\(\int \frac {1}{(d+e x)^{3/2} (a d e+(c d^2+a e^2) x+c d e x^2)^2} \, dx\) [2017]

Optimal result

Integrand size = 37, antiderivative size = 192 \[ \int \frac {1}{(d+e x)^{3/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {7 e}{5 \left (c d^2-a e^2\right )^2 (d+e x)^{5/2}}-\frac {1}{\left (c d^2-a e^2\right ) (a e+c d x) (d+e x)^{5/2}}-\frac {7 c d e}{3 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}-\frac {7 c^2 d^2 e}{\left (c d^2-a e^2\right )^4 \sqrt {d+e x}}+\frac {7 c^{5/2} d^{5/2} e \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{9/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {640, 44, 53, 65, 214} \[ \int \frac {1}{(d+e x)^{3/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {7 c^{5/2} d^{5/2} e \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{9/2}}-\frac {7 c^2 d^2 e}{\sqrt {d+e x} \left (c d^2-a e^2\right )^4}-\frac {7 c d e}{3 (d+e x)^{3/2} \left (c d^2-a e^2\right )^3}-\frac {1}{(d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)}-\frac {7 e}{5 (d+e x)^{5/2} \left (c d^2-a e^2\right )^2} \]

[In]

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

[Out]

(-7*e)/(5*(c*d^2 - a*e^2)^2*(d + e*x)^(5/2)) - 1/((c*d^2 - a*e^2)*(a*e + c*d*x)*(d + e*x)^(5/2)) - (7*c*d*e)/(
3*(c*d^2 - a*e^2)^3*(d + e*x)^(3/2)) - (7*c^2*d^2*e)/((c*d^2 - a*e^2)^4*Sqrt[d + e*x]) + (7*c^(5/2)*d^(5/2)*e*
ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(c*d^2 - a*e^2)^(9/2)

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

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

Rule 640

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{(a e+c d x)^2 (d+e x)^{7/2}} \, dx \\ & = -\frac {1}{\left (c d^2-a e^2\right ) (a e+c d x) (d+e x)^{5/2}}-\frac {(7 e) \int \frac {1}{(a e+c d x) (d+e x)^{7/2}} \, dx}{2 \left (c d^2-a e^2\right )} \\ & = -\frac {7 e}{5 \left (c d^2-a e^2\right )^2 (d+e x)^{5/2}}-\frac {1}{\left (c d^2-a e^2\right ) (a e+c d x) (d+e x)^{5/2}}-\frac {(7 c d e) \int \frac {1}{(a e+c d x) (d+e x)^{5/2}} \, dx}{2 \left (c d^2-a e^2\right )^2} \\ & = -\frac {7 e}{5 \left (c d^2-a e^2\right )^2 (d+e x)^{5/2}}-\frac {1}{\left (c d^2-a e^2\right ) (a e+c d x) (d+e x)^{5/2}}-\frac {7 c d e}{3 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}-\frac {\left (7 c^2 d^2 e\right ) \int \frac {1}{(a e+c d x) (d+e x)^{3/2}} \, dx}{2 \left (c d^2-a e^2\right )^3} \\ & = -\frac {7 e}{5 \left (c d^2-a e^2\right )^2 (d+e x)^{5/2}}-\frac {1}{\left (c d^2-a e^2\right ) (a e+c d x) (d+e x)^{5/2}}-\frac {7 c d e}{3 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}-\frac {7 c^2 d^2 e}{\left (c d^2-a e^2\right )^4 \sqrt {d+e x}}-\frac {\left (7 c^3 d^3 e\right ) \int \frac {1}{(a e+c d x) \sqrt {d+e x}} \, dx}{2 \left (c d^2-a e^2\right )^4} \\ & = -\frac {7 e}{5 \left (c d^2-a e^2\right )^2 (d+e x)^{5/2}}-\frac {1}{\left (c d^2-a e^2\right ) (a e+c d x) (d+e x)^{5/2}}-\frac {7 c d e}{3 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}-\frac {7 c^2 d^2 e}{\left (c d^2-a e^2\right )^4 \sqrt {d+e x}}-\frac {\left (7 c^3 d^3\right ) \text {Subst}\left (\int \frac {1}{-\frac {c d^2}{e}+a e+\frac {c d x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{\left (c d^2-a e^2\right )^4} \\ & = -\frac {7 e}{5 \left (c d^2-a e^2\right )^2 (d+e x)^{5/2}}-\frac {1}{\left (c d^2-a e^2\right ) (a e+c d x) (d+e x)^{5/2}}-\frac {7 c d e}{3 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}-\frac {7 c^2 d^2 e}{\left (c d^2-a e^2\right )^4 \sqrt {d+e x}}+\frac {7 c^{5/2} d^{5/2} e \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.04 \[ \int \frac {1}{(d+e x)^{3/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {-6 a^3 e^6+2 a^2 c d e^4 (16 d+7 e x)-2 a c^2 d^2 e^2 \left (58 d^2+84 d e x+35 e^2 x^2\right )-c^3 d^3 \left (15 d^3+161 d^2 e x+245 d e^2 x^2+105 e^3 x^3\right )}{15 \left (c d^2-a e^2\right )^4 (a e+c d x) (d+e x)^{5/2}}-\frac {7 c^{5/2} d^{5/2} e \arctan \left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{\left (-c d^2+a e^2\right )^{9/2}} \]

