\(\int \frac {d+e x}{(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}} \, dx\) [1969]

Optimal result

Integrand size = 35, antiderivative size = 118 \[ \int \frac {d+e x}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=-\frac {2 (d+e x)}{3 \left (c d^2-a e^2\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}+\frac {8 e \left (c d^2+a e^2+2 c d e x\right )}{3 \left (c d^2-a e^2\right )^3 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \]

[Out]

-2/3*(e*x+d)/(-a*e^2+c*d^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)+8/3*e*(2*c*d*e*x+a*e^2+c*d^2)/(-a*e^2+c*d^
2)^3/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {652, 627} \[ \int \frac {d+e x}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\frac {8 e \left (a e^2+c d^2+2 c d e x\right )}{3 \left (c d^2-a e^2\right )^3 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac {2 (d+e x)}{3 \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}} \]

[In]

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

[Out]

(-2*(d + e*x))/(3*(c*d^2 - a*e^2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)) + (8*e*(c*d^2 + a*e^2 + 2*c*d
*e*x))/(3*(c*d^2 - a*e^2)^3*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])

Rule 627

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[-2*((b + 2*c*x)/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2])), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 652

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)/((p + 1)*(b^2 - 4*a*c)))*(a + b*x + c*x^2)^(p + 1), x] - Dist[(2*p + 3)*((2*c*d - b*e)/((p + 1)*(b^2 -
4*a*c))), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 (d+e x)}{3 \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}}-\frac {(4 e) \int \frac {1}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx}{3 \left (c d^2-a e^2\right )} \\ & = -\frac {2 (d+e x)}{3 \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}}+\frac {8 e \left (c d^2+a e^2+2 c d e x\right )}{3 \left (c d^2-a e^2\right )^3 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.76 \[ \int \frac {d+e x}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\frac {2 (d+e x) \left (3 a^2 e^4+6 a c d e^2 (d+2 e x)+c^2 d^2 \left (-d^2+4 d e x+8 e^2 x^2\right )\right )}{3 \left (c d^2-a e^2\right )^3 ((a e+c d x) (d+e x))^{3/2}} \]

[In]

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

[Out]

(2*(d + e*x)*(3*a^2*e^4 + 6*a*c*d*e^2*(d + 2*e*x) + c^2*d^2*(-d^2 + 4*d*e*x + 8*e^2*x^2)))/(3*(c*d^2 - a*e^2)^
3*((a*e + c*d*x)*(d + e*x))^(3/2))

Maple [A] (verified)

Time = 2.74 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.24

method	result	size
gosper	\(-\frac {2 \left (c d x +a e \right ) \left (e x +d \right )^{2} \left (8 x^{2} c^{2} d^{2} e^{2}+12 x a c d \,e^{3}+4 x \,c^{2} d^{3} e +3 a^{2} e^{4}+6 a c \,d^{2} e^{2}-c^{2} d^{4}\right )}{3 \left (e^{6} a^{3}-3 d^{2} e^{4} a^{2} c +3 d^{4} e^{2} c^{2} a -c^{3} d^{6}\right ) \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {5}{2}}}\)	\(146\)
trager	\(-\frac {2 \left (8 x^{2} c^{2} d^{2} e^{2}+12 x a c d \,e^{3}+4 x \,c^{2} d^{3} e +3 a^{2} e^{4}+6 a c \,d^{2} e^{2}-c^{2} d^{4}\right ) \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}}{3 \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \left (c d x +a e \right )^{2} \left (e^{2} a -c \,d^{2}\right ) \left (e x +d \right )}\)	\(147\)
default	\(d \left (\frac {\frac {4}{3} x c d e +\frac {2}{3} e^{2} a +\frac {2}{3} c \,d^{2}}{\left (4 a c \,d^{2} e^{2}-\left (e^{2} a +c \,d^{2}\right )^{2}\right ) {\left (a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}\right )}^{\frac {3}{2}}}+\frac {16 c d e \left (2 x c d e +e^{2} a +c \,d^{2}\right )}{3 \left (4 a c \,d^{2} e^{2}-\left (e^{2} a +c \,d^{2}\right )^{2}\right )^{2} \sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}}\right )+e \left (-\frac {1}{3 c d e {\left (a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}\right )}^{\frac {3}{2}}}-\frac {\left (e^{2} a +c \,d^{2}\right ) \left (\frac {\frac {4}{3} x c d e +\frac {2}{3} e^{2} a +\frac {2}{3} c \,d^{2}}{\left (4 a c \,d^{2} e^{2}-\left (e^{2} a +c \,d^{2}\right )^{2}\right ) {\left (a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}\right )}^{\frac {3}{2}}}+\frac {16 c d e \left (2 x c d e +e^{2} a +c \,d^{2}\right )}{3 \left (4 a c \,d^{2} e^{2}-\left (e^{2} a +c \,d^{2}\right )^{2}\right )^{2} \sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}}\right )}{2 c d e}\right )\)	\(371\)

