Integrand size = 26, antiderivative size = 227 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^3} \, dx=\frac {154 c^2}{3 \left (b^2-4 a c\right )^3 d (b d+2 c d x)^{3/2}}-\frac {1}{2 \left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2}+\frac {11 c}{2 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )}-\frac {77 c^2 \arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{15/4} d^{5/2}}-\frac {77 c^2 \text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{15/4} d^{5/2}} \] Output:
154/3*c^2/(-4*a*c+b^2)^3/d/(2*c*d*x+b*d)^(3/2)-1/2/(-4*a*c+b^2)/d/(2*c*d*x +b*d)^(3/2)/(c*x^2+b*x+a)^2+11/2*c/(-4*a*c+b^2)^2/d/(2*c*d*x+b*d)^(3/2)/(c *x^2+b*x+a)-77*c^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)^(15/4)/d^(5/2)-77*c^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)^(15/4)/d^(5/2)
Result contains complex when optimal does not.
Time = 1.65 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.35 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^3} \, dx=\frac {\left (\frac {1}{6}+\frac {i}{6}\right ) c^2 \left (\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) (b+2 c x) \left (32 b^4-256 a b^2 c+512 a^2 c^2-121 b^2 (b+2 c x)^2+484 a c (b+2 c x)^2+77 (b+2 c x)^4\right )}{c^2 \left (b^2-4 a c\right )^3 (a+x (b+c x))^2}-\frac {231 i (b+2 c x)^{5/2} \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{15/4}}+\frac {231 i (b+2 c x)^{5/2} \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{15/4}}+\frac {231 i (b+2 c x)^{5/2} \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )}{\left (b^2-4 a c\right )^{15/4}}\right )}{(d (b+2 c x))^{5/2}} \] Input:
Integrate[1/((b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)^3),x]
Output:
((1/6 + I/6)*c^2*(((1/8 - I/8)*(b + 2*c*x)*(32*b^4 - 256*a*b^2*c + 512*a^2 *c^2 - 121*b^2*(b + 2*c*x)^2 + 484*a*c*(b + 2*c*x)^2 + 77*(b + 2*c*x)^4))/ (c^2*(b^2 - 4*a*c)^3*(a + x*(b + c*x))^2) - ((231*I)*(b + 2*c*x)^(5/2)*Arc Tan[1 - ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/(b^2 - 4*a*c)^(15/ 4) + ((231*I)*(b + 2*c*x)^(5/2)*ArcTan[1 + ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/(b^2 - 4*a*c)^(15/4) + ((231*I)*(b + 2*c*x)^(5/2)*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)^(15/4)))/(d*(b + 2*c*x))^(5/2)
Time = 0.45 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.17, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {1111, 1111, 1117, 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.
| \(\displaystyle \int \frac {1}{\left (a+b x+c x^2\right )^3 (b d+2 c d x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1111 |
| \(\displaystyle -\frac {11 c \int \frac {1}{(b d+2 c x d)^{5/2} \left (c x^2+b x+a\right )^2}dx}{2 \left (b^2-4 a c\right )}-\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2 (b d+2 c d x)^{3/2}}\) |
| \(\Big \downarrow \) 1111 |
| \(\displaystyle -\frac {11 c \left (-\frac {7 c \int \frac {1}{(b d+2 c x d)^{5/2} \left (c x^2+b x+a\right )}dx}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{3/2}}\right )}{2 \left (b^2-4 a c\right )}-\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2 (b d+2 c d x)^{3/2}}\) |
| \(\Big \downarrow \) 1117 |
| \(\displaystyle -\frac {11 c \left (-\frac {7 c \left (\frac {\int \frac {1}{\sqrt {b d+2 c x d} \left (c x^2+b x+a\right )}dx}{d^2 \left (b^2-4 a c\right )}+\frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{3/2}}\right )}{2 \left (b^2-4 a c\right )}-\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2 (b d+2 c d x)^{3/2}}\) |
| \(\Big \downarrow \) 1118 |
