Integrand size = 26, antiderivative size = 174 \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2} \, dx=-\frac {20 c}{\left (b^2-4 a c\right )^2 d \sqrt {b d+2 c d x}}-\frac {1}{\left (b^2-4 a c\right ) d \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )}-\frac {10 c \arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{9/4} d^{3/2}}+\frac {10 c \text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{9/4} d^{3/2}} \] Output:
-20*c/(-4*a*c+b^2)^2/d/(2*c*d*x+b*d)^(1/2)-1/(-4*a*c+b^2)/d/(2*c*d*x+b*d)^ (1/2)/(c*x^2+b*x+a)-10*c*arctan((2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1/4)/d^( 1/2))/(-4*a*c+b^2)^(9/4)/d^(3/2)+10*c*arctanh((2*c*d*x+b*d)^(1/2)/(-4*a*c+ b^2)^(1/4)/d^(1/2))/(-4*a*c+b^2)^(9/4)/d^(3/2)
Result contains complex when optimal does not.
Time = 0.85 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.49 \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2} \, dx=\frac {(1+i) c \left (-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) (b+2 c x) \left (b^2+20 b c x+4 c \left (4 a+5 c x^2\right )\right )}{c \left (b^2-4 a c\right )^2 (a+x (b+c x))}+\frac {5 (b+2 c x)^{3/2} \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{9/4}}-\frac {5 (b+2 c x)^{3/2} \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{9/4}}+\frac {5 (b+2 c x)^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )}{\left (b^2-4 a c\right )^{9/4}}\right )}{(d (b+2 c x))^{3/2}} \] Input:
Integrate[1/((b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)^2),x]
Output:
((1 + I)*c*(((-1/2 + I/2)*(b + 2*c*x)*(b^2 + 20*b*c*x + 4*c*(4*a + 5*c*x^2 )))/(c*(b^2 - 4*a*c)^2*(a + x*(b + c*x))) + (5*(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)^(9/4) - (5 *(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)^(9/4) + (5*(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)^(9/4)))/(d*(b + 2*c*x))^(3/2)
Time = 0.38 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.18, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {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.
| \(\displaystyle \int \frac {1}{\left (a+b x+c x^2\right )^2 (b d+2 c d x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1111 |
| \(\displaystyle -\frac {5 c \int \frac {1}{(b d+2 c x d)^{3/2} \left (c x^2+b x+a\right )}dx}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \sqrt {b d+2 c d x}}\) |
| \(\Big \downarrow \) 1117 |
| \(\displaystyle -\frac {5 c \left (\frac {\int \frac {\sqrt {b d+2 c x d}}{c x^2+b x+a}dx}{d^2 \left (b^2-4 a c\right )}+\frac {4}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \sqrt {b d+2 c d x}}\) |
| \(\Big \downarrow \) 1118 |
| \(\displaystyle -\frac {5 c \left (\frac {\int \frac {4 c d^2 \sqrt {b d+2 c x d}}{\left (4 a-\frac {b^2}{c}\right ) c d^2+(b d+2 c x d)^2}d(b d+2 c x d)}{2 c d^3 \left (b^2-4 a c\right )}+\frac {4}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \sqrt {b d+2 c d x}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {5 c \left (\frac {2 \int -\frac {\sqrt {b d+2 c x d}}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d(b d+2 c x d)}{d \left (b^2-4 a c\right )}+\frac {4}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \sqrt {b d+2 c d x}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {5 c \left (\frac {4}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}-\frac {2 \int \frac {\sqrt {b d+2 c x d}}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d(b d+2 c x d)}{d \left (b^2-4 a c\right )}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \sqrt {b d+2 c d x}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle -\frac {5 c \left (\frac {4}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}-\frac {4 \int \frac {b d+2 c x d}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d\sqrt {b d+2 c x d}}{d \left (b^2-4 a c\right )}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \sqrt {b d+2 c d x}}\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle -\frac {5 c \left (\frac {4}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}-\frac {4 \left (\frac {1}{2} \int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}-\frac {1}{2} \int \frac {1}{b d+2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}\right )}{d \left (b^2-4 a c\right )}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \sqrt {b d+2 c d x}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {5 c \left (\frac {4}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}-\frac {4 \left (\frac {1}{2} \int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}-\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 \sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d \left (b^2-4 a c\right )}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \sqrt {b d+2 c d x}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {5 c \left (\frac {4}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}-\frac {4 \left (\frac {\text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 \sqrt {d} \sqrt [4]{b^2-4 a c}}-\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 \sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d \left (b^2-4 a c\right )}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \sqrt {b d+2 c d x}}\) |
Input:
Int[1/((b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)^2),x]
Output:
-(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)
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]
Leaf count of result is larger than twice the leaf count of optimal. \(322\) vs. \(2(150)=300\).
