\(\int \frac {\sqrt {a d e+(c d^2+a e^2) x+c d e x^2}}{(d+e x)^4} \, dx\) [1915]

Optimal result

Integrand size = 37, antiderivative size = 111 \[ \int \frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{(d+e x)^4} \, dx=\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{5 \left (c d^2-a e^2\right ) (d+e x)^4}+\frac {4 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{15 \left (c d^2-a e^2\right )^2 (d+e x)^3} \]

[Out]

2/5*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/(-a*e^2+c*d^2)/(e*x+d)^4+4/15*c*d*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2
)^(3/2)/(-a*e^2+c*d^2)^2/(e*x+d)^3

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} \[ \int \frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{(d+e x)^4} \, dx=\frac {4 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{15 (d+e x)^3 \left (c d^2-a e^2\right )^2}+\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{5 (d+e x)^4 \left (c d^2-a e^2\right )} \]

[In]

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

[Out]

(2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(5*(c*d^2 - a*e^2)*(d + e*x)^4) + (4*c*d*(a*d*e + (c*d^2 + a
*e^2)*x + c*d*e*x^2)^(3/2))/(15*(c*d^2 - a*e^2)^2*(d + e*x)^3)

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 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{5 \left (c d^2-a e^2\right ) (d+e x)^4}+\frac {(2 c d) \int \frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{(d+e x)^3} \, dx}{5 \left (c d^2-a e^2\right )} \\ & = \frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{5 \left (c d^2-a e^2\right ) (d+e x)^4}+\frac {4 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{15 \left (c d^2-a e^2\right )^2 (d+e x)^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.73 \[ \int \frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{(d+e x)^4} \, dx=\frac {2 \sqrt {(a e+c d x) (d+e x)} \left (-3 a^2 e^3+a c d e (5 d-e x)+c^2 d^2 x (5 d+2 e x)\right )}{15 \left (c d^2-a e^2\right )^2 (d+e x)^3} \]

[In]

Integrate[Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]/(d + e*x)^4,x]

[Out]

(2*Sqrt[(a*e + c*d*x)*(d + e*x)]*(-3*a^2*e^3 + a*c*d*e*(5*d - e*x) + c^2*d^2*x*(5*d + 2*e*x)))/(15*(c*d^2 - a*
e^2)^2*(d + e*x)^3)

Maple [A] (verified)

Time = 3.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.81

method	result	size
gosper	\(-\frac {2 \left (c d x +a e \right ) \left (-2 x c d e +3 e^{2} a -5 c \,d^{2}\right ) \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}}{15 \left (e x +d \right )^{3} \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right )}\)	\(90\)
trager	\(-\frac {2 \left (-2 c^{2} d^{2} e \,x^{2}+a d \,e^{2} x c -5 c^{2} d^{3} x +3 a^{2} e^{3}-5 d^{2} e a c \right ) \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}}{15 \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \left (e x +d \right )^{3}}\)	\(109\)
default	\(\frac {-\frac {2 \left (c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{5 \left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )^{4}}+\frac {4 c d e \left (c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{15 \left (e^{2} a -c \,d^{2}\right )^{2} \left (x +\frac {d}{e}\right )^{3}}}{e^{4}}\)	\(131\)

[In]

int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/(e*x+d)^4,x,method=_RETURNVERBOSE)

[Out]

-2/15*(c*d*x+a*e)*(-2*c*d*e*x+3*a*e^2-5*c*d^2)*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(1/2)/(e*x+d)^3/(a^2*e^4-2*a*
c*d^2*e^2+c^2*d^4)

Fricas [A] (verification not implemented)

none

Time = 0.84 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.85 \[ \int \frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{(d+e x)^4} \, dx=\frac {2 \, {\left (2 \, c^{2} d^{2} e x^{2} + 5 \, a c d^{2} e - 3 \, a^{2} e^{3} + {\left (5 \, c^{2} d^{3} - a c d e^{2}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{15 \, {\left (c^{2} d^{7} - 2 \, a c d^{5} e^{2} + a^{2} d^{3} e^{4} + {\left (c^{2} d^{4} e^{3} - 2 \, a c d^{2} e^{5} + a^{2} e^{7}\right )} x^{3} + 3 \, {\left (c^{2} d^{5} e^{2} - 2 \, a c d^{3} e^{4} + a^{2} d e^{6}\right )} x^{2} + 3 \, {\left (c^{2} d^{6} e - 2 \, a c d^{4} e^{3} + a^{2} d^{2} e^{5}\right )} x\right )}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/(e*x+d)^4,x, algorithm="fricas")

[Out]

