\(\int \frac {(b d+2 c d x)^{5/2}}{(a+b x+c x^2)^2} \, dx\) [124]

Optimal result

Integrand size = 26, antiderivative size = 133 \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^2} \, dx=-\frac {d (b d+2 c d x)^{3/2}}{a+b x+c x^2}+\frac {6 c d^{5/2} \arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac {6 c d^{5/2} \text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\sqrt [4]{b^2-4 a c}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.80 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.73 \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^2} \, dx=(1+i) c (d (b+2 c x))^{5/2} \left (-\frac {\frac {1}{2}-\frac {i}{2}}{c (b+2 c x) (a+x (b+c x))}-\frac {3 \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\sqrt [4]{b^2-4 a c} (b+2 c x)^{5/2}}+\frac {3 \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\sqrt [4]{b^2-4 a c} (b+2 c x)^{5/2}}-\frac {3 \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )}{\sqrt [4]{b^2-4 a c} (b+2 c x)^{5/2}}\right ) \] Input:

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

Output:

(1 + I)*c*(d*(b + 2*c*x))^(5/2)*((-1/2 + I/2)/(c*(b + 2*c*x)*(a + x*(b + c 
*x))) - (3*ArcTan[1 - ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/((b^ 
2 - 4*a*c)^(1/4)*(b + 2*c*x)^(5/2)) + (3*ArcTan[1 + ((1 + I)*Sqrt[b + 2*c* 
x])/(b^2 - 4*a*c)^(1/4)])/((b^2 - 4*a*c)^(1/4)*(b + 2*c*x)^(5/2)) - (3*Arc 
Tanh[((1 + I)*(b^2 - 4*a*c)^(1/4)*Sqrt[b + 2*c*x])/(Sqrt[b^2 - 4*a*c] + I* 
(b + 2*c*x))])/((b^2 - 4*a*c)^(1/4)*(b + 2*c*x)^(5/2)))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {1110, 1118, 27, 25, 266, 827, 216, 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.

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

\(\Big \downarrow \) 1110

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

\(\Big \downarrow \) 1118

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 266

\(\displaystyle -12 c d^3 \int \frac {b d+2 c x d}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d\sqrt {b d+2 c x d}-\frac {d (b d+2 c d x)^{3/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 827

\(\displaystyle -12 c d^3 \left (\frac {1}{2} \int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}-\frac {1}{2} \int \frac {1}{b d+2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}\right )-\frac {d (b d+2 c d x)^{3/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 216

\(\displaystyle -12 c d^3 \left (\frac {1}{2} \int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}-\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 \sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-\frac {d (b d+2 c d x)^{3/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 219

\(\displaystyle -12 c d^3 \left (\frac {\text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 \sqrt {d} \sqrt [4]{b^2-4 a c}}-\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 \sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-\frac {d (b d+2 c d x)^{3/2}}{a+b x+c x^2}\)

Input:

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

Output:

-((d*(b*d + 2*c*d*x)^(3/2))/(a + b*x + c*x^2)) - 12*c*d^3*(-1/2*ArcTan[Sqr 
t[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])]/((b^2 - 4*a*c)^(1/4)*Sqrt[ 
d]) + ArcTanh[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])]/(2*(b^2 - 
 4*a*c)^(1/4)*Sqrt[d]))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

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 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 827

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ 
[a/b, 0]
 

rule 1110

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Sy 
mbol] :> Simp[d*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(b*(p + 1))), 
x] - Simp[d*e*((m - 1)/(b*(p + 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] && N 
eQ[m + 2*p + 3, 0] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]
 

rule 1118

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[1/e   Subst[Int[x^m*(a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, 
d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[2*c*d - b*e, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(276\) vs. \(2(111)=222\).

Time = 1.35 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.08

method	result	size
derivativedivides	\(16 c \,d^{3} \left (-\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{16 \left (a \,d^{2} c -\frac {b^{2} d^{2}}{4}+\frac {\left (2 c d x +b d \right )^{2}}{4}\right )}+\frac {3 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{32 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}\right )\)	\(277\)
default	\(16 c \,d^{3} \left (-\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{16 \left (a \,d^{2} c -\frac {b^{2} d^{2}}{4}+\frac {\left (2 c d x +b d \right )^{2}}{4}\right )}+\frac {3 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{32 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}\right )\)	\(277\)
pseudoelliptic	\(\frac {d^{2} \left (-2 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \left (2 c x +b \right )+3 \sqrt {2}\, c d \left (c \,x^{2}+b x +a \right ) \left (\ln \left (\frac {\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )\right )\right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \left (2 c \,x^{2}+2 b x +2 a \right )}\)	\(309\)

