3.1.48 \(\int \frac {(f+g x)^2 (a+b \sinh ^{-1}(c x))}{\sqrt {d+c^2 d x^2}} \, dx\) [48]

Optimal. Leaf size=258 \[ -\frac {2 b f g x \sqrt {1+c^2 x^2}}{c \sqrt {d+c^2 d x^2}}-\frac {b g^2 x^2 \sqrt {1+c^2 x^2}}{4 c \sqrt {d+c^2 d x^2}}+\frac {2 f g \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^2 \sqrt {d+c^2 d x^2}}+\frac {g^2 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 \sqrt {d+c^2 d x^2}}+\frac {f^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c \sqrt {d+c^2 d x^2}}-\frac {g^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c^3 \sqrt {d+c^2 d x^2}} \]

[Out]

2*f*g*(c^2*x^2+1)*(a+b*arcsinh(c*x))/c^2/(c^2*d*x^2+d)^(1/2)+1/2*g^2*x*(c^2*x^2+1)*(a+b*arcsinh(c*x))/c^2/(c^2
*d*x^2+d)^(1/2)-2*b*f*g*x*(c^2*x^2+1)^(1/2)/c/(c^2*d*x^2+d)^(1/2)-1/4*b*g^2*x^2*(c^2*x^2+1)^(1/2)/c/(c^2*d*x^2
+d)^(1/2)+1/2*f^2*(a+b*arcsinh(c*x))^2*(c^2*x^2+1)^(1/2)/b/c/(c^2*d*x^2+d)^(1/2)-1/4*g^2*(a+b*arcsinh(c*x))^2*
(c^2*x^2+1)^(1/2)/b/c^3/(c^2*d*x^2+d)^(1/2)

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Rubi [A]
time = 0.28, antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {5845, 5838, 5783, 5798, 8, 5812, 30} \begin {gather*} \frac {f^2 \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 {2 f 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 {g^2 x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 \sqrt {c^2 d x^2+d}}-\frac {g^2 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c^3 \sqrt {c^2 d x^2+d}}-\frac {2 b f g x \sqrt {c^2 x^2+1}}{c \sqrt {c^2 d x^2+d}}-\frac {b g^2 x^2 \sqrt {c^2 x^2+1}}{4 c \sqrt {c^2 d x^2+d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 5812

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

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

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Mathematica [A]
time = 0.36, size = 233, normalized size = 0.90 \begin {gather*} \frac {4 c \sqrt {d} g \left (-4 b c f x \sqrt {1+c^2 x^2}+a (4 f+g x) \left (1+c^2 x^2\right )\right )+4 b c \sqrt {d} g (4 f+g x) \left (1+c^2 x^2\right ) \sinh ^{-1}(c x)+2 b \sqrt {d} \left (2 c^2 f^2-g^2\right ) \sqrt {1+c^2 x^2} \sinh ^{-1}(c x)^2-b \sqrt {d} g^2 \sqrt {1+c^2 x^2} \cosh \left (2 \sinh ^{-1}(c x)\right )+4 a \left (2 c^2 f^2-g^2\right ) \sqrt {d+c^2 d x^2} \log \left (c d x+\sqrt {d} \sqrt {d+c^2 d x^2}\right )}{8 c^3 \sqrt {d} \sqrt {d+c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*c*Sqrt[d]*g*(-4*b*c*f*x*Sqrt[1 + c^2*x^2] + a*(4*f + g*x)*(1 + c^2*x^2)) + 4*b*c*Sqrt[d]*g*(4*f + g*x)*(1 +
c^2*x^2)*ArcSinh[c*x] + 2*b*Sqrt[d]*(2*c^2*f^2 - g^2)*Sqrt[1 + c^2*x^2]*ArcSinh[c*x]^2 - b*Sqrt[d]*g^2*Sqrt[1
+ c^2*x^2]*Cosh[2*ArcSinh[c*x]] + 4*a*(2*c^2*f^2 - g^2)*Sqrt[d + c^2*d*x^2]*Log[c*d*x + Sqrt[d]*Sqrt[d + c^2*
d*x^2]])/(8*c^3*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. \(485\) vs. \(2(230)=460\).
time = 6.14, size = 486, normalized size = 1.88

