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

Optimal. Leaf size=120 \[ \frac {b d x \sqrt {1+c^2 x^2}}{24 c^3}-\frac {b d x^3 \sqrt {1+c^2 x^2}}{36 c}-\frac {1}{36} b c d x^5 \sqrt {1+c^2 x^2}-\frac {b d \sinh ^{-1}(c x)}{24 c^4}+\frac {1}{4} d x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} c^2 d x^6 \left (a+b \sinh ^{-1}(c x)\right ) \]

[Out]

-1/24*b*d*arcsinh(c*x)/c^4+1/4*d*x^4*(a+b*arcsinh(c*x))+1/6*c^2*d*x^6*(a+b*arcsinh(c*x))+1/24*b*d*x*(c^2*x^2+1
)^(1/2)/c^3-1/36*b*d*x^3*(c^2*x^2+1)^(1/2)/c-1/36*b*c*d*x^5*(c^2*x^2+1)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.06, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {14, 5803, 12, 470, 327, 221} \begin {gather*} \frac {1}{6} c^2 d x^6 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{4} d x^4 \left (a+b \sinh ^{-1}(c x)\right )-\frac {b d \sinh ^{-1}(c x)}{24 c^4}-\frac {1}{36} b c d x^5 \sqrt {c^2 x^2+1}-\frac {b d x^3 \sqrt {c^2 x^2+1}}{36 c}+\frac {b d x \sqrt {c^2 x^2+1}}{24 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d*x*Sqrt[1 + c^2*x^2])/(24*c^3) - (b*d*x^3*Sqrt[1 + c^2*x^2])/(36*c) - (b*c*d*x^5*Sqrt[1 + c^2*x^2])/36 - (
b*d*ArcSinh[c*x])/(24*c^4) + (d*x^4*(a + b*ArcSinh[c*x]))/4 + (c^2*d*x^6*(a + b*ArcSinh[c*x]))/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 221

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

Rule 327

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

Rule 470

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

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 x^3 \left (d+c^2 d x^2\right ) \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac {1}{4} d x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} c^2 d x^6 \left (a+b \sinh ^{-1}(c x)\right )-(b c) \int \frac {d x^4 \left (3+2 c^2 x^2\right )}{12 \sqrt {1+c^2 x^2}} \, dx\\ &=\frac {1}{4} d x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} c^2 d x^6 \left (a+b \sinh ^{-1}(c x)\right )-\frac {1}{12} (b c d) \int \frac {x^4 \left (3+2 c^2 x^2\right )}{\sqrt {1+c^2 x^2}} \, dx\\ &=-\frac {1}{36} b c d x^5 \sqrt {1+c^2 x^2}+\frac {1}{4} d x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} c^2 d x^6 \left (a+b \sinh ^{-1}(c x)\right )-\frac {1}{9} (b c d) \int \frac {x^4}{\sqrt {1+c^2 x^2}} \, dx\\ &=-\frac {b d x^3 \sqrt {1+c^2 x^2}}{36 c}-\frac {1}{36} b c d x^5 \sqrt {1+c^2 x^2}+\frac {1}{4} d x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} c^2 d x^6 \left (a+b \sinh ^{-1}(c x)\right )+\frac {(b d) \int \frac {x^2}{\sqrt {1+c^2 x^2}} \, dx}{12 c}\\ &=\frac {b d x \sqrt {1+c^2 x^2}}{24 c^3}-\frac {b d x^3 \sqrt {1+c^2 x^2}}{36 c}-\frac {1}{36} b c d x^5 \sqrt {1+c^2 x^2}+\frac {1}{4} d x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} c^2 d x^6 \left (a+b \sinh ^{-1}(c x)\right )-\frac {(b d) \int \frac {1}{\sqrt {1+c^2 x^2}} \, dx}{24 c^3}\\ &=\frac {b d x \sqrt {1+c^2 x^2}}{24 c^3}-\frac {b d x^3 \sqrt {1+c^2 x^2}}{36 c}-\frac {1}{36} b c d x^5 \sqrt {1+c^2 x^2}-\frac {b d \sinh ^{-1}(c x)}{24 c^4}+\frac {1}{4} d x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} c^2 d x^6 \left (a+b \sinh ^{-1}(c x)\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.04, size = 88, normalized size = 0.73 \begin {gather*} \frac {d \left (6 a c^4 x^4 \left (3+2 c^2 x^2\right )+b c x \sqrt {1+c^2 x^2} \left (3-2 c^2 x^2-2 c^4 x^4\right )+3 b \left (-1+6 c^4 x^4+4 c^6 x^6\right ) \sinh ^{-1}(c x)\right )}{72 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(6*a*c^4*x^4*(3 + 2*c^2*x^2) + b*c*x*Sqrt[1 + c^2*x^2]*(3 - 2*c^2*x^2 - 2*c^4*x^4) + 3*b*(-1 + 6*c^4*x^4 +
4*c^6*x^6)*ArcSinh[c*x]))/(72*c^4)

