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

Optimal. Leaf size=215 \[ \frac{x^4 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^2 d}-\frac{4 x^2 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^4 d}+\frac{8 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^6 d}-\frac{b x^5 \sqrt{c^2 x^2+1}}{25 c \sqrt{c^2 d x^2+d}}+\frac{4 b x^3 \sqrt{c^2 x^2+1}}{45 c^3 \sqrt{c^2 d x^2+d}}-\frac{8 b x \sqrt{c^2 x^2+1}}{15 c^5 \sqrt{c^2 d x^2+d}} \]

[Out]

(-8*b*x*Sqrt[1 + c^2*x^2])/(15*c^5*Sqrt[d + c^2*d*x^2]) + (4*b*x^3*Sqrt[1 + c^2*x^2])/(45*c^3*Sqrt[d + c^2*d*x
^2]) - (b*x^5*Sqrt[1 + c^2*x^2])/(25*c*Sqrt[d + c^2*d*x^2]) + (8*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(15
*c^6*d) - (4*x^2*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(15*c^4*d) + (x^4*Sqrt[d + c^2*d*x^2]*(a + b*ArcSin
h[c*x]))/(5*c^2*d)

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Rubi [A]  time = 0.264369, antiderivative size = 215, 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 = {5758, 5717, 8, 30} \[ \frac{x^4 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^2 d}-\frac{4 x^2 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^4 d}+\frac{8 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^6 d}-\frac{b x^5 \sqrt{c^2 x^2+1}}{25 c \sqrt{c^2 d x^2+d}}+\frac{4 b x^3 \sqrt{c^2 x^2+1}}{45 c^3 \sqrt{c^2 d x^2+d}}-\frac{8 b x \sqrt{c^2 x^2+1}}{15 c^5 \sqrt{c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*b*x*Sqrt[1 + c^2*x^2])/(15*c^5*Sqrt[d + c^2*d*x^2]) + (4*b*x^3*Sqrt[1 + c^2*x^2])/(45*c^3*Sqrt[d + c^2*d*x
^2]) - (b*x^5*Sqrt[1 + c^2*x^2])/(25*c*Sqrt[d + c^2*d*x^2]) + (8*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(15
*c^6*d) - (4*x^2*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(15*c^4*d) + (x^4*Sqrt[d + c^2*d*x^2]*(a + b*ArcSin
h[c*x]))/(5*c^2*d)

Rule 5758

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[(f*(f*x)^(m - 1)*Sqrt[d + e*x^2]*(a + b*ArcSinh[c*x])^n)/(e*m), x] + (-Dist[(f^2*(m - 1))/(c^2*m), Int[((f*x)
^(m - 2)*(a + b*ArcSinh[c*x])^n)/Sqrt[d + e*x^2], x], x] - Dist[(b*f*n*Sqrt[1 + c^2*x^2])/(c*m*Sqrt[d + e*x^2]
), Int[(f*x)^(m - 1)*(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[m, 1] && IntegerQ[m]

Rule 5717

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

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

Rubi steps

\begin{align*} \int \frac{x^5 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{d+c^2 d x^2}} \, dx &=\frac{x^4 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^2 d}-\frac{4 \int \frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{d+c^2 d x^2}} \, dx}{5 c^2}-\frac{\left (b \sqrt{1+c^2 x^2}\right ) \int x^4 \, dx}{5 c \sqrt{d+c^2 d x^2}}\\ &=-\frac{b x^5 \sqrt{1+c^2 x^2}}{25 c \sqrt{d+c^2 d x^2}}-\frac{4 x^2 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^4 d}+\frac{x^4 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^2 d}+\frac{8 \int \frac{x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{d+c^2 d x^2}} \, dx}{15 c^4}+\frac{\left (4 b \sqrt{1+c^2 x^2}\right ) \int x^2 \, dx}{15 c^3 \sqrt{d+c^2 d x^2}}\\ &=\frac{4 b x^3 \sqrt{1+c^2 x^2}}{45 c^3 \sqrt{d+c^2 d x^2}}-\frac{b x^5 \sqrt{1+c^2 x^2}}{25 c \sqrt{d+c^2 d x^2}}+\frac{8 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^6 d}-\frac{4 x^2 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^4 d}+\frac{x^4 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^2 d}-\frac{\left (8 b \sqrt{1+c^2 x^2}\right ) \int 1 \, dx}{15 c^5 \sqrt{d+c^2 d x^2}}\\ &=-\frac{8 b x \sqrt{1+c^2 x^2}}{15 c^5 \sqrt{d+c^2 d x^2}}+\frac{4 b x^3 \sqrt{1+c^2 x^2}}{45 c^3 \sqrt{d+c^2 d x^2}}-\frac{b x^5 \sqrt{1+c^2 x^2}}{25 c \sqrt{d+c^2 d x^2}}+\frac{8 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^6 d}-\frac{4 x^2 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^4 d}+\frac{x^4 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^2 d}\\ \end{align*}

