3.269 \(\int (d+c^2 d x^2)^{3/2} (a+b \sinh ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=294 \[ \frac {d \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^3}{8 b c \sqrt {c^2 x^2+1}}+\frac {1}{4} x \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {3}{8} d x \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {b d \left (c^2 x^2+1\right )^{3/2} \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{8 c}-\frac {3 b c d x^2 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{8 \sqrt {c^2 x^2+1}}+\frac {15}{64} b^2 d x \sqrt {c^2 d x^2+d}+\frac {1}{32} b^2 d x \left (c^2 x^2+1\right ) \sqrt {c^2 d x^2+d}-\frac {9 b^2 d \sqrt {c^2 d x^2+d} \sinh ^{-1}(c x)}{64 c \sqrt {c^2 x^2+1}} \]

[Out]

1/4*x*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))^2+15/64*b^2*d*x*(c^2*d*x^2+d)^(1/2)+1/32*b^2*d*x*(c^2*x^2+1)*(c^2
*d*x^2+d)^(1/2)-1/8*b*d*(c^2*x^2+1)^(3/2)*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)/c+3/8*d*x*(a+b*arcsinh(c*x))^
2*(c^2*d*x^2+d)^(1/2)-9/64*b^2*d*arcsinh(c*x)*(c^2*d*x^2+d)^(1/2)/c/(c^2*x^2+1)^(1/2)-3/8*b*c*d*x^2*(a+b*arcsi
nh(c*x))*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)+1/8*d*(a+b*arcsinh(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 = 0.25, antiderivative size = 294, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {5684, 5682, 5675, 5661, 321, 215, 5717, 195} \[ \frac {d \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^3}{8 b c \sqrt {c^2 x^2+1}}+\frac {1}{4} x \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {3}{8} d x \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {b d \left (c^2 x^2+1\right )^{3/2} \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{8 c}-\frac {3 b c d x^2 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{8 \sqrt {c^2 x^2+1}}+\frac {15}{64} b^2 d x \sqrt {c^2 d x^2+d}+\frac {1}{32} b^2 d x \left (c^2 x^2+1\right ) \sqrt {c^2 d x^2+d}-\frac {9 b^2 d \sqrt {c^2 d x^2+d} \sinh ^{-1}(c x)}{64 c \sqrt {c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Int[(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x])^2,x]

[Out]

(15*b^2*d*x*Sqrt[d + c^2*d*x^2])/64 + (b^2*d*x*(1 + c^2*x^2)*Sqrt[d + c^2*d*x^2])/32 - (9*b^2*d*Sqrt[d + c^2*d
*x^2]*ArcSinh[c*x])/(64*c*Sqrt[1 + c^2*x^2]) - (3*b*c*d*x^2*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(8*Sqrt[
1 + c^2*x^2]) - (b*d*(1 + c^2*x^2)^(3/2)*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(8*c) + (3*d*x*Sqrt[d + c^2
*d*x^2]*(a + b*ArcSinh[c*x])^2)/8 + (x*(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x])^2)/4 + (d*Sqrt[d + c^2*d*x^2
]*(a + b*ArcSinh[c*x])^3)/(8*b*c*Sqrt[1 + c^2*x^2])

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 215

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 321

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 5661

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 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 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 5684

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(x*(d + e*x^2)^p*
(a + b*ArcSinh[c*x])^n)/(2*p + 1), x] + (Dist[(2*d*p)/(2*p + 1), Int[(d + e*x^2)^(p - 1)*(a + b*ArcSinh[c*x])^
n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/((2*p + 1)*(1 + c^2*x^2)^FracPart[p]), Int[x*(1
+ c^2*x^2)^(p - 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && Gt
Q[n, 0] && GtQ[p, 0]

