3.17.42 \(\int \frac {(d+e x)^3}{(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx\)

Optimal. Leaf size=180 \[ \frac {3 \sqrt {e} \left (c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{2 c^{5/2} d^{5/2}}+\frac {3 e \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c^2 d^2}-\frac {2 (d+e x)^2}{c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {668, 640, 621, 206} \begin {gather*} \frac {3 e \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c^2 d^2}+\frac {3 \sqrt {e} \left (c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{2 c^{5/2} d^{5/2}}-\frac {2 (d+e x)^2}{c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 640

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

Rule 668

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] - Dist[(e^2*(m + p))/(c*(p + 1)), Int[(d + e*x)^(m - 2)*(a + b*x +
c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &
& LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]

Rubi steps

\begin {align*} \int \frac {(d+e x)^3}{\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)^2}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {(3 e) \int \frac {d+e x}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{c d}\\ &=-\frac {2 (d+e x)^2}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {3 e \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^2 d^2}+\frac {\left (3 e \left (c d^2-a e^2\right )\right ) \int \frac {1}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{2 c^2 d^2}\\ &=-\frac {2 (d+e x)^2}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {3 e \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^2 d^2}+\frac {\left (3 e \left (c d^2-a e^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 c d e-x^2} \, dx,x,\frac {c d^2+a e^2+2 c d e x}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{c^2 d^2}\\ &=-\frac {2 (d+e x)^2}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {3 e \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^2 d^2}+\frac {3 \sqrt {e} \left (c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {c d^2+a e^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{2 c^{5/2} d^{5/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.05, size = 93, normalized size = 0.52 \begin {gather*} -\frac {2 (d+e x)^2 \, _2F_1\left (-\frac {3}{2},-\frac {1}{2};\frac {1}{2};\frac {e (a e+c d x)}{a e^2-c d^2}\right )}{c d \left (\frac {c d (d+e x)}{c d^2-a e^2}\right )^{3/2} \sqrt {(d+e x) (a e+c d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(d + e*x)^2*Hypergeometric2F1[-3/2, -1/2, 1/2, (e*(a*e + c*d*x))/(-(c*d^2) + a*e^2)])/(c*d*((c*d*(d + e*x)
)/(c*d^2 - a*e^2))^(3/2)*Sqrt[(a*e + c*d*x)*(d + e*x)])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 1.69, size = 302, normalized size = 1.68 \begin {gather*} -\frac {3 \sqrt {c d e} \left (c d^2-a e^2\right ) \log \left (a^2 e^4+8 c d e x \sqrt {c d e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}-2 a c d^2 e^2-4 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 )}{4 c^3 d^3}+\frac {3 \left (c d^2 \sqrt {e}-a e^{5/2}\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {e} \left (2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}-2 x \sqrt {c d e}\right )}{a e^2+c d^2}\right )}{2 c^{5/2} d^{5/2}}+\frac {\sqrt {a d e+a e^2 x+c d^2 x+c d e x^2} \left (3 a e^2-2 c d^2+c d e x\right )}{c^2 d^2 (a e+c d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.56, size = 447, normalized size = 2.48 \begin {gather*} \left [\frac {3 \, {\left (a c d^{2} e - a^{2} e^{3} + {\left (c^{2} d^{3} - a c d e^{2}\right )} x\right )} \sqrt {\frac {e}{c d}} \log \left (8 \, c^{2} d^{2} e^{2} x^{2} + c^{2} d^{4} + 6 \, a c d^{2} e^{2} + a^{2} e^{4} + 8 \, {\left (c^{2} d^{3} e + a c d e^{3}\right )} x + 4 \, {\left (2 \, c^{2} d^{2} e x + c^{2} d^{3} + a c d e^{2}\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {\frac {e}{c d}}\right ) + 4 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (c d e x - 2 \, c d^{2} + 3 \, a e^{2}\right )}}{4 \, {\left (c^{3} d^{3} x + a c^{2} d^{2} e\right )}}, -\frac {3 \, {\left (a c d^{2} e - a^{2} e^{3} + {\left (c^{2} d^{3} - a c d e^{2}\right )} x\right )} \sqrt {-\frac {e}{c d}} \arctan \left (\frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt {-\frac {e}{c d}}}{2 \, {\left (c d e^{2} x^{2} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x\right )}}\right ) - 2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (c d e x - 2 \, c d^{2} + 3 \, a e^{2}\right )}}{2 \, {\left (c^{3} d^{3} x + a c^{2} d^{2} e\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(3*(a*c*d^2*e - a^2*e^3 + (c^2*d^3 - a*c*d*e^2)*x)*sqrt(e/(c*d))*log(8*c^2*d^2*e^2*x^2 + c^2*d^4 + 6*a*c*
d^2*e^2 + a^2*e^4 + 8*(c^2*d^3*e + a*c*d*e^3)*x + 4*(2*c^2*d^2*e*x + c^2*d^3 + a*c*d*e^2)*sqrt(c*d*e*x^2 + a*d
*e + (c*d^2 + a*e^2)*x)*sqrt(e/(c*d))) + 4*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(c*d*e*x - 2*c*d^2 + 3*
a*e^2))/(c^3*d^3*x + a*c^2*d^2*e), -1/2*(3*(a*c*d^2*e - a^2*e^3 + (c^2*d^3 - a*c*d*e^2)*x)*sqrt(-e/(c*d))*arct
an(1/2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*c*d*e*x + c*d^2 + a*e^2)*sqrt(-e/(c*d))/(c*d*e^2*x^2 + a
*d*e^2 + (c*d^2*e + a*e^3)*x)) - 2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(c*d*e*x - 2*c*d^2 + 3*a*e^2))/
(c^3*d^3*x + a*c^2*d^2*e)]

