\(\int (c e+d e x)^{3/2} (a+b \text {arcsinh}(c+d x)) \, dx\) [230]

Optimal result

Integrand size = 23, antiderivative size = 261 \[ \int (c e+d e x)^{3/2} (a+b \text {arcsinh}(c+d x)) \, dx=-\frac {4 b (e (c+d x))^{3/2} \sqrt {1+(c+d x)^2}}{25 d}+\frac {12 b e \sqrt {e (c+d x)} \sqrt {1+(c+d x)^2}}{25 d (1+c+d x)}+\frac {2 (e (c+d x))^{5/2} (a+b \text {arcsinh}(c+d x))}{5 d e}-\frac {12 b e^{3/2} (1+c+d x) \sqrt {\frac {1+(c+d x)^2}{(1+c+d x)^2}} E\left (2 \arctan \left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{25 d \sqrt {1+(c+d x)^2}}+\frac {6 b e^{3/2} (1+c+d x) \sqrt {\frac {1+(c+d x)^2}{(1+c+d x)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right ),\frac {1}{2}\right )}{25 d \sqrt {1+(c+d x)^2}} \]

[Out]

2/5*(e*(d*x+c))^(5/2)*(a+b*arcsinh(d*x+c))/d/e-4/25*b*(e*(d*x+c))^(3/2)*(1+(d*x+c)^2)^(1/2)/d+12/25*b*e*(e*(d*
x+c))^(1/2)*(1+(d*x+c)^2)^(1/2)/d/(d*x+c+1)-12/25*b*e^(3/2)*(d*x+c+1)*(cos(2*arctan((e*(d*x+c))^(1/2)/e^(1/2))
)^2)^(1/2)/cos(2*arctan((e*(d*x+c))^(1/2)/e^(1/2)))*EllipticE(sin(2*arctan((e*(d*x+c))^(1/2)/e^(1/2))),1/2*2^(
1/2))*((1+(d*x+c)^2)/(d*x+c+1)^2)^(1/2)/d/(1+(d*x+c)^2)^(1/2)+6/25*b*e^(3/2)*(d*x+c+1)*(cos(2*arctan((e*(d*x+c
))^(1/2)/e^(1/2)))^2)^(1/2)/cos(2*arctan((e*(d*x+c))^(1/2)/e^(1/2)))*EllipticF(sin(2*arctan((e*(d*x+c))^(1/2)/
e^(1/2))),1/2*2^(1/2))*((1+(d*x+c)^2)/(d*x+c+1)^2)^(1/2)/d/(1+(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {5859, 5776, 327, 335, 311, 226, 1210} \[ \int (c e+d e x)^{3/2} (a+b \text {arcsinh}(c+d x)) \, dx=\frac {2 (e (c+d x))^{5/2} (a+b \text {arcsinh}(c+d x))}{5 d e}+\frac {6 b e^{3/2} (c+d x+1) \sqrt {\frac {(c+d x)^2+1}{(c+d x+1)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right ),\frac {1}{2}\right )}{25 d \sqrt {(c+d x)^2+1}}-\frac {12 b e^{3/2} (c+d x+1) \sqrt {\frac {(c+d x)^2+1}{(c+d x+1)^2}} E\left (2 \arctan \left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{25 d \sqrt {(c+d x)^2+1}}-\frac {4 b \sqrt {(c+d x)^2+1} (e (c+d x))^{3/2}}{25 d}+\frac {12 b e \sqrt {(c+d x)^2+1} \sqrt {e (c+d x)}}{25 d (c+d x+1)} \]

[In]

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

[Out]

(-4*b*(e*(c + d*x))^(3/2)*Sqrt[1 + (c + d*x)^2])/(25*d) + (12*b*e*Sqrt[e*(c + d*x)]*Sqrt[1 + (c + d*x)^2])/(25
*d*(1 + c + d*x)) + (2*(e*(c + d*x))^(5/2)*(a + b*ArcSinh[c + d*x]))/(5*d*e) - (12*b*e^(3/2)*(1 + c + d*x)*Sqr
t[(1 + (c + d*x)^2)/(1 + c + d*x)^2]*EllipticE[2*ArcTan[Sqrt[e*(c + d*x)]/Sqrt[e]], 1/2])/(25*d*Sqrt[1 + (c +
d*x)^2]) + (6*b*e^(3/2)*(1 + c + d*x)*Sqrt[(1 + (c + d*x)^2)/(1 + c + d*x)^2]*EllipticF[2*ArcTan[Sqrt[e*(c + d
*x)]/Sqrt[e]], 1/2])/(25*d*Sqrt[1 + (c + d*x)^2])

