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

Optimal. Leaf size=144 \[ -\frac{a+b \sinh ^{-1}(c x)}{c^4 d^2 \sqrt{c^2 d x^2+d}}+\frac{a+b \sinh ^{-1}(c x)}{3 c^4 d \left (c^2 d x^2+d\right )^{3/2}}-\frac{b x \sqrt{c^2 d x^2+d}}{6 c^3 d^3 \left (c^2 x^2+1\right )^{3/2}}+\frac{5 b \sqrt{c^2 d x^2+d} \tan ^{-1}(c x)}{6 c^4 d^3 \sqrt{c^2 x^2+1}} \]

[Out]

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

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Rubi [A]  time = 0.182464, antiderivative size = 149, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {5751, 5717, 203, 288} \[ -\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt{c^2 d x^2+d}}-\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d \left (c^2 d x^2+d\right )^{3/2}}-\frac{b x}{6 c^3 d^2 \sqrt{c^2 x^2+1} \sqrt{c^2 d x^2+d}}+\frac{5 b \sqrt{c^2 x^2+1} \tan ^{-1}(c x)}{6 c^4 d^2 \sqrt{c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

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

Mathematica [A]  time = 0.239703, size = 151, normalized size = 1.05 \[ \frac{5 b \sqrt{d \left (c^2 x^2+1\right )} \tan ^{-1}(c x)}{6 c^4 d^3 \sqrt{c^2 x^2+1}}-\frac{\sqrt{c^2 d x^2+d} \left (2 a \sqrt{c^2 x^2+1} \left (3 c^2 x^2+2\right )+b \left (c^3 x^3+c x\right )+2 b \sqrt{c^2 x^2+1} \left (3 c^2 x^2+2\right ) \sinh ^{-1}(c x)\right )}{6 c^4 d^3 \left (c^2 x^2+1\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.151, size = 262, normalized size = 1.8 \begin{align*} -{\frac{a{x}^{2}}{{c}^{2}d} \left ({c}^{2}d{x}^{2}+d \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,a}{3\,d{c}^{4}} \left ({c}^{2}d{x}^{2}+d \right ) ^{-{\frac{3}{2}}}}-{\frac{b{\it Arcsinh} \left ( cx \right ){x}^{2}}{{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}{c}^{2}}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{bx}{6\,{d}^{3}{c}^{3}}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) } \left ({c}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,b{\it Arcsinh} \left ( cx \right ) }{3\,{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}{c}^{4}}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{{\frac{5\,i}{6}}b}{{c}^{4}{d}^{3}}\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{{\frac{5\,i}{6}}b}{{c}^{4}{d}^{3}}\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}}}} \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)^(5/2),x)

[Out]

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

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Maxima [A]  time = 1.78168, size = 186, normalized size = 1.29 \begin{align*} -\frac{1}{6} \, b c{\left (\frac{x}{c^{6} d^{\frac{5}{2}} x^{2} + c^{4} d^{\frac{5}{2}}} - \frac{5 \, \arctan \left (c x\right )}{c^{5} d^{\frac{5}{2}}}\right )} - \frac{1}{3} \, b{\left (\frac{3 \, x^{2}}{{\left (c^{2} d x^{2} + d\right )}^{\frac{3}{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}{3} \, a{\left (\frac{3 \, x^{2}}{{\left (c^{2} d x^{2} + d\right )}^{\frac{3}{2}} c^{2} d} + \frac{2}{{\left (c^{2} d x^{2} + d\right )}^{\frac{3}{2}} c^{4} d}\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)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 3.23417, size = 416, normalized size = 2.89 \begin{align*} -\frac{5 \,{\left (b c^{4} x^{4} + 2 \, 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 ) + 4 \,{\left (3 \, 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 ) + 2 \,{\left (6 \, a c^{2} x^{2} + \sqrt{c^{2} x^{2} + 1} b c x + 4 \, a\right )} \sqrt{c^{2} d x^{2} + d}}{12 \,{\left (c^{8} d^{3} x^{4} + 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\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)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(5*(b*c^4*x^4 + 2*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)) + 4*(3*b*c^2*x^2 + 2*b)*sqrt(c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 + 1)) + 2*(6*a*c^2*x^2 + sqrt(
c^2*x^2 + 1)*b*c*x + 4*a)*sqrt(c^2*d*x^2 + d))/(c^8*d^3*x^4 + 2*c^6*d^3*x^2 + c^4*d^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**(5/2),x)

[Out]

Timed out

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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^{3}}{{\left (c^{2} d x^{2} + d\right )}^{\frac{5}{2}}}\,{d x} \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)^(5/2),x, algorithm="giac")

[Out]

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