Time = 1.31 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.86
| method | result | size |
| derivativedivides | \(16 c \,d^{3} \left (-\frac {\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{16 a \,d^{2} c -4 b^{2} d^{2}+4 \left (2 c d x +b d \right )^{2}}+\frac {5 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{32 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}}{d^{4} \left (4 a c -b^{2}\right )^{2}}-\frac {1}{d^{4} \left (4 a c -b^{2}\right )^{2} \sqrt {2 c d x +b d}}\right )\) | \(323\) |
| default | \(16 c \,d^{3} \left (-\frac {\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{16 a \,d^{2} c -4 b^{2} d^{2}+4 \left (2 c d x +b d \right )^{2}}+\frac {5 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{32 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}}{d^{4} \left (4 a c -b^{2}\right )^{2}}-\frac {1}{d^{4} \left (4 a c -b^{2}\right )^{2} \sqrt {2 c d x +b d}}\right )\) | \(323\) |
| pseudoelliptic | \(-\frac {5 \left (c \sqrt {2}\, \left (c \,x^{2}+b x +a \right ) \left (\ln \left (\frac {\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )\right ) \sqrt {d \left (2 c x +b \right )}+\frac {32 \left (\frac {5 c^{2} x^{2}}{4}+\left (\frac {5 b x}{4}+a \right ) c +\frac {b^{2}}{16}\right ) \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{5}\right )}{2 \sqrt {d \left (2 c x +b \right )}\, \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} d \left (4 a c -b^{2}\right )^{2} \left (c \,x^{2}+b x +a \right )}\) | \(342\) |
Input:
int(1/(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
16*c*d^3*(-1/d^4/(4*a*c-b^2)^2*(1/16*(2*c*d*x+b*d)^(3/2)/(a*d^2*c-1/4*b^2* d^2+1/4*(2*c*d*x+b*d)^2)+5/32/(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)))-1/d^4/(4*a*c-b^2)^2/(2*c*d*x+b*d)^(1/2))
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 1884, normalized size of antiderivative = 10.83 \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2} \, dx=\text {Too large to display} \] Input:
integrate(1/(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^2,x, algorithm="fricas")
Output:
(5*(2*(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*d^2*x^3 + 3*(b^5*c - 8*a*b^3*c^ 2 + 16*a^2*b*c^3)*d^2*x^2 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*d^2*x + (a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*d^2)*(c^4/((b^18 - 36*a*b^16*c + 576*a^2*b^1 4*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 3440 64*a^6*b^6*c^6 - 589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9) *d^6))^(1/4)*log(125*(b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2 - 2240*a^3*b^8 *c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^ 7*c^7)*d^5*(c^4/((b^18 - 36*a*b^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^ 3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824* a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^6))^(3/4) + 125*sqrt( 2*c*d*x + b*d)*c^3) + 5*(-2*I*(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*d^2*x^3 - 3*I*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*d^2*x^2 - I*(b^6 - 6*a*b^4*c + 32*a^3*c^3)*d^2*x - I*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*d^2)*(c^4/((b^ 18 - 36*a*b^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c ^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7*b^4*c^7 + 589824 *a^8*b^2*c^8 - 262144*a^9*c^9)*d^6))^(1/4)*log(125*I*(b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2 - 2240*a^3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^ 5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*d^5*(c^4/((b^18 - 36*a*b^16*c + 576 *a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^ 5 + 344064*a^6*b^6*c^6 - 589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 2621...
