\(\int \frac {a+b \text {arcsinh}(c+d x)}{(c e+d e x)^{3/2}} \, dx\) [233]

Optimal result

Integrand size = 23, antiderivative size = 106 \[ \int \frac {a+b \text {arcsinh}(c+d x)}{(c e+d e x)^{3/2}} \, dx=-\frac {2 (a+b \text {arcsinh}(c+d x))}{d e \sqrt {e (c+d x)}}+\frac {2 b (1+c+d x) \sqrt {\frac {1+(c+d x)^2}{(1+c+d x)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right ),\frac {1}{2}\right )}{d e^{3/2} \sqrt {1+(c+d x)^2}} \]

[Out]

-2*(a+b*arcsinh(d*x+c))/d/e/(e*(d*x+c))^(1/2)+2*b*(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/e^(3/2)/(1+(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {5859, 5776, 335, 226} \[ \int \frac {a+b \text {arcsinh}(c+d x)}{(c e+d e x)^{3/2}} \, dx=\frac {2 b (c+d x+1) \sqrt {\frac {(c+d x)^2+1}{(c+d x+1)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right ),\frac {1}{2}\right )}{d e^{3/2} \sqrt {(c+d x)^2+1}}-\frac {2 (a+b \text {arcsinh}(c+d x))}{d e \sqrt {e (c+d x)}} \]

[In]

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

[Out]

(-2*(a + b*ArcSinh[c + d*x]))/(d*e*Sqrt[e*(c + d*x)]) + (2*b*(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])/(d*e^(3/2)*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 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 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 \frac {a+b \text {arcsinh}(x)}{(e x)^{3/2}} \, dx,x,c+d x\right )}{d} \\ & = -\frac {2 (a+b \text {arcsinh}(c+d x))}{d e \sqrt {e (c+d x)}}+\frac {(2 b) \text {Subst}\left (\int \frac {1}{\sqrt {e x} \sqrt {1+x^2}} \, dx,x,c+d x\right )}{d e} \\ & = -\frac {2 (a+b \text {arcsinh}(c+d x))}{d e \sqrt {e (c+d x)}}+\frac {(4 b) \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^4}{e^2}}} \, dx,x,\sqrt {e (c+d x)}\right )}{d e^2} \\ & = -\frac {2 (a+b \text {arcsinh}(c+d x))}{d e \sqrt {e (c+d x)}}+\frac {2 b (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 )}{d e^{3/2} \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 = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.53 \[ \int \frac {a+b \text {arcsinh}(c+d x)}{(c e+d e x)^{3/2}} \, dx=-\frac {2 \left (a+b \text {arcsinh}(c+d x)-2 b (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-(c+d x)^2\right )\right )}{d e \sqrt {e (c+d x)}} \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.44 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.32

method	result	size
derivativedivides	\(\frac {-\frac {2 a}{\sqrt {d e x +c e}}+2 b \left (-\frac {\operatorname {arcsinh}\left (\frac {d e x +c e}{e}\right )}{\sqrt {d e x +c e}}+\frac {2 \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \operatorname {EllipticF}\left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )}{e \sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{d e}\)	\(140\)
default	\(\frac {-\frac {2 a}{\sqrt {d e x +c e}}+2 b \left (-\frac {\operatorname {arcsinh}\left (\frac {d e x +c e}{e}\right )}{\sqrt {d e x +c e}}+\frac {2 \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \operatorname {EllipticF}\left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )}{e \sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{d e}\)	\(140\)
parts	\(-\frac {2 a}{\sqrt {d e x +c e}\, d e}+\frac {2 b \left (-\frac {\operatorname {arcsinh}\left (\frac {d e x +c e}{e}\right )}{\sqrt {d e x +c e}}+\frac {2 \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \operatorname {EllipticF}\left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )}{e \sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{d e}\)	\(145\)

[In]

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

[Out]

2/d/e*(-a/(d*e*x+c*e)^(1/2)+b*(-1/(d*e*x+c*e)^(1/2)*arcsinh(1/e*(d*e*x+c*e))+2/e/(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)))

Fricas [C] (verification not implemented)

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

Time = 0.09 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.04 \[ \int \frac {a+b \text {arcsinh}(c+d x)}{(c e+d e x)^{3/2}} \, dx=-\frac {2 \, {\left (\sqrt {d e x + c e} b d^{2} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) + \sqrt {d e x + c e} a d^{2} - 2 \, \sqrt {d^{3} e} {\left (b d x + b c\right )} {\rm weierstrassPInverse}\left (-\frac {4}{d^{2}}, 0, \frac {d x + c}{d}\right )\right )}}{d^{4} e^{2} x + c d^{3} e^{2}} \]

[In]

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

[Out]

-2*(sqrt(d*e*x + c*e)*b*d^2*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) + sqrt(d*e*x + c*e)*a*d^2 - 2*sqr
t(d^3*e)*(b*d*x + b*c)*weierstrassPInverse(-4/d^2, 0, (d*x + c)/d))/(d^4*e^2*x + c*d^3*e^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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