\(\int \frac {1}{(b d+2 c d x)^3 (a+b x+c x^2)^{5/2}} \, dx\) [1258]

Optimal result

Integrand size = 26, antiderivative size = 176 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2}{3 \left (b^2-4 a c\right ) d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{3/2}}+\frac {40 c}{3 \left (b^2-4 a c\right )^2 d^3 (b+2 c x)^2 \sqrt {a+b x+c x^2}}+\frac {80 c^2 \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right )^3 d^3 (b+2 c x)^2}+\frac {40 c^{3/2} \arctan \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{7/2} d^3} \]

[Out]

-2/3/(-4*a*c+b^2)/d^3/(2*c*x+b)^2/(c*x^2+b*x+a)^(3/2)+40*c^(3/2)*arctan(2*c^(1/2)*(c*x^2+b*x+a)^(1/2)/(-4*a*c+
b^2)^(1/2))/(-4*a*c+b^2)^(7/2)/d^3+40/3*c/(-4*a*c+b^2)^2/d^3/(2*c*x+b)^2/(c*x^2+b*x+a)^(1/2)+80*c^2*(c*x^2+b*x
+a)^(1/2)/(-4*a*c+b^2)^3/d^3/(2*c*x+b)^2

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {701, 707, 702, 211} \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {40 c^{3/2} \arctan \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{d^3 \left (b^2-4 a c\right )^{7/2}}+\frac {80 c^2 \sqrt {a+b x+c x^2}}{d^3 \left (b^2-4 a c\right )^3 (b+2 c x)^2}+\frac {40 c}{3 d^3 \left (b^2-4 a c\right )^2 (b+2 c x)^2 \sqrt {a+b x+c x^2}}-\frac {2}{3 d^3 \left (b^2-4 a c\right ) (b+2 c x)^2 \left (a+b x+c x^2\right )^{3/2}} \]

[In]

Int[1/((b*d + 2*c*d*x)^3*(a + b*x + c*x^2)^(5/2)),x]

[Out]

-2/(3*(b^2 - 4*a*c)*d^3*(b + 2*c*x)^2*(a + b*x + c*x^2)^(3/2)) + (40*c)/(3*(b^2 - 4*a*c)^2*d^3*(b + 2*c*x)^2*S
qrt[a + b*x + c*x^2]) + (80*c^2*Sqrt[a + b*x + c*x^2])/((b^2 - 4*a*c)^3*d^3*(b + 2*c*x)^2) + (40*c^(3/2)*ArcTa
n[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]])/((b^2 - 4*a*c)^(7/2)*d^3)

Rule 211

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

Rule 701

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

Rule 702

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

Rule 707

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2}{3 \left (b^2-4 a c\right ) d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{3/2}}-\frac {(20 c) \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^{3/2}} \, dx}{3 \left (b^2-4 a c\right )} \\ & = -\frac {2}{3 \left (b^2-4 a c\right ) d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{3/2}}+\frac {40 c}{3 \left (b^2-4 a c\right )^2 d^3 (b+2 c x)^2 \sqrt {a+b x+c x^2}}+\frac {\left (80 c^2\right ) \int \frac {1}{(b d+2 c d x)^3 \sqrt {a+b x+c x^2}} \, dx}{\left (b^2-4 a c\right )^2} \\ & = -\frac {2}{3 \left (b^2-4 a c\right ) d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{3/2}}+\frac {40 c}{3 \left (b^2-4 a c\right )^2 d^3 (b+2 c x)^2 \sqrt {a+b x+c x^2}}+\frac {80 c^2 \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right )^3 d^3 (b+2 c x)^2}+\frac {\left (40 c^2\right ) \int \frac {1}{(b d+2 c d x) \sqrt {a+b x+c x^2}} \, dx}{\left (b^2-4 a c\right )^3 d^2} \\ & = -\frac {2}{3 \left (b^2-4 a c\right ) d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{3/2}}+\frac {40 c}{3 \left (b^2-4 a c\right )^2 d^3 (b+2 c x)^2 \sqrt {a+b x+c x^2}}+\frac {80 c^2 \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right )^3 d^3 (b+2 c x)^2}+\frac {\left (160 c^3\right ) \text {Subst}\left (\int \frac {1}{2 b^2 c d-8 a c^2 d+8 c^2 d x^2} \, dx,x,\sqrt {a+b x+c x^2}\right )}{\left (b^2-4 a c\right )^3 d^2} \\ & = -\frac {2}{3 \left (b^2-4 a c\right ) d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{3/2}}+\frac {40 c}{3 \left (b^2-4 a c\right )^2 d^3 (b+2 c x)^2 \sqrt {a+b x+c x^2}}+\frac {80 c^2 \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right )^3 d^3 (b+2 c x)^2}+\frac {40 c^{3/2} \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{7/2} d^3} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 10.04 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.35 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},2,-\frac {1}{2},\frac {4 c (a+x (b+c x))}{-b^2+4 a c}\right )}{3 \left (b^2-4 a c\right )^2 d^3 (a+x (b+c x))^{3/2}} \]