________________________________________________________________________________________

giac [B]  time = 0.63, size = 354, normalized size = 1.97 \begin {gather*} \frac {{\left (\frac {{\left (c^{3} d^{5} e^{3} - 2 \, a c^{2} d^{3} e^{5} + a^{2} c d e^{7}\right )} x}{c^{4} d^{6} e - 2 \, a c^{3} d^{4} e^{3} + a^{2} c^{2} d^{2} e^{5}} - \frac {c^{3} d^{6} e^{2} - 5 \, a c^{2} d^{4} e^{4} + 7 \, a^{2} c d^{2} e^{6} - 3 \, a^{3} e^{8}}{c^{4} d^{6} e - 2 \, a c^{3} d^{4} e^{3} + a^{2} c^{2} d^{2} e^{5}}\right )} x - \frac {2 \, c^{3} d^{7} e - 7 \, a c^{2} d^{5} e^{3} + 8 \, a^{2} c d^{3} e^{5} - 3 \, a^{3} d e^{7}}{c^{4} d^{6} e - 2 \, a c^{3} d^{4} e^{3} + a^{2} c^{2} d^{2} e^{5}}}{\sqrt {c d x^{2} e + a d e + {\left (c d^{2} + a e^{2}\right )} x}} - \frac {3 \, {\left (c d^{2} e - a e^{3}\right )} \sqrt {c d} e^{\left (-\frac {1}{2}\right )} \log \left ({\left | -\sqrt {c d} c d^{2} e^{\frac {1}{2}} - 2 \, {\left (\sqrt {c d} x e^{\frac {1}{2}} - \sqrt {c d x^{2} e + a d e + {\left (c d^{2} + a e^{2}\right )} x}\right )} c d e - \sqrt {c d} a e^{\frac {5}{2}} \right |}\right )}{2 \, c^{3} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((c^3*d^5*e^3 - 2*a*c^2*d^3*e^5 + a^2*c*d*e^7)*x/(c^4*d^6*e - 2*a*c^3*d^4*e^3 + a^2*c^2*d^2*e^5) - (c^3*d^6*e
^2 - 5*a*c^2*d^4*e^4 + 7*a^2*c*d^2*e^6 - 3*a^3*e^8)/(c^4*d^6*e - 2*a*c^3*d^4*e^3 + a^2*c^2*d^2*e^5))*x - (2*c^
3*d^7*e - 7*a*c^2*d^5*e^3 + 8*a^2*c*d^3*e^5 - 3*a^3*d*e^7)/(c^4*d^6*e - 2*a*c^3*d^4*e^3 + a^2*c^2*d^2*e^5))/sq
rt(c*d*x^2*e + a*d*e + (c*d^2 + a*e^2)*x) - 3/2*(c*d^2*e - a*e^3)*sqrt(c*d)*e^(-1/2)*log(abs(-sqrt(c*d)*c*d^2*
e^(1/2) - 2*(sqrt(c*d)*x*e^(1/2) - sqrt(c*d*x^2*e + a*d*e + (c*d^2 + a*e^2)*x))*c*d*e - sqrt(c*d)*a*e^(5/2)))/
(c^3*d^3)

