Integrand size = 26, antiderivative size = 176 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^2} \, dx=-\frac {28 c}{3 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{3/2}}-\frac {1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )}+\frac {14 c \arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{11/4} d^{5/2}}+\frac {14 c \text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{11/4} d^{5/2}} \] Output:
-28/3*c/(-4*a*c+b^2)^2/d/(2*c*d*x+b*d)^(3/2)-1/(-4*a*c+b^2)/d/(2*c*d*x+b*d )^(3/2)/(c*x^2+b*x+a)+14*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)^(11/4)/d^(5/2)+14*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)^(11/4)/d^(5/2)
Result contains complex when optimal does not.
Time = 0.99 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.52 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^2} \, dx=\frac {\left (\frac {1}{3}+\frac {i}{3}\right ) c \left (-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) (b+2 c x) \left (-4 b^2+16 a c+7 (b+2 c x)^2\right )}{c \left (b^2-4 a c\right )^2 (a+x (b+c x))}+\frac {21 i (b+2 c x)^{5/2} \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{11/4}}-\frac {21 i (b+2 c x)^{5/2} \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{11/4}}-\frac {21 i (b+2 c x)^{5/2} \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )}{\left (b^2-4 a c\right )^{11/4}}\right )}{(d (b+2 c x))^{5/2}} \] Input:
Integrate[1/((b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)^2),x]
Output:
((1/3 + I/3)*c*(((-1/2 + I/2)*(b + 2*c*x)*(-4*b^2 + 16*a*c + 7*(b + 2*c*x) ^2))/(c*(b^2 - 4*a*c)^2*(a + x*(b + c*x))) + ((21*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)^(11/ 4) - ((21*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)^(11/4) - ((21*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)^(11/4)))/(d*(b + 2*c*x))^(5/2)
Time = 0.38 (sec) , antiderivative size = 207, 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, 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.
| \(\displaystyle \int \frac {1}{\left (a+b x+c x^2\right )^2 (b d+2 c d x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1111 |
| \(\displaystyle -\frac {7 c \int \frac {1}{(b d+2 c x d)^{5/2} \left (c x^2+b x+a\right )}dx}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{3/2}}\) |
| \(\Big \downarrow \) 1117 |
| \(\displaystyle -\frac {7 c \left (\frac {\int \frac {1}{\sqrt {b d+2 c x d} \left (c x^2+b x+a\right )}dx}{d^2 \left (b^2-4 a c\right )}+\frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{3/2}}\) |
| \(\Big \downarrow \) 1118 |
| \(\displaystyle -\frac {7 c \left (\frac {\int \frac {4 c d^2}{\sqrt {b d+2 c x d} \left (\left (4 a-\frac {b^2}{c}\right ) c d^2+(b d+2 c x d)^2\right )}d(b d+2 c x d)}{2 c d^3 \left (b^2-4 a c\right )}+\frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {7 c \left (\frac {2 \int -\frac {1}{\sqrt {b d+2 c x d} \left (\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2\right )}d(b d+2 c x d)}{d \left (b^2-4 a c\right )}+\frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{3/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {7 c \left (\frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}-\frac {2 \int \frac {1}{\sqrt {b d+2 c x d} \left (\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2\right )}d(b d+2 c x d)}{d \left (b^2-4 a c\right )}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{3/2}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle -\frac {7 c \left (\frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}-\frac {4 \int \frac {1}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d\sqrt {b d+2 c x d}}{d \left (b^2-4 a c\right )}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{3/2}}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle -\frac {7 c \left (\frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}-\frac {4 \left (\frac {\int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}}{2 d \sqrt {b^2-4 a c}}+\frac {\int \frac {1}{b d+2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}}{2 d \sqrt {b^2-4 a c}}\right )}{d \left (b^2-4 a c\right )}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{3/2}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {7 c \left (\frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}-\frac {4 \left (\frac {\int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}}{2 d \sqrt {b^2-4 a c}}+\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 d^{3/2} \left (b^2-4 a c\right )^{3/4}}\right )}{d \left (b^2-4 a c\right )}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{3/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {7 c \left (\frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}-\frac {4 \left (\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 d^{3/2} \left (b^2-4 a c\right )^{3/4}}+\frac {\text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 d^{3/2} \left (b^2-4 a c\right )^{3/4}}\right )}{d \left (b^2-4 a c\right )}\right )}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{3/2}}\) |
Input:
Int[1/((b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)^2),x]
Output:
-(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[S qrt[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)
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]
Leaf count of result is larger than twice the leaf count of optimal. \(317\) vs. \(2(150)=300\).
