3.3.31 \(\int \sqrt {c e+d e x} (a+b \sinh ^{-1}(c+d x)) \, dx\) [231]

Optimal. Leaf size=142 \[ -\frac {4 b \sqrt {e (c+d x)} \sqrt {1+(c+d x)^2}}{9 d}+\frac {2 (e (c+d x))^{3/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d e}+\frac {2 b \sqrt {e} (1+c+d x) \sqrt {\frac {1+(c+d x)^2}{(1+c+d x)^2}} F\left (2 \text {ArcTan}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{9 d \sqrt {1+(c+d x)^2}} \]

[Out]

2/3*(e*(d*x+c))^(3/2)*(a+b*arcsinh(d*x+c))/d/e-4/9*b*(e*(d*x+c))^(1/2)*(1+(d*x+c)^2)^(1/2)/d+2/9*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*arc
tan((e*(d*x+c))^(1/2)/e^(1/2))),1/2*2^(1/2))*e^(1/2)*((1+(d*x+c)^2)/(d*x+c+1)^2)^(1/2)/d/(1+(d*x+c)^2)^(1/2)

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Rubi [A]
time = 0.09, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {5859, 5776, 327, 335, 226} \begin {gather*} \frac {2 (e (c+d x))^{3/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d e}+\frac {2 b \sqrt {e} (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 )}{9 d \sqrt {(c+d x)^2+1}}-\frac {4 b \sqrt {(c+d x)^2+1} \sqrt {e (c+d x)}}{9 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*b*Sqrt[e*(c + d*x)]*Sqrt[1 + (c + d*x)^2])/(9*d) + (2*(e*(c + d*x))^(3/2)*(a + b*ArcSinh[c + d*x]))/(3*d*e
) + (2*b*Sqrt[e]*(1 + c + d*x)*Sqrt[(1 + (c + d*x)^2)/(1 + c + d*x)^2]*EllipticF[2*ArcTan[Sqrt[e*(c + d*x)]/Sq
rt[e]], 1/2])/(9*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 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 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 \sqrt {c e+d e x} \left (a+b \sinh ^{-1}(c+d x)\right ) \, dx &=\frac {\text {Subst}\left (\int \sqrt {e x} \left (a+b \sinh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {2 (e (c+d x))^{3/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d e}-\frac {(2 b) \text {Subst}\left (\int \frac {(e x)^{3/2}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{3 d e}\\ &=-\frac {4 b \sqrt {e (c+d x)} \sqrt {1+(c+d x)^2}}{9 d}+\frac {2 (e (c+d x))^{3/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d e}+\frac {(2 b e) \text {Subst}\left (\int \frac {1}{\sqrt {e x} \sqrt {1+x^2}} \, dx,x,c+d x\right )}{9 d}\\ &=-\frac {4 b \sqrt {e (c+d x)} \sqrt {1+(c+d x)^2}}{9 d}+\frac {2 (e (c+d x))^{3/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{3 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 )}{9 d}\\ &=-\frac {4 b \sqrt {e (c+d x)} \sqrt {1+(c+d x)^2}}{9 d}+\frac {2 (e (c+d x))^{3/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d e}+\frac {2 b \sqrt {e} (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 )}{9 d \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 = 87, normalized size = 0.61 \begin {gather*} \frac {2 \sqrt {e (c+d x)} \left (3 a c+3 a d x-2 b \sqrt {1+(c+d x)^2}+3 b c \sinh ^{-1}(c+d x)+3 b d x \sinh ^{-1}(c+d x)+2 b \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-(c+d x)^2\right )\right )}{9 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[e*(c + d*x)]*(3*a*c + 3*a*d*x - 2*b*Sqrt[1 + (c + d*x)^2] + 3*b*c*ArcSinh[c + d*x] + 3*b*d*x*ArcSinh[c
+ d*x] + 2*b*Hypergeometric2F1[1/4, 1/2, 5/4, -(c + d*x)^2]))/(9*d)

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Maple [C] Result contains complex when optimal does not.
time = 3.02, size = 179, normalized size = 1.26

method	result	size
derivativedivides	\(\frac {\frac {2 a \left (d e x +c e \right )^{\frac {3}{2}}}{3}+2 b \left (\frac {\left (d e x +c e \right )^{\frac {3}{2}} \arcsinh \left (\frac {d e x +c e}{e}\right )}{3}-\frac {2 \left (\frac {e^{2} \sqrt {d e x +c e}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{3}-\frac {e^{2} \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \EllipticF \left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )}{3 \sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{3 e}\right )}{d e}\)	\(179\)
default	\(\frac {\frac {2 a \left (d e x +c e \right )^{\frac {3}{2}}}{3}+2 b \left (\frac {\left (d e x +c e \right )^{\frac {3}{2}} \arcsinh \left (\frac {d e x +c e}{e}\right )}{3}-\frac {2 \left (\frac {e^{2} \sqrt {d e x +c e}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{3}-\frac {e^{2} \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \EllipticF \left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )}{3 \sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{3 e}\right )}{d e}\)	\(179\)

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

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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]

2/3*(d*x*e + c*e)^(3/2)*a*e^(-1)/d + 1/18*(12*(d*x*e^(1/2) + c*e^(1/2))*sqrt(d*x + c)*log(d*x + c + sqrt(d^2*x
^2 + 2*c*d*x + c^2 + 1))/d - (3*(I*sqrt(2)*(log(1/2*I*sqrt(2)*(sqrt(2) + 2*sqrt(d*x + c)) + 1) - log(-1/2*I*sq
rt(2)*(sqrt(2) + 2*sqrt(d*x + c)) + 1)) - 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)) + sqrt(2)*log(d*x + sqrt(2)*sqrt(d*x + c) + c + 1) - sqrt(2)*
log(d*x - sqrt(2)*sqrt(d*x + c) + c + 1))*e^(1/2) + 8*e^(3/2*log(d*x + c) + 1/2))/d - 18*integrate(2/3*(d*x*e^
(1/2) + c*e^(1/2))*sqrt(d*x + c)/(d^3*x^3 + 3*c*d^2*x^2 + c^3 + (3*c^2*d + d)*x + (d^2*x^2 + 2*c*d*x + c^2 + 1
)^(3/2) + c), x))*b

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 159, normalized size = 1.12 \begin {gather*} \frac {2 \, {\left (3 \, {\left (b d^{3} x + b c d^{2}\right )} \sqrt {{\left (d x + c\right )} \cosh \left (1\right ) + {\left (d x + c\right )} \sinh \left (1\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) + 2 \, \sqrt {d^{3} \cosh \left (1\right ) + d^{3} \sinh \left (1\right )} b {\rm weierstrassPInverse}\left (-\frac {4}{d^{2}}, 0, \frac {d x + c}{d}\right ) + {\left (3 \, a d^{3} x + 3 \, a c d^{2} - 2 \, \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} b d^{2}\right )} \sqrt {{\left (d x + c\right )} \cosh \left (1\right ) + {\left (d x + c\right )} \sinh \left (1\right )}\right )}}{9 \, d^{3}} \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]

2/9*(3*(b*d^3*x + b*c*d^2)*sqrt((d*x + c)*cosh(1) + (d*x + c)*sinh(1))*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x +
c^2 + 1)) + 2*sqrt(d^3*cosh(1) + d^3*sinh(1))*b*weierstrassPInverse(-4/d^2, 0, (d*x + c)/d) + (3*a*d^3*x + 3*a
*c*d^2 - 2*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*b*d^2)*sqrt((d*x + c)*cosh(1) + (d*x + c)*sinh(1)))/d^3

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sqrt {c\,e+d\,e\,x}\,\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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