3.2.57 \(\int \frac {x^3 (a+b \sinh ^{-1}(c x))}{(d+c^2 d x^2)^{3/2}} \, dx\) [157]

Optimal. Leaf size=136 \[ -\frac {b x \sqrt {d+c^2 d x^2}}{c^3 d^2 \sqrt {1+c^2 x^2}}+\frac {a+b \sinh ^{-1}(c x)}{c^4 d \sqrt {d+c^2 d x^2}}+\frac {\sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d^2}-\frac {b \sqrt {d+c^2 d x^2} \text {ArcTan}(c x)}{c^4 d^2 \sqrt {1+c^2 x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.10, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {272, 45, 5804, 12, 396, 209} \begin {gather*} \frac {\sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d^2}+\frac {a+b \sinh ^{-1}(c x)}{c^4 d \sqrt {c^2 d x^2+d}}-\frac {b \text {ArcTan}(c x) \sqrt {c^2 d x^2+d}}{c^4 d^2 \sqrt {c^2 x^2+1}}-\frac {b x \sqrt {c^2 d x^2+d}}{c^3 d^2 \sqrt {c^2 x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2)^(3/2),x]

[Out]

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

Rule 12

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

Rule 45

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 209

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

Rule 272

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 396

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

Rule 5804

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[x
^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c*Simp[Sqrt[d + e*x^2]/Sqrt[1 + c^2*x^2]], Int[
SimplifyIntegrand[u/Sqrt[d + e*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IntegerQ[p -
1/2] && NeQ[p, -2^(-1)] && (IGtQ[(m + 1)/2, 0] || ILtQ[(m + 2*p + 3)/2, 0])

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.15, size = 143, normalized size = 1.05 \begin {gather*} \frac {\sqrt {d+c^2 d x^2} \left (a \sqrt {1+c^2 x^2} \left (2+c^2 x^2\right )-b \left (c x+c^3 x^3\right )+b \sqrt {1+c^2 x^2} \left (2+c^2 x^2\right ) \sinh ^{-1}(c x)\right )}{c^4 d^2 \left (1+c^2 x^2\right )^{3/2}}-\frac {b \sqrt {d \left (1+c^2 x^2\right )} \text {ArcTan}(c x)}{c^4 d^2 \sqrt {1+c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2)^(3/2),x]

[Out]

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

________________________________________________________________________________________

Maple [C] Result contains complex when optimal does not.
time = 2.61, size = 261, normalized size = 1.92

method	result	size
default	\(a \left (\frac {x^{2}}{c^{2} d \sqrt {c^{2} d \,x^{2}+d}}+\frac {2}{d \,c^{4} \sqrt {c^{2} d \,x^{2}+d}}\right )+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) x^{2}}{c^{2} d^{2} \left (c^{2} x^{2}+1\right )}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, x}{c^{3} d^{2} \sqrt {c^{2} x^{2}+1}}+\frac {2 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )}{c^{4} d^{2} \left (c^{2} x^{2}+1\right )}+\frac {i b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}-i\right )}{\sqrt {c^{2} x^{2}+1}\, c^{4} d^{2}}-\frac {i b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}+i\right )}{\sqrt {c^{2} x^{2}+1}\, c^{4} d^{2}}\)	\(261\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(3/2),x,method=_RETURNVERBOSE)

[Out]

a*(x^2/c^2/d/(c^2*d*x^2+d)^(1/2)+2/d/c^4/(c^2*d*x^2+d)^(1/2))+b*(d*(c^2*x^2+1))^(1/2)/c^2/d^2/(c^2*x^2+1)*arcs
inh(c*x)*x^2-b*(d*(c^2*x^2+1))^(1/2)/c^3/d^2/(c^2*x^2+1)^(1/2)*x+2*b*(d*(c^2*x^2+1))^(1/2)/c^4/d^2/(c^2*x^2+1)
*arcsinh(c*x)+I*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^4/d^2*ln(c*x+(c^2*x^2+1)^(1/2)-I)-I*b*(d*(c^2*x^2+
1))^(1/2)/(c^2*x^2+1)^(1/2)/c^4/d^2*ln(c*x+(c^2*x^2+1)^(1/2)+I)

________________________________________________________________________________________

Maxima [A]
time = 0.54, size = 119, normalized size = 0.88 \begin {gather*} -b c {\left (\frac {x}{c^{4} d^{\frac {3}{2}}} + \frac {\arctan \left (c x\right )}{c^{5} d^{\frac {3}{2}}}\right )} + b {\left (\frac {x^{2}}{\sqrt {c^{2} d x^{2} + d} c^{2} d} + \frac {2}{\sqrt {c^{2} d x^{2} + d} c^{4} d}\right )} \operatorname {arsinh}\left (c x\right ) + a {\left (\frac {x^{2}}{\sqrt {c^{2} d x^{2} + d} c^{2} d} + \frac {2}{\sqrt {c^{2} d x^{2} + d} c^{4} d}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(3/2),x, algorithm="maxima")

[Out]

-b*c*(x/(c^4*d^(3/2)) + arctan(c*x)/(c^5*d^(3/2))) + b*(x^2/(sqrt(c^2*d*x^2 + d)*c^2*d) + 2/(sqrt(c^2*d*x^2 +
d)*c^4*d))*arcsinh(c*x) + a*(x^2/(sqrt(c^2*d*x^2 + d)*c^2*d) + 2/(sqrt(c^2*d*x^2 + d)*c^4*d))

________________________________________________________________________________________

Fricas [A]
time = 0.41, size = 166, normalized size = 1.22 \begin {gather*} \frac {{\left (b c^{2} x^{2} + b\right )} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} + 1} c \sqrt {d} x}{c^{4} d x^{4} - d}\right ) + 2 \, {\left (b c^{2} x^{2} + 2 \, b\right )} \sqrt {c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + 2 \, {\left (a c^{2} x^{2} - \sqrt {c^{2} x^{2} + 1} b c x + 2 \, a\right )} \sqrt {c^{2} d x^{2} + d}}{2 \, {\left (c^{6} d^{2} x^{2} + c^{4} d^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(3/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )}{\left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*asinh(c*x))/(c**2*d*x**2+d)**(3/2),x)

[Out]

Integral(x**3*(a + b*asinh(c*x))/(d*(c**2*x**2 + 1))**(3/2), x)

________________________________________________________________________________________

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(x^3*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(3/2),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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{{\left (d\,c^2\,x^2+d\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a + b*asinh(c*x)))/(d + c^2*d*x^2)^(3/2),x)

[Out]

int((x^3*(a + b*asinh(c*x)))/(d + c^2*d*x^2)^(3/2), x)

________________________________________________________________________________________