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

Optimal. Leaf size=132 \[ \frac {c d \tan ^{-1}\left (\frac {\sqrt {g} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {c d f-a e g}}\right )}{g^{3/2} \sqrt {c d f-a e g}}-\frac {\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{g \sqrt {d+e x} (f+g x)} \]

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Rubi [A]  time = 0.16, antiderivative size = 132, 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 = {862, 874, 205} \begin {gather*} \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 )}{g^{3/2} \sqrt {c d f-a e g}}-\frac {\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{g \sqrt {d+e x} (f+g x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 862

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

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

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Mathematica [A]  time = 0.19, size = 110, normalized size = 0.83 \begin {gather*} \frac {\sqrt {(d+e x) (a e+c d x)} \left (\frac {c d \tan ^{-1}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c d f-a e g}}\right )}{\sqrt {a e+c d x} \sqrt {c d f-a e g}}-\frac {\sqrt {g}}{f+g x}\right )}{g^{3/2} \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [C]  time = 12.70, size = 1241, normalized size = 9.40 \begin {gather*} -\frac {i c^{3/2} \sqrt {d g-e f} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {c g d^2-2 c e f d-2 i \sqrt {c} \sqrt {e} \sqrt {d g-e f} \sqrt {c d f-a e g} \sqrt {d}+a e^2 g} \sqrt {d+e x}}{e \sqrt {g} \sqrt {-\frac {c (d+e x) d^2}{e}+\frac {c (d+e x)^2 d}{e}+a e (d+e x)}-\sqrt {c d e} \sqrt {g} (d+e x)}\right ) d^{3/2}}{g^{3/2} \sqrt {c d f-a e g} \sqrt {c g d^2-2 c e f d-2 i \sqrt {c} \sqrt {e} \sqrt {d g-e f} \sqrt {c d f-a e g} \sqrt {d}+a e^2 g}}+\frac {i c^{3/2} \sqrt {d g-e f} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {c g d^2-2 c e f d+2 i \sqrt {c} \sqrt {e} \sqrt {d g-e f} \sqrt {c d f-a e g} \sqrt {d}+a e^2 g} \sqrt {d+e x}}{e \sqrt {g} \sqrt {-\frac {c (d+e x) d^2}{e}+\frac {c (d+e x)^2 d}{e}+a e (d+e x)}-\sqrt {c d e} \sqrt {g} (d+e x)}\right ) d^{3/2}}{g^{3/2} \sqrt {c d f-a e g} \sqrt {c g d^2-2 c e f d+2 i \sqrt {c} \sqrt {e} \sqrt {d g-e f} \sqrt {c d f-a e g} \sqrt {d}+a e^2 g}}-\frac {c \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {c g d^2-2 c e f d-2 i \sqrt {c} \sqrt {e} \sqrt {d g-e f} \sqrt {c d f-a e g} \sqrt {d}+a e^2 g} \sqrt {d+e x}}{e \sqrt {g} \sqrt {-\frac {c (d+e x) d^2}{e}+\frac {c (d+e x)^2 d}{e}+a e (d+e x)}-\sqrt {c d e} \sqrt {g} (d+e x)}\right ) d}{g^{3/2} \sqrt {c g d^2-2 c e f d-2 i \sqrt {c} \sqrt {e} \sqrt {d g-e f} \sqrt {c d f-a e g} \sqrt {d}+a e^2 g}}-\frac {c \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {c g d^2-2 c e f d+2 i \sqrt {c} \sqrt {e} \sqrt {d g-e f} \sqrt {c d f-a e g} \sqrt {d}+a e^2 g} \sqrt {d+e x}}{e \sqrt {g} \sqrt {-\frac {c (d+e x) d^2}{e}+\frac {c (d+e x)^2 d}{e}+a e (d+e x)}-\sqrt {c d e} \sqrt {g} (d+e x)}\right ) d}{g^{3/2} \sqrt {c g d^2-2 c e f d+2 i \sqrt {c} \sqrt {e} \sqrt {d g-e f} \sqrt {c d f-a e g} \sqrt {d}+a e^2 g}}+\frac {\sqrt {-\frac {c (d+e x) d^2}{e}+\frac {c (d+e x)^2 d}{e}+a e (d+e x)} \left (-a e^3+c d^2 e-2 c d (d+e x) e\right )+\sqrt {c d e} \left (-2 c (d+e x) d^2+2 c (d+e x)^2 d+2 a e^2 (d+e x)\right )}{g \sqrt {d+e x} \left (-c d^2+2 c (d+e x) d+a e^2\right ) (e f-d g+g (d+e x))-2 \sqrt {c d e} g \sqrt {d+e x} (e f-d g+g (d+e x)) \sqrt {-\frac {c (d+e x) d^2}{e}+\frac {c (d+e x)^2 d}{e}+a e (d+e x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c*d*e]*(-2*c*d^2*(d + e*x) + 2*a*e^2*(d + e*x) + 2*c*d*(d + e*x)^2) + (c*d^2*e - a*e^3 - 2*c*d*e*(d + e*
x))*Sqrt[-((c*d^2*(d + e*x))/e) + a*e*(d + e*x) + (c*d*(d + e*x)^2)/e])/(g*Sqrt[d + e*x]*(-(c*d^2) + a*e^2 + 2
*c*d*(d + e*x))*(e*f - d*g + g*(d + e*x)) - 2*Sqrt[c*d*e]*g*Sqrt[d + e*x]*(e*f - d*g + g*(d + e*x))*Sqrt[-((c*
d^2*(d + e*x))/e) + a*e*(d + e*x) + (c*d*(d + e*x)^2)/e]) - (c*d*Sqrt[e]*ArcTanh[(Sqrt[e]*Sqrt[-2*c*d*e*f + c*
d^2*g + a*e^2*g - (2*I)*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[-(e*f) + d*g]*Sqrt[c*d*f - a*e*g]]*Sqrt[d + e*x])/(-(Sqrt
[c*d*e]*Sqrt[g]*(d + e*x)) + e*Sqrt[g]*Sqrt[-((c*d^2*(d + e*x))/e) + a*e*(d + e*x) + (c*d*(d + e*x)^2)/e])])/(
g^(3/2)*Sqrt[-2*c*d*e*f + c*d^2*g + a*e^2*g - (2*I)*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[-(e*f) + d*g]*Sqrt[c*d*f - a*
e*g]]) - (I*c^(3/2)*d^(3/2)*Sqrt[-(e*f) + d*g]*ArcTanh[(Sqrt[e]*Sqrt[-2*c*d*e*f + c*d^2*g + a*e^2*g - (2*I)*Sq
rt[c]*Sqrt[d]*Sqrt[e]*Sqrt[-(e*f) + d*g]*Sqrt[c*d*f - a*e*g]]*Sqrt[d + e*x])/(-(Sqrt[c*d*e]*Sqrt[g]*(d + e*x))
+ e*Sqrt[g]*Sqrt[-((c*d^2*(d + e*x))/e) + a*e*(d + e*x) + (c*d*(d + e*x)^2)/e])])/(g^(3/2)*Sqrt[c*d*f - a*e*g
]*Sqrt[-2*c*d*e*f + c*d^2*g + a*e^2*g - (2*I)*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[-(e*f) + d*g]*Sqrt[c*d*f - a*e*g]])
- (c*d*Sqrt[e]*ArcTanh[(Sqrt[e]*Sqrt[-2*c*d*e*f + c*d^2*g + a*e^2*g + (2*I)*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[-(e*
f) + d*g]*Sqrt[c*d*f - a*e*g]]*Sqrt[d + e*x])/(-(Sqrt[c*d*e]*Sqrt[g]*(d + e*x)) + e*Sqrt[g]*Sqrt[-((c*d^2*(d +
e*x))/e) + a*e*(d + e*x) + (c*d*(d + e*x)^2)/e])])/(g^(3/2)*Sqrt[-2*c*d*e*f + c*d^2*g + a*e^2*g + (2*I)*Sqrt[
c]*Sqrt[d]*Sqrt[e]*Sqrt[-(e*f) + d*g]*Sqrt[c*d*f - a*e*g]]) + (I*c^(3/2)*d^(3/2)*Sqrt[-(e*f) + d*g]*ArcTanh[(S
qrt[e]*Sqrt[-2*c*d*e*f + c*d^2*g + a*e^2*g + (2*I)*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[-(e*f) + d*g]*Sqrt[c*d*f - a*e
*g]]*Sqrt[d + e*x])/(-(Sqrt[c*d*e]*Sqrt[g]*(d + e*x)) + e*Sqrt[g]*Sqrt[-((c*d^2*(d + e*x))/e) + a*e*(d + e*x)
+ (c*d*(d + e*x)^2)/e])])/(g^(3/2)*Sqrt[c*d*f - a*e*g]*Sqrt[-2*c*d*e*f + c*d^2*g + a*e^2*g + (2*I)*Sqrt[c]*Sqr
t[d]*Sqrt[e]*Sqrt[-(e*f) + d*g]*Sqrt[c*d*f - a*e*g]])

