\(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}}{(d+e x)^{3/2}} \, dx\) [2049]

Optimal result

Integrand size = 39, antiderivative size = 109 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{3/2}} \, dx=\frac {4 \left (c d^2-a e^2\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{7/2}}{63 c^2 d^2 (d+e x)^{7/2}}+\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{7/2}}{9 c d (d+e x)^{5/2}} \]

[Out]

4/63*(-a*e^2+c*d^2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(7/2)/c^2/d^2/(e*x+d)^(7/2)+2/9*(a*d*e+(a*e^2+c*d^2)*x+c
*d*e*x^2)^(7/2)/c/d/(e*x+d)^(5/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {670, 662} \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{3/2}} \, dx=\frac {4 \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{7/2}}{63 c^2 d^2 (d+e x)^{7/2}}+\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{7/2}}{9 c d (d+e x)^{5/2}} \]

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(d + e*x)^(3/2),x]

[Out]

(4*(c*d^2 - a*e^2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(7/2))/(63*c^2*d^2*(d + e*x)^(7/2)) + (2*(a*d*e + (
c*d^2 + a*e^2)*x + c*d*e*x^2)^(7/2))/(9*c*d*(d + e*x)^(5/2))

Rule 662

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*
((a + b*x + c*x^2)^(p + 1)/(c*(p + 1))), 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 + p, 0]

Rule 670

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*
((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Dist[Simplify[m + p]*((2*c*d - b*e)/(c*(m + 2*p + 1))), In
t[(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] && E
qQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p], 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 )^{7/2}}{9 c d (d+e x)^{5/2}}+\frac {\left (2 \left (d^2-\frac {a e^2}{c}\right )\right ) \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{5/2}} \, dx}{9 d} \\ & = \frac {4 \left (c d^2-a e^2\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{7/2}}{63 c^2 d^2 (d+e x)^{7/2}}+\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{7/2}}{9 c d (d+e x)^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.60 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{3/2}} \, dx=\frac {2 (a e+c d x)^3 \sqrt {(a e+c d x) (d+e x)} \left (-2 a e^2+c d (9 d+7 e x)\right )}{63 c^2 d^2 \sqrt {d+e x}} \]

[In]

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(d + e*x)^(3/2),x]

[Out]

(2*(a*e + c*d*x)^3*Sqrt[(a*e + c*d*x)*(d + e*x)]*(-2*a*e^2 + c*d*(9*d + 7*e*x)))/(63*c^2*d^2*Sqrt[d + e*x])

Maple [A] (verified)

Time = 2.45 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.56

method	result	size
default	\(-\frac {2 \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}\, \left (c d x +a e \right )^{3} \left (-7 x c d e +2 e^{2} a -9 c \,d^{2}\right )}{63 \sqrt {e x +d}\, c^{2} d^{2}}\)	\(61\)
gosper	\(-\frac {2 \left (c d x +a e \right ) \left (-7 x c d e +2 e^{2} a -9 c \,d^{2}\right ) \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {5}{2}}}{63 c^{2} d^{2} \left (e x +d \right )^{\frac {5}{2}}}\)	\(69\)

[In]

int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2/63*((c*d*x+a*e)*(e*x+d))^(1/2)/(e*x+d)^(1/2)*(c*d*x+a*e)^3*(-7*c*d*e*x+2*a*e^2-9*c*d^2)/c^2/d^2

Fricas [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.55 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (7 \, c^{4} d^{4} e x^{4} + 9 \, a^{3} c d^{2} e^{3} - 2 \, a^{4} e^{5} + {\left (9 \, c^{4} d^{5} + 19 \, a c^{3} d^{3} e^{2}\right )} x^{3} + 3 \, {\left (9 \, a c^{3} d^{4} e + 5 \, a^{2} c^{2} d^{2} e^{3}\right )} x^{2} + {\left (27 \, a^{2} c^{2} d^{3} e^{2} + a^{3} c d e^{4}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{63 \, {\left (c^{2} d^{2} e x + c^{2} d^{3}\right )}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(3/2),x, algorithm="fricas")

[Out]

2/63*(7*c^4*d^4*e*x^4 + 9*a^3*c*d^2*e^3 - 2*a^4*e^5 + (9*c^4*d^5 + 19*a*c^3*d^3*e^2)*x^3 + 3*(9*a*c^3*d^4*e +
5*a^2*c^2*d^2*e^3)*x^2 + (27*a^2*c^2*d^3*e^2 + a^3*c*d*e^4)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqr
t(e*x + d)/(c^2*d^2*e*x + c^2*d^3)

Sympy [F]

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

[In]

integrate((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(5/2)/(e*x+d)**(3/2),x)

[Out]

Integral(((d + e*x)*(a*e + c*d*x))**(5/2)/(d + e*x)**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.21 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (7 \, c^{4} d^{4} e x^{4} + 9 \, a^{3} c d^{2} e^{3} - 2 \, a^{4} e^{5} + {\left (9 \, c^{4} d^{5} + 19 \, a c^{3} d^{3} e^{2}\right )} x^{3} + 3 \, {\left (9 \, a c^{3} d^{4} e + 5 \, a^{2} c^{2} d^{2} e^{3}\right )} x^{2} + {\left (27 \, a^{2} c^{2} d^{3} e^{2} + a^{3} c d e^{4}\right )} x\right )} \sqrt {c d x + a e}}{63 \, c^{2} d^{2}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(3/2),x, algorithm="maxima")

[Out]

2/63*(7*c^4*d^4*e*x^4 + 9*a^3*c*d^2*e^3 - 2*a^4*e^5 + (9*c^4*d^5 + 19*a*c^3*d^3*e^2)*x^3 + 3*(9*a*c^3*d^4*e +
5*a^2*c^2*d^2*e^3)*x^2 + (27*a^2*c^2*d^3*e^2 + a^3*c*d*e^4)*x)*sqrt(c*d*x + a*e)/(c^2*d^2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1138 vs. \(2 (97) = 194\).