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 \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx &=\frac {1}{4} x \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{4} (3 d) \int \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx-\frac {\left (b c d \sqrt {d+c^2 d x^2}\right ) \int x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{2 \sqrt {1+c^2 x^2}}\\ &=-\frac {b d \left (1+c^2 x^2\right )^{3/2} \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c}+\frac {3}{8} d x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{4} x \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {\left (3 d \sqrt {d+c^2 d x^2}\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt {1+c^2 x^2}} \, dx}{8 \sqrt {1+c^2 x^2}}+\frac {\left (b^2 d \sqrt {d+c^2 d x^2}\right ) \int \left (1+c^2 x^2\right )^{3/2} \, dx}{8 \sqrt {1+c^2 x^2}}-\frac {\left (3 b c d \sqrt {d+c^2 d x^2}\right ) \int x \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{4 \sqrt {1+c^2 x^2}}\\ &=\frac {1}{32} b^2 d x \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2}-\frac {3 b c d x^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 \sqrt {1+c^2 x^2}}-\frac {b d \left (1+c^2 x^2\right )^{3/2} \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c}+\frac {3}{8} d x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{4} x \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{8 b c \sqrt {1+c^2 x^2}}+\frac {\left (3 b^2 d \sqrt {d+c^2 d x^2}\right ) \int \sqrt {1+c^2 x^2} \, dx}{32 \sqrt {1+c^2 x^2}}+\frac {\left (3 b^2 c^2 d \sqrt {d+c^2 d x^2}\right ) \int \frac {x^2}{\sqrt {1+c^2 x^2}} \, dx}{8 \sqrt {1+c^2 x^2}}\\ &=\frac {15}{64} b^2 d x \sqrt {d+c^2 d x^2}+\frac {1}{32} b^2 d x \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2}-\frac {3 b c d x^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 \sqrt {1+c^2 x^2}}-\frac {b d \left (1+c^2 x^2\right )^{3/2} \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c}+\frac {3}{8} d x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{4} x \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{8 b c \sqrt {1+c^2 x^2}}+\frac {\left (3 b^2 d \sqrt {d+c^2 d x^2}\right ) \int \frac {1}{\sqrt {1+c^2 x^2}} \, dx}{64 \sqrt {1+c^2 x^2}}-\frac {\left (3 b^2 d \sqrt {d+c^2 d x^2}\right ) \int \frac {1}{\sqrt {1+c^2 x^2}} \, dx}{16 \sqrt {1+c^2 x^2}}\\ &=\frac {15}{64} b^2 d x \sqrt {d+c^2 d x^2}+\frac {1}{32} b^2 d x \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2}-\frac {9 b^2 d \sqrt {d+c^2 d x^2} \sinh ^{-1}(c x)}{64 c \sqrt {1+c^2 x^2}}-\frac {3 b c d x^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 \sqrt {1+c^2 x^2}}-\frac {b d \left (1+c^2 x^2\right )^{3/2} \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c}+\frac {3}{8} d x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{4} x \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{8 b c \sqrt {1+c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 1.91, size = 329, normalized size = 1.12 \[ \frac {288 a^2 d^{3/2} \sqrt {c^2 x^2+1} \log \left (\sqrt {d} \sqrt {c^2 d x^2+d}+c d x\right )+96 a^2 c d x \sqrt {c^2 x^2+1} \left (2 c^2 x^2+5\right ) \sqrt {c^2 d x^2+d}-192 a b d \sqrt {c^2 d x^2+d} \left (\cosh \left (2 \sinh ^{-1}(c x)\right )-2 \sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)+\sinh \left (2 \sinh ^{-1}(c x)\right )\right )\right )-12 a b d \sqrt {c^2 d x^2+d} \left (8 \sinh ^{-1}(c x)^2-4 \sinh \left (4 \sinh ^{-1}(c x)\right ) \sinh ^{-1}(c x)+\cosh \left (4 \sinh ^{-1}(c x)\right )\right )+32 b^2 d \sqrt {c^2 d x^2+d} \left (4 \sinh ^{-1}(c x)^3+\left (6 \sinh ^{-1}(c x)^2+3\right ) \sinh \left (2 \sinh ^{-1}(c x)\right )-6 \sinh ^{-1}(c x) \cosh \left (2 \sinh ^{-1}(c x)\right )\right )-b^2 d \sqrt {c^2 d x^2+d} \left (32 \sinh ^{-1}(c x)^3-3 \left (8 \sinh ^{-1}(c x)^2+1\right ) \sinh \left (4 \sinh ^{-1}(c x)\right )+12 \sinh ^{-1}(c x) \cosh \left (4 \sinh ^{-1}(c x)\right )\right )}{768 c \sqrt {c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x])^2,x]

