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

Optimal. Leaf size=210 \[ \frac{\sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{c^6 d^3}+\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{c^6 d^2 \sqrt{c^2 d x^2+d}}-\frac{a+b \sinh ^{-1}(c x)}{3 c^6 d \left (c^2 d x^2+d\right )^{3/2}}-\frac{b x \sqrt{c^2 d x^2+d}}{c^5 d^3 \sqrt{c^2 x^2+1}}+\frac{b x \sqrt{c^2 d x^2+d}}{6 c^5 d^3 \left (c^2 x^2+1\right )^{3/2}}-\frac{11 b \sqrt{c^2 d x^2+d} \tan ^{-1}(c x)}{6 c^6 d^3 \sqrt{c^2 x^2+1}} \]

[Out]

(b*x*Sqrt[d + c^2*d*x^2])/(6*c^5*d^3*(1 + c^2*x^2)^(3/2)) - (b*x*Sqrt[d + c^2*d*x^2])/(c^5*d^3*Sqrt[1 + c^2*x^
2]) - (a + b*ArcSinh[c*x])/(3*c^6*d*(d + c^2*d*x^2)^(3/2)) + (2*(a + b*ArcSinh[c*x]))/(c^6*d^2*Sqrt[d + c^2*d*
x^2]) + (Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(c^6*d^3) - (11*b*Sqrt[d + c^2*d*x^2]*ArcTan[c*x])/(6*c^6*d
^3*Sqrt[1 + c^2*x^2])

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Rubi [A]  time = 0.308664, antiderivative size = 225, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {5751, 5717, 8, 321, 203, 288} \[ -\frac{4 x^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt{c^2 d x^2+d}}+\frac{8 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^6 d^3}-\frac{x^4 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d \left (c^2 d x^2+d\right )^{3/2}}-\frac{b x^3}{6 c^3 d^2 \sqrt{c^2 x^2+1} \sqrt{c^2 d x^2+d}}-\frac{5 b x \sqrt{c^2 x^2+1}}{6 c^5 d^2 \sqrt{c^2 d x^2+d}}-\frac{11 b \sqrt{c^2 x^2+1} \tan ^{-1}(c x)}{6 c^6 d^2 \sqrt{c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*x^3)/(6*c^3*d^2*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2]) - (5*b*x*Sqrt[1 + c^2*x^2])/(6*c^5*d^2*Sqrt[d + c^2
*d*x^2]) - (x^4*(a + b*ArcSinh[c*x]))/(3*c^2*d*(d + c^2*d*x^2)^(3/2)) - (4*x^2*(a + b*ArcSinh[c*x]))/(3*c^4*d^
2*Sqrt[d + c^2*d*x^2]) + (8*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(3*c^6*d^3) - (11*b*Sqrt[1 + c^2*x^2]*Ar
cTan[c*x])/(6*c^6*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 8

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

Rule 321

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

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^5 \left (a+b \sinh ^{-1}(c x)\right )}{\left (d+c^2 d x^2\right )^{5/2}} \, dx &=-\frac{x^4 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}+\frac{4 \int \frac{x^3 \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^4}{\left (1+c^2 x^2\right )^2} \, dx}{3 c d^2 \sqrt{d+c^2 d x^2}}\\ &=-\frac{b x^3}{6 c^3 d^2 \sqrt{1+c^2 x^2} \sqrt{d+c^2 d x^2}}-\frac{x^4 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}-\frac{4 x^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt{d+c^2 d x^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^2}+\frac{\left (b \sqrt{1+c^2 x^2}\right ) \int \frac{x^2}{1+c^2 x^2} \, dx}{2 c^3 d^2 \sqrt{d+c^2 d x^2}}+\frac{\left (4 b \sqrt{1+c^2 x^2}\right ) \int \frac{x^2}{1+c^2 x^2} \, dx}{3 c^3 d^2 \sqrt{d+c^2 d x^2}}\\ &=-\frac{b x^3}{6 c^3 d^2 \sqrt{1+c^2 x^2} \sqrt{d+c^2 d x^2}}+\frac{11 b x \sqrt{1+c^2 x^2}}{6 c^5 d^2 \sqrt{d+c^2 d x^2}}-\frac{x^4 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}-\frac{4 x^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 d^2 \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^3}-\frac{\left (b \sqrt{1+c^2 x^2}\right ) \int \frac{1}{1+c^2 x^2} \, dx}{2 c^5 d^2 \sqrt{d+c^2 d x^2}}-\frac{\left (4 b \sqrt{1+c^2 x^2}\right ) \int \frac{1}{1+c^2 x^2} \, dx}{3 c^5 d^2 \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^2 \sqrt{d+c^2 d x^2}}\\ &=-\frac{b x^3}{6 c^3 d^2 \sqrt{1+c^2 x^2} \sqrt{d+c^2 d x^2}}-\frac{5 b x \sqrt{1+c^2 x^2}}{6 c^5 d^2 \sqrt{d+c^2 d x^2}}-\frac{x^4 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}-\frac{4 x^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 d^2 \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^3}-\frac{11 b \sqrt{1+c^2 x^2} \tan ^{-1}(c x)}{6 c^6 d^2 \sqrt{d+c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.271615, size = 154, normalized size = 0.73 \[ \frac{\sqrt{c^2 d x^2+d} \left (2 a \left (3 c^4 x^4+12 c^2 x^2+8\right )-b c x \sqrt{c^2 x^2+1} \left (6 c^2 x^2+5\right )+2 b \left (3 c^4 x^4+12 c^2 x^2+8\right ) \sinh ^{-1}(c x)\right )}{6 c^6 d^3 \left (c^2 x^2+1\right )^2}-\frac{11 b \sqrt{d \left (c^2 x^2+1\right )} \tan ^{-1}(c x)}{6 c^6 d^3 \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + c^2*d*x^2]*(-(b*c*x*Sqrt[1 + c^2*x^2]*(5 + 6*c^2*x^2)) + 2*a*(8 + 12*c^2*x^2 + 3*c^4*x^4) + 2*b*(8 +
12*c^2*x^2 + 3*c^4*x^4)*ArcSinh[c*x]))/(6*c^6*d^3*(1 + c^2*x^2)^2) - (11*b*Sqrt[d*(1 + c^2*x^2)]*ArcTan[c*x])
/(6*c^6*d^3*Sqrt[1 + c^2*x^2])