[In]

Integrate[1/((b*d + 2*c*d*x)^3*(a + b*x + c*x^2)^(5/2)),x]

[Out]

(-2*Hypergeometric2F1[-3/2, 2, -1/2, (4*c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])/(3*(b^2 - 4*a*c)^2*d^3*(a + x*(b
+ c*x))^(3/2))

Maple [A] (verified)

Time = 2.67 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.10

method	result	size
pseudoelliptic	\(\frac {\frac {5 \left (\frac {b^{4}}{60}-\frac {b^{3} c x}{3}-\frac {7 c \left (5 c \,x^{2}+a \right ) b^{2}}{15}+\left (-4 c^{3} x^{3}-\frac {8}{3} a \,c^{2} x \right ) b -2 c^{4} x^{4}-\frac {8 x^{2} c^{3} a}{3}-\frac {2 a^{2} c^{2}}{5}\right ) \sqrt {4 c^{2} a -b^{2} c}}{8}+\frac {5 c^{2} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} \left (2 c x +b \right )^{2} \operatorname {arctanh}\left (\frac {2 c \sqrt {c \,x^{2}+b x +a}}{\sqrt {4 c^{2} a -b^{2} c}}\right )}{8}}{\sqrt {4 c^{2} a -b^{2} c}\, \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} d^{3} \left (2 c x +b \right )^{2} \left (-\frac {b^{2}}{4}+a c \right )^{3}}\)	\(194\)
default	\(\frac {-\frac {2 c}{\left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )^{2} \left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}-\frac {10 c^{2} \left (\frac {4 c}{3 \left (4 a c -b^{2}\right ) \left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}+\frac {4 c \left (\frac {4 c}{\left (4 a c -b^{2}\right ) \sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}}-\frac {8 c \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{\left (4 a c -b^{2}\right ) \sqrt {\frac {4 a c -b^{2}}{c}}}\right )}{4 a c -b^{2}}\right )}{4 a c -b^{2}}}{8 d^{3} c^{3}}\)	\(296\)

[In]

int(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

5/8*((1/60*b^4-1/3*b^3*c*x-7/15*c*(5*c*x^2+a)*b^2+(-4*c^3*x^3-8/3*a*c^2*x)*b-2*c^4*x^4-8/3*x^2*c^3*a-2/5*a^2*c
^2)*(4*a*c^2-b^2*c)^(1/2)+c^2*(c*x^2+b*x+a)^(3/2)*(2*c*x+b)^2*arctanh(2*c*(c*x^2+b*x+a)^(1/2)/(4*a*c^2-b^2*c)^
(1/2)))/(4*a*c^2-b^2*c)^(1/2)/(c*x^2+b*x+a)^(3/2)/d^3/(2*c*x+b)^2/(-1/4*b^2+a*c)^3

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 628 vs. \(2 (156) = 312\).

