*a*c^2/d/x*(c^2*d*x^2+d)^(7/2)+5/3*a*c^4*d
*x*(c^2*d*x^2+d)^(3/2)+5/2*a*c^4*d^2*x*(c^2*d*x^2+d)^(1/2)+5/2*a*c^4*d^3*ln(x*c^2*d/(c^2*d)^(1/2)+(c^2*d*x^2+d
)^(1/2))/(c^2*d)^(1/2)-1/8*b*(d*(c^2*x^2+1))^(1/2)*d^2*c^3/(c^2*x^2+1)^(1/2)+147*b*(d*(c^2*x^2+1))^(1/2)*d^2/(
63*c^4*x^4+15*c^2*x^2+1)*x^4/(c^2*x^2+1)^(1/2)*arcsinh(c*x)*c^7+35*b*(d*(c^2*x^2+1))^(1/2)*d^2/(63*c^4*x^4+15*
c^2*x^2+1)*x^2/(c^2*x^2+1)^(1/2)*arcsinh(c*x)*c^5-147*b*(d*(c^2*x^2+1))^(1/2)*d^2/(63*c^4*x^4+15*c^2*x^2+1)*x^
5/(c^2*x^2+1)*arcsinh(c*x)*c^8-203*b*(d*(c^2*x^2+1))^(1/2)*d^2/(63*c^4*x^4+15*c^2*x^2+1)*x^3/(c^2*x^2+1)*arcsi
nh(c*x)*c^6-190/3*b*(d*(c^2*x^2+1))^(1/2)*d^2/(63*c^4*x^4+15*c^2*x^2+1)*x/(c^2*x^2+1)*arcsinh(c*x)*c^4-23/3*b*
(d*(c^2*x^2+1))^(1/2)*d^2/(63*c^4*x^4+15*c^2*x^2+1)/x/(c^2*x^2+1)*arcsinh(c*x)*c^2-1/3*b*(d*(c^2*x^2+1))^(1/2)
*d^2/(63*c^4*x^4+15*c^2*x^2+1)/x^3/(c^2*x^2+1)*arcsinh(c*x)+1/2*b*(d*(c^2*x^2+1))^(1/2)*d^2*c^6/(c^2*x^2+1)*ar
csinh(c*x)*x^3+1/2*b*(d*(c^2*x^2+1))^(1/2)*d^2*c^4/(c^2*x^2+1)*arcsinh(c*x)*x-49/6*b*(d*(c^2*x^2+1))^(1/2)*d^2
/(63*c^4*x^4+15*c^2*x^2+1)*x^5/(c^2*x^2+1)*c^8-28/3*b*(d*(c^2*x^2+1))^(1/2)*d^2/(63*c^4*x^4+15*c^2*x^2+1)*x^3/
(c^2*x^2+1)*c^6-7/6*b*(d*(c^2*x^2+1))^(1/2)*d^2/(63*c^4*x^4+15*c^2*x^2+1)*x/(c^2*x^2+1)*c^4-21/2*b*(d*(c^2*x^2
+1))^(1/2)*d^2/(63*c^4*x^4+15*c^2*x^2+1)*x^2/(c^2*x^2+1)^(1/2)*c^5-1/6*b*(d*(c^2*x^2+1))^(1/2)*d^2/(63*c^4*x^4
+15*c^2*x^2+1)/x^2/(c^2*x^2+1)^(1/2)*c+7/3*b*(d*(c^2*x^2+1))^(1/2)*d^2/(63*c^4*x^4+15*c^2*x^2+1)/(c^2*x^2+1)^(
1/2)*arcsinh(c*x)*c^3-5/2*b*(d*(c^2*x^2+1))^(1/2)*d^2/(63*c^4*x^4+15*c^2*x^2+1)/(c^2*x^2+1)^(1/2)*c^3-1/4*b*(d
*(c^2*x^2+1))^(1/2)*d^2*c^5/(c^2*x^2+1)^(1/2)*x^2+49/6*b*(d*(c^2*x^2+1))^(1/2)*d^2/(63*c^4*x^4+15*c^2*x^2+1)*x
^3*c^6+7/6*b*(d*(c^2*x^2+1))^(1/2)*d^2/(63*c^4*x^4+15*c^2*x^2+1)*x*c^4+5/4*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)
^(1/2)*arcsinh(c*x)^2*d^2*c^3+7/3*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)*ln((c*x+(c^2*x^2+1)^(1/2))^2-1)*d^
2*c^3-14/3*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)*arcsinh(c*x)*d^2*c^3

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))/x^4,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (d\,c^2\,x^2+d\right )}^{5/2}}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*asinh(c*x))*(d + c^2*d*x^2)^(5/2))/x^4,x)

[Out]

int(((a + b*asinh(c*x))*(d + c^2*d*x^2)^(5/2))/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {5}{2}} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c**2*d*x**2+d)**(5/2)*(a+b*asinh(c*x))/x**4,x)

[Out]

Integral((d*(c**2*x**2 + 1))**(5/2)*(a + b*asinh(c*x))/x**4, x)

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