3.1.41 \(\int (f+g x) (d+c^2 d x^2)^{3/2} (a+b \sinh ^{-1}(c x)) \, dx\) [41]

Optimal. Leaf size=353 \[ -\frac {b d g x \sqrt {d+c^2 d x^2}}{5 c \sqrt {1+c^2 x^2}}-\frac {5 b c d f x^2 \sqrt {d+c^2 d x^2}}{16 \sqrt {1+c^2 x^2}}-\frac {2 b c d g x^3 \sqrt {d+c^2 d x^2}}{15 \sqrt {1+c^2 x^2}}-\frac {b c^3 d f x^4 \sqrt {d+c^2 d x^2}}{16 \sqrt {1+c^2 x^2}}-\frac {b c^3 d g x^5 \sqrt {d+c^2 d x^2}}{25 \sqrt {1+c^2 x^2}}+\frac {3}{8} d f x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{4} d f x \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {d g \left (1+c^2 x^2\right )^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^2}+\frac {3 d f \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c \sqrt {1+c^2 x^2}} \]

[Out]

3/8*d*f*x*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)+1/4*d*f*x*(c^2*x^2+1)*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)+
1/5*d*g*(c^2*x^2+1)^2*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)/c^2-1/5*b*d*g*x*(c^2*d*x^2+d)^(1/2)/c/(c^2*x^2+1)
^(1/2)-5/16*b*c*d*f*x^2*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-2/15*b*c*d*g*x^3*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)
^(1/2)-1/16*b*c^3*d*f*x^4*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-1/25*b*c^3*d*g*x^5*(c^2*d*x^2+d)^(1/2)/(c^2*x^
2+1)^(1/2)+3/16*d*f*(a+b*arcsinh(c*x))^2*(c^2*d*x^2+d)^(1/2)/b/c/(c^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.23, antiderivative size = 353, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {5845, 5838, 5786, 5785, 5783, 30, 14, 5798, 200} \begin {gather*} \frac {3}{8} d f x \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{4} d f x \left (c^2 x^2+1\right ) \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac {3 d f \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c \sqrt {c^2 x^2+1}}+\frac {d g \left (c^2 x^2+1\right )^2 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^2}-\frac {5 b c d f x^2 \sqrt {c^2 d x^2+d}}{16 \sqrt {c^2 x^2+1}}-\frac {b d g x \sqrt {c^2 d x^2+d}}{5 c \sqrt {c^2 x^2+1}}-\frac {2 b c d g x^3 \sqrt {c^2 d x^2+d}}{15 \sqrt {c^2 x^2+1}}-\frac {b c^3 d f x^4 \sqrt {c^2 d x^2+d}}{16 \sqrt {c^2 x^2+1}}-\frac {b c^3 d g x^5 \sqrt {c^2 d x^2+d}}{25 \sqrt {c^2 x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*(b*d*g*x*Sqrt[d + c^2*d*x^2])/(c*Sqrt[1 + c^2*x^2]) - (5*b*c*d*f*x^2*Sqrt[d + c^2*d*x^2])/(16*Sqrt[1 + c^
2*x^2]) - (2*b*c*d*g*x^3*Sqrt[d + c^2*d*x^2])/(15*Sqrt[1 + c^2*x^2]) - (b*c^3*d*f*x^4*Sqrt[d + c^2*d*x^2])/(16
*Sqrt[1 + c^2*x^2]) - (b*c^3*d*g*x^5*Sqrt[d + c^2*d*x^2])/(25*Sqrt[1 + c^2*x^2]) + (3*d*f*x*Sqrt[d + c^2*d*x^2
]*(a + b*ArcSinh[c*x]))/8 + (d*f*x*(1 + c^2*x^2)*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/4 + (d*g*(1 + c^2*x
^2)^2*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(5*c^2) + (3*d*f*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^2)/(
16*b*c*Sqrt[1 + c^2*x^2])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

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

Rule 200

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

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]

Rule 5786

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/(2*p + 1))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p], Int[x*(1 + c^2*x^2)^(p - 1/2)*(a + b*A
rcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && GtQ[p, 0]

