\(\int \frac {(b d+2 c d x)^{15/2}}{(a+b x+c x^2)^2} \, dx\) [119]

Optimal result

Integrand size = 26, antiderivative size = 212 \[ \int \frac {(b d+2 c d x)^{15/2}}{\left (a+b x+c x^2\right )^2} \, dx=52 c \left (b^2-4 a c\right )^2 d^7 \sqrt {b d+2 c d x}+\frac {52}{5} c \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{5/2}+\frac {52}{9} c d^3 (b d+2 c d x)^{9/2}-\frac {d (b d+2 c d x)^{13/2}}{a+b x+c x^2}-26 c \left (b^2-4 a c\right )^{9/4} d^{15/2} \arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-26 c \left (b^2-4 a c\right )^{9/4} d^{15/2} \text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ) \] Output:

52*c*(-4*a*c+b^2)^2*d^7*(2*c*d*x+b*d)^(1/2)+52/5*c*(-4*a*c+b^2)*d^5*(2*c*d 
*x+b*d)^(5/2)+52/9*c*d^3*(2*c*d*x+b*d)^(9/2)-d*(2*c*d*x+b*d)^(13/2)/(c*x^2 
+b*x+a)-26*c*(-4*a*c+b^2)^(9/4)*d^(15/2)*arctan((2*c*d*x+b*d)^(1/2)/(-4*a* 
c+b^2)^(1/4)/d^(1/2))-26*c*(-4*a*c+b^2)^(9/4)*d^(15/2)*arctanh((2*c*d*x+b* 
d)^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.39 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.66 \[ \int \frac {(b d+2 c d x)^{15/2}}{\left (a+b x+c x^2\right )^2} \, dx=\left (\frac {1}{45}+\frac {i}{45}\right ) c (d (b+2 c x))^{15/2} \left (\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (-585 b^6+7020 a b^4 c-28080 a^2 b^2 c^2+37440 a^3 c^3+468 b^4 (b+2 c x)^2-3744 a b^2 c (b+2 c x)^2+7488 a^2 c^2 (b+2 c x)^2+52 b^2 (b+2 c x)^4-208 a c (b+2 c x)^4+20 (b+2 c x)^6\right )}{c (b+2 c x)^7 (a+x (b+c x))}-\frac {585 i \left (b^2-4 a c\right )^{9/4} \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{(b+2 c x)^{15/2}}+\frac {585 i \left (b^2-4 a c\right )^{9/4} \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{(b+2 c x)^{15/2}}+\frac {585 i \left (b^2-4 a c\right )^{9/4} \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )}{(b+2 c x)^{15/2}}\right ) \] Input:

Integrate[(b*d + 2*c*d*x)^(15/2)/(a + b*x + c*x^2)^2,x]
 

Output:

(1/45 + I/45)*c*(d*(b + 2*c*x))^(15/2)*(((1/2 - I/2)*(-585*b^6 + 7020*a*b^ 
4*c - 28080*a^2*b^2*c^2 + 37440*a^3*c^3 + 468*b^4*(b + 2*c*x)^2 - 3744*a*b 
^2*c*(b + 2*c*x)^2 + 7488*a^2*c^2*(b + 2*c*x)^2 + 52*b^2*(b + 2*c*x)^4 - 2 
08*a*c*(b + 2*c*x)^4 + 20*(b + 2*c*x)^6))/(c*(b + 2*c*x)^7*(a + x*(b + c*x 
))) - ((585*I)*(b^2 - 4*a*c)^(9/4)*ArcTan[1 - ((1 + I)*Sqrt[b + 2*c*x])/(b 
^2 - 4*a*c)^(1/4)])/(b + 2*c*x)^(15/2) + ((585*I)*(b^2 - 4*a*c)^(9/4)*ArcT 
an[1 + ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/(b + 2*c*x)^(15/2) 
+ ((585*I)*(b^2 - 4*a*c)^(9/4)*ArcTanh[((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*c*x)^(15/2))
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.10, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {1110, 1116, 1116, 1116, 1118, 27, 25, 266, 756, 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.

