Integrand size = 26, antiderivative size = 131 \[ \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^2} \, dx=-\frac {d \sqrt {b d+2 c d x}}{a+b x+c x^2}-\frac {2 c d^{3/2} \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{3/4}}-\frac {2 c d^{3/2} \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{3/4}} \]
-2*c*d^(3/2)*arctan((d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))/(-4*a* c+b^2)^(3/4)-2*c*d^(3/2)*arctanh((d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^ (1/2))/(-4*a*c+b^2)^(3/4)-d*(2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)
Result contains complex when optimal does not.
Time = 0.56 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.76 \[ \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^2} \, dx=-\frac {i (d (b+2 c x))^{3/2} \left ((1+i) c (a+x (b+c x)) \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-i \left (\left (b^2-4 a c\right )^{3/4} \sqrt {b+2 c x}+(1-i) c (a+x (b+c x)) \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )\right )-(1+i) c (a+x (b+c x)) \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )\right )}{\left (b^2-4 a c\right )^{3/4} (b+2 c x)^{3/2} (a+x (b+c x))} \]
((-I)*(d*(b + 2*c*x))^(3/2)*((1 + I)*c*(a + x*(b + c*x))*ArcTan[1 - ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)] - I*((b^2 - 4*a*c)^(3/4)*Sqrt[b + 2*c*x] + (1 - I)*c*(a + x*(b + c*x))*ArcTan[1 + ((1 + I)*Sqrt[b + 2*c*x]) /(b^2 - 4*a*c)^(1/4)]) - (1 + I)*c*(a + x*(b + c*x))*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 - 4*a*c)^(3/4)*(b + 2*c*x)^(3/2)*(a + x*(b + c*x)))
Time = 0.30 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.08, 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, 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.
| \(\displaystyle \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 1110 |
| \(\displaystyle c d^2 \int \frac {1}{\sqrt {b d+2 c x d} \left (c x^2+b x+a\right )}dx-\frac {d \sqrt {b d+2 c d x}}{a+b x+c x^2}\) |
| \(\Big \downarrow \) 1118 |
| \(\displaystyle \frac {1}{2} d \int \frac {4 c d^2}{\sqrt {b d+2 c x d} \left (\left (4 a-\frac {b^2}{c}\right ) c d^2+(b d+2 c x d)^2\right )}d(b d+2 c x d)-\frac {d \sqrt {b d+2 c d x}}{a+b x+c x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 2 c d^3 \int -\frac {1}{\sqrt {b d+2 c x d} \left (\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2\right )}d(b d+2 c x d)-\frac {d \sqrt {b d+2 c d x}}{a+b x+c x^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -2 c d^3 \int \frac {1}{\sqrt {b d+2 c x d} \left (\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2\right )}d(b d+2 c x d)-\frac {d \sqrt {b d+2 c d x}}{a+b x+c x^2}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle -4 c d^3 \int \frac {1}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d\sqrt {b d+2 c x d}-\frac {d \sqrt {b d+2 c d x}}{a+b x+c x^2}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle -4 c d^3 \left (\frac {\int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}}{2 d \sqrt {b^2-4 a c}}+\frac {\int \frac {1}{b d+2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}}{2 d \sqrt {b^2-4 a c}}\right )-\frac {d \sqrt {b d+2 c d x}}{a+b x+c x^2}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -4 c d^3 \left (\frac {\int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}}{2 d \sqrt {b^2-4 a c}}+\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 d^{3/2} \left (b^2-4 a c\right )^{3/4}}\right )-\frac {d \sqrt {b d+2 c d x}}{a+b x+c x^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -4 c d^3 \left (\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 d^{3/2} \left (b^2-4 a c\right )^{3/4}}+\frac {\text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 d^{3/2} \left (b^2-4 a c\right )^{3/4}}\right )-\frac {d \sqrt {b d+2 c d x}}{a+b x+c x^2}\) |
-((d*Sqrt[b*d + 2*c*d*x])/(a + b*x + c*x^2)) - 4*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)))
3.14.1.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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])
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]
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]
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]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(276\) vs. \(2(109)=218\).