[In]

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

[Out]

(-6*a^3*e^6 + 2*a^2*c*d*e^4*(16*d + 7*e*x) - 2*a*c^2*d^2*e^2*(58*d^2 + 84*d*e*x + 35*e^2*x^2) - c^3*d^3*(15*d^
3 + 161*d^2*e*x + 245*d*e^2*x^2 + 105*e^3*x^3))/(15*(c*d^2 - a*e^2)^4*(a*e + c*d*x)*(d + e*x)^(5/2)) - (7*c^(5
/2)*d^(5/2)*e*ArcTan[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[-(c*d^2) + a*e^2]])/(-(c*d^2) + a*e^2)^(9/2)

Maple [A] (verified)

Time = 2.82 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(2 e \left (-\frac {1}{5 \left (e^{2} a -c \,d^{2}\right )^{2} \left (e x +d \right )^{\frac {5}{2}}}-\frac {3 c^{2} d^{2}}{\left (e^{2} a -c \,d^{2}\right )^{4} \sqrt {e x +d}}+\frac {2 c d}{3 \left (e^{2} a -c \,d^{2}\right )^{3} \left (e x +d \right )^{\frac {3}{2}}}-\frac {c^{3} d^{3} \left (\frac {\sqrt {e x +d}}{2 c d \left (e x +d \right )+2 e^{2} a -2 c \,d^{2}}+\frac {7 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{2 \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{\left (e^{2} a -c \,d^{2}\right )^{4}}\right )\)	\(183\)
default	\(2 e \left (-\frac {1}{5 \left (e^{2} a -c \,d^{2}\right )^{2} \left (e x +d \right )^{\frac {5}{2}}}-\frac {3 c^{2} d^{2}}{\left (e^{2} a -c \,d^{2}\right )^{4} \sqrt {e x +d}}+\frac {2 c d}{3 \left (e^{2} a -c \,d^{2}\right )^{3} \left (e x +d \right )^{\frac {3}{2}}}-\frac {c^{3} d^{3} \left (\frac {\sqrt {e x +d}}{2 c d \left (e x +d \right )+2 e^{2} a -2 c \,d^{2}}+\frac {7 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{2 \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{\left (e^{2} a -c \,d^{2}\right )^{4}}\right )\)	\(183\)
pseudoelliptic	\(-\frac {2 \left (\frac {35 c^{3} d^{3} e \left (e x +d \right )^{\frac {5}{2}} \left (c d x +a e \right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{2}+\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}\, \left (\frac {5 \left (7 e^{3} x^{3}+\frac {49}{3} d \,e^{2} x^{2}+\frac {161}{15} d^{2} e x +d^{3}\right ) d^{3} c^{3}}{2}+\frac {58 \left (\frac {35}{58} x^{2} e^{2}+\frac {42}{29} d e x +d^{2}\right ) e^{2} d^{2} a \,c^{2}}{3}-\frac {16 \left (\frac {7 e x}{16}+d \right ) e^{4} d \,a^{2} c}{3}+e^{6} a^{3}\right )\right )}{5 \left (e x +d \right )^{\frac {5}{2}} \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}\, \left (e^{2} a -c \,d^{2}\right )^{4} \left (c d x +a e \right )}\)	\(211\)

[In]

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

[Out]

2*e*(-1/5/(a*e^2-c*d^2)^2/(e*x+d)^(5/2)-3/(a*e^2-c*d^2)^4*c^2*d^2/(e*x+d)^(1/2)+2/3/(a*e^2-c*d^2)^3*c*d/(e*x+d
)^(3/2)-1/(a*e^2-c*d^2)^4*c^3*d^3*(1/2*(e*x+d)^(1/2)/(c*d*(e*x+d)+e^2*a-c*d^2)+7/2/((a*e^2-c*d^2)*c*d)^(1/2)*a
rctan(c*d*(e*x+d)^(1/2)/((a*e^2-c*d^2)*c*d)^(1/2))))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 658 vs. \(2 (166) = 332\).

