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

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

[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])

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Rubi [A]  time = 0.217123, antiderivative size = 120, normalized size of antiderivative = 1., 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 = {5835, 5821, 5675, 5717, 8} \[ \frac{f \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c \sqrt{c^2 d x^2+d}}+\frac{g \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^2 \sqrt{c^2 d x^2+d}}-\frac{b g x \sqrt{c^2 x^2+1}}{c \sqrt{c^2 d x^2+d}} \]

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 5835

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol]
:> Dist[(d^IntPart[p]*(d + e*x^2)^FracPart[p])/(1 + c^2*x^2)^FracPart[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]

Rule 5821

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 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 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]

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*}

Mathematica [A]  time = 0.253685, size = 158, normalized size = 1.32 \[ \frac{2 \sqrt{d} g \left (a c^2 x^2+a-b c x \sqrt{c^2 x^2+1}\right )+2 a c f \sqrt{c^2 d x^2+d} \log \left (\sqrt{d} \sqrt{c^2 d x^2+d}+c d x\right )+b c \sqrt{d} f \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)^2+2 b \sqrt{d} g \left (c^2 x^2+1\right ) \sinh ^{-1}(c x)}{2 c^2 \sqrt{d} \sqrt{c^2 d x^2+d}} \]

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 [A]  time = 0.178, size = 209, normalized size = 1.7 \begin{align*}{\frac{ag}{{c}^{2}d}\sqrt{{c}^{2}d{x}^{2}+d}}+{af\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{bf \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{2\,cd}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{bg{\it Arcsinh} \left ( cx \right ){x}^{2}}{d \left ({c}^{2}{x}^{2}+1 \right ) }\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{bgx}{cd}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{bg{\it Arcsinh} \left ( cx \right ) }{{c}^{2}d \left ({c}^{2}{x}^{2}+1 \right ) }\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }} \end{align*}

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)

[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)+1/2*b*(d*(c^2*x^
2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c/d*f*arcsinh(c*x)^2+b*(d*(c^2*x^2+1))^(1/2)*g/d/(c^2*x^2+1)*arcsinh(c*x)*x^2-b*
(d*(c^2*x^2+1))^(1/2)*g/c/d/(c^2*x^2+1)^(1/2)*x+b*(d*(c^2*x^2+1))^(1/2)*g/c^2/d/(c^2*x^2+1)*arcsinh(c*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((g*x+f)*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(1/2),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{a g x + a f +{\left (b g x + b f\right )} \operatorname{arsinh}\left (c x\right )}{\sqrt{c^{2} d x^{2} + d}}, x\right ) \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (g x + f\right )}{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}}{\sqrt{c^{2} d x^{2} + d}}\,{d x} \end{align*}

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)