\(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^3}{(d+e x)^{7/2}} \, dx\) [1996]

Optimal result

Integrand size = 37, antiderivative size = 115 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{7/2}} \, dx=-\frac {2 \left (c d^2-a e^2\right )^3 \sqrt {d+e x}}{e^4}+\frac {2 c d \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}{e^4}-\frac {6 c^2 d^2 \left (c d^2-a e^2\right ) (d+e x)^{5/2}}{5 e^4}+\frac {2 c^3 d^3 (d+e x)^{7/2}}{7 e^4} \]

[Out]

2*c*d*(-a*e^2+c*d^2)^2*(e*x+d)^(3/2)/e^4-6/5*c^2*d^2*(-a*e^2+c*d^2)*(e*x+d)^(5/2)/e^4+2/7*c^3*d^3*(e*x+d)^(7/2
)/e^4-2*(-a*e^2+c*d^2)^3*(e*x+d)^(1/2)/e^4

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 115, 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} \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{7/2}} \, dx=-\frac {6 c^2 d^2 (d+e x)^{5/2} \left (c d^2-a e^2\right )}{5 e^4}+\frac {2 c d (d+e x)^{3/2} \left (c d^2-a e^2\right )^2}{e^4}-\frac {2 \sqrt {d+e x} \left (c d^2-a e^2\right )^3}{e^4}+\frac {2 c^3 d^3 (d+e x)^{7/2}}{7 e^4} \]

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3/(d + e*x)^(7/2),x]

[Out]

(-2*(c*d^2 - a*e^2)^3*Sqrt[d + e*x])/e^4 + (2*c*d*(c*d^2 - a*e^2)^2*(d + e*x)^(3/2))/e^4 - (6*c^2*d^2*(c*d^2 -
a*e^2)*(d + e*x)^(5/2))/(5*e^4) + (2*c^3*d^3*(d + e*x)^(7/2))/(7*e^4)

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)^3}{\sqrt {d+e x}} \, dx \\ & = \int \left (\frac {\left (-c d^2+a e^2\right )^3}{e^3 \sqrt {d+e x}}+\frac {3 c d \left (c d^2-a e^2\right )^2 \sqrt {d+e x}}{e^3}-\frac {3 c^2 d^2 \left (c d^2-a e^2\right ) (d+e x)^{3/2}}{e^3}+\frac {c^3 d^3 (d+e x)^{5/2}}{e^3}\right ) \, dx \\ & = -\frac {2 \left (c d^2-a e^2\right )^3 \sqrt {d+e x}}{e^4}+\frac {2 c d \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}{e^4}-\frac {6 c^2 d^2 \left (c d^2-a e^2\right ) (d+e x)^{5/2}}{5 e^4}+\frac {2 c^3 d^3 (d+e x)^{7/2}}{7 e^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{7/2}} \, dx=\frac {2 \sqrt {d+e x} \left (35 a^3 e^6+35 a^2 c d e^4 (-2 d+e x)+7 a c^2 d^2 e^2 \left (8 d^2-4 d e x+3 e^2 x^2\right )+c^3 d^3 \left (-16 d^3+8 d^2 e x-6 d e^2 x^2+5 e^3 x^3\right )\right )}{35 e^4} \]

[In]

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3/(d + e*x)^(7/2),x]

[Out]

(2*Sqrt[d + e*x]*(35*a^3*e^6 + 35*a^2*c*d*e^4*(-2*d + e*x) + 7*a*c^2*d^2*e^2*(8*d^2 - 4*d*e*x + 3*e^2*x^2) + c
^3*d^3*(-16*d^3 + 8*d^2*e*x - 6*d*e^2*x^2 + 5*e^3*x^3)))/(35*e^4)

Maple [A] (verified)

