\(\int \frac {1}{(d+e x)^{3/2} (a+b x+c x^2)} \, dx\) [537]

Optimal result

Integrand size = 22, antiderivative size = 310 \[ \int \frac {1}{(d+e x)^{3/2} \left (a+b x+c x^2\right )} \, dx=-\frac {2 e}{\left (c d^2-b d e+a e^2\right ) \sqrt {d+e x}}-\frac {\sqrt {2} \sqrt {c} \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d+e x}}{\sqrt {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}}\right )}{\sqrt {b^2-4 a c} \sqrt {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e} \left (c d^2-b d e+a e^2\right )}+\frac {\sqrt {2} \sqrt {c} \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d+e x}}{\sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}}\right )}{\sqrt {b^2-4 a c} \sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e} \left (c d^2-b d e+a e^2\right )} \] Output:

-2*e/(a*e^2-b*d*e+c*d^2)/(e*x+d)^(1/2)-2^(1/2)*c^(1/2)*(2*c*d-(b+(-4*a*c+b 
^2)^(1/2))*e)*arctanh(2^(1/2)*c^(1/2)*(e*x+d)^(1/2)/(2*c*d-(b-(-4*a*c+b^2) 
^(1/2))*e)^(1/2))/(-4*a*c+b^2)^(1/2)/(2*c*d-(b-(-4*a*c+b^2)^(1/2))*e)^(1/2 
)/(a*e^2-b*d*e+c*d^2)+2^(1/2)*c^(1/2)*(2*c*d-(b-(-4*a*c+b^2)^(1/2))*e)*arc 
tanh(2^(1/2)*c^(1/2)*(e*x+d)^(1/2)/(2*c*d-(b+(-4*a*c+b^2)^(1/2))*e)^(1/2)) 
/(-4*a*c+b^2)^(1/2)/(2*c*d-(b+(-4*a*c+b^2)^(1/2))*e)^(1/2)/(a*e^2-b*d*e+c* 
d^2)
 

Mathematica [A] (verified)

Time = 0.94 (sec) , antiderivative size = 305, normalized size of antiderivative = 0.98 \[ \int \frac {1}{(d+e x)^{3/2} \left (a+b x+c x^2\right )} \, dx=-\frac {2 e}{\left (c d^2-b d e+a e^2\right ) \sqrt {d+e x}}+\frac {\sqrt {2} \sqrt {c} \left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} \sqrt {d+e x}}{\sqrt {-2 c d+b e-\sqrt {b^2-4 a c} e}}\right )}{\sqrt {b^2-4 a c} \sqrt {-2 c d+\left (b-\sqrt {b^2-4 a c}\right ) e} \left (-c d^2+e (b d-a e)\right )}+\frac {\sqrt {2} \sqrt {c} \left (2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} \sqrt {d+e x}}{\sqrt {-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e}}\right )}{\sqrt {b^2-4 a c} \sqrt {-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e} \left (-c d^2+e (b d-a e)\right )} \] Input:

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

Output:

(-2*e)/((c*d^2 - b*d*e + a*e^2)*Sqrt[d + e*x]) + (Sqrt[2]*Sqrt[c]*(-2*c*d 
+ (b + Sqrt[b^2 - 4*a*c])*e)*ArcTan[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[- 
2*c*d + b*e - Sqrt[b^2 - 4*a*c]*e]])/(Sqrt[b^2 - 4*a*c]*Sqrt[-2*c*d + (b - 
 Sqrt[b^2 - 4*a*c])*e]*(-(c*d^2) + e*(b*d - a*e))) + (Sqrt[2]*Sqrt[c]*(2*c 
*d + (-b + Sqrt[b^2 - 4*a*c])*e)*ArcTan[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sq 
rt[-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e]])/(Sqrt[b^2 - 4*a*c]*Sqrt[-2*c*d + 
(b + Sqrt[b^2 - 4*a*c])*e]*(-(c*d^2) + e*(b*d - a*e)))
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 290, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1147, 1197, 27, 1480, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 1147

\(\displaystyle \frac {\int \frac {c d-b e-c e x}{\sqrt {d+e x} \left (c x^2+b x+a\right )}dx}{a e^2-b d e+c d^2}-\frac {2 e}{\sqrt {d+e x} \left (a e^2-b d e+c d^2\right )}\)

