3.483 \(\int \frac{1}{x (d+e x) (a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx\)

Optimal. Leaf size=271 \[ \frac{2 \left (7 a^2 c d^2 e^4-3 a^3 e^6+a c^2 d^4 e^2+c d e x \left (3 c d^2-a e^2\right ) \left (3 a e^2+c d^2\right )+3 c^3 d^6\right )}{3 a d^2 e \left (c d^2-a e^2\right )^3 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac{\tanh ^{-1}\left (\frac{x \left (a e^2+c d^2\right )+2 a d e}{2 \sqrt{a} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{a^{3/2} d^{5/2} e^{3/2}}-\frac{2 e (a e+c d x)}{3 d \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}} \]

[Out]

(-2*e*(a*e + c*d*x))/(3*d*(c*d^2 - a*e^2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)) + (2*(3*c^3*d^6 + a*c
^2*d^4*e^2 + 7*a^2*c*d^2*e^4 - 3*a^3*e^6 + c*d*e*(3*c*d^2 - a*e^2)*(c*d^2 + 3*a*e^2)*x))/(3*a*d^2*e*(c*d^2 - a
*e^2)^3*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - ArcTanh[(2*a*d*e + (c*d^2 + a*e^2)*x)/(2*Sqrt[a]*Sqrt[d
]*Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])]/(a^(3/2)*d^(5/2)*e^(3/2))

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Rubi [A]  time = 0.337703, antiderivative size = 271, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {851, 822, 12, 724, 206} \[ \frac{2 \left (7 a^2 c d^2 e^4-3 a^3 e^6+a c^2 d^4 e^2+c d e x \left (3 c d^2-a e^2\right ) \left (3 a e^2+c d^2\right )+3 c^3 d^6\right )}{3 a d^2 e \left (c d^2-a e^2\right )^3 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac{\tanh ^{-1}\left (\frac{x \left (a e^2+c d^2\right )+2 a d e}{2 \sqrt{a} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{a^{3/2} d^{5/2} e^{3/2}}-\frac{2 e (a e+c d x)}{3 d \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*e*(a*e + c*d*x))/(3*d*(c*d^2 - a*e^2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)) + (2*(3*c^3*d^6 + a*c
^2*d^4*e^2 + 7*a^2*c*d^2*e^4 - 3*a^3*e^6 + c*d*e*(3*c*d^2 - a*e^2)*(c*d^2 + 3*a*e^2)*x))/(3*a*d^2*e*(c*d^2 - a
*e^2)^3*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - ArcTanh[(2*a*d*e + (c*d^2 + a*e^2)*x)/(2*Sqrt[a]*Sqrt[d
]*Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])]/(a^(3/2)*d^(5/2)*e^(3/2))

Rule 851

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :>
Int[((f + g*x)^n*(a + b*x + c*x^2)^(m + p))/(a/d + (c*x)/e)^m, x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] &&
NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[m, 0] && In
tegerQ[n] && (LtQ[n, 0] || GtQ[p, 0])

Rule 822

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 724

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

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])

