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

Optimal. Leaf size=227 \[ \frac{1}{2} f x \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac{f \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c \sqrt{c^2 x^2+1}}+\frac{g \left (c^2 x^2+1\right ) \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2}-\frac{b c f x^2 \sqrt{c^2 d x^2+d}}{4 \sqrt{c^2 x^2+1}}-\frac{b c g x^3 \sqrt{c^2 d x^2+d}}{9 \sqrt{c^2 x^2+1}}-\frac{b g x \sqrt{c^2 d x^2+d}}{3 c \sqrt{c^2 x^2+1}} \]

[Out]

-(b*g*x*Sqrt[d + c^2*d*x^2])/(3*c*Sqrt[1 + c^2*x^2]) - (b*c*f*x^2*Sqrt[d + c^2*d*x^2])/(4*Sqrt[1 + c^2*x^2]) -
(b*c*g*x^3*Sqrt[d + c^2*d*x^2])/(9*Sqrt[1 + c^2*x^2]) + (f*x*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/2 + (g
*(1 + c^2*x^2)*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(3*c^2) + (f*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])
^2)/(4*b*c*Sqrt[1 + c^2*x^2])

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Rubi [A]  time = 0.250144, antiderivative size = 227, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {5835, 5821, 5682, 5675, 30, 5717} \[ \frac{1}{2} f x \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac{f \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c \sqrt{c^2 x^2+1}}+\frac{g \left (c^2 x^2+1\right ) \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2}-\frac{b c f x^2 \sqrt{c^2 d x^2+d}}{4 \sqrt{c^2 x^2+1}}-\frac{b c g x^3 \sqrt{c^2 d x^2+d}}{9 \sqrt{c^2 x^2+1}}-\frac{b g x \sqrt{c^2 d x^2+d}}{3 c \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*g*x*Sqrt[d + c^2*d*x^2])/(3*c*Sqrt[1 + c^2*x^2]) - (b*c*f*x^2*Sqrt[d + c^2*d*x^2])/(4*Sqrt[1 + c^2*x^2]) -
(b*c*g*x^3*Sqrt[d + c^2*d*x^2])/(9*Sqrt[1 + c^2*x^2]) + (f*x*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/2 + (g
*(1 + c^2*x^2)*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(3*c^2) + (f*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])
^2)/(4*b*c*Sqrt[1 + c^2*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 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 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]

Rubi steps

\begin{align*} \int (f+g x) \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac{\sqrt{d+c^2 d x^2} \int (f+g x) \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\sqrt{1+c^2 x^2}}\\ &=\frac{\sqrt{d+c^2 d x^2} \int \left (f \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )+g x \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )\right ) \, dx}{\sqrt{1+c^2 x^2}}\\ &=\frac{\left (f \sqrt{d+c^2 d x^2}\right ) \int \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\sqrt{1+c^2 x^2}}+\frac{\left (g \sqrt{d+c^2 d x^2}\right ) \int x \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\sqrt{1+c^2 x^2}}\\ &=\frac{1}{2} f x \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{g \left (1+c^2 x^2\right ) \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2}+\frac{\left (f \sqrt{d+c^2 d x^2}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\sqrt{1+c^2 x^2}} \, dx}{2 \sqrt{1+c^2 x^2}}-\frac{\left (b c f \sqrt{d+c^2 d x^2}\right ) \int x \, dx}{2 \sqrt{1+c^2 x^2}}-\frac{\left (b g \sqrt{d+c^2 d x^2}\right ) \int \left (1+c^2 x^2\right ) \, dx}{3 c \sqrt{1+c^2 x^2}}\\ &=-\frac{b g x \sqrt{d+c^2 d x^2}}{3 c \sqrt{1+c^2 x^2}}-\frac{b c f x^2 \sqrt{d+c^2 d x^2}}{4 \sqrt{1+c^2 x^2}}-\frac{b c g x^3 \sqrt{d+c^2 d x^2}}{9 \sqrt{1+c^2 x^2}}+\frac{1}{2} f x \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{g \left (1+c^2 x^2\right ) \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2}+\frac{f \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c \sqrt{1+c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 1.16786, size = 208, normalized size = 0.92 \[ \frac{1}{6} a \sqrt{c^2 d x^2+d} \left (\frac{2 g}{c^2}+x (3 f+2 g x)\right )+\frac{a \sqrt{d} f \log \left (\sqrt{d} \sqrt{c^2 d x^2+d}+c d x\right )}{2 c}+\frac{b f \sqrt{c^2 d x^2+d} \left (2 \sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)+\sinh \left (2 \sinh ^{-1}(c x)\right )\right )-\cosh \left (2 \sinh ^{-1}(c x)\right )\right )}{8 c \sqrt{c^2 x^2+1}}-\frac{b g \sqrt{c^2 d x^2+d} \left (c^3 x^3-3 \left (c^2 x^2+1\right )^{3/2} \sinh ^{-1}(c x)+3 c x\right )}{9 c^2 \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[d + c^2*d*x^2]*((2*g)/c^2 + x*(3*f + 2*g*x)))/6 - (b*g*Sqrt[d + c^2*d*x^2]*(3*c*x + c^3*x^3 - 3*(1 + c
^2*x^2)^(3/2)*ArcSinh[c*x]))/(9*c^2*Sqrt[1 + c^2*x^2]) + (a*Sqrt[d]*f*Log[c*d*x + Sqrt[d]*Sqrt[d + c^2*d*x^2]]
)/(2*c) + (b*f*Sqrt[d + c^2*d*x^2]*(-Cosh[2*ArcSinh[c*x]] + 2*ArcSinh[c*x]*(ArcSinh[c*x] + Sinh[2*ArcSinh[c*x]
])))/(8*c*Sqrt[1 + c^2*x^2])