\(\Big \downarrow \) 1197

\(\displaystyle \frac {2 \int \frac {e (2 c d-b e-c (d+e x))}{c d^2-b e d+a e^2+c (d+e x)^2-(2 c d-b e) (d+e x)}d\sqrt {d+e x}}{a e^2-b d e+c d^2}-\frac {2 e}{\sqrt {d+e x} \left (a e^2-b d e+c d^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 e \int \frac {2 c d-b e-c (d+e x)}{c d^2-b e d+a e^2+c (d+e x)^2-(2 c d-b e) (d+e x)}d\sqrt {d+e x}}{a e^2-b d e+c d^2}-\frac {2 e}{\sqrt {d+e x} \left (a e^2-b d e+c d^2\right )}\)

\(\Big \downarrow \) 1480

\(\displaystyle \frac {2 e \left (\frac {c \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right ) \int \frac {1}{\frac {1}{2} \left (\left (b-\sqrt {b^2-4 a c}\right ) e-2 c d\right )+c (d+e x)}d\sqrt {d+e x}}{2 e \sqrt {b^2-4 a c}}-\frac {1}{2} c \left (\frac {2 c d-b e}{e \sqrt {b^2-4 a c}}+1\right ) \int \frac {1}{\frac {1}{2} \left (\left (b+\sqrt {b^2-4 a c}\right ) e-2 c d\right )+c (d+e x)}d\sqrt {d+e x}\right )}{a e^2-b d e+c d^2}-\frac {2 e}{\sqrt {d+e x} \left (a e^2-b d e+c d^2\right )}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 e \left (\frac {\sqrt {c} \left (\frac {2 c d-b e}{e \sqrt {b^2-4 a c}}+1\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d+e x}}{\sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}\right )}{\sqrt {2} \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}-\frac {\sqrt {c} \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d+e x}}{\sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}\right )}{\sqrt {2} e \sqrt {b^2-4 a c} \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}\right )}{a e^2-b d e+c d^2}-\frac {2 e}{\sqrt {d+e x} \left (a e^2-b d e+c d^2\right )}\)

Input:

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

Output:

(-2*e)/((c*d^2 - b*d*e + a*e^2)*Sqrt[d + e*x]) + (2*e*(-((Sqrt[c]*(2*c*d - 
 (b + Sqrt[b^2 - 4*a*c])*e)*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2 
*c*d - (b - Sqrt[b^2 - 4*a*c])*e]])/(Sqrt[2]*Sqrt[b^2 - 4*a*c]*e*Sqrt[2*c* 
d - (b - Sqrt[b^2 - 4*a*c])*e])) + (Sqrt[c]*(1 + (2*c*d - b*e)/(Sqrt[b^2 - 
 4*a*c]*e))*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - (b + Sqrt 
[b^2 - 4*a*c])*e]])/(Sqrt[2]*Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e])))/(c 
*d^2 - b*d*e + a*e^2)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 221

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 1147

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

rule 1197

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)), x_Symbol] :> Simp[2   Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - 
b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; Fr 
eeQ[{a, b, c, d, e, f, g}, x]
 

rule 1480

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

Maple [A] (verified)

Time = 1.20 (sec) , antiderivative size = 297, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(2 e \left (-\frac {1}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {e x +d}}+\frac {4 c \left (\frac {\left (b e -2 c d -\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\right ) \sqrt {2}\, \arctan \left (\frac {\sqrt {e x +d}\, c \sqrt {2}}{\sqrt {\left (b e -2 c d +\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\right ) c}}\right )}{8 \sqrt {-e^{2} \left (4 a c -b^{2}\right )}\, \sqrt {\left (b e -2 c d +\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\right ) c}}-\frac {\left (-b e +2 c d -\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {e x +d}\, c \sqrt {2}}{\sqrt {\left (-b e +2 c d +\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\right ) c}}\right )}{8 \sqrt {-e^{2} \left (4 a c -b^{2}\right )}\, \sqrt {\left (-b e +2 c d +\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\right ) c}}\right )}{a \,e^{2}-b d e +c \,d^{2}}\right )\)	\(297\)
default	\(2 e \left (-\frac {1}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {e x +d}}+\frac {4 c \left (\frac {\left (b e -2 c d -\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\right ) \sqrt {2}\, \arctan \left (\frac {\sqrt {e x +d}\, c \sqrt {2}}{\sqrt {\left (b e -2 c d +\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\right ) c}}\right )}{8 \sqrt {-e^{2} \left (4 a c -b^{2}\right )}\, \sqrt {\left (b e -2 c d +\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\right ) c}}-\frac {\left (-b e +2 c d -\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {e x +d}\, c \sqrt {2}}{\sqrt {\left (-b e +2 c d +\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\right ) c}}\right )}{8 \sqrt {-e^{2} \left (4 a c -b^{2}\right )}\, \sqrt {\left (-b e +2 c d +\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\right ) c}}\right )}{a \,e^{2}-b d e +c \,d^{2}}\right )\)	\(297\)
pseudoelliptic	\(-\frac {2 e \left (-\frac {\sqrt {\left (b e -2 c d +\sqrt {-4 e^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) c}\, \sqrt {2}\, \left (b e -2 c d +\sqrt {-4 e^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) c \sqrt {e x +d}\, \operatorname {arctanh}\left (\frac {\sqrt {e x +d}\, c \sqrt {2}}{\sqrt {\left (-b e +2 c d +\sqrt {-4 e^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) c}}\right )}{2}+\left (\frac {c \sqrt {2}\, \sqrt {e x +d}\, \left (-b e +2 c d +\sqrt {-4 e^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) \arctan \left (\frac {\sqrt {e x +d}\, c \sqrt {2}}{\sqrt {\left (b e -2 c d +\sqrt {-4 e^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) c}}\right )}{2}+\sqrt {\left (b e -2 c d +\sqrt {-4 e^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) c}\, \sqrt {-4 e^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) \sqrt {\left (-b e +2 c d +\sqrt {-4 e^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) c}\right )}{\sqrt {\left (-b e +2 c d +\sqrt {-4 e^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) c}\, \sqrt {e x +d}\, \sqrt {\left (b e -2 c d +\sqrt {-4 e^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) c}\, \sqrt {-4 e^{2} \left (a c -\frac {b^{2}}{4}\right )}\, \left (a \,e^{2}-b d e +c \,d^{2}\right )}\)	\(365\)

