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

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

[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 = 0.19, 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 = {872, 874, 205} \[ \frac {\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} (f+g x) (c d f-a e g)}+\frac {c d \tan ^{-1}\left (\frac {\sqrt {g} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {c d f-a e g}}\right )}{\sqrt {g} (c d f-a e g)^{3/2}} \]

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 205

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

Rule 872

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 874

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 ) \operatorname {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 = 0.11, size = 136, normalized size = 0.97 \[ \frac {\sqrt {d+e x} \left (\sqrt {g} (a e+c d x) \sqrt {c d f-a e g}+c d (f+g x) \sqrt {a e+c d x} \tan ^{-1}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c d f-a e g}}\right )\right )}{\sqrt {g} (f+g x) \sqrt {(d+e x) (a e+c d x)} (c d f-a e g)^{3/2}} \]

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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fricas [B]  time = 1.27, size = 703, normalized size = 5.02 \[ \left [\frac {{\left (c d e g x^{2} + c d^{2} f + {\left (c d e f + c d^{2} g\right )} x\right )} \sqrt {-c d f g + a e g^{2}} \log \left (-\frac {c d e g x^{2} - c d^{2} f + 2 \, a d e g - {\left (c d e f - {\left (c d^{2} + 2 \, a e^{2}\right )} g\right )} x + 2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {-c d f g + a e g^{2}} \sqrt {e x + d}}{e g x^{2} + d f + {\left (e f + d g\right )} x}\right ) + 2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (c d f g - a e g^{2}\right )} \sqrt {e x + d}}{2 \, {\left (c^{2} d^{3} f^{3} g - 2 \, a c d^{2} e f^{2} g^{2} + a^{2} d e^{2} f g^{3} + {\left (c^{2} d^{2} e f^{2} g^{2} - 2 \, a c d e^{2} f g^{3} + a^{2} e^{3} g^{4}\right )} x^{2} + {\left (c^{2} d^{2} e f^{3} g + a^{2} d e^{2} g^{4} + {\left (c^{2} d^{3} - 2 \, a c d e^{2}\right )} f^{2} g^{2} - {\left (2 \, a c d^{2} e - a^{2} e^{3}\right )} f g^{3}\right )} x\right )}}, -\frac {{\left (c d e g x^{2} + c d^{2} f + {\left (c d e f + c d^{2} g\right )} x\right )} \sqrt {c d f g - a e g^{2}} \arctan \left (\frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {c d f g - a e g^{2}} \sqrt {e x + d}}{c d e g x^{2} + a d e g + {\left (c d^{2} + a e^{2}\right )} g x}\right ) - \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (c d f g - a e g^{2}\right )} \sqrt {e x + d}}{c^{2} d^{3} f^{3} g - 2 \, a c d^{2} e f^{2} g^{2} + a^{2} d e^{2} f g^{3} + {\left (c^{2} d^{2} e f^{2} g^{2} - 2 \, a c d e^{2} f g^{3} + a^{2} e^{3} g^{4}\right )} x^{2} + {\left (c^{2} d^{2} e f^{3} g + a^{2} d e^{2} g^{4} + {\left (c^{2} d^{3} - 2 \, a c d e^{2}\right )} f^{2} g^{2} - {\left (2 \, a c d^{2} e - a^{2} e^{3}\right )} f g^{3}\right )} x}\right ] \]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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maple [A]  time = 0.02, size = 168, normalized size = 1.20 \[ \frac {\sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}\, \left (c d g x \arctanh \left (\frac {\sqrt {c d x +a e}\, g}{\sqrt {\left (a e g -c d f \right ) g}}\right )+c d f \arctanh \left (\frac {\sqrt {c d x +a e}\, g}{\sqrt {\left (a e g -c d f \right ) g}}\right )-\sqrt {c d x +a e}\, \sqrt {\left (a e g -c d f \right ) g}\right )}{\sqrt {e x +d}\, \sqrt {c d x +a e}\, \left (a e g -c d f \right ) \left (g x +f \right ) \sqrt {\left (a e g -c d f \right ) g}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(1/2)*(arctanh((c*d*x+a*e)^(1/2)/((a*e*g-c*d*f)*g)^(1/2)*g)*x*c*d*g+arctanh(
(c*d*x+a*e)^(1/2)/((a*e*g-c*d*f)*g)^(1/2)*g)*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 \[ \int \frac {\sqrt {e x + d}}{\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (g x + f\right )}^{2}}\,{d x} \]

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

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

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

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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