\(\int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^5} \, dx\) [159]

Optimal result

Integrand size = 26, antiderivative size = 133 \[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^5} \, dx=-\frac {\sqrt {a+b x+c x^2}}{8 c d^5 (b+2 c x)^4}+\frac {\sqrt {a+b x+c x^2}}{16 c \left (b^2-4 a c\right ) d^5 (b+2 c x)^2}+\frac {\arctan \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{32 c^{3/2} \left (b^2-4 a c\right )^{3/2} d^5} \] Output:

-1/8*(c*x^2+b*x+a)^(1/2)/c/d^5/(2*c*x+b)^4+1/16*(c*x^2+b*x+a)^(1/2)/c/(-4* 
a*c+b^2)/d^5/(2*c*x+b)^2+1/32*arctan(2*c^(1/2)*(c*x^2+b*x+a)^(1/2)/(-4*a*c 
+b^2)^(1/2))/c^(3/2)/(-4*a*c+b^2)^(3/2)/d^5
 

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 = 0.47 \[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^5} \, dx=\frac {2 (a+x (b+c x))^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},3,\frac {5}{2},\frac {4 c (a+x (b+c x))}{-b^2+4 a c}\right )}{3 \left (b^2-4 a c\right )^3 d^5} \] Input:

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

Output:

(2*(a + x*(b + c*x))^(3/2)*Hypergeometric2F1[3/2, 3, 5/2, (4*c*(a + x*(b + 
 c*x)))/(-b^2 + 4*a*c)])/(3*(b^2 - 4*a*c)^3*d^5)
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1108, 27, 1117, 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.

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

\(\Big \downarrow \) 1108

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1117

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

\(\Big \downarrow \) 1112

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

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

rule 1117

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> 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] + Simp[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] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[m, -1] & 
& (IntegerQ[2*p] || (IntegerQ[m] && RationalQ[p]) || IntegerQ[(m + 2*p + 3) 
/2])
 

Maple [A] (verified)

Time = 1.03 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.02

method	result	size
pseudoelliptic	\(\frac {-16 \sqrt {c \,x^{2}+b x +a}\, \left (\frac {c^{2} x^{2}}{2}+\left (\frac {b x}{2}+a \right ) c -\frac {b^{2}}{8}\right ) \sqrt {4 a \,c^{2}-b^{2} c}+\left (2 c x +b \right )^{4} \operatorname {arctanh}\left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, c}{\sqrt {4 a \,c^{2}-b^{2} c}}\right )}{128 \sqrt {4 a \,c^{2}-b^{2} c}\, \left (2 c x +b \right )^{4} \left (a c -\frac {b^{2}}{4}\right ) d^{5} c}\)	\(135\)
default	\(\frac {-\frac {c \left (c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}{\left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )^{4}}-\frac {c^{2} \left (-\frac {2 c \left (c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}{\left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )^{2}}+\frac {2 c^{2} \left (\frac {\sqrt {4 c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{c}}}{2}-\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{2 c \sqrt {\frac {4 a c -b^{2}}{c}}}\right )}{4 a c -b^{2}}\right )}{4 a c -b^{2}}}{32 d^{5} c^{5}}\)	\(295\)

Input:

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

Output:

1/128/(4*a*c^2-b^2*c)^(1/2)*(-16*(c*x^2+b*x+a)^(1/2)*(1/2*c^2*x^2+(1/2*b*x 
+a)*c-1/8*b^2)*(4*a*c^2-b^2*c)^(1/2)+(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)))/(2*c*x+b)^4/(a*c-1/4*b^2)/d^5/c
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 342 vs. \(2 (113) = 226\).