Input:

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

Output:

2*e*(-1/(a*e^2-b*d*e+c*d^2)/(e*x+d)^(1/2)+4/(a*e^2-b*d*e+c*d^2)*c*(1/8*(b* 
e-2*c*d-(-e^2*(4*a*c-b^2))^(1/2))/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((b*e-2 
*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2^(1/2)/((b 
*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))-1/8*(-b*e+2*c*d-(-e^2*(4*a*c- 
b^2))^(1/2))/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^ 
2))^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4* 
a*c-b^2))^(1/2))*c)^(1/2))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 11293 vs. \(2 (264) = 528\).

Time = 0.48 (sec) , antiderivative size = 11293, normalized size of antiderivative = 36.43 \[ \int \frac {1}{(d+e x)^{3/2} \left (a+b x+c x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1492 vs. \(2 (264) = 528\).

Time = 0.46 (sec) , antiderivative size = 1492, normalized size of antiderivative = 4.81 \[ \int \frac {1}{(d+e x)^{3/2} \left (a+b x+c x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

-2*e/((c*d^2 - b*d*e + a*e^2)*sqrt(e*x + d)) - 1/4*((c*d^2*e - b*d*e^2 + a 
*e^3)^2*sqrt(-4*c^2*d + 2*(b*c + sqrt(b^2 - 4*a*c)*c)*e)*(b^2 - 4*a*c)*e - 
 2*(2*sqrt(b^2 - 4*a*c)*c^2*d^3*e - 3*sqrt(b^2 - 4*a*c)*b*c*d^2*e^2 - sqrt 
(b^2 - 4*a*c)*a*b*e^4 + (b^2 + 2*a*c)*sqrt(b^2 - 4*a*c)*d*e^3)*sqrt(-4*c^2 
*d + 2*(b*c + sqrt(b^2 - 4*a*c)*c)*e)*abs(-c*d^2*e + b*d*e^2 - a*e^3) + (4 
*c^4*d^6*e - 12*b*c^3*d^5*e^2 + a^2*b^2*e^7 + (13*b^2*c^2 + 8*a*c^3)*d^4*e 
^3 - 2*(3*b^3*c + 8*a*b*c^2)*d^3*e^4 + (b^4 + 10*a*b^2*c + 4*a^2*c^2)*d^2* 
e^5 - 2*(a*b^3 + 2*a^2*b*c)*d*e^6)*sqrt(-4*c^2*d + 2*(b*c + sqrt(b^2 - 4*a 
*c)*c)*e))*arctan(2*sqrt(1/2)*sqrt(e*x + d)/sqrt(-(2*c^2*d^3 - 3*b*c*d^2*e 
 + b^2*d*e^2 + 2*a*c*d*e^2 - a*b*e^3 + sqrt((2*c^2*d^3 - 3*b*c*d^2*e + b^2 
*d*e^2 + 2*a*c*d*e^2 - a*b*e^3)^2 - 4*(c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2 
 + 2*a*c*d^2*e^2 - 2*a*b*d*e^3 + a^2*e^4)*(c^2*d^2 - b*c*d*e + a*c*e^2)))/ 
(c^2*d^2 - b*c*d*e + a*c*e^2)))/((sqrt(b^2 - 4*a*c)*c^3*d^6 - 3*sqrt(b^2 - 
 4*a*c)*b*c^2*d^5*e - 3*sqrt(b^2 - 4*a*c)*a^2*b*d*e^5 + sqrt(b^2 - 4*a*c)* 
a^3*e^6 + 3*(b^2*c + a*c^2)*sqrt(b^2 - 4*a*c)*d^4*e^2 - (b^3 + 6*a*b*c)*sq 
rt(b^2 - 4*a*c)*d^3*e^3 + 3*(a*b^2 + a^2*c)*sqrt(b^2 - 4*a*c)*d^2*e^4)*abs 
(-c*d^2*e + b*d*e^2 - a*e^3)*abs(c)) + 1/4*((c*d^2*e - b*d*e^2 + a*e^3)^2* 
sqrt(-4*c^2*d + 2*(b*c - sqrt(b^2 - 4*a*c)*c)*e)*(b^2 - 4*a*c)*e + 2*(2*sq 
rt(b^2 - 4*a*c)*c^2*d^3*e - 3*sqrt(b^2 - 4*a*c)*b*c*d^2*e^2 - sqrt(b^2 - 4 
*a*c)*a*b*e^4 + (b^2 + 2*a*c)*sqrt(b^2 - 4*a*c)*d*e^3)*sqrt(-4*c^2*d + ...
 

