\(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^2}{(d+e x)^{11/2}} \, dx\) [1990]

Optimal result

Integrand size = 37, antiderivative size = 81 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{11/2}} \, dx=-\frac {2 \left (c d^2-a e^2\right )^2}{5 e^3 (d+e x)^{5/2}}+\frac {4 c d \left (c d^2-a e^2\right )}{3 e^3 (d+e x)^{3/2}}-\frac {2 c^2 d^2}{e^3 \sqrt {d+e x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {640, 45} \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{11/2}} \, dx=\frac {4 c d \left (c d^2-a e^2\right )}{3 e^3 (d+e x)^{3/2}}-\frac {2 \left (c d^2-a e^2\right )^2}{5 e^3 (d+e x)^{5/2}}-\frac {2 c^2 d^2}{e^3 \sqrt {d+e x}} \]

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2/(d + e*x)^(11/2),x]

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 640

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {(a e+c d x)^2}{(d+e x)^{7/2}} \, dx \\ & = \int \left (\frac {\left (-c d^2+a e^2\right )^2}{e^2 (d+e x)^{7/2}}-\frac {2 c d \left (c d^2-a e^2\right )}{e^2 (d+e x)^{5/2}}+\frac {c^2 d^2}{e^2 (d+e x)^{3/2}}\right ) \, dx \\ & = -\frac {2 \left (c d^2-a e^2\right )^2}{5 e^3 (d+e x)^{5/2}}+\frac {4 c d \left (c d^2-a e^2\right )}{3 e^3 (d+e x)^{3/2}}-\frac {2 c^2 d^2}{e^3 \sqrt {d+e x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{11/2}} \, dx=-\frac {2 \left (3 a^2 e^4+2 a c d e^2 (2 d+5 e x)+c^2 d^2 \left (8 d^2+20 d e x+15 e^2 x^2\right )\right )}{15 e^3 (d+e x)^{5/2}} \]

[In]

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2/(d + e*x)^(11/2),x]

[Out]

(-2*(3*a^2*e^4 + 2*a*c*d*e^2*(2*d + 5*e*x) + c^2*d^2*(8*d^2 + 20*d*e*x + 15*e^2*x^2)))/(15*e^3*(d + e*x)^(5/2)
)

Maple [A] (verified)

Time = 2.71 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.80

method	result	size
pseudoelliptic	\(-\frac {2 \left (a^{2} e^{4}+\frac {10 x a c d \,e^{3}}{3}+\frac {4 c \left (\frac {15 c \,x^{2}}{4}+a \right ) d^{2} e^{2}}{3}+\frac {20 x \,c^{2} d^{3} e}{3}+\frac {8 c^{2} d^{4}}{3}\right )}{5 \left (e x +d \right )^{\frac {5}{2}} e^{3}}\)	\(65\)
gosper	\(-\frac {2 \left (15 x^{2} c^{2} d^{2} e^{2}+10 x a c d \,e^{3}+20 x \,c^{2} d^{3} e +3 a^{2} e^{4}+4 a c \,d^{2} e^{2}+8 c^{2} d^{4}\right )}{15 \left (e x +d \right )^{\frac {5}{2}} e^{3}}\)	\(73\)
trager	\(-\frac {2 \left (15 x^{2} c^{2} d^{2} e^{2}+10 x a c d \,e^{3}+20 x \,c^{2} d^{3} e +3 a^{2} e^{4}+4 a c \,d^{2} e^{2}+8 c^{2} d^{4}\right )}{15 \left (e x +d \right )^{\frac {5}{2}} e^{3}}\)	\(73\)
derivativedivides	\(\frac {-\frac {2 \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right )}{5 \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 c^{2} d^{2}}{\sqrt {e x +d}}-\frac {4 c d \left (e^{2} a -c \,d^{2}\right )}{3 \left (e x +d \right )^{\frac {3}{2}}}}{e^{3}}\)	\(79\)
default	\(\frac {-\frac {2 \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right )}{5 \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 c^{2} d^{2}}{\sqrt {e x +d}}-\frac {4 c d \left (e^{2} a -c \,d^{2}\right )}{3 \left (e x +d \right )^{\frac {3}{2}}}}{e^{3}}\)	\(79\)

[In]

int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^(11/2),x,method=_RETURNVERBOSE)

[Out]

-2/5*(a^2*e^4+10/3*x*a*c*d*e^3+4/3*c*(15/4*c*x^2+a)*d^2*e^2+20/3*x*c^2*d^3*e+8/3*c^2*d^4)/(e*x+d)^(5/2)/e^3

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.30 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{11/2}} \, dx=-\frac {2 \, {\left (15 \, c^{2} d^{2} e^{2} x^{2} + 8 \, c^{2} d^{4} + 4 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4} + 10 \, {\left (2 \, c^{2} d^{3} e + a c d e^{3}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^(11/2),x, algorithm="fricas")

[Out]

-2/15*(15*c^2*d^2*e^2*x^2 + 8*c^2*d^4 + 4*a*c*d^2*e^2 + 3*a^2*e^4 + 10*(2*c^2*d^3*e + a*c*d*e^3)*x)*sqrt(e*x +
d)/(e^6*x^3 + 3*d*e^5*x^2 + 3*d^2*e^4*x + d^3*e^3)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 388 vs. \(2 (75) = 150\).

