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

Optimal. Leaf size=542 \[ \frac {d 5^{-n-1} e^{-\frac {5 a}{b}} \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \Gamma \left (n+1,-\frac {5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 c^2 \sqrt {c^2 x^2+1}}+\frac {d 3^{-n} e^{-\frac {3 a}{b}} \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \Gamma \left (n+1,-\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 c^2 \sqrt {c^2 x^2+1}}+\frac {d e^{-\frac {a}{b}} \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \Gamma \left (n+1,-\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{16 c^2 \sqrt {c^2 x^2+1}}+\frac {d e^{a/b} \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^n \left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \Gamma \left (n+1,\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{16 c^2 \sqrt {c^2 x^2+1}}+\frac {d 3^{-n} e^{\frac {3 a}{b}} \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^n \left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \Gamma \left (n+1,\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 c^2 \sqrt {c^2 x^2+1}}+\frac {d 5^{-n-1} e^{\frac {5 a}{b}} \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^n \left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \Gamma \left (n+1,\frac {5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 c^2 \sqrt {c^2 x^2+1}} \]

[Out]

1/32*5^(-1-n)*d*(a+b*arcsinh(c*x))^n*GAMMA(1+n,-5*(a+b*arcsinh(c*x))/b)*(c^2*d*x^2+d)^(1/2)/c^2/exp(5*a/b)/(((
-a-b*arcsinh(c*x))/b)^n)/(c^2*x^2+1)^(1/2)+1/32*d*(a+b*arcsinh(c*x))^n*GAMMA(1+n,-3*(a+b*arcsinh(c*x))/b)*(c^2
*d*x^2+d)^(1/2)/(3^n)/c^2/exp(3*a/b)/(((-a-b*arcsinh(c*x))/b)^n)/(c^2*x^2+1)^(1/2)+1/16*d*(a+b*arcsinh(c*x))^n
*GAMMA(1+n,(-a-b*arcsinh(c*x))/b)*(c^2*d*x^2+d)^(1/2)/c^2/exp(a/b)/(((-a-b*arcsinh(c*x))/b)^n)/(c^2*x^2+1)^(1/
2)+1/16*d*exp(a/b)*(a+b*arcsinh(c*x))^n*GAMMA(1+n,(a+b*arcsinh(c*x))/b)*(c^2*d*x^2+d)^(1/2)/c^2/(((a+b*arcsinh
(c*x))/b)^n)/(c^2*x^2+1)^(1/2)+1/32*d*exp(3*a/b)*(a+b*arcsinh(c*x))^n*GAMMA(1+n,3*(a+b*arcsinh(c*x))/b)*(c^2*d
*x^2+d)^(1/2)/(3^n)/c^2/(((a+b*arcsinh(c*x))/b)^n)/(c^2*x^2+1)^(1/2)+1/32*5^(-1-n)*d*exp(5*a/b)*(a+b*arcsinh(c
*x))^n*GAMMA(1+n,5*(a+b*arcsinh(c*x))/b)*(c^2*d*x^2+d)^(1/2)/c^2/(((a+b*arcsinh(c*x))/b)^n)/(c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.63, antiderivative size = 542, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {5782, 5779, 5448, 3308, 2181} \[ \frac {d 5^{-n-1} e^{-\frac {5 a}{b}} \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \text {Gamma}\left (n+1,-\frac {5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 c^2 \sqrt {c^2 x^2+1}}+\frac {d 3^{-n} e^{-\frac {3 a}{b}} \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \text {Gamma}\left (n+1,-\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 c^2 \sqrt {c^2 x^2+1}}+\frac {d e^{-\frac {a}{b}} \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \text {Gamma}\left (n+1,-\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{16 c^2 \sqrt {c^2 x^2+1}}+\frac {d e^{a/b} \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^n \left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \text {Gamma}\left (n+1,\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{16 c^2 \sqrt {c^2 x^2+1}}+\frac {d 3^{-n} e^{\frac {3 a}{b}} \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^n \left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \text {Gamma}\left (n+1,\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 c^2 \sqrt {c^2 x^2+1}}+\frac {d 5^{-n-1} e^{\frac {5 a}{b}} \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^n \left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \text {Gamma}\left (n+1,\frac {5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 c^2 \sqrt {c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5^(-1 - n)*d*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^n*Gamma[1 + n, (-5*(a + b*ArcSinh[c*x]))/b])/(32*c^2*E^
((5*a)/b)*Sqrt[1 + c^2*x^2]*(-((a + b*ArcSinh[c*x])/b))^n) + (d*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^n*Gam
ma[1 + n, (-3*(a + b*ArcSinh[c*x]))/b])/(32*3^n*c^2*E^((3*a)/b)*Sqrt[1 + c^2*x^2]*(-((a + b*ArcSinh[c*x])/b))^
n) + (d*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^n*Gamma[1 + n, -((a + b*ArcSinh[c*x])/b)])/(16*c^2*E^(a/b)*Sq
rt[1 + c^2*x^2]*(-((a + b*ArcSinh[c*x])/b))^n) + (d*E^(a/b)*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^n*Gamma[1
+ n, (a + b*ArcSinh[c*x])/b])/(16*c^2*Sqrt[1 + c^2*x^2]*((a + b*ArcSinh[c*x])/b)^n) + (d*E^((3*a)/b)*Sqrt[d +
c^2*d*x^2]*(a + b*ArcSinh[c*x])^n*Gamma[1 + n, (3*(a + b*ArcSinh[c*x]))/b])/(32*3^n*c^2*Sqrt[1 + c^2*x^2]*((a
+ b*ArcSinh[c*x])/b)^n) + (5^(-1 - n)*d*E^((5*a)/b)*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^n*Gamma[1 + n, (
5*(a + b*ArcSinh[c*x]))/b])/(32*c^2*Sqrt[1 + c^2*x^2]*((a + b*ArcSinh[c*x])/b)^n)