\(\displaystyle \int \frac {(b d+2 c d x)^{15/2}}{\left (a+b x+c x^2\right )^2} \, dx\)

\(\Big \downarrow \) 1110

\(\displaystyle 13 c d^2 \int \frac {(b d+2 c x d)^{11/2}}{c x^2+b x+a}dx-\frac {d (b d+2 c d x)^{13/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 1116

\(\displaystyle 13 c d^2 \left (d^2 \left (b^2-4 a c\right ) \int \frac {(b d+2 c x d)^{7/2}}{c x^2+b x+a}dx+\frac {4}{9} d (b d+2 c d x)^{9/2}\right )-\frac {d (b d+2 c d x)^{13/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 1116

\(\displaystyle 13 c d^2 \left (d^2 \left (b^2-4 a c\right ) \left (d^2 \left (b^2-4 a c\right ) \int \frac {(b d+2 c x d)^{3/2}}{c x^2+b x+a}dx+\frac {4}{5} d (b d+2 c d x)^{5/2}\right )+\frac {4}{9} d (b d+2 c d x)^{9/2}\right )-\frac {d (b d+2 c d x)^{13/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 1116

\(\displaystyle 13 c d^2 \left (d^2 \left (b^2-4 a c\right ) \left (d^2 \left (b^2-4 a c\right ) \left (d^2 \left (b^2-4 a c\right ) \int \frac {1}{\sqrt {b d+2 c x d} \left (c x^2+b x+a\right )}dx+4 d \sqrt {b d+2 c d x}\right )+\frac {4}{5} d (b d+2 c d x)^{5/2}\right )+\frac {4}{9} d (b d+2 c d x)^{9/2}\right )-\frac {d (b d+2 c d x)^{13/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 1118

\(\displaystyle 13 c d^2 \left (d^2 \left (b^2-4 a c\right ) \left (d^2 \left (b^2-4 a c\right ) \left (\frac {d \left (b^2-4 a c\right ) \int \frac {4 c d^2}{\sqrt {b d+2 c x d} \left (\left (4 a-\frac {b^2}{c}\right ) c d^2+(b d+2 c x d)^2\right )}d(b d+2 c x d)}{2 c}+4 d \sqrt {b d+2 c d x}\right )+\frac {4}{5} d (b d+2 c d x)^{5/2}\right )+\frac {4}{9} d (b d+2 c d x)^{9/2}\right )-\frac {d (b d+2 c d x)^{13/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 27

\(\displaystyle 13 c d^2 \left (d^2 \left (b^2-4 a c\right ) \left (d^2 \left (b^2-4 a c\right ) \left (2 d^3 \left (b^2-4 a c\right ) \int -\frac {1}{\sqrt {b d+2 c x d} \left (\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2\right )}d(b d+2 c x d)+4 d \sqrt {b d+2 c d x}\right )+\frac {4}{5} d (b d+2 c d x)^{5/2}\right )+\frac {4}{9} d (b d+2 c d x)^{9/2}\right )-\frac {d (b d+2 c d x)^{13/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 25

\(\displaystyle 13 c d^2 \left (d^2 \left (b^2-4 a c\right ) \left (d^2 \left (b^2-4 a c\right ) \left (4 d \sqrt {b d+2 c d x}-2 d^3 \left (b^2-4 a c\right ) \int \frac {1}{\sqrt {b d+2 c x d} \left (\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2\right )}d(b d+2 c x d)\right )+\frac {4}{5} d (b d+2 c d x)^{5/2}\right )+\frac {4}{9} d (b d+2 c d x)^{9/2}\right )-\frac {d (b d+2 c d x)^{13/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 266

\(\displaystyle 13 c d^2 \left (d^2 \left (b^2-4 a c\right ) \left (d^2 \left (b^2-4 a c\right ) \left (4 d \sqrt {b d+2 c d x}-4 d^3 \left (b^2-4 a c\right ) \int \frac {1}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d\sqrt {b d+2 c x d}\right )+\frac {4}{5} d (b d+2 c d x)^{5/2}\right )+\frac {4}{9} d (b d+2 c d x)^{9/2}\right )-\frac {d (b d+2 c d x)^{13/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 756

