\(\int \frac {(d+e x)^{5/2}}{\sqrt {a d e+(c d^2+a e^2) x+c d e x^2}} \, dx\) [2058]

Optimal result

Integrand size = 39, antiderivative size = 171 \[ \int \frac {(d+e x)^{5/2}}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {16 \left (c d^2-a e^2\right )^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{15 c^3 d^3 \sqrt {d+e x}}+\frac {8 \left (c d^2-a e^2\right ) \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{15 c^2 d^2}+\frac {2 (d+e x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c d} \]

[Out]

2/5*(e*x+d)^(3/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/c/d+16/15*(-a*e^2+c*d^2)^2*(a*d*e+(a*e^2+c*d^2)*x+c*
d*e*x^2)^(1/2)/c^3/d^3/(e*x+d)^(1/2)+8/15*(-a*e^2+c*d^2)*(e*x+d)^(1/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)
/c^2/d^2

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {670, 662} \[ \int \frac {(d+e x)^{5/2}}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {16 \left (c d^2-a e^2\right )^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{15 c^3 d^3 \sqrt {d+e x}}+\frac {8 \sqrt {d+e x} \left (c d^2-a e^2\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{15 c^2 d^2}+\frac {2 (d+e x)^{3/2} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{5 c d} \]

[In]

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

[Out]

(16*(c*d^2 - a*e^2)^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(15*c^3*d^3*Sqrt[d + e*x]) + (8*(c*d^2 - a*
e^2)*Sqrt[d + e*x]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(15*c^2*d^2) + (2*(d + e*x)^(3/2)*Sqrt[a*d*e +
(c*d^2 + a*e^2)*x + c*d*e*x^2])/(5*c*d)

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} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c d}+\frac {\left (4 \left (d^2-\frac {a e^2}{c}\right )\right ) \int \frac {(d+e x)^{3/2}}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{5 d} \\ & = \frac {8 \left (c d^2-a e^2\right ) \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{15 c^2 d^2}+\frac {2 (d+e x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c d}+\frac {\left (8 \left (d^2-\frac {a e^2}{c}\right )^2\right ) \int \frac {\sqrt {d+e x}}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{15 d^2} \\ & = \frac {16 \left (c d^2-a e^2\right )^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{15 c^3 d^3 \sqrt {d+e x}}+\frac {8 \left (c d^2-a e^2\right ) \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{15 c^2 d^2}+\frac {2 (d+e x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.51 \[ \int \frac {(d+e x)^{5/2}}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \sqrt {(a e+c d x) (d+e x)} \left (8 a^2 e^4-4 a c d e^2 (5 d+e x)+c^2 d^2 \left (15 d^2+10 d e x+3 e^2 x^2\right )\right )}{15 c^3 d^3 \sqrt {d+e x}} \]

[In]

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

[Out]

(2*Sqrt[(a*e + c*d*x)*(d + e*x)]*(8*a^2*e^4 - 4*a*c*d*e^2*(5*d + e*x) + c^2*d^2*(15*d^2 + 10*d*e*x + 3*e^2*x^2
)))/(15*c^3*d^3*Sqrt[d + e*x])

Maple [A] (verified)

Time = 2.95 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.54

method	result	size
default	\(\frac {2 \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}\, \left (3 x^{2} c^{2} d^{2} e^{2}-4 x a c d \,e^{3}+10 x \,c^{2} d^{3} e +8 a^{2} e^{4}-20 a c \,d^{2} e^{2}+15 c^{2} d^{4}\right )}{15 \sqrt {e x +d}\, c^{3} d^{3}}\)	\(92\)
gosper	\(\frac {2 \left (c d x +a e \right ) \left (3 x^{2} c^{2} d^{2} e^{2}-4 x a c d \,e^{3}+10 x \,c^{2} d^{3} e +8 a^{2} e^{4}-20 a c \,d^{2} e^{2}+15 c^{2} d^{4}\right ) \sqrt {e x +d}}{15 c^{3} d^{3} \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}}\)	\(110\)

[In]

int((e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/15/(e*x+d)^(1/2)*((c*d*x+a*e)*(e*x+d))^(1/2)*(3*c^2*d^2*e^2*x^2-4*a*c*d*e^3*x+10*c^2*d^3*e*x+8*a^2*e^4-20*a*
c*d^2*e^2+15*c^2*d^4)/c^3/d^3