Mathematica [A]  time = 0.171129, size = 119, normalized size = 0.55 \[ \frac{15 a \left (3 c^6 x^6-c^4 x^4+4 c^2 x^2+8\right )+b c x \sqrt{c^2 x^2+1} \left (-9 c^4 x^4+20 c^2 x^2-120\right )+15 b \left (3 c^6 x^6-c^4 x^4+4 c^2 x^2+8\right ) \sinh ^{-1}(c x)}{225 c^6 \sqrt{c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c*x*Sqrt[1 + c^2*x^2]*(-120 + 20*c^2*x^2 - 9*c^4*x^4) + 15*a*(8 + 4*c^2*x^2 - c^4*x^4 + 3*c^6*x^6) + 15*b*(
8 + 4*c^2*x^2 - c^4*x^4 + 3*c^6*x^6)*ArcSinh[c*x])/(225*c^6*Sqrt[d + c^2*d*x^2])

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Maple [B]  time = 0.214, size = 625, normalized size = 2.9 \begin{align*} a \left ({\frac{{x}^{4}}{5\,{c}^{2}d}\sqrt{{c}^{2}d{x}^{2}+d}}-{\frac{4}{5\,{c}^{2}} \left ({\frac{{x}^{2}}{3\,{c}^{2}d}\sqrt{{c}^{2}d{x}^{2}+d}}-{\frac{2}{3\,d{c}^{4}}\sqrt{{c}^{2}d{x}^{2}+d}} \right ) } \right ) +b \left ({\frac{-1+5\,{\it Arcsinh} \left ( cx \right ) }{800\,d{c}^{6} \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{-5+15\,{\it Arcsinh} \left ( cx \right ) }{288\,d{c}^{6} \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{-5+5\,{\it Arcsinh} \left ( cx \right ) }{16\,d{c}^{6} \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{5+5\,{\it Arcsinh} \left ( cx \right ) }{16\,d{c}^{6} \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{5+15\,{\it Arcsinh} \left ( cx \right ) }{288\,d{c}^{6} \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\,d{c}^{6} \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^5*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(1/2),x)

[Out]

a*(1/5*x^4/c^2/d*(c^2*d*x^2+d)^(1/2)-4/5/c^2*(1/3*x^2/c^2/d*(c^2*d*x^2+d)^(1/2)-2/3/d/c^4*(c^2*d*x^2+d)^(1/2))
)+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^6/d/(c^2*x^2+1)-5/288*(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^6/d/(c^2*x^2+1
)+5/16*(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^6/d/(c^2*x^2+1)+5/16*(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^6/d/(c^2*x^2+1)-5/288*(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^6/d/(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*arcsinh(c*x))/c^6/d/(c^2*x^2+1))

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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^5*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.5436, size = 355, normalized size = 1.65 \begin{align*} \frac{15 \,{\left (3 \, b c^{6} x^{6} - b c^{4} x^{4} + 4 \, b c^{2} x^{2} + 8 \, 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} - 15 \, a c^{4} x^{4} + 60 \, a c^{2} x^{2} -{\left (9 \, b c^{5} x^{5} - 20 \, b c^{3} x^{3} + 120 \, b c x\right )} \sqrt{c^{2} x^{2} + 1} + 120 \, a\right )} \sqrt{c^{2} d x^{2} + d}}{225 \,{\left (c^{8} d x^{2} + c^{6} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(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 - b*c^4*x^4 + 4*b*c^2*x^2 + 8*b)*sqrt(c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 + 1)) + (45
*a*c^6*x^6 - 15*a*c^4*x^4 + 60*a*c^2*x^2 - (9*b*c^5*x^5 - 20*b*c^3*x^3 + 120*b*c*x)*sqrt(c^2*x^2 + 1) + 120*a)
*sqrt(c^2*d*x^2 + d))/(c^8*d*x^2 + c^6*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5} \left (a + b \operatorname{asinh}{\left (c x \right )}\right )}{\sqrt{d \left (c^{2} x^{2} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )} x^{5}}{\sqrt{c^{2} d x^{2} + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)*x^5/sqrt(c^2*d*x^2 + d), x)