Time = 2.14 (sec) , antiderivative size = 1287, normalized size of antiderivative = 7.31 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

[-2/3*(30*(4*c^5*x^6 + 12*b*c^4*x^5 + a^2*b^2*c + (13*b^2*c^3 + 8*a*c^4)*x^4 + 2*(3*b^3*c^2 + 8*a*b*c^3)*x^3 +
(b^4*c + 10*a*b^2*c^2 + 4*a^2*c^3)*x^2 + 2*(a*b^3*c + 2*a^2*b*c^2)*x)*sqrt(-c/(b^2 - 4*a*c))*log(-(4*c^2*x^2
+ 4*b*c*x - b^2 + 8*a*c - 4*sqrt(c*x^2 + b*x + a)*(b^2 - 4*a*c)*sqrt(-c/(b^2 - 4*a*c)))/(4*c^2*x^2 + 4*b*c*x +
b^2)) - (120*c^4*x^4 + 240*b*c^3*x^3 - b^4 + 28*a*b^2*c + 24*a^2*c^2 + 20*(7*b^2*c^2 + 8*a*c^3)*x^2 + 20*(b^3
*c + 8*a*b*c^2)*x)*sqrt(c*x^2 + b*x + a))/(4*(b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^3*x^6 +
12*(b^7*c^3 - 12*a*b^5*c^4 + 48*a^2*b^3*c^5 - 64*a^3*b*c^6)*d^3*x^5 + (13*b^8*c^2 - 148*a*b^6*c^3 + 528*a^2*b^
4*c^4 - 448*a^3*b^2*c^5 - 512*a^4*c^6)*d^3*x^4 + 2*(3*b^9*c - 28*a*b^7*c^2 + 48*a^2*b^5*c^3 + 192*a^3*b^3*c^4
- 512*a^4*b*c^5)*d^3*x^3 + (b^10 - 2*a*b^8*c - 68*a^2*b^6*c^2 + 368*a^3*b^4*c^3 - 448*a^4*b^2*c^4 - 256*a^5*c^
5)*d^3*x^2 + 2*(a*b^9 - 10*a^2*b^7*c + 24*a^3*b^5*c^2 + 32*a^4*b^3*c^3 - 128*a^5*b*c^4)*d^3*x + (a^2*b^8 - 12*
a^3*b^6*c + 48*a^4*b^4*c^2 - 64*a^5*b^2*c^3)*d^3), 2/3*(60*(4*c^5*x^6 + 12*b*c^4*x^5 + a^2*b^2*c + (13*b^2*c^3
+ 8*a*c^4)*x^4 + 2*(3*b^3*c^2 + 8*a*b*c^3)*x^3 + (b^4*c + 10*a*b^2*c^2 + 4*a^2*c^3)*x^2 + 2*(a*b^3*c + 2*a^2*
b*c^2)*x)*sqrt(c/(b^2 - 4*a*c))*arctan(-1/2*sqrt(c*x^2 + b*x + a)*(b^2 - 4*a*c)*sqrt(c/(b^2 - 4*a*c))/(c^2*x^2
+ b*c*x + a*c)) + (120*c^4*x^4 + 240*b*c^3*x^3 - b^4 + 28*a*b^2*c + 24*a^2*c^2 + 20*(7*b^2*c^2 + 8*a*c^3)*x^2
+ 20*(b^3*c + 8*a*b*c^2)*x)*sqrt(c*x^2 + b*x + a))/(4*(b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*
d^3*x^6 + 12*(b^7*c^3 - 12*a*b^5*c^4 + 48*a^2*b^3*c^5 - 64*a^3*b*c^6)*d^3*x^5 + (13*b^8*c^2 - 148*a*b^6*c^3 +
528*a^2*b^4*c^4 - 448*a^3*b^2*c^5 - 512*a^4*c^6)*d^3*x^4 + 2*(3*b^9*c - 28*a*b^7*c^2 + 48*a^2*b^5*c^3 + 192*a^
3*b^3*c^4 - 512*a^4*b*c^5)*d^3*x^3 + (b^10 - 2*a*b^8*c - 68*a^2*b^6*c^2 + 368*a^3*b^4*c^3 - 448*a^4*b^2*c^4 -
256*a^5*c^5)*d^3*x^2 + 2*(a*b^9 - 10*a^2*b^7*c + 24*a^3*b^5*c^2 + 32*a^4*b^3*c^3 - 128*a^5*b*c^4)*d^3*x + (a^2
*b^8 - 12*a^3*b^6*c + 48*a^4*b^4*c^2 - 64*a^5*b^2*c^3)*d^3)]

