\(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^2}{(d+e x)^{7/2}} \, dx\) [1988]

Optimal result

Integrand size = 37, antiderivative size = 79 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{7/2}} \, dx=-\frac {2 \left (c d^2-a e^2\right )^2}{e^3 \sqrt {d+e x}}-\frac {4 c d \left (c d^2-a e^2\right ) \sqrt {d+e x}}{e^3}+\frac {2 c^2 d^2 (d+e x)^{3/2}}{3 e^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 79, 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} \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{7/2}} \, dx=-\frac {4 c d \sqrt {d+e x} \left (c d^2-a e^2\right )}{e^3}-\frac {2 \left (c d^2-a e^2\right )^2}{e^3 \sqrt {d+e x}}+\frac {2 c^2 d^2 (d+e x)^{3/2}}{3 e^3} \]

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2/(d + e*x)^(7/2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.82 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{7/2}} \, dx=\frac {2 \left (-3 a^2 e^4+6 a c d e^2 (2 d+e x)+c^2 d^2 \left (-8 d^2-4 d e x+e^2 x^2\right )\right )}{3 e^3 \sqrt {d+e x}} \]

[In]

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2/(d + e*x)^(7/2),x]

[Out]

(2*(-3*a^2*e^4 + 6*a*c*d*e^2*(2*d + e*x) + c^2*d^2*(-8*d^2 - 4*d*e*x + e^2*x^2)))/(3*e^3*Sqrt[d + e*x])

Maple [A] (verified)

Time = 3.06 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.82

method	result	size
pseudoelliptic	\(-\frac {2 \left (a^{2} e^{4}-2 x a c d \,e^{3}-4 \left (\frac {c \,x^{2}}{12}+a \right ) c \,d^{2} e^{2}+\frac {4 x \,c^{2} d^{3} e}{3}+\frac {8 c^{2} d^{4}}{3}\right )}{\sqrt {e x +d}\, e^{3}}\)	\(65\)
risch	\(\frac {2 c d \left (x c d e +6 e^{2} a -5 c \,d^{2}\right ) \sqrt {e x +d}}{3 e^{3}}-\frac {2 \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right )}{e^{3} \sqrt {e x +d}}\)	\(71\)
gosper	\(-\frac {2 \left (-x^{2} c^{2} d^{2} e^{2}-6 x a c d \,e^{3}+4 x \,c^{2} d^{3} e +3 a^{2} e^{4}-12 a c \,d^{2} e^{2}+8 c^{2} d^{4}\right )}{3 \sqrt {e x +d}\, e^{3}}\)	\(73\)
trager	\(-\frac {2 \left (-x^{2} c^{2} d^{2} e^{2}-6 x a c d \,e^{3}+4 x \,c^{2} d^{3} e +3 a^{2} e^{4}-12 a c \,d^{2} e^{2}+8 c^{2} d^{4}\right )}{3 \sqrt {e x +d}\, e^{3}}\)	\(73\)
derivativedivides	\(\frac {\frac {2 c^{2} d^{2} \left (e x +d \right )^{\frac {3}{2}}}{3}+4 a c d \,e^{2} \sqrt {e x +d}-4 c^{2} d^{3} \sqrt {e x +d}-\frac {2 \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right )}{\sqrt {e x +d}}}{e^{3}}\)	\(86\)
default	\(\frac {\frac {2 c^{2} d^{2} \left (e x +d \right )^{\frac {3}{2}}}{3}+4 a c d \,e^{2} \sqrt {e x +d}-4 c^{2} d^{3} \sqrt {e x +d}-\frac {2 \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right )}{\sqrt {e x +d}}}{e^{3}}\)	\(86\)

[In]

int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^(7/2),x,method=_RETURNVERBOSE)

[Out]

-2/(e*x+d)^(1/2)*(a^2*e^4-2*x*a*c*d*e^3-4*(1/12*c*x^2+a)*c*d^2*e^2+4/3*x*c^2*d^3*e+8/3*c^2*d^4)/e^3

