\(\int \frac {(a+b x+c x^2)^{5/2}}{(b d+2 c d x)^2} \, dx\) [183]

Optimal result

Integrand size = 26, antiderivative size = 153 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^2} \, dx=-\frac {15 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{256 c^3 d^2}+\frac {5 (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{32 c^2 d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{2 c d^2 (b+2 c x)}+\frac {15 \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{512 c^{7/2} d^2} \] Output:

-15/256*(-4*a*c+b^2)*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c^3/d^2+5/32*(2*c*x+b)* 
(c*x^2+b*x+a)^(3/2)/c^2/d^2-1/2*(c*x^2+b*x+a)^(5/2)/c/d^2/(2*c*x+b)+15/512 
*(-4*a*c+b^2)^2*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(7/2) 
/d^2
 

Mathematica [A] (verified)

Time = 2.37 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.93 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^2} \, dx=\frac {\frac {\sqrt {c} \sqrt {a+x (b+c x)} \left (-15 b^4-20 b^3 c x+4 b^2 c \left (25 a+3 c x^2\right )+16 b c^2 x \left (9 a+4 c x^2\right )+16 c^2 \left (-8 a^2+9 a c x^2+2 c^2 x^4\right )\right )}{b+2 c x}-\frac {8 \left (b^2-4 a c\right )^{5/2} \arctan \left (\frac {\sqrt {c} \sqrt {b^2-4 a c} x}{\sqrt {a} (b+2 c x)-b \sqrt {a+x (b+c x)}}\right )}{b}+15 \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )-\frac {8 \left (-b^2+4 a c\right )^{5/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {-b^2+4 a c} x}{\sqrt {a} (b+2 c x)-b \sqrt {a+x (b+c x)}}\right )}{b}}{256 c^{7/2} d^2} \] Input:

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

Output:

((Sqrt[c]*Sqrt[a + x*(b + c*x)]*(-15*b^4 - 20*b^3*c*x + 4*b^2*c*(25*a + 3* 
c*x^2) + 16*b*c^2*x*(9*a + 4*c*x^2) + 16*c^2*(-8*a^2 + 9*a*c*x^2 + 2*c^2*x 
^4)))/(b + 2*c*x) - (8*(b^2 - 4*a*c)^(5/2)*ArcTan[(Sqrt[c]*Sqrt[b^2 - 4*a* 
c]*x)/(Sqrt[a]*(b + 2*c*x) - b*Sqrt[a + x*(b + c*x)])])/b + 15*(b^2 - 4*a* 
c)^2*ArcTanh[(Sqrt[c]*x)/(-Sqrt[a] + Sqrt[a + x*(b + c*x)])] - (8*(-b^2 + 
4*a*c)^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[-b^2 + 4*a*c]*x)/(Sqrt[a]*(b + 2*c*x) - 
 b*Sqrt[a + x*(b + c*x)])])/b)/(256*c^(7/2)*d^2)
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1108, 1087, 1087, 1092, 219}

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.

\(\displaystyle \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^2} \, dx\)

\(\Big \downarrow \) 1108

\(\displaystyle \frac {5 \int \left (c x^2+b x+a\right )^{3/2}dx}{4 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{2 c d^2 (b+2 c x)}\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {5 \left (\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \int \sqrt {c x^2+b x+a}dx}{16 c}\right )}{4 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{2 c d^2 (b+2 c x)}\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {5 \left (\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{8 c}\right )}{16 c}\right )}{4 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{2 c d^2 (b+2 c x)}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {5 \left (\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{4 c}\right )}{16 c}\right )}{4 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{2 c d^2 (b+2 c x)}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {5 \left (\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{3/2}}\right )}{16 c}\right )}{4 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{2 c d^2 (b+2 c x)}\)

Input:

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

Output:

-1/2*(a + b*x + c*x^2)^(5/2)/(c*d^2*(b + 2*c*x)) + (5*(((b + 2*c*x)*(a + b 
*x + c*x^2)^(3/2))/(8*c) - (3*(b^2 - 4*a*c)*(((b + 2*c*x)*Sqrt[a + b*x + c 
*x^2])/(4*c) - ((b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x 
+ c*x^2])])/(8*c^(3/2))))/(16*c)))/(4*c*d^2)
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1087