Sympy [F]

\[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {\int \frac {1}{a^{2} b^{3} \sqrt {a + b x + c x^{2}} + 6 a^{2} b^{2} c x \sqrt {a + b x + c x^{2}} + 12 a^{2} b c^{2} x^{2} \sqrt {a + b x + c x^{2}} + 8 a^{2} c^{3} x^{3} \sqrt {a + b x + c x^{2}} + 2 a b^{4} x \sqrt {a + b x + c x^{2}} + 14 a b^{3} c x^{2} \sqrt {a + b x + c x^{2}} + 36 a b^{2} c^{2} x^{3} \sqrt {a + b x + c x^{2}} + 40 a b c^{3} x^{4} \sqrt {a + b x + c x^{2}} + 16 a c^{4} x^{5} \sqrt {a + b x + c x^{2}} + b^{5} x^{2} \sqrt {a + b x + c x^{2}} + 8 b^{4} c x^{3} \sqrt {a + b x + c x^{2}} + 25 b^{3} c^{2} x^{4} \sqrt {a + b x + c x^{2}} + 38 b^{2} c^{3} x^{5} \sqrt {a + b x + c x^{2}} + 28 b c^{4} x^{6} \sqrt {a + b x + c x^{2}} + 8 c^{5} x^{7} \sqrt {a + b x + c x^{2}}}\, dx}{d^{3}} \]

[In]

integrate(1/(2*c*d*x+b*d)**3/(c*x**2+b*x+a)**(5/2),x)

[Out]

Integral(1/(a**2*b**3*sqrt(a + b*x + c*x**2) + 6*a**2*b**2*c*x*sqrt(a + b*x + c*x**2) + 12*a**2*b*c**2*x**2*sq
rt(a + b*x + c*x**2) + 8*a**2*c**3*x**3*sqrt(a + b*x + c*x**2) + 2*a*b**4*x*sqrt(a + b*x + c*x**2) + 14*a*b**3
*c*x**2*sqrt(a + b*x + c*x**2) + 36*a*b**2*c**2*x**3*sqrt(a + b*x + c*x**2) + 40*a*b*c**3*x**4*sqrt(a + b*x +
c*x**2) + 16*a*c**4*x**5*sqrt(a + b*x + c*x**2) + b**5*x**2*sqrt(a + b*x + c*x**2) + 8*b**4*c*x**3*sqrt(a + b*
x + c*x**2) + 25*b**3*c**2*x**4*sqrt(a + b*x + c*x**2) + 38*b**2*c**3*x**5*sqrt(a + b*x + c*x**2) + 28*b*c**4*
x**6*sqrt(a + b*x + c*x**2) + 8*c**5*x**7*sqrt(a + b*x + c*x**2)), x)/d**3

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(5/2),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(4*a*c-b^2>0)', see `assume?` f
or more deta

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1443 vs. \(2 (156) = 312\).

Time = 0.43 (sec) , antiderivative size = 1443, normalized size of antiderivative = 8.20 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