\[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2} \, dx=\int \frac {1}{\left (d \left (b + 2 c x\right )\right )^{\frac {3}{2}} \left (a + b x + c x^{2}\right )^{2}}\, dx \] Input:
integrate(1/(2*c*d*x+b*d)**(3/2)/(c*x**2+b*x+a)**2,x)
Output:
Integral(1/((d*(b + 2*c*x))**(3/2)*(a + b*x + c*x**2)**2), x)
Exception generated. \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(2*c*d*x+b*d)^(3/2)/(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
Leaf count of result is larger than twice the leaf count of optimal. 650 vs. \(2 (150) = 300\).
Time = 0.16 (sec) , antiderivative size = 650, normalized size of antiderivative = 3.74 \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2} \, dx=\frac {5 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c \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^{6} d^{3} - 12 \, a b^{4} c d^{3} + 48 \, a^{2} b^{2} c^{2} d^{3} - 64 \, a^{3} c^{3} d^{3}} + \frac {5 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c \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^{6} d^{3} - 12 \, a b^{4} c d^{3} + 48 \, a^{2} b^{2} c^{2} d^{3} - 64 \, a^{3} c^{3} d^{3}} - \frac {5 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c \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^{6} d^{3} - 12 \, \sqrt {2} a b^{4} c d^{3} + 48 \, \sqrt {2} a^{2} b^{2} c^{2} d^{3} - 64 \, \sqrt {2} a^{3} c^{3} d^{3}} + \frac {5 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c \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^{6} d^{3} - 12 \, \sqrt {2} a b^{4} c d^{3} + 48 \, \sqrt {2} a^{2} b^{2} c^{2} d^{3} - 64 \, \sqrt {2} a^{3} c^{3} d^{3}} - \frac {4 \, {\left (4 \, b^{2} c d^{2} - 16 \, a c^{2} d^{2} - 5 \, {\left (2 \, c d x + b d\right )}^{2} c\right )}}{{\left (b^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d\right )} {\left (\sqrt {2 \, c d x + b d} b^{2} d^{2} - 4 \, \sqrt {2 \, c d x + b d} a c d^{2} - {\left (2 \, c d x + b d\right )}^{\frac {5}{2}}\right )}} \] Input:
integrate(1/(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^2,x, algorithm="giac")
Output:
5*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*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^6*d^3 - 12*a*b^4*c*d^3 + 48*a^2*b^2*c^2*d^3 - 64*a^3*c^3*d^3) + 5* sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*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^6*d^3 - 12*a*b^4*c*d^3 + 48*a^2*b^2*c^2*d^3 - 64*a^3*c^3*d^3) - 5*( -b^2*d^2 + 4*a*c*d^2)^(3/4)*c*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^ 6*d^3 - 12*sqrt(2)*a*b^4*c*d^3 + 48*sqrt(2)*a^2*b^2*c^2*d^3 - 64*sqrt(2)*a ^3*c^3*d^3) + 5*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*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^6*d^3 - 12*sqrt(2)*a*b^4*c*d^3 + 48*sqrt(2)*a^2*b^2*c^2*d ^3 - 64*sqrt(2)*a^3*c^3*d^3) - 4*(4*b^2*c*d^2 - 16*a*c^2*d^2 - 5*(2*c*d*x + b*d)^2*c)/((b^4*d - 8*a*b^2*c*d + 16*a^2*c^2*d)*(sqrt(2*c*d*x + b*d)*b^2 *d^2 - 4*sqrt(2*c*d*x + b*d)*a*c*d^2 - (2*c*d*x + b*d)^(5/2)))
