\(\int \frac {(a+b x+c x^2)^{3/2}}{(b d+2 c d x)^5} \, dx\) [171]

Optimal result

Integrand size = 26, antiderivative size = 123 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^5} \, dx=-\frac {3 \sqrt {a+b x+c x^2}}{64 c^2 d^5 (b+2 c x)^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{8 c d^5 (b+2 c x)^4}+\frac {3 \arctan \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{128 c^{5/2} \sqrt {b^2-4 a c} d^5} \] Output:

-3/64*(c*x^2+b*x+a)^(1/2)/c^2/d^5/(2*c*x+b)^2-1/8*(c*x^2+b*x+a)^(3/2)/c/d^ 
5/(2*c*x+b)^4+3/128*arctan(2*c^(1/2)*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)^(1/2 
))/c^(5/2)/(-4*a*c+b^2)^(1/2)/d^5
 

Mathematica [A] (verified)

Time = 10.21 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.32 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^5} \, dx=\frac {-2 c \left (8 a^2 c+a \left (3 b^2+28 b c x+28 c^2 x^2\right )+x \left (3 b^3+23 b^2 c x+40 b c^2 x^2+20 c^3 x^3\right )\right )-3 (b+2 c x)^4 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}} \text {arctanh}\left (2 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}\right )}{128 c^3 d^5 (b+2 c x)^4 \sqrt {a+x (b+c x)}} \] Input:

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

Output:

(-2*c*(8*a^2*c + a*(3*b^2 + 28*b*c*x + 28*c^2*x^2) + x*(3*b^3 + 23*b^2*c*x 
 + 40*b*c^2*x^2 + 20*c^3*x^3)) - 3*(b + 2*c*x)^4*Sqrt[(c*(a + x*(b + c*x)) 
)/(-b^2 + 4*a*c)]*ArcTanh[2*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]])/( 
128*c^3*d^5*(b + 2*c*x)^4*Sqrt[a + x*(b + c*x)])
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.04, 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, 1112, 218}

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.

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

\(\Big \downarrow \) 1108

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1108

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

\(\Big \downarrow \) 1112

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

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

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 218

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

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]
 

rule 1112

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symb 
ol] :> Simp[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] && EqQ[2*c*d - b*e, 0]
 

Maple [A] (verified)

Time = 1.01 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.01

method	result	size
pseudoelliptic	\(-\frac {\frac {3 \left (2 c x +b \right )^{4} \operatorname {arctanh}\left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, c}{\sqrt {4 a \,c^{2}-b^{2} c}}\right )}{16}+\sqrt {c \,x^{2}+b x +a}\, \sqrt {4 a \,c^{2}-b^{2} c}\, \left (\frac {5 c^{2} x^{2}}{2}+\left (\frac {5 b x}{2}+a \right ) c +\frac {3 b^{2}}{8}\right )}{8 \sqrt {4 a \,c^{2}-b^{2} c}\, c^{2} \left (2 c x +b \right )^{4} d^{5}}\)	\(124\)
default	\(\frac {-\frac {c \left (c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}\right )^{\frac {5}{2}}}{\left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )^{4}}+\frac {c^{2} \left (-\frac {2 c \left (c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}\right )^{\frac {5}{2}}}{\left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )^{2}}+\frac {6 c^{2} \left (\frac {\left (c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}{3}+\frac {\left (4 a c -b^{2}\right ) \left (\frac {\sqrt {4 c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{c}}}{2}-\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{2 c \sqrt {\frac {4 a c -b^{2}}{c}}}\right )}{4 c}\right )}{4 a c -b^{2}}\right )}{4 a c -b^{2}}}{32 d^{5} c^{5}}\)	\(342\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 289 vs. \(2 (103) = 206\).

