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

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

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.160971, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {266, 43, 5734, 12} \[ \frac{\left (c^2 d x^2+d\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^4 d^2}-\frac{\left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 d}-\frac{b c x^5 \sqrt{c^2 d x^2+d}}{25 \sqrt{c^2 x^2+1}}-\frac{b x^3 \sqrt{c^2 d x^2+d}}{45 c \sqrt{c^2 x^2+1}}+\frac{2 b x \sqrt{c^2 d x^2+d}}{15 c^3 \sqrt{c^2 x^2+1}} \]

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 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 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 5734

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

Rule 12

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

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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [A]  time = 0.117843, size = 120, normalized size = 0.69 \[ \frac{\sqrt{c^2 d x^2+d} \left (15 a \left (3 c^2 x^2-2\right ) \left (c^2 x^2+1\right )^2+b c x \left (-9 c^4 x^4-5 c^2 x^2+30\right ) \sqrt{c^2 x^2+1}+15 b \left (3 c^2 x^2-2\right ) \left (c^2 x^2+1\right )^2 \sinh ^{-1}(c x)\right )}{225 c^4 \left (c^2 x^2+1\right )} \]

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))

________________________________________________________________________________________

Maple [B]  time = 0.219, size = 578, normalized size = 3.3 \begin{align*} a \left ({\frac{{x}^{2}}{5\,{c}^{2}d} \left ({c}^{2}d{x}^{2}+d \right ) ^{{\frac{3}{2}}}}-{\frac{2}{15\,d{c}^{4}} \left ({c}^{2}d{x}^{2}+d \right ) ^{{\frac{3}{2}}}} \right ) +b \left ({\frac{-1+5\,{\it Arcsinh} \left ( cx \right ) }{800\,{c}^{4} \left ({c}^{2}{x}^{2}+1 \right ) }\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) } \left ( 16\,{c}^{6}{x}^{6}+16\,{c}^{5}{x}^{5}\sqrt{{c}^{2}{x}^{2}+1}+28\,{c}^{4}{x}^{4}+20\,{c}^{3}{x}^{3}\sqrt{{c}^{2}{x}^{2}+1}+13\,{c}^{2}{x}^{2}+5\,cx\sqrt{{c}^{2}{x}^{2}+1}+1 \right ) }-{\frac{-1+3\,{\it Arcsinh} \left ( cx \right ) }{288\,{c}^{4} \left ({c}^{2}{x}^{2}+1 \right ) }\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) } \left ( 4\,{c}^{4}{x}^{4}+4\,{c}^{3}{x}^{3}\sqrt{{c}^{2}{x}^{2}+1}+5\,{c}^{2}{x}^{2}+3\,cx\sqrt{{c}^{2}{x}^{2}+1}+1 \right ) }-{\frac{-1+{\it Arcsinh} \left ( cx \right ) }{16\,{c}^{4} \left ({c}^{2}{x}^{2}+1 \right ) }\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) } \left ({c}^{2}{x}^{2}+cx\sqrt{{c}^{2}{x}^{2}+1}+1 \right ) }-{\frac{1+{\it Arcsinh} \left ( cx \right ) }{16\,{c}^{4} \left ({c}^{2}{x}^{2}+1 \right ) }\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) } \left ({c}^{2}{x}^{2}-cx\sqrt{{c}^{2}{x}^{2}+1}+1 \right ) }-{\frac{1+3\,{\it Arcsinh} \left ( cx \right ) }{288\,{c}^{4} \left ({c}^{2}{x}^{2}+1 \right ) }\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) } \left ( 4\,{c}^{4}{x}^{4}-4\,{c}^{3}{x}^{3}\sqrt{{c}^{2}{x}^{2}+1}+5\,{c}^{2}{x}^{2}-3\,cx\sqrt{{c}^{2}{x}^{2}+1}+1 \right ) }+{\frac{1+5\,{\it Arcsinh} \left ( cx \right ) }{800\,{c}^{4} \left ({c}^{2}{x}^{2}+1 \right ) }\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) } \left ( 16\,{c}^{6}{x}^{6}-16\,{c}^{5}{x}^{5}\sqrt{{c}^{2}{x}^{2}+1}+28\,{c}^{4}{x}^{4}-20\,{c}^{3}{x}^{3}\sqrt{{c}^{2}{x}^{2}+1}+13\,{c}^{2}{x}^{2}-5\,cx\sqrt{{c}^{2}{x}^{2}+1}+1 \right ) } \right ) \end{align*}

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)

[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*c^6*x^
6+16*c^5*x^5*(c^2*x^2+1)^(1/2)+28*c^4*x^4+20*c^3*x^3*(c^2*x^2+1)^(1/2)+13*c^2*x^2+5*c*x*(c^2*x^2+1)^(1/2)+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^3*x^3*(c^2*x^2+1)^(1/2)+5*c^2*x^
2+3*c*x*(c^2*x^2+1)^(1/2)+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*x*(c^2*
x^2+1)^(1/2)+1)*(-1+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*x*(c^2*x^2+1)^(1/2)+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^3*x^3*(c^2*x^2+1)^(1/2)+5*c^2*x^2
-3*c*x*(c^2*x^2+1)^(1/2)+1)*(1+3*arcsinh(c*x))/c^4/(c^2*x^2+1)+1/800*(d*(c^2*x^2+1))^(1/2)*(16*c^6*x^6-16*c^5*
x^5*(c^2*x^2+1)^(1/2)+28*c^4*x^4-20*c^3*x^3*(c^2*x^2+1)^(1/2)+13*c^2*x^2-5*c*x*(c^2*x^2+1)^(1/2)+1)*(1+5*arcsi
nh(c*x))/c^4/(c^2*x^2+1))

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.65268, size = 346, normalized size = 1.98 \begin{align*} \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{align*}

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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

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: NotImplementedError