Integrand size = 26, antiderivative size = 225 \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^3} \, dx=\frac {90 c^2}{\left (b^2-4 a c\right )^3 d \sqrt {b d+2 c d x}}-\frac {1}{2 \left (b^2-4 a c\right ) d \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2}+\frac {9 c}{2 \left (b^2-4 a c\right )^2 d \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )}+\frac {45 c^2 \arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{13/4} d^{3/2}}-\frac {45 c^2 \text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{13/4} d^{3/2}} \] Output:
90*c^2/(-4*a*c+b^2)^3/d/(2*c*d*x+b*d)^(1/2)-1/2/(-4*a*c+b^2)/d/(2*c*d*x+b* d)^(1/2)/(c*x^2+b*x+a)^2+9/2*c/(-4*a*c+b^2)^2/d/(2*c*d*x+b*d)^(1/2)/(c*x^2 +b*x+a)+45*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)^(13/4)/d^(3/2)-45*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)^(13/4)/d^(3/2)
Result contains complex when optimal does not.
Time = 1.49 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.33 \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^3} \, dx=\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) c^2 \left (\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) (b+2 c x) \left (32 b^4-256 a b^2 c+512 a^2 c^2-81 b^2 (b+2 c x)^2+324 a c (b+2 c x)^2+45 (b+2 c x)^4\right )}{c^2 \left (b^2-4 a c\right )^3 (a+x (b+c x))^2}-\frac {45 (b+2 c x)^{3/2} \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{13/4}}+\frac {45 (b+2 c x)^{3/2} \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{13/4}}-\frac {45 (b+2 c x)^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )}{\left (b^2-4 a c\right )^{13/4}}\right )}{(d (b+2 c x))^{3/2}} \] Input:
Integrate[1/((b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)^3),x]
Output:
((1/2 + I/2)*c^2*(((1/8 - I/8)*(b + 2*c*x)*(32*b^4 - 256*a*b^2*c + 512*a^2 *c^2 - 81*b^2*(b + 2*c*x)^2 + 324*a*c*(b + 2*c*x)^2 + 45*(b + 2*c*x)^4))/( c^2*(b^2 - 4*a*c)^3*(a + x*(b + c*x))^2) - (45*(b + 2*c*x)^(3/2)*ArcTan[1 - ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/(b^2 - 4*a*c)^(13/4) + ( 45*(b + 2*c*x)^(3/2)*ArcTan[1 + ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1 /4)])/(b^2 - 4*a*c)^(13/4) - (45*(b + 2*c*x)^(3/2)*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)^(13/4)))/(d*(b + 2*c*x))^(3/2)
Time = 0.44 (sec) , antiderivative size = 263, 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, 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.
| \(\displaystyle \int \frac {1}{\left (a+b x+c x^2\right )^3 (b d+2 c d x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1111 |
| \(\displaystyle -\frac {9 c \int \frac {1}{(b d+2 c x d)^{3/2} \left (c x^2+b x+a\right )^2}dx}{2 \left (b^2-4 a c\right )}-\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2 \sqrt {b d+2 c d x}}\) |
| \(\Big \downarrow \) 1111 |
| \(\displaystyle -\frac {9 c \left (-\frac {5 c \int \frac {1}{(b d+2 c x d)^{3/2} \left (c x^2+b x+a\right )}dx}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \sqrt {b d+2 c d x}}\right )}{2 \left (b^2-4 a c\right )}-\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2 \sqrt {b d+2 c d x}}\) |
| \(\Big \downarrow \) 1117 |
| \(\displaystyle -\frac {9 c \left (-\frac {5 c \left (\frac {\int \frac {\sqrt {b d+2 c x d}}{c x^2+b x+a}dx}{d^2 \left (b^2-4 a c\right )}+\frac {4}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \sqrt {b d+2 c d x}}\right )}{2 \left (b^2-4 a c\right )}-\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2 \sqrt {b d+2 c d x}}\) |
| \(\Big \downarrow \) 1118 |