Time = 0.50 (sec) , antiderivative size = 615, normalized size of antiderivative = 5.00 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^5} \, dx=\left [-\frac {3 \, {\left (16 \, c^{4} x^{4} + 32 \, b c^{3} x^{3} + 24 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x + b^{4}\right )} \sqrt {-b^{2} c + 4 \, a c^{2}} \log \left (-\frac {4 \, c^{2} x^{2} + 4 \, b c x - b^{2} + 8 \, a c - 4 \, \sqrt {-b^{2} c + 4 \, a c^{2}} \sqrt {c x^{2} + b x + a}}{4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}\right ) + 4 \, {\left (3 \, b^{4} c - 4 \, a b^{2} c^{2} - 32 \, a^{2} c^{3} + 20 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{2} + 20 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{256 \, {\left (16 \, {\left (b^{2} c^{7} - 4 \, a c^{8}\right )} d^{5} x^{4} + 32 \, {\left (b^{3} c^{6} - 4 \, a b c^{7}\right )} d^{5} x^{3} + 24 \, {\left (b^{4} c^{5} - 4 \, a b^{2} c^{6}\right )} d^{5} x^{2} + 8 \, {\left (b^{5} c^{4} - 4 \, a b^{3} c^{5}\right )} d^{5} x + {\left (b^{6} c^{3} - 4 \, a b^{4} c^{4}\right )} d^{5}\right )}}, -\frac {3 \, {\left (16 \, c^{4} x^{4} + 32 \, b c^{3} x^{3} + 24 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x + b^{4}\right )} \sqrt {b^{2} c - 4 \, a c^{2}} \arctan \left (-\frac {2 \, \sqrt {b^{2} c - 4 \, a c^{2}} \sqrt {c x^{2} + b x + a}}{b^{2} - 4 \, a c}\right ) + 2 \, {\left (3 \, b^{4} c - 4 \, a b^{2} c^{2} - 32 \, a^{2} c^{3} + 20 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{2} + 20 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{128 \, {\left (16 \, {\left (b^{2} c^{7} - 4 \, a c^{8}\right )} d^{5} x^{4} + 32 \, {\left (b^{3} c^{6} - 4 \, a b c^{7}\right )} d^{5} x^{3} + 24 \, {\left (b^{4} c^{5} - 4 \, a b^{2} c^{6}\right )} d^{5} x^{2} + 8 \, {\left (b^{5} c^{4} - 4 \, a b^{3} c^{5}\right )} d^{5} x + {\left (b^{6} c^{3} - 4 \, a b^{4} c^{4}\right )} d^{5}\right )}}\right ] \] Input:

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

Output:

[-1/256*(3*(16*c^4*x^4 + 32*b*c^3*x^3 + 24*b^2*c^2*x^2 + 8*b^3*c*x + b^4)* 
sqrt(-b^2*c + 4*a*c^2)*log(-(4*c^2*x^2 + 4*b*c*x - b^2 + 8*a*c - 4*sqrt(-b 
^2*c + 4*a*c^2)*sqrt(c*x^2 + b*x + a))/(4*c^2*x^2 + 4*b*c*x + b^2)) + 4*(3 
*b^4*c - 4*a*b^2*c^2 - 32*a^2*c^3 + 20*(b^2*c^3 - 4*a*c^4)*x^2 + 20*(b^3*c 
^2 - 4*a*b*c^3)*x)*sqrt(c*x^2 + b*x + a))/(16*(b^2*c^7 - 4*a*c^8)*d^5*x^4 
+ 32*(b^3*c^6 - 4*a*b*c^7)*d^5*x^3 + 24*(b^4*c^5 - 4*a*b^2*c^6)*d^5*x^2 + 
8*(b^5*c^4 - 4*a*b^3*c^5)*d^5*x + (b^6*c^3 - 4*a*b^4*c^4)*d^5), -1/128*(3* 
(16*c^4*x^4 + 32*b*c^3*x^3 + 24*b^2*c^2*x^2 + 8*b^3*c*x + b^4)*sqrt(b^2*c 
- 4*a*c^2)*arctan(-2*sqrt(b^2*c - 4*a*c^2)*sqrt(c*x^2 + b*x + a)/(b^2 - 4* 
a*c)) + 2*(3*b^4*c - 4*a*b^2*c^2 - 32*a^2*c^3 + 20*(b^2*c^3 - 4*a*c^4)*x^2 
 + 20*(b^3*c^2 - 4*a*b*c^3)*x)*sqrt(c*x^2 + b*x + a))/(16*(b^2*c^7 - 4*a*c 
^8)*d^5*x^4 + 32*(b^3*c^6 - 4*a*b*c^7)*d^5*x^3 + 24*(b^4*c^5 - 4*a*b^2*c^6 
)*d^5*x^2 + 8*(b^5*c^4 - 4*a*b^3*c^5)*d^5*x + (b^6*c^3 - 4*a*b^4*c^4)*d^5) 
]
 