Time = 5.37 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.52 \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2} \, dx=\frac {\frac {16\,c\,d}{4\,a\,c-b^2}+\frac {20\,c\,{\left (b\,d+2\,c\,d\,x\right )}^2}{d\,{\left (4\,a\,c-b^2\right )}^2}}{\sqrt {b\,d+2\,c\,d\,x}\,\left (b^2\,d^2-4\,a\,c\,d^2\right )-{\left (b\,d+2\,c\,d\,x\right )}^{5/2}}-\frac {10\,c\,\mathrm {atan}\left (\frac {b^4\,\sqrt {b\,d+2\,c\,d\,x}+16\,a^2\,c^2\,\sqrt {b\,d+2\,c\,d\,x}-8\,a\,b^2\,c\,\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{9/4}}\right )}{d^{3/2}\,{\left (b^2-4\,a\,c\right )}^{9/4}}-\frac {c\,\mathrm {atan}\left (\frac {b^4\,\sqrt {b\,d+2\,c\,d\,x}\,1{}\mathrm {i}+a^2\,c^2\,\sqrt {b\,d+2\,c\,d\,x}\,16{}\mathrm {i}-a\,b^2\,c\,\sqrt {b\,d+2\,c\,d\,x}\,8{}\mathrm {i}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{9/4}}\right )\,10{}\mathrm {i}}{d^{3/2}\,{\left (b^2-4\,a\,c\right )}^{9/4}} \] Input:
int(1/((b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)^2),x)
Output:
((16*c*d)/(4*a*c - b^2) + (20*c*(b*d + 2*c*d*x)^2)/(d*(4*a*c - b^2)^2))/(( b*d + 2*c*d*x)^(1/2)*(b^2*d^2 - 4*a*c*d^2) - (b*d + 2*c*d*x)^(5/2)) - (10* c*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))))/(d^(3/2)*(b^ 2 - 4*a*c)^(9/4)) - (c*atan((b^4*(b*d + 2*c*d*x)^(1/2)*1i + a^2*c^2*(b*d + 2*c*d*x)^(1/2)*16i - a*b^2*c*(b*d + 2*c*d*x)^(1/2)*8i)/(d^(1/2)*(b^2 - 4* a*c)^(9/4)))*10i)/(d^(3/2)*(b^2 - 4*a*c)^(9/4))
Time = 0.22 (sec) , antiderivative size = 1012, normalized size of antiderivative = 5.82 \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:
int(1/(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^2,x)
Output:
(sqrt(d)*(10*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 + 10*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*x + 10*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** 2*x**2 - 10*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 - 10*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 *x - 10*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**2 *x**2 - 5*sqrt(b + 2*c*x)*(4*a*c - b**2)**(3/4)*sqrt(2)*log( - sqrt(b + 2* c*x)*(4*a*c - b**2)**(1/4)*sqrt(2) + sqrt(4*a*c - b**2) + b + 2*c*x)*a*c - 5*sqrt(b + 2*c*x)*(4*a*c - b**2)**(3/4)*sqrt(2)*log( - sqrt(b + 2*c*x)*(4 *a*c - b**2)**(1/4)*sqrt(2) + sqrt(4*a*c - b**2) + b + 2*c*x)*b*c*x - 5*sq rt(b + 2*c*x)*(4*a*c - b**2)**(3/4)*sqrt(2)*log( - sqrt(b + 2*c*x)*(4*a*c - b**2)**(1/4)*sqrt(2) + sqrt(4*a*c - b**2) + b + 2*c*x)*c**2*x**2 + 5*sqr t(b + 2*c*x)*(4*a*c - b**2)**(3/4)*sqrt(2)*log(sqrt(b + 2*c*x)*(4*a*c - b* *2)**(1/4)*sqrt(2) + sqrt(4*a*c - b**2) + b + 2*c*x)*a*c + 5*sqrt(b + 2...