Time = 1.36 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.81
| method | result | size |
| pseudoelliptic | \(\frac {8 c \,d^{3} \left (-\frac {2}{3 d^{4} \left (d \left (2 c x +b \right )\right )^{\frac {3}{2}}}-\frac {\sqrt {d \left (2 c x +b \right )}}{8 c \,d^{6} \left (c \,x^{2}+b x +a \right )}-\frac {7 \ln \left (\frac {\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}{\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}\right ) \sqrt {2}}{16 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}} d^{4}}-\frac {7 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}+1\right ) \sqrt {2}}{8 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}} d^{4}}-\frac {7 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}-1\right ) \sqrt {2}}{8 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}} d^{4}}\right )}{\left (4 a c -b^{2}\right )^{2}}\) | \(318\) |
| derivativedivides | \(16 c \,d^{3} \left (-\frac {1}{3 d^{4} \left (4 a c -b^{2}\right )^{2} \left (2 c d x +b d \right )^{\frac {3}{2}}}-\frac {\frac {\sqrt {2 c d x +b d}}{16 a \,d^{2} c -4 b^{2} d^{2}+4 \left (2 c d x +b d \right )^{2}}+\frac {7 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{32 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {3}{4}}}}{d^{4} \left (4 a c -b^{2}\right )^{2}}\right )\) | \(323\) |
| default | \(16 c \,d^{3} \left (-\frac {1}{3 d^{4} \left (4 a c -b^{2}\right )^{2} \left (2 c d x +b d \right )^{\frac {3}{2}}}-\frac {\frac {\sqrt {2 c d x +b d}}{16 a \,d^{2} c -4 b^{2} d^{2}+4 \left (2 c d x +b d \right )^{2}}+\frac {7 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{32 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {3}{4}}}}{d^{4} \left (4 a c -b^{2}\right )^{2}}\right )\) | \(323\) |
Input:
int(1/(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
8*c*d^3/(4*a*c-b^2)^2*(-2/3/d^4/(d*(2*c*x+b))^(3/2)-1/8*(d*(2*c*x+b))^(1/2 )/c/d^6/(c*x^2+b*x+a)-7/16*ln(((d^2*(4*a*c-b^2))^(1/4)*(d*(2*c*x+b))^(1/2) *2^(1/2)+(d^2*(4*a*c-b^2))^(1/2)+d*(2*c*x+b))/((d^2*(4*a*c-b^2))^(1/2)-(d^ 2*(4*a*c-b^2))^(1/4)*(d*(2*c*x+b))^(1/2)*2^(1/2)+d*(2*c*x+b)))/(d^2*(4*a*c -b^2))^(3/4)*2^(1/2)/d^4-7/8*arctan(2^(1/2)/(d^2*(4*a*c-b^2))^(1/4)*(d*(2* c*x+b))^(1/2)+1)/(d^2*(4*a*c-b^2))^(3/4)*2^(1/2)/d^4-7/8*arctan(2^(1/2)/(d ^2*(4*a*c-b^2))^(1/4)*(d*(2*c*x+b))^(1/2)-1)/(d^2*(4*a*c-b^2))^(3/4)*2^(1/ 2)/d^4)
Result contains complex when optimal does not.
