\(\int \sqrt {d+e x} \sqrt {a d e+(c d^2+a e^2) x+c d e x^2} \, dx\) [2030]

Optimal result

Integrand size = 39, antiderivative size = 109 \[ \int \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {4 \left (c d^2-a e^2\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{15 c^2 d^2 (d+e x)^{3/2}}+\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{5 c d \sqrt {d+e x}} \]

[Out]

4/15*(-a*e^2+c*d^2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/c^2/d^2/(e*x+d)^(3/2)+2/5*(a*d*e+(a*e^2+c*d^2)*x+c
*d*e*x^2)^(3/2)/c/d/(e*x+d)^(1/2)

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} \[ \int \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {4 \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{15 c^2 d^2 (d+e x)^{3/2}}+\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{5 c d \sqrt {d+e x}} \]

[In]

Int[Sqrt[d + e*x]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^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)^(3/2))/(15*c^2*d^2*(d + e*x)^(3/2)) + (2*(a*d*e + (
c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(5*c*d*Sqrt[d + e*x])

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 )^{3/2}}{5 c d \sqrt {d+e x}}+\frac {\left (2 \left (d^2-\frac {a e^2}{c}\right )\right ) \int \frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {d+e x}} \, dx}{5 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 )^{3/2}}{15 c^2 d^2 (d+e x)^{3/2}}+\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{5 c d \sqrt {d+e x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.50 \[ \int \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {2 ((a e+c d x) (d+e x))^{3/2} \left (-2 a e^2+c d (5 d+3 e x)\right )}{15 c^2 d^2 (d+e x)^{3/2}} \]

[In]

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

[Out]

(2*((a*e + c*d*x)*(d + e*x))^(3/2)*(-2*a*e^2 + c*d*(5*d + 3*e*x)))/(15*c^2*d^2*(d + e*x)^(3/2))

Maple [A] (verified)

Time = 2.74 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.54

method	result	size
default	\(-\frac {2 \left (c d x +a e \right ) \left (-3 x c d e +2 e^{2} a -5 c \,d^{2}\right ) \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}}{15 c^{2} d^{2} \sqrt {e x +d}}\)	\(59\)
gosper	\(-\frac {2 \left (c d x +a e \right ) \left (-3 x c d e +2 e^{2} a -5 c \,d^{2}\right ) \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}}{15 c^{2} d^{2} \sqrt {e x +d}}\)	\(69\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.94 \[ \int \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {2 \, {\left (3 \, c^{2} d^{2} e x^{2} + 5 \, a c d^{2} e - 2 \, a^{2} e^{3} + {\left (5 \, c^{2} d^{3} + a c d e^{2}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{15 \, {\left (c^{2} d^{2} e x + c^{2} d^{3}\right )}} \]

[In]

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

[Out]

2/15*(3*c^2*d^2*e*x^2 + 5*a*c*d^2*e - 2*a^2*e^3 + (5*c^2*d^3 + a*c*d*e^2)*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]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.76 \[ \int \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {2 \, {\left (3 \, c^{2} d^{2} e x^{2} + 5 \, a c d^{2} e - 2 \, a^{2} e^{3} + {\left (5 \, c^{2} d^{3} + a c d e^{2}\right )} x\right )} \sqrt {c d x + a e} {\left (e x + d\right )}}{15 \, {\left (c^{2} d^{2} e x + c^{2} d^{3}\right )}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 252 vs. \(2 (97) = 194\).

Time = 0.28 (sec) , antiderivative size = 252, normalized size of antiderivative = 2.31 \[ \int \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {2 \, {\left (\frac {5 \, d {\left (\frac {\sqrt {-c d^{2} e + a e^{3}} c d^{2} - \sqrt {-c d^{2} e + a e^{3}} a e^{2}}{c d} + \frac {{\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}}}{c d e}\right )} {\left | e \right |}}{e^{2}} - \frac {{\left (\frac {3 \, \sqrt {-c d^{2} e + a e^{3}} c^{2} d^{4} - \sqrt {-c d^{2} e + a e^{3}} a c d^{2} e^{2} - 2 \, \sqrt {-c d^{2} e + a e^{3}} a^{2} e^{4}}{c^{2} d^{2}} + \frac {5 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} a e^{3} - 3 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {5}{2}}}{c^{2} d^{2} e^{2}}\right )} {\left | e \right |}}{e^{2}}\right )}}{15 \, e} \]

[In]

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

[Out]

2/15*(5*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^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 = 10.13 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.11 \[ \int \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (\frac {2\,x^2\,\sqrt {d+e\,x}}{5}-\frac {\left (4\,a^2\,e^3-10\,a\,c\,d^2\,e\right )\,\sqrt {d+e\,x}}{15\,c^2\,d^2\,e}+\frac {x\,\left (10\,c^2\,d^3+2\,a\,c\,d\,e^2\right )\,\sqrt {d+e\,x}}{15\,c^2\,d^2\,e}\right )}{x+\frac {d}{e}} \]

[In]

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

[Out]

((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)*((2*x^2*(d + e*x)^(1/2))/5 - ((4*a^2*e^3 - 10*a*c*d^2*e)*(d + e
*x)^(1/2))/(15*c^2*d^2*e) + (x*(10*c^2*d^3 + 2*a*c*d*e^2)*(d + e*x)^(1/2))/(15*c^2*d^2*e)))/(x + d/e)