3.155 \(\int \frac{x^5 (a+b \sinh ^{-1}(c x))}{(d+c^2 d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=212 \[ \frac{\left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^6 d^3}-\frac{2 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{c^6 d^2}-\frac{a+b \sinh ^{-1}(c x)}{c^6 d \sqrt{c^2 d x^2+d}}-\frac{b x^3 \sqrt{c^2 d x^2+d}}{9 c^3 d^2 \sqrt{c^2 x^2+1}}+\frac{5 b x \sqrt{c^2 d x^2+d}}{3 c^5 d^2 \sqrt{c^2 x^2+1}}+\frac{b \sqrt{c^2 d x^2+d} \tan ^{-1}(c x)}{c^6 d^2 \sqrt{c^2 x^2+1}} \]

[Out]

(5*b*x*Sqrt[d + c^2*d*x^2])/(3*c^5*d^2*Sqrt[1 + c^2*x^2]) - (b*x^3*Sqrt[d + c^2*d*x^2])/(9*c^3*d^2*Sqrt[1 + c^
2*x^2]) - (a + b*ArcSinh[c*x])/(c^6*d*Sqrt[d + c^2*d*x^2]) - (2*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(c^6
*d^2) + ((d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/(3*c^6*d^3) + (b*Sqrt[d + c^2*d*x^2]*ArcTan[c*x])/(c^6*d^
2*Sqrt[1 + c^2*x^2])

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Rubi [A]  time = 0.293835, antiderivative size = 220, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {5751, 5758, 5717, 8, 30, 302, 203} \[ \frac{4 x^2 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 d^2}-\frac{8 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^6 d^2}-\frac{x^4 \left (a+b \sinh ^{-1}(c x)\right )}{c^2 d \sqrt{c^2 d x^2+d}}-\frac{b x^3 \sqrt{c^2 x^2+1}}{9 c^3 d \sqrt{c^2 d x^2+d}}+\frac{5 b x \sqrt{c^2 x^2+1}}{3 c^5 d \sqrt{c^2 d x^2+d}}+\frac{b \sqrt{c^2 x^2+1} \tan ^{-1}(c x)}{c^6 d \sqrt{c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2)^(3/2),x]

[Out]

(5*b*x*Sqrt[1 + c^2*x^2])/(3*c^5*d*Sqrt[d + c^2*d*x^2]) - (b*x^3*Sqrt[1 + c^2*x^2])/(9*c^3*d*Sqrt[d + c^2*d*x^
2]) - (x^4*(a + b*ArcSinh[c*x]))/(c^2*d*Sqrt[d + c^2*d*x^2]) - (8*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(3
*c^6*d^2) + (4*x^2*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(3*c^4*d^2) + (b*Sqrt[1 + c^2*x^2]*ArcTan[c*x])/(
c^6*d*Sqrt[d + c^2*d*x^2])

Rule 5751

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

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]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 203

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

Rubi steps

\begin{align*} \int \frac{x^5 \left (a+b \sinh ^{-1}(c x)\right )}{\left (d+c^2 d x^2\right )^{3/2}} \, dx &=-\frac{x^4 \left (a+b \sinh ^{-1}(c x)\right )}{c^2 d \sqrt{d+c^2 d x^2}}+\frac{4 \int \frac{x^3 \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^4}{1+c^2 x^2} \, dx}{c d \sqrt{d+c^2 d x^2}}\\ &=-\frac{x^4 \left (a+b \sinh ^{-1}(c x)\right )}{c^2 d \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 )}{3 c^4 d^2}-\frac{8 \int \frac{x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{d+c^2 d x^2}} \, dx}{3 c^4 d}-\frac{\left (4 b \sqrt{1+c^2 x^2}\right ) \int x^2 \, dx}{3 c^3 d \sqrt{d+c^2 d x^2}}+\frac{\left (b \sqrt{1+c^2 x^2}\right ) \int \left (-\frac{1}{c^4}+\frac{x^2}{c^2}+\frac{1}{c^4 \left (1+c^2 x^2\right )}\right ) \, dx}{c d \sqrt{d+c^2 d x^2}}\\ &=-\frac{b x \sqrt{1+c^2 x^2}}{c^5 d \sqrt{d+c^2 d x^2}}-\frac{b x^3 \sqrt{1+c^2 x^2}}{9 c^3 d \sqrt{d+c^2 d x^2}}-\frac{x^4 \left (a+b \sinh ^{-1}(c x)\right )}{c^2 d \sqrt{d+c^2 d x^2}}-\frac{8 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^6 d^2}+\frac{4 x^2 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 d^2}+\frac{\left (b \sqrt{1+c^2 x^2}\right ) \int \frac{1}{1+c^2 x^2} \, dx}{c^5 d \sqrt{d+c^2 d x^2}}+\frac{\left (8 b \sqrt{1+c^2 x^2}\right ) \int 1 \, dx}{3 c^5 d \sqrt{d+c^2 d x^2}}\\ &=\frac{5 b x \sqrt{1+c^2 x^2}}{3 c^5 d \sqrt{d+c^2 d x^2}}-\frac{b x^3 \sqrt{1+c^2 x^2}}{9 c^3 d \sqrt{d+c^2 d x^2}}-\frac{x^4 \left (a+b \sinh ^{-1}(c x)\right )}{c^2 d \sqrt{d+c^2 d x^2}}-\frac{8 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^6 d^2}+\frac{4 x^2 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 d^2}+\frac{b \sqrt{1+c^2 x^2} \tan ^{-1}(c x)}{c^6 d \sqrt{d+c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.264626, size = 148, normalized size = 0.7 \[ \frac{\sqrt{c^2 d x^2+d} \left (3 a \left (c^4 x^4-4 c^2 x^2-8\right )+b c x \sqrt{c^2 x^2+1} \left (15-c^2 x^2\right )+3 b \left (c^4 x^4-4 c^2 x^2-8\right ) \sinh ^{-1}(c x)\right )}{9 c^6 d^2 \left (c^2 x^2+1\right )}+\frac{b \sqrt{d \left (c^2 x^2+1\right )} \tan ^{-1}(c x)}{c^6 d^2 \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2)^(3/2),x]

