3.3.33 \(\int \frac {a+b \sinh ^{-1}(c+d x)}{(c e+d e x)^{3/2}} \, dx\) [233]

Optimal. Leaf size=106 \[ -\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{d e \sqrt {e (c+d x)}}+\frac {2 b (1+c+d x) \sqrt {\frac {1+(c+d x)^2}{(1+c+d x)^2}} F\left (2 \text {ArcTan}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{d e^{3/2} \sqrt {1+(c+d x)^2}} \]

[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]
time = 0.08, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {5859, 5776, 335, 226} \begin {gather*} \frac {2 b (c+d x+1) \sqrt {\frac {(c+d x)^2+1}{(c+d x+1)^2}} F\left (2 \text {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 \left (a+b \sinh ^{-1}(c+d x)\right )}{d e \sqrt {e (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \frac {a+b \sinh ^{-1}(c+d x)}{(c e+d e x)^{3/2}} \, dx &=\frac {\text {Subst}\left (\int \frac {a+b \sinh ^{-1}(x)}{(e x)^{3/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{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 \left (a+b \sinh ^{-1}(c+d x)\right )}{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 \left (a+b \sinh ^{-1}(c+d x)\right )}{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}} F\left (2 \tan ^{-1}\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] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.02, size = 56, normalized size = 0.53 \begin {gather*} -\frac {2 \left (a+b \sinh ^{-1}(c+d x)-2 b (c+d x) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-(c+d x)^2\right )\right )}{d e \sqrt {e (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[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] Result contains complex when optimal does not.
time = 3.24, size = 140, normalized size = 1.32

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

Verification of antiderivative is not currently implemented for this CAS.

[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((d*e*x+c*e)/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)/((d*e*x+c*e)^2/e^2+1)^(1/2)*EllipticF((d*e*x+c*e)^(1/2)*(I/e)^(1/2),I)))

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b*(4*e^(-3/2)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))/(sqrt(d*x + c)*d) + (I*sqrt(2)*(log(1/2*I*
sqrt(2)*(sqrt(2) + 2*sqrt(d*x + c)) + 1) - log(-1/2*I*sqrt(2)*(sqrt(2) + 2*sqrt(d*x + c)) + 1))*e^(-3/2) - I*s
qrt(2)*(log(1/2*I*sqrt(2)*(sqrt(2) - 2*sqrt(d*x + c)) + 1) - log(-1/2*I*sqrt(2)*(sqrt(2) - 2*sqrt(d*x + c)) +
1))*e^(-3/2) + sqrt(2)*e^(-3/2)*log(d*x + sqrt(2)*sqrt(d*x + c) + c + 1) - sqrt(2)*e^(-3/2)*log(d*x - sqrt(2)*
sqrt(d*x + c) + c + 1))/d - 4*integrate(1/((d^2*x^2*e^(3/2) + 2*c*d*x*e^(3/2) + (c^2 + 1)*e^(3/2))*sqrt(d^2*x^
2 + 2*c*d*x + c^2 + 1)*sqrt(d*x + c) + (d^3*x^3*e^(3/2) + 3*c*d^2*x^2*e^(3/2) + (3*c^2*d + d)*x*e^(3/2) + (c^3
+ c)*e^(3/2))*sqrt(d*x + c)), x)) - 2*a*e^(-1)/(sqrt(d*x*e + c*e)*d)

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 169, normalized size = 1.59 \begin {gather*} -\frac {2 \, {\left (\sqrt {{\left (d x + c\right )} \cosh \left (1\right ) + {\left (d x + c\right )} \sinh \left (1\right )} b d^{2} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) + \sqrt {{\left (d x + c\right )} \cosh \left (1\right ) + {\left (d x + c\right )} \sinh \left (1\right )} a d^{2} - 2 \, \sqrt {d^{3} \cosh \left (1\right ) + d^{3} \sinh \left (1\right )} {\left (b d x + b c\right )} {\rm weierstrassPInverse}\left (-\frac {4}{d^{2}}, 0, \frac {d x + c}{d}\right )\right )}}{{\left (d^{4} x + c d^{3}\right )} \cosh \left (1\right )^{2} + 2 \, {\left (d^{4} x + c d^{3}\right )} \cosh \left (1\right ) \sinh \left (1\right ) + {\left (d^{4} x + c d^{3}\right )} \sinh \left (1\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(sqrt((d*x + c)*cosh(1) + (d*x + c)*sinh(1))*b*d^2*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) + sqrt(
(d*x + c)*cosh(1) + (d*x + c)*sinh(1))*a*d^2 - 2*sqrt(d^3*cosh(1) + d^3*sinh(1))*(b*d*x + b*c)*weierstrassPInv
erse(-4/d^2, 0, (d*x + c)/d))/((d^4*x + c*d^3)*cosh(1)^2 + 2*(d^4*x + c*d^3)*cosh(1)*sinh(1) + (d^4*x + c*d^3)
*sinh(1)^2)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \operatorname {asinh}{\left (c + d x \right )}}{\left (e \left (c + d x\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Giac [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((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]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,\mathrm {asinh}\left (c+d\,x\right )}{{\left (c\,e+d\,e\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________