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{7/2}} \, dx=\frac {2 \, {\left (c^{2} d^{2} e^{2} x^{2} - 8 \, c^{2} d^{4} + 12 \, a c d^{2} e^{2} - 3 \, a^{2} e^{4} - 2 \, {\left (2 \, c^{2} d^{3} e - 3 \, a c d e^{3}\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (e^{4} x + d e^{3}\right )}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^(7/2),x, algorithm="fricas")

[Out]

2/3*(c^2*d^2*e^2*x^2 - 8*c^2*d^4 + 12*a*c*d^2*e^2 - 3*a^2*e^4 - 2*(2*c^2*d^3*e - 3*a*c*d*e^3)*x)*sqrt(e*x + d)
/(e^4*x + d*e^3)

Sympy [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.68 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{7/2}} \, dx=\begin {cases} - \frac {2 a^{2} e}{\sqrt {d + e x}} + \frac {8 a c d^{2}}{e \sqrt {d + e x}} + \frac {4 a c d x}{\sqrt {d + e x}} - \frac {16 c^{2} d^{4}}{3 e^{3} \sqrt {d + e x}} - \frac {8 c^{2} d^{3} x}{3 e^{2} \sqrt {d + e x}} + \frac {2 c^{2} d^{2} x^{2}}{3 e \sqrt {d + e x}} & \text {for}\: e \neq 0 \\\frac {c^{2} \sqrt {d} x^{3}}{3} & \text {otherwise} \end {cases} \]

[In]

integrate((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**2/(e*x+d)**(7/2),x)

[Out]

Piecewise((-2*a**2*e/sqrt(d + e*x) + 8*a*c*d**2/(e*sqrt(d + e*x)) + 4*a*c*d*x/sqrt(d + e*x) - 16*c**2*d**4/(3*
e**3*sqrt(d + e*x)) - 8*c**2*d**3*x/(3*e**2*sqrt(d + e*x)) + 2*c**2*d**2*x**2/(3*e*sqrt(d + e*x)), Ne(e, 0)),
(c**2*sqrt(d)*x**3/3, True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{7/2}} \, dx=\frac {2 \, {\left (\frac {{\left (e x + d\right )}^{\frac {3}{2}} c^{2} d^{2} - 6 \, {\left (c^{2} d^{3} - a c d e^{2}\right )} \sqrt {e x + d}}{e^{2}} - \frac {3 \, {\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )}}{\sqrt {e x + d} e^{2}}\right )}}{3 \, e} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^(7/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.19 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{7/2}} \, dx=-\frac {2 \, {\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )}}{\sqrt {e x + d} e^{3}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} c^{2} d^{2} e^{6} - 6 \, \sqrt {e x + d} c^{2} d^{3} e^{6} + 6 \, \sqrt {e x + d} a c d e^{8}\right )}}{3 \, e^{9}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^(7/2),x, algorithm="giac")

[Out]

-2*(c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)/(sqrt(e*x + d)*e^3) + 2/3*((e*x + d)^(3/2)*c^2*d^2*e^6 - 6*sqrt(e*x + d
)*c^2*d^3*e^6 + 6*sqrt(e*x + d)*a*c*d*e^8)/e^9

Mupad [B] (verification not implemented)

Time = 9.76 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{7/2}} \, dx=-\frac {6\,a^2\,e^4+6\,c^2\,d^4-2\,c^2\,d^2\,{\left (d+e\,x\right )}^2+12\,c^2\,d^3\,\left (d+e\,x\right )-12\,a\,c\,d^2\,e^2-12\,a\,c\,d\,e^2\,\left (d+e\,x\right )}{3\,e^3\,\sqrt {d+e\,x}} \]

[In]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^2/(d + e*x)^(7/2),x)

[Out]

-(6*a^2*e^4 + 6*c^2*d^4 - 2*c^2*d^2*(d + e*x)^2 + 12*c^2*d^3*(d + e*x) - 12*a*c*d^2*e^2 - 12*a*c*d*e^2*(d + e*
x))/(3*e^3*(d + e*x)^(1/2))