\(\int \frac {\sqrt {d+e x}}{(a d e+(c d^2+a e^2) x+c d e x^2)^3} \, dx\) [2025]

Optimal result

Integrand size = 37, antiderivative size = 208 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {35 e^2}{12 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}-\frac {1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{3/2}}+\frac {7 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{3/2}}+\frac {35 c d e^2}{4 \left (c d^2-a e^2\right )^4 \sqrt {d+e x}}-\frac {35 c^{3/2} d^{3/2} e^2 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 \left (c d^2-a e^2\right )^{9/2}} \]

[Out]

35/12*e^2/(-a*e^2+c*d^2)^3/(e*x+d)^(3/2)-1/2/(-a*e^2+c*d^2)/(c*d*x+a*e)^2/(e*x+d)^(3/2)+7/4*e/(-a*e^2+c*d^2)^2
/(c*d*x+a*e)/(e*x+d)^(3/2)-35/4*c^(3/2)*d^(3/2)*e^2*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)+35/4*c*d*e^2/(-a*e^2+c*d^2)^4/(e*x+d)^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 208, 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} \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=-\frac {35 c^{3/2} d^{3/2} e^2 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 \left (c d^2-a e^2\right )^{9/2}}+\frac {35 c d e^2}{4 \sqrt {d+e x} \left (c d^2-a e^2\right )^4}+\frac {35 e^2}{12 (d+e x)^{3/2} \left (c d^2-a e^2\right )^3}+\frac {7 e}{4 (d+e x)^{3/2} \left (c d^2-a e^2\right )^2 (a e+c d x)}-\frac {1}{2 (d+e x)^{3/2} \left (c d^2-a e^2\right ) (a e+c d x)^2} \]

[In]

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

[Out]

(35*e^2)/(12*(c*d^2 - a*e^2)^3*(d + e*x)^(3/2)) - 1/(2*(c*d^2 - a*e^2)*(a*e + c*d*x)^2*(d + e*x)^(3/2)) + (7*e
)/(4*(c*d^2 - a*e^2)^2*(a*e + c*d*x)*(d + e*x)^(3/2)) + (35*c*d*e^2)/(4*(c*d^2 - a*e^2)^4*Sqrt[d + e*x]) - (35
*c^(3/2)*d^(3/2)*e^2*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(4*(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)^3 (d+e x)^{5/2}} \, dx \\ & = -\frac {1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{3/2}}-\frac {(7 e) \int \frac {1}{(a e+c d x)^2 (d+e x)^{5/2}} \, dx}{4 \left (c d^2-a e^2\right )} \\ & = -\frac {1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{3/2}}+\frac {7 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{3/2}}+\frac {\left (35 e^2\right ) \int \frac {1}{(a e+c d x) (d+e x)^{5/2}} \, dx}{8 \left (c d^2-a e^2\right )^2} \\ & = \frac {35 e^2}{12 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}-\frac {1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{3/2}}+\frac {7 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{3/2}}+\frac {\left (35 c d e^2\right ) \int \frac {1}{(a e+c d x) (d+e x)^{3/2}} \, dx}{8 \left (c d^2-a e^2\right )^3} \\ & = \frac {35 e^2}{12 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}-\frac {1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{3/2}}+\frac {7 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{3/2}}+\frac {35 c d e^2}{4 \left (c d^2-a e^2\right )^4 \sqrt {d+e x}}+\frac {\left (35 c^2 d^2 e^2\right ) \int \frac {1}{(a e+c d x) \sqrt {d+e x}} \, dx}{8 \left (c d^2-a e^2\right )^4} \\ & = \frac {35 e^2}{12 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}-\frac {1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{3/2}}+\frac {7 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{3/2}}+\frac {35 c d e^2}{4 \left (c d^2-a e^2\right )^4 \sqrt {d+e x}}+\frac {\left (35 c^2 d^2 e\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 )}{4 \left (c d^2-a e^2\right )^4} \\ & = \frac {35 e^2}{12 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}-\frac {1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{3/2}}+\frac {7 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{3/2}}+\frac {35 c d e^2}{4 \left (c d^2-a e^2\right )^4 \sqrt {d+e x}}-\frac {35 c^{3/2} d^{3/2} e^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 \left (c d^2-a e^2\right )^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.97 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {-8 a^3 e^6+8 a^2 c d e^4 (10 d+7 e x)+a c^2 d^2 e^2 \left (39 d^2+238 d e x+175 e^2 x^2\right )+c^3 d^3 \left (-6 d^3+21 d^2 e x+140 d e^2 x^2+105 e^3 x^3\right )}{12 \left (c d^2-a e^2\right )^4 (a e+c d x)^2 (d+e x)^{3/2}}+\frac {35 c^{3/2} d^{3/2} e^2 \arctan \left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{4 \left (-c d^2+a e^2\right )^{9/2}} \]

