Integrand size = 20, antiderivative size = 272 \[ \int \frac {1}{(d+e x)^3 \left (a+b x+c x^2\right )} \, dx=-\frac {e}{2 \left (c d^2-b d e+a e^2\right ) (d+e x)^2}-\frac {e (2 c d-b e)}{\left (c d^2-b d e+a e^2\right )^2 (d+e x)}-\frac {(2 c d-b e) \left (c^2 d^2+b^2 e^2-c e (b d+3 a e)\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (c d^2-b d e+a e^2\right )^3}+\frac {e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right ) \log (d+e x)}{\left (c d^2-b d e+a e^2\right )^3}-\frac {e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right ) \log \left (a+b x+c x^2\right )}{2 \left (c d^2-b d e+a e^2\right )^3} \] Output:
-1/2*e/(a*e^2-b*d*e+c*d^2)/(e*x+d)^2-e*(-b*e+2*c*d)/(a*e^2-b*d*e+c*d^2)^2/ (e*x+d)-(-b*e+2*c*d)*(c^2*d^2+b^2*e^2-c*e*(3*a*e+b*d))*arctanh((2*c*x+b)/( -4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2)/(a*e^2-b*d*e+c*d^2)^3+e*(3*c^2*d^2+b ^2*e^2-c*e*(a*e+3*b*d))*ln(e*x+d)/(a*e^2-b*d*e+c*d^2)^3-1/2*e*(3*c^2*d^2+b ^2*e^2-c*e*(a*e+3*b*d))*ln(c*x^2+b*x+a)/(a*e^2-b*d*e+c*d^2)^3
Time = 0.20 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(d+e x)^3 \left (a+b x+c x^2\right )} \, dx=-\frac {e}{2 \left (c d^2+e (-b d+a e)\right ) (d+e x)^2}+\frac {e (-2 c d+b e)}{\left (c d^2+e (-b d+a e)\right )^2 (d+e x)}+\frac {(-2 c d+b e) \left (c^2 d^2+b^2 e^2-c e (b d+3 a e)\right ) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c} \left (-c d^2+e (b d-a e)\right )^3}+\frac {e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right ) \log (d+e x)}{\left (c d^2+e (-b d+a e)\right )^3}+\frac {e \left (-3 c^2 d^2-b^2 e^2+c e (3 b d+a e)\right ) \log (a+x (b+c x))}{2 \left (c d^2+e (-b d+a e)\right )^3} \] Input:
Integrate[1/((d + e*x)^3*(a + b*x + c*x^2)),x]
Output:
-1/2*e/((c*d^2 + e*(-(b*d) + a*e))*(d + e*x)^2) + (e*(-2*c*d + b*e))/((c*d ^2 + e*(-(b*d) + a*e))^2*(d + e*x)) + ((-2*c*d + b*e)*(c^2*d^2 + b^2*e^2 - c*e*(b*d + 3*a*e))*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/(Sqrt[-b^2 + 4 *a*c]*(-(c*d^2) + e*(b*d - a*e))^3) + (e*(3*c^2*d^2 + b^2*e^2 - c*e*(3*b*d + a*e))*Log[d + e*x])/(c*d^2 + e*(-(b*d) + a*e))^3 + (e*(-3*c^2*d^2 - b^2 *e^2 + c*e*(3*b*d + a*e))*Log[a + x*(b + c*x)])/(2*(c*d^2 + e*(-(b*d) + a* e))^3)
Time = 0.60 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.07, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {1145, 1200, 2009}
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.