[Out]

(96*a^2*c*d*x*Sqrt[1 + c^2*x^2]*(5 + 2*c^2*x^2)*Sqrt[d + c^2*d*x^2] + 288*a^2*d^(3/2)*Sqrt[1 + c^2*x^2]*Log[c*
d*x + Sqrt[d]*Sqrt[d + c^2*d*x^2]] + 32*b^2*d*Sqrt[d + c^2*d*x^2]*(4*ArcSinh[c*x]^3 - 6*ArcSinh[c*x]*Cosh[2*Ar
cSinh[c*x]] + (3 + 6*ArcSinh[c*x]^2)*Sinh[2*ArcSinh[c*x]]) - 192*a*b*d*Sqrt[d + c^2*d*x^2]*(Cosh[2*ArcSinh[c*x
]] - 2*ArcSinh[c*x]*(ArcSinh[c*x] + Sinh[2*ArcSinh[c*x]])) - 12*a*b*d*Sqrt[d + c^2*d*x^2]*(8*ArcSinh[c*x]^2 +
Cosh[4*ArcSinh[c*x]] - 4*ArcSinh[c*x]*Sinh[4*ArcSinh[c*x]]) - b^2*d*Sqrt[d + c^2*d*x^2]*(32*ArcSinh[c*x]^3 + 1
2*ArcSinh[c*x]*Cosh[4*ArcSinh[c*x]] - 3*(1 + 8*ArcSinh[c*x]^2)*Sinh[4*ArcSinh[c*x]]))/(768*c*Sqrt[1 + c^2*x^2]
)

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fricas [F]  time = 0.79, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{2} c^{2} d x^{2} + a^{2} d + {\left (b^{2} c^{2} d x^{2} + b^{2} d\right )} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, {\left (a b c^{2} d x^{2} + a b d\right )} \operatorname {arsinh}\left (c x\right )\right )} \sqrt {c^{2} d x^{2} + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))^2,x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [B]  time = 0.28, size = 709, normalized size = 2.41 \[ \frac {x \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}} a^{2}}{4}+\frac {3 a^{2} d x \sqrt {c^{2} d \,x^{2}+d}}{8}+\frac {3 a^{2} d^{2} \ln \left (\frac {x \,c^{2} d}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{8 \sqrt {c^{2} d}}+\frac {5 b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}\, d \arcsinh \left (c x \right )^{2} x}{8 \left (c^{2} x^{2}+1\right )}-\frac {17 b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}\, d \arcsinh \left (c x \right )}{64 c \sqrt {c^{2} x^{2}+1}}+\frac {b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}\, d \,c^{4} \arcsinh \left (c x \right )^{2} x^{5}}{4 c^{2} x^{2}+4}+\frac {7 b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}\, d \,c^{2} \arcsinh \left (c x \right )^{2} x^{3}}{8 \left (c^{2} x^{2}+1\right )}+\frac {b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}\, d \,c^{4} x^{5}}{32 c^{2} x^{2}+32}+\frac {19 b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}\, d \,c^{2} x^{3}}{64 \left (c^{2} x^{2}+1\right )}+\frac {17 b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}\, d x}{64 \left (c^{2} x^{2}+1\right )}+\frac {b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )^{3} d}{8 \sqrt {c^{2} x^{2}+1}\, c}-\frac {b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}\, d \,c^{3} \arcsinh \left (c x \right ) x^{4}}{8 \sqrt {c^{2} x^{2}+1}}-\frac {5 b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}\, d c \arcsinh \left (c x \right ) x^{2}}{8 \sqrt {c^{2} x^{2}+1}}+\frac {7 a b \sqrt {d \left (c^{2} x^{2}+1\right )}\, d \,c^{2} \arcsinh \left (c x \right ) x^{3}}{4 \left (c^{2} x^{2}+1\right )}-\frac {5 a b \sqrt {d \left (c^{2} x^{2}+1\right )}\, d c \,x^{2}}{8 \sqrt {c^{2} x^{2}+1}}+\frac {5 a b \sqrt {d \left (c^{2} x^{2}+1\right )}\, d \arcsinh \left (c x \right ) x}{4 \left (c^{2} x^{2}+1\right )}+\frac {a b \sqrt {d \left (c^{2} x^{2}+1\right )}\, d \,c^{4} \arcsinh \left (c x \right ) x^{5}}{2 c^{2} x^{2}+2}-\frac {a b \sqrt {d \left (c^{2} x^{2}+1\right )}\, d \,c^{3} x^{4}}{8 \sqrt {c^{2} x^{2}+1}}+\frac {3 a b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )^{2} d}{8 \sqrt {c^{2} x^{2}+1}\, c}-\frac {17 a b \sqrt {d \left (c^{2} x^{2}+1\right )}\, d}{64 c \sqrt {c^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))^2,x)