[In]

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

[Out]

(-8*a^3*e^6 + 8*a^2*c*d*e^4*(10*d + 7*e*x) + a*c^2*d^2*e^2*(39*d^2 + 238*d*e*x + 175*e^2*x^2) + c^3*d^3*(-6*d^
3 + 21*d^2*e*x + 140*d*e^2*x^2 + 105*e^3*x^3))/(12*(c*d^2 - a*e^2)^4*(a*e + c*d*x)^2*(d + e*x)^(3/2)) + (35*c^
(3/2)*d^(3/2)*e^2*ArcTan[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[-(c*d^2) + a*e^2]])/(4*(-(c*d^2) + a*e^2)^(9/2))

Maple [A] (verified)

Time = 2.56 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.87

method	result	size
derivativedivides	\(2 e^{2} \left (-\frac {1}{3 \left (e^{2} a -c \,d^{2}\right )^{3} \left (e x +d \right )^{\frac {3}{2}}}+\frac {3 c d}{\left (e^{2} a -c \,d^{2}\right )^{4} \sqrt {e x +d}}+\frac {c^{2} d^{2} \left (\frac {\frac {11 c d \left (e x +d \right )^{\frac {3}{2}}}{8}+\left (\frac {13 e^{2} a}{8}-\frac {13 c \,d^{2}}{8}\right ) \sqrt {e x +d}}{\left (c d \left (e x +d \right )+e^{2} a -c \,d^{2}\right )^{2}}+\frac {35 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{8 \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{\left (e^{2} a -c \,d^{2}\right )^{4}}\right )\)	\(180\)
default	\(2 e^{2} \left (-\frac {1}{3 \left (e^{2} a -c \,d^{2}\right )^{3} \left (e x +d \right )^{\frac {3}{2}}}+\frac {3 c d}{\left (e^{2} a -c \,d^{2}\right )^{4} \sqrt {e x +d}}+\frac {c^{2} d^{2} \left (\frac {\frac {11 c d \left (e x +d \right )^{\frac {3}{2}}}{8}+\left (\frac {13 e^{2} a}{8}-\frac {13 c \,d^{2}}{8}\right ) \sqrt {e x +d}}{\left (c d \left (e x +d \right )+e^{2} a -c \,d^{2}\right )^{2}}+\frac {35 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{8 \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{\left (e^{2} a -c \,d^{2}\right )^{4}}\right )\)	\(180\)
pseudoelliptic	\(-\frac {2 \left (-\frac {105 c^{2} d^{2} e^{2} \left (e x +d \right )^{\frac {3}{2}} \left (c d x +a e \right )^{2} \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{8}+\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}\, \left (\frac {3 \left (-\frac {35}{2} e^{3} x^{3}-\frac {70}{3} d \,e^{2} x^{2}-\frac {7}{2} d^{2} e x +d^{3}\right ) d^{3} c^{3}}{4}-\frac {39 e^{2} \left (\frac {175}{39} x^{2} e^{2}+\frac {238}{39} d e x +d^{2}\right ) d^{2} a \,c^{2}}{8}-10 \left (\frac {7 e x}{10}+d \right ) e^{4} d \,a^{2} c +e^{6} a^{3}\right )\right )}{3 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}\, \left (e^{2} a -c \,d^{2}\right )^{4} \left (c d x +a e \right )^{2}}\)	\(215\)

[In]

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

[Out]

2*e^2*(-1/3/(a*e^2-c*d^2)^3/(e*x+d)^(3/2)+3/(a*e^2-c*d^2)^4*c*d/(e*x+d)^(1/2)+1/(a*e^2-c*d^2)^4*c^2*d^2*((11/8
*c*d*(e*x+d)^(3/2)+(13/8*e^2*a-13/8*c*d^2)*(e*x+d)^(1/2))/(c*d*(e*x+d)+e^2*a-c*d^2)^2+35/8/((a*e^2-c*d^2)*c*d)
^(1/2)*arctan(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. 669 vs. \(2 (176) = 352\).