\(\displaystyle \int \frac {1}{(d+e x)^3 \left (a+b x+c x^2\right )} \, dx\) |
\(\Big \downarrow \) 1145 |
\(\displaystyle \frac {\int \frac {c d-b e-c e x}{(d+e x)^2 \left (c x^2+b x+a\right )}dx}{a e^2-b d e+c d^2}-\frac {e}{2 (d+e x)^2 \left (a e^2-b d e+c d^2\right )}\) |
\(\Big \downarrow \) 1200 |
\(\displaystyle \frac {\int \left (\frac {\left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right ) e^2}{\left (c d^2-b e d+a e^2\right )^2 (d+e x)}-\frac {(b e-2 c d) e^2}{\left (c d^2-b e d+a e^2\right ) (d+e x)^2}+\frac {c^3 d^3-3 c^2 e (b d+a e) d-b^3 e^3+b c e^2 (3 b d+2 a e)-c e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right ) x}{\left (c d^2-b e d+a e^2\right )^2 \left (c x^2+b x+a\right )}\right )dx}{a e^2-b d e+c d^2}-\frac {e}{2 (d+e x)^2 \left (a e^2-b d e+c d^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {(2 c d-b e) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \left (-c e (3 a e+b d)+b^2 e^2+c^2 d^2\right )}{\sqrt {b^2-4 a c} \left (a e^2-b d e+c d^2\right )^2}-\frac {e \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right ) \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )^2}+\frac {e \log (d+e x) \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )}{\left (a e^2-b d e+c d^2\right )^2}-\frac {e (2 c d-b e)}{(d+e x) \left (a e^2-b d e+c d^2\right )}}{a e^2-b d e+c d^2}-\frac {e}{2 (d+e x)^2 \left (a e^2-b d e+c d^2\right )}\) |
Input:
Int[1/((d + e*x)^3*(a + b*x + c*x^2)),x]
Output:
-1/2*e/((c*d^2 - b*d*e + a*e^2)*(d + e*x)^2) + (-((e*(2*c*d - b*e))/((c*d^ 2 - b*d*e + a*e^2)*(d + e*x))) - ((2*c*d - b*e)*(c^2*d^2 + b^2*e^2 - c*e*( b*d + 3*a*e))*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(Sqrt[b^2 - 4*a*c]*( c*d^2 - b*d*e + a*e^2)^2) + (e*(3*c^2*d^2 + b^2*e^2 - c*e*(3*b*d + a*e))*L og[d + e*x])/(c*d^2 - b*d*e + a*e^2)^2 - (e*(3*c^2*d^2 + b^2*e^2 - c*e*(3* b*d + a*e))*Log[a + b*x + c*x^2])/(2*(c*d^2 - b*d*e + a*e^2)^2))/(c*d^2 - b*d*e + a*e^2)
Int[((d_.) + (e_.)*(x_))^(m_)/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[e*((d + e*x)^(m + 1)/((m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Simp [1/(c*d^2 - b*d*e + a*e^2) Int[(d + e*x)^(m + 1)*(Simp[c*d - b*e - c*e*x, x]/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && ILtQ[m, -1]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* (x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In tegersQ[n]
Time = 1.15 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.19
method | result | size |
default | \(-\frac {e}{2 \left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (e x +d \right )^{2}}+\frac {e \left (b e -2 c d \right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{2} \left (e x +d \right )}-\frac {e \left (a c \,e^{2}-b^{2} e^{2}+3 b c d e -3 c^{2} d^{2}\right ) \ln \left (e x +d \right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{3}}+\frac {\frac {\left (a \,c^{2} e^{3}-b^{2} c \,e^{3}+3 d \,e^{2} b \,c^{2}-3 d^{2} e \,c^{3}\right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (2 a b c \,e^{3}-3 d \,e^{2} a \,c^{2}-b^{3} e^{3}+3 d \,e^{2} b^{2} c -3 d^{2} e b \,c^{2}+d^{3} c^{3}-\frac {\left (a \,c^{2} e^{3}-b^{2} c \,e^{3}+3 d \,e^{2} b \,c^{2}-3 d^{2} e \,c^{3}\right ) b}{2 c}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{3}}\) | \(325\) |
risch | \(\text {Expression too large to display}\) | \(2086\) |
Input:
int(1/(e*x+d)^3/(c*x^2+b*x+a),x,method=_RETURNVERBOSE)
Output:
-1/2*e/(a*e^2-b*d*e+c*d^2)/(e*x+d)^2+e*(b*e-2*c*d)/(a*e^2-b*d*e+c*d^2)^2/( e*x+d)-e*(a*c*e^2-b^2*e^2+3*b*c*d*e-3*c^2*d^2)/(a*e^2-b*d*e+c*d^2)^3*ln(e* x+d)+1/(a*e^2-b*d*e+c*d^2)^3*(1/2*(a*c^2*e^3-b^2*c*e^3+3*b*c^2*d*e^2-3*c^3 *d^2*e)/c*ln(c*x^2+b*x+a)+2*(2*a*b*c*e^3-3*d*e^2*a*c^2-b^3*e^3+3*d*e^2*b^2 *c-3*d^2*e*b*c^2+d^3*c^3-1/2*(a*c^2*e^3-b^2*c*e^3+3*b*c^2*d*e^2-3*c^3*d^2* e)*b/c)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 1386 vs. \(2 (264) = 528\).