[Out]

1/4*x*(c^2*d*x^2+d)^(3/2)*a^2+3/8*a^2*d*x*(c^2*d*x^2+d)^(1/2)+3/8*a^2*d^2*ln(x*c^2*d/(c^2*d)^(1/2)+(c^2*d*x^2+
d)^(1/2))/(c^2*d)^(1/2)+5/8*b^2*(d*(c^2*x^2+1))^(1/2)*d/(c^2*x^2+1)*arcsinh(c*x)^2*x-17/64*b^2*(d*(c^2*x^2+1))
^(1/2)*d/c/(c^2*x^2+1)^(1/2)*arcsinh(c*x)+1/4*b^2*(d*(c^2*x^2+1))^(1/2)*d*c^4/(c^2*x^2+1)*arcsinh(c*x)^2*x^5+7
/8*b^2*(d*(c^2*x^2+1))^(1/2)*d*c^2/(c^2*x^2+1)*arcsinh(c*x)^2*x^3+1/32*b^2*(d*(c^2*x^2+1))^(1/2)*d*c^4/(c^2*x^
2+1)*x^5+19/64*b^2*(d*(c^2*x^2+1))^(1/2)*d*c^2/(c^2*x^2+1)*x^3+17/64*b^2*(d*(c^2*x^2+1))^(1/2)*d/(c^2*x^2+1)*x
+1/8*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c*arcsinh(c*x)^3*d-1/8*b^2*(d*(c^2*x^2+1))^(1/2)*d*c^3/(c^2*x
^2+1)^(1/2)*arcsinh(c*x)*x^4-5/8*b^2*(d*(c^2*x^2+1))^(1/2)*d*c/(c^2*x^2+1)^(1/2)*arcsinh(c*x)*x^2+7/4*a*b*(d*(
c^2*x^2+1))^(1/2)*d*c^2/(c^2*x^2+1)*arcsinh(c*x)*x^3-5/8*a*b*(d*(c^2*x^2+1))^(1/2)*d*c/(c^2*x^2+1)^(1/2)*x^2+5
/4*a*b*(d*(c^2*x^2+1))^(1/2)*d/(c^2*x^2+1)*arcsinh(c*x)*x+1/2*a*b*(d*(c^2*x^2+1))^(1/2)*d*c^4/(c^2*x^2+1)*arcs
inh(c*x)*x^5-1/8*a*b*(d*(c^2*x^2+1))^(1/2)*d*c^3/(c^2*x^2+1)^(1/2)*x^4+3/8*a*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+
1)^(1/2)/c*arcsinh(c*x)^2*d-17/64*a*b*(d*(c^2*x^2+1))^(1/2)*d/c/(c^2*x^2+1)^(1/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))^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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,{\left (d\,c^2\,x^2+d\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c*x))^2*(d + c^2*d*x^2)^(3/2),x)

[Out]

int((a + b*asinh(c*x))^2*(d + c^2*d*x^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c**2*d*x**2+d)**(3/2)*(a+b*asinh(c*x))**2,x)

[Out]

Integral((d*(c**2*x**2 + 1))**(3/2)*(a + b*asinh(c*x))**2, x)

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