Rule 226

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 311

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

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 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 1210

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a +
c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

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 5859

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + f*(x/d))^m*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int (e x)^{3/2} (a+b \text {arcsinh}(x)) \, dx,x,c+d x\right )}{d} \\ & = \frac {2 (e (c+d x))^{5/2} (a+b \text {arcsinh}(c+d x))}{5 d e}-\frac {(2 b) \text {Subst}\left (\int \frac {(e x)^{5/2}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{5 d e} \\ & = -\frac {4 b (e (c+d x))^{3/2} \sqrt {1+(c+d x)^2}}{25 d}+\frac {2 (e (c+d x))^{5/2} (a+b \text {arcsinh}(c+d x))}{5 d e}+\frac {(6 b e) \text {Subst}\left (\int \frac {\sqrt {e x}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{25 d} \\ & = -\frac {4 b (e (c+d x))^{3/2} \sqrt {1+(c+d x)^2}}{25 d}+\frac {2 (e (c+d x))^{5/2} (a+b \text {arcsinh}(c+d x))}{5 d e}+\frac {(12 b) \text {Subst}\left (\int \frac {x^2}{\sqrt {1+\frac {x^4}{e^2}}} \, dx,x,\sqrt {e (c+d x)}\right )}{25 d} \\ & = -\frac {4 b (e (c+d x))^{3/2} \sqrt {1+(c+d x)^2}}{25 d}+\frac {2 (e (c+d x))^{5/2} (a+b \text {arcsinh}(c+d x))}{5 d e}+\frac {(12 b e) \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^4}{e^2}}} \, dx,x,\sqrt {e (c+d x)}\right )}{25 d}-\frac {(12 b e) \text {Subst}\left (\int \frac {1-\frac {x^2}{e}}{\sqrt {1+\frac {x^4}{e^2}}} \, dx,x,\sqrt {e (c+d x)}\right )}{25 d} \\ & = -\frac {4 b (e (c+d x))^{3/2} \sqrt {1+(c+d x)^2}}{25 d}+\frac {12 b e \sqrt {e (c+d x)} \sqrt {1+(c+d x)^2}}{25 d (1+c+d x)}+\frac {2 (e (c+d x))^{5/2} (a+b \text {arcsinh}(c+d x))}{5 d e}-\frac {12 b e^{3/2} (1+c+d x) \sqrt {\frac {1+(c+d x)^2}{(1+c+d x)^2}} E\left (2 \arctan \left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{25 d \sqrt {1+(c+d x)^2}}+\frac {6 b e^{3/2} (1+c+d x) \sqrt {\frac {1+(c+d x)^2}{(1+c+d x)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right ),\frac {1}{2}\right )}{25 d \sqrt {1+(c+d x)^2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.33 \[ \int (c e+d e x)^{3/2} (a+b \text {arcsinh}(c+d x)) \, dx=\frac {2 (e (c+d x))^{3/2} \left (5 a c+5 a d x-2 b \sqrt {1+(c+d x)^2}+5 b c \text {arcsinh}(c+d x)+5 b d x \text {arcsinh}(c+d x)+2 b \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-(c+d x)^2\right )\right )}{25 d} \]

[In]

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

[Out]