Time = 3.10 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.11
| method | result | size |
| derivativedivides | \(16 c \,d^{3} \left (-\frac {\sqrt {2 c d x +b d}}{16 \left (a c \,d^{2}-\frac {b^{2} d^{2}}{4}+\frac {\left (2 c d x +b d \right )^{2}}{4}\right )}+\frac {\sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{32 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}\right )\) | \(277\) |
| default | \(16 c \,d^{3} \left (-\frac {\sqrt {2 c d x +b d}}{16 \left (a c \,d^{2}-\frac {b^{2} d^{2}}{4}+\frac {\left (2 c d x +b d \right )^{2}}{4}\right )}+\frac {\sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{32 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}\right )\) | \(277\) |
| pseudoelliptic | \(\frac {d \left (-2 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}} \sqrt {d \left (2 c x +b \right )}+\sqrt {2}\, d^{2} \left (c \,x^{2}+b x +a \right ) c \left (2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )+\ln \left (\frac {\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}{\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}\right )-2 \arctan \left (\frac {-\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )\right )\right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}} \left (2 c \,x^{2}+2 b x +2 a \right )}\) | \(301\) |
16*c*d^3*(-1/16*(2*c*d*x+b*d)^(1/2)/(a*c*d^2-1/4*b^2*d^2+1/4*(2*c*d*x+b*d) ^2)+1/32/(4*a*c*d^2-b^2*d^2)^(3/4)*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)))
Result contains complex when optimal does not.
Time = 0.28 (sec) , antiderivative size = 516, normalized size of antiderivative = 3.94 \[ \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^2} \, dx=-\frac {\left (\frac {c^{4} d^{6}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac {1}{4}} {\left (c x^{2} + b x + a\right )} \log \left (\sqrt {2 \, c d x + b d} c d + \left (\frac {c^{4} d^{6}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac {1}{4}} {\left (b^{2} - 4 \, a c\right )}\right ) - \left (\frac {c^{4} d^{6}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac {1}{4}} {\left (c x^{2} + b x + a\right )} \log \left (\sqrt {2 \, c d x + b d} c d - \left (\frac {c^{4} d^{6}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac {1}{4}} {\left (b^{2} - 4 \, a c\right )}\right ) - \left (\frac {c^{4} d^{6}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac {1}{4}} {\left (-i \, c x^{2} - i \, b x - i \, a\right )} \log \left (\sqrt {2 \, c d x + b d} c d + \left (\frac {c^{4} d^{6}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac {1}{4}} {\left (i \, b^{2} - 4 i \, a c\right )}\right ) - \left (\frac {c^{4} d^{6}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac {1}{4}} {\left (i \, c x^{2} + i \, b x + i \, a\right )} \log \left (\sqrt {2 \, c d x + b d} c d + \left (\frac {c^{4} d^{6}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac {1}{4}} {\left (-i \, b^{2} + 4 i \, a c\right )}\right ) + \sqrt {2 \, c d x + b d} d}{c x^{2} + b x + a} \]
-((c^4*d^6/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))^(1/4)*(c*x^2 + b*x + a)*log(sqrt(2*c*d*x + b*d)*c*d + (c^4*d^6/(b^6 - 12*a*b^4*c + 48*a ^2*b^2*c^2 - 64*a^3*c^3))^(1/4)*(b^2 - 4*a*c)) - (c^4*d^6/(b^6 - 12*a*b^4* c + 48*a^2*b^2*c^2 - 64*a^3*c^3))^(1/4)*(c*x^2 + b*x + a)*log(sqrt(2*c*d*x + b*d)*c*d - (c^4*d^6/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))^( 1/4)*(b^2 - 4*a*c)) - (c^4*d^6/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3 *c^3))^(1/4)*(-I*c*x^2 - I*b*x - I*a)*log(sqrt(2*c*d*x + b*d)*c*d + (c^4*d ^6/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))^(1/4)*(I*b^2 - 4*I*a* c)) - (c^4*d^6/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))^(1/4)*(I* c*x^2 + I*b*x + I*a)*log(sqrt(2*c*d*x + b*d)*c*d + (c^4*d^6/(b^6 - 12*a*b^ 4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))^(1/4)*(-I*b^2 + 4*I*a*c)) + sqrt(2*c*d *x + b*d)*d)/(c*x^2 + b*x + a)
Timed out. \[ \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^2} \, dx=\text {Exception raised: ValueError} \]
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
Leaf count of result is larger than twice the leaf count of optimal. 438 vs. \(2 (109) = 218\).