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) 
*((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* 
p + 1)))   Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && 
GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
 

rule 1092

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[I 
nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a 
, b, c}, x]
 

rule 1108

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

Maple [A] (verified)

Time = 1.13 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.71

method	result	size
risch	\(\frac {\left (16 c^{3} x^{3}+24 b \,c^{2} x^{2}+72 a \,c^{2} x -6 b^{2} c x +36 a b c -7 b^{3}\right ) \sqrt {c \,x^{2}+b x +a}}{256 c^{3} d^{2}}+\frac {-\frac {\left (512 a^{3} c^{3}-384 a^{2} b^{2} c^{2}+96 a \,b^{4} c -8 b^{6}\right ) \sqrt {c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}}}{c \left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )}+\frac {15 b^{4} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}+240 a^{2} c^{\frac {3}{2}} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )-120 \sqrt {c}\, a \,b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{512 c^{3} d^{2}}\)	\(262\)
default	\(\frac {-\frac {4 c \left (c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}\right )^{\frac {7}{2}}}{\left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )}+\frac {24 c^{2} \left (\frac {\left (x +\frac {b}{2 c}\right ) \left (c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}\right )^{\frac {5}{2}}}{6}+\frac {5 \left (4 a c -b^{2}\right ) \left (\frac {\left (x +\frac {b}{2 c}\right ) \left (c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}{4}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (x +\frac {b}{2 c}\right ) \sqrt {c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}}}{2}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\sqrt {c}\, \left (x +\frac {b}{2 c}\right )+\sqrt {c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{24 c}\right )}{4 a c -b^{2}}}{4 d^{2} c^{2}}\)	\(294\)

Input:

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

Output:

1/256*(16*c^3*x^3+24*b*c^2*x^2+72*a*c^2*x-6*b^2*c*x+36*a*b*c-7*b^3)/c^3*(c 
*x^2+b*x+a)^(1/2)/d^2+1/512/c^3*(-(512*a^3*c^3-384*a^2*b^2*c^2+96*a*b^4*c- 
8*b^6)/c/(4*a*c-b^2)/(x+1/2*b/c)*(c*(x+1/2*b/c)^2+1/4*(4*a*c-b^2)/c)^(1/2) 
+15*b^4*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)+240*a^2*c^(3/2 
)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-120*c^(1/2)*a*b^2*ln((1/2*b+ 
c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2)))/d^2
 

Fricas [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 427, normalized size of antiderivative = 2.79 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^2} \, dx=\left [\frac {15 \, {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2} + 2 \, {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (32 \, c^{5} x^{4} + 64 \, b c^{4} x^{3} - 15 \, b^{4} c + 100 \, a b^{2} c^{2} - 128 \, a^{2} c^{3} + 12 \, {\left (b^{2} c^{3} + 12 \, a c^{4}\right )} x^{2} - 4 \, {\left (5 \, b^{3} c^{2} - 36 \, a b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{1024 \, {\left (2 \, c^{5} d^{2} x + b c^{4} d^{2}\right )}}, -\frac {15 \, {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2} + 2 \, {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - 2 \, {\left (32 \, c^{5} x^{4} + 64 \, b c^{4} x^{3} - 15 \, b^{4} c + 100 \, a b^{2} c^{2} - 128 \, a^{2} c^{3} + 12 \, {\left (b^{2} c^{3} + 12 \, a c^{4}\right )} x^{2} - 4 \, {\left (5 \, b^{3} c^{2} - 36 \, a b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{512 \, {\left (2 \, c^{5} d^{2} x + b c^{4} d^{2}\right )}}\right ] \] Input:

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

Output:

[1/1024*(15*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2 + 2*(b^4*c - 8*a*b^2*c^2 + 16* 
a^2*c^3)*x)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + 
a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*(32*c^5*x^4 + 64*b*c^4*x^3 - 15*b^4*c 
+ 100*a*b^2*c^2 - 128*a^2*c^3 + 12*(b^2*c^3 + 12*a*c^4)*x^2 - 4*(5*b^3*c^2 
 - 36*a*b*c^3)*x)*sqrt(c*x^2 + b*x + a))/(2*c^5*d^2*x + b*c^4*d^2), -1/512 
*(15*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2 + 2*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3 
)*x)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x 
^2 + b*c*x + a*c)) - 2*(32*c^5*x^4 + 64*b*c^4*x^3 - 15*b^4*c + 100*a*b^2*c 
^2 - 128*a^2*c^3 + 12*(b^2*c^3 + 12*a*c^4)*x^2 - 4*(5*b^3*c^2 - 36*a*b*c^3 
)*x)*sqrt(c*x^2 + b*x + a))/(2*c^5*d^2*x + b*c^4*d^2)]
 

