Integrand size = 26, antiderivative size = 180 \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^3} \, dx=-\frac {d (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {3 c d (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {3 c^2 d^{5/2} \arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{5/4}}+\frac {3 c^2 d^{5/2} \text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{5/4}} \] Output:
-1/2*d*(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^2-3/2*c*d*(2*c*d*x+b*d)^(3/2)/(-4 *a*c+b^2)/(c*x^2+b*x+a)-3*c^2*d^(5/2)*arctan((2*c*d*x+b*d)^(1/2)/(-4*a*c+b ^2)^(1/4)/d^(1/2))/(-4*a*c+b^2)^(5/4)+3*c^2*d^(5/2)*arctanh((2*c*d*x+b*d)^ (1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))/(-4*a*c+b^2)^(5/4)
Result contains complex when optimal does not.
Time = 1.89 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.48 \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^3} \, dx=\left (\frac {1}{2}+\frac {i}{2}\right ) c^2 (d (b+2 c x))^{5/2} \left (-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (b^2+3 b c x+c \left (-a+3 c x^2\right )\right )}{c^2 \left (b^2-4 a c\right ) (b+2 c x) (a+x (b+c x))^2}+\frac {3 \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/4} (b+2 c x)^{5/2}}-\frac {3 \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/4} (b+2 c x)^{5/2}}+\frac {3 \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )}{\left (b^2-4 a c\right )^{5/4} (b+2 c x)^{5/2}}\right ) \] Input:
Integrate[(b*d + 2*c*d*x)^(5/2)/(a + b*x + c*x^2)^3,x]
Output:
(1/2 + I/2)*c^2*(d*(b + 2*c*x))^(5/2)*(((-1/2 + I/2)*(b^2 + 3*b*c*x + c*(- a + 3*c*x^2)))/(c^2*(b^2 - 4*a*c)*(b + 2*c*x)*(a + x*(b + c*x))^2) + (3*Ar cTan[1 - ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/((b^2 - 4*a*c)^(5 /4)*(b + 2*c*x)^(5/2)) - (3*ArcTan[1 + ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4* a*c)^(1/4)])/((b^2 - 4*a*c)^(5/4)*(b + 2*c*x)^(5/2)) + (3*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)^(5/4)*(b + 2*c*x)^(5/2)))
Time = 0.37 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {1110, 1111, 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.
| \(\displaystyle \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 1110 |
| \(\displaystyle \frac {3}{2} c d^2 \int \frac {\sqrt {b d+2 c x d}}{\left (c x^2+b x+a\right )^2}dx-\frac {d (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )^2}\) |
| \(\Big \downarrow \) 1111 |
| \(\displaystyle \frac {3}{2} c d^2 \left (-\frac {c \int \frac {\sqrt {b d+2 c x d}}{c x^2+b x+a}dx}{b^2-4 a c}-\frac {(b d+2 c d x)^{3/2}}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\right )-\frac {d (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )^2}\) |
| \(\Big \downarrow \) 1118 |
| \(\displaystyle \frac {3}{2} c d^2 \left (-\frac {\int \frac {4 c d^2 \sqrt {b d+2 c x d}}{\left (4 a-\frac {b^2}{c}\right ) c d^2+(b d+2 c x d)^2}d(b d+2 c x d)}{2 d \left (b^2-4 a c\right )}-\frac {(b d+2 c d x)^{3/2}}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\right )-\frac {d (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3}{2} c d^2 \left (-\frac {2 c d \int -\frac {\sqrt {b d+2 c x d}}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d(b d+2 c x d)}{b^2-4 a c}-\frac {(b d+2 c d x)^{3/2}}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\right )-\frac {d (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {3}{2} c d^2 \left (\frac {2 c d \int \frac {\sqrt {b d+2 c x d}}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d(b d+2 c x d)}{b^2-4 a c}-\frac {(b d+2 c d x)^{3/2}}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\right )-\frac {d (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )^2}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {3}{2} c d^2 \left (\frac {4 c d \int \frac {b d+2 c x d}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d\sqrt {b d+2 c x d}}{b^2-4 a c}-\frac {(b d+2 c d x)^{3/2}}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\right )-\frac {d (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )^2}\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle \frac {3}{2} c d^2 \left (\frac {4 c d \left (\frac {1}{2} \int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}-\frac {1}{2} \int \frac {1}{b d+2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}\right )}{b^2-4 a c}-\frac {(b d+2 c d x)^{3/2}}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\right )-\frac {d (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )^2}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {3}{2} c d^2 \left (\frac {4 c d \left (\frac {1}{2} \int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}-\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 \sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{b^2-4 a c}-\frac {(b d+2 c d x)^{3/2}}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\right )-\frac {d (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {3}{2} c d^2 \left (\frac {4 c d \left (\frac {\text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 \sqrt {d} \sqrt [4]{b^2-4 a c}}-\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 \sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{b^2-4 a c}-\frac {(b d+2 c d x)^{3/2}}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\right )-\frac {d (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )^2}\) |
Input:
Int[(b*d + 2*c*d*x)^(5/2)/(a + b*x + c*x^2)^3,x]
Output:
-1/2*(d*(b*d + 2*c*d*x)^(3/2))/(a + b*x + c*x^2)^2 + (3*c*d^2*(-((b*d + 2* c*d*x)^(3/2)/((b^2 - 4*a*c)*d*(a + b*x + c*x^2))) + (4*c*d*(-1/2*ArcTan[Sq rt[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)))/2
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*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[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. \(323\) vs. \(2(152)=304\).