| \(\displaystyle -\frac {11 c \left (-\frac {7 c \left (\frac {\int \frac {4 c d^2}{\sqrt {b d+2 c x d} \left (\left (4 a-\frac {b^2}{c}\right ) c d^2+(b d+2 c x d)^2\right )}d(b d+2 c x d)}{2 c d^3 \left (b^2-4 a c\right )}+\frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{3/2}}\right )}{2 \left (b^2-4 a c\right )}-\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2 (b d+2 c d x)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {11 c \left (-\frac {7 c \left (\frac {2 \int -\frac {1}{\sqrt {b d+2 c x d} \left (\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2\right )}d(b d+2 c x d)}{d \left (b^2-4 a c\right )}+\frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{3/2}}\right )}{2 \left (b^2-4 a c\right )}-\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2 (b d+2 c d x)^{3/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {11 c \left (-\frac {7 c \left (\frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}-\frac {2 \int \frac {1}{\sqrt {b d+2 c x d} \left (\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2\right )}d(b d+2 c x d)}{d \left (b^2-4 a c\right )}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{3/2}}\right )}{2 \left (b^2-4 a c\right )}-\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2 (b d+2 c d x)^{3/2}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle -\frac {11 c \left (-\frac {7 c \left (\frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}-\frac {4 \int \frac {1}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d\sqrt {b d+2 c x d}}{d \left (b^2-4 a c\right )}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{3/2}}\right )}{2 \left (b^2-4 a c\right )}-\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2 (b d+2 c d x)^{3/2}}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle -\frac {11 c \left (-\frac {7 c \left (\frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}-\frac {4 \left (\frac {\int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}}{2 d \sqrt {b^2-4 a c}}+\frac {\int \frac {1}{b d+2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}}{2 d \sqrt {b^2-4 a c}}\right )}{d \left (b^2-4 a c\right )}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{3/2}}\right )}{2 \left (b^2-4 a c\right )}-\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2 (b d+2 c d x)^{3/2}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {11 c \left (-\frac {7 c \left (\frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}-\frac {4 \left (\frac {\int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}}{2 d \sqrt {b^2-4 a c}}+\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 d^{3/2} \left (b^2-4 a c\right )^{3/4}}\right )}{d \left (b^2-4 a c\right )}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{3/2}}\right )}{2 \left (b^2-4 a c\right )}-\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2 (b d+2 c d x)^{3/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {11 c \left (-\frac {7 c \left (\frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}-\frac {4 \left (\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 d^{3/2} \left (b^2-4 a c\right )^{3/4}}+\frac {\text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 d^{3/2} \left (b^2-4 a c\right )^{3/4}}\right )}{d \left (b^2-4 a c\right )}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{3/2}}\right )}{2 \left (b^2-4 a c\right )}-\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2 (b d+2 c d x)^{3/2}}\) |
Input:
Int[1/((b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)^3),x]
Output:
-1/2*1/((b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)^2) - (11*c *(-(1/((b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2))) - (7*c*(4 /(3*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(3/2)) - (4*(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))))/((b^2 - 4*a*c)*d)))/(b^2 - 4*a*c)))/(2*(b^2 - 4*a*c))
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_S ymbol] :> Simp[2*c*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(e*(p + 1)* (b^2 - 4*a*c))), x] - Simp[2*c*e*((m + 2*p + 3)/(e*(p + 1)*(b^2 - 4*a*c))) Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e , m}, x] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[p, -1] && !G tQ[m, 1] && RationalQ[m] && IntegerQ[2*p]
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])
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]