(2*(e*(c + d*x))^(3/2)*(5*a*c + 5*a*d*x - 2*b*Sqrt[1 + (c + d*x)^2] + 5*b*c*ArcSinh[c + d*x] + 5*b*d*x*ArcSinh
[c + d*x] + 2*b*Hypergeometric2F1[1/2, 3/4, 7/4, -(c + d*x)^2]))/(25*d)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.91 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(\frac {\frac {2 \left (d e x +c e \right )^{\frac {5}{2}} a}{5}+2 b \left (\frac {\left (d e x +c e \right )^{\frac {5}{2}} \operatorname {arcsinh}\left (\frac {d e x +c e}{e}\right )}{5}-\frac {2 \left (\frac {e^{2} \left (d e x +c e \right )^{\frac {3}{2}} \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{5}-\frac {3 i e^{3} \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \left (\operatorname {EllipticF}\left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )-\operatorname {EllipticE}\left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )\right )}{5 \sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{5 e}\right )}{d e}\)	\(205\)
default	\(\frac {\frac {2 \left (d e x +c e \right )^{\frac {5}{2}} a}{5}+2 b \left (\frac {\left (d e x +c e \right )^{\frac {5}{2}} \operatorname {arcsinh}\left (\frac {d e x +c e}{e}\right )}{5}-\frac {2 \left (\frac {e^{2} \left (d e x +c e \right )^{\frac {3}{2}} \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{5}-\frac {3 i e^{3} \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \left (\operatorname {EllipticF}\left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )-\operatorname {EllipticE}\left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )\right )}{5 \sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{5 e}\right )}{d e}\)	\(205\)
parts	\(\frac {2 a \left (d e x +c e \right )^{\frac {5}{2}}}{5 d e}+\frac {2 b \left (\frac {\left (d e x +c e \right )^{\frac {5}{2}} \operatorname {arcsinh}\left (\frac {d e x +c e}{e}\right )}{5}-\frac {2 \left (\frac {e^{2} \left (d e x +c e \right )^{\frac {3}{2}} \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{5}-\frac {3 i e^{3} \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \left (\operatorname {EllipticF}\left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )-\operatorname {EllipticE}\left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )\right )}{5 \sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{5 e}\right )}{d e}\)	\(210\)

[In]

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

[Out]

2/d/e*(1/5*(d*e*x+c*e)^(5/2)*a+b*(1/5*(d*e*x+c*e)^(5/2)*arcsinh(1/e*(d*e*x+c*e))-2/5/e*(1/5*e^2*(d*e*x+c*e)^(3
/2)*(1/e^2*(d*e*x+c*e)^2+1)^(1/2)-3/5*I*e^3/(I/e)^(1/2)*(1-I/e*(d*e*x+c*e))^(1/2)*(1+I/e*(d*e*x+c*e))^(1/2)/(1
/e^2*(d*e*x+c*e)^2+1)^(1/2)*(EllipticF((d*e*x+c*e)^(1/2)*(I/e)^(1/2),I)-EllipticE((d*e*x+c*e)^(1/2)*(I/e)^(1/2
),I)))))

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.71 \[ \int (c e+d e x)^{3/2} (a+b \text {arcsinh}(c+d x)) \, dx=-\frac {2 \, {\left (6 \, \sqrt {d^{3} e} b e {\rm weierstrassZeta}\left (-\frac {4}{d^{2}}, 0, {\rm weierstrassPInverse}\left (-\frac {4}{d^{2}}, 0, \frac {d x + c}{d}\right )\right ) - 5 \, {\left (b d^{3} e x^{2} + 2 \, b c d^{2} e x + b c^{2} d e\right )} \sqrt {d e x + c e} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) + 2 \, {\left (b d^{2} e x + b c d e\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} \sqrt {d e x + c e} - 5 \, {\left (a d^{3} e x^{2} + 2 \, a c d^{2} e x + a c^{2} d e\right )} \sqrt {d e x + c e}\right )}}{25 \, d^{2}} \]

[In]

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

[Out]

-2/25*(6*sqrt(d^3*e)*b*e*weierstrassZeta(-4/d^2, 0, weierstrassPInverse(-4/d^2, 0, (d*x + c)/d)) - 5*(b*d^3*e*
x^2 + 2*b*c*d^2*e*x + b*c^2*d*e)*sqrt(d*e*x + c*e)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) + 2*(b*d^2
*e*x + b*c*d*e)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*sqrt(d*e*x + c*e) - 5*(a*d^3*e*x^2 + 2*a*c*d^2*e*x + a*c^2*d
*e)*sqrt(d*e*x + c*e))/d^2

Sympy [F]

\[ \int (c e+d e x)^{3/2} (a+b \text {arcsinh}(c+d x)) \, dx=\int \left (e \left (c + d x\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asinh}{\left (c + d x \right )}\right )\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int (c e+d e x)^{3/2} (a+b \text {arcsinh}(c+d x)) \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [F]

\[ \int (c e+d e x)^{3/2} (a+b \text {arcsinh}(c+d x)) \, dx=\int { {\left (d e x + c e\right )}^{\frac {3}{2}} {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (c e+d e x)^{3/2} (a+b \text {arcsinh}(c+d x)) \, dx=\int {\left (c\,e+d\,e\,x\right )}^{3/2}\,\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right ) \,d x \]

[In]

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

[Out]

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