Time = 0.32 (sec) , antiderivative size = 1138, normalized size of antiderivative = 10.44 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{3/2}} \, dx=\text {Too large to display} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(3/2),x, algorithm="giac")

[Out]

2/315*(105*a^2*d*((sqrt(-c*d^2*e + a*e^3)*c*d^2 - sqrt(-c*d^2*e + a*e^3)*a*e^2)/(c*d) + ((e*x + d)*c*d*e - c*d
^2*e + a*e^3)^(3/2)/(c*d*e))*abs(e) + 6*a*c*d*((15*sqrt(-c*d^2*e + a*e^3)*c^3*d^6 - 3*sqrt(-c*d^2*e + a*e^3)*a
*c^2*d^4*e^2 - 4*sqrt(-c*d^2*e + a*e^3)*a^2*c*d^2*e^4 - 8*sqrt(-c*d^2*e + a*e^3)*a^3*e^6)/(c^3*d^3*e^2) + (35*
((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*a^2*e^6 - 42*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(5/2)*a*e^3 + 15*((
e*x + d)*c*d*e - c*d^2*e + a*e^3)^(7/2))/(c^3*d^3*e^5))*abs(e) + 3*c^2*d^3*((15*sqrt(-c*d^2*e + a*e^3)*c^3*d^6
- 3*sqrt(-c*d^2*e + a*e^3)*a*c^2*d^4*e^2 - 4*sqrt(-c*d^2*e + a*e^3)*a^2*c*d^2*e^4 - 8*sqrt(-c*d^2*e + a*e^3)*
a^3*e^6)/(c^3*d^3*e^2) + (35*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*a^2*e^6 - 42*((e*x + d)*c*d*e - c*d^2*e
+ a*e^3)^(5/2)*a*e^3 + 15*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(7/2))/(c^3*d^3*e^5))*abs(e)/e^2 - c^2*d^2*((35
*sqrt(-c*d^2*e + a*e^3)*c^4*d^8 - 5*sqrt(-c*d^2*e + a*e^3)*a*c^3*d^6*e^2 - 6*sqrt(-c*d^2*e + a*e^3)*a^2*c^2*d^
4*e^4 - 8*sqrt(-c*d^2*e + a*e^3)*a^3*c*d^2*e^6 - 16*sqrt(-c*d^2*e + a*e^3)*a^4*e^8)/(c^4*d^4*e^3) + (105*((e*x
+ d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*a^3*e^9 - 189*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(5/2)*a^2*e^6 + 135*((e
*x + d)*c*d*e - c*d^2*e + a*e^3)^(7/2)*a*e^3 - 35*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(9/2))/(c^4*d^4*e^7))*ab
s(e)/e - 21*a^2*((3*sqrt(-c*d^2*e + a*e^3)*c^2*d^4 - sqrt(-c*d^2*e + a*e^3)*a*c*d^2*e^2 - 2*sqrt(-c*d^2*e + a*
e^3)*a^2*e^4)/(c^2*d^2) + (5*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*a*e^3 - 3*((e*x + d)*c*d*e - c*d^2*e +
a*e^3)^(5/2))/(c^2*d^2*e^2))*abs(e) - 42*a*c*d^2*((3*sqrt(-c*d^2*e + a*e^3)*c^2*d^4 - sqrt(-c*d^2*e + a*e^3)*a
*c*d^2*e^2 - 2*sqrt(-c*d^2*e + a*e^3)*a^2*e^4)/(c^2*d^2) + (5*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*a*e^3
- 3*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(5/2))/(c^2*d^2*e^2))*abs(e)/e^2)/e

Mupad [B] (verification not implemented)

Time = 10.54 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.43 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{3/2}} \, dx=\frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (\frac {x^3\,\left (18\,c^4\,d^5+38\,a\,c^3\,d^3\,e^2\right )}{63\,c^2\,d^2}-\frac {4\,a^4\,e^5-18\,a^3\,c\,d^2\,e^3}{63\,c^2\,d^2}+\frac {2\,c^2\,d^2\,e\,x^4}{9}+\frac {2\,a\,e\,x^2\,\left (9\,c\,d^2+5\,a\,e^2\right )}{21}+\frac {2\,a^2\,e^2\,x\,\left (27\,c\,d^2+a\,e^2\right )}{63\,c\,d}\right )}{\sqrt {d+e\,x}} \]

[In]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2)/(d + e*x)^(3/2),x)

[Out]

((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)*((x^3*(18*c^4*d^5 + 38*a*c^3*d^3*e^2))/(63*c^2*d^2) - (4*a^4*e^
5 - 18*a^3*c*d^2*e^3)/(63*c^2*d^2) + (2*c^2*d^2*e*x^4)/9 + (2*a*e*x^2*(5*a*e^2 + 9*c*d^2))/21 + (2*a^2*e^2*x*(
a*e^2 + 27*c*d^2))/(63*c*d)))/(d + e*x)^(1/2)