Integrand size = 26, antiderivative size = 152 \[ \int \frac {(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^2} \, dx=\frac {28}{3} c d^3 (b d+2 c d x)^{3/2}-\frac {d (b d+2 c d x)^{7/2}}{a+b x+c x^2}+14 c \left (b^2-4 a c\right )^{3/4} d^{9/2} \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-14 c \left (b^2-4 a c\right )^{3/4} d^{9/2} \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ) \]
28/3*c*d^3*(2*c*d*x+b*d)^(3/2)-d*(2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)+14*c*(- 4*a*c+b^2)^(3/4)*d^(9/2)*arctan((d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^( 1/2))-14*c*(-4*a*c+b^2)^(3/4)*d^(9/2)*arctanh((d*(2*c*x+b))^(1/2)/(-4*a*c+ b^2)^(1/4)/d^(1/2))
Result contains complex when optimal does not.
Time = 0.68 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.67 \[ \int \frac {(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^2} \, dx=\left (\frac {1}{3}+\frac {i}{3}\right ) c (d (b+2 c x))^{9/2} \left (\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (-7 b^2+28 a c+4 (b+2 c x)^2\right )}{c (b+2 c x)^3 (a+x (b+c x))}-\frac {21 \left (b^2-4 a c\right )^{3/4} \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{(b+2 c x)^{9/2}}+\frac {21 \left (b^2-4 a c\right )^{3/4} \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{(b+2 c x)^{9/2}}-\frac {21 \left (b^2-4 a c\right )^{3/4} \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )}{(b+2 c x)^{9/2}}\right ) \]
(1/3 + I/3)*c*(d*(b + 2*c*x))^(9/2)*(((1/2 - I/2)*(-7*b^2 + 28*a*c + 4*(b + 2*c*x)^2))/(c*(b + 2*c*x)^3*(a + x*(b + c*x))) - (21*(b^2 - 4*a*c)^(3/4) *ArcTan[1 - ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/(b + 2*c*x)^(9 /2) + (21*(b^2 - 4*a*c)^(3/4)*ArcTan[1 + ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/(b + 2*c*x)^(9/2) - (21*(b^2 - 4*a*c)^(3/4)*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)^(9/2))
Time = 0.36 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.14, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {1110, 1116, 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.
| \(\displaystyle \int \frac {(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 1110 |
| \(\displaystyle 7 c d^2 \int \frac {(b d+2 c x d)^{5/2}}{c x^2+b x+a}dx-\frac {d (b d+2 c d x)^{7/2}}{a+b x+c x^2}\) |
| \(\Big \downarrow \) 1116 |
| \(\displaystyle 7 c d^2 \left (d^2 \left (b^2-4 a c\right ) \int \frac {\sqrt {b d+2 c x d}}{c x^2+b x+a}dx+\frac {4}{3} d (b d+2 c d x)^{3/2}\right )-\frac {d (b d+2 c d x)^{7/2}}{a+b x+c x^2}\) |
| \(\Big \downarrow \) 1118 |
| \(\displaystyle 7 c d^2 \left (\frac {d \left (b^2-4 a c\right ) \int \frac {4 c d^2 \sqrt {b d+2 c x d}}{\left (4 a-\frac {b^2}{c}\right ) c d^2+(b d+2 c x d)^2}d(b d+2 c x d)}{2 c}+\frac {4}{3} d (b d+2 c d x)^{3/2}\right )-\frac {d (b d+2 c d x)^{7/2}}{a+b x+c x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 7 c d^2 \left (2 d^3 \left (b^2-4 a c\right ) \int -\frac {\sqrt {b d+2 c x d}}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d(b d+2 c x d)+\frac {4}{3} d (b d+2 c d x)^{3/2}\right )-\frac {d (b d+2 c d x)^{7/2}}{a+b x+c x^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle 7 c d^2 \left (\frac {4}{3} d (b d+2 c d x)^{3/2}-2 d^3 \left (b^2-4 a c\right ) \int \frac {\sqrt {b d+2 c x d}}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d(b d+2 c x d)\right )-\frac {d (b d+2 c d x)^{7/2}}{a+b x+c x^2}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle 7 c d^2 \left (\frac {4}{3} d (b d+2 c d x)^{3/2}-4 d^3 \left (b^2-4 a c\right ) \int \frac {b d+2 c x d}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d\sqrt {b d+2 c x d}\right )-\frac {d (b d+2 c d x)^{7/2}}{a+b x+c x^2}\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle 7 c d^2 \left (\frac {4}{3} d (b d+2 c d x)^{3/2}-4 d^3 \left (b^2-4 a c\right ) \left (\frac {1}{2} \int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}-\frac {1}{2} \int \frac {1}{b d+2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}\right )\right )-\frac {d (b d+2 c d x)^{7/2}}{a+b x+c x^2}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle 7 c d^2 \left (\frac {4}{3} d (b d+2 c d x)^{3/2}-4 d^3 \left (b^2-4 a c\right ) \left (\frac {1}{2} \int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}-\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 \sqrt {d} \sqrt [4]{b^2-4 a c}}\right )\right )-\frac {d (b d+2 c d x)^{7/2}}{a+b x+c x^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle 7 c d^2 \left (\frac {4}{3} d (b d+2 c d x)^{3/2}-4 d^3 \left (b^2-4 a c\right ) \left (\frac {\text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 \sqrt {d} \sqrt [4]{b^2-4 a c}}-\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 \sqrt {d} \sqrt [4]{b^2-4 a c}}\right )\right )-\frac {d (b d+2 c d x)^{7/2}}{a+b x+c x^2}\) |
-((d*(b*d + 2*c*d*x)^(7/2))/(a + b*x + c*x^2)) + 7*c*d^2*((4*d*(b*d + 2*c* d*x)^(3/2))/3 - 4*(b^2 - 4*a*c)*d^3*(-1/2*ArcTan[Sqrt[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]) ))
3.13.98.3.1 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[(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]
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[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])
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. \(314\) vs. \(2(126)=252\).
Time = 2.68 (sec) , antiderivative size = 315, normalized size of antiderivative = 2.07
| method | result | size |
| derivativedivides | \(16 c \,d^{3} \left (\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{3}-d^{2} \left (\frac {\left (-\frac {a c}{4}+\frac {b^{2}}{16}\right ) \left (2 c d x +b d \right )^{\frac {3}{2}}}{a c \,d^{2}-\frac {b^{2} d^{2}}{4}+\frac {\left (2 c d x +b d \right )^{2}}{4}}+\frac {\left (7 a c -\frac {7 b^{2}}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}\right )\right )\) | \(315\) |
| default | \(16 c \,d^{3} \left (\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{3}-d^{2} \left (\frac {\left (-\frac {a c}{4}+\frac {b^{2}}{16}\right ) \left (2 c d x +b d \right )^{\frac {3}{2}}}{a c \,d^{2}-\frac {b^{2} d^{2}}{4}+\frac {\left (2 c d x +b d \right )^{2}}{4}}+\frac {\left (7 a c -\frac {7 b^{2}}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}\right )\right )\) | \(315\) |
| pseudoelliptic | \(\frac {56 \left (\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \left (\frac {4 c^{2} x^{2}}{7}+\left (\frac {4 b x}{7}+a \right ) c -\frac {3 b^{2}}{28}\right ) \left (d \left (2 c x +b \right )\right )^{\frac {3}{2}}-\frac {3 \sqrt {2}\, c \,d^{2} \left (c \,x^{2}+b x +a \right ) \left (-\frac {b^{2}}{4}+a c \right ) \left (2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )+\ln \left (\frac {\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}\right )-2 \arctan \left (\frac {-\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )\right )}{2}\right ) d^{3}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \left (6 c \,x^{2}+6 b x +6 a \right )}\) | \(335\) |
16*c*d^3*(1/3*(2*c*d*x+b*d)^(3/2)-d^2*((-1/4*a*c+1/16*b^2)*(2*c*d*x+b*d)^( 3/2)/(a*c*d^2-1/4*b^2*d^2+1/4*(2*c*d*x+b*d)^2)+1/8*(7*a*c-7/4*b^2)/(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))))
Result contains complex when optimal does not.