Time = 25.42 (sec) , antiderivative size = 2793, normalized size of antiderivative = 10.27 \[ \int \frac {1}{(d+e x)^3 \left (a+b x+c x^2\right )} \, dx=\text {Too large to display} \] Input:
integrate(1/(e*x+d)^3/(c*x^2+b*x+a),x, algorithm="fricas")
Output:
[-1/2*(5*(b^2*c^2 - 4*a*c^3)*d^4*e - 8*(b^3*c - 4*a*b*c^2)*d^3*e^2 + 3*(b^ 4 - 2*a*b^2*c - 8*a^2*c^2)*d^2*e^3 - 4*(a*b^3 - 4*a^2*b*c)*d*e^4 + (a^2*b^ 2 - 4*a^3*c)*e^5 - (2*c^3*d^5 - 3*b*c^2*d^4*e + 3*(b^2*c - 2*a*c^2)*d^3*e^ 2 - (b^3 - 3*a*b*c)*d^2*e^3 + (2*c^3*d^3*e^2 - 3*b*c^2*d^2*e^3 + 3*(b^2*c - 2*a*c^2)*d*e^4 - (b^3 - 3*a*b*c)*e^5)*x^2 + 2*(2*c^3*d^4*e - 3*b*c^2*d^3 *e^2 + 3*(b^2*c - 2*a*c^2)*d^2*e^3 - (b^3 - 3*a*b*c)*d*e^4)*x)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^2 + 2*b*c*x + b^2 - 2*a*c - sqrt(b^2 - 4*a*c)*(2*c*x + b))/(c*x^2 + b*x + a)) + 2*(2*(b^2*c^2 - 4*a*c^3)*d^3*e^2 - 3*(b^3*c - 4* a*b*c^2)*d^2*e^3 + (b^4 - 2*a*b^2*c - 8*a^2*c^2)*d*e^4 - (a*b^3 - 4*a^2*b* c)*e^5)*x + (3*(b^2*c^2 - 4*a*c^3)*d^4*e - 3*(b^3*c - 4*a*b*c^2)*d^3*e^2 + (b^4 - 5*a*b^2*c + 4*a^2*c^2)*d^2*e^3 + (3*(b^2*c^2 - 4*a*c^3)*d^2*e^3 - 3*(b^3*c - 4*a*b*c^2)*d*e^4 + (b^4 - 5*a*b^2*c + 4*a^2*c^2)*e^5)*x^2 + 2*( 3*(b^2*c^2 - 4*a*c^3)*d^3*e^2 - 3*(b^3*c - 4*a*b*c^2)*d^2*e^3 + (b^4 - 5*a *b^2*c + 4*a^2*c^2)*d*e^4)*x)*log(c*x^2 + b*x + a) - 2*(3*(b^2*c^2 - 4*a*c ^3)*d^4*e - 3*(b^3*c - 4*a*b*c^2)*d^3*e^2 + (b^4 - 5*a*b^2*c + 4*a^2*c^2)* d^2*e^3 + (3*(b^2*c^2 - 4*a*c^3)*d^2*e^3 - 3*(b^3*c - 4*a*b*c^2)*d*e^4 + ( b^4 - 5*a*b^2*c + 4*a^2*c^2)*e^5)*x^2 + 2*(3*(b^2*c^2 - 4*a*c^3)*d^3*e^2 - 3*(b^3*c - 4*a*b*c^2)*d^2*e^3 + (b^4 - 5*a*b^2*c + 4*a^2*c^2)*d*e^4)*x)*l og(e*x + d))/((b^2*c^3 - 4*a*c^4)*d^8 - 3*(b^3*c^2 - 4*a*b*c^3)*d^7*e + 3* (b^4*c - 3*a*b^2*c^2 - 4*a^2*c^3)*d^6*e^2 - (b^5 + 2*a*b^3*c - 24*a^2*b...