Time = 1.50 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.80
| method | result | size |
| derivativedivides | \(64 c^{2} d^{5} \left (\frac {\frac {3 \left (2 c d x +b d \right )^{\frac {7}{2}}}{32 d^{2} \left (4 a c -b^{2}\right )}-\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{32}}{\left (\left (2 c d x +b d \right )^{2}+4 a \,d^{2} c -b^{2} d^{2}\right )^{2}}+\frac {3 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{256 d^{2} \left (4 a c -b^{2}\right ) \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}\right )\) | \(324\) |
| default | \(64 c^{2} d^{5} \left (\frac {\frac {3 \left (2 c d x +b d \right )^{\frac {7}{2}}}{32 d^{2} \left (4 a c -b^{2}\right )}-\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{32}}{\left (\left (2 c d x +b d \right )^{2}+4 a \,d^{2} c -b^{2} d^{2}\right )^{2}}+\frac {3 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{256 d^{2} \left (4 a c -b^{2}\right ) \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}\right )\) | \(324\) |
| pseudoelliptic | \(\frac {3 d \left (\frac {\sqrt {2}\, c^{2} d^{2} \left (c \,x^{2}+b x +a \right )^{2} \ln \left (\frac {\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}\right )}{2}+\sqrt {2}\, c^{2} d^{2} \left (c \,x^{2}+b x +a \right )^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}-1\right )+\sqrt {2}\, c^{2} d^{2} \left (c \,x^{2}+b x +a \right )^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}+1\right )-\frac {\left (d \left (2 c x +b \right )\right )^{\frac {3}{2}} \left (-3 c^{2} x^{2}+\left (-3 b x +a \right ) c -b^{2}\right ) \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{3}\right )}{8 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \left (a c -\frac {b^{2}}{4}\right ) \left (c \,x^{2}+b x +a \right )^{2}}\) | \(342\) |
Input:
int((2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
64*c^2*d^5*(16*(3/512/d^2/(4*a*c-b^2)*(2*c*d*x+b*d)^(7/2)-1/512*(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+3/256/d^2/(4*a*c-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*arct an(-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 = 0.11 (sec) , antiderivative size = 1296, normalized size of antiderivative = 7.20 \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^3} \, dx=\text {Too large to display} \] Input:
integrate((2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^3,x, algorithm="fricas")
Output:
1/2*(3*(c^8*d^10/(b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5))^(1/4)*((b^2*c^2 - 4*a*c^3)*x^4 + a^2*b^2 - 4*a^3*c + 2*(b^3*c - 4*a*b*c^2)*x^3 + (b^4 - 2*a*b^2*c - 8*a^2*c^2)*x^2 + 2*(a*b^3 - 4*a^2*b*c)*x)*log(27*sqrt(2*c*d*x + b*d)*c^6*d^7 + 27*(c^8*d ^10/(b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2* c^4 - 1024*a^5*c^5))^(3/4)*(b^8 - 16*a*b^6*c + 96*a^2*b^4*c^2 - 256*a^3*b^ 2*c^3 + 256*a^4*c^4)) - 3*(c^8*d^10/(b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5))^(1/4)*((b^2*c^2 - 4*a *c^3)*x^4 + a^2*b^2 - 4*a^3*c + 2*(b^3*c - 4*a*b*c^2)*x^3 + (b^4 - 2*a*b^2 *c - 8*a^2*c^2)*x^2 + 2*(a*b^3 - 4*a^2*b*c)*x)*log(27*sqrt(2*c*d*x + b*d)* c^6*d^7 - 27*(c^8*d^10/(b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4* c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5))^(3/4)*(b^8 - 16*a*b^6*c + 96*a^2*b ^4*c^2 - 256*a^3*b^2*c^3 + 256*a^4*c^4)) + 3*(c^8*d^10/(b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5))^(1 /4)*(I*(b^2*c^2 - 4*a*c^3)*x^4 + I*a^2*b^2 - 4*I*a^3*c + 2*I*(b^3*c - 4*a* b*c^2)*x^3 + I*(b^4 - 2*a*b^2*c - 8*a^2*c^2)*x^2 + 2*I*(a*b^3 - 4*a^2*b*c) *x)*log(27*sqrt(2*c*d*x + b*d)*c^6*d^7 - 27*(c^8*d^10/(b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5))^(3/ 4)*(I*b^8 - 16*I*a*b^6*c + 96*I*a^2*b^4*c^2 - 256*I*a^3*b^2*c^3 + 256*I*a^ 4*c^4)) + 3*(c^8*d^10/(b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^...