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 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 (f+g x) \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac {\left (d \sqrt {d+c^2 d x^2}\right ) \int (f+g x) \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\sqrt {1+c^2 x^2}}\\ &=\frac {\left (d \sqrt {d+c^2 d x^2}\right ) \int \left (f \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+g x \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )\right ) \, dx}{\sqrt {1+c^2 x^2}}\\ &=\frac {\left (d f \sqrt {d+c^2 d x^2}\right ) \int \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\sqrt {1+c^2 x^2}}+\frac {\left (d g \sqrt {d+c^2 d x^2}\right ) \int x \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\sqrt {1+c^2 x^2}}\\ &=\frac {1}{4} d f x \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {d g \left (1+c^2 x^2\right )^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^2}+\frac {\left (3 d 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}{4 \sqrt {1+c^2 x^2}}-\frac {\left (b c d f \sqrt {d+c^2 d x^2}\right ) \int x \left (1+c^2 x^2\right ) \, dx}{4 \sqrt {1+c^2 x^2}}-\frac {\left (b d g \sqrt {d+c^2 d x^2}\right ) \int \left (1+c^2 x^2\right )^2 \, dx}{5 c \sqrt {1+c^2 x^2}}\\ &=\frac {3}{8} d f x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{4} d f x \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {d g \left (1+c^2 x^2\right )^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^2}+\frac {\left (3 d f \sqrt {d+c^2 d x^2}\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}} \, dx}{8 \sqrt {1+c^2 x^2}}-\frac {\left (b c d f \sqrt {d+c^2 d x^2}\right ) \int \left (x+c^2 x^3\right ) \, dx}{4 \sqrt {1+c^2 x^2}}-\frac {\left (3 b c d f \sqrt {d+c^2 d x^2}\right ) \int x \, dx}{8 \sqrt {1+c^2 x^2}}-\frac {\left (b d g \sqrt {d+c^2 d x^2}\right ) \int \left (1+2 c^2 x^2+c^4 x^4\right ) \, dx}{5 c \sqrt {1+c^2 x^2}}\\ &=-\frac {b d g x \sqrt {d+c^2 d x^2}}{5 c \sqrt {1+c^2 x^2}}-\frac {5 b c d f x^2 \sqrt {d+c^2 d x^2}}{16 \sqrt {1+c^2 x^2}}-\frac {2 b c d g x^3 \sqrt {d+c^2 d x^2}}{15 \sqrt {1+c^2 x^2}}-\frac {b c^3 d f x^4 \sqrt {d+c^2 d x^2}}{16 \sqrt {1+c^2 x^2}}-\frac {b c^3 d g x^5 \sqrt {d+c^2 d x^2}}{25 \sqrt {1+c^2 x^2}}+\frac {3}{8} d f x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{4} d f x \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {d g \left (1+c^2 x^2\right )^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^2}+\frac {3 d f \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c \sqrt {1+c^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.82, size = 392, normalized size = 1.11 \begin {gather*} \frac {-640 b c d g x \left (3+c^2 x^2\right ) \sqrt {d+c^2 d x^2}-128 b c^3 d g x^3 \left (5+3 c^2 x^2\right ) \sqrt {d+c^2 d x^2}+240 a d \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2} \left (8 g \left (1+c^2 x^2\right )^2+5 c^2 f x \left (5+2 c^2 x^2\right )\right )+3200 b d g \left (1+c^2 x^2\right )^{3/2} \sqrt {d+c^2 d x^2} \sinh ^{-1}(c x)+640 b d g \left (1+c^2 x^2\right )^{3/2} \left (-2+3 c^2 x^2\right ) \sqrt {d+c^2 d x^2} \sinh ^{-1}(c x)-1200 b c d f \sqrt {d+c^2 d x^2} \cosh \left (2 \sinh ^{-1}(c x)\right )+3600 a c d^{3/2} f \sqrt {1+c^2 x^2} \log \left (c d x+\sqrt {d} \sqrt {d+c^2 d x^2}\right )+2400 b c d f \sqrt {d+c^2 d x^2} \sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)+\sinh \left (2 \sinh ^{-1}(c x)\right )\right )-75 b c d f \sqrt {d+c^2 d x^2} \left (8 \sinh ^{-1}(c x)^2+\cosh \left (4 \sinh ^{-1}(c x)\right )-4 \sinh ^{-1}(c x) \sinh \left (4 \sinh ^{-1}(c x)\right )\right )}{9600 c^2 \sqrt {1+c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-640*b*c*d*g*x*(3 + c^2*x^2)*Sqrt[d + c^2*d*x^2] - 128*b*c^3*d*g*x^3*(5 + 3*c^2*x^2)*Sqrt[d + c^2*d*x^2] + 24
0*a*d*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2]*(8*g*(1 + c^2*x^2)^2 + 5*c^2*f*x*(5 + 2*c^2*x^2)) + 3200*b*d*g*(1
+ c^2*x^2)^(3/2)*Sqrt[d + c^2*d*x^2]*ArcSinh[c*x] + 640*b*d*g*(1 + c^2*x^2)^(3/2)*(-2 + 3*c^2*x^2)*Sqrt[d + c^
2*d*x^2]*ArcSinh[c*x] - 1200*b*c*d*f*Sqrt[d + c^2*d*x^2]*Cosh[2*ArcSinh[c*x]] + 3600*a*c*d^(3/2)*f*Sqrt[1 + c^
2*x^2]*Log[c*d*x + Sqrt[d]*Sqrt[d + c^2*d*x^2]] + 2400*b*c*d*f*Sqrt[d + c^2*d*x^2]*ArcSinh[c*x]*(ArcSinh[c*x]
+ Sinh[2*ArcSinh[c*x]]) - 75*b*c*d*f*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]]))/(9600*c^2*Sqrt[1 + c^2*x^2])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1064\) vs. \(2(305)=610\).
time = 3.16, size = 1065, normalized size = 3.02