method	result	size
default	\(\frac {a \,g^{2} x \sqrt {c^{2} d \,x^{2}+d}}{2 c^{2} d}-\frac {a \,g^{2} \ln \left (\frac {x \,c^{2} d}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{2 c^{2} \sqrt {c^{2} d}}+\frac {2 a f g \sqrt {c^{2} d \,x^{2}+d}}{c^{2} d}+\frac {a \,f^{2} \ln \left (\frac {x \,c^{2} d}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{\sqrt {c^{2} d}}+b \left (\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )^{2} \left (2 c^{2} f^{2}-g^{2}\right )}{4 \sqrt {c^{2} x^{2}+1}\, c^{3} d}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 c^{3} x^{3}+2 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+2 c x +\sqrt {c^{2} x^{2}+1}\right ) g^{2} \left (2 \arcsinh \left (c x \right )-1\right )}{16 c^{3} d \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (c^{2} x^{2}+\sqrt {c^{2} x^{2}+1}\, c x +1\right ) f g \left (\arcsinh \left (c x \right )-1\right )}{c^{2} d \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (c^{2} x^{2}-\sqrt {c^{2} x^{2}+1}\, c x +1\right ) f g \left (\arcsinh \left (c x \right )+1\right )}{c^{2} d \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 c^{3} x^{3}-2 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+2 c x -\sqrt {c^{2} x^{2}+1}\right ) g^{2} \left (2 \arcsinh \left (c x \right )+1\right )}{16 c^{3} d \left (c^{2} x^{2}+1\right )}\right )\)	\(486\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*g^2*x/c^2/d*(c^2*d*x^2+d)^(1/2)-1/2*a*g^2/c^2*ln(x*c^2*d/(c^2*d)^(1/2)+(c^2*d*x^2+d)^(1/2))/(c^2*d)^(1/2
)+2*a*f*g/c^2/d*(c^2*d*x^2+d)^(1/2)+a*f^2*ln(x*c^2*d/(c^2*d)^(1/2)+(c^2*d*x^2+d)^(1/2))/(c^2*d)^(1/2)+b*(1/4*(
d*(c^2*x^2+1))^(1/2)*arcsinh(c*x)^2*(2*c^2*f^2-g^2)/(c^2*x^2+1)^(1/2)/c^3/d+1/16*(d*(c^2*x^2+1))^(1/2)*(2*c^3*
x^3+2*c^2*x^2*(c^2*x^2+1)^(1/2)+2*c*x+(c^2*x^2+1)^(1/2))*g^2*(2*arcsinh(c*x)-1)/c^3/d/(c^2*x^2+1)+(d*(c^2*x^2+
1))^(1/2)*(c^2*x^2+(c^2*x^2+1)^(1/2)*c*x+1)*f*g*(arcsinh(c*x)-1)/c^2/d/(c^2*x^2+1)+(d*(c^2*x^2+1))^(1/2)*(c^2*
x^2-(c^2*x^2+1)^(1/2)*c*x+1)*f*g*(arcsinh(c*x)+1)/c^2/d/(c^2*x^2+1)+1/16*(d*(c^2*x^2+1))^(1/2)*(2*c^3*x^3-2*c^
2*x^2*(c^2*x^2+1)^(1/2)+2*c*x-(c^2*x^2+1)^(1/2))*g^2*(2*arcsinh(c*x)+1)/c^3/d/(c^2*x^2+1))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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

[Out]

integral((a*g^2*x^2 + 2*a*f*g*x + a*f^2 + (b*g^2*x^2 + 2*b*f*g*x + b*f^2)*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 )^{2}}{\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)**2*(a+b*asinh(c*x))/(c**2*d*x**2+d)**(1/2),x)

[Out]

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

[Out]

integrate((g*x + f)^2*(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 = -0.00 \begin {gather*} \int \frac {{\left (f+g\,x\right )}^2\,\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)^2*(a + b*asinh(c*x)))/(d + c^2*d*x^2)^(1/2),x)

[Out]

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

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