[Out]

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

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Maple [C]  time = 0.206, size = 362, normalized size = 1.7 \begin{align*}{\frac{a{x}^{4}}{3\,{c}^{2}d}{\frac{1}{\sqrt{{c}^{2}d{x}^{2}+d}}}}-{\frac{4\,a{x}^{2}}{3\,d{c}^{4}}{\frac{1}{\sqrt{{c}^{2}d{x}^{2}+d}}}}-{\frac{8\,a}{3\,d{c}^{6}}{\frac{1}{\sqrt{{c}^{2}d{x}^{2}+d}}}}+{\frac{b{\it Arcsinh} \left ( cx \right ){x}^{4}}{3\,{c}^{2}{d}^{2} \left ({c}^{2}{x}^{2}+1 \right ) }\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{b{x}^{3}}{9\,{c}^{3}{d}^{2}}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}-{\frac{4\,b{\it Arcsinh} \left ( cx \right ){x}^{2}}{3\,{c}^{4}{d}^{2} \left ({c}^{2}{x}^{2}+1 \right ) }\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{5\,bx}{3\,{c}^{5}{d}^{2}}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{ib}{{d}^{2}{c}^{6}}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }\ln \left ( cx+\sqrt{{c}^{2}{x}^{2}+1}+i \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}-{\frac{ib}{{d}^{2}{c}^{6}}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }\ln \left ( cx+\sqrt{{c}^{2}{x}^{2}+1}-i \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}-{\frac{8\,b{\it Arcsinh} \left ( cx \right ) }{3\,{d}^{2}{c}^{6} \left ({c}^{2}{x}^{2}+1 \right ) }\sqrt{d \left ({c}^{2}{x}^{2}+1 \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)^(3/2),x)

[Out]

1/3*a*x^4/c^2/d/(c^2*d*x^2+d)^(1/2)-4/3*a/c^4*x^2/d/(c^2*d*x^2+d)^(1/2)-8/3*a/c^6/d/(c^2*d*x^2+d)^(1/2)+1/3*b*
(d*(c^2*x^2+1))^(1/2)/c^2/d^2/(c^2*x^2+1)*arcsinh(c*x)*x^4-1/9*b*(d*(c^2*x^2+1))^(1/2)/c^3/d^2/(c^2*x^2+1)^(1/
2)*x^3-4/3*b*(d*(c^2*x^2+1))^(1/2)/c^4/d^2/(c^2*x^2+1)*arcsinh(c*x)*x^2+5/3*b*(d*(c^2*x^2+1))^(1/2)/c^5/d^2/(c
^2*x^2+1)^(1/2)*x+I*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^6/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^6/d^2*ln(c*x+(c^2*x^2+1)^(1/2)-I)-8/3*b*(d*(c^2*x^2+1))^(1/2)/c^6/d^2/(c^2*x
^2+1)*arcsinh(c*x)

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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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.61838, size = 436, normalized size = 2.06 \begin{align*} -\frac{9 \,{\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 ) - 6 \,{\left (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 ) - 2 \,{\left (3 \, a c^{4} x^{4} - 12 \, a c^{2} x^{2} -{\left (b c^{3} x^{3} - 15 \, b c x\right )} \sqrt{c^{2} x^{2} + 1} - 24 \, a\right )} \sqrt{c^{2} d x^{2} + d}}{18 \,{\left (c^{8} d^{2} x^{2} + c^{6} d^{2}\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)^(3/2),x, algorithm="fricas")

[Out]

-1/18*(9*(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)) -
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)) - 2*(3*a*c^4*x^4 - 12*a*c^
2*x^2 - (b*c^3*x^3 - 15*b*c*x)*sqrt(c^2*x^2 + 1) - 24*a)*sqrt(c^2*d*x^2 + d))/(c^8*d^2*x^2 + c^6*d^2)

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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 )}{\left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac{3}{2}}}\, 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)**(3/2),x)

[Out]

Integral(x**5*(a + b*asinh(c*x))/(d*(c**2*x**2 + 1))**(3/2), 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}}{{\left (c^{2} d x^{2} + d\right )}^{\frac{3}{2}}}\,{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)^(3/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)*x^5/(c^2*d*x^2 + d)^(3/2), x)