Time = 0.46 (sec) , antiderivative size = 1331, normalized size of antiderivative = 6.93 \[ \int \frac {1}{(d+e x)^{3/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[1/30*(105*(c^3*d^3*e^4*x^4 + a*c^2*d^5*e^2 + (3*c^3*d^4*e^3 + a*c^2*d^2*e^5)*x^3 + 3*(c^3*d^5*e^2 + a*c^2*d^3
*e^4)*x^2 + (c^3*d^6*e + 3*a*c^2*d^4*e^3)*x)*sqrt(c*d/(c*d^2 - a*e^2))*log((c*d*e*x + 2*c*d^2 - a*e^2 + 2*(c*d
^2 - a*e^2)*sqrt(e*x + d)*sqrt(c*d/(c*d^2 - a*e^2)))/(c*d*x + a*e)) - 2*(105*c^3*d^3*e^3*x^3 + 15*c^3*d^6 + 11
6*a*c^2*d^4*e^2 - 32*a^2*c*d^2*e^4 + 6*a^3*e^6 + 35*(7*c^3*d^4*e^2 + 2*a*c^2*d^2*e^4)*x^2 + 7*(23*c^3*d^5*e +
24*a*c^2*d^3*e^3 - 2*a^2*c*d*e^5)*x)*sqrt(e*x + d))/(a*c^4*d^11*e - 4*a^2*c^3*d^9*e^3 + 6*a^3*c^2*d^7*e^5 - 4*
a^4*c*d^5*e^7 + a^5*d^3*e^9 + (c^5*d^9*e^3 - 4*a*c^4*d^7*e^5 + 6*a^2*c^3*d^5*e^7 - 4*a^3*c^2*d^3*e^9 + a^4*c*d
*e^11)*x^4 + (3*c^5*d^10*e^2 - 11*a*c^4*d^8*e^4 + 14*a^2*c^3*d^6*e^6 - 6*a^3*c^2*d^4*e^8 - a^4*c*d^2*e^10 + a^
5*e^12)*x^3 + 3*(c^5*d^11*e - 3*a*c^4*d^9*e^3 + 2*a^2*c^3*d^7*e^5 + 2*a^3*c^2*d^5*e^7 - 3*a^4*c*d^3*e^9 + a^5*
d*e^11)*x^2 + (c^5*d^12 - a*c^4*d^10*e^2 - 6*a^2*c^3*d^8*e^4 + 14*a^3*c^2*d^6*e^6 - 11*a^4*c*d^4*e^8 + 3*a^5*d
^2*e^10)*x), 1/15*(105*(c^3*d^3*e^4*x^4 + a*c^2*d^5*e^2 + (3*c^3*d^4*e^3 + a*c^2*d^2*e^5)*x^3 + 3*(c^3*d^5*e^2
+ a*c^2*d^3*e^4)*x^2 + (c^3*d^6*e + 3*a*c^2*d^4*e^3)*x)*sqrt(-c*d/(c*d^2 - a*e^2))*arctan(-(c*d^2 - a*e^2)*sq
rt(e*x + d)*sqrt(-c*d/(c*d^2 - a*e^2))/(c*d*e*x + c*d^2)) - (105*c^3*d^3*e^3*x^3 + 15*c^3*d^6 + 116*a*c^2*d^4*
e^2 - 32*a^2*c*d^2*e^4 + 6*a^3*e^6 + 35*(7*c^3*d^4*e^2 + 2*a*c^2*d^2*e^4)*x^2 + 7*(23*c^3*d^5*e + 24*a*c^2*d^3
*e^3 - 2*a^2*c*d*e^5)*x)*sqrt(e*x + d))/(a*c^4*d^11*e - 4*a^2*c^3*d^9*e^3 + 6*a^3*c^2*d^7*e^5 - 4*a^4*c*d^5*e^
7 + a^5*d^3*e^9 + (c^5*d^9*e^3 - 4*a*c^4*d^7*e^5 + 6*a^2*c^3*d^5*e^7 - 4*a^3*c^2*d^3*e^9 + a^4*c*d*e^11)*x^4 +
(3*c^5*d^10*e^2 - 11*a*c^4*d^8*e^4 + 14*a^2*c^3*d^6*e^6 - 6*a^3*c^2*d^4*e^8 - a^4*c*d^2*e^10 + a^5*e^12)*x^3
+ 3*(c^5*d^11*e - 3*a*c^4*d^9*e^3 + 2*a^2*c^3*d^7*e^5 + 2*a^3*c^2*d^5*e^7 - 3*a^4*c*d^3*e^9 + a^5*d*e^11)*x^2
+ (c^5*d^12 - a*c^4*d^10*e^2 - 6*a^2*c^3*d^8*e^4 + 14*a^3*c^2*d^6*e^6 - 11*a^4*c*d^4*e^8 + 3*a^5*d^2*e^10)*x)]

Sympy [F]

\[ \int \frac {1}{(d+e x)^{3/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\int \frac {1}{\left (d + e x\right )^{\frac {7}{2}} \left (a e + c d x\right )^{2}}\, dx \]

[In]

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

[Out]

Integral(1/((d + e*x)**(7/2)*(a*e + c*d*x)**2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(d+e x)^{3/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(e*x+d)^(3/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 340 vs. \(2 (166) = 332\).