Timed out. \[ \int \frac {1}{(d+e x)^3 \left (a+b x+c x^2\right )} \, dx=\text {Timed out} \] Input:
integrate(1/(e*x+d)**3/(c*x**2+b*x+a),x)
Output:
Timed out
Exception generated. \[ \int \frac {1}{(d+e x)^3 \left (a+b x+c x^2\right )} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(e*x+d)^3/(c*x^2+b*x+a),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. 631 vs. \(2 (264) = 528\).
Time = 0.37 (sec) , antiderivative size = 631, normalized size of antiderivative = 2.32 \[ \int \frac {1}{(d+e x)^3 \left (a+b x+c x^2\right )} \, dx=-\frac {{\left (3 \, c^{2} d^{2} e - 3 \, b c d e^{2} + b^{2} e^{3} - a c e^{3}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (c^{3} d^{6} - 3 \, b c^{2} d^{5} e + 3 \, b^{2} c d^{4} e^{2} + 3 \, a c^{2} d^{4} e^{2} - b^{3} d^{3} e^{3} - 6 \, a b c d^{3} e^{3} + 3 \, a b^{2} d^{2} e^{4} + 3 \, a^{2} c d^{2} e^{4} - 3 \, a^{2} b d e^{5} + a^{3} e^{6}\right )}} + \frac {{\left (3 \, c^{2} d^{2} e^{2} - 3 \, b c d e^{3} + b^{2} e^{4} - a c e^{4}\right )} \log \left ({\left | e x + d \right |}\right )}{c^{3} d^{6} e - 3 \, b c^{2} d^{5} e^{2} + 3 \, b^{2} c d^{4} e^{3} + 3 \, a c^{2} d^{4} e^{3} - b^{3} d^{3} e^{4} - 6 \, a b c d^{3} e^{4} + 3 \, a b^{2} d^{2} e^{5} + 3 \, a^{2} c d^{2} e^{5} - 3 \, a^{2} b d e^{6} + a^{3} e^{7}} + \frac {{\left (2 \, c^{3} d^{3} - 3 \, b c^{2} d^{2} e + 3 \, b^{2} c d e^{2} - 6 \, a c^{2} d e^{2} - b^{3} e^{3} + 3 \, a b c e^{3}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (c^{3} d^{6} - 3 \, b c^{2} d^{5} e + 3 \, b^{2} c d^{4} e^{2} + 3 \, a c^{2} d^{4} e^{2} - b^{3} d^{3} e^{3} - 6 \, a b c d^{3} e^{3} + 3 \, a b^{2} d^{2} e^{4} + 3 \, a^{2} c d^{2} e^{4} - 3 \, a^{2} b d e^{5} + a^{3} e^{6}\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {5 \, c^{2} d^{4} e - 8 \, b c d^{3} e^{2} + 3 \, b^{2} d^{2} e^{3} + 6 \, a c d^{2} e^{3} - 4 \, a b d e^{4} + a^{2} e^{5} + 2 \, {\left (2 \, c^{2} d^{3} e^{2} - 3 \, b c d^{2} e^{3} + b^{2} d e^{4} + 2 \, a c d e^{4} - a b e^{5}\right )} x}{2 \, {\left (c d^{2} - b d e + a e^{2}\right )}^{3} {\left (e x + d\right )}^{2}} \] Input:
integrate(1/(e*x+d)^3/(c*x^2+b*x+a),x, algorithm="giac")
Output:
-1/2*(3*c^2*d^2*e - 3*b*c*d*e^2 + b^2*e^3 - a*c*e^3)*log(c*x^2 + b*x + a)/ (c^3*d^6 - 3*b*c^2*d^5*e + 3*b^2*c*d^4*e^2 + 3*a*c^2*d^4*e^2 - b^3*d^3*e^3 - 6*a*b*c*d^3*e^3 + 3*a*b^2*d^2*e^4 + 3*a^2*c*d^2*e^4 - 3*a^2*b*d*e^5 + a ^3*e^6) + (3*c^2*d^2*e^2 - 3*b*c*d*e^3 + b^2*e^4 - a*c*e^4)*log(abs(e*x + d))/(c^3*d^6*e - 3*b*c^2*d^5*e^2 + 3*b^2*c*d^4*e^3 + 3*a*c^2*d^4*e^3 - b^3 *d^3*e^4 - 6*a*b*c*d^3*e^4 + 3*a*b^2*d^2*e^5 + 3*a^2*c*d^2*e^5 - 3*a^2*b*d *e^6 + a^3*e^7) + (2*c^3*d^3 - 3*b*c^2*d^2*e + 3*b^2*c*d*e^2 - 6*a*c^2*d*e ^2 - b^3*e^3 + 3*a*b*c*e^3)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((c^3*d ^6 - 3*b*c^2*d^5*e + 3*b^2*c*d^4*e^2 + 3*a*c^2*d^4*e^2 - b^3*d^3*e^3 - 6*a *b*c*d^3*e^3 + 3*a*b^2*d^2*e^4 + 3*a^2*c*d^2*e^4 - 3*a^2*b*d*e^5 + a^3*e^6 )*sqrt(-b^2 + 4*a*c)) - 1/2*(5*c^2*d^4*e - 8*b*c*d^3*e^2 + 3*b^2*d^2*e^3 + 6*a*c*d^2*e^3 - 4*a*b*d*e^4 + a^2*e^5 + 2*(2*c^2*d^3*e^2 - 3*b*c*d^2*e^3 + b^2*d*e^4 + 2*a*c*d*e^4 - a*b*e^5)*x)/((c*d^2 - b*d*e + a*e^2)^3*(e*x + d)^2)
Time = 9.97 (sec) , antiderivative size = 3506, normalized size of antiderivative = 12.89 \[ \int \frac {1}{(d+e x)^3 \left (a+b x+c x^2\right )} \, dx=\text {Too large to display} \] Input:
int(1/((d + e*x)^3*(a + b*x + c*x^2)),x)
Output:
(log((27*d*e^6*(b^2 - 4*a*c)^(7/2))/16 + (9*e^7*x*(b^2 - 4*a*c)^(7/2))/16 - 8*a*b^6*e^7 + 4*b*c^6*d^7 - 8*b^7*e^7*x + 8*c^7*d^7*x + 4*c^6*d^7*(b^2 - 4*a*c)^(1/2) + 72*a^4*c^3*e^7 + (57*b^2*e^7*x*(b^2 - 4*a*c)^(5/2))/16 + ( 51*b^4*e^7*x*(b^2 - 4*a*c)^(3/2))/16 + (11*b^6*e^7*x*(b^2 - 4*a*c)^(1/2))/ 16 + 60*a^2*b^4*c*e^7 + 8*b^2*c^5*d^6*e + 4*b^6*c*d^2*e^5 + (75*c^2*d^3*e^ 4*(b^2 - 4*a*c)^(5/2))/4 + 25*c^4*d^5*e^2*(b^2 - 4*a*c)^(3/2) - 132*a^3*b^ 2*c^2*e^7 + 408*a^2*c^5*d^4*e^3 - 456*a^3*c^4*d^2*e^5 - 20*b^3*c^4*d^5*e^2 + 28*b^4*c^3*d^4*e^3 - 16*b^5*c^2*d^3*e^4 + (9*a*b*e^7*(b^2 - 4*a*c)^(5/2 ))/4 - 88*a*c^6*d^6*e + (9*a*b^3*e^7*(b^2 - 4*a*c)^(3/2))/2 + (5*a*b^5*e^7 *(b^2 - 4*a*c)^(1/2))/4 + (111*b^2*d*e^6*(b^2 - 4*a*c)^(5/2))/16 - (79*b^4 *d*e^6*(b^2 - 4*a*c)^(3/2))/16 - (59*b^6*d*e^6*(b^2 - 4*a*c)^(1/2))/16 + 4 0*a*b^5*c*d*e^6 + (23*b^2*c^2*d^3*e^4*(b^2 - 4*a*c)^(3/2))/2 - 45*b^2*c^4* d^5*e^2*(b^2 - 4*a*c)^(1/2) + 65*b^3*c^3*d^4*e^3*(b^2 - 4*a*c)^(1/2) - (18 5*b^4*c^2*d^3*e^4*(b^2 - 4*a*c)^(1/2))/4 + 64*a*b^5*c*e^7*x - 28*b*c^6*d^6 *e*x + 48*b^6*c*d*e^6*x + 504*a^2*b^2*c^3*d^2*e^5 - 21*b*c*d^2*e^5*(b^2 - 4*a*c)^(5/2) + 8*b*c^5*d^6*e*(b^2 - 4*a*c)^(1/2) + 44*c^6*d^6*e*x*(b^2 - 4 *a*c)^(1/2) + 164*a*b*c^5*d^5*e^2 + 348*a^3*b*c^3*d*e^6 + 108*a^3*b*c^3*e^ 7*x - 200*a*c^6*d^5*e^2*x - 216*a^3*c^4*d*e^6*x - 37*b*c^3*d^4*e^3*(b^2 - 4*a*c)^(3/2) + 7*b^3*c*d^2*e^5*(b^2 - 4*a*c)^(3/2) + 18*b^5*c*d^2*e^5*(b^2 - 4*a*c)^(1/2) + (57*c^2*d^2*e^5*x*(b^2 - 4*a*c)^(5/2))/4 + 51*c^4*d^4...
Time = 0.20 (sec) , antiderivative size = 2708, normalized size of antiderivative = 9.96 \[ \int \frac {1}{(d+e x)^3 \left (a+b x+c x^2\right )} \, dx =\text {Too large to display} \] Input:
int(1/(e*x+d)^3/(c*x^2+b*x+a),x)
Output:
(6*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*b*c*d**3*e**3 + 12*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*b*c*d**2*e **4*x + 6*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*b*c*d* e**5*x**2 - 12*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*c **2*d**4*e**2 - 24*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2)) *a*c**2*d**3*e**3*x - 12*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*c**2*d**2*e**4*x**2 - 2*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt( 4*a*c - b**2))*b**3*d**3*e**3 - 4*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt (4*a*c - b**2))*b**3*d**2*e**4*x - 2*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/s qrt(4*a*c - b**2))*b**3*d*e**5*x**2 + 6*sqrt(4*a*c - b**2)*atan((b + 2*c*x )/sqrt(4*a*c - b**2))*b**2*c*d**4*e**2 + 12*sqrt(4*a*c - b**2)*atan((b + 2 *c*x)/sqrt(4*a*c - b**2))*b**2*c*d**3*e**3*x + 6*sqrt(4*a*c - b**2)*atan(( b + 2*c*x)/sqrt(4*a*c - b**2))*b**2*c*d**2*e**4*x**2 - 6*sqrt(4*a*c - b**2 )*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b*c**2*d**5*e - 12*sqrt(4*a*c - b** 2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b*c**2*d**4*e**2*x - 6*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b*c**2*d**3*e**3*x**2 + 4*sqr t(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*c**3*d**6 + 8*sqrt(4* a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*c**3*d**5*e*x + 4*sqrt(4* a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*c**3*d**4*e**2*x**2 + 4*l og(a + b*x + c*x**2)*a**2*c**2*d**3*e**3 + 8*log(a + b*x + c*x**2)*a**2...