Time = 0.12 (sec) , antiderivative size = 2171, normalized size of antiderivative = 12.34 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^2} \, dx=\text {Too large to display} \] Input:
integrate(1/(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^2,x, algorithm="fricas")
Output:
1/3*(21*(4*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^3*x^4 + 8*(b^5*c^2 - 8*a *b^3*c^3 + 16*a^2*b*c^4)*d^3*x^3 + (5*b^6*c - 36*a*b^4*c^2 + 48*a^2*b^2*c^ 3 + 64*a^3*c^4)*d^3*x^2 + (b^7 - 4*a*b^5*c - 16*a^2*b^3*c^2 + 64*a^3*b*c^3 )*d^3*x + (a*b^6 - 8*a^2*b^4*c + 16*a^3*b^2*c^2)*d^3)*(c^4/((b^22 - 44*a*b ^20*c + 880*a^2*b^18*c^2 - 10560*a^3*b^16*c^3 + 84480*a^4*b^14*c^4 - 47308 8*a^5*b^12*c^5 + 1892352*a^6*b^10*c^6 - 5406720*a^7*b^8*c^7 + 10813440*a^8 *b^6*c^8 - 14417920*a^9*b^4*c^9 + 11534336*a^10*b^2*c^10 - 4194304*a^11*c^ 11)*d^10))^(1/4)*log(7*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^ 3*(c^4/((b^22 - 44*a*b^20*c + 880*a^2*b^18*c^2 - 10560*a^3*b^16*c^3 + 8448 0*a^4*b^14*c^4 - 473088*a^5*b^12*c^5 + 1892352*a^6*b^10*c^6 - 5406720*a^7* b^8*c^7 + 10813440*a^8*b^6*c^8 - 14417920*a^9*b^4*c^9 + 11534336*a^10*b^2* c^10 - 4194304*a^11*c^11)*d^10))^(1/4) + 7*sqrt(2*c*d*x + b*d)*c) - 21*(-4 *I*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^3*x^4 - 8*I*(b^5*c^2 - 8*a*b^3*c ^3 + 16*a^2*b*c^4)*d^3*x^3 - I*(5*b^6*c - 36*a*b^4*c^2 + 48*a^2*b^2*c^3 + 64*a^3*c^4)*d^3*x^2 - I*(b^7 - 4*a*b^5*c - 16*a^2*b^3*c^2 + 64*a^3*b*c^3)* d^3*x - I*(a*b^6 - 8*a^2*b^4*c + 16*a^3*b^2*c^2)*d^3)*(c^4/((b^22 - 44*a*b ^20*c + 880*a^2*b^18*c^2 - 10560*a^3*b^16*c^3 + 84480*a^4*b^14*c^4 - 47308 8*a^5*b^12*c^5 + 1892352*a^6*b^10*c^6 - 5406720*a^7*b^8*c^7 + 10813440*a^8 *b^6*c^8 - 14417920*a^9*b^4*c^9 + 11534336*a^10*b^2*c^10 - 4194304*a^11*c^ 11)*d^10))^(1/4)*log(7*I*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^...
Timed out. \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^2} \, dx=\text {Timed out} \] Input:
integrate(1/(2*c*d*x+b*d)**(5/2)/(c*x**2+b*x+a)**2,x)
Output:
Timed out
Exception generated. \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(2*c*d*x+b*d)^(5/2)/(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. 646 vs. \(2 (150) = 300\).