Mupad [B] (verification not implemented)

Time = 8.76 (sec) , antiderivative size = 23975, normalized size of antiderivative = 77.34 \[ \int \frac {1}{(d+e x)^{3/2} \left (a+b x+c x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

atan((((d + e*x)^(1/2)*(16*c^8*d^8*e^2 - 16*a^4*c^4*e^10 + 32*a*c^7*d^6*e^ 
4 - 64*b*c^7*d^7*e^3 + 8*a^3*b^2*c^3*e^10 - 32*a^3*c^5*d^2*e^8 + 104*b^2*c 
^6*d^6*e^4 - 88*b^3*c^5*d^5*e^5 + 40*b^4*c^4*d^4*e^6 - 8*b^5*c^3*d^3*e^7 + 
 24*a^2*b^2*c^4*d^2*e^8 - 96*a*b*c^6*d^5*e^5 + 32*a^3*b*c^4*d*e^9 + 120*a* 
b^2*c^5*d^4*e^6 - 80*a*b^3*c^4*d^3*e^7 + 24*a*b^4*c^3*d^2*e^8 - 24*a^2*b^3 
*c^3*d*e^9) + (-(b^5*e^3 + 8*a*c^4*d^3 - 2*b^2*c^3*d^3 + b^2*e^3*(-(4*a*c 
- b^2)^3)^(1/2) + 12*a^2*b*c^2*e^3 - 24*a^2*c^3*d*e^2 + 3*b^3*c^2*d^2*e + 
3*c^2*d^2*e*(-(4*a*c - b^2)^3)^(1/2) - 7*a*b^3*c*e^3 - a*c*e^3*(-(4*a*c - 
b^2)^3)^(1/2) - 3*b^4*c*d*e^2 - 12*a*b*c^3*d^2*e - 3*b*c*d*e^2*(-(4*a*c - 
b^2)^3)^(1/2) + 18*a*b^2*c^2*d*e^2)/(2*(16*a^2*c^5*d^6 + a^3*b^4*e^6 + 16* 
a^5*c^2*e^6 + b^4*c^3*d^6 - b^7*d^3*e^3 - 8*a*b^2*c^4*d^6 - 8*a^4*b^2*c*e^ 
6 + 3*a*b^6*d^2*e^4 - 3*a^2*b^5*d*e^5 - 3*b^5*c^2*d^5*e + 3*b^6*c*d^4*e^2 
+ 48*a^3*c^4*d^4*e^2 + 48*a^4*c^3*d^2*e^4 + 24*a^2*b^2*c^3*d^4*e^2 + 32*a^ 
2*b^3*c^2*d^3*e^3 + 24*a^3*b^2*c^2*d^2*e^4 + 24*a*b^3*c^3*d^5*e + 2*a*b^5* 
c*d^3*e^3 - 48*a^2*b*c^4*d^5*e + 24*a^3*b^3*c*d*e^5 - 48*a^4*b*c^2*d*e^5 - 
 21*a*b^4*c^2*d^4*e^2 - 21*a^2*b^4*c*d^2*e^4 - 96*a^3*b*c^3*d^3*e^3)))^(1/ 
2)*((d + e*x)^(1/2)*(-(b^5*e^3 + 8*a*c^4*d^3 - 2*b^2*c^3*d^3 + b^2*e^3*(-( 
4*a*c - b^2)^3)^(1/2) + 12*a^2*b*c^2*e^3 - 24*a^2*c^3*d*e^2 + 3*b^3*c^2*d^ 
2*e + 3*c^2*d^2*e*(-(4*a*c - b^2)^3)^(1/2) - 7*a*b^3*c*e^3 - a*c*e^3*(-(4* 
a*c - b^2)^3)^(1/2) - 3*b^4*c*d*e^2 - 12*a*b*c^3*d^2*e - 3*b*c*d*e^2*(-...
 

