\(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^2}{(d+e x)^{3/2}} \, dx\) [1986]

Optimal result

Integrand size = 37, antiderivative size = 83 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{3/2}} \, dx=\frac {2 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}{3 e^3}-\frac {4 c d \left (c d^2-a e^2\right ) (d+e x)^{5/2}}{5 e^3}+\frac {2 c^2 d^2 (d+e x)^{7/2}}{7 e^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 83, 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} \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{3/2}} \, dx=-\frac {4 c d (d+e x)^{5/2} \left (c d^2-a e^2\right )}{5 e^3}+\frac {2 (d+e x)^{3/2} \left (c d^2-a e^2\right )^2}{3 e^3}+\frac {2 c^2 d^2 (d+e x)^{7/2}}{7 e^3} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.81 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{3/2}} \, dx=\frac {2 (d+e x)^{3/2} \left (35 a^2 e^4+14 a c d e^2 (-2 d+3 e x)+c^2 d^2 \left (8 d^2-12 d e x+15 e^2 x^2\right )\right )}{105 e^3} \]

[In]

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

[Out]

(2*(d + e*x)^(3/2)*(35*a^2*e^4 + 14*a*c*d*e^2*(-2*d + 3*e*x) + c^2*d^2*(8*d^2 - 12*d*e*x + 15*e^2*x^2)))/(105*
e^3)

Maple [A] (verified)

Time = 2.82 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.78

method	result	size
pseudoelliptic	\(\frac {2 \left (a^{2} e^{4}+\frac {6 x a c d \,e^{3}}{5}-\frac {4 \left (-\frac {15 c \,x^{2}}{28}+a \right ) c \,d^{2} e^{2}}{5}-\frac {12 x \,c^{2} d^{3} e}{35}+\frac {8 c^{2} d^{4}}{35}\right ) \left (e x +d \right )^{\frac {3}{2}}}{3 e^{3}}\)	\(65\)
derivativedivides	\(\frac {\frac {2 c^{2} d^{2} \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {4 \left (e^{2} a -c \,d^{2}\right ) c d \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (e^{2} a -c \,d^{2}\right )^{2} \left (e x +d \right )^{\frac {3}{2}}}{3}}{e^{3}}\)	\(68\)
default	\(\frac {\frac {2 c^{2} d^{2} \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {4 \left (e^{2} a -c \,d^{2}\right ) c d \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (e^{2} a -c \,d^{2}\right )^{2} \left (e x +d \right )^{\frac {3}{2}}}{3}}{e^{3}}\)	\(68\)
gosper	\(\frac {2 \left (e x +d \right )^{\frac {3}{2}} \left (15 x^{2} c^{2} d^{2} e^{2}+42 x a c d \,e^{3}-12 x \,c^{2} d^{3} e +35 a^{2} e^{4}-28 a c \,d^{2} e^{2}+8 c^{2} d^{4}\right )}{105 e^{3}}\)	\(73\)
trager	\(\frac {2 \left (15 c^{2} d^{2} e^{3} x^{3}+42 a c d \,e^{4} x^{2}+3 c^{2} d^{3} e^{2} x^{2}+35 a^{2} e^{5} x +14 a c \,d^{2} e^{3} x -4 c^{2} d^{4} e x +35 a^{2} d \,e^{4}-28 a \,d^{3} e^{2} c +8 d^{5} c^{2}\right ) \sqrt {e x +d}}{105 e^{3}}\)	\(110\)
risch	\(\frac {2 \left (15 c^{2} d^{2} e^{3} x^{3}+42 a c d \,e^{4} x^{2}+3 c^{2} d^{3} e^{2} x^{2}+35 a^{2} e^{5} x +14 a c \,d^{2} e^{3} x -4 c^{2} d^{4} e x +35 a^{2} d \,e^{4}-28 a \,d^{3} e^{2} c +8 d^{5} c^{2}\right ) \sqrt {e x +d}}{105 e^{3}}\)	\(110\)

[In]

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

[Out]

2/3*(a^2*e^4+6/5*x*a*c*d*e^3-4/5*(-15/28*c*x^2+a)*c*d^2*e^2-12/35*x*c^2*d^3*e+8/35*c^2*d^4)*(e*x+d)^(3/2)/e^3

