3.8.15 \(\int \frac {\sqrt {d+e x}}{\sqrt {f+g x} \sqrt {a d e+(c d^2+a e^2) x+c d e x^2}} \, dx\) [715]

Optimal. Leaf size=105 \[ \frac {2 \sqrt {a e+c d x} \sqrt {d+e x} \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c} \sqrt {d} \sqrt {f+g x}}\right )}{\sqrt {c} \sqrt {d} \sqrt {g} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \]

[Out]

2*arctanh(g^(1/2)*(c*d*x+a*e)^(1/2)/c^(1/2)/d^(1/2)/(g*x+f)^(1/2))*(c*d*x+a*e)^(1/2)*(e*x+d)^(1/2)/c^(1/2)/d^(
1/2)/g^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {905, 65, 223, 212} \begin {gather*} \frac {2 \sqrt {d+e x} \sqrt {a e+c d x} \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c} \sqrt {d} \sqrt {f+g x}}\right )}{\sqrt {c} \sqrt {d} \sqrt {g} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[d + e*x]/(Sqrt[f + g*x]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]),x]

[Out]

(2*Sqrt[a*e + c*d*x]*Sqrt[d + e*x]*ArcTanh[(Sqrt[g]*Sqrt[a*e + c*d*x])/(Sqrt[c]*Sqrt[d]*Sqrt[f + g*x])])/(Sqrt
[c]*Sqrt[d]*Sqrt[g]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 905

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :>
Dist[(a + b*x + c*x^2)^FracPart[p]/((d + e*x)^FracPart[p]*(a/d + (c*x)/e)^FracPart[p]), Int[(d + e*x)^(m + p)*
(f + g*x)^n*(a/d + (c/e)*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2
- 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] &&  !IGtQ[m, 0] &&  !IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\sqrt {d+e x}}{\sqrt {f+g x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx &=\frac {\left (\sqrt {a e+c d x} \sqrt {d+e x}\right ) \int \frac {1}{\sqrt {a e+c d x} \sqrt {f+g x}} \, dx}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=\frac {\left (2 \sqrt {a e+c d x} \sqrt {d+e x}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {f-\frac {a e g}{c d}+\frac {g x^2}{c d}}} \, dx,x,\sqrt {a e+c d x}\right )}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=\frac {\left (2 \sqrt {a e+c d x} \sqrt {d+e x}\right ) \text {Subst}\left (\int \frac {1}{1-\frac {g x^2}{c d}} \, dx,x,\frac {\sqrt {a e+c d x}}{\sqrt {f+g x}}\right )}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=\frac {2 \sqrt {a e+c d x} \sqrt {d+e x} \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c} \sqrt {d} \sqrt {f+g x}}\right )}{\sqrt {c} \sqrt {d} \sqrt {g} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 94, normalized size = 0.90 \begin {gather*} \frac {2 \sqrt {a e+c d x} \sqrt {d+e x} \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c} \sqrt {d} \sqrt {f+g x}}\right )}{\sqrt {c} \sqrt {d} \sqrt {g} \sqrt {(a e+c d x) (d+e x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d + e*x]/(Sqrt[f + g*x]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]),x]

[Out]

(2*Sqrt[a*e + c*d*x]*Sqrt[d + e*x]*ArcTanh[(Sqrt[g]*Sqrt[a*e + c*d*x])/(Sqrt[c]*Sqrt[d]*Sqrt[f + g*x])])/(Sqrt
[c]*Sqrt[d]*Sqrt[g]*Sqrt[(a*e + c*d*x)*(d + e*x)])

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Maple [A]
time = 0.14, size = 102, normalized size = 0.97

method	result	size
default	\(\frac {\sqrt {g x +f}\, \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}\, \ln \left (\frac {2 c d g x +a e g +c d f +2 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}}{2 \sqrt {d g c}}\right )}{\sqrt {e x +d}\, \sqrt {d g c}\, \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}}\)	\(102\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(e*x+d)^(1/2)*(g*x+f)^(1/2)*((c*d*x+a*e)*(e*x+d))^(1/2)*ln(1/2*(2*c*d*g*x+a*e*g+c*d*f+2*((g*x+f)*(c*d*x+a*e)
)^(1/2)*(d*g*c)^(1/2))/(d*g*c)^(1/2))/(d*g*c)^(1/2)/((g*x+f)*(c*d*x+a*e))^(1/2)

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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((e*x+d)^(1/2)/(g*x+f)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(x*e + d)/(sqrt(c*d*x^2*e + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(g*x + f)), x)

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Fricas [A]
time = 2.62, size = 349, normalized size = 3.32 \begin {gather*} \left [\frac {\sqrt {c d g} \log \left (-\frac {8 \, c^{2} d^{3} g^{2} x^{2} + 8 \, c^{2} d^{3} f g x + c^{2} d^{3} f^{2} + a^{2} g^{2} x e^{3} + 4 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} {\left (2 \, c d g x + c d f + a g e\right )} \sqrt {c d g} \sqrt {g x + f} \sqrt {x e + d} + {\left (8 \, a c d g^{2} x^{2} + 6 \, a c d f g x + a^{2} d g^{2}\right )} e^{2} + {\left (8 \, c^{2} d^{2} g^{2} x^{3} + 8 \, c^{2} d^{2} f g x^{2} + 6 \, a c d^{2} f g + {\left (c^{2} d^{2} f^{2} + 8 \, a c d^{2} g^{2}\right )} x\right )} e}{x e + d}\right )}{2 \, c d g}, -\frac {\sqrt {-c d g} \arctan \left (\frac {2 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {-c d g} \sqrt {g x + f} \sqrt {x e + d}}{2 \, c d^{2} g x + c d^{2} f + a g x e^{2} + {\left (2 \, c d g x^{2} + c d f x + a d g\right )} e}\right )}{c d g}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(1/2)/(g*x+f)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="fricas")

[Out]

[1/2*sqrt(c*d*g)*log(-(8*c^2*d^3*g^2*x^2 + 8*c^2*d^3*f*g*x + c^2*d^3*f^2 + a^2*g^2*x*e^3 + 4*sqrt(c*d^2*x + a*
x*e^2 + (c*d*x^2 + a*d)*e)*(2*c*d*g*x + c*d*f + a*g*e)*sqrt(c*d*g)*sqrt(g*x + f)*sqrt(x*e + d) + (8*a*c*d*g^2*
x^2 + 6*a*c*d*f*g*x + a^2*d*g^2)*e^2 + (8*c^2*d^2*g^2*x^3 + 8*c^2*d^2*f*g*x^2 + 6*a*c*d^2*f*g + (c^2*d^2*f^2 +
8*a*c*d^2*g^2)*x)*e)/(x*e + d))/(c*d*g), -sqrt(-c*d*g)*arctan(2*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*s
qrt(-c*d*g)*sqrt(g*x + f)*sqrt(x*e + d)/(2*c*d^2*g*x + c*d^2*f + a*g*x*e^2 + (2*c*d*g*x^2 + c*d*f*x + a*d*g)*e
))/(c*d*g)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {d + e x}}{\sqrt {\left (d + e x\right ) \left (a e + c d x\right )} \sqrt {f + g x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(1/2)/(g*x+f)**(1/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(1/2),x)

[Out]

Integral(sqrt(d + e*x)/(sqrt((d + e*x)*(a*e + c*d*x))*sqrt(f + g*x)), x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(1/2)/(g*x+f)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {d+e\,x}}{\sqrt {f+g\,x}\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(1/2)/((f + g*x)^(1/2)*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)),x)

[Out]

int((d + e*x)^(1/2)/((f + g*x)^(1/2)*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)), x)

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