3.1.49 \(\int \frac {(f+g x) (a+b \sinh ^{-1}(c x))}{\sqrt {d+c^2 d x^2}} \, dx\) [49]

Optimal. Leaf size=120 \[ -\frac {b g x \sqrt {1+c^2 x^2}}{c \sqrt {d+c^2 d x^2}}+\frac {g \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^2 \sqrt {d+c^2 d x^2}}+\frac {f \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c \sqrt {d+c^2 d x^2}} \]

[Out]

g*(c^2*x^2+1)*(a+b*arcsinh(c*x))/c^2/(c^2*d*x^2+d)^(1/2)-b*g*x*(c^2*x^2+1)^(1/2)/c/(c^2*d*x^2+d)^(1/2)+1/2*f*(
a+b*arcsinh(c*x))^2*(c^2*x^2+1)^(1/2)/b/c/(c^2*d*x^2+d)^(1/2)

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Rubi [A]
time = 0.15, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {5845, 5838, 5783, 5798, 8} \begin {gather*} \frac {f \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c \sqrt {c^2 d x^2+d}}+\frac {g \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^2 \sqrt {c^2 d x^2+d}}-\frac {b g x \sqrt {c^2 x^2+1}}{c \sqrt {c^2 d x^2+d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*g*x*Sqrt[1 + c^2*x^2])/(c*Sqrt[d + c^2*d*x^2])) + (g*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/(c^2*Sqrt[d + c^
2*d*x^2]) + (f*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^2)/(2*b*c*Sqrt[d + c^2*d*x^2])

Rule 8

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

Rule 5783

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

Rule 5798

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/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)
^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 5838

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol]
:> Int[ExpandIntegrand[(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g
}, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ[n, 0] && ((EqQ[n, 1] && GtQ[p,
-1]) || GtQ[p, 0] || EqQ[m, 1] || (EqQ[m, 2] && LtQ[p, -2]))

Rule 5845

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol]
:> Dist[Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p], Int[(f + g*x)^m*(1 + c^2*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] /;
FreeQ[{a, b, c, d, e, f, g, n}, x] && EqQ[e, c^2*d] && IntegerQ[m] && IntegerQ[p - 1/2] &&  !GtQ[d, 0]

Rubi steps

\begin {align*} \int \frac {(f+g x) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {d+c^2 d x^2}} \, dx &=\frac {\sqrt {1+c^2 x^2} \int \frac {(f+g x) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx}{\sqrt {d+c^2 d x^2}}\\ &=\frac {\sqrt {1+c^2 x^2} \int \left (\frac {f \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}}+\frac {g x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}}\right ) \, dx}{\sqrt {d+c^2 d x^2}}\\ &=\frac {\left (f \sqrt {1+c^2 x^2}\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}} \, dx}{\sqrt {d+c^2 d x^2}}+\frac {\left (g \sqrt {1+c^2 x^2}\right ) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx}{\sqrt {d+c^2 d x^2}}\\ &=\frac {g \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^2 \sqrt {d+c^2 d x^2}}+\frac {f \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c \sqrt {d+c^2 d x^2}}-\frac {\left (b g \sqrt {1+c^2 x^2}\right ) \int 1 \, dx}{c \sqrt {d+c^2 d x^2}}\\ &=-\frac {b g x \sqrt {1+c^2 x^2}}{c \sqrt {d+c^2 d x^2}}+\frac {g \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^2 \sqrt {d+c^2 d x^2}}+\frac {f \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c \sqrt {d+c^2 d x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 158, normalized size = 1.32 \begin {gather*} \frac {2 \sqrt {d} g \left (a+a c^2 x^2-b c x \sqrt {1+c^2 x^2}\right )+2 b \sqrt {d} g \left (1+c^2 x^2\right ) \sinh ^{-1}(c x)+b c \sqrt {d} f \sqrt {1+c^2 x^2} \sinh ^{-1}(c x)^2+2 a c f \sqrt {d+c^2 d x^2} \log \left (c d x+\sqrt {d} \sqrt {d+c^2 d x^2}\right )}{2 c^2 \sqrt {d} \sqrt {d+c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d]*g*(a + a*c^2*x^2 - b*c*x*Sqrt[1 + c^2*x^2]) + 2*b*Sqrt[d]*g*(1 + c^2*x^2)*ArcSinh[c*x] + b*c*Sqrt[d
]*f*Sqrt[1 + c^2*x^2]*ArcSinh[c*x]^2 + 2*a*c*f*Sqrt[d + c^2*d*x^2]*Log[c*d*x + Sqrt[d]*Sqrt[d + c^2*d*x^2]])/(
2*c^2*Sqrt[d]*Sqrt[d + c^2*d*x^2])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(226\) vs. \(2(108)=216\).
time = 4.54, size = 227, normalized size = 1.89

