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

Optimal. Leaf size=257 \[ \frac{15}{8} c^2 d^2 x \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac{15 c d^2 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b \sqrt{c^2 x^2+1}}+\frac{5}{4} c^2 d x \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{\left (c^2 d x^2+d\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{x}-\frac{b c^5 d^2 x^4 \sqrt{c^2 d x^2+d}}{16 \sqrt{c^2 x^2+1}}-\frac{9 b c^3 d^2 x^2 \sqrt{c^2 d x^2+d}}{16 \sqrt{c^2 x^2+1}}+\frac{b c d^2 \log (x) \sqrt{c^2 d x^2+d}}{\sqrt{c^2 x^2+1}} \]

[Out]

(-9*b*c^3*d^2*x^2*Sqrt[d + c^2*d*x^2])/(16*Sqrt[1 + c^2*x^2]) - (b*c^5*d^2*x^4*Sqrt[d + c^2*d*x^2])/(16*Sqrt[1
+ c^2*x^2]) + (15*c^2*d^2*x*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/8 + (5*c^2*d*x*(d + c^2*d*x^2)^(3/2)*(a
+ b*ArcSinh[c*x]))/4 - ((d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/x + (15*c*d^2*Sqrt[d + c^2*d*x^2]*(a + b*
ArcSinh[c*x])^2)/(16*b*Sqrt[1 + c^2*x^2]) + (b*c*d^2*Sqrt[d + c^2*d*x^2]*Log[x])/Sqrt[1 + c^2*x^2]

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Rubi [A]  time = 0.23772, antiderivative size = 257, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {5739, 5684, 5682, 5675, 30, 14, 266, 43} \[ \frac{15}{8} c^2 d^2 x \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac{15 c d^2 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b \sqrt{c^2 x^2+1}}+\frac{5}{4} c^2 d x \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{\left (c^2 d x^2+d\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{x}-\frac{b c^5 d^2 x^4 \sqrt{c^2 d x^2+d}}{16 \sqrt{c^2 x^2+1}}-\frac{9 b c^3 d^2 x^2 \sqrt{c^2 d x^2+d}}{16 \sqrt{c^2 x^2+1}}+\frac{b c d^2 \log (x) \sqrt{c^2 d x^2+d}}{\sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*b*c^3*d^2*x^2*Sqrt[d + c^2*d*x^2])/(16*Sqrt[1 + c^2*x^2]) - (b*c^5*d^2*x^4*Sqrt[d + c^2*d*x^2])/(16*Sqrt[1
+ c^2*x^2]) + (15*c^2*d^2*x*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/8 + (5*c^2*d*x*(d + c^2*d*x^2)^(3/2)*(a
+ b*ArcSinh[c*x]))/4 - ((d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/x + (15*c*d^2*Sqrt[d + c^2*d*x^2]*(a + b*
ArcSinh[c*x])^2)/(16*b*Sqrt[1 + c^2*x^2]) + (b*c*d^2*Sqrt[d + c^2*d*x^2]*Log[x])/Sqrt[1 + c^2*x^2]

Rule 5739

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

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 266

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{x^2} \, dx &=-\frac{\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{x}+\left (5 c^2 d\right ) \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 \frac{\left (1+c^2 x^2\right )^2}{x} \, dx}{\sqrt{1+c^2 x^2}}\\ &=\frac{5}{4} c^2 d x \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{x}+\frac{1}{4} \left (15 c^2 d^2\right ) \int \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx+\frac{\left (b c d^2 \sqrt{d+c^2 d x^2}\right ) \operatorname{Subst}\left (\int \frac{\left (1+c^2 x\right )^2}{x} \, dx,x,x^2\right )}{2 \sqrt{1+c^2 x^2}}-\frac{\left (5 b c^3 d^2 \sqrt{d+c^2 d x^2}\right ) \int x \left (1+c^2 x^2\right ) \, dx}{4 \sqrt{1+c^2 x^2}}\\ &=\frac{15}{8} c^2 d^2 x \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5}{4} c^2 d x \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{x}+\frac{\left (b c d^2 \sqrt{d+c^2 d x^2}\right ) \operatorname{Subst}\left (\int \left (2 c^2+\frac{1}{x}+c^4 x\right ) \, dx,x,x^2\right )}{2 \sqrt{1+c^2 x^2}}+\frac{\left (15 c^2 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}{8 \sqrt{1+c^2 x^2}}-\frac{\left (5 b c^3 d^2 \sqrt{d+c^2 d x^2}\right ) \int \left (x+c^2 x^3\right ) \, dx}{4 \sqrt{1+c^2 x^2}}-\frac{\left (15 b c^3 d^2 \sqrt{d+c^2 d x^2}\right ) \int x \, dx}{8 \sqrt{1+c^2 x^2}}\\ &=-\frac{9 b c^3 d^2 x^2 \sqrt{d+c^2 d x^2}}{16 \sqrt{1+c^2 x^2}}-\frac{b c^5 d^2 x^4 \sqrt{d+c^2 d x^2}}{16 \sqrt{1+c^2 x^2}}+\frac{15}{8} c^2 d^2 x \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5}{4} c^2 d x \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{x}+\frac{15 c d^2 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b \sqrt{1+c^2 x^2}}+\frac{b c d^2 \sqrt{d+c^2 d x^2} \log (x)}{\sqrt{1+c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 1.34322, size = 270, normalized size = 1.05 \[ \frac{1}{128} d^2 \left (\frac{16 a \left (2 c^4 x^4+9 c^2 x^2-8\right ) \sqrt{c^2 d x^2+d}}{x}+240 a c \sqrt{d} \log \left (\sqrt{d} \sqrt{c^2 d x^2+d}+c d x\right )+\frac{64 b \sqrt{c^2 d x^2+d} \left (-2 \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)+2 c x \log (c x)+c x \sinh ^{-1}(c x)^2\right )}{x \sqrt{c^2 x^2+1}}+\frac{32 b c \sqrt{c^2 d x^2+d} \left (2 \sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)+\sinh \left (2 \sinh ^{-1}(c x)\right )\right )-\cosh \left (2 \sinh ^{-1}(c x)\right )\right )}{\sqrt{c^2 x^2+1}}-\frac{b c \sqrt{c^2 d x^2+d} \left (8 \sinh ^{-1}(c x)^2-4 \sinh \left (4 \sinh ^{-1}(c x)\right ) \sinh ^{-1}(c x)+\cosh \left (4 \sinh ^{-1}(c x)\right )\right )}{\sqrt{c^2 x^2+1}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*((16*a*Sqrt[d + c^2*d*x^2]*(-8 + 9*c^2*x^2 + 2*c^4*x^4))/x + (64*b*Sqrt[d + c^2*d*x^2]*(-2*Sqrt[1 + c^2*x
^2]*ArcSinh[c*x] + c*x*ArcSinh[c*x]^2 + 2*c*x*Log[c*x]))/(x*Sqrt[1 + c^2*x^2]) + 240*a*c*Sqrt[d]*Log[c*d*x + S
qrt[d]*Sqrt[d + c^2*d*x^2]] + (32*b*c*Sqrt[d + c^2*d*x^2]*(-Cosh[2*ArcSinh[c*x]] + 2*ArcSinh[c*x]*(ArcSinh[c*x
] + Sinh[2*ArcSinh[c*x]])))/Sqrt[1 + c^2*x^2] - (b*c*Sqrt[d + c^2*d*x^2]*(8*ArcSinh[c*x]^2 + Cosh[4*ArcSinh[c*
x]] - 4*ArcSinh[c*x]*Sinh[4*ArcSinh[c*x]]))/Sqrt[1 + c^2*x^2]))/128