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.31 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (15 \, c^{2} d^{2} e^{3} x^{3} + 8 \, c^{2} d^{5} - 28 \, a c d^{3} e^{2} + 35 \, a^{2} d e^{4} + 3 \, {\left (c^{2} d^{3} e^{2} + 14 \, a c d e^{4}\right )} x^{2} - {\left (4 \, c^{2} d^{4} e - 14 \, a c d^{2} e^{3} - 35 \, a^{2} e^{5}\right )} x\right )} \sqrt {e x + d}}{105 \, e^{3}} \]

[In]

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

[Out]

2/105*(15*c^2*d^2*e^3*x^3 + 8*c^2*d^5 - 28*a*c*d^3*e^2 + 35*a^2*d*e^4 + 3*(c^2*d^3*e^2 + 14*a*c*d*e^4)*x^2 - (
4*c^2*d^4*e - 14*a*c*d^2*e^3 - 35*a^2*e^5)*x)*sqrt(e*x + d)/e^3

Sympy [A] (verification not implemented)

Time = 0.93 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.33 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {c^{2} d^{2} \left (d + e x\right )^{\frac {7}{2}}}{7 e^{2}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \cdot \left (2 a c d e^{2} - 2 c^{2} d^{3}\right )}{5 e^{2}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (a^{2} e^{4} - 2 a c d^{2} e^{2} + c^{2} d^{4}\right )}{3 e^{2}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {c^{2} d^{\frac {5}{2}} x^{3}}{3} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(c**2*d**2*(d + e*x)**(7/2)/(7*e**2) + (d + e*x)**(5/2)*(2*a*c*d*e**2 - 2*c**2*d**3)/(5*e**2) + (
d + e*x)**(3/2)*(a**2*e**4 - 2*a*c*d**2*e**2 + c**2*d**4)/(3*e**2))/e, Ne(e, 0)), (c**2*d**(5/2)*x**3/3, True)
)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (15 \, {\left (e x + d\right )}^{\frac {7}{2}} c^{2} d^{2} - 42 \, {\left (c^{2} d^{3} - a c d e^{2}\right )} {\left (e x + d\right )}^{\frac {5}{2}} + 35 \, {\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} {\left (e x + d\right )}^{\frac {3}{2}}\right )}}{105 \, e^{3}} \]

[In]

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

[Out]

2/105*(15*(e*x + d)^(7/2)*c^2*d^2 - 42*(c^2*d^3 - a*c*d*e^2)*(e*x + d)^(5/2) + 35*(c^2*d^4 - 2*a*c*d^2*e^2 + a
^2*e^4)*(e*x + d)^(3/2))/e^3

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 208 vs. \(2 (71) = 142\).

Time = 0.27 (sec) , antiderivative size = 208, normalized size of antiderivative = 2.51 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (105 \, \sqrt {e x + d} a^{2} d e^{2} + 70 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a c d^{2} + 35 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a^{2} e^{2} + 14 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a c d + \frac {7 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} c^{2} d^{3}}{e^{2}} + \frac {3 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} c^{2} d^{2}}{e^{2}}\right )}}{105 \, e} \]

[In]

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

[Out]

2/105*(105*sqrt(e*x + d)*a^2*d*e^2 + 70*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*a*c*d^2 + 35*((e*x + d)^(3/2) -
3*sqrt(e*x + d)*d)*a^2*e^2 + 14*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*a*c*d + 7*(3
*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*c^2*d^3/e^2 + 3*(5*(e*x + d)^(7/2) - 21*(e*x +
d)^(5/2)*d + 35*(e*x + d)^(3/2)*d^2 - 35*sqrt(e*x + d)*d^3)*c^2*d^2/e^2)/e

Mupad [B] (verification not implemented)

Time = 9.71 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{3/2}} \, dx=\frac {2\,{\left (d+e\,x\right )}^{3/2}\,\left (35\,a^2\,e^4+35\,c^2\,d^4+15\,c^2\,d^2\,{\left (d+e\,x\right )}^2-42\,c^2\,d^3\,\left (d+e\,x\right )-70\,a\,c\,d^2\,e^2+42\,a\,c\,d\,e^2\,\left (d+e\,x\right )\right )}{105\,e^3} \]

[In]

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

[Out]

(2*(d + e*x)^(3/2)*(35*a^2*e^4 + 35*c^2*d^4 + 15*c^2*d^2*(d + e*x)^2 - 42*c^2*d^3*(d + e*x) - 70*a*c*d^2*e^2 +
42*a*c*d*e^2*(d + e*x)))/(105*e^3)