Time = 3.06 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {\frac {2 c^{3} d^{3} \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {6 \left (e^{2} a -c \,d^{2}\right ) c^{2} d^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}+2 \left (e^{2} a -c \,d^{2}\right )^{2} c d \left (e x +d \right )^{\frac {3}{2}}+2 \left (e^{2} a -c \,d^{2}\right )^{3} \sqrt {e x +d}}{e^{4}}\)	\(95\)
default	\(\frac {\frac {2 c^{3} d^{3} \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {6 \left (e^{2} a -c \,d^{2}\right ) c^{2} d^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}+2 \left (e^{2} a -c \,d^{2}\right )^{2} c d \left (e x +d \right )^{\frac {3}{2}}+2 \left (e^{2} a -c \,d^{2}\right )^{3} \sqrt {e x +d}}{e^{4}}\)	\(95\)
pseudoelliptic	\(\frac {2 \sqrt {e x +d}\, \left (-\frac {16 \left (-\frac {5}{16} e^{3} x^{3}+\frac {3}{8} d \,e^{2} x^{2}-\frac {1}{2} d^{2} e x +d^{3}\right ) d^{3} c^{3}}{35}+\frac {8 e^{2} d^{2} a \left (\frac {3}{8} x^{2} e^{2}-\frac {1}{2} d e x +d^{2}\right ) c^{2}}{5}-2 \left (-\frac {e x}{2}+d \right ) e^{4} d \,a^{2} c +e^{6} a^{3}\right )}{e^{4}}\)	\(102\)
gosper	\(\frac {2 \sqrt {e x +d}\, \left (5 x^{3} c^{3} d^{3} e^{3}+21 x^{2} a \,c^{2} d^{2} e^{4}-6 x^{2} c^{3} d^{4} e^{2}+35 x \,a^{2} c d \,e^{5}-28 x a \,c^{2} d^{3} e^{3}+8 x \,c^{3} d^{5} e +35 e^{6} a^{3}-70 d^{2} e^{4} a^{2} c +56 d^{4} e^{2} c^{2} a -16 c^{3} d^{6}\right )}{35 e^{4}}\)	\(131\)
trager	\(\frac {2 \sqrt {e x +d}\, \left (5 x^{3} c^{3} d^{3} e^{3}+21 x^{2} a \,c^{2} d^{2} e^{4}-6 x^{2} c^{3} d^{4} e^{2}+35 x \,a^{2} c d \,e^{5}-28 x a \,c^{2} d^{3} e^{3}+8 x \,c^{3} d^{5} e +35 e^{6} a^{3}-70 d^{2} e^{4} a^{2} c +56 d^{4} e^{2} c^{2} a -16 c^{3} d^{6}\right )}{35 e^{4}}\)	\(131\)
risch	\(\frac {2 \sqrt {e x +d}\, \left (5 x^{3} c^{3} d^{3} e^{3}+21 x^{2} a \,c^{2} d^{2} e^{4}-6 x^{2} c^{3} d^{4} e^{2}+35 x \,a^{2} c d \,e^{5}-28 x a \,c^{2} d^{3} e^{3}+8 x \,c^{3} d^{5} e +35 e^{6} a^{3}-70 d^{2} e^{4} a^{2} c +56 d^{4} e^{2} c^{2} a -16 c^{3} d^{6}\right )}{35 e^{4}}\)	\(131\)

[In]

int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3/(e*x+d)^(7/2),x,method=_RETURNVERBOSE)

[Out]

2/e^4*(1/7*c^3*d^3*(e*x+d)^(7/2)+3/5*(a*e^2-c*d^2)*c^2*d^2*(e*x+d)^(5/2)+(a*e^2-c*d^2)^2*c*d*(e*x+d)^(3/2)+(a*
e^2-c*d^2)^3*(e*x+d)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.13 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{7/2}} \, dx=\frac {2 \, {\left (5 \, c^{3} d^{3} e^{3} x^{3} - 16 \, c^{3} d^{6} + 56 \, a c^{2} d^{4} e^{2} - 70 \, a^{2} c d^{2} e^{4} + 35 \, a^{3} e^{6} - 3 \, {\left (2 \, c^{3} d^{4} e^{2} - 7 \, a c^{2} d^{2} e^{4}\right )} x^{2} + {\left (8 \, c^{3} d^{5} e - 28 \, a c^{2} d^{3} e^{3} + 35 \, a^{2} c d e^{5}\right )} x\right )} \sqrt {e x + d}}{35 \, e^{4}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3/(e*x+d)^(7/2),x, algorithm="fricas")

[Out]

2/35*(5*c^3*d^3*e^3*x^3 - 16*c^3*d^6 + 56*a*c^2*d^4*e^2 - 70*a^2*c*d^2*e^4 + 35*a^3*e^6 - 3*(2*c^3*d^4*e^2 - 7
*a*c^2*d^2*e^4)*x^2 + (8*c^3*d^5*e - 28*a*c^2*d^3*e^3 + 35*a^2*c*d*e^5)*x)*sqrt(e*x + d)/e^4

