\(\int \frac {\sqrt {d+e x}}{(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx\) [2068]

Optimal result

Integrand size = 39, antiderivative size = 139 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=-\frac {2 \sqrt {d+e x}}{\left (c d^2-a e^2\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {2 \sqrt {e} \arctan \left (\frac {\sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {c d^2-a e^2} \sqrt {d+e x}}\right )}{\left (c d^2-a e^2\right )^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {680, 674, 211} \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=-\frac {2 \sqrt {e} \arctan \left (\frac {\sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{3/2}}-\frac {2 \sqrt {d+e x}}{\left (c d^2-a e^2\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \]

[In]

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

[Out]

(-2*Sqrt[d + e*x])/((c*d^2 - a*e^2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - (2*Sqrt[e]*ArcTan[(Sqrt[e]*
Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(Sqrt[c*d^2 - a*e^2]*Sqrt[d + e*x])])/(c*d^2 - a*e^2)^(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 674

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(
2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^
2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 680

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(2*c*d - b*e)*(d + e
*x)^m*((a + b*x + c*x^2)^(p + 1)/(e*(p + 1)*(b^2 - 4*a*c))), x] - Dist[(2*c*d - b*e)*((m + 2*p + 2)/((p + 1)*(
b^2 - 4*a*c))), Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^
2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && LtQ[0, m, 1] && IntegerQ[2*p]

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.83 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=-\frac {2 \sqrt {d+e x} \left (\sqrt {c d^2-a e^2}+\sqrt {e} \sqrt {a e+c d x} \arctan \left (\frac {\sqrt {e} \sqrt {a e+c d x}}{\sqrt {c d^2-a e^2}}\right )\right )}{\left (c d^2-a e^2\right )^{3/2} \sqrt {(a e+c d x) (d+e x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.65 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.91

method	result	size
default	\(-\frac {2 \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}\, \left (e \,\operatorname {arctanh}\left (\frac {e \sqrt {c d x +a e}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) e}}\right ) \sqrt {c d x +a e}-\sqrt {\left (e^{2} a -c \,d^{2}\right ) e}\right )}{\sqrt {e x +d}\, \left (c d x +a e \right ) \left (e^{2} a -c \,d^{2}\right ) \sqrt {\left (e^{2} a -c \,d^{2}\right ) e}}\)	\(126\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 500, normalized size of antiderivative = 3.60 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=\left [-\frac {{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )} \sqrt {-\frac {e}{c d^{2} - a e^{2}}} \log \left (-\frac {c d e^{2} x^{2} + 2 \, a e^{3} x - c d^{3} + 2 \, a d e^{2} + 2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (c d^{2} - a e^{2}\right )} \sqrt {e x + d} \sqrt {-\frac {e}{c d^{2} - a e^{2}}}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + 2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{a c d^{3} e - a^{2} d e^{3} + {\left (c^{2} d^{3} e - a c d e^{3}\right )} x^{2} + {\left (c^{2} d^{4} - a^{2} e^{4}\right )} x}, -\frac {2 \, {\left ({\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )} \sqrt {\frac {e}{c d^{2} - a e^{2}}} \arctan \left (-\frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (c d^{2} - a e^{2}\right )} \sqrt {e x + d} \sqrt {\frac {e}{c d^{2} - a e^{2}}}}{c d e^{2} x^{2} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x}\right ) + \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}\right )}}{a c d^{3} e - a^{2} d e^{3} + {\left (c^{2} d^{3} e - a c d e^{3}\right )} x^{2} + {\left (c^{2} d^{4} - a^{2} e^{4}\right )} x}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

integrate((e*x+d)**(1/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(3/2),x)

[Out]

Integral(sqrt(d + e*x)/((d + e*x)*(a*e + c*d*x))**(3/2), x)

Maxima [F]

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

[In]

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

[Out]

integrate(sqrt(e*x + d)/(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(3/2), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 276 vs. \(2 (123) = 246\).

Time = 0.30 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.99 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=-2 \, e {\left (\frac {e \arctan \left (\frac {\sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}}}{\sqrt {c d^{2} e - a e^{3}}}\right )}{\sqrt {c d^{2} e - a e^{3}} {\left (c d^{2} {\left | e \right |} - a e^{2} {\left | e \right |}\right )}} + \frac {e}{\sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} {\left (c d^{2} {\left | e \right |} - a e^{2} {\left | e \right |}\right )}}\right )} + \frac {2 \, {\left (\sqrt {-c d^{2} e + a e^{3}} e^{2} \arctan \left (\frac {\sqrt {-c d^{2} e + a e^{3}}}{\sqrt {c d^{2} e - a e^{3}}}\right ) + \sqrt {c d^{2} e - a e^{3}} e^{2}\right )}}{\sqrt {c d^{2} e - a e^{3}} \sqrt {-c d^{2} e + a e^{3}} c d^{2} {\left | e \right |} - \sqrt {c d^{2} e - a e^{3}} \sqrt {-c d^{2} e + a e^{3}} a e^{2} {\left | e \right |}} \]

[In]

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

[Out]

-2*e*(e*arctan(sqrt((e*x + d)*c*d*e - c*d^2*e + a*e^3)/sqrt(c*d^2*e - a*e^3))/(sqrt(c*d^2*e - a*e^3)*(c*d^2*ab
s(e) - a*e^2*abs(e))) + e/(sqrt((e*x + d)*c*d*e - c*d^2*e + a*e^3)*(c*d^2*abs(e) - a*e^2*abs(e)))) + 2*(sqrt(-
c*d^2*e + a*e^3)*e^2*arctan(sqrt(-c*d^2*e + a*e^3)/sqrt(c*d^2*e - a*e^3)) + sqrt(c*d^2*e - a*e^3)*e^2)/(sqrt(c
*d^2*e - a*e^3)*sqrt(-c*d^2*e + a*e^3)*c*d^2*abs(e) - sqrt(c*d^2*e - a*e^3)*sqrt(-c*d^2*e + a*e^3)*a*e^2*abs(e
))

Mupad [F(-1)]

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

[In]

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

[Out]

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