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

Optimal. Leaf size=254 \[ \frac{5}{16} d^2 x \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5 d^2 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{32 b c \sqrt{c^2 x^2+1}}+\frac{1}{6} x \left (c^2 d x^2+d\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5}{24} d x \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{5 b c^3 d^2 x^4 \sqrt{c^2 d x^2+d}}{96 \sqrt{c^2 x^2+1}}-\frac{25 b c d^2 x^2 \sqrt{c^2 d x^2+d}}{96 \sqrt{c^2 x^2+1}}-\frac{b d^2 \left (c^2 x^2+1\right )^{5/2} \sqrt{c^2 d x^2+d}}{36 c} \]

[Out]

(-25*b*c*d^2*x^2*Sqrt[d + c^2*d*x^2])/(96*Sqrt[1 + c^2*x^2]) - (5*b*c^3*d^2*x^4*Sqrt[d + c^2*d*x^2])/(96*Sqrt[
1 + c^2*x^2]) - (b*d^2*(1 + c^2*x^2)^(5/2)*Sqrt[d + c^2*d*x^2])/(36*c) + (5*d^2*x*Sqrt[d + c^2*d*x^2]*(a + b*A
rcSinh[c*x]))/16 + (5*d*x*(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/24 + (x*(d + c^2*d*x^2)^(5/2)*(a + b*Arc
Sinh[c*x]))/6 + (5*d^2*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^2)/(32*b*c*Sqrt[1 + c^2*x^2])

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Rubi [A]  time = 0.159522, antiderivative size = 254, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {5684, 5682, 5675, 30, 14, 261} \[ \frac{5}{16} d^2 x \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5 d^2 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{32 b c \sqrt{c^2 x^2+1}}+\frac{1}{6} x \left (c^2 d x^2+d\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5}{24} d x \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{5 b c^3 d^2 x^4 \sqrt{c^2 d x^2+d}}{96 \sqrt{c^2 x^2+1}}-\frac{25 b c d^2 x^2 \sqrt{c^2 d x^2+d}}{96 \sqrt{c^2 x^2+1}}-\frac{b d^2 \left (c^2 x^2+1\right )^{5/2} \sqrt{c^2 d x^2+d}}{36 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-25*b*c*d^2*x^2*Sqrt[d + c^2*d*x^2])/(96*Sqrt[1 + c^2*x^2]) - (5*b*c^3*d^2*x^4*Sqrt[d + c^2*d*x^2])/(96*Sqrt[
1 + c^2*x^2]) - (b*d^2*(1 + c^2*x^2)^(5/2)*Sqrt[d + c^2*d*x^2])/(36*c) + (5*d^2*x*Sqrt[d + c^2*d*x^2]*(a + b*A
rcSinh[c*x]))/16 + (5*d*x*(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/24 + (x*(d + c^2*d*x^2)^(5/2)*(a + b*Arc
Sinh[c*x]))/6 + (5*d^2*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^2)/(32*b*c*Sqrt[1 + c^2*x^2])