Sympy [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^2} \, dx=\frac {\int \frac {a^{2} \sqrt {a + b x + c x^{2}}}{b^{2} + 4 b c x + 4 c^{2} x^{2}}\, dx + \int \frac {b^{2} x^{2} \sqrt {a + b x + c x^{2}}}{b^{2} + 4 b c x + 4 c^{2} x^{2}}\, dx + \int \frac {c^{2} x^{4} \sqrt {a + b x + c x^{2}}}{b^{2} + 4 b c x + 4 c^{2} x^{2}}\, dx + \int \frac {2 a b x \sqrt {a + b x + c x^{2}}}{b^{2} + 4 b c x + 4 c^{2} x^{2}}\, dx + \int \frac {2 a c x^{2} \sqrt {a + b x + c x^{2}}}{b^{2} + 4 b c x + 4 c^{2} x^{2}}\, dx + \int \frac {2 b c x^{3} \sqrt {a + b x + c x^{2}}}{b^{2} + 4 b c x + 4 c^{2} x^{2}}\, dx}{d^{2}} \] Input:

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

Output:

(Integral(a**2*sqrt(a + b*x + c*x**2)/(b**2 + 4*b*c*x + 4*c**2*x**2), x) + 
 Integral(b**2*x**2*sqrt(a + b*x + c*x**2)/(b**2 + 4*b*c*x + 4*c**2*x**2), 
 x) + Integral(c**2*x**4*sqrt(a + b*x + c*x**2)/(b**2 + 4*b*c*x + 4*c**2*x 
**2), x) + Integral(2*a*b*x*sqrt(a + b*x + c*x**2)/(b**2 + 4*b*c*x + 4*c** 
2*x**2), x) + Integral(2*a*c*x**2*sqrt(a + b*x + c*x**2)/(b**2 + 4*b*c*x + 
 4*c**2*x**2), x) + Integral(2*b*c*x**3*sqrt(a + b*x + c*x**2)/(b**2 + 4*b 
*c*x + 4*c**2*x**2), x))/d**2
 

Maxima [F(-2)]

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

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for 
 more deta
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 814 vs. \(2 (131) = 262\).

