3.3.32 \(\int \frac {a+b \sinh ^{-1}(c+d x)}{\sqrt {c e+d e x}} \, dx\) [232]

Optimal. Leaf size=223 \[ -\frac {4 b \sqrt {e (c+d x)} \sqrt {1+(c+d x)^2}}{d e (1+c+d x)}+\frac {2 \sqrt {e (c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right )}{d e}+\frac {4 b (1+c+d x) \sqrt {\frac {1+(c+d x)^2}{(1+c+d x)^2}} E\left (2 \text {ArcTan}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{d \sqrt {e} \sqrt {1+(c+d x)^2}}-\frac {2 b (1+c+d x) \sqrt {\frac {1+(c+d x)^2}{(1+c+d x)^2}} F\left (2 \text {ArcTan}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{d \sqrt {e} \sqrt {1+(c+d x)^2}} \]

[Out]

2*(a+b*arcsinh(d*x+c))*(e*(d*x+c))^(1/2)/d/e-4*b*(e*(d*x+c))^(1/2)*(1+(d*x+c)^2)^(1/2)/d/e/(d*x+c+1)+4*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)))*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/e^(1/2)/(1+(d*x+c)^2)^(
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^(1/2)/(
1+(d*x+c)^2)^(1/2)

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Rubi [A]
time = 0.15, antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {5859, 5776, 335, 311, 226, 1210} \begin {gather*} \frac {2 \sqrt {e (c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right )}{d e}-\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 \sqrt {e} \sqrt {(c+d x)^2+1}}+\frac {4 b (c+d x+1) \sqrt {\frac {(c+d x)^2+1}{(c+d x+1)^2}} E\left (2 \text {ArcTan}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{d \sqrt {e} \sqrt {(c+d x)^2+1}}-\frac {4 b \sqrt {(c+d x)^2+1} \sqrt {e (c+d x)}}{d e (c+d x+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*b*Sqrt[e*(c + d*x)]*Sqrt[1 + (c + d*x)^2])/(d*e*(1 + c + d*x)) + (2*Sqrt[e*(c + d*x)]*(a + b*ArcSinh[c + d
*x]))/(d*e) + (4*b*(1 + c + d*x)*Sqrt[(1 + (c + d*x)^2)/(1 + c + d*x)^2]*EllipticE[2*ArcTan[Sqrt[e*(c + d*x)]/
Sqrt[e]], 1/2])/(d*Sqrt[e]*Sqrt[1 + (c + d*x)^2]) - (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*Sqrt[e]*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 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*} \int \frac {a+b \sinh ^{-1}(c+d x)}{\sqrt {c e+d e x}} \, dx &=\frac {\text {Subst}\left (\int \frac {a+b \sinh ^{-1}(x)}{\sqrt {e x}} \, dx,x,c+d x\right )}{d}\\ &=\frac {2 \sqrt {e (c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right )}{d e}-\frac {(2 b) \text {Subst}\left (\int \frac {\sqrt {e x}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{d e}\\ &=\frac {2 \sqrt {e (c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right )}{d e}-\frac {(4 b) \text {Subst}\left (\int \frac {x^2}{\sqrt {1+\frac {x^4}{e^2}}} \, dx,x,\sqrt {e (c+d x)}\right )}{d e^2}\\ &=\frac {2 \sqrt {e (c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right )}{d e}-\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}+\frac {(4 b) \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 )}{d e}\\ &=-\frac {4 b \sqrt {e (c+d x)} \sqrt {1+(c+d x)^2}}{d e (1+c+d x)}+\frac {2 \sqrt {e (c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right )}{d e}+\frac {4 b (1+c+d x) \sqrt {\frac {1+(c+d x)^2}{(1+c+d x)^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{d \sqrt {e} \sqrt {1+(c+d x)^2}}-\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 \sqrt {e} \sqrt {1+(c+d x)^2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.02, size = 61, normalized size = 0.27 \begin {gather*} -\frac {2 \sqrt {e (c+d x)} \left (-3 \left (a+b \sinh ^{-1}(c+d x)\right )+2 b (c+d x) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-(c+d x)^2\right )\right )}{3 d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 3.25, size = 161, normalized size = 0.72

method	result	size
derivativedivides	\(\frac {2 \sqrt {d e x +c e}\, a +2 b \left (\sqrt {d e x +c e}\, \arcsinh \left (\frac {d e x +c e}{e}\right )-\frac {2 i \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \left (\EllipticF \left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )-\EllipticE \left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )\right )}{\sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{d e}\)	\(161\)
default	\(\frac {2 \sqrt {d e x +c e}\, a +2 b \left (\sqrt {d e x +c e}\, \arcsinh \left (\frac {d e x +c e}{e}\right )-\frac {2 i \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \left (\EllipticF \left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )-\EllipticE \left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )\right )}{\sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{d e}\)	\(161\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(d*x+c))/(d*e*x+c*e)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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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)^(1/2),x, algorithm="maxima")

[Out]

1/2*(4*(d*x*e^(1/2) + c*e^(1/2))*e^(-1)*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^(1/2) - 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^(1/2) - sqrt(2)*e^(1/2)*log(d*x + sqrt(2)*sqrt(d*x + c) + c + 1) + sqrt(2)*e^(1/2)*
log(d*x - sqrt(2)*sqrt(d*x + c) + c + 1))*e^(-1) + 8*e^(1/2*log(d*x + c) - 1/2))/d - 2*integrate(2*(d*x*e^(1/2
) + c*e^(1/2))/((d^3*x^3*e + 3*c*d^2*x^2*e + (3*c^2*d + d)*x*e + (c^3 + c)*e + (d^2*x^2*e + 2*c*d*x*e + (c^2 +
1)*e)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))*sqrt(d*x + c)), x))*b + 2*sqrt(d*x*e + c*e)*a*e^(-1)/d

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.13, size = 128, normalized size = 0.57 \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 \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 \, \sqrt {d^{3} \cosh \left (1\right ) + d^{3} \sinh \left (1\right )} b {\rm weierstrassZeta}\left (-\frac {4}{d^{2}}, 0, {\rm weierstrassPInverse}\left (-\frac {4}{d^{2}}, 0, \frac {d x + c}{d}\right )\right )\right )}}{d^{2} \cosh \left (1\right ) + d^{2} \sinh \left (1\right )} \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)^(1/2),x, algorithm="fricas")

[Out]

2*(sqrt((d*x + c)*cosh(1) + (d*x + c)*sinh(1))*b*d*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*sqrt(d^3*cosh(1) + d^3*sinh(1))*b*weierstrassZeta(-4/d^2, 0, weier
strassPInverse(-4/d^2, 0, (d*x + c)/d)))/(d^2*cosh(1) + d^2*sinh(1))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \operatorname {asinh}{\left (c + d x \right )}}{\sqrt {e \left (c + d x\right )}}\, 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)**(1/2),x)

[Out]

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

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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)^(1/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(d*x + c) + a)/sqrt(d*e*x + c*e), x)

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

[Out]

int((a + b*asinh(c + d*x))/(c*e + d*e*x)^(1/2), x)

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