Time = 0.51 (sec) , antiderivative size = 721, normalized size of antiderivative = 5.42 \[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^5} \, dx=\left [\frac {{\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 (b^{4} c - 12 \, a b^{2} c^{2} + 32 \, a^{2} c^{3} - 4 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{2} - 4 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{64 \, {\left (16 \, {\left (b^{4} c^{6} - 8 \, a b^{2} c^{7} + 16 \, a^{2} c^{8}\right )} d^{5} x^{4} + 32 \, {\left (b^{5} c^{5} - 8 \, a b^{3} c^{6} + 16 \, a^{2} b c^{7}\right )} d^{5} x^{3} + 24 \, {\left (b^{6} c^{4} - 8 \, a b^{4} c^{5} + 16 \, a^{2} b^{2} c^{6}\right )} d^{5} x^{2} + 8 \, {\left (b^{7} c^{3} - 8 \, a b^{5} c^{4} + 16 \, a^{2} b^{3} c^{5}\right )} d^{5} x + {\left (b^{8} c^{2} - 8 \, a b^{6} c^{3} + 16 \, a^{2} b^{4} c^{4}\right )} d^{5}\right )}}, -\frac {{\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 (b^{4} c - 12 \, a b^{2} c^{2} + 32 \, a^{2} c^{3} - 4 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{2} - 4 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{32 \, {\left (16 \, {\left (b^{4} c^{6} - 8 \, a b^{2} c^{7} + 16 \, a^{2} c^{8}\right )} d^{5} x^{4} + 32 \, {\left (b^{5} c^{5} - 8 \, a b^{3} c^{6} + 16 \, a^{2} b c^{7}\right )} d^{5} x^{3} + 24 \, {\left (b^{6} c^{4} - 8 \, a b^{4} c^{5} + 16 \, a^{2} b^{2} c^{6}\right )} d^{5} x^{2} + 8 \, {\left (b^{7} c^{3} - 8 \, a b^{5} c^{4} + 16 \, a^{2} b^{3} c^{5}\right )} d^{5} x + {\left (b^{8} c^{2} - 8 \, a b^{6} c^{3} + 16 \, a^{2} b^{4} c^{4}\right )} d^{5}\right )}}\right ] \] Input:

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

Output:

[1/64*((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*(b^4*c 
 - 12*a*b^2*c^2 + 32*a^2*c^3 - 4*(b^2*c^3 - 4*a*c^4)*x^2 - 4*(b^3*c^2 - 4* 
a*b*c^3)*x)*sqrt(c*x^2 + b*x + a))/(16*(b^4*c^6 - 8*a*b^2*c^7 + 16*a^2*c^8 
)*d^5*x^4 + 32*(b^5*c^5 - 8*a*b^3*c^6 + 16*a^2*b*c^7)*d^5*x^3 + 24*(b^6*c^ 
4 - 8*a*b^4*c^5 + 16*a^2*b^2*c^6)*d^5*x^2 + 8*(b^7*c^3 - 8*a*b^5*c^4 + 16* 
a^2*b^3*c^5)*d^5*x + (b^8*c^2 - 8*a*b^6*c^3 + 16*a^2*b^4*c^4)*d^5), -1/32* 
((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*(b^4*c - 12*a*b^2*c^2 + 32*a^2*c^3 - 4*(b^2*c^3 - 4*a*c^4)*x^2 
- 4*(b^3*c^2 - 4*a*b*c^3)*x)*sqrt(c*x^2 + b*x + a))/(16*(b^4*c^6 - 8*a*b^2 
*c^7 + 16*a^2*c^8)*d^5*x^4 + 32*(b^5*c^5 - 8*a*b^3*c^6 + 16*a^2*b*c^7)*d^5 
*x^3 + 24*(b^6*c^4 - 8*a*b^4*c^5 + 16*a^2*b^2*c^6)*d^5*x^2 + 8*(b^7*c^3 - 
8*a*b^5*c^4 + 16*a^2*b^3*c^5)*d^5*x + (b^8*c^2 - 8*a*b^6*c^3 + 16*a^2*b^4* 
c^4)*d^5)]
 

Sympy [F]

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

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

Output:

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

integrate((c*x^2+b*x+a)^(1/2)/(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. \[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^5} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.66 (sec) , antiderivative size = 994, normalized size of antiderivative = 7.47 \[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^5} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 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 - 8*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 - 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**2*c**2*x**2 - 32*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 - 16*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 + 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 + 8* 
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 + 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**2*c**2*x**2 + 32*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 + 16*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 - 64*sqrt(a + b*x + c*x**2)*a**2*c**3 + 24*sqrt(a + b*x 
 + c*x**2)*a*b**2*c**2 - 32*sqrt(a + b*x + c*x**2)*a*b*c**3*x - 32*sqrt...