method	result	size
default	\(\frac {a g \left (c^{2} d \,x^{2}+d \right )^{\frac {5}{2}}}{5 c^{2} d}+\frac {a f x \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{4}+\frac {3 a f d x \sqrt {c^{2} d \,x^{2}+d}}{8}+\frac {3 a f \,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}}+b \left (\frac {3 \sqrt {d \left (c^{2} x^{2}+1\right )}\, f \arcsinh \left (c x \right )^{2} d}{16 \sqrt {c^{2} x^{2}+1}\, c}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (16 x^{6} c^{6}+16 \sqrt {c^{2} x^{2}+1}\, x^{5} c^{5}+28 c^{4} x^{4}+20 \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}+13 c^{2} x^{2}+5 \sqrt {c^{2} x^{2}+1}\, c x +1\right ) g \left (-1+5 \arcsinh \left (c x \right )\right ) d}{800 c^{2} \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (8 c^{5} x^{5}+8 \sqrt {c^{2} x^{2}+1}\, c^{4} x^{4}+12 c^{3} x^{3}+8 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+4 c x +\sqrt {c^{2} x^{2}+1}\right ) f \left (-1+4 \arcsinh \left (c x \right )\right ) d}{256 c \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (4 c^{4} x^{4}+4 \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}+5 c^{2} x^{2}+3 \sqrt {c^{2} x^{2}+1}\, c x +1\right ) g \left (-1+3 \arcsinh \left (c x \right )\right ) d}{96 c^{2} \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 ) f \left (2 \arcsinh \left (c x \right )-1\right ) d}{16 c \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 ) g \left (\arcsinh \left (c x \right )-1\right ) d}{16 c^{2} \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 ) g \left (\arcsinh \left (c x \right )+1\right ) d}{16 c^{2} \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 ) f \left (2 \arcsinh \left (c x \right )+1\right ) d}{16 c \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (4 c^{4} x^{4}-4 \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}+5 c^{2} x^{2}-3 \sqrt {c^{2} x^{2}+1}\, c x +1\right ) g \left (1+3 \arcsinh \left (c x \right )\right ) d}{96 c^{2} \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (8 c^{5} x^{5}-8 \sqrt {c^{2} x^{2}+1}\, c^{4} x^{4}+12 c^{3} x^{3}-8 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+4 c x -\sqrt {c^{2} x^{2}+1}\right ) f \left (1+4 \arcsinh \left (c x \right )\right ) d}{256 c \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (16 x^{6} c^{6}-16 \sqrt {c^{2} x^{2}+1}\, x^{5} c^{5}+28 c^{4} x^{4}-20 \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}+13 c^{2} x^{2}-5 \sqrt {c^{2} x^{2}+1}\, c x +1\right ) g \left (1+5 \arcsinh \left (c x \right )\right ) d}{800 c^{2} \left (c^{2} x^{2}+1\right )}\right )\)	\(1065\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x)),x,method=_RETURNVERBOSE)

