Integrand size = 17, antiderivative size = 496 \[ \int \frac {1}{x^2 \left (a+b (c+d x)^4\right )} \, dx=-\frac {1}{\left (a+b c^4\right ) x}-\frac {\sqrt {b} c \left (a-b c^4\right ) d \arctan \left (\frac {\sqrt {b} (c+d x)^2}{\sqrt {a}}\right )}{\sqrt {a} \left (a+b c^4\right )^2}+\frac {\sqrt [4]{b} \left (\sqrt {a} \left (a-3 b c^4\right )+\sqrt {b} c^2 \left (3 a-b c^4\right )\right ) d \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} \left (a+b c^4\right )^2}-\frac {\sqrt [4]{b} \left (\sqrt {a} \left (a-3 b c^4\right )+\sqrt {b} c^2 \left (3 a-b c^4\right )\right ) d \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} \left (a+b c^4\right )^2}-\frac {4 b c^3 d \log (x)}{\left (a+b c^4\right )^2}-\frac {\sqrt [4]{b} \left (\sqrt {a} \left (a-3 b c^4\right )-\sqrt {b} c^2 \left (3 a-b c^4\right )\right ) d \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} (c+d x)+\sqrt {b} (c+d x)^2\right )}{4 \sqrt {2} a^{3/4} \left (a+b c^4\right )^2}+\frac {\sqrt [4]{b} \left (\sqrt {a} \left (a-3 b c^4\right )-\sqrt {b} c^2 \left (3 a-b c^4\right )\right ) d \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} (c+d x)+\sqrt {b} (c+d x)^2\right )}{4 \sqrt {2} a^{3/4} \left (a+b c^4\right )^2}+\frac {b c^3 d \log \left (a+b (c+d x)^4\right )}{\left (a+b c^4\right )^2} \]
-1/(b*c^4+a)/x-4*b*c^3*d*ln(x)/(b*c^4+a)^2+b*c^3*d*ln(a+b*(d*x+c)^4)/(b*c^ 4+a)^2-c*(-b*c^4+a)*d*arctan((d*x+c)^2*b^(1/2)/a^(1/2))*b^(1/2)/(b*c^4+a)^ 2/a^(1/2)-1/8*b^(1/4)*d*ln(-a^(1/4)*b^(1/4)*(d*x+c)*2^(1/2)+a^(1/2)+(d*x+c )^2*b^(1/2))*((-3*b*c^4+a)*a^(1/2)-c^2*(-b*c^4+3*a)*b^(1/2))/a^(3/4)/(b*c^ 4+a)^2*2^(1/2)+1/8*b^(1/4)*d*ln(a^(1/4)*b^(1/4)*(d*x+c)*2^(1/2)+a^(1/2)+(d *x+c)^2*b^(1/2))*((-3*b*c^4+a)*a^(1/2)-c^2*(-b*c^4+3*a)*b^(1/2))/a^(3/4)/( b*c^4+a)^2*2^(1/2)-1/4*b^(1/4)*d*arctan(-1+b^(1/4)*(d*x+c)*2^(1/2)/a^(1/4) )*((-3*b*c^4+a)*a^(1/2)+c^2*(-b*c^4+3*a)*b^(1/2))/a^(3/4)/(b*c^4+a)^2*2^(1 /2)-1/4*b^(1/4)*d*arctan(1+b^(1/4)*(d*x+c)*2^(1/2)/a^(1/4))*((-3*b*c^4+a)* a^(1/2)+c^2*(-b*c^4+3*a)*b^(1/2))/a^(3/4)/(b*c^4+a)^2*2^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.09 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.48 \[ \int \frac {1}{x^2 \left (a+b (c+d x)^4\right )} \, dx=\frac {-4 \left (a+b c^4+4 b c^3 d x \log (x)\right )+d x \text {RootSum}\left [a+b c^4+4 b c^3 d \text {$\#$1}+6 b c^2 d^2 \text {$\#$1}^2+4 b c d^3 \text {$\#$1}^3+b d^4 \text {$\#$1}^4\&,\frac {-6 a c^2 \log (x-\text {$\#$1})+10 b c^6 \log (x-\text {$\#$1})-4 a c d \log (x-\text {$\#$1}) \text {$\#$1}+20 b c^5 d \log (x-\text {$\#$1}) \text {$\#$1}-a d^2 \log (x-\text {$\#$1}) \text {$\#$1}^2+15 b c^4 d^2 \log (x-\text {$\#$1}) \text {$\#$1}^2+4 b c^3 d^3 \log (x-\text {$\#$1}) \text {$\#$1}^3}{c^3+3 c^2 d \text {$\#$1}+3 c d^2 \text {$\#$1}^2+d^3 \text {$\#$1}^3}\&\right ]}{4 \left (a+b c^4\right )^2 x} \]