Sympy [A] (verification not implemented)

Time = 1.15 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.54 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{7/2}} \, dx=\begin {cases} \frac {2 \left (\frac {c^{3} d^{3} \left (d + e x\right )^{\frac {7}{2}}}{7 e^{3}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \cdot \left (3 a c^{2} d^{2} e^{2} - 3 c^{3} d^{4}\right )}{5 e^{3}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \cdot \left (3 a^{2} c d e^{4} - 6 a c^{2} d^{3} e^{2} + 3 c^{3} d^{5}\right )}{3 e^{3}} + \frac {\sqrt {d + e x} \left (a^{3} e^{6} - 3 a^{2} c d^{2} e^{4} + 3 a c^{2} d^{4} e^{2} - c^{3} d^{6}\right )}{e^{3}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {c^{3} d^{\frac {5}{2}} x^{4}}{4} & \text {otherwise} \end {cases} \]

[In]

integrate((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**3/(e*x+d)**(7/2),x)

[Out]

Piecewise((2*(c**3*d**3*(d + e*x)**(7/2)/(7*e**3) + (d + e*x)**(5/2)*(3*a*c**2*d**2*e**2 - 3*c**3*d**4)/(5*e**
3) + (d + e*x)**(3/2)*(3*a**2*c*d*e**4 - 6*a*c**2*d**3*e**2 + 3*c**3*d**5)/(3*e**3) + sqrt(d + e*x)*(a**3*e**6
- 3*a**2*c*d**2*e**4 + 3*a*c**2*d**4*e**2 - c**3*d**6)/e**3)/e, Ne(e, 0)), (c**3*d**(5/2)*x**4/4, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.19 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{7/2}} \, dx=\frac {2 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} c^{3} d^{3} - 21 \, {\left (c^{3} d^{4} - a c^{2} d^{2} e^{2}\right )} {\left (e x + d\right )}^{\frac {5}{2}} + 35 \, {\left (c^{3} d^{5} - 2 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4}\right )} {\left (e x + d\right )}^{\frac {3}{2}} - 35 \, {\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} \sqrt {e x + d}\right )}}{35 \, e^{4}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3/(e*x+d)^(7/2),x, algorithm="maxima")

[Out]

2/35*(5*(e*x + d)^(7/2)*c^3*d^3 - 21*(c^3*d^4 - a*c^2*d^2*e^2)*(e*x + d)^(5/2) + 35*(c^3*d^5 - 2*a*c^2*d^3*e^2
+ a^2*c*d*e^4)*(e*x + d)^(3/2) - 35*(c^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)*sqrt(e*x + d))/e^
4

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.26 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{7/2}} \, dx=\frac {2 \, {\left (35 \, \sqrt {e x + d} a^{3} e^{3} + 35 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a^{2} c d e + \frac {7 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a c^{2} d^{2}}{e} + \frac {{\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} c^{3} d^{3}}{e^{3}}\right )}}{35 \, e} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3/(e*x+d)^(7/2),x, algorithm="giac")

[Out]

2/35*(35*sqrt(e*x + d)*a^3*e^3 + 35*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*a^2*c*d*e + 7*(3*(e*x + d)^(5/2) - 1
0*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*a*c^2*d^2/e + (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^3*d^3/e^3)/e

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{7/2}} \, dx=\frac {2\,{\left (a\,e^2-c\,d^2\right )}^3\,\sqrt {d+e\,x}}{e^4}-\frac {\left (6\,c^3\,d^4-6\,a\,c^2\,d^2\,e^2\right )\,{\left (d+e\,x\right )}^{5/2}}{5\,e^4}+\frac {2\,c^3\,d^3\,{\left (d+e\,x\right )}^{7/2}}{7\,e^4}+\frac {2\,c\,d\,{\left (a\,e^2-c\,d^2\right )}^2\,{\left (d+e\,x\right )}^{3/2}}{e^4} \]

[In]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^3/(d + e*x)^(7/2),x)

[Out]

(2*(a*e^2 - c*d^2)^3*(d + e*x)^(1/2))/e^4 - ((6*c^3*d^4 - 6*a*c^2*d^2*e^2)*(d + e*x)^(5/2))/(5*e^4) + (2*c^3*d
^3*(d + e*x)^(7/2))/(7*e^4) + (2*c*d*(a*e^2 - c*d^2)^2*(d + e*x)^(3/2))/e^4