Reduce [B] (verification not implemented)

Time = 2.25 (sec) , antiderivative size = 4661, normalized size of antiderivative = 15.04 \[ \int \frac {1}{(d+e x)^{3/2} \left (a+b x+c x^2\right )} \, dx =\text {Too large to display} \] Input:

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

Output:

(4*sqrt(d + e*x)*sqrt(2*sqrt(c)*sqrt(a*e**2 - b*d*e + c*d**2) + b*e - 2*c* 
d)*sqrt(a*e**2 - b*d*e + c*d**2)*atan((sqrt(2*sqrt(c)*sqrt(a*e**2 - b*d*e 
+ c*d**2) - b*e + 2*c*d) - 2*sqrt(c)*sqrt(d + e*x))/sqrt(2*sqrt(c)*sqrt(a* 
e**2 - b*d*e + c*d**2) + b*e - 2*c*d))*a*c*e**2 - 2*sqrt(d + e*x)*sqrt(2*s 
qrt(c)*sqrt(a*e**2 - b*d*e + c*d**2) + b*e - 2*c*d)*sqrt(a*e**2 - b*d*e + 
c*d**2)*atan((sqrt(2*sqrt(c)*sqrt(a*e**2 - b*d*e + c*d**2) - b*e + 2*c*d) 
- 2*sqrt(c)*sqrt(d + e*x))/sqrt(2*sqrt(c)*sqrt(a*e**2 - b*d*e + c*d**2) + 
b*e - 2*c*d))*b**2*e**2 + 4*sqrt(d + e*x)*sqrt(2*sqrt(c)*sqrt(a*e**2 - b*d 
*e + c*d**2) + b*e - 2*c*d)*sqrt(a*e**2 - b*d*e + c*d**2)*atan((sqrt(2*sqr 
t(c)*sqrt(a*e**2 - b*d*e + c*d**2) - b*e + 2*c*d) - 2*sqrt(c)*sqrt(d + e*x 
))/sqrt(2*sqrt(c)*sqrt(a*e**2 - b*d*e + c*d**2) + b*e - 2*c*d))*b*c*d*e - 
4*sqrt(d + e*x)*sqrt(2*sqrt(c)*sqrt(a*e**2 - b*d*e + c*d**2) + b*e - 2*c*d 
)*sqrt(a*e**2 - b*d*e + c*d**2)*atan((sqrt(2*sqrt(c)*sqrt(a*e**2 - b*d*e + 
 c*d**2) - b*e + 2*c*d) - 2*sqrt(c)*sqrt(d + e*x))/sqrt(2*sqrt(c)*sqrt(a*e 
**2 - b*d*e + c*d**2) + b*e - 2*c*d))*c**2*d**2 + 2*sqrt(c)*sqrt(d + e*x)* 
sqrt(2*sqrt(c)*sqrt(a*e**2 - b*d*e + c*d**2) + b*e - 2*c*d)*atan((sqrt(2*s 
qrt(c)*sqrt(a*e**2 - b*d*e + c*d**2) - b*e + 2*c*d) - 2*sqrt(c)*sqrt(d + e 
*x))/sqrt(2*sqrt(c)*sqrt(a*e**2 - b*d*e + c*d**2) + b*e - 2*c*d))*a*b*e**3 
 - 4*sqrt(c)*sqrt(d + e*x)*sqrt(2*sqrt(c)*sqrt(a*e**2 - b*d*e + c*d**2) + 
b*e - 2*c*d)*atan((sqrt(2*sqrt(c)*sqrt(a*e**2 - b*d*e + c*d**2) - b*e +...