| \(\displaystyle -\frac {9 c \left (-\frac {5 c \left (\frac {\int \frac {4 c d^2 \sqrt {b d+2 c x d}}{\left (4 a-\frac {b^2}{c}\right ) c d^2+(b d+2 c x d)^2}d(b d+2 c x d)}{2 c d^3 \left (b^2-4 a c\right )}+\frac {4}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \sqrt {b d+2 c d x}}\right )}{2 \left (b^2-4 a c\right )}-\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2 \sqrt {b d+2 c d x}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {9 c \left (-\frac {5 c \left (\frac {2 \int -\frac {\sqrt {b d+2 c x d}}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d(b d+2 c x d)}{d \left (b^2-4 a c\right )}+\frac {4}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \sqrt {b d+2 c d x}}\right )}{2 \left (b^2-4 a c\right )}-\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2 \sqrt {b d+2 c d x}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {9 c \left (-\frac {5 c \left (\frac {4}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}-\frac {2 \int \frac {\sqrt {b d+2 c x d}}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d(b d+2 c x d)}{d \left (b^2-4 a c\right )}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \sqrt {b d+2 c d x}}\right )}{2 \left (b^2-4 a c\right )}-\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2 \sqrt {b d+2 c d x}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle -\frac {9 c \left (-\frac {5 c \left (\frac {4}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}-\frac {4 \int \frac {b d+2 c x d}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d\sqrt {b d+2 c x d}}{d \left (b^2-4 a c\right )}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \sqrt {b d+2 c d x}}\right )}{2 \left (b^2-4 a c\right )}-\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2 \sqrt {b d+2 c d x}}\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle -\frac {9 c \left (-\frac {5 c \left (\frac {4}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}-\frac {4 \left (\frac {1}{2} \int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}-\frac {1}{2} \int \frac {1}{b d+2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}\right )}{d \left (b^2-4 a c\right )}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \sqrt {b d+2 c d x}}\right )}{2 \left (b^2-4 a c\right )}-\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2 \sqrt {b d+2 c d x}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {9 c \left (-\frac {5 c \left (\frac {4}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}-\frac {4 \left (\frac {1}{2} \int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}-\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 \sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d \left (b^2-4 a c\right )}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \sqrt {b d+2 c d x}}\right )}{2 \left (b^2-4 a c\right )}-\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2 \sqrt {b d+2 c d x}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {9 c \left (-\frac {5 c \left (\frac {4}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}-\frac {4 \left (\frac {\text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 \sqrt {d} \sqrt [4]{b^2-4 a c}}-\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 \sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d \left (b^2-4 a c\right )}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \sqrt {b d+2 c d x}}\right )}{2 \left (b^2-4 a c\right )}-\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2 \sqrt {b d+2 c d x}}\) |
Input:
Int[1/((b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)^3),x]
Output:
-1/2*1/((b^2 - 4*a*c)*d*Sqrt[b*d + 2*c*d*x]*(a + b*x + c*x^2)^2) - (9*c*(- (1/((b^2 - 4*a*c)*d*Sqrt[b*d + 2*c*d*x]*(a + b*x + c*x^2))) - (5*c*(4/((b^ 2 - 4*a*c)*d*Sqrt[b*d + 2*c*d*x]) - (4*(-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])))/((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[(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_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 = 1.37 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.58