80*c^2*arctan(-(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*c + b*sqrt(c))/sqrt(b^2*c - 4*a*c^2))/((b^6*d^3 - 12*a*b
^4*c*d^3 + 48*a^2*b^2*c^2*d^3 - 64*a^3*c^3*d^3)*sqrt(b^2*c - 4*a*c^2)) + 2/3*(24*((b^22*c^2*d^9 - 44*a*b^20*c^
3*d^9 + 880*a^2*b^18*c^4*d^9 - 10560*a^3*b^16*c^5*d^9 + 84480*a^4*b^14*c^6*d^9 - 473088*a^5*b^12*c^7*d^9 + 189
2352*a^6*b^10*c^8*d^9 - 5406720*a^7*b^8*c^9*d^9 + 10813440*a^8*b^6*c^10*d^9 - 14417920*a^9*b^4*c^11*d^9 + 1153
4336*a^10*b^2*c^12*d^9 - 4194304*a^11*c^13*d^9)*x/(b^28*d^12 - 56*a*b^26*c*d^12 + 1456*a^2*b^24*c^2*d^12 - 232
96*a^3*b^22*c^3*d^12 + 256256*a^4*b^20*c^4*d^12 - 2050048*a^5*b^18*c^5*d^12 + 12300288*a^6*b^16*c^6*d^12 - 562
29888*a^7*b^14*c^7*d^12 + 196804608*a^8*b^12*c^8*d^12 - 524812288*a^9*b^10*c^9*d^12 + 1049624576*a^10*b^8*c^10
*d^12 - 1526726656*a^11*b^6*c^11*d^12 + 1526726656*a^12*b^4*c^12*d^12 - 939524096*a^13*b^2*c^13*d^12 + 2684354
56*a^14*c^14*d^12) + (b^23*c*d^9 - 44*a*b^21*c^2*d^9 + 880*a^2*b^19*c^3*d^9 - 10560*a^3*b^17*c^4*d^9 + 84480*a
^4*b^15*c^5*d^9 - 473088*a^5*b^13*c^6*d^9 + 1892352*a^6*b^11*c^7*d^9 - 5406720*a^7*b^9*c^8*d^9 + 10813440*a^8*
b^7*c^9*d^9 - 14417920*a^9*b^5*c^10*d^9 + 11534336*a^10*b^3*c^11*d^9 - 4194304*a^11*b*c^12*d^9)/(b^28*d^12 - 5
6*a*b^26*c*d^12 + 1456*a^2*b^24*c^2*d^12 - 23296*a^3*b^22*c^3*d^12 + 256256*a^4*b^20*c^4*d^12 - 2050048*a^5*b^
18*c^5*d^12 + 12300288*a^6*b^16*c^6*d^12 - 56229888*a^7*b^14*c^7*d^12 + 196804608*a^8*b^12*c^8*d^12 - 52481228
8*a^9*b^10*c^9*d^12 + 1049624576*a^10*b^8*c^10*d^12 - 1526726656*a^11*b^6*c^11*d^12 + 1526726656*a^12*b^4*c^12
*d^12 - 939524096*a^13*b^2*c^13*d^12 + 268435456*a^14*c^14*d^12))*x - (b^24*d^9 - 72*a*b^22*c*d^9 + 2112*a^2*b
^20*c^2*d^9 - 35200*a^3*b^18*c^3*d^9 + 380160*a^4*b^16*c^4*d^9 - 2838528*a^5*b^14*c^5*d^9 + 15138816*a^6*b^12*
c^6*d^9 - 58392576*a^7*b^10*c^7*d^9 + 162201600*a^8*b^8*c^8*d^9 - 317194240*a^9*b^6*c^9*d^9 + 415236096*a^10*b
^4*c^10*d^9 - 327155712*a^11*b^2*c^11*d^9 + 117440512*a^12*c^12*d^9)/(b^28*d^12 - 56*a*b^26*c*d^12 + 1456*a^2*
b^24*c^2*d^12 - 23296*a^3*b^22*c^3*d^12 + 256256*a^4*b^20*c^4*d^12 - 2050048*a^5*b^18*c^5*d^12 + 12300288*a^6*
b^16*c^6*d^12 - 56229888*a^7*b^14*c^7*d^12 + 196804608*a^8*b^12*c^8*d^12 - 524812288*a^9*b^10*c^9*d^12 + 10496
24576*a^10*b^8*c^10*d^12 - 1526726656*a^11*b^6*c^11*d^12 + 1526726656*a^12*b^4*c^12*d^12 - 939524096*a^13*b^2*
c^13*d^12 + 268435456*a^14*c^14*d^12))/(c*x^2 + b*x + a)^(3/2) - 16*(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*c
^3 + 3*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*b*c^(5/2) + (sqrt(c)*x - sqrt(c*x^2 + b*x + a))*b^2*c^2 + 2*(sqrt
(c)*x - sqrt(c*x^2 + b*x + a))*a*c^3 + a*b*c^(5/2))/((b^6*d^3 - 12*a*b^4*c*d^3 + 48*a^2*b^2*c^2*d^3 - 64*a^3*c
^3*d^3)*(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*c + 2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*b*sqrt(c) + b^2 - 2
*a*c)^2)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^{5/2}} \, dx=\int \frac {1}{{\left (b\,d+2\,c\,d\,x\right )}^3\,{\left (c\,x^2+b\,x+a\right )}^{5/2}} \,d x \]

[In]

int(1/((b*d + 2*c*d*x)^3*(a + b*x + c*x^2)^(5/2)),x)

[Out]

int(1/((b*d + 2*c*d*x)^3*(a + b*x + c*x^2)^(5/2)), x)