Time = 0.29 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.77 \[ \int \frac {1}{(d+e x)^{3/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {7 \, c^{3} d^{3} e \arctan \left (\frac {\sqrt {e x + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{{\left (c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}\right )} \sqrt {-c^{2} d^{3} + a c d e^{2}}} - \frac {\sqrt {e x + d} c^{3} d^{3} e}{{\left (c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}\right )} {\left ({\left (e x + d\right )} c d - c d^{2} + a e^{2}\right )}} - \frac {2 \, {\left (45 \, {\left (e x + d\right )}^{2} c^{2} d^{2} e + 10 \, {\left (e x + d\right )} c^{2} d^{3} e + 3 \, c^{2} d^{4} e - 10 \, {\left (e x + d\right )} a c d e^{3} - 6 \, a c d^{2} e^{3} + 3 \, a^{2} e^{5}\right )}}{15 \, {\left (c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}\right )} {\left (e x + d\right )}^{\frac {5}{2}}} \]

[In]

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

[Out]

-7*c^3*d^3*e*arctan(sqrt(e*x + d)*c*d/sqrt(-c^2*d^3 + a*c*d*e^2))/((c^4*d^8 - 4*a*c^3*d^6*e^2 + 6*a^2*c^2*d^4*
e^4 - 4*a^3*c*d^2*e^6 + a^4*e^8)*sqrt(-c^2*d^3 + a*c*d*e^2)) - sqrt(e*x + d)*c^3*d^3*e/((c^4*d^8 - 4*a*c^3*d^6
*e^2 + 6*a^2*c^2*d^4*e^4 - 4*a^3*c*d^2*e^6 + a^4*e^8)*((e*x + d)*c*d - c*d^2 + a*e^2)) - 2/15*(45*(e*x + d)^2*
c^2*d^2*e + 10*(e*x + d)*c^2*d^3*e + 3*c^2*d^4*e - 10*(e*x + d)*a*c*d*e^3 - 6*a*c*d^2*e^3 + 3*a^2*e^5)/((c^4*d
^8 - 4*a*c^3*d^6*e^2 + 6*a^2*c^2*d^4*e^4 - 4*a^3*c*d^2*e^6 + a^4*e^8)*(e*x + d)^(5/2))

Mupad [B] (verification not implemented)

Time = 10.13 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.27 \[ \int \frac {1}{(d+e x)^{3/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {\frac {2\,e}{5\,\left (a\,e^2-c\,d^2\right )}-\frac {14\,c\,d\,e\,\left (d+e\,x\right )}{15\,{\left (a\,e^2-c\,d^2\right )}^2}+\frac {14\,c^2\,d^2\,e\,{\left (d+e\,x\right )}^2}{3\,{\left (a\,e^2-c\,d^2\right )}^3}+\frac {7\,c^3\,d^3\,e\,{\left (d+e\,x\right )}^3}{{\left (a\,e^2-c\,d^2\right )}^4}}{\left (a\,e^2-c\,d^2\right )\,{\left (d+e\,x\right )}^{5/2}+c\,d\,{\left (d+e\,x\right )}^{7/2}}-\frac {7\,c^{5/2}\,d^{5/2}\,e\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,\sqrt {d+e\,x}\,\left (a^4\,e^8-4\,a^3\,c\,d^2\,e^6+6\,a^2\,c^2\,d^4\,e^4-4\,a\,c^3\,d^6\,e^2+c^4\,d^8\right )}{{\left (a\,e^2-c\,d^2\right )}^{9/2}}\right )}{{\left (a\,e^2-c\,d^2\right )}^{9/2}} \]

[In]

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

[Out]

- ((2*e)/(5*(a*e^2 - c*d^2)) - (14*c*d*e*(d + e*x))/(15*(a*e^2 - c*d^2)^2) + (14*c^2*d^2*e*(d + e*x)^2)/(3*(a*
e^2 - c*d^2)^3) + (7*c^3*d^3*e*(d + e*x)^3)/(a*e^2 - c*d^2)^4)/((a*e^2 - c*d^2)*(d + e*x)^(5/2) + c*d*(d + e*x
)^(7/2)) - (7*c^(5/2)*d^(5/2)*e*atan((c^(1/2)*d^(1/2)*(d + e*x)^(1/2)*(a^4*e^8 + c^4*d^8 - 4*a*c^3*d^6*e^2 - 4
*a^3*c*d^2*e^6 + 6*a^2*c^2*d^4*e^4))/(a*e^2 - c*d^2)^(9/2)))/(a*e^2 - c*d^2)^(9/2)