Time = 0.40 (sec) , antiderivative size = 1353, normalized size of antiderivative = 6.50 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[1/24*(105*(c^3*d^3*e^4*x^4 + a^2*c*d^3*e^4 + 2*(c^3*d^4*e^3 + a*c^2*d^2*e^5)*x^3 + (c^3*d^5*e^2 + 4*a*c^2*d^3
*e^4 + a^2*c*d*e^6)*x^2 + 2*(a*c^2*d^4*e^3 + a^2*c*d^2*e^5)*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
- 6*c^3*d^6 + 39*a*c^2*d^4*e^2 + 80*a^2*c*d^2*e^4 - 8*a^3*e^6 + 35*(4*c^3*d^4*e^2 + 5*a*c^2*d^2*e^4)*x^2 + 7*
(3*c^3*d^5*e + 34*a*c^2*d^3*e^3 + 8*a^2*c*d*e^5)*x)*sqrt(e*x + d))/(a^2*c^4*d^10*e^2 - 4*a^3*c^3*d^8*e^4 + 6*a
^4*c^2*d^6*e^6 - 4*a^5*c*d^4*e^8 + a^6*d^2*e^10 + (c^6*d^10*e^2 - 4*a*c^5*d^8*e^4 + 6*a^2*c^4*d^6*e^6 - 4*a^3*
c^3*d^4*e^8 + a^4*c^2*d^2*e^10)*x^4 + 2*(c^6*d^11*e - 3*a*c^5*d^9*e^3 + 2*a^2*c^4*d^7*e^5 + 2*a^3*c^3*d^5*e^7
- 3*a^4*c^2*d^3*e^9 + a^5*c*d*e^11)*x^3 + (c^6*d^12 - 9*a^2*c^4*d^8*e^4 + 16*a^3*c^3*d^6*e^6 - 9*a^4*c^2*d^4*e
^8 + a^6*e^12)*x^2 + 2*(a*c^5*d^11*e - 3*a^2*c^4*d^9*e^3 + 2*a^3*c^3*d^7*e^5 + 2*a^4*c^2*d^5*e^7 - 3*a^5*c*d^3
*e^9 + a^6*d*e^11)*x), -1/12*(105*(c^3*d^3*e^4*x^4 + a^2*c*d^3*e^4 + 2*(c^3*d^4*e^3 + a*c^2*d^2*e^5)*x^3 + (c^
3*d^5*e^2 + 4*a*c^2*d^3*e^4 + a^2*c*d*e^6)*x^2 + 2*(a*c^2*d^4*e^3 + a^2*c*d^2*e^5)*x)*sqrt(-c*d/(c*d^2 - a*e^2
))*arctan(-(c*d^2 - a*e^2)*sqrt(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
- 6*c^3*d^6 + 39*a*c^2*d^4*e^2 + 80*a^2*c*d^2*e^4 - 8*a^3*e^6 + 35*(4*c^3*d^4*e^2 + 5*a*c^2*d^2*e^4)*x^2 + 7*(
3*c^3*d^5*e + 34*a*c^2*d^3*e^3 + 8*a^2*c*d*e^5)*x)*sqrt(e*x + d))/(a^2*c^4*d^10*e^2 - 4*a^3*c^3*d^8*e^4 + 6*a^
4*c^2*d^6*e^6 - 4*a^5*c*d^4*e^8 + a^6*d^2*e^10 + (c^6*d^10*e^2 - 4*a*c^5*d^8*e^4 + 6*a^2*c^4*d^6*e^6 - 4*a^3*c
^3*d^4*e^8 + a^4*c^2*d^2*e^10)*x^4 + 2*(c^6*d^11*e - 3*a*c^5*d^9*e^3 + 2*a^2*c^4*d^7*e^5 + 2*a^3*c^3*d^5*e^7 -
3*a^4*c^2*d^3*e^9 + a^5*c*d*e^11)*x^3 + (c^6*d^12 - 9*a^2*c^4*d^8*e^4 + 16*a^3*c^3*d^6*e^6 - 9*a^4*c^2*d^4*e^
8 + a^6*e^12)*x^2 + 2*(a*c^5*d^11*e - 3*a^2*c^4*d^9*e^3 + 2*a^3*c^3*d^7*e^5 + 2*a^4*c^2*d^5*e^7 - 3*a^5*c*d^3*
e^9 + a^6*d*e^11)*x)]

