3.7.62 \(\int \frac {\sqrt {d+e x}}{(f+g x)^2 \sqrt {a d e+(c d^2+a e^2) x+c d e x^2}} \, dx\) [662]

Optimal. Leaf size=140 \[ \frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{(c d f-a e g) \sqrt {d+e x} (f+g x)}+\frac {c d \tan ^{-1}\left (\frac {\sqrt {g} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {c d f-a e g} \sqrt {d+e x}}\right )}{\sqrt {g} (c d f-a e g)^{3/2}} \]

[Out]

c*d*arctan(g^(1/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/(-a*e*g+c*d*f)^(1/2)/(e*x+d)^(1/2))/(-a*e*g+c*d*f)^
(3/2)/g^(1/2)+(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/(-a*e*g+c*d*f)/(g*x+f)/(e*x+d)^(1/2)

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Rubi [A]
time = 0.12, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {886, 888, 211} \begin {gather*} \frac {c d \text {ArcTan}\left (\frac {\sqrt {g} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {c d f-a e g}}\right )}{\sqrt {g} (c d f-a e g)^{3/2}}+\frac {\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} (f+g x) (c d f-a e g)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]/((c*d*f - a*e*g)*Sqrt[d + e*x]*(f + g*x)) + (c*d*ArcTan[(Sqrt[g]*S
qrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(Sqrt[c*d*f - a*e*g]*Sqrt[d + e*x])])/(Sqrt[g]*(c*d*f - a*e*g)^(3/
2))

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 886

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :>
Simp[(-e^2)*(d + e*x)^(m - 1)*(f + g*x)^(n + 1)*((a + b*x + c*x^2)^(p + 1)/((n + 1)*(c*e*f + c*d*g - b*e*g))),
x] - Dist[c*e*((m - n - 2)/((n + 1)*(c*e*f + c*d*g - b*e*g))), Int[(d + e*x)^m*(f + g*x)^(n + 1)*(a + b*x + c
*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, 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] && EqQ[m + p, 0] && LtQ[n, -1] && IntegerQ[2*p]

Rule 888

Int[Sqrt[(d_) + (e_.)*(x_)]/(((f_.) + (g_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[
2*e^2, Subst[Int[1/(c*(e*f + d*g) - b*e*g + e^2*g*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; Fre
eQ[{a, b, c, d, e, f, g}, 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]

Rubi steps

\begin {align*} \int \frac {\sqrt {d+e x}}{(f+g x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx &=\frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{(c d f-a e g) \sqrt {d+e x} (f+g x)}+\frac {(c d) \int \frac {\sqrt {d+e x}}{(f+g x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{2 (c d f-a e g)}\\ &=\frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{(c d f-a e g) \sqrt {d+e x} (f+g x)}+\frac {\left (c d e^2\right ) \text {Subst}\left (\int \frac {1}{-e \left (c d^2+a e^2\right ) g+c d e (e f+d g)+e^2 g x^2} \, dx,x,\frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {d+e x}}\right )}{c d f-a e g}\\ &=\frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{(c d f-a e g) \sqrt {d+e x} (f+g x)}+\frac {c d \tan ^{-1}\left (\frac {\sqrt {g} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {c d f-a e g} \sqrt {d+e x}}\right )}{\sqrt {g} (c d f-a e g)^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.27, size = 136, normalized size = 0.97 \begin {gather*} \frac {\sqrt {d+e x} \left (\sqrt {g} \sqrt {c d f-a e g} (a e+c d x)+c d \sqrt {a e+c d x} (f+g x) \tan ^{-1}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c d f-a e g}}\right )\right )}{\sqrt {g} (c d f-a e g)^{3/2} \sqrt {(a e+c d x) (d+e x)} (f+g x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + e*x]*(Sqrt[g]*Sqrt[c*d*f - a*e*g]*(a*e + c*d*x) + c*d*Sqrt[a*e + c*d*x]*(f + g*x)*ArcTan[(Sqrt[g]*Sq
rt[a*e + c*d*x])/Sqrt[c*d*f - a*e*g]]))/(Sqrt[g]*(c*d*f - a*e*g)^(3/2)*Sqrt[(a*e + c*d*x)*(d + e*x)]*(f + g*x)
)

