\(\int \frac {1}{(d+e x)^2 \sqrt {a d e+(c d^2+a e^2) x+c d e x^2}} \, dx\) [1951]

Optimal result

Integrand size = 37, antiderivative size = 111 \[ \int \frac {1}{(d+e x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 \left (c d^2-a e^2\right ) (d+e x)^2}+\frac {4 c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 \left (c d^2-a e^2\right )^2 (d+e x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {672, 664} \[ \int \frac {1}{(d+e x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {4 c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 (d+e x) \left (c d^2-a e^2\right )^2}+\frac {2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 (d+e x)^2 \left (c d^2-a e^2\right )} \]

[In]

Int[1/((d + e*x)^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])/(3*(c*d^2 - a*e^2)*(d + e*x)^2) + (4*c*d*Sqrt[a*d*e + (c*d^2 +
a*e^2)*x + c*d*e*x^2])/(3*(c*d^2 - a*e^2)^2*(d + e*x))

Rule 664

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

Rule 672

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-e)*(d + e*x)^m*((a
+ b*x + c*x^2)^(p + 1)/((m + p + 1)*(2*c*d - b*e))), x] + Dist[c*(Simplify[m + 2*p + 2]/((m + p + 1)*(2*c*d -
b*e))), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a
*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 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}}{3 \left (c d^2-a e^2\right ) (d+e x)^2}+\frac {(2 c d) \int \frac {1}{(d+e x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{3 \left (c d^2-a e^2\right )} \\ & = \frac {2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 \left (c d^2-a e^2\right ) (d+e x)^2}+\frac {4 c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 \left (c d^2-a e^2\right )^2 (d+e x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.55 \[ \int \frac {1}{(d+e x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \sqrt {(a e+c d x) (d+e x)} \left (-a e^2+c d (3 d+2 e x)\right )}{3 \left (c d^2-a e^2\right )^2 (d+e x)^2} \]

[In]

Integrate[1/((d + e*x)^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)]*(-(a*e^2) + c*d*(3*d + 2*e*x)))/(3*(c*d^2 - a*e^2)^2*(d + e*x)^2)

Maple [A] (verified)

Time = 2.78 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.73

method	result	size
trager	\(-\frac {2 \left (-2 x c d e +e^{2} a -3 c \,d^{2}\right ) \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}}{3 \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \left (e x +d \right )^{2}}\)	\(81\)
gosper	\(-\frac {2 \left (c d x +a e \right ) \left (-2 x c d e +e^{2} a -3 c \,d^{2}\right )}{3 \left (e x +d \right ) \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}}\)	\(89\)
default	\(\frac {-\frac {2 \sqrt {c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}}{3 \left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )^{2}}+\frac {4 c d e \sqrt {c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}}{3 \left (e^{2} a -c \,d^{2}\right )^{2} \left (x +\frac {d}{e}\right )}}{e^{2}}\)	\(131\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.46 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.26 \[ \int \frac {1}{(d+e x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d e x + 3 \, c d^{2} - a e^{2}\right )}}{3 \, {\left (c^{2} d^{6} - 2 \, a c d^{4} e^{2} + a^{2} d^{2} e^{4} + {\left (c^{2} d^{4} e^{2} - 2 \, a c d^{2} e^{4} + a^{2} e^{6}\right )} x^{2} + 2 \, {\left (c^{2} d^{5} e - 2 \, a c d^{3} e^{3} + a^{2} d e^{5}\right )} x\right )}} \]

[In]

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

[Out]

2/3*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*c*d*e*x + 3*c*d^2 - a*e^2)/(c^2*d^6 - 2*a*c*d^4*e^2 + a^2*d
^2*e^4 + (c^2*d^4*e^2 - 2*a*c*d^2*e^4 + a^2*e^6)*x^2 + 2*(c^2*d^5*e - 2*a*c*d^3*e^3 + a^2*d*e^5)*x)

Sympy [F]

\[ \int \frac {1}{(d+e x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\int \frac {1}{\sqrt {\left (d + e x\right ) \left (a e + c d x\right )} \left (d + e x\right )^{2}}\, dx \]

[In]

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

[Out]

Integral(1/(sqrt((d + e*x)*(a*e + c*d*x))*(d + e*x)**2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(d+e x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e*(a*e^2-c*d^2)>0)', see `assu
me?` for mor

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.59 \[ \int \frac {1}{(d+e x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=-\frac {4 \, \sqrt {c d e} c d \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}{3 \, {\left (c^{2} d^{4} e - 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )}} + \frac {2 \, {\left (3 \, \sqrt {c d e - \frac {c d^{2} e}{e x + d} + \frac {a e^{3}}{e x + d}} c d e - {\left (c d e - \frac {c d^{2} e}{e x + d} + \frac {a e^{3}}{e x + d}\right )}^{\frac {3}{2}}\right )}}{3 \, {\left (c d^{2} e^{2} - a e^{4}\right )} {\left (c d^{2} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - a e^{2} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )\right )}} \]

[In]

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

[Out]

-4/3*sqrt(c*d*e)*c*d*sgn(1/(e*x + d))*sgn(e)/(c^2*d^4*e - 2*a*c*d^2*e^3 + a^2*e^5) + 2/3*(3*sqrt(c*d*e - c*d^2
*e/(e*x + d) + a*e^3/(e*x + d))*c*d*e - (c*d*e - c*d^2*e/(e*x + d) + a*e^3/(e*x + d))^(3/2))/((c*d^2*e^2 - a*e
^4)*(c*d^2*sgn(1/(e*x + d))*sgn(e) - a*e^2*sgn(1/(e*x + d))*sgn(e)))

Mupad [B] (verification not implemented)

Time = 10.24 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.62 \[ \int \frac {1}{(d+e x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2\,\left (3\,c\,d^2+2\,c\,x\,d\,e-a\,e^2\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{3\,{\left (a\,e^2-c\,d^2\right )}^2\,{\left (d+e\,x\right )}^2} \]

[In]

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

[Out]

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