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 376, normalized size of antiderivative = 2.83 \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^2} \, dx=-\frac {3 \, \left (\frac {c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac {1}{4}} {\left (c x^{2} + b x + a\right )} \log \left (27 \, \sqrt {2 \, c d x + b d} c^{3} d^{7} + 27 \, \left (\frac {c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac {3}{4}} {\left (b^{2} - 4 \, a c\right )}\right ) - 3 \, \left (\frac {c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac {1}{4}} {\left (c x^{2} + b x + a\right )} \log \left (27 \, \sqrt {2 \, c d x + b d} c^{3} d^{7} - 27 \, \left (\frac {c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac {3}{4}} {\left (b^{2} - 4 \, a c\right )}\right ) - 3 \, \left (\frac {c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac {1}{4}} {\left (i \, c x^{2} + i \, b x + i \, a\right )} \log \left (27 \, \sqrt {2 \, c d x + b d} c^{3} d^{7} + 27 \, \left (\frac {c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac {3}{4}} {\left (i \, b^{2} - 4 i \, a c\right )}\right ) - 3 \, \left (\frac {c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac {1}{4}} {\left (-i \, c x^{2} - i \, b x - i \, a\right )} \log \left (27 \, \sqrt {2 \, c d x + b d} c^{3} d^{7} + 27 \, \left (\frac {c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac {3}{4}} {\left (-i \, b^{2} + 4 i \, a c\right )}\right ) + {\left (2 \, c d^{2} x + b d^{2}\right )} \sqrt {2 \, c d x + b d}}{c x^{2} + b x + a} \] Input:

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

Output:

-(3*(c^4*d^10/(b^2 - 4*a*c))^(1/4)*(c*x^2 + b*x + a)*log(27*sqrt(2*c*d*x + 
 b*d)*c^3*d^7 + 27*(c^4*d^10/(b^2 - 4*a*c))^(3/4)*(b^2 - 4*a*c)) - 3*(c^4* 
d^10/(b^2 - 4*a*c))^(1/4)*(c*x^2 + b*x + a)*log(27*sqrt(2*c*d*x + b*d)*c^3 
*d^7 - 27*(c^4*d^10/(b^2 - 4*a*c))^(3/4)*(b^2 - 4*a*c)) - 3*(c^4*d^10/(b^2 
 - 4*a*c))^(1/4)*(I*c*x^2 + I*b*x + I*a)*log(27*sqrt(2*c*d*x + b*d)*c^3*d^ 
7 + 27*(c^4*d^10/(b^2 - 4*a*c))^(3/4)*(I*b^2 - 4*I*a*c)) - 3*(c^4*d^10/(b^ 
2 - 4*a*c))^(1/4)*(-I*c*x^2 - I*b*x - I*a)*log(27*sqrt(2*c*d*x + b*d)*c^3* 
d^7 + 27*(c^4*d^10/(b^2 - 4*a*c))^(3/4)*(-I*b^2 + 4*I*a*c)) + (2*c*d^2*x + 
 b*d^2)*sqrt(2*c*d*x + b*d))/(c*x^2 + b*x + a)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^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. 439 vs. \(2 (111) = 222\).

Time = 0.15 (sec) , antiderivative size = 439, normalized size of antiderivative = 3.30 \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^2} \, dx=\frac {4 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} c d^{3}}{b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}} - \frac {3 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c d \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{2} - 4 \, a c} - \frac {3 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c d \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{2} - 4 \, a c} + \frac {3 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c d \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{2} - 4 \, \sqrt {2} a c} - \frac {3 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c d \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{2} - 4 \, \sqrt {2} a c} \] Input:

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

Output:

4*(2*c*d*x + b*d)^(3/2)*c*d^3/(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2) - 
3*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*d*arctan(1/2*sqrt(2)*(sqrt(2)*(-b 
^2*d^2 + 4*a*c*d^2)^(1/4) + 2*sqrt(2*c*d*x + b*d))/(-b^2*d^2 + 4*a*c*d^2)^ 
(1/4))/(b^2 - 4*a*c) - 3*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*d*arctan(- 
1/2*sqrt(2)*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4) - 2*sqrt(2*c*d*x + b*d)) 
/(-b^2*d^2 + 4*a*c*d^2)^(1/4))/(b^2 - 4*a*c) + 3*(-b^2*d^2 + 4*a*c*d^2)^(3 
/4)*c*d*log(2*c*d*x + b*d + sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*sqrt(2*c* 
d*x + b*d) + sqrt(-b^2*d^2 + 4*a*c*d^2))/(sqrt(2)*b^2 - 4*sqrt(2)*a*c) - 3 
*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*d*log(2*c*d*x + b*d - sqrt(2)*(-b^2*d^2 + 
4*a*c*d^2)^(1/4)*sqrt(2*c*d*x + b*d) + sqrt(-b^2*d^2 + 4*a*c*d^2))/(sqrt(2 
)*b^2 - 4*sqrt(2)*a*c)
 