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Maple [B]  time = 0.185, size = 506, normalized size = 2. \begin{align*} -{\frac{a}{dx} \left ({c}^{2}d{x}^{2}+d \right ) ^{{\frac{7}{2}}}}+a{c}^{2}x \left ({c}^{2}d{x}^{2}+d \right ) ^{{\frac{5}{2}}}+{\frac{5\,a{c}^{2}dx}{4} \left ({c}^{2}d{x}^{2}+d \right ) ^{{\frac{3}{2}}}}+{\frac{15\,a{c}^{2}{d}^{2}x}{8}\sqrt{{c}^{2}d{x}^{2}+d}}+{\frac{15\,a{c}^{2}{d}^{3}}{8}\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{15\,b \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}c{d}^{2}}{16}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{11\,b{c}^{4}{d}^{2}{\it Arcsinh} \left ( cx \right ){x}^{3}}{8\,{c}^{2}{x}^{2}+8}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{9\,b{c}^{3}{d}^{2}{x}^{2}}{16}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{{d}^{2}b{c}^{2}{\it Arcsinh} \left ( cx \right ) x}{8\,{c}^{2}{x}^{2}+8}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{bc{d}^{2}{\it Arcsinh} \left ( cx \right ) \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}^{2}}{x \left ({c}^{2}{x}^{2}+1 \right ) }\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{b{c}^{6}{d}^{2}{\it Arcsinh} \left ( cx \right ){x}^{5}}{4\,{c}^{2}{x}^{2}+4}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{b{c}^{5}{d}^{2}{x}^{4}}{16}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{bc{d}^{2}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }\ln \left ( \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2}-1 \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}-{\frac{33\,bc{d}^{2}}{128}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \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^2,x)

[Out]

-a/d/x*(c^2*d*x^2+d)^(7/2)+a*c^2*x*(c^2*d*x^2+d)^(5/2)+5/4*a*c^2*d*x*(c^2*d*x^2+d)^(3/2)+15/8*a*c^2*d^2*x*(c^2
*d*x^2+d)^(1/2)+15/8*a*c^2*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)+15/16*b*(d*(c^2*x^2
+1))^(1/2)/(c^2*x^2+1)^(1/2)*arcsinh(c*x)^2*c*d^2+11/8*b*(d*(c^2*x^2+1))^(1/2)*c^4*d^2/(c^2*x^2+1)*arcsinh(c*x
)*x^3-9/16*b*(d*(c^2*x^2+1))^(1/2)*c^3*d^2/(c^2*x^2+1)^(1/2)*x^2+1/8*b*(d*(c^2*x^2+1))^(1/2)*c^2*d^2/(c^2*x^2+
1)*arcsinh(c*x)*x-b*(d*(c^2*x^2+1))^(1/2)*c*d^2/(c^2*x^2+1)^(1/2)*arcsinh(c*x)-b*(d*(c^2*x^2+1))^(1/2)*arcsinh
(c*x)*d^2/x/(c^2*x^2+1)+1/4*b*(d*(c^2*x^2+1))^(1/2)*c^6*d^2/(c^2*x^2+1)*arcsinh(c*x)*x^5-1/16*b*(d*(c^2*x^2+1)
)^(1/2)*c^5*d^2/(c^2*x^2+1)^(1/2)*x^4+b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)*ln((c*x+(c^2*x^2+1)^(1/2))^2-1
)*c*d^2-33/128*b*(d*(c^2*x^2+1))^(1/2)*c*d^2/(c^2*x^2+1)^(1/2)

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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^2,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 (\frac{{\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^{2}}, 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^2,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^2, 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**2,x)

[Out]

Timed out

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

[Out]

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