Time = 0.30 (sec) , antiderivative size = 438, normalized size of antiderivative = 3.34 \[ \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^2} \, dx=\frac {4 \, \sqrt {2 \, c d x + b d} c d^{3}}{b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}} - \frac {\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c d \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{2} - 4 \, a c} - \frac {\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c d \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{2} - 4 \, a c} - \frac {{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c d \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{2} - 4 \, \sqrt {2} a c} + \frac {{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c d \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{2} - 4 \, \sqrt {2} a c} \]
4*sqrt(2*c*d*x + b*d)*c*d^3/(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2) - sq rt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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) - sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c*d*arctan(-1/2*sq rt(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) - (-b^2*d^2 + 4*a*c*d^2)^(1/4)*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) + (-b^2*d^2 + 4*a*c*d^2)^(1/4)*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)
Time = 9.22 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.72 \[ \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^2} \, dx=-\frac {4\,c\,d^3\,\sqrt {b\,d+2\,c\,d\,x}}{{\left (b\,d+2\,c\,d\,x\right )}^2-b^2\,d^2+4\,a\,c\,d^2}-\frac {2\,c\,d^{3/2}\,\mathrm {atan}\left (\frac {128\,c^3\,d^{15/2}\,\sqrt {b\,d+2\,c\,d\,x}}{\left (\frac {128\,b^2\,c^3\,d^8}{{\left (b^2-4\,a\,c\right )}^{3/2}}-\frac {512\,a\,c^4\,d^8}{{\left (b^2-4\,a\,c\right )}^{3/2}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{3/4}}-\frac {2\,c\,d^{3/2}\,\mathrm {atanh}\left (\frac {128\,c^3\,d^{15/2}\,\sqrt {b\,d+2\,c\,d\,x}}{\left (\frac {128\,b^2\,c^3\,d^8}{{\left (b^2-4\,a\,c\right )}^{3/2}}-\frac {512\,a\,c^4\,d^8}{{\left (b^2-4\,a\,c\right )}^{3/2}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{3/4}} \]
- (4*c*d^3*(b*d + 2*c*d*x)^(1/2))/((b*d + 2*c*d*x)^2 - b^2*d^2 + 4*a*c*d^2 ) - (2*c*d^(3/2)*atan((128*c^3*d^(15/2)*(b*d + 2*c*d*x)^(1/2))/(((128*b^2* c^3*d^8)/(b^2 - 4*a*c)^(3/2) - (512*a*c^4*d^8)/(b^2 - 4*a*c)^(3/2))*(b^2 - 4*a*c)^(3/4))))/(b^2 - 4*a*c)^(3/4) - (2*c*d^(3/2)*atanh((128*c^3*d^(15/2 )*(b*d + 2*c*d*x)^(1/2))/(((128*b^2*c^3*d^8)/(b^2 - 4*a*c)^(3/2) - (512*a* c^4*d^8)/(b^2 - 4*a*c)^(3/2))*(b^2 - 4*a*c)^(3/4))))/(b^2 - 4*a*c)^(3/4)