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.68 \[ \int \frac {(d+e x)^{5/2}}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \, {\left (3 \, c^{2} d^{2} e^{2} x^{2} + 15 \, c^{2} d^{4} - 20 \, a c d^{2} e^{2} + 8 \, a^{2} e^{4} + 2 \, {\left (5 \, c^{2} d^{3} e - 2 \, a c d e^{3}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{15 \, {\left (c^{3} d^{3} e x + c^{3} d^{4}\right )}} \]

[In]

integrate((e*x+d)^(5/2)/(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^2*x^2 + 15*c^2*d^4 - 20*a*c*d^2*e^2 + 8*a^2*e^4 + 2*(5*c^2*d^3*e - 2*a*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^3*d^3*e*x + c^3*d^4)

Sympy [F]

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

[In]

integrate((e*x+d)**(5/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(1/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.71 \[ \int \frac {(d+e x)^{5/2}}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \, {\left (3 \, c^{3} d^{3} e^{2} x^{3} + 15 \, a c^{2} d^{4} e - 20 \, a^{2} c d^{2} e^{3} + 8 \, a^{3} e^{5} + {\left (10 \, c^{3} d^{4} e - a c^{2} d^{2} e^{3}\right )} x^{2} + {\left (15 \, c^{3} d^{5} - 10 \, a c^{2} d^{3} e^{2} + 4 \, a^{2} c d e^{4}\right )} x\right )}}{15 \, \sqrt {c d x + a e} c^{3} d^{3}} \]

[In]

integrate((e*x+d)^(5/2)/(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^3*d^3*e^2*x^3 + 15*a*c^2*d^4*e - 20*a^2*c*d^2*e^3 + 8*a^3*e^5 + (10*c^3*d^4*e - a*c^2*d^2*e^3)*x^2 +
(15*c^3*d^5 - 10*a*c^2*d^3*e^2 + 4*a^2*c*d*e^4)*x)/(sqrt(c*d*x + a*e)*c^3*d^3)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.44 \[ \int \frac {(d+e x)^{5/2}}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \, e {\left (\frac {15 \, {\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}}}{c^{3} d^{3} e} - \frac {8 \, {\left (\sqrt {-c d^{2} e + a e^{3}} c^{2} d^{4} - 2 \, \sqrt {-c d^{2} e + a e^{3}} a c d^{2} e^{2} + \sqrt {-c d^{2} e + a e^{3}} a^{2} e^{4}\right )}}{c^{3} d^{3} e} + \frac {10 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} c d^{2} e - 10 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} a e^{3} + 3 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {5}{2}}}{c^{3} d^{3} e^{3}}\right )}}{15 \, {\left | e \right |}} \]

[In]

integrate((e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="giac")

[Out]

2/15*e*(15*(c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*sqrt((e*x + d)*c*d*e - c*d^2*e + a*e^3)/(c^3*d^3*e) - 8*(sqrt(-
c*d^2*e + a*e^3)*c^2*d^4 - 2*sqrt(-c*d^2*e + a*e^3)*a*c*d^2*e^2 + sqrt(-c*d^2*e + a*e^3)*a^2*e^4)/(c^3*d^3*e)
+ (10*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*c*d^2*e - 10*((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^3*d^3*e^3))/abs(e)

Mupad [B] (verification not implemented)

Time = 10.44 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.77 \[ \int \frac {(d+e x)^{5/2}}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (\frac {2\,e\,x^2\,\sqrt {d+e\,x}}{5\,c\,d}-\frac {4\,x\,\left (2\,a\,e^2-5\,c\,d^2\right )\,\sqrt {d+e\,x}}{15\,c^2\,d^2}+\frac {\sqrt {d+e\,x}\,\left (16\,a^2\,e^4-40\,a\,c\,d^2\,e^2+30\,c^2\,d^4\right )}{15\,c^3\,d^3\,e}\right )}{x+\frac {d}{e}} \]

[In]

int((d + e*x)^(5/2)/(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*e*x^2*(d + e*x)^(1/2))/(5*c*d) - (4*x*(2*a*e^2 - 5*c*d^2)*(
d + e*x)^(1/2))/(15*c^2*d^2) + ((d + e*x)^(1/2)*(16*a^2*e^4 + 30*c^2*d^4 - 40*a*c*d^2*e^2))/(15*c^3*d^3*e)))/(
x + d/e)