Mupad [B] (verification not implemented)

Time = 5.27 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.98 \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^2} \, dx=\frac {6\,c\,d^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{1/4}}-\frac {4\,c\,d^3\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}}{{\left (b\,d+2\,c\,d\,x\right )}^2-b^2\,d^2+4\,a\,c\,d^2}-\frac {6\,c\,d^{5/2}\,\mathrm {atanh}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{1/4}} \] Input:

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

Output:

(6*c*d^(5/2)*atan((b*d + 2*c*d*x)^(1/2)/(d^(1/2)*(b^2 - 4*a*c)^(1/4))))/(b 
^2 - 4*a*c)^(1/4) - (4*c*d^3*(b*d + 2*c*d*x)^(3/2))/((b*d + 2*c*d*x)^2 - b 
^2*d^2 + 4*a*c*d^2) - (6*c*d^(5/2)*atanh((b*d + 2*c*d*x)^(1/2)/(d^(1/2)*(b 
^2 - 4*a*c)^(1/4))))/(b^2 - 4*a*c)^(1/4)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 841, normalized size of antiderivative = 6.32 \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(sqrt(d)*d**2*( - 6*(4*a*c - b**2)**(3/4)*sqrt(2)*atan(((4*a*c - b**2)**(1 
/4)*sqrt(2) - 2*sqrt(b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2)))*a*c - 6* 
(4*a*c - b**2)**(3/4)*sqrt(2)*atan(((4*a*c - b**2)**(1/4)*sqrt(2) - 2*sqrt 
(b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2)))*b*c*x - 6*(4*a*c - b**2)**(3 
/4)*sqrt(2)*atan(((4*a*c - b**2)**(1/4)*sqrt(2) - 2*sqrt(b + 2*c*x))/((4*a 
*c - b**2)**(1/4)*sqrt(2)))*c**2*x**2 + 6*(4*a*c - b**2)**(3/4)*sqrt(2)*at 
an(((4*a*c - b**2)**(1/4)*sqrt(2) + 2*sqrt(b + 2*c*x))/((4*a*c - b**2)**(1 
/4)*sqrt(2)))*a*c + 6*(4*a*c - b**2)**(3/4)*sqrt(2)*atan(((4*a*c - b**2)** 
(1/4)*sqrt(2) + 2*sqrt(b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2)))*b*c*x 
+ 6*(4*a*c - b**2)**(3/4)*sqrt(2)*atan(((4*a*c - b**2)**(1/4)*sqrt(2) + 2* 
sqrt(b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2)))*c**2*x**2 + 3*(4*a*c - b 
**2)**(3/4)*sqrt(2)*log( - sqrt(b + 2*c*x)*(4*a*c - b**2)**(1/4)*sqrt(2) + 
 sqrt(4*a*c - b**2) + b + 2*c*x)*a*c + 3*(4*a*c - b**2)**(3/4)*sqrt(2)*log 
( - sqrt(b + 2*c*x)*(4*a*c - b**2)**(1/4)*sqrt(2) + sqrt(4*a*c - b**2) + b 
 + 2*c*x)*b*c*x + 3*(4*a*c - b**2)**(3/4)*sqrt(2)*log( - sqrt(b + 2*c*x)*( 
4*a*c - b**2)**(1/4)*sqrt(2) + sqrt(4*a*c - b**2) + b + 2*c*x)*c**2*x**2 - 
 3*(4*a*c - b**2)**(3/4)*sqrt(2)*log(sqrt(b + 2*c*x)*(4*a*c - b**2)**(1/4) 
*sqrt(2) + sqrt(4*a*c - b**2) + b + 2*c*x)*a*c - 3*(4*a*c - b**2)**(3/4)*s 
qrt(2)*log(sqrt(b + 2*c*x)*(4*a*c - b**2)**(1/4)*sqrt(2) + sqrt(4*a*c - b* 
*2) + b + 2*c*x)*b*c*x - 3*(4*a*c - b**2)**(3/4)*sqrt(2)*log(sqrt(b + 2...