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Maple [C]  time = 0.194, size = 394, normalized size = 1.9 \begin{align*}{\frac{a{x}^{4}}{{c}^{2}d} \left ({c}^{2}d{x}^{2}+d \right ) ^{-{\frac{3}{2}}}}+4\,{\frac{a{x}^{2}}{d{c}^{4} \left ({c}^{2}d{x}^{2}+d \right ) ^{3/2}}}+{\frac{8\,a}{3\,d{c}^{6}} \left ({c}^{2}d{x}^{2}+d \right ) ^{-{\frac{3}{2}}}}+{\frac{b{\it Arcsinh} \left ( cx \right ){x}^{2}}{{c}^{4}{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) }\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{bx}{{c}^{5}{d}^{3}}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{b{\it Arcsinh} \left ( cx \right ) }{{d}^{3}{c}^{6} \left ({c}^{2}{x}^{2}+1 \right ) }\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}+2\,{\frac{b\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\it Arcsinh} \left ( cx \right ){x}^{2}}{{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}{c}^{4}}}+{\frac{bx}{6\,{c}^{5}{d}^{3}}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) } \left ({c}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}+{\frac{5\,b{\it Arcsinh} \left ( cx \right ) }{3\,{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}{c}^{6}}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{{\frac{11\,i}{6}}b}{{d}^{3}{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{{\frac{11\,i}{6}}b}{{d}^{3}{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}}}} \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)^(5/2),x)

[Out]

a*x^4/c^2/d/(c^2*d*x^2+d)^(3/2)+4*a/c^4*x^2/d/(c^2*d*x^2+d)^(3/2)+8/3*a/c^6/d/(c^2*d*x^2+d)^(3/2)+b*(d*(c^2*x^
2+1))^(1/2)/c^4/d^3/(c^2*x^2+1)*arcsinh(c*x)*x^2-b*(d*(c^2*x^2+1))^(1/2)/c^5/d^3/(c^2*x^2+1)^(1/2)*x+b*(d*(c^2
*x^2+1))^(1/2)/c^6/d^3/(c^2*x^2+1)*arcsinh(c*x)+2*b*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^2/c^4*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^5*x+5/3*b*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^2/c^6*a
rcsinh(c*x)+11/6*I*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^6/d^3*ln(c*x+(c^2*x^2+1)^(1/2)-I)-11/6*I*b*(d*(
c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^6/d^3*ln(c*x+(c^2*x^2+1)^(1/2)+I)

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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)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.3151, size = 483, normalized size = 2.3 \begin{align*} \frac{11 \,{\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^{4} x^{4} + 12 \, 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 (6 \, a c^{4} x^{4} + 24 \, a c^{2} x^{2} -{\left (6 \, b c^{3} x^{3} + 5 \, b c x\right )} \sqrt{c^{2} x^{2} + 1} + 16 \, a\right )} \sqrt{c^{2} d x^{2} + d}}{12 \,{\left (c^{10} d^{3} x^{4} + 2 \, c^{8} d^{3} x^{2} + c^{6} d^{3}\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)^(5/2),x, algorithm="fricas")

[Out]

1/12*(11*(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^4*x^4 + 12*b*c^2*x^2 + 8*b)*sqrt(c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 + 1)) + 2*(6*a*
c^4*x^4 + 24*a*c^2*x^2 - (6*b*c^3*x^3 + 5*b*c*x)*sqrt(c^2*x^2 + 1) + 16*a)*sqrt(c^2*d*x^2 + d))/(c^10*d^3*x^4
+ 2*c^8*d^3*x^2 + c^6*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**5*(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^{5}}{{\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^5*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(5/2),x, algorithm="giac")

[Out]

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