\(\int \frac {(a+b x+c x^2)^{3/2}}{(b d+2 c d x)^4} \, dx\) [170]

Optimal result

Integrand size = 26, antiderivative size = 107 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^4} \, dx=-\frac {\sqrt {a+b x+c x^2}}{8 c^2 d^4 (b+2 c x)}-\frac {\left (a+b x+c x^2\right )^{3/2}}{6 c d^4 (b+2 c x)^3}+\frac {\text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{5/2} d^4} \] Output:

-1/8*(c*x^2+b*x+a)^(1/2)/c^2/d^4/(2*c*x+b)-1/6*(c*x^2+b*x+a)^(3/2)/c/d^4/( 
2*c*x+b)^3+1/16*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(5/2) 
/d^4
 

Mathematica [C] (verified)

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

Time = 10.07 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^4} \, dx=\frac {\left (b^2-4 a c\right ) \sqrt {a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{2},-\frac {1}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{48 c^2 d^4 (b+2 c x)^3 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \] Input:

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

Output:

((b^2 - 4*a*c)*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-3/2, -3/2, -1/2, ( 
b + 2*c*x)^2/(b^2 - 4*a*c)])/(48*c^2*d^4*(b + 2*c*x)^3*Sqrt[(c*(a + x*(b + 
 c*x)))/(-b^2 + 4*a*c)])
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 112, 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, 27, 1108, 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.

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

\(\Big \downarrow \) 1108

\(\displaystyle \frac {\int \frac {\sqrt {c x^2+b x+a}}{d^2 (b+2 c x)^2}dx}{4 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{6 c d^4 (b+2 c x)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sqrt {c x^2+b x+a}}{(b+2 c x)^2}dx}{4 c d^4}-\frac {\left (a+b x+c x^2\right )^{3/2}}{6 c d^4 (b+2 c x)^3}\)

\(\Big \downarrow \) 1108

\(\displaystyle \frac {\frac {\int \frac {1}{\sqrt {c x^2+b x+a}}dx}{4 c}-\frac {\sqrt {a+b x+c x^2}}{2 c (b+2 c x)}}{4 c d^4}-\frac {\left (a+b x+c x^2\right )^{3/2}}{6 c d^4 (b+2 c x)^3}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {\frac {\int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{2 c}-\frac {\sqrt {a+b x+c x^2}}{2 c (b+2 c x)}}{4 c d^4}-\frac {\left (a+b x+c x^2\right )^{3/2}}{6 c d^4 (b+2 c x)^3}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{4 c^{3/2}}-\frac {\sqrt {a+b x+c x^2}}{2 c (b+2 c x)}}{4 c d^4}-\frac {\left (a+b x+c x^2\right )^{3/2}}{6 c d^4 (b+2 c x)^3}\)

Input:

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

Output:

-1/6*(a + b*x + c*x^2)^(3/2)/(c*d^4*(b + 2*c*x)^3) + (-1/2*Sqrt[a + b*x + 
c*x^2]/(c*(b + 2*c*x)) + ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x 
^2])]/(4*c^(3/2)))/(4*c*d^4)
 

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 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 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(310\) vs. \(2(89)=178\).

Time = 1.18 (sec) , antiderivative size = 311, normalized size of antiderivative = 2.91

method	result	size
default	\(\frac {-\frac {4 c \left (c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}\right )^{\frac {5}{2}}}{3 \left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )^{3}}+\frac {8 c^{2} \left (-\frac {4 c \left (c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}\right )^{\frac {5}{2}}}{\left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )}+\frac {16 c^{2} \left (\frac {\left (x +\frac {b}{2 c}\right ) \left (c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}{4}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (x +\frac {b}{2 c}\right ) \sqrt {c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}}}{2}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\sqrt {c}\, \left (x +\frac {b}{2 c}\right )+\sqrt {c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{4 a c -b^{2}}\right )}{3 \left (4 a c -b^{2}\right )}}{16 d^{4} c^{4}}\)	\(311\)

Input:

int((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^4,x,method=_RETURNVERBOSE)
 

Output:

1/16/d^4/c^4*(-4/3/(4*a*c-b^2)*c/(x+1/2*b/c)^3*(c*(x+1/2*b/c)^2+1/4*(4*a*c 
-b^2)/c)^(5/2)+8/3*c^2/(4*a*c-b^2)*(-4/(4*a*c-b^2)*c/(x+1/2*b/c)*(c*(x+1/2 
*b/c)^2+1/4*(4*a*c-b^2)/c)^(5/2)+16*c^2/(4*a*c-b^2)*(1/4*(x+1/2*b/c)*(c*(x 
+1/2*b/c)^2+1/4*(4*a*c-b^2)/c)^(3/2)+3/16*(4*a*c-b^2)/c*(1/2*(x+1/2*b/c)*( 
c*(x+1/2*b/c)^2+1/4*(4*a*c-b^2)/c)^(1/2)+1/8*(4*a*c-b^2)/c^(3/2)*ln(c^(1/2 
)*(x+1/2*b/c)+(c*(x+1/2*b/c)^2+1/4*(4*a*c-b^2)/c)^(1/2))))))
 

Fricas [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 345, normalized size of antiderivative = 3.22 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^4} \, dx=\left [\frac {3 \, {\left (8 \, c^{3} x^{3} + 12 \, b c^{2} x^{2} + 6 \, b^{2} c x + b^{3}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (16 \, c^{3} x^{2} + 16 \, b c^{2} x + 3 \, b^{2} c + 4 \, a c^{2}\right )} \sqrt {c x^{2} + b x + a}}{96 \, {\left (8 \, c^{6} d^{4} x^{3} + 12 \, b c^{5} d^{4} x^{2} + 6 \, b^{2} c^{4} d^{4} x + b^{3} c^{3} d^{4}\right )}}, -\frac {3 \, {\left (8 \, c^{3} x^{3} + 12 \, b c^{2} x^{2} + 6 \, b^{2} c x + b^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (16 \, c^{3} x^{2} + 16 \, b c^{2} x + 3 \, b^{2} c + 4 \, a c^{2}\right )} \sqrt {c x^{2} + b x + a}}{48 \, {\left (8 \, c^{6} d^{4} x^{3} + 12 \, b c^{5} d^{4} x^{2} + 6 \, b^{2} c^{4} d^{4} x + b^{3} c^{3} d^{4}\right )}}\right ] \] Input:

integrate((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^4,x, algorithm="fricas")
 

Output:

[1/96*(3*(8*c^3*x^3 + 12*b*c^2*x^2 + 6*b^2*c*x + b^3)*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*(16*c^3*x^2 + 16*b*c^2*x + 3*b^2*c + 4*a*c^2)*sqrt(c*x^2 + b*x + a))/( 
8*c^6*d^4*x^3 + 12*b*c^5*d^4*x^2 + 6*b^2*c^4*d^4*x + b^3*c^3*d^4), -1/48*( 
3*(8*c^3*x^3 + 12*b*c^2*x^2 + 6*b^2*c*x + b^3)*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*(16*c^3*x 
^2 + 16*b*c^2*x + 3*b^2*c + 4*a*c^2)*sqrt(c*x^2 + b*x + a))/(8*c^6*d^4*x^3 
 + 12*b*c^5*d^4*x^2 + 6*b^2*c^4*d^4*x + b^3*c^3*d^4)]
 

Sympy [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^4} \, dx=\frac {\int \frac {a \sqrt {a + b x + c x^{2}}}{b^{4} + 8 b^{3} c x + 24 b^{2} c^{2} x^{2} + 32 b c^{3} x^{3} + 16 c^{4} x^{4}}\, dx + \int \frac {b x \sqrt {a + b x + c x^{2}}}{b^{4} + 8 b^{3} c x + 24 b^{2} c^{2} x^{2} + 32 b c^{3} x^{3} + 16 c^{4} x^{4}}\, dx + \int \frac {c x^{2} \sqrt {a + b x + c x^{2}}}{b^{4} + 8 b^{3} c x + 24 b^{2} c^{2} x^{2} + 32 b c^{3} x^{3} + 16 c^{4} x^{4}}\, dx}{d^{4}} \] Input:

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

Output:

(Integral(a*sqrt(a + b*x + c*x**2)/(b**4 + 8*b**3*c*x + 24*b**2*c**2*x**2 
+ 32*b*c**3*x**3 + 16*c**4*x**4), x) + Integral(b*x*sqrt(a + b*x + c*x**2) 
/(b**4 + 8*b**3*c*x + 24*b**2*c**2*x**2 + 32*b*c**3*x**3 + 16*c**4*x**4), 
x) + Integral(c*x**2*sqrt(a + b*x + c*x**2)/(b**4 + 8*b**3*c*x + 24*b**2*c 
**2*x**2 + 32*b*c**3*x**3 + 16*c**4*x**4), x))/d**4
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^4} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^4,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. 431 vs. \(2 (89) = 178\).