[Out]

1/5*a*g/c^2/d*(c^2*d*x^2+d)^(5/2)+1/4*a*f*x*(c^2*d*x^2+d)^(3/2)+3/8*a*f*d*x*(c^2*d*x^2+d)^(1/2)+3/8*a*f*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)+b*(3/16*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c*f*
arcsinh(c*x)^2*d+1/800*(d*(c^2*x^2+1))^(1/2)*(16*x^6*c^6+16*(c^2*x^2+1)^(1/2)*x^5*c^5+28*c^4*x^4+20*(c^2*x^2+1
)^(1/2)*x^3*c^3+13*c^2*x^2+5*(c^2*x^2+1)^(1/2)*c*x+1)*g*(-1+5*arcsinh(c*x))*d/c^2/(c^2*x^2+1)+1/256*(d*(c^2*x^
2+1))^(1/2)*(8*c^5*x^5+8*(c^2*x^2+1)^(1/2)*c^4*x^4+12*c^3*x^3+8*c^2*x^2*(c^2*x^2+1)^(1/2)+4*c*x+(c^2*x^2+1)^(1
/2))*f*(-1+4*arcsinh(c*x))*d/c/(c^2*x^2+1)+1/96*(d*(c^2*x^2+1))^(1/2)*(4*c^4*x^4+4*(c^2*x^2+1)^(1/2)*x^3*c^3+5
*c^2*x^2+3*(c^2*x^2+1)^(1/2)*c*x+1)*g*(-1+3*arcsinh(c*x))*d/c^2/(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))*f*(2*arcsinh(c*x)-1)*d/c/(c^2*x^2+1)+1/16*(d*(c^2*x^2
+1))^(1/2)*(c^2*x^2+(c^2*x^2+1)^(1/2)*c*x+1)*g*(arcsinh(c*x)-1)*d/c^2/(c^2*x^2+1)+1/16*(d*(c^2*x^2+1))^(1/2)*(
c^2*x^2-(c^2*x^2+1)^(1/2)*c*x+1)*g*(arcsinh(c*x)+1)*d/c^2/(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))*f*(2*arcsinh(c*x)+1)*d/c/(c^2*x^2+1)+1/96*(d*(c^2*x^2+1))^(
1/2)*(4*c^4*x^4-4*(c^2*x^2+1)^(1/2)*x^3*c^3+5*c^2*x^2-3*(c^2*x^2+1)^(1/2)*c*x+1)*g*(1+3*arcsinh(c*x))*d/c^2/(c
^2*x^2+1)+1/256*(d*(c^2*x^2+1))^(1/2)*(8*c^5*x^5-8*(c^2*x^2+1)^(1/2)*c^4*x^4+12*c^3*x^3-8*c^2*x^2*(c^2*x^2+1)^
(1/2)+4*c*x-(c^2*x^2+1)^(1/2))*f*(1+4*arcsinh(c*x))*d/c/(c^2*x^2+1)+1/800*(d*(c^2*x^2+1))^(1/2)*(16*x^6*c^6-16
*(c^2*x^2+1)^(1/2)*x^5*c^5+28*c^4*x^4-20*(c^2*x^2+1)^(1/2)*x^3*c^3+13*c^2*x^2-5*(c^2*x^2+1)^(1/2)*c*x+1)*g*(1+
5*arcsinh(c*x))*d/c^2/(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)*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x)),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)*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x)),x, algorithm="fricas")

[Out]

integral((a*c^2*d*g*x^3 + a*c^2*d*f*x^2 + a*d*g*x + a*d*f + (b*c^2*d*g*x^3 + b*c^2*d*f*x^2 + b*d*g*x + b*d*f)*
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 \left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asinh}{\left (c x \right )}\right ) \left (f + g x\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:sym2poly/r2sym(co
nst 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 = -0.00 \begin {gather*} \int \left (f+g\,x\right )\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (d\,c^2\,x^2+d\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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