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Maple [B]  time = 0.27, size = 423, normalized size = 1.9 \begin{align*}{\frac{ag}{3\,{c}^{2}d} \left ({c}^{2}d{x}^{2}+d \right ) ^{{\frac{3}{2}}}}+{\frac{afx}{2}\sqrt{{c}^{2}d{x}^{2}+d}}+{\frac{afd}{2}\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{bg{\it Arcsinh} \left ( cx \right ) }{3\,{c}^{2} \left ({c}^{2}{x}^{2}+1 \right ) }\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{bf}{8\,c}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{bg{c}^{2}{\it Arcsinh} \left ( cx \right ){x}^{4}}{3\,{c}^{2}{x}^{2}+3}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{bgc{x}^{3}}{9}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{2\,bg{\it Arcsinh} \left ( cx \right ){x}^{2}}{3\,{c}^{2}{x}^{2}+3}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{bgx}{3\,c}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{bf{c}^{2}{\it Arcsinh} \left ( cx \right ){x}^{3}}{2\,{c}^{2}{x}^{2}+2}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{bcf{x}^{2}}{4}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{bf{\it Arcsinh} \left ( cx \right ) x}{2\,{c}^{2}{x}^{2}+2}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{bf \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{4\,c}\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((g*x+f)*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2),x)

[Out]

1/3*a*g/c^2/d*(c^2*d*x^2+d)^(3/2)+1/2*a*f*x*(c^2*d*x^2+d)^(1/2)+1/2*a*f*d*ln(x*c^2*d/(c^2*d)^(1/2)+(c^2*d*x^2+
d)^(1/2))/(c^2*d)^(1/2)+1/3*b*(d*(c^2*x^2+1))^(1/2)*g/c^2/(c^2*x^2+1)*arcsinh(c*x)-1/8*b*(d*(c^2*x^2+1))^(1/2)
*f/c/(c^2*x^2+1)^(1/2)+1/3*b*(d*(c^2*x^2+1))^(1/2)*g*c^2/(c^2*x^2+1)*arcsinh(c*x)*x^4-1/9*b*(d*(c^2*x^2+1))^(1
/2)*g*c/(c^2*x^2+1)^(1/2)*x^3+2/3*b*(d*(c^2*x^2+1))^(1/2)*g/(c^2*x^2+1)*arcsinh(c*x)*x^2-1/3*b*(d*(c^2*x^2+1))
^(1/2)*g/c/(c^2*x^2+1)^(1/2)*x+1/2*b*(d*(c^2*x^2+1))^(1/2)*f*c^2/(c^2*x^2+1)*arcsinh(c*x)*x^3-1/4*b*(d*(c^2*x^
2+1))^(1/2)*f*c/(c^2*x^2+1)^(1/2)*x^2+1/2*b*(d*(c^2*x^2+1))^(1/2)*f/(c^2*x^2+1)*arcsinh(c*x)*x+1/4*b*(d*(c^2*x
^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c*f*arcsinh(c*x)^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((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 (\sqrt{c^{2} d x^{2} + d}{\left (a g x + a f +{\left (b g x + b f\right )} \operatorname{arsinh}\left (c x\right )\right )}, 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(sqrt(c^2*d*x^2 + d)*(a*g*x + a*f + (b*g*x + b*f)*arcsinh(c*x)), x)

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{c^{2} d x^{2} + d}{\left (g x + f\right )}{\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((g*x+f)*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

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