Time = 1.05 (sec) , antiderivative size = 388, normalized size of antiderivative = 4.79 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{11/2}} \, dx=\begin {cases} - \frac {6 a^{2} e^{4}}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} - \frac {8 a c d^{2} e^{2}}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} - \frac {20 a c d e^{3} x}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} - \frac {16 c^{2} d^{4}}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} - \frac {40 c^{2} d^{3} e x}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} - \frac {30 c^{2} d^{2} e^{2} x^{2}}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} & \text {for}\: e \neq 0 \\\frac {c^{2} x^{3}}{3 d^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**2/(e*x+d)**(11/2),x)

[Out]

Piecewise((-6*a**2*e**4/(15*d**2*e**3*sqrt(d + e*x) + 30*d*e**4*x*sqrt(d + e*x) + 15*e**5*x**2*sqrt(d + e*x))
- 8*a*c*d**2*e**2/(15*d**2*e**3*sqrt(d + e*x) + 30*d*e**4*x*sqrt(d + e*x) + 15*e**5*x**2*sqrt(d + e*x)) - 20*a
*c*d*e**3*x/(15*d**2*e**3*sqrt(d + e*x) + 30*d*e**4*x*sqrt(d + e*x) + 15*e**5*x**2*sqrt(d + e*x)) - 16*c**2*d*
*4/(15*d**2*e**3*sqrt(d + e*x) + 30*d*e**4*x*sqrt(d + e*x) + 15*e**5*x**2*sqrt(d + e*x)) - 40*c**2*d**3*e*x/(1
5*d**2*e**3*sqrt(d + e*x) + 30*d*e**4*x*sqrt(d + e*x) + 15*e**5*x**2*sqrt(d + e*x)) - 30*c**2*d**2*e**2*x**2/(
15*d**2*e**3*sqrt(d + e*x) + 30*d*e**4*x*sqrt(d + e*x) + 15*e**5*x**2*sqrt(d + e*x)), Ne(e, 0)), (c**2*x**3/(3
*d**(3/2)), True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{11/2}} \, dx=-\frac {2 \, {\left (15 \, {\left (e x + d\right )}^{2} c^{2} d^{2} + 3 \, c^{2} d^{4} - 6 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4} - 10 \, {\left (c^{2} d^{3} - a c d e^{2}\right )} {\left (e x + d\right )}\right )}}{15 \, {\left (e x + d\right )}^{\frac {5}{2}} e^{3}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^(11/2),x, algorithm="maxima")

[Out]

-2/15*(15*(e*x + d)^2*c^2*d^2 + 3*c^2*d^4 - 6*a*c*d^2*e^2 + 3*a^2*e^4 - 10*(c^2*d^3 - a*c*d*e^2)*(e*x + d))/((
e*x + d)^(5/2)*e^3)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{11/2}} \, dx=-\frac {2 \, {\left (15 \, {\left (e x + d\right )}^{2} c^{2} d^{2} - 10 \, {\left (e x + d\right )} c^{2} d^{3} + 3 \, c^{2} d^{4} + 10 \, {\left (e x + d\right )} a c d e^{2} - 6 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4}\right )}}{15 \, {\left (e x + d\right )}^{\frac {5}{2}} e^{3}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^(11/2),x, algorithm="giac")

[Out]

-2/15*(15*(e*x + d)^2*c^2*d^2 - 10*(e*x + d)*c^2*d^3 + 3*c^2*d^4 + 10*(e*x + d)*a*c*d*e^2 - 6*a*c*d^2*e^2 + 3*
a^2*e^4)/((e*x + d)^(5/2)*e^3)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{11/2}} \, dx=-\frac {\frac {2\,a^2\,e^4}{5}+\frac {2\,c^2\,d^4}{5}-\left (\frac {4\,c^2\,d^3}{3}-\frac {4\,a\,c\,d\,e^2}{3}\right )\,\left (d+e\,x\right )+2\,c^2\,d^2\,{\left (d+e\,x\right )}^2-\frac {4\,a\,c\,d^2\,e^2}{5}}{e^3\,{\left (d+e\,x\right )}^{5/2}} \]

[In]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^2/(d + e*x)^(11/2),x)

[Out]

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