Rubi steps

\begin{align*} \int \frac{1}{x (d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx &=\int \frac{a e+c d x}{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 e (a e+c d x)}{3 d \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{2 \int \frac{-\frac{3}{2} a e \left (c d^2-a e^2\right )^2+2 a c d e^2 \left (c d^2-a e^2\right ) x}{x \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx}{3 a d e \left (c d^2-a e^2\right )^2}\\ &=-\frac{2 e (a e+c d x)}{3 d \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{2 \left (3 c^3 d^6+a c^2 d^4 e^2+7 a^2 c d^2 e^4-3 a^3 e^6+c d e \left (3 c d^2-a e^2\right ) \left (c d^2+3 a e^2\right ) x\right )}{3 a d^2 e \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}}+\frac{4 \int \frac{3 a e \left (c d^2-a e^2\right )^4}{4 x \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{3 a^2 d^2 e^2 \left (c d^2-a e^2\right )^4}\\ &=-\frac{2 e (a e+c d x)}{3 d \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{2 \left (3 c^3 d^6+a c^2 d^4 e^2+7 a^2 c d^2 e^4-3 a^3 e^6+c d e \left (3 c d^2-a e^2\right ) \left (c d^2+3 a e^2\right ) x\right )}{3 a d^2 e \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}}+\frac{\int \frac{1}{x \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{a d^2 e}\\ &=-\frac{2 e (a e+c d x)}{3 d \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{2 \left (3 c^3 d^6+a c^2 d^4 e^2+7 a^2 c d^2 e^4-3 a^3 e^6+c d e \left (3 c d^2-a e^2\right ) \left (c d^2+3 a e^2\right ) x\right )}{3 a d^2 e \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}}-\frac{2 \operatorname{Subst}\left (\int \frac{1}{4 a d e-x^2} \, dx,x,\frac{2 a d e-\left (-c d^2-a e^2\right ) x}{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{a d^2 e}\\ &=-\frac{2 e (a e+c d x)}{3 d \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{2 \left (3 c^3 d^6+a c^2 d^4 e^2+7 a^2 c d^2 e^4-3 a^3 e^6+c d e \left (3 c d^2-a e^2\right ) \left (c d^2+3 a e^2\right ) x\right )}{3 a d^2 e \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}}-\frac{\tanh ^{-1}\left (\frac{2 a d e+\left (c d^2+a e^2\right ) x}{2 \sqrt{a} \sqrt{d} \sqrt{e} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{a^{3/2} d^{5/2} e^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.444473, size = 262, normalized size = 0.97 \[ \frac{2 \left (-\frac{(d+e x) (a e+c d x)^{3/2} \left (\sqrt{a} \sqrt{d} \sqrt{e} \left (3 a^2 e^5-8 a c d^2 e^3-3 c^2 d^4 e\right ) \sqrt{a e+c d x}+3 \sqrt{d+e x} \left (c d^2-a e^2\right )^3 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a e+c d x}}{\sqrt{a} \sqrt{e} \sqrt{d+e x}}\right )\right )}{3 \sqrt{a} d^{5/2} \sqrt{e} \left (c d^2-a e^2\right )^2}+\frac{\left (a e^3+3 c d^2 e\right ) (a e+c d x)^2}{3 c d^3-3 a d e^2}+c d (a e+c d x)\right )}{a e \left (c d^2-a e^2\right ) ((d+e x) (a e+c d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.059, size = 682, normalized size = 2.5 \begin{align*}{\frac{1}{a{d}^{2}e}{\frac{1}{\sqrt{ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+cde{x}^{2}}}}}-2\,{\frac{{e}^{2}xc}{d \left ( -{a}^{2}{e}^{4}+2\,ac{d}^{2}{e}^{2}-{c}^{2}{d}^{4} \right ) \sqrt{ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+cde{x}^{2}}}}-2\,{\frac{dx{c}^{2}}{a \left ( -{a}^{2}{e}^{4}+2\,ac{d}^{2}{e}^{2}-{c}^{2}{d}^{4} \right ) \sqrt{ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+cde{x}^{2}}}}-{\frac{a{e}^{3}}{{d}^{2} \left ( -{a}^{2}{e}^{4}+2\,ac{d}^{2}{e}^{2}-{c}^{2}{d}^{4} \right ) }{\frac{1}{\sqrt{ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+cde{x}^{2}}}}}-2\,{\frac{ce}{ \left ( -{a}^{2}{e}^{4}+2\,ac{d}^{2}{e}^{2}-{c}^{2}{d}^{4} \right ) \sqrt{ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+cde{x}^{2}}}}-{\frac{{c}^{2}{d}^{2}}{ae \left ( -{a}^{2}{e}^{4}+2\,ac{d}^{2}{e}^{2}-{c}^{2}{d}^{4} \right ) }{\frac{1}{\sqrt{ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+cde{x}^{2}}}}}-{\frac{1}{a{d}^{2}e}\ln \left ({\frac{1}{x} \left ( 2\,ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+2\,\sqrt{ade}\sqrt{ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+cde{x}^{2}} \right ) } \right ){\frac{1}{\sqrt{ade}}}}+{\frac{2}{3\,d \left ( a{e}^{2}-c{d}^{2} \right ) } \left ({\frac{d}{e}}+x \right ) ^{-1}{\frac{1}{\sqrt{cde \left ({\frac{d}{e}}+x \right ) ^{2}+ \left ( a{e}^{2}-c{d}^{2} \right ) \left ({\frac{d}{e}}+x \right ) }}}}-{\frac{16\,d{e}^{2}{c}^{2}x}{3\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{3}}{\frac{1}{\sqrt{cde \left ({\frac{d}{e}}+x \right ) ^{2}+ \left ( a{e}^{2}-c{d}^{2} \right ) \left ({\frac{d}{e}}+x \right ) }}}}-{\frac{8\,ac{e}^{3}}{3\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{3}}{\frac{1}{\sqrt{cde \left ({\frac{d}{e}}+x \right ) ^{2}+ \left ( a{e}^{2}-c{d}^{2} \right ) \left ({\frac{d}{e}}+x \right ) }}}}-{\frac{8\,{c}^{2}{d}^{2}e}{3\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{3}}{\frac{1}{\sqrt{cde \left ({\frac{d}{e}}+x \right ) ^{2}+ \left ( a{e}^{2}-c{d}^{2} \right ) \left ({\frac{d}{e}}+x \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(e*x+d)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x)