________________________________________________________________________________________

maple [B]  time = 0.06, size = 1047, normalized size = 5.82 \begin {gather*} -\frac {3 a^{3} e^{7} x}{2 \left (-a^{2} e^{4}+2 a c \,d^{2} e^{2}-c^{2} d^{4}\right ) \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, c^{2} d^{2}}+\frac {5 a^{2} e^{5} x}{2 \left (-a^{2} e^{4}+2 a c \,d^{2} e^{2}-c^{2} d^{4}\right ) \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, c}-\frac {a \,d^{2} e^{3} x}{2 \left (-a^{2} e^{4}+2 a c \,d^{2} e^{2}-c^{2} d^{4}\right ) \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}}-\frac {9 c \,d^{4} e x}{2 \left (-a^{2} e^{4}+2 a c \,d^{2} e^{2}-c^{2} d^{4}\right ) \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}}-\frac {3 a^{4} e^{8}}{4 \left (-a^{2} e^{4}+2 a c \,d^{2} e^{2}-c^{2} d^{4}\right ) \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, c^{3} d^{3}}+\frac {a^{3} e^{6}}{2 \left (-a^{2} e^{4}+2 a c \,d^{2} e^{2}-c^{2} d^{4}\right ) \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, c^{2} d}+\frac {a^{2} d \,e^{4}}{\left (-a^{2} e^{4}+2 a c \,d^{2} e^{2}-c^{2} d^{4}\right ) \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, c}-\frac {5 a \,d^{3} e^{2}}{2 \left (-a^{2} e^{4}+2 a c \,d^{2} e^{2}-c^{2} d^{4}\right ) \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}}-\frac {9 c \,d^{5}}{4 \left (-a^{2} e^{4}+2 a c \,d^{2} e^{2}-c^{2} d^{4}\right ) \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}}+\frac {2 \left (2 c d e x +a \,e^{2}+c \,d^{2}\right ) d^{3}}{\left (4 a c \,d^{2} e^{2}-\left (a \,e^{2}+c \,d^{2}\right )^{2}\right ) \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}}+\frac {e^{2} x^{2}}{\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, c d}+\frac {3 a \,e^{3} x}{2 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, c^{2} d^{2}}-\frac {3 a \,e^{3} \ln \left (\frac {c d e x +\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}}{\sqrt {c d e}}+\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\right )}{2 \sqrt {c d e}\, c^{2} d^{2}}-\frac {3 e x}{2 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, c}+\frac {3 e \ln \left (\frac {c d e x +\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}}{\sqrt {c d e}}+\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\right )}{2 \sqrt {c d e}\, c}-\frac {3 a^{2} e^{4}}{4 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, c^{3} d^{3}}+\frac {2 a \,e^{2}}{\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, c^{2} d}-\frac {9 d}{4 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^4/c*d/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*a^2-3/2*e^3/c^2/d^2*ln((c*d*e
*x+1/2*a*e^2+1/2*c*d^2)/(c*d*e)^(1/2)+(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2))/(c*d*e)^(1/2)*a-9/2*e*c*d^4/(-a
^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*x-3/4*e^8/c^3/d^3/(-a^2*e^4+2*a*c*d^2*e^
2-c^2*d^4)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*a^4+1/2*e^6/c^2/d/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(c*d*e*x
^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*a^3-1/2*e^3*d^2/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2
)*x)^(1/2)*x*a+3/2*e^3/c^2/d^2*x/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*a+5/2*e^5/c/(-a^2*e^4+2*a*c*d^2*e^2-c
^2*d^4)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*x*a^2-3/2*e^7/c^2/d^2/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(c*d*e*
x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*x*a^3+e^2*x^2/c/d/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)-3/4*e^4/c^3/d^3/(c*
d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*a^2+2*e^2/c^2/d/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*a-5/2*e^2*d^3/(-a
^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*a-3/2*e/c*x/(c*d*e*x^2+a*d*e+(a*e^2+c*d^
2)*x)^(1/2)-9/4*c*d^5/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)+3/2*e/c*ln((c*d
*e*x+1/2*a*e^2+1/2*c*d^2)/(c*d*e)^(1/2)+(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2))/(c*d*e)^(1/2)+2*d^3*(2*c*d*e*
x+a*e^2+c*d^2)/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)-9/4/c*d/(c*d*e*x^2+a*d*
e+(a*e^2+c*d^2)*x)^(1/2)

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),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 details)Is a*e^2-c*d^2 zero or nonzero?

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^3}{{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{3}}{\left (\left (d + e x\right ) \left (a e + c d x\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________