Timed out. \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^3} \, dx=\text {Timed out} \] Input:
integrate((2*c*d*x+b*d)**(5/2)/(c*x**2+b*x+a)**3,x)
Output:
Timed out
Exception generated. \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((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. 567 vs. \(2 (152) = 304\).
Time = 0.16 (sec) , antiderivative size = 567, normalized size of antiderivative = 3.15 \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^3} \, dx=\frac {3 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} d \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{\sqrt {2} b^{4} - 8 \, \sqrt {2} a b^{2} c + 16 \, \sqrt {2} a^{2} c^{2}} + \frac {3 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} d \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{\sqrt {2} b^{4} - 8 \, \sqrt {2} a b^{2} c + 16 \, \sqrt {2} a^{2} c^{2}} - \frac {3 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} d \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{2 \, {\left (\sqrt {2} b^{4} - 8 \, \sqrt {2} a b^{2} c + 16 \, \sqrt {2} a^{2} c^{2}\right )}} + \frac {3 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} d \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{2 \, {\left (\sqrt {2} b^{4} - 8 \, \sqrt {2} a b^{2} c + 16 \, \sqrt {2} a^{2} c^{2}\right )}} - \frac {2 \, {\left ({\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{2} c^{2} d^{5} - 4 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} a c^{3} d^{5} + 3 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} c^{2} d^{3}\right )}}{{\left (b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}\right )}^{2} {\left (b^{2} - 4 \, a c\right )}} \] Input:
integrate((2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^3,x, algorithm="giac")
Output:
3*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*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)) /(sqrt(2)*b^4 - 8*sqrt(2)*a*b^2*c + 16*sqrt(2)*a^2*c^2) + 3*(-b^2*d^2 + 4* a*c*d^2)^(3/4)*c^2*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))/(sqrt(2)*b^4 - 8*sqrt(2)*a*b^2*c + 16*sqrt(2)*a^2*c^2) - 3/2*(-b^2*d^2 + 4*a*c*d^2)^(3/4 )*c^2*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^4 - 8*sqrt(2)*a*b^2*c + 16*sqrt(2)*a^2*c^2) + 3/2*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*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^4 - 8*sqrt(2)*a*b^2*c + 16*sqrt(2)*a^2*c^ 2) - 2*((2*c*d*x + b*d)^(3/2)*b^2*c^2*d^5 - 4*(2*c*d*x + b*d)^(3/2)*a*c^3* d^5 + 3*(2*c*d*x + b*d)^(7/2)*c^2*d^3)/((b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2)^2*(b^2 - 4*a*c))
Time = 0.08 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.39 \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^3} \, dx=\frac {3\,c^2\,d^{5/2}\,\mathrm {atanh}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{9/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{5/4}}-\frac {3\,c^2\,d^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{9/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{5/4}}-\frac {2\,c^2\,d^5\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}-\frac {6\,c^2\,d^3\,{\left (b\,d+2\,c\,d\,x\right )}^{7/2}}{4\,a\,c-b^2}}{{\left (b\,d+2\,c\,d\,x\right )}^4-{\left (b\,d+2\,c\,d\,x\right )}^2\,\left (2\,b^2\,d^2-8\,a\,c\,d^2\right )+b^4\,d^4+16\,a^2\,c^2\,d^4-8\,a\,b^2\,c\,d^4} \] Input:
int((b*d + 2*c*d*x)^(5/2)/(a + b*x + c*x^2)^3,x)
Output:
(3*c^2*d^(5/2)*atanh(((b*d + 2*c*d*x)^(1/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c) )/(d^(1/2)*(b^2 - 4*a*c)^(9/4))))/(b^2 - 4*a*c)^(5/4) - (3*c^2*d^(5/2)*ata n(((b*d + 2*c*d*x)^(1/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c))/(d^(1/2)*(b^2 - 4 *a*c)^(9/4))))/(b^2 - 4*a*c)^(5/4) - (2*c^2*d^5*(b*d + 2*c*d*x)^(3/2) - (6 *c^2*d^3*(b*d + 2*c*d*x)^(7/2))/(4*a*c - b^2))/((b*d + 2*c*d*x)^4 - (b*d + 2*c*d*x)^2*(2*b^2*d^2 - 8*a*c*d^2) + b^4*d^4 + 16*a^2*c^2*d^4 - 8*a*b^2*c *d^4)
Time = 0.21 (sec) , antiderivative size = 1866, normalized size of antiderivative = 10.37 \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^3} \, dx =\text {Too large to display} \] Input:
int((2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^3,x)
Output:
(sqrt(d)*d**2*( - 6*(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 - 12*(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 - 12*(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 - 6*(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**2*c**2*x**2 - 12*(4*a*c - b**2)**(3/4)*sqr t(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 - 6*(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)*s qrt(2)))*c**4*x**4 + 6*(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 + 12*(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 + 12*(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 + 6*(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**2*c**2*x**2 + 12*(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*...