[Out]

1/d^2/a/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-2/d*e^2/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2
)*x+c*d*e*x^2)^(1/2)*x*c-2*d/a/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*x*c^2-
1/d^2*a*e^3/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-2*e/(-a^2*e^4+2*a*c*d^2*e
^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*c-d^2/a/e/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c
*d^2)*x+c*d*e*x^2)^(1/2)*c^2-1/d^2/a/e/(a*d*e)^(1/2)*ln((2*a*d*e+(a*e^2+c*d^2)*x+2*(a*d*e)^(1/2)*(a*d*e+(a*e^2
+c*d^2)*x+c*d*e*x^2)^(1/2))/x)+2/3/d/(a*e^2-c*d^2)/(d/e+x)/(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)-16/3*
d*e^2*c^2/(a*e^2-c*d^2)^3/(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)*x-8/3*e^3*c/(a*e^2-c*d^2)^3/(c*d*e*(d/
e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)*a-8/3*d^2*e*c^2/(a*e^2-c*d^2)^3/(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1
/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac{3}{2}}{\left (e x + d\right )} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(e*x+d)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(3/2)*(e*x + d)*x), x)

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Fricas [B]  time = 39.1122, size = 2923, normalized size = 10.79 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(e*x+d)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="fricas")

[Out]