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Maple [A]
time = 0.15, size = 158, normalized size = 1.13

method	result	size
default	\(\frac {\sqrt {\left (c d x +a e \right ) \left (e x +d \right )}\, \left (\arctanh \left (\frac {g \sqrt {c d x +a e}}{\sqrt {\left (a e g -c d f \right ) g}}\right ) c d g x +\arctanh \left (\frac {g \sqrt {c d x +a e}}{\sqrt {\left (a e g -c d f \right ) g}}\right ) c d f -\sqrt {c d x +a e}\, \sqrt {\left (a e g -c d f \right ) g}\right )}{\sqrt {e x +d}\, \sqrt {c d x +a e}\, \left (a e g -c d f \right ) \left (g x +f \right ) \sqrt {\left (a e g -c d f \right ) g}}\)	\(158\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 344 vs. \(2 (131) = 262\).
time = 3.94, size = 727, normalized size = 5.19 \begin {gather*} \left [\frac {{\left (c d^{2} g x + c d^{2} f + {\left (c d g x^{2} + c d f x\right )} e\right )} \sqrt {-c d f g + a g^{2} e} \log \left (-\frac {c d^{2} g x - c d^{2} f + 2 \, a g x e^{2} + {\left (c d g x^{2} - c d f x + 2 \, a d g\right )} e + 2 \, \sqrt {-c d f g + a g^{2} e} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {x e + d}}{d g x + d f + {\left (g x^{2} + f x\right )} e}\right ) + 2 \, {\left (c d f g - a g^{2} e\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {x e + d}}{2 \, {\left (c^{2} d^{3} f^{2} g^{2} x + c^{2} d^{3} f^{3} g + {\left (a^{2} g^{4} x^{2} + a^{2} f g^{3} x\right )} e^{3} - {\left (2 \, a c d f g^{3} x^{2} - a^{2} d f g^{3} + {\left (2 \, a c d f^{2} g^{2} - a^{2} d g^{4}\right )} x\right )} e^{2} + {\left (c^{2} d^{2} f^{2} g^{2} x^{2} - 2 \, a c d^{2} f^{2} g^{2} + {\left (c^{2} d^{2} f^{3} g - 2 \, a c d^{2} f g^{3}\right )} x\right )} e\right )}}, -\frac {{\left (c d^{2} g x + c d^{2} f + {\left (c d g x^{2} + c d f x\right )} e\right )} \sqrt {c d f g - a g^{2} e} \arctan \left (\frac {\sqrt {c d f g - a g^{2} e} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {x e + d}}{c d^{2} g x + a g x e^{2} + {\left (c d g x^{2} + a d g\right )} e}\right ) - {\left (c d f g - a g^{2} e\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {x e + d}}{c^{2} d^{3} f^{2} g^{2} x + c^{2} d^{3} f^{3} g + {\left (a^{2} g^{4} x^{2} + a^{2} f g^{3} x\right )} e^{3} - {\left (2 \, a c d f g^{3} x^{2} - a^{2} d f g^{3} + {\left (2 \, a c d f^{2} g^{2} - a^{2} d g^{4}\right )} x\right )} e^{2} + {\left (c^{2} d^{2} f^{2} g^{2} x^{2} - 2 \, a c d^{2} f^{2} g^{2} + {\left (c^{2} d^{2} f^{3} g - 2 \, a c d^{2} f g^{3}\right )} x\right )} e}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )} \left (f + g x\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(1/2)/(g*x+f)**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))*(f + g*x)**2), 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)^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}}{{\left (f+g\,x\right )}^2\,\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)^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)^2*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)), x)

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