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

Optimal. Leaf size=139 \[ -\frac {2 \sqrt {d+e x}}{\left (c d^2-a e^2\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {2 \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {c d^2-a e^2} \sqrt {d+e x}}\right )}{\left (c d^2-a e^2\right )^{3/2}} \]

[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]
time = 0.06, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {680, 674, 211} \begin {gather*} -\frac {2 \sqrt {e} \text {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}} \end {gather*}

Antiderivative was successfully verified.

[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*} \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 {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]
time = 0.08, size = 115, normalized size = 0.83 \begin {gather*} -\frac {2 \sqrt {d+e x} \left (\sqrt {c d^2-a e^2}+\sqrt {e} \sqrt {a e+c d x} \tan ^{-1}\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)}} \end {gather*}

Antiderivative was successfully verified.

[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]
time = 0.73, size = 126, normalized size = 0.91

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

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

________________________________________________________________________________________

Fricas [A]
time = 2.16, size = 474, normalized size = 3.41 \begin {gather*} \left [-\frac {{\left (c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e\right )} \sqrt {-\frac {e}{c d^{2} - a e^{2}}} \log \left (\frac {c d^{3} - 2 \, a x e^{3} - 2 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} {\left (c d^{2} - a e^{2}\right )} \sqrt {x e + d} \sqrt {-\frac {e}{c d^{2} - a e^{2}}} - {\left (c d x^{2} + 2 \, a d\right )} e^{2}}{x^{2} e^{2} + 2 \, d x e + d^{2}}\right ) + 2 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {x e + d}}{c^{2} d^{4} x - a^{2} x e^{4} - {\left (a c d x^{2} + a^{2} d\right )} e^{3} + {\left (c^{2} d^{3} x^{2} + a c d^{3}\right )} e}, -\frac {2 \, {\left (\frac {{\left (c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e\right )} \arctan \left (-\frac {\sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {c d^{2} - a e^{2}} \sqrt {x e + d} e^{\frac {1}{2}}}{c d^{2} x e + a x e^{3} + {\left (c d x^{2} + a d\right )} e^{2}}\right ) e^{\frac {1}{2}}}{\sqrt {c d^{2} - a e^{2}}} + \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {x e + d}\right )}}{c^{2} d^{4} x - a^{2} x e^{4} - {\left (a c d x^{2} + a^{2} d\right )} e^{3} + {\left (c^{2} d^{3} x^{2} + a c d^{3}\right )} e}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*sqrt(-e/(c*d^2 - a*e^2))*log((c*d^3 - 2*a*x*e^3 - 2*sqrt(c*d^2*x +
a*x*e^2 + (c*d*x^2 + a*d)*e)*(c*d^2 - a*e^2)*sqrt(x*e + d)*sqrt(-e/(c*d^2 - a*e^2)) - (c*d*x^2 + 2*a*d)*e^2)/(
x^2*e^2 + 2*d*x*e + d^2)) + 2*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*sqrt(x*e + d))/(c^2*d^4*x - a^2*x*e^
4 - (a*c*d*x^2 + a^2*d)*e^3 + (c^2*d^3*x^2 + a*c*d^3)*e), -2*((c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*arctan(-
sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*sqrt(c*d^2 - a*e^2)*sqrt(x*e + d)*e^(1/2)/(c*d^2*x*e + a*x*e^3 + (
c*d*x^2 + a*d)*e^2))*e^(1/2)/sqrt(c*d^2 - a*e^2) + sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*sqrt(x*e + d))/
(c^2*d^4*x - a^2*x*e^4 - (a*c*d*x^2 + a^2*d)*e^3 + (c^2*d^3*x^2 + a*c*d^3)*e)]

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {d + e x}}{\left (\left (d + e x\right ) \left (a e + c d x\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 263 vs. \(2 (122) = 244\).
time = 0.96, size = 263, normalized size = 1.89 \begin {gather*} -\frac {2 \, \arctan \left (\frac {\sqrt {{\left (x e + d\right )} c d e - c d^{2} e + a e^{3}}}{\sqrt {c d^{2} e - a e^{3}}}\right ) e}{\sqrt {c d^{2} e - a e^{3}} {\left (c d^{2} - a e^{2}\right )}} + \frac {2 \, {\left (\sqrt {-c d^{2} e + a e^{3}} \arctan \left (\frac {\sqrt {-c d^{2} e + a e^{3}}}{\sqrt {c d^{2} e - a e^{3}}}\right ) e + \sqrt {c d^{2} e - a e^{3}} e\right )}}{\sqrt {c d^{2} e - a e^{3}} \sqrt {-c d^{2} e + a e^{3}} c d^{2} - \sqrt {c d^{2} e - a e^{3}} \sqrt {-c d^{2} e + a e^{3}} a e^{2}} - \frac {2 \, e}{\sqrt {{\left (x e + d\right )} c d e - c d^{2} e + a e^{3}} {\left (c d^{2} - a e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________