Time = 0.28 (sec) , antiderivative size = 659, normalized size of antiderivative = 4.34 \[ \int \frac {(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^2} \, dx=-\frac {21 \, \left ({\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{18}\right )^{\frac {1}{4}} {\left (c x^{2} + b x + a\right )} \log \left (343 \, {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} \sqrt {2 \, c d x + b d} d^{13} + 343 \, \left ({\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{18}\right )^{\frac {3}{4}}\right ) + 21 \, \left ({\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{18}\right )^{\frac {1}{4}} {\left (-i \, c x^{2} - i \, b x - i \, a\right )} \log \left (343 \, {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} \sqrt {2 \, c d x + b d} d^{13} + 343 i \, \left ({\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{18}\right )^{\frac {3}{4}}\right ) + 21 \, \left ({\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{18}\right )^{\frac {1}{4}} {\left (i \, c x^{2} + i \, b x + i \, a\right )} \log \left (343 \, {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} \sqrt {2 \, c d x + b d} d^{13} - 343 i \, \left ({\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{18}\right )^{\frac {3}{4}}\right ) - 21 \, \left ({\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{18}\right )^{\frac {1}{4}} {\left (c x^{2} + b x + a\right )} \log \left (343 \, {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} \sqrt {2 \, c d x + b d} d^{13} - 343 \, \left ({\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{18}\right )^{\frac {3}{4}}\right ) - {\left (32 \, c^{3} d^{4} x^{3} + 48 \, b c^{2} d^{4} x^{2} + 2 \, {\left (5 \, b^{2} c + 28 \, a c^{2}\right )} d^{4} x - {\left (3 \, b^{3} - 28 \, a b c\right )} d^{4}\right )} \sqrt {2 \, c d x + b d}}{3 \, {\left (c x^{2} + b x + a\right )}} \]
-1/3*(21*((b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)^(1/ 4)*(c*x^2 + b*x + a)*log(343*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*sqrt(2*c *d*x + b*d)*d^13 + 343*((b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3* c^7)*d^18)^(3/4)) + 21*((b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3* c^7)*d^18)^(1/4)*(-I*c*x^2 - I*b*x - I*a)*log(343*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*sqrt(2*c*d*x + b*d)*d^13 + 343*I*((b^6*c^4 - 12*a*b^4*c^5 + 4 8*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)^(3/4)) + 21*((b^6*c^4 - 12*a*b^4*c^5 + 4 8*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)^(1/4)*(I*c*x^2 + I*b*x + I*a)*log(343*(b ^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*sqrt(2*c*d*x + b*d)*d^13 - 343*I*((b^6* c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)^(3/4)) - 21*((b^6* c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)^(1/4)*(c*x^2 + b*x + a)*log(343*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*sqrt(2*c*d*x + b*d)*d^1 3 - 343*((b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)^(3/4 )) - (32*c^3*d^4*x^3 + 48*b*c^2*d^4*x^2 + 2*(5*b^2*c + 28*a*c^2)*d^4*x - ( 3*b^3 - 28*a*b*c)*d^4)*sqrt(2*c*d*x + b*d))/(c*x^2 + b*x + a)
Timed out. \[ \int \frac {(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^2} \, dx=\text {Exception raised: ValueError} \]
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. 441 vs. \(2 (126) = 252\).
Time = 0.32 (sec) , antiderivative size = 441, normalized size of antiderivative = 2.90 \[ \int \frac {(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^2} \, dx=-7 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c d^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right ) - 7 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c d^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right ) + \frac {7}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c d^{3} \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right ) - \frac {7}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c d^{3} \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right ) + \frac {16}{3} \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} c d^{3} + \frac {4 \, {\left ({\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{2} c d^{5} - 4 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} a c^{2} d^{5}\right )}}{b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}} \]
-7*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*d^3*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)) - 7*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*d^3*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)) + 7/2*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*d^3*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)) - 7/2*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c *d^3*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)) + 16/3*(2*c*d*x + b*d)^(3/2)*c*d^3 + 4*((2*c*d*x + b*d)^(3/2)*b^2*c*d^5 - 4*(2*c*d*x + b*d)^(3/2)*a*c^2*d^5)/( b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2)
Time = 9.30 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.09 \[ \int \frac {(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^2} \, dx=\frac {16\,c\,d^3\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}}{3}+\frac {{\left (b\,d+2\,c\,d\,x\right )}^{3/2}\,\left (16\,a\,c^2\,d^5-4\,b^2\,c\,d^5\right )}{{\left (b\,d+2\,c\,d\,x\right )}^2-b^2\,d^2+4\,a\,c\,d^2}+14\,c\,d^{9/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4}+c\,d^{9/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,1{}\mathrm {i}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4}\,14{}\mathrm {i} \]
(16*c*d^3*(b*d + 2*c*d*x)^(3/2))/3 + ((b*d + 2*c*d*x)^(3/2)*(16*a*c^2*d^5 - 4*b^2*c*d^5))/((b*d + 2*c*d*x)^2 - b^2*d^2 + 4*a*c*d^2) + 14*c*d^(9/2)*a tan((b*d + 2*c*d*x)^(1/2)/(d^(1/2)*(b^2 - 4*a*c)^(1/4)))*(b^2 - 4*a*c)^(3/ 4) + c*d^(9/2)*atan(((b*d + 2*c*d*x)^(1/2)*1i)/(d^(1/2)*(b^2 - 4*a*c)^(1/4 )))*(b^2 - 4*a*c)^(3/4)*14i