Time = 0.66 (sec) , antiderivative size = 814, normalized size of antiderivative = 5.32 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^2} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/512*d^2*(15*(b^4*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d) - 8*a*b^2*c*sgn(1 
/(2*c*d*x + b*d))*sgn(c)*sgn(d) + 16*a^2*c^2*sgn(1/(2*c*d*x + b*d))*sgn(c) 
*sgn(d))*arctan(sqrt(-b^2*c*d^2/(2*c*d*x + b*d)^2 + 4*a*c^2*d^2/(2*c*d*x + 
 b*d)^2 + c)/sqrt(-c))/(sqrt(-c)*c^3*d^4*abs(c)) + 8*(sqrt(-b^2*c*d^2/(2*c 
*d*x + b*d)^2 + 4*a*c^2*d^2/(2*c*d*x + b*d)^2 + c)*b^4*sgn(1/(2*c*d*x + b* 
d))*sgn(c)*sgn(d) - 8*sqrt(-b^2*c*d^2/(2*c*d*x + b*d)^2 + 4*a*c^2*d^2/(2*c 
*d*x + b*d)^2 + c)*a*b^2*c*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d) + 16*sqrt( 
-b^2*c*d^2/(2*c*d*x + b*d)^2 + 4*a*c^2*d^2/(2*c*d*x + b*d)^2 + c)*a^2*c^2* 
sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d))/(c^4*d^4*abs(c)) - (9*(-b^2*c*d^2/(2 
*c*d*x + b*d)^2 + 4*a*c^2*d^2/(2*c*d*x + b*d)^2 + c)^(3/2)*b^4*sgn(1/(2*c* 
d*x + b*d))*sgn(c)*sgn(d) - 72*(-b^2*c*d^2/(2*c*d*x + b*d)^2 + 4*a*c^2*d^2 
/(2*c*d*x + b*d)^2 + c)^(3/2)*a*b^2*c*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d) 
 - 7*sqrt(-b^2*c*d^2/(2*c*d*x + b*d)^2 + 4*a*c^2*d^2/(2*c*d*x + b*d)^2 + c 
)*b^4*c*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d) + 144*(-b^2*c*d^2/(2*c*d*x + 
b*d)^2 + 4*a*c^2*d^2/(2*c*d*x + b*d)^2 + c)^(3/2)*a^2*c^2*sgn(1/(2*c*d*x + 
 b*d))*sgn(c)*sgn(d) + 56*sqrt(-b^2*c*d^2/(2*c*d*x + b*d)^2 + 4*a*c^2*d^2/ 
(2*c*d*x + b*d)^2 + c)*a*b^2*c^2*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d) - 11 
2*sqrt(-b^2*c*d^2/(2*c*d*x + b*d)^2 + 4*a*c^2*d^2/(2*c*d*x + b*d)^2 + c)*a 
^2*c^3*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d))/((b^2*c*d^2/(2*c*d*x + b*d)^2 
 - 4*a*c^2*d^2/(2*c*d*x + b*d)^2)^2*c^3*d^4*abs(c)))*abs(c)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 527, normalized size of antiderivative = 3.44 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^2} \, dx=\frac {-256 \sqrt {c \,x^{2}+b x +a}\, a^{2} c^{3}+200 \sqrt {c \,x^{2}+b x +a}\, a \,b^{2} c^{2}+288 \sqrt {c \,x^{2}+b x +a}\, a b \,c^{3} x +288 \sqrt {c \,x^{2}+b x +a}\, a \,c^{4} x^{2}-30 \sqrt {c \,x^{2}+b x +a}\, b^{4} c -40 \sqrt {c \,x^{2}+b x +a}\, b^{3} c^{2} x +24 \sqrt {c \,x^{2}+b x +a}\, b^{2} c^{3} x^{2}+128 \sqrt {c \,x^{2}+b x +a}\, b \,c^{4} x^{3}+64 \sqrt {c \,x^{2}+b x +a}\, c^{5} x^{4}+240 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a^{2} b \,c^{2}+480 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a^{2} c^{3} x -120 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a \,b^{3} c -240 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a \,b^{2} c^{2} x +15 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) b^{5}+30 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) b^{4} c x -160 \sqrt {c}\, a^{2} b \,c^{2}-320 \sqrt {c}\, a^{2} c^{3} x +80 \sqrt {c}\, a \,b^{3} c +160 \sqrt {c}\, a \,b^{2} c^{2} x -10 \sqrt {c}\, b^{5}-20 \sqrt {c}\, b^{4} c x}{512 c^{4} d^{2} \left (2 c x +b \right )} \] Input:

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

Output:

( - 256*sqrt(a + b*x + c*x**2)*a**2*c**3 + 200*sqrt(a + b*x + c*x**2)*a*b* 
*2*c**2 + 288*sqrt(a + b*x + c*x**2)*a*b*c**3*x + 288*sqrt(a + b*x + c*x** 
2)*a*c**4*x**2 - 30*sqrt(a + b*x + c*x**2)*b**4*c - 40*sqrt(a + b*x + c*x* 
*2)*b**3*c**2*x + 24*sqrt(a + b*x + c*x**2)*b**2*c**3*x**2 + 128*sqrt(a + 
b*x + c*x**2)*b*c**4*x**3 + 64*sqrt(a + b*x + c*x**2)*c**5*x**4 + 240*sqrt 
(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2)) 
*a**2*b*c**2 + 480*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c 
*x)/sqrt(4*a*c - b**2))*a**2*c**3*x - 120*sqrt(c)*log((2*sqrt(c)*sqrt(a + 
b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*a*b**3*c - 240*sqrt(c)*log( 
(2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*a*b**2* 
c**2*x + 15*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqr 
t(4*a*c - b**2))*b**5 + 30*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + 
 b + 2*c*x)/sqrt(4*a*c - b**2))*b**4*c*x - 160*sqrt(c)*a**2*b*c**2 - 320*s 
qrt(c)*a**2*c**3*x + 80*sqrt(c)*a*b**3*c + 160*sqrt(c)*a*b**2*c**2*x - 10* 
sqrt(c)*b**5 - 20*sqrt(c)*b**4*c*x)/(512*c**4*d**2*(b + 2*c*x))