3.3.69 \(\int \frac {(1+a^2+2 a b x+b^2 x^2)^{3/2}}{\sinh ^{-1}(a+b x)} \, dx\) [269]

Optimal. Leaf size=47 \[ \frac {\text {Chi}\left (2 \sinh ^{-1}(a+b x)\right )}{2 b}+\frac {\text {Chi}\left (4 \sinh ^{-1}(a+b x)\right )}{8 b}+\frac {3 \log \left (\sinh ^{-1}(a+b x)\right )}{8 b} \]

[Out]

1/2*Chi(2*arcsinh(b*x+a))/b+1/8*Chi(4*arcsinh(b*x+a))/b+3/8*ln(arcsinh(b*x+a))/b

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Rubi [A]
time = 0.10, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {5860, 5791, 3393, 3382} \begin {gather*} \frac {\text {Chi}\left (2 \sinh ^{-1}(a+b x)\right )}{2 b}+\frac {\text {Chi}\left (4 \sinh ^{-1}(a+b x)\right )}{8 b}+\frac {3 \log \left (\sinh ^{-1}(a+b x)\right )}{8 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + a^2 + 2*a*b*x + b^2*x^2)^(3/2)/ArcSinh[a + b*x],x]

[Out]

CoshIntegral[2*ArcSinh[a + b*x]]/(2*b) + CoshIntegral[4*ArcSinh[a + b*x]]/(8*b) + (3*Log[ArcSinh[a + b*x]])/(8
*b)

Rule 3382

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[c*f*(fz/d)
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rule 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 5791

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

Rule 5860

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)^(p_.), x_Symbol] :> D
ist[1/d, Subst[Int[(C/d^2 + (C/d^2)*x^2)^p*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, A,
B, C, n, p}, x] && EqQ[B*(1 + c^2) - 2*A*c*d, 0] && EqQ[2*c*C - B*d, 0]

Rubi steps

\begin {align*} \int \frac {\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}}{\sinh ^{-1}(a+b x)} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (1+x^2\right )^{3/2}}{\sinh ^{-1}(x)} \, dx,x,a+b x\right )}{b}\\ &=\frac {\text {Subst}\left (\int \frac {\cosh ^4(x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=\frac {\text {Subst}\left (\int \left (\frac {3}{8 x}+\frac {\cosh (2 x)}{2 x}+\frac {\cosh (4 x)}{8 x}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=\frac {3 \log \left (\sinh ^{-1}(a+b x)\right )}{8 b}+\frac {\text {Subst}\left (\int \frac {\cosh (4 x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b}+\frac {\text {Subst}\left (\int \frac {\cosh (2 x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b}\\ &=\frac {\text {Chi}\left (2 \sinh ^{-1}(a+b x)\right )}{2 b}+\frac {\text {Chi}\left (4 \sinh ^{-1}(a+b x)\right )}{8 b}+\frac {3 \log \left (\sinh ^{-1}(a+b x)\right )}{8 b}\\ \end {align*}

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Mathematica [A]
time = 0.18, size = 37, normalized size = 0.79 \begin {gather*} \frac {4 \text {Chi}\left (2 \sinh ^{-1}(a+b x)\right )+\text {Chi}\left (4 \sinh ^{-1}(a+b x)\right )+3 \log \left (\sinh ^{-1}(a+b x)\right )}{8 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + a^2 + 2*a*b*x + b^2*x^2)^(3/2)/ArcSinh[a + b*x],x]

[Out]

(4*CoshIntegral[2*ArcSinh[a + b*x]] + CoshIntegral[4*ArcSinh[a + b*x]] + 3*Log[ArcSinh[a + b*x]])/(8*b)

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Maple [A]
time = 3.96, size = 36, normalized size = 0.77

method	result	size
default	\(\frac {3 \ln \left (\arcsinh \left (b x +a \right )\right )+4 \hyperbolicCosineIntegral \left (2 \arcsinh \left (b x +a \right )\right )+\hyperbolicCosineIntegral \left (4 \arcsinh \left (b x +a \right )\right )}{8 b}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^2+2*a*b*x+a^2+1)^(3/2)/arcsinh(b*x+a),x,method=_RETURNVERBOSE)

[Out]

1/8*(3*ln(arcsinh(b*x+a))+4*Chi(2*arcsinh(b*x+a))+Chi(4*arcsinh(b*x+a)))/b

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2+1)^(3/2)/arcsinh(b*x+a),x, algorithm="maxima")

[Out]

integrate((b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)/arcsinh(b*x + a), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2+1)^(3/2)/arcsinh(b*x+a),x, algorithm="fricas")

[Out]

integral((b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)/arcsinh(b*x + a), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a^{2} + 2 a b x + b^{2} x^{2} + 1\right )^{\frac {3}{2}}}{\operatorname {asinh}{\left (a + b x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**2+2*a*b*x+a**2+1)**(3/2)/asinh(b*x+a),x)

[Out]

Integral((a**2 + 2*a*b*x + b**2*x**2 + 1)**(3/2)/asinh(a + b*x), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2+1)^(3/2)/arcsinh(b*x+a),x, algorithm="giac")

[Out]

integrate((b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)/arcsinh(b*x + a), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}^{3/2}}{\mathrm {asinh}\left (a+b\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^2 + 2*a*b*x + 1)^(3/2)/asinh(a + b*x),x)

[Out]

int((a^2 + b^2*x^2 + 2*a*b*x + 1)^(3/2)/asinh(a + b*x), x)

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