method	result	size
default	\(\frac {a g \sqrt {c^{2} d \,x^{2}+d}}{c^{2} d}+\frac {a f \ln \left (\frac {x \,c^{2} d}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{\sqrt {c^{2} d}}+b \left (\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, f \arcsinh \left (c x \right )^{2}}{2 \sqrt {c^{2} x^{2}+1}\, c d}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (c^{2} x^{2}+\sqrt {c^{2} x^{2}+1}\, c x +1\right ) g \left (\arcsinh \left (c x \right )-1\right )}{2 c^{2} d \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (c^{2} x^{2}-\sqrt {c^{2} x^{2}+1}\, c x +1\right ) g \left (\arcsinh \left (c x \right )+1\right )}{2 c^{2} d \left (c^{2} x^{2}+1\right )}\right )\)	\(227\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(1/2),x,method=_RETURNVERBOSE)

[Out]

a*g/c^2/d*(c^2*d*x^2+d)^(1/2)+a*f*ln(x*c^2*d/(c^2*d)^(1/2)+(c^2*d*x^2+d)^(1/2))/(c^2*d)^(1/2)+b*(1/2*(d*(c^2*x
^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c/d*f*arcsinh(c*x)^2+1/2*(d*(c^2*x^2+1))^(1/2)*(c^2*x^2+(c^2*x^2+1)^(1/2)*c*x+1
)*g*(arcsinh(c*x)-1)/c^2/d/(c^2*x^2+1)+1/2*(d*(c^2*x^2+1))^(1/2)*(c^2*x^2-(c^2*x^2+1)^(1/2)*c*x+1)*g*(arcsinh(
c*x)+1)/c^2/d/(c^2*x^2+1))

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Maxima [A]
time = 0.27, size = 87, normalized size = 0.72 \begin {gather*} \frac {b f \operatorname {arsinh}\left (c x\right )^{2}}{2 \, c \sqrt {d}} - \frac {b g x}{c \sqrt {d}} + \frac {a f \operatorname {arsinh}\left (c x\right )}{c \sqrt {d}} + \frac {\sqrt {c^{2} d x^{2} + d} b g \operatorname {arsinh}\left (c x\right )}{c^{2} d} + \frac {\sqrt {c^{2} d x^{2} + d} a g}{c^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

1/2*b*f*arcsinh(c*x)^2/(c*sqrt(d)) - b*g*x/(c*sqrt(d)) + a*f*arcsinh(c*x)/(c*sqrt(d)) + sqrt(c^2*d*x^2 + d)*b*
g*arcsinh(c*x)/(c^2*d) + sqrt(c^2*d*x^2 + d)*a*g/(c^2*d)

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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((g*x+f)*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

integral((a*g*x + a*f + (b*g*x + b*f)*arcsinh(c*x))/sqrt(c^2*d*x^2 + d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(a+b*asinh(c*x))/(c**2*d*x**2+d)**(1/2),x)

[Out]

Integral((a + b*asinh(c*x))*(f + g*x)/sqrt(d*(c**2*x**2 + 1)), 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((g*x+f)*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

integrate((g*x + f)*(b*arcsinh(c*x) + a)/sqrt(c^2*d*x^2 + d), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (f+g\,x\right )\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{\sqrt {d\,c^2\,x^2+d}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((f + g*x)*(a + b*asinh(c*x)))/(d + c^2*d*x^2)^(1/2),x)

[Out]

int(((f + g*x)*(a + b*asinh(c*x)))/(d + c^2*d*x^2)^(1/2), x)

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