\(\int \frac {\sqrt {d+e x}}{(f+g x)^{3/2} \sqrt {a d e+(c d^2+a e^2) x+c d e x^2}} \, dx\) [716]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 48, antiderivative size = 61 \[ \int \frac {\sqrt {d+e x}}{(f+g x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{(c d f-a e g) \sqrt {d+e x} \sqrt {f+g x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {874} \[ \int \frac {\sqrt {d+e x}}{(f+g x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {f+g x} (c d f-a e g)} \]

[In]

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

[Out]

(2*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]*Sqrt[f + g*x])

Rule 874

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] /; FreeQ[{a, b, c, d, e, f, g, m, n, 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] && EqQ[m - n - 2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \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} \sqrt {f+g x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.82 \[ \int \frac {\sqrt {d+e x}}{(f+g x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \sqrt {(a e+c d x) (d+e x)}}{(c d f-a e g) \sqrt {d+e x} \sqrt {f+g x}} \]

[In]

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

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.74

method result size
default \(-\frac {2 \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}}{\sqrt {e x +d}\, \sqrt {g x +f}\, \left (a e g -c d f \right )}\) \(45\)
gosper \(-\frac {2 \left (c d x +a e \right ) \sqrt {e x +d}}{\sqrt {g x +f}\, \left (a e g -c d f \right ) \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}}\) \(63\)

[In]

int((e*x+d)^(1/2)/(g*x+f)^(3/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 114 vs. \(2 (55) = 110\).

Time = 0.35 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.87 \[ \int \frac {\sqrt {d+e x}}{(f+g x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d} \sqrt {g x + f}}{c d^{2} f^{2} - a d e f g + {\left (c d e f g - a e^{2} g^{2}\right )} x^{2} + {\left (c d e f^{2} - a d e g^{2} + {\left (c d^{2} - a e^{2}\right )} f g\right )} x} \]

[In]

integrate((e*x+d)^(1/2)/(g*x+f)^(3/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\sqrt {d+e x}}{(f+g x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\int \frac {\sqrt {d + e x}}{\sqrt {\left (d + e x\right ) \left (a e + c d x\right )} \left (f + g x\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate((e*x+d)**(1/2)/(g*x+f)**(3/2)/(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)**(3/2)), x)

Maxima [F]

\[ \int \frac {\sqrt {d+e x}}{(f+g x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\int { \frac {\sqrt {e x + d}}{\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (g x + f\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate((e*x+d)^(1/2)/(g*x+f)^(3/2)/(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)^(3/2)), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (55) = 110\).

Time = 0.31 (sec) , antiderivative size = 194, normalized size of antiderivative = 3.18 \[ \int \frac {\sqrt {d+e x}}{(f+g x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=-\frac {2 \, \sqrt {c d g} e^{2} g}{c d e^{2} f g {\left | g \right |} - c d^{2} e g^{2} {\left | g \right |} - \sqrt {-c d^{2} e g^{2} + a e^{3} g^{2}} \sqrt {e^{2} f - d e g} \sqrt {c d g} {\left | g \right |}} + \frac {4 \, \sqrt {c d g} e^{2} g}{{\left (c d e^{2} f g - a e^{3} g^{2} + {\left (\sqrt {e^{2} f + {\left (e x + d\right )} e g - d e g} \sqrt {c d g} - \sqrt {-c d e^{2} f g + a e^{3} g^{2} + {\left (e^{2} f + {\left (e x + d\right )} e g - d e g\right )} c d g}\right )}^{2}\right )} {\left | g \right |}} \]

[In]

integrate((e*x+d)^(1/2)/(g*x+f)^(3/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="giac")

[Out]

-2*sqrt(c*d*g)*e^2*g/(c*d*e^2*f*g*abs(g) - c*d^2*e*g^2*abs(g) - sqrt(-c*d^2*e*g^2 + a*e^3*g^2)*sqrt(e^2*f - d*
e*g)*sqrt(c*d*g)*abs(g)) + 4*sqrt(c*d*g)*e^2*g/((c*d*e^2*f*g - a*e^3*g^2 + (sqrt(e^2*f + (e*x + d)*e*g - d*e*g
)*sqrt(c*d*g) - sqrt(-c*d*e^2*f*g + a*e^3*g^2 + (e^2*f + (e*x + d)*e*g - d*e*g)*c*d*g))^2)*abs(g))

Mupad [B] (verification not implemented)

Time = 13.99 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.64 \[ \int \frac {\sqrt {d+e x}}{(f+g x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=-\frac {2\,\sqrt {d+e\,x}\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{\left (x\,\sqrt {f+g\,x}-\frac {\sqrt {f+g\,x}\,\left (c\,d^2\,f-a\,d\,e\,g\right )}{a\,e^2\,g-c\,d\,e\,f}\right )\,\left (a\,e^2\,g-c\,d\,e\,f\right )} \]

[In]

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

[Out]

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