2/15*(2*c^2*d^2*e*x^2 + 5*a*c*d^2*e - 3*a^2*e^3 + (5*c^2*d^3 - a*c*d*e^2)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 +
a*e^2)*x)/(c^2*d^7 - 2*a*c*d^5*e^2 + a^2*d^3*e^4 + (c^2*d^4*e^3 - 2*a*c*d^2*e^5 + a^2*e^7)*x^3 + 3*(c^2*d^5*e
^2 - 2*a*c*d^3*e^4 + a^2*d*e^6)*x^2 + 3*(c^2*d^6*e - 2*a*c*d^4*e^3 + a^2*d^2*e^5)*x)

Sympy [F]

\[ \int \frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{(d+e x)^4} \, dx=\int \frac {\sqrt {\left (d + e x\right ) \left (a e + c d x\right )}}{\left (d + e x\right )^{4}}\, dx \]

[In]

integrate((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(1/2)/(e*x+d)**4,x)

[Out]

Integral(sqrt((d + e*x)*(a*e + c*d*x))/(d + e*x)**4, x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{(d+e x)^4} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/(e*x+d)^4,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 [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{(d+e x)^4} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/(e*x+d)^4,x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{%%%{1,[0,0,3]%%%},[6]%%%}+%%%{%%{[%%%{-6,[0,1,2]%%%},0]:
[1,0,%%%{-1

Mupad [B] (verification not implemented)

Time = 10.57 (sec) , antiderivative size = 562, normalized size of antiderivative = 5.06 \[ \int \frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{(d+e x)^4} \, dx=\frac {\left (\frac {4\,c^2\,d^3}{5\,\left (a\,e^2-c\,d^2\right )\,\left (3\,a\,e^3-3\,c\,d^2\,e\right )}-\frac {4\,a\,c\,d\,e^2}{5\,\left (a\,e^2-c\,d^2\right )\,\left (3\,a\,e^3-3\,c\,d^2\,e\right )}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{{\left (d+e\,x\right )}^2}-\frac {\left (\frac {2\,a\,e^2}{5\,a\,e^3-5\,c\,d^2\,e}-\frac {2\,c\,d^2}{5\,a\,e^3-5\,c\,d^2\,e}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{{\left (d+e\,x\right )}^3}+\frac {\left (\frac {4\,c^3\,d^4+4\,a\,c^2\,d^2\,e^2}{15\,e\,{\left (a\,e^2-c\,d^2\right )}^3}-\frac {8\,c^3\,d^4}{15\,e\,{\left (a\,e^2-c\,d^2\right )}^3}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{d+e\,x}+\frac {\left (\frac {2\,c^2\,d^3+2\,a\,c\,d\,e^2}{5\,\left (a\,e^2-c\,d^2\right )\,\left (3\,a\,e^3-3\,c\,d^2\,e\right )}-\frac {4\,c^2\,d^3}{5\,\left (a\,e^2-c\,d^2\right )\,\left (3\,a\,e^3-3\,c\,d^2\,e\right )}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{{\left (d+e\,x\right )}^2}+\frac {\left (\frac {8\,c^3\,d^4}{15\,e\,{\left (a\,e^2-c\,d^2\right )}^3}-\frac {8\,a\,c^2\,d^2\,e}{15\,{\left (a\,e^2-c\,d^2\right )}^3}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{d+e\,x}+\frac {8\,c^2\,d^2\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{15\,e\,{\left (a\,e^2-c\,d^2\right )}^2\,\left (d+e\,x\right )} \]

[In]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)/(d + e*x)^4,x)

[Out]

(((4*c^2*d^3)/(5*(a*e^2 - c*d^2)*(3*a*e^3 - 3*c*d^2*e)) - (4*a*c*d*e^2)/(5*(a*e^2 - c*d^2)*(3*a*e^3 - 3*c*d^2*
e)))*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2))/(d + e*x)^2 - (((2*a*e^2)/(5*a*e^3 - 5*c*d^2*e) - (2*c*d^2
)/(5*a*e^3 - 5*c*d^2*e))*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2))/(d + e*x)^3 + (((4*c^3*d^4 + 4*a*c^2*d
^2*e^2)/(15*e*(a*e^2 - c*d^2)^3) - (8*c^3*d^4)/(15*e*(a*e^2 - c*d^2)^3))*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^
2)^(1/2))/(d + e*x) + (((2*c^2*d^3 + 2*a*c*d*e^2)/(5*(a*e^2 - c*d^2)*(3*a*e^3 - 3*c*d^2*e)) - (4*c^2*d^3)/(5*(
a*e^2 - c*d^2)*(3*a*e^3 - 3*c*d^2*e)))*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2))/(d + e*x)^2 + (((8*c^3*d
^4)/(15*e*(a*e^2 - c*d^2)^3) - (8*a*c^2*d^2*e)/(15*(a*e^2 - c*d^2)^3))*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)
^(1/2))/(d + e*x) + (8*c^2*d^2*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2))/(15*e*(a*e^2 - c*d^2)^2*(d + e*x
))