________________________________________________________________________________________

Maple [A]
time = 1.71, size = 115, normalized size = 0.96

method	result	size
derivativedivides	\(\frac {d a \left (\frac {\left (c^{2} x^{2}+1\right )^{3}}{6}-\frac {\left (c^{2} x^{2}+1\right )^{2}}{4}\right )+b d \left (\frac {\arcsinh \left (c x \right ) c^{6} x^{6}}{6}+\frac {\arcsinh \left (c x \right ) c^{4} x^{4}}{4}-\frac {\arcsinh \left (c x \right )}{24}-\frac {c x \left (c^{2} x^{2}+1\right )^{\frac {5}{2}}}{36}+\frac {c x \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}{36}+\frac {\sqrt {c^{2} x^{2}+1}\, c x}{24}\right )}{c^{4}}\)	\(115\)
default	\(\frac {d a \left (\frac {\left (c^{2} x^{2}+1\right )^{3}}{6}-\frac {\left (c^{2} x^{2}+1\right )^{2}}{4}\right )+b d \left (\frac {\arcsinh \left (c x \right ) c^{6} x^{6}}{6}+\frac {\arcsinh \left (c x \right ) c^{4} x^{4}}{4}-\frac {\arcsinh \left (c x \right )}{24}-\frac {c x \left (c^{2} x^{2}+1\right )^{\frac {5}{2}}}{36}+\frac {c x \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}{36}+\frac {\sqrt {c^{2} x^{2}+1}\, c x}{24}\right )}{c^{4}}\)	\(115\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^4*(d*a*(1/6*(c^2*x^2+1)^3-1/4*(c^2*x^2+1)^2)+b*d*(1/6*arcsinh(c*x)*c^6*x^6+1/4*arcsinh(c*x)*c^4*x^4-1/24*a
rcsinh(c*x)-1/36*c*x*(c^2*x^2+1)^(5/2)+1/36*c*x*(c^2*x^2+1)^(3/2)+1/24*(c^2*x^2+1)^(1/2)*c*x))

________________________________________________________________________________________

Maxima [A]
time = 0.25, size = 166, normalized size = 1.38 \begin {gather*} \frac {1}{6} \, a c^{2} d x^{6} + \frac {1}{4} \, a d x^{4} + \frac {1}{288} \, {\left (48 \, x^{6} \operatorname {arsinh}\left (c x\right ) - {\left (\frac {8 \, \sqrt {c^{2} x^{2} + 1} x^{5}}{c^{2}} - \frac {10 \, \sqrt {c^{2} x^{2} + 1} x^{3}}{c^{4}} + \frac {15 \, \sqrt {c^{2} x^{2} + 1} x}{c^{6}} - \frac {15 \, \operatorname {arsinh}\left (c x\right )}{c^{7}}\right )} c\right )} b c^{2} d + \frac {1}{32} \, {\left (8 \, x^{4} \operatorname {arsinh}\left (c x\right ) - {\left (\frac {2 \, \sqrt {c^{2} x^{2} + 1} x^{3}}{c^{2}} - \frac {3 \, \sqrt {c^{2} x^{2} + 1} x}{c^{4}} + \frac {3 \, \operatorname {arsinh}\left (c x\right )}{c^{5}}\right )} c\right )} b d \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a*c^2*d*x^6 + 1/4*a*d*x^4 + 1/288*(48*x^6*arcsinh(c*x) - (8*sqrt(c^2*x^2 + 1)*x^5/c^2 - 10*sqrt(c^2*x^2 +
1)*x^3/c^4 + 15*sqrt(c^2*x^2 + 1)*x/c^6 - 15*arcsinh(c*x)/c^7)*c)*b*c^2*d + 1/32*(8*x^4*arcsinh(c*x) - (2*sqrt
(c^2*x^2 + 1)*x^3/c^2 - 3*sqrt(c^2*x^2 + 1)*x/c^4 + 3*arcsinh(c*x)/c^5)*c)*b*d

________________________________________________________________________________________

Fricas [A]
time = 0.36, size = 109, normalized size = 0.91 \begin {gather*} \frac {12 \, a c^{6} d x^{6} + 18 \, a c^{4} d x^{4} + 3 \, {\left (4 \, b c^{6} d x^{6} + 6 \, b c^{4} d x^{4} - b d\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - {\left (2 \, b c^{5} d x^{5} + 2 \, b c^{3} d x^{3} - 3 \, b c d x\right )} \sqrt {c^{2} x^{2} + 1}}{72 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/72*(12*a*c^6*d*x^6 + 18*a*c^4*d*x^4 + 3*(4*b*c^6*d*x^6 + 6*b*c^4*d*x^4 - b*d)*log(c*x + sqrt(c^2*x^2 + 1)) -
(2*b*c^5*d*x^5 + 2*b*c^3*d*x^3 - 3*b*c*d*x)*sqrt(c^2*x^2 + 1))/c^4

________________________________________________________________________________________

Sympy [A]
time = 0.50, size = 138, normalized size = 1.15 \begin {gather*} \begin {cases} \frac {a c^{2} d x^{6}}{6} + \frac {a d x^{4}}{4} + \frac {b c^{2} d x^{6} \operatorname {asinh}{\left (c x \right )}}{6} - \frac {b c d x^{5} \sqrt {c^{2} x^{2} + 1}}{36} + \frac {b d x^{4} \operatorname {asinh}{\left (c x \right )}}{4} - \frac {b d x^{3} \sqrt {c^{2} x^{2} + 1}}{36 c} + \frac {b d x \sqrt {c^{2} x^{2} + 1}}{24 c^{3}} - \frac {b d \operatorname {asinh}{\left (c x \right )}}{24 c^{4}} & \text {for}\: c \neq 0 \\\frac {a d x^{4}}{4} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*c**2*d*x**6/6 + a*d*x**4/4 + b*c**2*d*x**6*asinh(c*x)/6 - b*c*d*x**5*sqrt(c**2*x**2 + 1)/36 + b*d
*x**4*asinh(c*x)/4 - b*d*x**3*sqrt(c**2*x**2 + 1)/(36*c) + b*d*x*sqrt(c**2*x**2 + 1)/(24*c**3) - b*d*asinh(c*x
)/(24*c**4), Ne(c, 0)), (a*d*x**4/4, True))

________________________________________________________________________________________

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*(c^2*d*x^2+d)*(a+b*arcsinh(c*x)),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 x^3\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,\left (d\,c^2\,x^2+d\right ) \,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),x)

[Out]

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

________________________________________________________________________________________