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

Optimal. Leaf size=124 \[ \frac {2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{g \sqrt {d+e x}}-\frac {2 \sqrt {c d f-a e g} \tan ^{-1}\left (\frac {\sqrt {g} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {c d f-a e g}}\right )}{g^{3/2}} \]

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Rubi [A]  time = 0.19, antiderivative size = 124, 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 = {864, 874, 205} \begin {gather*} \frac {2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{g \sqrt {d+e x}}-\frac {2 \sqrt {c d f-a e g} \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}} \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)),x]

[Out]

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

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 864

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*(m - n - 1)), x] - Dist[(m*(c*e*f + c*d*g - b*e*g
))/(e^2*g*(m - n - 1)), Int[(d + e*x)^(m + 1)*(f + g*x)^n*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c,
d, e, f, g, 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] && EqQ[m + p, 0] && GtQ[p, 0] && NeQ[m - n - 1, 0] &&  !IGtQ[n, 0] &&  !(IntegerQ[n + p] && LtQ[n + p + 2,
0]) && RationalQ[n]

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

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Mathematica [A]  time = 0.13, size = 101, normalized size = 0.81 \begin {gather*} \frac {2 \sqrt {(d+e x) (a e+c d x)} \left (\sqrt {g}-\frac {\sqrt {c d f-a e g} \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}}\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)),x]

[Out]

(2*Sqrt[(a*e + c*d*x)*(d + e*x)]*(Sqrt[g] - (Sqrt[c*d*f - a*e*g]*ArcTan[(Sqrt[g]*Sqrt[a*e + c*d*x])/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 = 4.70, size = 931, normalized size = 7.51 \begin {gather*} -\frac {2 \left (-a 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} e^{3/2}+c d f \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 {e}-i \sqrt {c} \sqrt {d} \sqrt {d g-e f} \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}\right ) \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 )}{\left (c d^2-a e^2\right ) g^{5/2}}-\frac {2 \left (-a 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} e^{3/2}+c d f \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 {e}+i \sqrt {c} \sqrt {d} \sqrt {d g-e f} \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}\right ) \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 )}{\left (c d^2-a e^2\right ) g^{5/2}}+\frac {2 \sqrt {-\frac {c (d+e x) d^2}{e}+\frac {c (d+e x)^2 d}{e}+a e (d+e x)}}{g \sqrt {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)),x]

[Out]

(2*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]) - (2*(c*d*Sqrt[e]*f*S
qrt[-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]] - a
*e^(3/2)*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]] - I*Sqrt[c]*Sqrt[d]*Sqrt[-(e*f) + d*g]*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]])*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])])/((c*d^2
- a*e^2)*g^(5/2)) - (2*(c*d*Sqrt[e]*f*Sqrt[-2*c*d*e*f + c*d^2*g + a*e^2*g + (2*I)*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqr
t[-(e*f) + d*g]*Sqrt[c*d*f - a*e*g]] - a*e^(3/2)*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]] + I*Sqrt[c]*Sqrt[d]*Sqrt[-(e*f) + d*g]*Sqrt[c*d*f - a*e*g]*Sq
rt[-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]])*Arc
Tanh[(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])])/((c*d^2 - a*e^2)*g^(5/2))

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fricas [A]  time = 0.44, size = 318, normalized size = 2.56 \begin {gather*} \left [\frac {{\left (e x + d\right )} \sqrt {-\frac {c d f - a e g}{g}} \log \left (-\frac {c d e g x^{2} - c d^{2} f + 2 \, a d e g - 2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d} g \sqrt {-\frac {c d f - a e g}{g}} - {\left (c d e f - {\left (c d^{2} + 2 \, a e^{2}\right )} g\right )} x}{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} \sqrt {e x + d}}{e g x + d g}, \frac {2 \, {\left ({\left (e x + d\right )} \sqrt {\frac {c d f - a e g}{g}} \arctan \left (\frac {\sqrt {e x + d} \sqrt {\frac {c d f - a e g}{g}}}{\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}\right ) + \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}\right )}}{e g x + d g}\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)/(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

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

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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)/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.02, size = 153, normalized size = 1.23 \begin {gather*} -\frac {2 \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}\, \left (a e g \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}\, \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)/(e*x+d)^(1/2),x)

[Out]

-2*(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)*a*e*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)/g/((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 )}}\,{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)/(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)), 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 )\,\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)*(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)*(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 )}\, 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)/(e*x+d)**(1/2),x)

[Out]

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

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