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fricas [B]  time = 0.45, size = 562, normalized size = 4.26 \begin {gather*} \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 d^{2} f^{2} g^{2} - a d e f g^{3} + {\left (c d e f g^{3} - a e^{2} g^{4}\right )} x^{2} + {\left (c d e f^{2} g^{2} - a d e g^{4} + {\left (c d^{2} - a e^{2}\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 d^{2} f^{2} g^{2} - a d e f g^{3} + {\left (c d e f g^{3} - a e^{2} g^{4}\right )} x^{2} + {\left (c d e f^{2} g^{2} - a d e g^{4} + {\left (c d^{2} - a e^{2}\right )} f g^{3}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/(g*x+f)^2/(e*x+d)^(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*d^2*f^2*g^2 - a*d*e*f*g^3 + (c*d*e*f*g^3 - a*e^2*g^4)*x^2 + (c*d*e*f^2*g^2 -
a*d*e*g^4 + (c*d^2 - a*e^2)*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*d^2*f^2*g^2 - a*d*e*f*g^3 + (c*d*e*f*g^3 - a*e^2*g^4)*x^2 + (c*d*e*f^2*g^2 - a*d*e*g^4 + (c*d^2 - a*e^2)*f*
g^3)*x)]

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

[Out]

Timed out

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maple [A]  time = 0.02, size = 161, normalized size = 1.22 \begin {gather*} \frac {\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 {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}}{\sqrt {e x +d}\, \sqrt {c d x +a e}\, \left (g x +f \right ) \sqrt {\left (a e g -c d f \right ) g}\, g} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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