Time = 0.44 (sec) , antiderivative size = 431, normalized size of antiderivative = 4.03 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^4} \, dx=-\frac {\log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{16 \, c^{\frac {5}{2}} d^{4}} - \frac {12 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{4} b^{2} c^{2} - 48 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{4} a c^{3} + 24 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} b^{3} c^{\frac {3}{2}} - 96 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} a b c^{\frac {5}{2}} + 21 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} b^{4} c - 96 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} a b^{2} c^{2} + 48 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} a^{2} c^{3} + 9 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} b^{5} \sqrt {c} - 48 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} a b^{3} c^{\frac {3}{2}} + 48 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} a^{2} b c^{\frac {5}{2}} + 2 \, b^{6} - 15 \, a b^{4} c + 36 \, a^{2} b^{2} c^{2} - 32 \, a^{3} c^{3}}{24 \, {\left (2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} c + 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} b \sqrt {c} + b^{2} - 2 \, a c\right )}^{3} c^{\frac {5}{2}} d^{4}} \] Input:

integrate((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^4,x, algorithm="giac")
 

Output:

-1/16*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/(c^(5/2) 
*d^4) - 1/24*(12*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*b^2*c^2 - 48*(sqrt( 
c)*x - sqrt(c*x^2 + b*x + a))^4*a*c^3 + 24*(sqrt(c)*x - sqrt(c*x^2 + b*x + 
 a))^3*b^3*c^(3/2) - 96*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a*b*c^(5/2) 
+ 21*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*b^4*c - 96*(sqrt(c)*x - sqrt(c* 
x^2 + b*x + a))^2*a*b^2*c^2 + 48*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a^2 
*c^3 + 9*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*b^5*sqrt(c) - 48*(sqrt(c)*x - 
 sqrt(c*x^2 + b*x + a))*a*b^3*c^(3/2) + 48*(sqrt(c)*x - sqrt(c*x^2 + b*x + 
 a))*a^2*b*c^(5/2) + 2*b^6 - 15*a*b^4*c + 36*a^2*b^2*c^2 - 32*a^3*c^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)^3*c^(5/2)*d^4)
 

Mupad [F(-1)]

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

int((a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^4,x)
 

Output:

int((a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^4, x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 291, normalized size of antiderivative = 2.72 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^4} \, dx=\frac {-8 \sqrt {c \,x^{2}+b x +a}\, a \,c^{2}-6 \sqrt {c \,x^{2}+b x +a}\, b^{2} c -32 \sqrt {c \,x^{2}+b x +a}\, b \,c^{2} x -32 \sqrt {c \,x^{2}+b x +a}\, c^{3} x^{2}+3 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) b^{3}+18 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) b^{2} c x +36 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) b \,c^{2} x^{2}+24 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) c^{3} x^{3}}{48 c^{3} d^{4} \left (8 c^{3} x^{3}+12 b \,c^{2} x^{2}+6 b^{2} c x +b^{3}\right )} \] Input:

int((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^4,x)
 

Output:

( - 8*sqrt(a + b*x + c*x**2)*a*c**2 - 6*sqrt(a + b*x + c*x**2)*b**2*c - 32 
*sqrt(a + b*x + c*x**2)*b*c**2*x - 32*sqrt(a + b*x + c*x**2)*c**3*x**2 + 3 
*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b 
**2))*b**3 + 18*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x) 
/sqrt(4*a*c - b**2))*b**2*c*x + 36*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c 
*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*b*c**2*x**2 + 24*sqrt(c)*log((2*sq 
rt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*c**3*x**3)/( 
48*c**3*d**4*(b**3 + 6*b**2*c*x + 12*b*c**2*x**2 + 8*c**3*x**3))