[In]

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

[Out]

-2/3*(c*d*x+a*e)*(e*x+d)^2*(8*c^2*d^2*e^2*x^2+12*a*c*d*e^3*x+4*c^2*d^3*e*x+3*a^2*e^4+6*a*c*d^2*e^2-c^2*d^4)/(a
^3*e^6-3*a^2*c*d^2*e^4+3*a*c^2*d^4*e^2-c^3*d^6)/(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(5/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 316 vs. \(2 (110) = 220\).

Time = 2.03 (sec) , antiderivative size = 316, normalized size of antiderivative = 2.68 \[ \int \frac {d+e x}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\frac {2 \, {\left (8 \, c^{2} d^{2} e^{2} x^{2} - c^{2} d^{4} + 6 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4} + 4 \, {\left (c^{2} d^{3} e + 3 \, a c d e^{3}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{3 \, {\left (a^{2} c^{3} d^{7} e^{2} - 3 \, a^{3} c^{2} d^{5} e^{4} + 3 \, a^{4} c d^{3} e^{6} - a^{5} d e^{8} + {\left (c^{5} d^{8} e - 3 \, a c^{4} d^{6} e^{3} + 3 \, a^{2} c^{3} d^{4} e^{5} - a^{3} c^{2} d^{2} e^{7}\right )} x^{3} + {\left (c^{5} d^{9} - a c^{4} d^{7} e^{2} - 3 \, a^{2} c^{3} d^{5} e^{4} + 5 \, a^{3} c^{2} d^{3} e^{6} - 2 \, a^{4} c d e^{8}\right )} x^{2} + {\left (2 \, a c^{4} d^{8} e - 5 \, a^{2} c^{3} d^{6} e^{3} + 3 \, a^{3} c^{2} d^{4} e^{5} + a^{4} c d^{2} e^{7} - a^{5} e^{9}\right )} x\right )}} \]

[In]

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

[Out]

2/3*(8*c^2*d^2*e^2*x^2 - c^2*d^4 + 6*a*c*d^2*e^2 + 3*a^2*e^4 + 4*(c^2*d^3*e + 3*a*c*d*e^3)*x)*sqrt(c*d*e*x^2 +
a*d*e + (c*d^2 + a*e^2)*x)/(a^2*c^3*d^7*e^2 - 3*a^3*c^2*d^5*e^4 + 3*a^4*c*d^3*e^6 - a^5*d*e^8 + (c^5*d^8*e -
3*a*c^4*d^6*e^3 + 3*a^2*c^3*d^4*e^5 - a^3*c^2*d^2*e^7)*x^3 + (c^5*d^9 - a*c^4*d^7*e^2 - 3*a^2*c^3*d^5*e^4 + 5*
a^3*c^2*d^3*e^6 - 2*a^4*c*d*e^8)*x^2 + (2*a*c^4*d^8*e - 5*a^2*c^3*d^6*e^3 + 3*a^3*c^2*d^4*e^5 + a^4*c*d^2*e^7
- a^5*e^9)*x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {d+e x}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e^2-c*d^2>0)', see `assume?`
for more de

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.39 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.02 \[ \int \frac {d+e x}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=-\frac {2\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (3\,a^2\,e^4+6\,a\,c\,d^2\,e^2+12\,a\,c\,d\,e^3\,x-c^2\,d^4+4\,c^2\,d^3\,e\,x+8\,c^2\,d^2\,e^2\,x^2\right )}{3\,{\left (a\,e+c\,d\,x\right )}^2\,{\left (a\,e^2-c\,d^2\right )}^3\,\left (d+e\,x\right )} \]

[In]

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

[Out]

-(2*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)*(3*a^2*e^4 - c^2*d^4 + 8*c^2*d^2*e^2*x^2 + 6*a*c*d^2*e^2 + 4
*c^2*d^3*e*x + 12*a*c*d*e^3*x))/(3*(a*e + c*d*x)^2*(a*e^2 - c*d^2)^3*(d + e*x))