3.3.61 \(\int \sqrt {d+c^2 d x^2} (a+b \sinh ^{-1}(c x))^2 \, dx\) [261]

Optimal. Leaf size=184 \[ \frac {1}{4} b^2 x \sqrt {d+c^2 d x^2}-\frac {b^2 \sqrt {d+c^2 d x^2} \sinh ^{-1}(c x)}{4 c \sqrt {1+c^2 x^2}}-\frac {b c x^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt {1+c^2 x^2}}+\frac {1}{2} x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {\sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b c \sqrt {1+c^2 x^2}} \]

[Out]

1/4*b^2*x*(c^2*d*x^2+d)^(1/2)+1/2*x*(a+b*arcsinh(c*x))^2*(c^2*d*x^2+d)^(1/2)-1/4*b^2*arcsinh(c*x)*(c^2*d*x^2+d
)^(1/2)/c/(c^2*x^2+1)^(1/2)-1/2*b*c*x^2*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)+1/6*(a+b*arcs
inh(c*x))^3*(c^2*d*x^2+d)^(1/2)/b/c/(c^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5785, 5783, 5776, 327, 221} \begin {gather*} \frac {\sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b c \sqrt {c^2 x^2+1}}+\frac {1}{2} x \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {b c x^2 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt {c^2 x^2+1}}+\frac {1}{4} b^2 x \sqrt {c^2 d x^2+d}-\frac {b^2 \sqrt {c^2 d x^2+d} \sinh ^{-1}(c x)}{4 c \sqrt {c^2 x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*x*Sqrt[d + c^2*d*x^2])/4 - (b^2*Sqrt[d + c^2*d*x^2]*ArcSinh[c*x])/(4*c*Sqrt[1 + c^2*x^2]) - (b*c*x^2*Sqrt
[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(2*Sqrt[1 + c^2*x^2]) + (x*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^2)/2
+ (Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^3)/(6*b*c*Sqrt[1 + c^2*x^2])

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 5776

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcS
inh[c*x])^n/(d*(m + 1))), x] - Dist[b*c*(n/(d*(m + 1))), Int[(d*x)^(m + 1)*((a + b*ArcSinh[c*x])^(n - 1)/Sqrt[
1 + c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && 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 5785

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[(1/2)*Simp[Sqrt[d + e*x^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/2)*Simp[Sqrt[d + e*x^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]

Rubi steps

\begin {align*} \int \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx &=\frac {1}{2} x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {\sqrt {d+c^2 d x^2} \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt {1+c^2 x^2}} \, dx}{2 \sqrt {1+c^2 x^2}}-\frac {\left (b c \sqrt {d+c^2 d x^2}\right ) \int x \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\sqrt {1+c^2 x^2}}\\ &=-\frac {b c x^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt {1+c^2 x^2}}+\frac {1}{2} x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {\sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b c \sqrt {1+c^2 x^2}}+\frac {\left (b^2 c^2 \sqrt {d+c^2 d x^2}\right ) \int \frac {x^2}{\sqrt {1+c^2 x^2}} \, dx}{2 \sqrt {1+c^2 x^2}}\\ &=\frac {1}{4} b^2 x \sqrt {d+c^2 d x^2}-\frac {b c x^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt {1+c^2 x^2}}+\frac {1}{2} x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {\sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b c \sqrt {1+c^2 x^2}}-\frac {\left (b^2 \sqrt {d+c^2 d x^2}\right ) \int \frac {1}{\sqrt {1+c^2 x^2}} \, dx}{4 \sqrt {1+c^2 x^2}}\\ &=\frac {1}{4} b^2 x \sqrt {d+c^2 d x^2}-\frac {b^2 \sqrt {d+c^2 d x^2} \sinh ^{-1}(c x)}{4 c \sqrt {1+c^2 x^2}}-\frac {b c x^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt {1+c^2 x^2}}+\frac {1}{2} x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {\sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b c \sqrt {1+c^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.74, size = 200, normalized size = 1.09 \begin {gather*} \frac {1}{24} \left (12 a^2 x \sqrt {d+c^2 d x^2}+\frac {12 a^2 \sqrt {d} \log \left (c d x+\sqrt {d} \sqrt {d+c^2 d x^2}\right )}{c}+\frac {b^2 \sqrt {d+c^2 d x^2} \left (4 \sinh ^{-1}(c x)^3-6 \sinh ^{-1}(c x) \cosh \left (2 \sinh ^{-1}(c x)\right )+\left (3+6 \sinh ^{-1}(c x)^2\right ) \sinh \left (2 \sinh ^{-1}(c x)\right )\right )}{c \sqrt {1+c^2 x^2}}+\frac {6 a b \sqrt {d+c^2 d x^2} \left (-\cosh \left (2 \sinh ^{-1}(c x)\right )+2 \sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)+\sinh \left (2 \sinh ^{-1}(c x)\right )\right )\right )}{c \sqrt {1+c^2 x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(12*a^2*x*Sqrt[d + c^2*d*x^2] + (12*a^2*Sqrt[d]*Log[c*d*x + Sqrt[d]*Sqrt[d + c^2*d*x^2]])/c + (b^2*Sqrt[d + c^
2*d*x^2]*(4*ArcSinh[c*x]^3 - 6*ArcSinh[c*x]*Cosh[2*ArcSinh[c*x]] + (3 + 6*ArcSinh[c*x]^2)*Sinh[2*ArcSinh[c*x]]
))/(c*Sqrt[1 + c^2*x^2]) + (6*a*b*Sqrt[d + c^2*d*x^2]*(-Cosh[2*ArcSinh[c*x]] + 2*ArcSinh[c*x]*(ArcSinh[c*x] +
Sinh[2*ArcSinh[c*x]])))/(c*Sqrt[1 + c^2*x^2]))/24