Sympy [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^5} \, dx=\frac {\int \frac {a \sqrt {a + b x + c x^{2}}}{b^{5} + 10 b^{4} c x + 40 b^{3} c^{2} x^{2} + 80 b^{2} c^{3} x^{3} + 80 b c^{4} x^{4} + 32 c^{5} x^{5}}\, dx + \int \frac {b x \sqrt {a + b x + c x^{2}}}{b^{5} + 10 b^{4} c x + 40 b^{3} c^{2} x^{2} + 80 b^{2} c^{3} x^{3} + 80 b c^{4} x^{4} + 32 c^{5} x^{5}}\, dx + \int \frac {c x^{2} \sqrt {a + b x + c x^{2}}}{b^{5} + 10 b^{4} c x + 40 b^{3} c^{2} x^{2} + 80 b^{2} c^{3} x^{3} + 80 b c^{4} x^{4} + 32 c^{5} x^{5}}\, dx}{d^{5}} \] Input:

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

Output:

(Integral(a*sqrt(a + b*x + c*x**2)/(b**5 + 10*b**4*c*x + 40*b**3*c**2*x**2 
 + 80*b**2*c**3*x**3 + 80*b*c**4*x**4 + 32*c**5*x**5), x) + Integral(b*x*s 
qrt(a + b*x + c*x**2)/(b**5 + 10*b**4*c*x + 40*b**3*c**2*x**2 + 80*b**2*c* 
*3*x**3 + 80*b*c**4*x**4 + 32*c**5*x**5), x) + Integral(c*x**2*sqrt(a + b* 
x + c*x**2)/(b**5 + 10*b**4*c*x + 40*b**3*c**2*x**2 + 80*b**2*c**3*x**3 + 
80*b*c**4*x**4 + 32*c**5*x**5), x))/d**5
 

Maxima [F(-2)]

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

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

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.69 (sec) , antiderivative size = 932, normalized size of antiderivative = 7.58 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^5} \, dx =\text {Too large to display} \] Input:

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

Output:

(3*sqrt(c)*sqrt(4*a*c - b**2)*log(( - sqrt(4*a*c - b**2) + 2*sqrt(c)*sqrt( 
a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*b**4 + 24*sqrt(c)*sqrt( 
4*a*c - b**2)*log(( - sqrt(4*a*c - b**2) + 2*sqrt(c)*sqrt(a + b*x + c*x**2 
) + b + 2*c*x)/sqrt(4*a*c - b**2))*b**3*c*x + 72*sqrt(c)*sqrt(4*a*c - b**2 
)*log(( - sqrt(4*a*c - b**2) + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c* 
x)/sqrt(4*a*c - b**2))*b**2*c**2*x**2 + 96*sqrt(c)*sqrt(4*a*c - b**2)*log( 
( - sqrt(4*a*c - b**2) + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqr 
t(4*a*c - b**2))*b*c**3*x**3 + 48*sqrt(c)*sqrt(4*a*c - b**2)*log(( - sqrt( 
4*a*c - b**2) + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - 
 b**2))*c**4*x**4 - 3*sqrt(c)*sqrt(4*a*c - b**2)*log((sqrt(4*a*c - b**2) + 
 2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*b**4 - 
24*sqrt(c)*sqrt(4*a*c - b**2)*log((sqrt(4*a*c - b**2) + 2*sqrt(c)*sqrt(a + 
 b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*b**3*c*x - 72*sqrt(c)*sqrt 
(4*a*c - b**2)*log((sqrt(4*a*c - b**2) + 2*sqrt(c)*sqrt(a + b*x + c*x**2) 
+ b + 2*c*x)/sqrt(4*a*c - b**2))*b**2*c**2*x**2 - 96*sqrt(c)*sqrt(4*a*c - 
b**2)*log((sqrt(4*a*c - b**2) + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c 
*x)/sqrt(4*a*c - b**2))*b*c**3*x**3 - 48*sqrt(c)*sqrt(4*a*c - b**2)*log((s 
qrt(4*a*c - b**2) + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a 
*c - b**2))*c**4*x**4 - 64*sqrt(a + b*x + c*x**2)*a**2*c**3 - 8*sqrt(a + b 
*x + c*x**2)*a*b**2*c**2 - 160*sqrt(a + b*x + c*x**2)*a*b*c**3*x - 160*...