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

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

[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)

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Rubi [A]  time = 0.11, 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 = {5865, 5661, 329, 220} \[ \frac {2 b (c+d x+1) \sqrt {\frac {(c+d x)^2+1}{(c+d x+1)^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{d e^{3/2} \sqrt {(c+d x)^2+1}}-\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{d e \sqrt {e (c+d x)}} \]

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 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 329

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 5661

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 5865

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 {\operatorname {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) \operatorname {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) \operatorname {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*}

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Mathematica [C]  time = 0.03, size = 56, normalized size = 0.53 \[ -\frac {2 \left (a-2 b (c+d x) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-(c+d x)^2\right )+b \sinh ^{-1}(c+d x)\right )}{d e \sqrt {e (c+d x)}} \]

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)])

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d e x + c e} {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}}{d^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}}, x\right ) \]

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]

integral(sqrt(d*e*x + c*e)*(b*arcsinh(d*x + c) + a)/(d^2*e^2*x^2 + 2*c*d*e^2*x + c^2*e^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arsinh}\left (d x + c\right ) + a}{{\left (d e x + c e\right )}^{\frac {3}{2}}}\,{d x} \]

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)

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maple [C]  time = 0.01, size = 140, normalized size = 1.32 \[ \frac {-\frac {2 a}{\sqrt {d e x +c e}}+2 b \left (-\frac {\arcsinh \left (\frac {d e x +c e}{e}\right )}{\sqrt {d e x +c e}}+\frac {2 \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \EllipticF \left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )}{e \sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{d e} \]

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)

[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)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, b {\left (\frac {\frac {i \, \sqrt {2} {\left (\log \left (\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {d x + c}\right )} + 1\right ) - \log \left (-\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {d x + c}\right )} + 1\right )\right )}}{e^{\frac {3}{2}}} - \frac {i \, \sqrt {2} {\left (\log \left (\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {d x + c}\right )} + 1\right ) - \log \left (-\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {d x + c}\right )} + 1\right )\right )}}{e^{\frac {3}{2}}} + \frac {\sqrt {2} \log \left (d x + \sqrt {2} \sqrt {d x + c} + c + 1\right )}{e^{\frac {3}{2}}} - \frac {\sqrt {2} \log \left (d x - \sqrt {2} \sqrt {d x + c} + c + 1\right )}{e^{\frac {3}{2}}}}{d} + \frac {4 \, \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )}{\sqrt {d x + c} d e^{\frac {3}{2}}} - 4 \, \int \frac {1}{{\left (d^{2} e^{\frac {3}{2}} x^{2} + 2 \, c d e^{\frac {3}{2}} x + c^{2} e^{\frac {3}{2}} + e^{\frac {3}{2}}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} \sqrt {d x + c} + {\left (d^{3} e^{\frac {3}{2}} x^{3} + 3 \, c d^{2} e^{\frac {3}{2}} x^{2} + c^{3} e^{\frac {3}{2}} + c e^{\frac {3}{2}} + {\left (3 \, c^{2} d e^{\frac {3}{2}} + d e^{\frac {3}{2}}\right )} x\right )} \sqrt {d x + c}}\,{d x}\right )} - \frac {2 \, a}{\sqrt {d e x + c e} d e} \]

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

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

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)

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

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)

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