Sympy [F]

\[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\int \frac {1}{\left (d + e x\right )^{\frac {5}{2}} \left (a e + c d x\right )^{3}}\, dx \]

[In]

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

[Out]

Integral(1/((d + e*x)**(5/2)*(a*e + c*d*x)**3), x)

Maxima [F(-2)]

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

[In]

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

none

Time = 0.28 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.63 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {35 \, c^{2} d^{2} e^{2} \arctan \left (\frac {\sqrt {e x + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{4 \, {\left (c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}\right )} \sqrt {-c^{2} d^{3} + a c d e^{2}}} + \frac {2 \, {\left (9 \, {\left (e x + d\right )} c d e^{2} + c d^{2} e^{2} - a e^{4}\right )}}{3 \, {\left (c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}\right )} {\left (e x + d\right )}^{\frac {3}{2}}} + \frac {11 \, {\left (e x + d\right )}^{\frac {3}{2}} c^{3} d^{3} e^{2} - 13 \, \sqrt {e x + d} c^{3} d^{4} e^{2} + 13 \, \sqrt {e x + d} a c^{2} d^{2} e^{4}}{4 \, {\left (c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}\right )} {\left ({\left (e x + d\right )} c d - c d^{2} + a e^{2}\right )}^{2}} \]

[In]

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

[Out]

35/4*c^2*d^2*e^2*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)) + 2/3*(9*(e*x + d)*c*d*e^2 + c*d^2*e^2 - a*e^
4)/((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)^(3/2)) + 1/4*(11*(e*
x + d)^(3/2)*c^3*d^3*e^2 - 13*sqrt(e*x + d)*c^3*d^4*e^2 + 13*sqrt(e*x + d)*a*c^2*d^2*e^4)/((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)

Mupad [B] (verification not implemented)

Time = 10.12 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.42 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {\frac {14\,c\,d\,e^2\,\left (d+e\,x\right )}{3\,{\left (a\,e^2-c\,d^2\right )}^2}-\frac {2\,e^2}{3\,\left (a\,e^2-c\,d^2\right )}+\frac {175\,c^2\,d^2\,e^2\,{\left (d+e\,x\right )}^2}{12\,{\left (a\,e^2-c\,d^2\right )}^3}+\frac {35\,c^3\,d^3\,e^2\,{\left (d+e\,x\right )}^3}{4\,{\left (a\,e^2-c\,d^2\right )}^4}}{{\left (d+e\,x\right )}^{3/2}\,\left (a^2\,e^4-2\,a\,c\,d^2\,e^2+c^2\,d^4\right )-\left (2\,c^2\,d^3-2\,a\,c\,d\,e^2\right )\,{\left (d+e\,x\right )}^{5/2}+c^2\,d^2\,{\left (d+e\,x\right )}^{7/2}}+\frac {35\,c^{3/2}\,d^{3/2}\,e^2\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,\sqrt {d+e\,x}\,\left (a^4\,e^8-4\,a^3\,c\,d^2\,e^6+6\,a^2\,c^2\,d^4\,e^4-4\,a\,c^3\,d^6\,e^2+c^4\,d^8\right )}{{\left (a\,e^2-c\,d^2\right )}^{9/2}}\right )}{4\,{\left (a\,e^2-c\,d^2\right )}^{9/2}} \]

[In]

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

[Out]

((14*c*d*e^2*(d + e*x))/(3*(a*e^2 - c*d^2)^2) - (2*e^2)/(3*(a*e^2 - c*d^2)) + (175*c^2*d^2*e^2*(d + e*x)^2)/(1
2*(a*e^2 - c*d^2)^3) + (35*c^3*d^3*e^2*(d + e*x)^3)/(4*(a*e^2 - c*d^2)^4))/((d + e*x)^(3/2)*(a^2*e^4 + c^2*d^4
- 2*a*c*d^2*e^2) - (2*c^2*d^3 - 2*a*c*d*e^2)*(d + e*x)^(5/2) + c^2*d^2*(d + e*x)^(7/2)) + (35*c^(3/2)*d^(3/2)
*e^2*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)))/(4*(a*e^2 - c*d^2)^(9/2))