Rule 5684

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

Rule 5682

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

Rule 5675

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

Rule 30

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

\begin{align*} \int \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac{1}{6} x \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{6} (5 d) \int \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx-\frac{\left (b c d^2 \sqrt{d+c^2 d x^2}\right ) \int x \left (1+c^2 x^2\right )^2 \, dx}{6 \sqrt{1+c^2 x^2}}\\ &=-\frac{b d^2 \left (1+c^2 x^2\right )^{5/2} \sqrt{d+c^2 d x^2}}{36 c}+\frac{5}{24} d x \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{6} x \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{8} \left (5 d^2\right ) \int \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx-\frac{\left (5 b c d^2 \sqrt{d+c^2 d x^2}\right ) \int x \left (1+c^2 x^2\right ) \, dx}{24 \sqrt{1+c^2 x^2}}\\ &=-\frac{b d^2 \left (1+c^2 x^2\right )^{5/2} \sqrt{d+c^2 d x^2}}{36 c}+\frac{5}{16} d^2 x \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5}{24} d x \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{6} x \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{\left (5 d^2 \sqrt{d+c^2 d x^2}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\sqrt{1+c^2 x^2}} \, dx}{16 \sqrt{1+c^2 x^2}}-\frac{\left (5 b c d^2 \sqrt{d+c^2 d x^2}\right ) \int \left (x+c^2 x^3\right ) \, dx}{24 \sqrt{1+c^2 x^2}}-\frac{\left (5 b c d^2 \sqrt{d+c^2 d x^2}\right ) \int x \, dx}{16 \sqrt{1+c^2 x^2}}\\ &=-\frac{25 b c d^2 x^2 \sqrt{d+c^2 d x^2}}{96 \sqrt{1+c^2 x^2}}-\frac{5 b c^3 d^2 x^4 \sqrt{d+c^2 d x^2}}{96 \sqrt{1+c^2 x^2}}-\frac{b d^2 \left (1+c^2 x^2\right )^{5/2} \sqrt{d+c^2 d x^2}}{36 c}+\frac{5}{16} d^2 x \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5}{24} d x \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{6} x \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5 d^2 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{32 b c \sqrt{1+c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.690686, size = 317, normalized size = 1.25 \[ \frac{d^2 \left (384 a c^5 x^5 \sqrt{c^2 x^2+1} \sqrt{c^2 d x^2+d}+1248 a c^3 x^3 \sqrt{c^2 x^2+1} \sqrt{c^2 d x^2+d}+1584 a c x \sqrt{c^2 x^2+1} \sqrt{c^2 d x^2+d}+720 a \sqrt{d} \sqrt{c^2 x^2+1} \log \left (\sqrt{d} \sqrt{c^2 d x^2+d}+c d x\right )+360 b \sqrt{c^2 d x^2+d} \sinh ^{-1}(c x)^2+12 b \sqrt{c^2 d x^2+d} \sinh ^{-1}(c x) \left (45 \sinh \left (2 \sinh ^{-1}(c x)\right )+9 \sinh \left (4 \sinh ^{-1}(c x)\right )+\sinh \left (6 \sinh ^{-1}(c x)\right )\right )-270 b \sqrt{c^2 d x^2+d} \cosh \left (2 \sinh ^{-1}(c x)\right )-27 b \sqrt{c^2 d x^2+d} \cosh \left (4 \sinh ^{-1}(c x)\right )-2 b \sqrt{c^2 d x^2+d} \cosh \left (6 \sinh ^{-1}(c x)\right )\right )}{2304 c \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(1584*a*c*x*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2] + 1248*a*c^3*x^3*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2]
+ 384*a*c^5*x^5*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2] + 360*b*Sqrt[d + c^2*d*x^2]*ArcSinh[c*x]^2 - 270*b*Sqrt[
d + c^2*d*x^2]*Cosh[2*ArcSinh[c*x]] - 27*b*Sqrt[d + c^2*d*x^2]*Cosh[4*ArcSinh[c*x]] - 2*b*Sqrt[d + c^2*d*x^2]*
Cosh[6*ArcSinh[c*x]] + 720*a*Sqrt[d]*Sqrt[1 + c^2*x^2]*Log[c*d*x + Sqrt[d]*Sqrt[d + c^2*d*x^2]] + 12*b*Sqrt[d
+ c^2*d*x^2]*ArcSinh[c*x]*(45*Sinh[2*ArcSinh[c*x]] + 9*Sinh[4*ArcSinh[c*x]] + Sinh[6*ArcSinh[c*x]])))/(2304*c*
Sqrt[1 + c^2*x^2])

