3.2.20 \(\int x^3 \sqrt {d+c^2 d x^2} (a+b \sinh ^{-1}(c x)) \, dx\) [120]

Optimal. Leaf size=175 \[ \frac {2 b x \sqrt {d+c^2 d x^2}}{15 c^3 \sqrt {1+c^2 x^2}}-\frac {b x^3 \sqrt {d+c^2 d x^2}}{45 c \sqrt {1+c^2 x^2}}-\frac {b c x^5 \sqrt {d+c^2 d x^2}}{25 \sqrt {1+c^2 x^2}}-\frac {\left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 d}+\frac {\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^4 d^2} \]

[Out]

-1/3*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))/c^4/d+1/5*(c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))/c^4/d^2+2/15*b*x*
(c^2*d*x^2+d)^(1/2)/c^3/(c^2*x^2+1)^(1/2)-1/45*b*x^3*(c^2*d*x^2+d)^(1/2)/c/(c^2*x^2+1)^(1/2)-1/25*b*c*x^5*(c^2
*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 45, 5804, 12} \begin {gather*} \frac {\left (c^2 d x^2+d\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^4 d^2}-\frac {\left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 d}-\frac {b c x^5 \sqrt {c^2 d x^2+d}}{25 \sqrt {c^2 x^2+1}}-\frac {b x^3 \sqrt {c^2 d x^2+d}}{45 c \sqrt {c^2 x^2+1}}+\frac {2 b x \sqrt {c^2 d x^2+d}}{15 c^3 \sqrt {c^2 x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*x*Sqrt[d + c^2*d*x^2])/(15*c^3*Sqrt[1 + c^2*x^2]) - (b*x^3*Sqrt[d + c^2*d*x^2])/(45*c*Sqrt[1 + c^2*x^2])
- (b*c*x^5*Sqrt[d + c^2*d*x^2])/(25*Sqrt[1 + c^2*x^2]) - ((d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/(3*c^4*d
) + ((d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/(5*c^4*d^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 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 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 x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=-\frac {\left (b c \sqrt {d+c^2 d x^2}\right ) \int \frac {-2+c^2 x^2+3 c^4 x^4}{15 c^4} \, dx}{\sqrt {1+c^2 x^2}}+\left (a+b \sinh ^{-1}(c x)\right ) \int x^3 \sqrt {d+c^2 d x^2} \, dx\\ &=-\frac {\left (b \sqrt {d+c^2 d x^2}\right ) \int \left (-2+c^2 x^2+3 c^4 x^4\right ) \, dx}{15 c^3 \sqrt {1+c^2 x^2}}+\frac {1}{2} \left (a+b \sinh ^{-1}(c x)\right ) \text {Subst}\left (\int x \sqrt {d+c^2 d x} \, dx,x,x^2\right )\\ &=\frac {2 b x \sqrt {d+c^2 d x^2}}{15 c^3 \sqrt {1+c^2 x^2}}-\frac {b x^3 \sqrt {d+c^2 d x^2}}{45 c \sqrt {1+c^2 x^2}}-\frac {b c x^5 \sqrt {d+c^2 d x^2}}{25 \sqrt {1+c^2 x^2}}+\frac {1}{2} \left (a+b \sinh ^{-1}(c x)\right ) \text {Subst}\left (\int \left (-\frac {\sqrt {d+c^2 d x}}{c^2}+\frac {\left (d+c^2 d x\right )^{3/2}}{c^2 d}\right ) \, dx,x,x^2\right )\\ &=\frac {2 b x \sqrt {d+c^2 d x^2}}{15 c^3 \sqrt {1+c^2 x^2}}-\frac {b x^3 \sqrt {d+c^2 d x^2}}{45 c \sqrt {1+c^2 x^2}}-\frac {b c x^5 \sqrt {d+c^2 d x^2}}{25 \sqrt {1+c^2 x^2}}-\frac {\left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 d}+\frac {\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^4 d^2}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 120, normalized size = 0.69 \begin {gather*} \frac {\sqrt {d+c^2 d x^2} \left (15 a \left (1+c^2 x^2\right )^2 \left (-2+3 c^2 x^2\right )+b c x \sqrt {1+c^2 x^2} \left (30-5 c^2 x^2-9 c^4 x^4\right )+15 b \left (1+c^2 x^2\right )^2 \left (-2+3 c^2 x^2\right ) \sinh ^{-1}(c x)\right )}{225 c^4 \left (1+c^2 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + c^2*d*x^2]*(15*a*(1 + c^2*x^2)^2*(-2 + 3*c^2*x^2) + b*c*x*Sqrt[1 + c^2*x^2]*(30 - 5*c^2*x^2 - 9*c^4*
x^4) + 15*b*(1 + c^2*x^2)^2*(-2 + 3*c^2*x^2)*ArcSinh[c*x]))/(225*c^4*(1 + c^2*x^2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(577\) vs. \(2(149)=298\).
time = 1.80, size = 578, normalized size = 3.30

