\(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}}{\sqrt {d+e x}} \, dx\) [2039]

Optimal result

Integrand size = 39, antiderivative size = 109 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{\sqrt {d+e x}} \, dx=\frac {4 \left (c d^2-a e^2\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{35 c^2 d^2 (d+e x)^{5/2}}+\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{7 c d (d+e x)^{3/2}} \]

[Out]

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

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} \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{\sqrt {d+e x}} \, dx=\frac {4 \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{35 c^2 d^2 (d+e x)^{5/2}}+\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{7 c d (d+e x)^{3/2}} \]

[In]

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

[Out]

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

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 )^{5/2}}{7 c d (d+e x)^{3/2}}+\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 )^{3/2}}{(d+e x)^{3/2}} \, dx}{7 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 )^{5/2}}{35 c^2 d^2 (d+e x)^{5/2}}+\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{7 c d (d+e x)^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.50 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{\sqrt {d+e x}} \, dx=\frac {2 ((a e+c d x) (d+e x))^{5/2} \left (-2 a e^2+c d (7 d+5 e x)\right )}{35 c^2 d^2 (d+e x)^{5/2}} \]

[In]

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

[Out]

(2*((a*e + c*d*x)*(d + e*x))^(5/2)*(-2*a*e^2 + c*d*(7*d + 5*e*x)))/(35*c^2*d^2*(d + e*x)^(5/2))

Maple [A] (verified)

Time = 2.49 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.56

method	result	size
default	\(-\frac {2 \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}\, \left (c d x +a e \right )^{2} \left (-5 x c d e +2 e^{2} a -7 c \,d^{2}\right )}{35 \sqrt {e x +d}\, c^{2} d^{2}}\)	\(61\)
gosper	\(-\frac {2 \left (c d x +a e \right ) \left (-5 x c d e +2 e^{2} a -7 c \,d^{2}\right ) \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {3}{2}}}{35 c^{2} d^{2} \left (e x +d \right )^{\frac {3}{2}}}\)	\(69\)

[In]

int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/(e*x+d)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.24 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (5 \, c^{3} d^{3} e x^{3} + 7 \, a^{2} c d^{2} e^{2} - 2 \, a^{3} e^{4} + {\left (7 \, c^{3} d^{4} + 8 \, a c^{2} d^{2} e^{2}\right )} x^{2} + {\left (14 \, a c^{2} d^{3} e + a^{2} c d e^{3}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{35 \, {\left (c^{2} d^{2} e x + c^{2} d^{3}\right )}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

2/35*(5*c^3*d^3*e*x^3 + 7*a^2*c*d^2*e^2 - 2*a^3*e^4 + (7*c^3*d^4 + 8*a*c^2*d^2*e^2)*x^2 + (14*a*c^2*d^3*e + a^
2*c*d*e^3)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d)/(c^2*d^2*e*x + c^2*d^3)

Sympy [F]

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

[In]

integrate((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(3/2)/(e*x+d)**(1/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (5 \, c^{3} d^{3} e x^{3} + 7 \, a^{2} c d^{2} e^{2} - 2 \, a^{3} e^{4} + {\left (7 \, c^{3} d^{4} + 8 \, a c^{2} d^{2} e^{2}\right )} x^{2} + {\left (14 \, a c^{2} d^{3} e + a^{2} c d e^{3}\right )} x\right )} \sqrt {c d x + a e}}{35 \, c^{2} d^{2}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/(e*x+d)^(1/2),x, algorithm="maxima")

[Out]

2/35*(5*c^3*d^3*e*x^3 + 7*a^2*c*d^2*e^2 - 2*a^3*e^4 + (7*c^3*d^4 + 8*a*c^2*d^2*e^2)*x^2 + (14*a*c^2*d^3*e + a^
2*c*d*e^3)*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. 631 vs. \(2 (97) = 194\).

Time = 0.31 (sec) , antiderivative size = 631, normalized size of antiderivative = 5.79 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (\frac {35 \, a d {\left (\frac {\sqrt {-c d^{2} e + a e^{3}} c d^{2} - \sqrt {-c d^{2} e + a e^{3}} a e^{2}}{c d} + \frac {{\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}}}{c d e}\right )} {\left | e \right |}}{e} + \frac {c d {\left (\frac {15 \, \sqrt {-c d^{2} e + a e^{3}} c^{3} d^{6} - 3 \, \sqrt {-c d^{2} e + a e^{3}} a c^{2} d^{4} e^{2} - 4 \, \sqrt {-c d^{2} e + a e^{3}} a^{2} c d^{2} e^{4} - 8 \, \sqrt {-c d^{2} e + a e^{3}} a^{3} e^{6}}{c^{3} d^{3} e^{2}} + \frac {35 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} a^{2} e^{6} - 42 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {5}{2}} a e^{3} + 15 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {7}{2}}}{c^{3} d^{3} e^{5}}\right )} {\left | e \right |}}{e} - \frac {7 \, c d^{2} {\left (\frac {3 \, \sqrt {-c d^{2} e + a e^{3}} c^{2} d^{4} - \sqrt {-c d^{2} e + a e^{3}} a c d^{2} e^{2} - 2 \, \sqrt {-c d^{2} e + a e^{3}} a^{2} e^{4}}{c^{2} d^{2}} + \frac {5 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} a e^{3} - 3 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {5}{2}}}{c^{2} d^{2} e^{2}}\right )} {\left | e \right |}}{e^{3}} - \frac {7 \, a {\left (\frac {3 \, \sqrt {-c d^{2} e + a e^{3}} c^{2} d^{4} - \sqrt {-c d^{2} e + a e^{3}} a c d^{2} e^{2} - 2 \, \sqrt {-c d^{2} e + a e^{3}} a^{2} e^{4}}{c^{2} d^{2}} + \frac {5 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} a e^{3} - 3 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {5}{2}}}{c^{2} d^{2} e^{2}}\right )} {\left | e \right |}}{e}\right )}}{105 \, e} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

2/105*(35*a*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)/e + 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)/e - 7*c*d^2*((3*sqrt(-c*d^2*e + a*e^3)*c^2*d^4 - sqr
t(-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^3 - 7*a*((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)/e

Mupad [B] (verification not implemented)

Time = 10.49 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.17 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{\sqrt {d+e x}} \, dx=\frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (\frac {x^2\,\left (14\,c^3\,d^4+16\,a\,c^2\,d^2\,e^2\right )}{35\,c^2\,d^2}-\frac {4\,a^3\,e^4-14\,a^2\,c\,d^2\,e^2}{35\,c^2\,d^2}+\frac {2\,c\,d\,e\,x^3}{7}+\frac {2\,a\,e\,x\,\left (14\,c\,d^2+a\,e^2\right )}{35\,c\,d}\right )}{\sqrt {d+e\,x}} \]

[In]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(3/2)/(d + e*x)^(1/2),x)

[Out]

((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)*((x^2*(14*c^3*d^4 + 16*a*c^2*d^2*e^2))/(35*c^2*d^2) - (4*a^3*e^
4 - 14*a^2*c*d^2*e^2)/(35*c^2*d^2) + (2*c*d*e*x^3)/7 + (2*a*e*x*(a*e^2 + 14*c*d^2))/(35*c*d)))/(d + e*x)^(1/2)