[1/6*(3*(a*c^3*d^8*e - 3*a^2*c^2*d^6*e^3 + 3*a^3*c*d^4*e^5 - a^4*d^2*e^7 + (c^4*d^7*e^2 - 3*a*c^3*d^5*e^4 + 3*
a^2*c^2*d^3*e^6 - a^3*c*d*e^8)*x^3 + (2*c^4*d^8*e - 5*a*c^3*d^6*e^3 + 3*a^2*c^2*d^4*e^5 + a^3*c*d^2*e^7 - a^4*
e^9)*x^2 + (c^4*d^9 - a*c^3*d^7*e^2 - 3*a^2*c^2*d^5*e^4 + 5*a^3*c*d^3*e^6 - 2*a^4*d*e^8)*x)*sqrt(a*d*e)*log((8
*a^2*d^2*e^2 + (c^2*d^4 + 6*a*c*d^2*e^2 + a^2*e^4)*x^2 - 4*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*a*d*
e + (c*d^2 + a*e^2)*x)*sqrt(a*d*e) + 8*(a*c*d^3*e + a^2*d*e^3)*x)/x^2) + 4*(3*a*c^3*d^8*e + 9*a^3*c*d^4*e^5 -
4*a^4*d^2*e^7 + (3*a*c^3*d^6*e^3 + 8*a^2*c^2*d^4*e^5 - 3*a^3*c*d^2*e^7)*x^2 + (6*a*c^3*d^7*e^2 + 9*a^2*c^2*d^5
*e^4 + 4*a^3*c*d^3*e^6 - 3*a^4*d*e^8)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(a^3*c^3*d^11*e^3 - 3*a^
4*c^2*d^9*e^5 + 3*a^5*c*d^7*e^7 - a^6*d^5*e^9 + (a^2*c^4*d^10*e^4 - 3*a^3*c^3*d^8*e^6 + 3*a^4*c^2*d^6*e^8 - a^
5*c*d^4*e^10)*x^3 + (2*a^2*c^4*d^11*e^3 - 5*a^3*c^3*d^9*e^5 + 3*a^4*c^2*d^7*e^7 + a^5*c*d^5*e^9 - a^6*d^3*e^11
)*x^2 + (a^2*c^4*d^12*e^2 - a^3*c^3*d^10*e^4 - 3*a^4*c^2*d^8*e^6 + 5*a^5*c*d^6*e^8 - 2*a^6*d^4*e^10)*x), 1/3*(
3*(a*c^3*d^8*e - 3*a^2*c^2*d^6*e^3 + 3*a^3*c*d^4*e^5 - a^4*d^2*e^7 + (c^4*d^7*e^2 - 3*a*c^3*d^5*e^4 + 3*a^2*c^
2*d^3*e^6 - a^3*c*d*e^8)*x^3 + (2*c^4*d^8*e - 5*a*c^3*d^6*e^3 + 3*a^2*c^2*d^4*e^5 + a^3*c*d^2*e^7 - a^4*e^9)*x
^2 + (c^4*d^9 - a*c^3*d^7*e^2 - 3*a^2*c^2*d^5*e^4 + 5*a^3*c*d^3*e^6 - 2*a^4*d*e^8)*x)*sqrt(-a*d*e)*arctan(1/2*
sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*a*d*e + (c*d^2 + a*e^2)*x)*sqrt(-a*d*e)/(a*c*d^2*e^2*x^2 + a^2*
d^2*e^2 + (a*c*d^3*e + a^2*d*e^3)*x)) + 2*(3*a*c^3*d^8*e + 9*a^3*c*d^4*e^5 - 4*a^4*d^2*e^7 + (3*a*c^3*d^6*e^3
+ 8*a^2*c^2*d^4*e^5 - 3*a^3*c*d^2*e^7)*x^2 + (6*a*c^3*d^7*e^2 + 9*a^2*c^2*d^5*e^4 + 4*a^3*c*d^3*e^6 - 3*a^4*d*
e^8)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(a^3*c^3*d^11*e^3 - 3*a^4*c^2*d^9*e^5 + 3*a^5*c*d^7*e^7 -
a^6*d^5*e^9 + (a^2*c^4*d^10*e^4 - 3*a^3*c^3*d^8*e^6 + 3*a^4*c^2*d^6*e^8 - a^5*c*d^4*e^10)*x^3 + (2*a^2*c^4*d^
11*e^3 - 5*a^3*c^3*d^9*e^5 + 3*a^4*c^2*d^7*e^7 + a^5*c*d^5*e^9 - a^6*d^3*e^11)*x^2 + (a^2*c^4*d^12*e^2 - a^3*c
^3*d^10*e^4 - 3*a^4*c^2*d^8*e^6 + 5*a^5*c*d^6*e^8 - 2*a^6*d^4*e^10)*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \left (\left (d + e x\right ) \left (a e + c d x\right )\right )^{\frac{3}{2}} \left (d + e x\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(e*x+d)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(3/2),x)

[Out]

Integral(1/(x*((d + e*x)*(a*e + c*d*x))**(3/2)*(d + e*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \left [\mathit{undef}, \mathit{undef}, \mathit{undef}, 1\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(e*x+d)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="giac")

[Out]

[undef, undef, undef, 1]