method	result	size
default	\(a \left (\frac {x^{2} \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{5 c^{2} d}-\frac {2 \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{15 d \,c^{4}}\right )+b \left (\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (16 x^{6} c^{6}+16 \sqrt {c^{2} x^{2}+1}\, x^{5} c^{5}+28 c^{4} x^{4}+20 \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}+13 c^{2} x^{2}+5 \sqrt {c^{2} x^{2}+1}\, c x +1\right ) \left (-1+5 \arcsinh \left (c x \right )\right )}{800 c^{4} \left (c^{2} x^{2}+1\right )}-\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (4 c^{4} x^{4}+4 \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}+5 c^{2} x^{2}+3 \sqrt {c^{2} x^{2}+1}\, c x +1\right ) \left (-1+3 \arcsinh \left (c x \right )\right )}{288 c^{4} \left (c^{2} x^{2}+1\right )}-\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (c^{2} x^{2}+\sqrt {c^{2} x^{2}+1}\, c x +1\right ) \left (\arcsinh \left (c x \right )-1\right )}{16 c^{4} \left (c^{2} x^{2}+1\right )}-\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (c^{2} x^{2}-\sqrt {c^{2} x^{2}+1}\, c x +1\right ) \left (1+\arcsinh \left (c x \right )\right )}{16 c^{4} \left (c^{2} x^{2}+1\right )}-\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (4 c^{4} x^{4}-4 \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}+5 c^{2} x^{2}-3 \sqrt {c^{2} x^{2}+1}\, c x +1\right ) \left (1+3 \arcsinh \left (c x \right )\right )}{288 c^{4} \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (16 x^{6} c^{6}-16 \sqrt {c^{2} x^{2}+1}\, x^{5} c^{5}+28 c^{4} x^{4}-20 \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}+13 c^{2} x^{2}-5 \sqrt {c^{2} x^{2}+1}\, c x +1\right ) \left (1+5 \arcsinh \left (c x \right )\right )}{800 c^{4} \left (c^{2} x^{2}+1\right )}\right )\)	\(578\)

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)^(1/2),x,method=_RETURNVERBOSE)

[Out]