\(\displaystyle 13 c d^2 \left (d^2 \left (b^2-4 a c\right ) \left (d^2 \left (b^2-4 a c\right ) \left (4 d \sqrt {b d+2 c d x}-4 d^3 \left (b^2-4 a c\right ) \left (\frac {\int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}}{2 d \sqrt {b^2-4 a c}}+\frac {\int \frac {1}{b d+2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}}{2 d \sqrt {b^2-4 a c}}\right )\right )+\frac {4}{5} d (b d+2 c d x)^{5/2}\right )+\frac {4}{9} d (b d+2 c d x)^{9/2}\right )-\frac {d (b d+2 c d x)^{13/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 216

\(\displaystyle 13 c d^2 \left (d^2 \left (b^2-4 a c\right ) \left (d^2 \left (b^2-4 a c\right ) \left (4 d \sqrt {b d+2 c d x}-4 d^3 \left (b^2-4 a c\right ) \left (\frac {\int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}}{2 d \sqrt {b^2-4 a c}}+\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 d^{3/2} \left (b^2-4 a c\right )^{3/4}}\right )\right )+\frac {4}{5} d (b d+2 c d x)^{5/2}\right )+\frac {4}{9} d (b d+2 c d x)^{9/2}\right )-\frac {d (b d+2 c d x)^{13/2}}{a+b x+c x^2}\)

\(\Big \downarrow \) 219

\(\displaystyle 13 c d^2 \left (d^2 \left (b^2-4 a c\right ) \left (d^2 \left (b^2-4 a c\right ) \left (4 d \sqrt {b d+2 c d x}-4 d^3 \left (b^2-4 a c\right ) \left (\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 d^{3/2} \left (b^2-4 a c\right )^{3/4}}+\frac {\text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 d^{3/2} \left (b^2-4 a c\right )^{3/4}}\right )\right )+\frac {4}{5} d (b d+2 c d x)^{5/2}\right )+\frac {4}{9} d (b d+2 c d x)^{9/2}\right )-\frac {d (b d+2 c d x)^{13/2}}{a+b x+c x^2}\)

Input:

Int[(b*d + 2*c*d*x)^(15/2)/(a + b*x + c*x^2)^2,x]
 

Output:

-((d*(b*d + 2*c*d*x)^(13/2))/(a + b*x + c*x^2)) + 13*c*d^2*((4*d*(b*d + 2* 
c*d*x)^(9/2))/9 + (b^2 - 4*a*c)*d^2*((4*d*(b*d + 2*c*d*x)^(5/2))/5 + (b^2 
- 4*a*c)*d^2*(4*d*Sqrt[b*d + 2*c*d*x] - 4*(b^2 - 4*a*c)*d^3*(ArcTan[Sqrt[b 
*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])]/(2*(b^2 - 4*a*c)^(3/4)*d^(3/2 
)) + ArcTanh[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])]/(2*(b^2 - 
4*a*c)^(3/4)*d^(3/2))))))
 

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 756

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   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 1116

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[2*d*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(b*(m + 2*p 
 + 1))), x] + Simp[d^2*(m - 1)*((b^2 - 4*a*c)/(b^2*(m + 2*p + 1)))   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] && GtQ[m, 1] && NeQ[m + 2*p 
+ 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] && RationalQ[p]) || OddQ[m])
 

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. \(406\) vs. \(2(180)=360\).