| method | result | size |
| derivativedivides | \(64 d^{5} c^{2} \left (-\frac {1}{d^{6} \left (4 a c -b^{2}\right )^{3} \sqrt {2 c d x +b d}}-\frac {\frac {\frac {13 \left (2 c d x +b d \right )^{\frac {7}{2}}}{32}+16 \left (\frac {17}{128} a \,d^{2} c -\frac {17}{512} b^{2} d^{2}\right ) \left (2 c d x +b d \right )^{\frac {3}{2}}}{\left (\left (2 c d x +b d \right )^{2}+4 a \,d^{2} c -b^{2} d^{2}\right )^{2}}+\frac {45 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{256 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}}{d^{6} \left (4 a c -b^{2}\right )^{3}}\right )\) | \(355\) |
| default | \(64 d^{5} c^{2} \left (-\frac {1}{d^{6} \left (4 a c -b^{2}\right )^{3} \sqrt {2 c d x +b d}}-\frac {\frac {\frac {13 \left (2 c d x +b d \right )^{\frac {7}{2}}}{32}+16 \left (\frac {17}{128} a \,d^{2} c -\frac {17}{512} b^{2} d^{2}\right ) \left (2 c d x +b d \right )^{\frac {3}{2}}}{\left (\left (2 c d x +b d \right )^{2}+4 a \,d^{2} c -b^{2} d^{2}\right )^{2}}+\frac {45 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{256 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}}{d^{6} \left (4 a c -b^{2}\right )^{3}}\right )\) | \(355\) |
| pseudoelliptic | \(\frac {32 c^{2} d^{5} \left (-\frac {2}{d^{6} \sqrt {d \left (2 c x +b \right )}}-\frac {13 \left (d \left (2 c x +b \right )\right )^{\frac {7}{2}}}{256 c^{2} d^{10} \left (c \,x^{2}+b x +a \right )^{2}}-\frac {17 \left (d \left (2 c x +b \right )\right )^{\frac {3}{2}} a}{64 c \,d^{8} \left (c \,x^{2}+b x +a \right )^{2}}+\frac {17 \left (d \left (2 c x +b \right )\right )^{\frac {3}{2}} b^{2}}{256 c^{2} d^{8} \left (c \,x^{2}+b x +a \right )^{2}}-\frac {45 \ln \left (\frac {\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}\right ) \sqrt {2}}{128 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} d^{6}}-\frac {45 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}+1\right ) \sqrt {2}}{64 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} d^{6}}-\frac {45 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}-1\right ) \sqrt {2}}{64 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} d^{6}}\right )}{\left (4 a c -b^{2}\right )^{3}}\) | \(384\) |
Input:
int(1/(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
64*d^5*c^2*(-1/d^6/(4*a*c-b^2)^3/(2*c*d*x+b*d)^(1/2)-1/d^6/(4*a*c-b^2)^3*( 16*(13/512*(2*c*d*x+b*d)^(7/2)+(17/128*a*d^2*c-17/512*b^2*d^2)*(2*c*d*x+b* d)^(3/2))/((2*c*d*x+b*d)^2+4*a*d^2*c-b^2*d^2)^2+45/256/(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 = 0.13 (sec) , antiderivative size = 3252, normalized size of antiderivative = 14.45 \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^3} \, dx=\text {Too large to display} \] Input:
integrate(1/(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^3,x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^3} \, dx=\text {Timed out} \] Input:
integrate(1/(2*c*d*x+b*d)**(3/2)/(c*x**2+b*x+a)**3,x)
Output:
Timed out
Exception generated. \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(2*c*d*x+b*d)^(3/2)/(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 = 3.61 \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^3} \, dx =\text {Too large to display} \] Input:
integrate(1/(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^3,x, algorithm="giac")
Output:
-45*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*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) - 45*(-b^2*d^2 + 4* a*c*d^2)^(3/4)*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) + 45/2*(-b^2*d^2 + 4*a*c*d^2)^(3/4)* 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) - 45/2*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*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 *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)*sqrt(2*c*d* x + b*d)) - 2*(17*(2*c*d*x + b*d)^(3/2)*b^2*c^2*d^2 - 68*(2*c*d*x + b*d)^( 3/2)*a*c^3*d^2 - 13*(2*c*d*x + b*d)^(7/2)*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 = 5.50 (sec) , antiderivative size = 421, normalized size of antiderivative = 1.87 \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^3} \, dx=\frac {45\,c^2\,\mathrm {atan}\left (\frac {b^6\,\sqrt {b\,d+2\,c\,d\,x}-64\,a^3\,c^3\,\sqrt {b\,d+2\,c\,d\,x}+48\,a^2\,b^2\,c^2\,\sqrt {b\,d+2\,c\,d\,x}-12\,a\,b^4\,c\,\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{13/4}}\right )}{d^{3/2}\,{\left (b^2-4\,a\,c\right )}^{13/4}}-\frac {\frac {64\,c^2\,d^3}{4\,a\,c-b^2}-\frac {90\,c^2\,{\left (b\,d+2\,c\,d\,x\right )}^4}{-64\,d\,a^3\,c^3+48\,d\,a^2\,b^2\,c^2-12\,d\,a\,b^4\,c+d\,b^6}+\frac {162\,c^2\,d\,{\left (b\,d+2\,c\,d\,x\right )}^2}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}}{\sqrt {b\,d+2\,c\,d\,x}\,\left (16\,a^2\,c^2\,d^4-8\,a\,b^2\,c\,d^4+b^4\,d^4\right )-{\left (b\,d+2\,c\,d\,x\right )}^{5/2}\,\left (2\,b^2\,d^2-8\,a\,c\,d^2\right )+{\left (b\,d+2\,c\,d\,x\right )}^{9/2}}+\frac {c^2\,\mathrm {atan}\left (\frac {b^6\,\sqrt {b\,d+2\,c\,d\,x}\,1{}\mathrm {i}-a^3\,c^3\,\sqrt {b\,d+2\,c\,d\,x}\,64{}\mathrm {i}+a^2\,b^2\,c^2\,\sqrt {b\,d+2\,c\,d\,x}\,48{}\mathrm {i}-a\,b^4\,c\,\sqrt {b\,d+2\,c\,d\,x}\,12{}\mathrm {i}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{13/4}}\right )\,45{}\mathrm {i}}{d^{3/2}\,{\left (b^2-4\,a\,c\right )}^{13/4}} \] Input:
int(1/((b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)^3),x)
Output:
(45*c^2*atan((b^6*(b*d + 2*c*d*x)^(1/2) - 64*a^3*c^3*(b*d + 2*c*d*x)^(1/2) + 48*a^2*b^2*c^2*(b*d + 2*c*d*x)^(1/2) - 12*a*b^4*c*(b*d + 2*c*d*x)^(1/2) )/(d^(1/2)*(b^2 - 4*a*c)^(13/4))))/(d^(3/2)*(b^2 - 4*a*c)^(13/4)) - ((64*c ^2*d^3)/(4*a*c - b^2) - (90*c^2*(b*d + 2*c*d*x)^4)/(b^6*d - 64*a^3*c^3*d + 48*a^2*b^2*c^2*d - 12*a*b^4*c*d) + (162*c^2*d*(b*d + 2*c*d*x)^2)/(b^4 + 1 6*a^2*c^2 - 8*a*b^2*c))/((b*d + 2*c*d*x)^(1/2)*(b^4*d^4 + 16*a^2*c^2*d^4 - 8*a*b^2*c*d^4) - (b*d + 2*c*d*x)^(5/2)*(2*b^2*d^2 - 8*a*c*d^2) + (b*d + 2 *c*d*x)^(9/2)) + (c^2*atan((b^6*(b*d + 2*c*d*x)^(1/2)*1i - a^3*c^3*(b*d + 2*c*d*x)^(1/2)*64i + a^2*b^2*c^2*(b*d + 2*c*d*x)^(1/2)*48i - a*b^4*c*(b*d + 2*c*d*x)^(1/2)*12i)/(d^(1/2)*(b^2 - 4*a*c)^(13/4)))*45i)/(d^(3/2)*(b^2 - 4*a*c)^(13/4))
Time = 0.27 (sec) , antiderivative size = 2160, normalized size of antiderivative = 9.60 \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^3} \, dx =\text {Too large to display} \] Input:
int(1/(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^3,x)
Output:
(sqrt(d)*(90*sqrt(b + 2*c*x)*(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**2*c**2 + 180*sqrt(b + 2*c*x)*(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*b*c**2*x + 180*sqrt(b + 2*c*x)*(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**3*x**2 + 90*sqrt(b + 2*c*x)*(4*a*c - b**2)**(3/4)*sqrt(2)*a tan(((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**2*c**2*x**2 + 180*sqrt(b + 2*c*x)*(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**3*x**3 + 90*sqrt(b + 2*c*x)*(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**4*x**4 - 90*sqrt(b + 2*c*x)*(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**2*c**2 - 180*sqrt(b + 2*c*x)*(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*b*c**2*x - 180*sqrt(b + 2*c*x)*(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**3*x**2 - 90*sqrt(b + 2*c*x)* (4*a*c - b**2)**(3/4)*sqrt(2)*atan(((4*a*c - b**2)**(1/4)*sqrt(2) + 2*s...