Rule 2181

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(F^(g*(e - (c*f)/d))*(c +
d*x)^FracPart[m]*Gamma[m + 1, (-((f*g*Log[F])/d))*(c + d*x)])/(d*(-((f*g*Log[F])/d))^(IntPart[m] + 1)*(-((f*g*
Log[F]*(c + d*x))/d))^FracPart[m]), x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rule 3308

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/E^(I*(e + f*x))
, x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e, f, m}, x]

Rule 5448

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rule 5779

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c^
(m + 1), Subst[Int[(a + b*x)^n*Sinh[x]^m*Cosh[x]^(2*p + 1), x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e,
n}, x] && EqQ[e, c^2*d] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

Rule 5782

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(d^IntPa
rt[p]*(d + e*x^2)^FracPart[p])/(1 + c^2*x^2)^FracPart[p], Int[x^m*(1 + c^2*x^2)^p*(a + b*ArcSinh[c*x])^n, x],
x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] &&  !(Integer
Q[p] || GtQ[d, 0])

Rubi steps

\begin {align*} \int x \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^n \, dx &=\frac {\left (d \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 )^n \, dx}{\sqrt {1+c^2 x^2}}\\ &=\frac {\left (d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int (a+b x)^n \cosh ^4(x) \sinh (x) \, dx,x,\sinh ^{-1}(c x)\right )}{c^2 \sqrt {1+c^2 x^2}}\\ &=\frac {\left (d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{8} (a+b x)^n \sinh (x)+\frac {3}{16} (a+b x)^n \sinh (3 x)+\frac {1}{16} (a+b x)^n \sinh (5 x)\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^2 \sqrt {1+c^2 x^2}}\\ &=\frac {\left (d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int (a+b x)^n \sinh (5 x) \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^2 \sqrt {1+c^2 x^2}}+\frac {\left (d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int (a+b x)^n \sinh (x) \, dx,x,\sinh ^{-1}(c x)\right )}{8 c^2 \sqrt {1+c^2 x^2}}+\frac {\left (3 d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int (a+b x)^n \sinh (3 x) \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^2 \sqrt {1+c^2 x^2}}\\ &=-\frac {\left (d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int e^{-5 x} (a+b x)^n \, dx,x,\sinh ^{-1}(c x)\right )}{32 c^2 \sqrt {1+c^2 x^2}}+\frac {\left (d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int e^{5 x} (a+b x)^n \, dx,x,\sinh ^{-1}(c x)\right )}{32 c^2 \sqrt {1+c^2 x^2}}-\frac {\left (d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int e^{-x} (a+b x)^n \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^2 \sqrt {1+c^2 x^2}}+\frac {\left (d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int e^x (a+b x)^n \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^2 \sqrt {1+c^2 x^2}}-\frac {\left (3 d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int e^{-3 x} (a+b x)^n \, dx,x,\sinh ^{-1}(c x)\right )}{32 c^2 \sqrt {1+c^2 x^2}}+\frac {\left (3 d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int e^{3 x} (a+b x)^n \, dx,x,\sinh ^{-1}(c x)\right )}{32 c^2 \sqrt {1+c^2 x^2}}\\ &=\frac {5^{-1-n} d e^{-\frac {5 a}{b}} \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \Gamma \left (1+n,-\frac {5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 c^2 \sqrt {1+c^2 x^2}}+\frac {3^{-n} d e^{-\frac {3 a}{b}} \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \Gamma \left (1+n,-\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 c^2 \sqrt {1+c^2 x^2}}+\frac {d e^{-\frac {a}{b}} \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \Gamma \left (1+n,-\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{16 c^2 \sqrt {1+c^2 x^2}}+\frac {d e^{a/b} \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^n \left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \Gamma \left (1+n,\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{16 c^2 \sqrt {1+c^2 x^2}}+\frac {3^{-n} d e^{\frac {3 a}{b}} \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^n \left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \Gamma \left (1+n,\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 c^2 \sqrt {1+c^2 x^2}}+\frac {5^{-1-n} d e^{\frac {5 a}{b}} \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^n \left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \Gamma \left (1+n,\frac {5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 c^2 \sqrt {1+c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 1.79, size = 390, normalized size = 0.72 \[ \frac {d^2 15^{-n-1} e^{-\frac {5 a}{b}} \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^n \left (-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{b^2}\right )^{-2 n} \left (3 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )^n \left (3^n \left (-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{b^2}\right )^n \Gamma \left (n+1,-\frac {5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )+5^{n+1} e^{\frac {2 a}{b}} \left (-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{b^2}\right )^n \Gamma \left (n+1,-\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )+2\ 3^n 5^{n+1} e^{\frac {4 a}{b}} \left (-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{b^2}\right )^n \Gamma \left (n+1,-\frac {a+b \sinh ^{-1}(c x)}{b}\right )+5^{n+1} e^{\frac {8 a}{b}} \left (-\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{2 n} \Gamma \left (n+1,\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )+3^n e^{\frac {10 a}{b}} \left (-\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{2 n} \Gamma \left (n+1,\frac {5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )\right )+2\ 15^{n+1} e^{\frac {6 a}{b}} \left (-\frac {a+b \sinh ^{-1}(c x)}{b}\right )^n \left (-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{b^2}\right )^n \Gamma \left (n+1,\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )}{32 c^2 \sqrt {c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15^(-1 - n)*d^2*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^n*(2*15^(1 + n)*E^((6*a)/b)*(-((a + b*ArcSinh[c*x])/b)
)^n*(-((a + b*ArcSinh[c*x])^2/b^2))^n*Gamma[1 + n, a/b + ArcSinh[c*x]] + 3*(a/b + ArcSinh[c*x])^n*(3^n*(-((a +
b*ArcSinh[c*x])^2/b^2))^n*Gamma[1 + n, (-5*(a + b*ArcSinh[c*x]))/b] + 5^(1 + n)*E^((2*a)/b)*(-((a + b*ArcSinh
[c*x])^2/b^2))^n*Gamma[1 + n, (-3*(a + b*ArcSinh[c*x]))/b] + 2*3^n*5^(1 + n)*E^((4*a)/b)*(-((a + b*ArcSinh[c*x
])^2/b^2))^n*Gamma[1 + n, -((a + b*ArcSinh[c*x])/b)] + 5^(1 + n)*E^((8*a)/b)*(-((a + b*ArcSinh[c*x])/b))^(2*n)
*Gamma[1 + n, (3*(a + b*ArcSinh[c*x]))/b] + 3^n*E^((10*a)/b)*(-((a + b*ArcSinh[c*x])/b))^(2*n)*Gamma[1 + n, (5
*(a + b*ArcSinh[c*x]))/b])))/(32*c^2*E^((5*a)/b)*Sqrt[d + c^2*d*x^2]*(-((a + b*ArcSinh[c*x])^2/b^2))^(2*n))

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (c^{2} d x^{3} + d x\right )} \sqrt {c^{2} d x^{2} + d} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c^2*d*x^3 + d*x)*sqrt(c^2*d*x^2 + d)*(b*arcsinh(c*x) + a)^n, 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(x*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))^n,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 [F]  time = 0.15, size = 0, normalized size = 0.00 \[ \int x \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \arcsinh \left (c x \right )\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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