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Maple [A]  time = 0.161, size = 427, normalized size = 1.7 \begin{align*}{\frac{ax}{6} \left ({c}^{2}d{x}^{2}+d \right ) ^{{\frac{5}{2}}}}+{\frac{5\,adx}{24} \left ({c}^{2}d{x}^{2}+d \right ) ^{{\frac{3}{2}}}}+{\frac{5\,a{d}^{2}x}{16}\sqrt{{c}^{2}d{x}^{2}+d}}+{\frac{5\,a{d}^{3}}{16}\ln \left ({{c}^{2}dx{\frac{1}{\sqrt{{c}^{2}d}}}}+\sqrt{{c}^{2}d{x}^{2}+d} \right ){\frac{1}{\sqrt{{c}^{2}d}}}}-{\frac{299\,b{d}^{2}}{2304\,c}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{5\,b \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}{d}^{2}}{32\,c}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{b{d}^{2}{c}^{6}{\it Arcsinh} \left ( cx \right ){x}^{7}}{6\,{c}^{2}{x}^{2}+6}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{b{d}^{2}{c}^{5}{x}^{6}}{36}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{17\,b{d}^{2}{c}^{4}{\it Arcsinh} \left ( cx \right ){x}^{5}}{24\,{c}^{2}{x}^{2}+24}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{13\,b{d}^{2}{c}^{3}{x}^{4}}{96}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{59\,{d}^{2}b{c}^{2}{\it Arcsinh} \left ( cx \right ){x}^{3}}{48\,{c}^{2}{x}^{2}+48}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{11\,b{d}^{2}c{x}^{2}}{32}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{11\,b{d}^{2}{\it Arcsinh} \left ( cx \right ) x}{16\,{c}^{2}{x}^{2}+16}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x*(c^2*d*x^2+d)^(5/2)*a+5/24*a*d*x*(c^2*d*x^2+d)^(3/2)+5/16*a*d^2*x*(c^2*d*x^2+d)^(1/2)+5/16*a*d^3*ln(x*c^
2*d/(c^2*d)^(1/2)+(c^2*d*x^2+d)^(1/2))/(c^2*d)^(1/2)-299/2304*b*(d*(c^2*x^2+1))^(1/2)*d^2/c/(c^2*x^2+1)^(1/2)+
5/32*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c*arcsinh(c*x)^2*d^2+1/6*b*(d*(c^2*x^2+1))^(1/2)*d^2*c^6/(c^2*x
^2+1)*arcsinh(c*x)*x^7-1/36*b*(d*(c^2*x^2+1))^(1/2)*d^2*c^5/(c^2*x^2+1)^(1/2)*x^6+17/24*b*(d*(c^2*x^2+1))^(1/2
)*d^2*c^4/(c^2*x^2+1)*arcsinh(c*x)*x^5-13/96*b*(d*(c^2*x^2+1))^(1/2)*d^2*c^3/(c^2*x^2+1)^(1/2)*x^4+59/48*b*(d*
(c^2*x^2+1))^(1/2)*d^2*c^2/(c^2*x^2+1)*arcsinh(c*x)*x^3-11/32*b*(d*(c^2*x^2+1))^(1/2)*d^2*c/(c^2*x^2+1)^(1/2)*
x^2+11/16*b*(d*(c^2*x^2+1))^(1/2)*d^2/(c^2*x^2+1)*arcsinh(c*x)*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((c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a c^{4} d^{2} x^{4} + 2 \, a c^{2} d^{2} x^{2} + a d^{2} +{\left (b c^{4} d^{2} x^{4} + 2 \, b c^{2} d^{2} x^{2} + b d^{2}\right )} \operatorname{arsinh}\left (c x\right )\right )} \sqrt{c^{2} d x^{2} + d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*c^4*d^2*x^4 + 2*a*c^2*d^2*x^2 + a*d^2 + (b*c^4*d^2*x^4 + 2*b*c^2*d^2*x^2 + b*d^2)*arcsinh(c*x))*sq
rt(c^2*d*x^2 + d), x)

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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