Time = 1.43 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.56
| method | result | size |
| derivativedivides | \(64 d^{5} c^{2} \left (-\frac {1}{3 d^{6} \left (4 a c -b^{2}\right )^{3} \left (2 c d x +b d \right )^{\frac {3}{2}}}-\frac {\frac {\frac {15 \left (2 c d x +b d \right )^{\frac {5}{2}}}{32}+16 \left (\frac {19}{128} a \,d^{2} c -\frac {19}{512} b^{2} d^{2}\right ) \sqrt {2 c d x +b d}}{\left (\left (2 c d x +b d \right )^{2}+4 a \,d^{2} c -b^{2} d^{2}\right )^{2}}+\frac {77 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{256 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {3}{4}}}}{d^{6} \left (4 a c -b^{2}\right )^{3}}\right )\) | \(355\) |
| default | \(64 d^{5} c^{2} \left (-\frac {1}{3 d^{6} \left (4 a c -b^{2}\right )^{3} \left (2 c d x +b d \right )^{\frac {3}{2}}}-\frac {\frac {\frac {15 \left (2 c d x +b d \right )^{\frac {5}{2}}}{32}+16 \left (\frac {19}{128} a \,d^{2} c -\frac {19}{512} b^{2} d^{2}\right ) \sqrt {2 c d x +b d}}{\left (\left (2 c d x +b d \right )^{2}+4 a \,d^{2} c -b^{2} d^{2}\right )^{2}}+\frac {77 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{256 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {3}{4}}}}{d^{6} \left (4 a c -b^{2}\right )^{3}}\right )\) | \(355\) |
| pseudoelliptic | \(\frac {32 c^{2} d^{5} \left (-\frac {2}{3 d^{6} \left (d \left (2 c x +b \right )\right )^{\frac {3}{2}}}-\frac {15 \left (d \left (2 c x +b \right )\right )^{\frac {5}{2}}}{256 c^{2} d^{10} \left (c \,x^{2}+b x +a \right )^{2}}-\frac {19 \sqrt {d \left (2 c x +b \right )}\, a}{64 c \,d^{8} \left (c \,x^{2}+b x +a \right )^{2}}+\frac {19 \sqrt {d \left (2 c x +b \right )}\, b^{2}}{256 c^{2} d^{8} \left (c \,x^{2}+b x +a \right )^{2}}-\frac {77 \ln \left (\frac {\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}{\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}\right ) \sqrt {2}}{128 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}} d^{6}}-\frac {77 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}+1\right ) \sqrt {2}}{64 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}} d^{6}}-\frac {77 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}-1\right ) \sqrt {2}}{64 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}} d^{6}}\right )}{\left (4 a c -b^{2}\right )^{3}}\) | \(384\) |
Input:
int(1/(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
64*d^5*c^2*(-1/3/d^6/(4*a*c-b^2)^3/(2*c*d*x+b*d)^(3/2)-1/d^6/(4*a*c-b^2)^3 *(16*(15/512*(2*c*d*x+b*d)^(5/2)+(19/128*a*d^2*c-19/512*b^2*d^2)*(2*c*d*x+ b*d)^(1/2))/((2*c*d*x+b*d)^2+4*a*d^2*c-b^2*d^2)^2+77/256/(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 = 0.14 (sec) , antiderivative size = 3517, normalized size of antiderivative = 15.49 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^3} \, dx=\text {Too large to display} \] Input:
integrate(1/(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^3,x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^3} \, dx=\text {Timed out} \] Input:
integrate(1/(2*c*d*x+b*d)**(5/2)/(c*x**2+b*x+a)**3,x)
Output:
Timed out
Exception generated. \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^3,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
Leaf count of result is larger than twice the leaf count of optimal. 813 vs. \(2 (195) = 390\).
Time = 0.16 (sec) , antiderivative size = 813, normalized size of antiderivative = 3.58 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^3} \, dx =\text {Too large to display} \] Input:
integrate(1/(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^3,x, algorithm="giac")
Output:
-77*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c^2*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)) /(sqrt(2)*b^8*d^3 - 16*sqrt(2)*a*b^6*c*d^3 + 96*sqrt(2)*a^2*b^4*c^2*d^3 - 256*sqrt(2)*a^3*b^2*c^3*d^3 + 256*sqrt(2)*a^4*c^4*d^3) - 77*(-b^2*d^2 + 4* a*c*d^2)^(1/4)*c^2*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))/(sqrt(2)*b^8*d^3 - 16*sqrt(2)*a*b^6*c*d^3 + 96*sqrt(2)*a^2*b^4*c^2*d^3 - 256*sqrt(2)*a^3*b ^2*c^3*d^3 + 256*sqrt(2)*a^4*c^4*d^3) - 77/2*(-b^2*d^2 + 4*a*c*d^2)^(1/4)* c^2*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^8*d^3 - 16*sqrt(2)*a*b^6*c *d^3 + 96*sqrt(2)*a^2*b^4*c^2*d^3 - 256*sqrt(2)*a^3*b^2*c^3*d^3 + 256*sqrt (2)*a^4*c^4*d^3) + 77/2*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c^2*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^8*d^3 - 16*sqrt(2)*a*b^6*c*d^3 + 96*sqrt(2)*a^2 *b^4*c^2*d^3 - 256*sqrt(2)*a^3*b^2*c^3*d^3 + 256*sqrt(2)*a^4*c^4*d^3) + 64 /3*c^2/((b^6*d - 12*a*b^4*c*d + 48*a^2*b^2*c^2*d - 64*a^3*c^3*d)*(2*c*d*x + b*d)^(3/2)) - 2*(19*sqrt(2*c*d*x + b*d)*b^2*c^2*d^2 - 76*sqrt(2*c*d*x + b*d)*a*c^3*d^2 - 15*(2*c*d*x + b*d)^(5/2)*c^2)/((b^6*d - 12*a*b^4*c*d + 48 *a^2*b^2*c^2*d - 64*a^3*c^3*d)*(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2)^2 )
Time = 5.97 (sec) , antiderivative size = 3227, normalized size of antiderivative = 14.22 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^3} \, dx=\text {Too large to display} \] Input:
int(1/((b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)^3),x)
Output:
- ((64*c^2*d^3)/(3*(4*a*c - b^2)) - (154*c^2*(b*d + 2*c*d*x)^4)/(3*(b^6*d - 64*a^3*c^3*d + 48*a^2*b^2*c^2*d - 12*a*b^4*c*d)) + (242*c^2*d*(b*d + 2*c *d*x)^2)/(3*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)))/((b*d + 2*c*d*x)^(3/2)*(b^4*d ^4 + 16*a^2*c^2*d^4 - 8*a*b^2*c*d^4) - (b*d + 2*c*d*x)^(7/2)*(2*b^2*d^2 - 8*a*c*d^2) + (b*d + 2*c*d*x)^(11/2)) - (c^2*atan(((c^2*((b*d + 2*c*d*x)^(1 /2)*(24868028416*a^9*c^13*d^3 - 94864*b^18*c^4*d^3 + 3415104*a*b^16*c^5*d^ 3 - 54641664*a^2*b^14*c^6*d^3 + 509988864*a^3*b^12*c^7*d^3 - 3059933184*a^ 4*b^10*c^8*d^3 + 12239732736*a^5*b^8*c^9*d^3 - 32639287296*a^6*b^6*c^10*d^ 3 + 55953063936*a^7*b^4*c^11*d^3 - 55953063936*a^8*b^2*c^12*d^3) - (77*c^2 *(165356240896*a^13*c^15*d^6 - 2464*b^26*c^2*d^6 + 128128*a*b^24*c^3*d^6 - 3075072*a^2*b^22*c^4*d^6 + 45101056*a^3*b^20*c^5*d^6 - 451010560*a^4*b^18 *c^6*d^6 + 3247276032*a^5*b^16*c^7*d^6 - 17318805504*a^6*b^14*c^8*d^6 + 69 275222016*a^7*b^12*c^9*d^6 - 207825666048*a^8*b^10*c^10*d^6 + 461834813440 *a^9*b^8*c^11*d^6 - 738935701504*a^10*b^6*c^12*d^6 + 806111674368*a^11*b^4 *c^13*d^6 - 537407782912*a^12*b^2*c^14*d^6))/(2*d^(5/2)*(b^2 - 4*a*c)^(15/ 4)))*77i)/(2*d^(5/2)*(b^2 - 4*a*c)^(15/4)) + (c^2*((b*d + 2*c*d*x)^(1/2)*( 24868028416*a^9*c^13*d^3 - 94864*b^18*c^4*d^3 + 3415104*a*b^16*c^5*d^3 - 5 4641664*a^2*b^14*c^6*d^3 + 509988864*a^3*b^12*c^7*d^3 - 3059933184*a^4*b^1 0*c^8*d^3 + 12239732736*a^5*b^8*c^9*d^3 - 32639287296*a^6*b^6*c^10*d^3 + 5 5953063936*a^7*b^4*c^11*d^3 - 55953063936*a^8*b^2*c^12*d^3) + (77*c^2*(...
Time = 0.30 (sec) , antiderivative size = 3095, normalized size of antiderivative = 13.63 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^3} \, dx =\text {Too large to display} \] Input:
int(1/(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^3,x)
Output:
(sqrt(d)*(462*sqrt(b + 2*c*x)*(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**2 + 924*sqrt(b + 2*c*x)*(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)*sq rt(2)))*a**2*c**3*x + 924*sqrt(b + 2*c*x)*(4*a*c - b**2)**(1/4)*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*b**2*c**2*x + 2772*sqrt(b + 2*c*x)*(4*a*c - b**2)**(1/4)*s qrt(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*c**3*x**2 + 1848*sqrt(b + 2*c*x)*(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*c**4*x**3 + 462*sqrt(b + 2*c*x)*(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**3*c**2*x**2 + 1848*sqrt(b + 2*c*x )*(4*a*c - b**2)**(1/4)*sqrt(2)*atan(((4*a*c - b**2)**(1/4)*sqrt(2) - 2*sq rt(b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2)))*b**2*c**3*x**3 + 2310*sqrt (b + 2*c*x)*(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*c**4*x**4 + 92 4*sqrt(b + 2*c*x)*(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)))*c**5*x**5 - 462*sqrt(b + 2*c*x)*(4*a*c - b**2)**(1/4)*sqrt(2)*atan(((4*a*c - b**2...