\(\int \frac {(d+e x)^{5/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx\) [2066]

Optimal result

Integrand size = 39, antiderivative size = 105 \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=-\frac {4 \left (c d^2-a e^2\right ) \sqrt {d+e x}}{c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {2 (d+e x)^{3/2}}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 105, 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} \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=\frac {2 (d+e x)^{3/2}}{c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac {4 \sqrt {d+e x} \left (c d^2-a e^2\right )}{c^2 d^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.49 \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=-\frac {2 \sqrt {d+e x} \left (-2 a e^2+c d (d-e x)\right )}{c^2 d^2 \sqrt {(a e+c d x) (d+e x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.79 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.57

method	result	size
default	\(\frac {2 \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}\, \left (x c d e +2 e^{2} a -c \,d^{2}\right )}{\sqrt {e x +d}\, \left (c d x +a e \right ) c^{2} d^{2}}\)	\(60\)
gosper	\(\frac {2 \left (c d x +a e \right ) \left (x c d e +2 e^{2} a -c \,d^{2}\right ) \left (e x +d \right )^{\frac {3}{2}}}{c^{2} d^{2} \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {3}{2}}}\)	\(68\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.93 \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=\frac {2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (c d e x - c d^{2} + 2 \, a e^{2}\right )} \sqrt {e x + d}}{c^{3} d^{3} e x^{2} + a c^{2} d^{3} e + {\left (c^{3} d^{4} + a c^{2} d^{2} e^{2}\right )} x} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.34 \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (c d e x - c d^{2} + 2 \, a e^{2}\right )}}{\sqrt {c d x + a e} c^{2} d^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.32 \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (\frac {\sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} e}{c d {\left | e \right |}} - \frac {c d^{2} e^{2} - a e^{4}}{\sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} c d {\left | e \right |}}\right )}}{c d} + \frac {4 \, {\left (c d^{2} e^{2} - a e^{4}\right )}}{\sqrt {-c d^{2} e + a e^{3}} c^{2} d^{2} {\left | e \right |}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.60 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.10 \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=\frac {\left (\frac {2\,x\,\sqrt {d+e\,x}}{c^2\,d^2}+\frac {\left (4\,a\,e^2-2\,c\,d^2\right )\,\sqrt {d+e\,x}}{c^3\,d^3\,e}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{\frac {a}{c}+x^2+\frac {x\,\left (c^3\,d^4+a\,c^2\,d^2\,e^2\right )}{c^3\,d^3\,e}} \]

[In]

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

[Out]

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