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(479\) vs. \(2(158)=316\).
time = 1.67, size = 480, normalized size = 2.61

method	result	size
default	\(\frac {a^{2} x \sqrt {c^{2} d \,x^{2}+d}}{2}+\frac {a^{2} d \ln \left (\frac {x \,c^{2} d}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{2 \sqrt {c^{2} d}}+b^{2} \left (\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )^{3}}{6 \sqrt {c^{2} x^{2}+1}\, c}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 c^{3} x^{3}+2 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+2 c x +\sqrt {c^{2} x^{2}+1}\right ) \left (2 \arcsinh \left (c x \right )^{2}-2 \arcsinh \left (c x \right )+1\right )}{16 c \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 c^{3} x^{3}-2 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+2 c x -\sqrt {c^{2} x^{2}+1}\right ) \left (2 \arcsinh \left (c x \right )^{2}+2 \arcsinh \left (c x \right )+1\right )}{16 c \left (c^{2} x^{2}+1\right )}\right )+2 a b \left (\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )^{2}}{4 \sqrt {c^{2} x^{2}+1}\, c}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 c^{3} x^{3}+2 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+2 c x +\sqrt {c^{2} x^{2}+1}\right ) \left (-1+2 \arcsinh \left (c x \right )\right )}{16 c \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 c^{3} x^{3}-2 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+2 c x -\sqrt {c^{2} x^{2}+1}\right ) \left (1+2 \arcsinh \left (c x \right )\right )}{16 c \left (c^{2} x^{2}+1\right )}\right )\)	\(480\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*x*(c^2*d*x^2+d)^(1/2)+1/2*a^2*d*ln(x*c^2*d/(c^2*d)^(1/2)+(c^2*d*x^2+d)^(1/2))/(c^2*d)^(1/2)+b^2*(1/6*(
d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c*arcsinh(c*x)^3+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))*(2*arcsinh(c*x)^2-2*arcsinh(c*x)+1)/c/(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))*(2*arcsinh(c*x)^2+2*arcsinh(c*x)+1)/c/(c^2
*x^2+1))+2*a*b*(1/4*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c*arcsinh(c*x)^2+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))*(-1+2*arcsinh(c*x))/c/(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))*(1+2*arcsinh(c*x))/c/(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((a+b*arcsinh(c*x))^2*(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((a+b*arcsinh(c*x))^2*(c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(d*(c**2*x**2 + 1))*(a + b*asinh(c*x))**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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