Time = 0.13 (sec) , antiderivative size = 646, normalized size of antiderivative = 3.67 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^2} \, dx=\frac {7 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{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 {7 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{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 {7 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{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 {7 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{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 \, \sqrt {2 \, c d x + b d} c}{{\left (b^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d\right )} {\left (b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}\right )}} - \frac {16 \, c}{3 \, {\left (b^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d\right )} {\left (2 \, c d x + b d\right )}^{\frac {3}{2}}} \] Input:
integrate(1/(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^2,x, algorithm="giac")
Output:
7*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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) + 7* sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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) + 7*( -b^2*d^2 + 4*a*c*d^2)^(1/4)*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) - 7*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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*sqrt(2*c*d*x + b*d)*c/((b^4*d - 8*a*b^2*c *d + 16*a^2*c^2*d)*(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2)) - 16/3*c/((b ^4*d - 8*a*b^2*c*d + 16*a^2*c^2*d)*(2*c*d*x + b*d)^(3/2))
Time = 5.33 (sec) , antiderivative size = 2280, normalized size of antiderivative = 12.95 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^2} \, dx=\text {Too large to display} \] Input:
int(1/((b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)^2),x)
Output:
((16*c*d)/(3*(4*a*c - b^2)) + (28*c*(b*d + 2*c*d*x)^2)/(3*d*(4*a*c - b^2)^ 2))/((b*d + 2*c*d*x)^(3/2)*(b^2*d^2 - 4*a*c*d^2) - (b*d + 2*c*d*x)^(7/2)) + (c*atan(((c*((b*d + 2*c*d*x)^(1/2)*(12845056*a^6*c^8*d^3 + 3136*b^12*c^2 *d^3 - 75264*a*b^10*c^3*d^3 + 752640*a^2*b^8*c^4*d^3 - 4014080*a^3*b^6*c^5 *d^3 + 12042240*a^4*b^4*c^6*d^3 - 19267584*a^5*b^2*c^7*d^3) - (7*c*(448*b^ 18*c*d^6 - 117440512*a^9*c^10*d^6 - 16128*a*b^16*c^2*d^6 + 258048*a^2*b^14 *c^3*d^6 - 2408448*a^3*b^12*c^4*d^6 + 14450688*a^4*b^10*c^5*d^6 - 57802752 *a^5*b^8*c^6*d^6 + 154140672*a^6*b^6*c^7*d^6 - 264241152*a^7*b^4*c^8*d^6 + 264241152*a^8*b^2*c^9*d^6))/(d^(5/2)*(b^2 - 4*a*c)^(11/4)))*7i)/(d^(5/2)* (b^2 - 4*a*c)^(11/4)) + (c*((b*d + 2*c*d*x)^(1/2)*(12845056*a^6*c^8*d^3 + 3136*b^12*c^2*d^3 - 75264*a*b^10*c^3*d^3 + 752640*a^2*b^8*c^4*d^3 - 401408 0*a^3*b^6*c^5*d^3 + 12042240*a^4*b^4*c^6*d^3 - 19267584*a^5*b^2*c^7*d^3) + (7*c*(448*b^18*c*d^6 - 117440512*a^9*c^10*d^6 - 16128*a*b^16*c^2*d^6 + 25 8048*a^2*b^14*c^3*d^6 - 2408448*a^3*b^12*c^4*d^6 + 14450688*a^4*b^10*c^5*d ^6 - 57802752*a^5*b^8*c^6*d^6 + 154140672*a^6*b^6*c^7*d^6 - 264241152*a^7* b^4*c^8*d^6 + 264241152*a^8*b^2*c^9*d^6))/(d^(5/2)*(b^2 - 4*a*c)^(11/4)))* 7i)/(d^(5/2)*(b^2 - 4*a*c)^(11/4)))/((7*c*((b*d + 2*c*d*x)^(1/2)*(12845056 *a^6*c^8*d^3 + 3136*b^12*c^2*d^3 - 75264*a*b^10*c^3*d^3 + 752640*a^2*b^8*c ^4*d^3 - 4014080*a^3*b^6*c^5*d^3 + 12042240*a^4*b^4*c^6*d^3 - 19267584*a^5 *b^2*c^7*d^3) - (7*c*(448*b^18*c*d^6 - 117440512*a^9*c^10*d^6 - 16128*a...
Time = 0.23 (sec) , antiderivative size = 1651, normalized size of antiderivative = 9.38 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:
int(1/(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^2,x)
Output:
(sqrt(d)*(42*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*b*c + 84*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**2*x + 42*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**2*c*x + 126*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**2*x**2 + 84*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**3*x**3 - 42*sqrt(b + 2*c*x)*(4*a*c - b**2)**(1/4)*sqrt(2)*ata n(((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 - 84*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**2*x - 42*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)))*b**2*c*x - 126*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**2*x**2 - 84*sqrt(b + 2*c*x)*(4*a*c - b**2)**(1/4)*sq rt(2)*atan(((4*a*c - b**2)**(1/4)*sqrt(2) + 2*sqrt(b + 2*c*x))/((4*a*c ...