a*(1/5*x^2*(c^2*d*x^2+d)^(3/2)/c^2/d-2/15/d/c^4*(c^2*d*x^2+d)^(3/2))+b*(1/800*(d*(c^2*x^2+1))^(1/2)*(16*x^6*c^
6+16*(c^2*x^2+1)^(1/2)*x^5*c^5+28*c^4*x^4+20*(c^2*x^2+1)^(1/2)*x^3*c^3+13*c^2*x^2+5*(c^2*x^2+1)^(1/2)*c*x+1)*(
-1+5*arcsinh(c*x))/c^4/(c^2*x^2+1)-1/288*(d*(c^2*x^2+1))^(1/2)*(4*c^4*x^4+4*(c^2*x^2+1)^(1/2)*x^3*c^3+5*c^2*x^
2+3*(c^2*x^2+1)^(1/2)*c*x+1)*(-1+3*arcsinh(c*x))/c^4/(c^2*x^2+1)-1/16*(d*(c^2*x^2+1))^(1/2)*(c^2*x^2+(c^2*x^2+
1)^(1/2)*c*x+1)*(arcsinh(c*x)-1)/c^4/(c^2*x^2+1)-1/16*(d*(c^2*x^2+1))^(1/2)*(c^2*x^2-(c^2*x^2+1)^(1/2)*c*x+1)*
(1+arcsinh(c*x))/c^4/(c^2*x^2+1)-1/288*(d*(c^2*x^2+1))^(1/2)*(4*c^4*x^4-4*(c^2*x^2+1)^(1/2)*x^3*c^3+5*c^2*x^2-
3*(c^2*x^2+1)^(1/2)*c*x+1)*(1+3*arcsinh(c*x))/c^4/(c^2*x^2+1)+1/800*(d*(c^2*x^2+1))^(1/2)*(16*x^6*c^6-16*(c^2*
x^2+1)^(1/2)*x^5*c^5+28*c^4*x^4-20*(c^2*x^2+1)^(1/2)*x^3*c^3+13*c^2*x^2-5*(c^2*x^2+1)^(1/2)*c*x+1)*(1+5*arcsin
h(c*x))/c^4/(c^2*x^2+1))

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Maxima [A]
time = 0.28, size = 134, normalized size = 0.77 \begin {gather*} \frac {1}{15} \, b {\left (\frac {3 \, {\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} x^{2}}{c^{2} d} - \frac {2 \, {\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}}}{c^{4} d}\right )} \operatorname {arsinh}\left (c x\right ) + \frac {1}{15} \, a {\left (\frac {3 \, {\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} x^{2}}{c^{2} d} - \frac {2 \, {\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}}}{c^{4} d}\right )} - \frac {{\left (9 \, c^{4} \sqrt {d} x^{5} + 5 \, c^{2} \sqrt {d} x^{3} - 30 \, \sqrt {d} x\right )} b}{225 \, c^{3}} \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)^(1/2),x, algorithm="maxima")

[Out]

1/15*b*(3*(c^2*d*x^2 + d)^(3/2)*x^2/(c^2*d) - 2*(c^2*d*x^2 + d)^(3/2)/(c^4*d))*arcsinh(c*x) + 1/15*a*(3*(c^2*d
*x^2 + d)^(3/2)*x^2/(c^2*d) - 2*(c^2*d*x^2 + d)^(3/2)/(c^4*d)) - 1/225*(9*c^4*sqrt(d)*x^5 + 5*c^2*sqrt(d)*x^3
- 30*sqrt(d)*x)*b/c^3

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Fricas [A]
time = 0.36, size = 158, normalized size = 0.90 \begin {gather*} \frac {15 \, {\left (3 \, b c^{6} x^{6} + 4 \, b c^{4} x^{4} - 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 ) + {\left (45 \, a c^{6} x^{6} + 60 \, a c^{4} x^{4} - 15 \, a c^{2} x^{2} - {\left (9 \, b c^{5} x^{5} + 5 \, b c^{3} x^{3} - 30 \, b c x\right )} \sqrt {c^{2} x^{2} + 1} - 30 \, a\right )} \sqrt {c^{2} d x^{2} + d}}{225 \, {\left (c^{6} x^{2} + c^{4}\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)^(1/2),x, algorithm="fricas")

[Out]

1/225*(15*(3*b*c^6*x^6 + 4*b*c^4*x^4 - b*c^2*x^2 - 2*b)*sqrt(c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 + 1)) + (45
*a*c^6*x^6 + 60*a*c^4*x^4 - 15*a*c^2*x^2 - (9*b*c^5*x^5 + 5*b*c^3*x^3 - 30*b*c*x)*sqrt(c^2*x^2 + 1) - 30*a)*sq
rt(c^2*d*x^2 + d))/(c^6*x^2 + c^4)

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

[Out]

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

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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 )\,\sqrt {d\,c^2\,x^2+d} \,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)^(1/2),x)

[Out]

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

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