Time = 1.71 (sec) , antiderivative size = 407, normalized size of antiderivative = 1.92

method	result	size
pseudoelliptic	\(\frac {832 d^{3} \left (-\frac {2 \left (a c -\frac {b^{2}}{4}\right ) d^{2} \left (c \,x^{2}+b x +a \right ) \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}} c \left (d \left (2 c x +b \right )\right )^{\frac {5}{2}}}{65}+\frac {\left (d \left (2 c x +b \right )\right )^{\frac {9}{2}} \left (c \,x^{2}+b x +a \right ) c \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}}}{468}+\frac {d^{4} \left (4 a c -b^{2}\right )^{2} \left (2 \left (\frac {12 c^{2} x^{2}}{13}+\left (\frac {12 b x}{13}+a \right ) c -\frac {b^{2}}{52}\right ) \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}} \sqrt {d \left (2 c x +b \right )}-c \,d^{2} \sqrt {2}\, \left (a c -\frac {b^{2}}{4}\right ) \left (c \,x^{2}+b x +a \right ) \left (\ln \left (\frac {\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}{\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}-1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}+1\right )\right )\right )}{32}\right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}} \left (c \,x^{2}+b x +a \right )}\)	\(407\)
derivativedivides	\(16 c \,d^{3} \left (48 a^{2} c^{2} d^{4} \sqrt {2 c d x +b d}-24 a \,b^{2} c \,d^{4} \sqrt {2 c d x +b d}-\frac {8 a c \,d^{2} \left (2 c d x +b d \right )^{\frac {5}{2}}}{5}+3 b^{4} d^{4} \sqrt {2 c d x +b d}+\frac {2 b^{2} d^{2} \left (2 c d x +b d \right )^{\frac {5}{2}}}{5}+\frac {\left (2 c d x +b d \right )^{\frac {9}{2}}}{9}-d^{6} \left (\frac {\left (-4 a^{3} c^{3}+3 a^{2} b^{2} c^{2}-\frac {3}{4} a \,b^{4} c +\frac {1}{16} b^{6}\right ) \sqrt {2 c d x +b d}}{a \,d^{2} c -\frac {b^{2} d^{2}}{4}+\frac {\left (2 c d x +b d \right )^{2}}{4}}+\frac {13 \left (16 a^{3} c^{3}-12 a^{2} b^{2} c^{2}+3 a \,b^{4} c -\frac {1}{4} b^{6}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {3}{4}}}\right )\right )\)	\(458\)
default	\(16 c \,d^{3} \left (48 a^{2} c^{2} d^{4} \sqrt {2 c d x +b d}-24 a \,b^{2} c \,d^{4} \sqrt {2 c d x +b d}-\frac {8 a c \,d^{2} \left (2 c d x +b d \right )^{\frac {5}{2}}}{5}+3 b^{4} d^{4} \sqrt {2 c d x +b d}+\frac {2 b^{2} d^{2} \left (2 c d x +b d \right )^{\frac {5}{2}}}{5}+\frac {\left (2 c d x +b d \right )^{\frac {9}{2}}}{9}-d^{6} \left (\frac {\left (-4 a^{3} c^{3}+3 a^{2} b^{2} c^{2}-\frac {3}{4} a \,b^{4} c +\frac {1}{16} b^{6}\right ) \sqrt {2 c d x +b d}}{a \,d^{2} c -\frac {b^{2} d^{2}}{4}+\frac {\left (2 c d x +b d \right )^{2}}{4}}+\frac {13 \left (16 a^{3} c^{3}-12 a^{2} b^{2} c^{2}+3 a \,b^{4} c -\frac {1}{4} b^{6}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {3}{4}}}\right )\right )\)	\(458\)

Input:

int((2*c*d*x+b*d)^(15/2)/(c*x^2+b*x+a)^2,x,method=_RETURNVERBOSE)
 

Output:

832*d^3*(-2/65*(a*c-1/4*b^2)*d^2*(c*x^2+b*x+a)*(d^2*(4*a*c-b^2))^(3/4)*c*( 
d*(2*c*x+b))^(5/2)+1/468*(d*(2*c*x+b))^(9/2)*(c*x^2+b*x+a)*c*(d^2*(4*a*c-b 
^2))^(3/4)+1/32*d^4*(4*a*c-b^2)^2*(2*(12/13*c^2*x^2+(12/13*b*x+a)*c-1/52*b 
^2)*(d^2*(4*a*c-b^2))^(3/4)*(d*(2*c*x+b))^(1/2)-c*d^2*2^(1/2)*(a*c-1/4*b^2 
)*(c*x^2+b*x+a)*(ln(((d^2*(4*a*c-b^2))^(1/4)*(d*(2*c*x+b))^(1/2)*2^(1/2)+( 
d^2*(4*a*c-b^2))^(1/2)+d*(2*c*x+b))/((d^2*(4*a*c-b^2))^(1/2)-(d^2*(4*a*c-b 
^2))^(1/4)*(d*(2*c*x+b))^(1/2)*2^(1/2)+d*(2*c*x+b)))+2*arctan(2^(1/2)/(d^2 
*(4*a*c-b^2))^(1/4)*(d*(2*c*x+b))^(1/2)-1)+2*arctan(2^(1/2)/(d^2*(4*a*c-b^ 
2))^(1/4)*(d*(2*c*x+b))^(1/2)+1))))/(d^2*(4*a*c-b^2))^(3/4)/(c*x^2+b*x+a)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.17 (sec) , antiderivative size = 1293, normalized size of antiderivative = 6.10 \[ \int \frac {(b d+2 c d x)^{15/2}}{\left (a+b x+c x^2\right )^2} \, dx=\text {Too large to display} \] Input:

integrate((2*c*d*x+b*d)^(15/2)/(c*x^2+b*x+a)^2,x, algorithm="fricas")
 

Output:

-1/45*(585*((b^18*c^4 - 36*a*b^16*c^5 + 576*a^2*b^14*c^6 - 5376*a^3*b^12*c 
^7 + 32256*a^4*b^10*c^8 - 129024*a^5*b^8*c^9 + 344064*a^6*b^6*c^10 - 58982 
4*a^7*b^4*c^11 + 589824*a^8*b^2*c^12 - 262144*a^9*c^13)*d^30)^(1/4)*(c*x^2 
 + b*x + a)*log(13*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*sqrt(2*c*d*x + b*d)* 
d^7 + 13*((b^18*c^4 - 36*a*b^16*c^5 + 576*a^2*b^14*c^6 - 5376*a^3*b^12*c^7 
 + 32256*a^4*b^10*c^8 - 129024*a^5*b^8*c^9 + 344064*a^6*b^6*c^10 - 589824* 
a^7*b^4*c^11 + 589824*a^8*b^2*c^12 - 262144*a^9*c^13)*d^30)^(1/4)) + 585*( 
(b^18*c^4 - 36*a*b^16*c^5 + 576*a^2*b^14*c^6 - 5376*a^3*b^12*c^7 + 32256*a 
^4*b^10*c^8 - 129024*a^5*b^8*c^9 + 344064*a^6*b^6*c^10 - 589824*a^7*b^4*c^ 
11 + 589824*a^8*b^2*c^12 - 262144*a^9*c^13)*d^30)^(1/4)*(I*c*x^2 + I*b*x + 
 I*a)*log(13*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*sqrt(2*c*d*x + b*d)*d^7 + 
13*I*((b^18*c^4 - 36*a*b^16*c^5 + 576*a^2*b^14*c^6 - 5376*a^3*b^12*c^7 + 3 
2256*a^4*b^10*c^8 - 129024*a^5*b^8*c^9 + 344064*a^6*b^6*c^10 - 589824*a^7* 
b^4*c^11 + 589824*a^8*b^2*c^12 - 262144*a^9*c^13)*d^30)^(1/4)) + 585*((b^1 
8*c^4 - 36*a*b^16*c^5 + 576*a^2*b^14*c^6 - 5376*a^3*b^12*c^7 + 32256*a^4*b 
^10*c^8 - 129024*a^5*b^8*c^9 + 344064*a^6*b^6*c^10 - 589824*a^7*b^4*c^11 + 
 589824*a^8*b^2*c^12 - 262144*a^9*c^13)*d^30)^(1/4)*(-I*c*x^2 - I*b*x - I* 
a)*log(13*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*sqrt(2*c*d*x + b*d)*d^7 - 13* 
I*((b^18*c^4 - 36*a*b^16*c^5 + 576*a^2*b^14*c^6 - 5376*a^3*b^12*c^7 + 3225 
6*a^4*b^10*c^8 - 129024*a^5*b^8*c^9 + 344064*a^6*b^6*c^10 - 589824*a^7*...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(b d+2 c d x)^{15/2}}{\left (a+b x+c x^2\right )^2} \, dx=\text {Timed out} \] Input:

integrate((2*c*d*x+b*d)**(15/2)/(c*x**2+b*x+a)**2,x)
 

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(b d+2 c d x)^{15/2}}{\left (a+b x+c x^2\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((2*c*d*x+b*d)^(15/2)/(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. 876 vs. \(2 (180) = 360\).

Time = 0.16 (sec) , antiderivative size = 876, normalized size of antiderivative = 4.13 \[ \int \frac {(b d+2 c d x)^{15/2}}{\left (a+b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:

integrate((2*c*d*x+b*d)^(15/2)/(c*x^2+b*x+a)^2,x, algorithm="giac")
 

Output:

48*sqrt(2*c*d*x + b*d)*b^4*c*d^7 - 384*sqrt(2*c*d*x + b*d)*a*b^2*c^2*d^7 + 
 768*sqrt(2*c*d*x + b*d)*a^2*c^3*d^7 + 32/5*(2*c*d*x + b*d)^(5/2)*b^2*c*d^ 
5 - 128/5*(2*c*d*x + b*d)^(5/2)*a*c^2*d^5 + 16/9*(2*c*d*x + b*d)^(9/2)*c*d 
^3 - 13*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*b^4*c*d^7 - 8*sqrt(2)*(-b^2* 
d^2 + 4*a*c*d^2)^(1/4)*a*b^2*c^2*d^7 + 16*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^( 
1/4)*a^2*c^3*d^7)*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)) - 13*(sqrt(2)*(-b^ 
2*d^2 + 4*a*c*d^2)^(1/4)*b^4*c*d^7 - 8*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4 
)*a*b^2*c^2*d^7 + 16*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*a^2*c^3*d^7)*arc 
tan(-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)) - 13/2*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2) 
^(1/4)*b^4*c*d^7 - 8*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*a*b^2*c^2*d^7 + 
16*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*a^2*c^3*d^7)*log(2*c*d*x + b*d + s 
qrt(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)) + 13/2*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*b^4*c*d^7 - 8*sqr 
t(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*a*b^2*c^2*d^7 + 16*sqrt(2)*(-b^2*d^2 + 4 
*a*c*d^2)^(1/4)*a^2*c^3*d^7)*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)) + 4*(sqrt(2* 
c*d*x + b*d)*b^6*c*d^9 - 12*sqrt(2*c*d*x + b*d)*a*b^4*c^2*d^9 + 48*sqrt(2* 
c*d*x + b*d)*a^2*b^2*c^3*d^9 - 64*sqrt(2*c*d*x + b*d)*a^3*c^4*d^9)/(b^2...
 

Mupad [B] (verification not implemented)

Time = 5.36 (sec) , antiderivative size = 1060, normalized size of antiderivative = 5.00 \[ \int \frac {(b d+2 c d x)^{15/2}}{\left (a+b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:

int((b*d + 2*c*d*x)^(15/2)/(a + b*x + c*x^2)^2,x)
 

Output:

(16*c*d^3*(b*d + 2*c*d*x)^(9/2))/9 - ((b*d + 2*c*d*x)^(1/2)*(4*b^6*c*d^9 - 
 256*a^3*c^4*d^9 - 48*a*b^4*c^2*d^9 + 192*a^2*b^2*c^3*d^9))/((b*d + 2*c*d* 
x)^2 - b^2*d^2 + 4*a*c*d^2) + 48*c*d^7*(b*d + 2*c*d*x)^(1/2)*(4*a*c - b^2) 
^2 - 26*c*d^(15/2)*atan((b^4*(b*d + 2*c*d*x)^(1/2) + 16*a^2*c^2*(b*d + 2*c 
*d*x)^(1/2) - 8*a*b^2*c*(b*d + 2*c*d*x)^(1/2))/(d^(1/2)*(b^2 - 4*a*c)^(9/4 
)))*(b^2 - 4*a*c)^(9/4) - c*d^(15/2)*atan((c*d^(15/2)*(b^2 - 4*a*c)^(9/4)* 
((b*d + 2*c*d*x)^(1/2)*(44302336*a^6*c^8*d^18 + 10816*b^12*c^2*d^18 - 2595 
84*a*b^10*c^3*d^18 + 2595840*a^2*b^8*c^4*d^18 - 13844480*a^3*b^6*c^5*d^18 
+ 41533440*a^4*b^4*c^6*d^18 - 66453504*a^5*b^2*c^7*d^18) - 13*c*d^(15/2)*( 
b^2 - 4*a*c)^(9/4)*(832*b^8*c*d^11 + 212992*a^4*c^5*d^11 - 13312*a*b^6*c^2 
*d^11 + 79872*a^2*b^4*c^3*d^11 - 212992*a^3*b^2*c^4*d^11))*13i + c*d^(15/2 
)*(b^2 - 4*a*c)^(9/4)*((b*d + 2*c*d*x)^(1/2)*(44302336*a^6*c^8*d^18 + 1081 
6*b^12*c^2*d^18 - 259584*a*b^10*c^3*d^18 + 2595840*a^2*b^8*c^4*d^18 - 1384 
4480*a^3*b^6*c^5*d^18 + 41533440*a^4*b^4*c^6*d^18 - 66453504*a^5*b^2*c^7*d 
^18) + 13*c*d^(15/2)*(b^2 - 4*a*c)^(9/4)*(832*b^8*c*d^11 + 212992*a^4*c^5* 
d^11 - 13312*a*b^6*c^2*d^11 + 79872*a^2*b^4*c^3*d^11 - 212992*a^3*b^2*c^4* 
d^11))*13i)/(13*c*d^(15/2)*(b^2 - 4*a*c)^(9/4)*((b*d + 2*c*d*x)^(1/2)*(443 
02336*a^6*c^8*d^18 + 10816*b^12*c^2*d^18 - 259584*a*b^10*c^3*d^18 + 259584 
0*a^2*b^8*c^4*d^18 - 13844480*a^3*b^6*c^5*d^18 + 41533440*a^4*b^4*c^6*d^18 
 - 66453504*a^5*b^2*c^7*d^18) - 13*c*d^(15/2)*(b^2 - 4*a*c)^(9/4)*(832*...
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 2640, normalized size of antiderivative = 12.45 \[ \int \frac {(b d+2 c d x)^{15/2}}{\left (a+b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:

int((2*c*d*x+b*d)^(15/2)/(c*x^2+b*x+a)^2,x)
 

Output:

(sqrt(d)*d**7*(18720*(4*a*c - b**2)**(1/4)*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**3*c* 
*3 - 9360*(4*a*c - b**2)**(1/4)*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**2*b**2*c**2 + 1 
8720*(4*a*c - b**2)**(1/4)*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**2*b*c**3*x + 18720*( 
4*a*c - b**2)**(1/4)*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**2*c**4*x**2 + 1170*(4*a*c 
- b**2)**(1/4)*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*b**4*c - 9360*(4*a*c - b**2)**(1/ 
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*b**3*c**2*x - 9360*(4*a*c - b**2)**(1/4)*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*b**2*c**3*x**2 + 1170*(4*a*c - b**2)**(1/4)*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**5*c*x + 1170*(4*a*c - b**2)**(1/4)*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**4*c**2*x**2 - 18720*(4*a*c - b**2)**(1/4)*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 
**3*c**3 + 9360*(4*a*c - b**2)**(1/4)*sqrt(2)*atan(((4*a*c - b**2)**(1/...