(-4*(a + b*c^4 + 4*b*c^3*d*x*Log[x]) + d*x*RootSum[a + b*c^4 + 4*b*c^3*d*# 1 + 6*b*c^2*d^2*#1^2 + 4*b*c*d^3*#1^3 + b*d^4*#1^4 & , (-6*a*c^2*Log[x - # 1] + 10*b*c^6*Log[x - #1] - 4*a*c*d*Log[x - #1]*#1 + 20*b*c^5*d*Log[x - #1 ]*#1 - a*d^2*Log[x - #1]*#1^2 + 15*b*c^4*d^2*Log[x - #1]*#1^2 + 4*b*c^3*d^ 3*Log[x - #1]*#1^3)/(c^3 + 3*c^2*d*#1 + 3*c*d^2*#1^2 + d^3*#1^3) & ])/(4*( a + b*c^4)^2*x)
Time = 0.96 (sec) , antiderivative size = 497, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {896, 7276, 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.
| \(\displaystyle \int \frac {1}{x^2 \left (a+b (c+d x)^4\right )} \, dx\) |
| \(\Big \downarrow \) 896 |
| \(\displaystyle d \int \frac {1}{d^2 x^2 \left (b (c+d x)^4+a\right )}d(c+d x)\) |
| \(\Big \downarrow \) 7276 |
| \(\displaystyle d \int \left (-\frac {4 b c^3}{\left (b c^4+a\right )^2 d x}+\frac {b \left (4 b c^3 (c+d x)^3-\left (a-3 b c^4\right ) (c+d x)^2-2 c \left (a-b c^4\right ) (c+d x)-c^2 \left (3 a-b c^4\right )\right )}{\left (b c^4+a\right )^2 \left (b (c+d x)^4+a\right )}+\frac {1}{\left (b c^4+a\right ) d^2 x^2}\right )d(c+d x)\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle d \left (\frac {\sqrt [4]{b} \left (\sqrt {a} \left (a-3 b c^4\right )+\sqrt {b} c^2 \left (3 a-b c^4\right )\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} \left (a+b c^4\right )^2}-\frac {\sqrt [4]{b} \left (\sqrt {a} \left (a-3 b c^4\right )+\sqrt {b} c^2 \left (3 a-b c^4\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}+1\right )}{2 \sqrt {2} a^{3/4} \left (a+b c^4\right )^2}-\frac {\sqrt [4]{b} \left (\sqrt {a} \left (a-3 b c^4\right )-\sqrt {b} c^2 \left (3 a-b c^4\right )\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} (c+d x)+\sqrt {a}+\sqrt {b} (c+d x)^2\right )}{4 \sqrt {2} a^{3/4} \left (a+b c^4\right )^2}+\frac {\sqrt [4]{b} \left (\sqrt {a} \left (a-3 b c^4\right )-\sqrt {b} c^2 \left (3 a-b c^4\right )\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} (c+d x)+\sqrt {a}+\sqrt {b} (c+d x)^2\right )}{4 \sqrt {2} a^{3/4} \left (a+b c^4\right )^2}-\frac {\sqrt {b} c \left (a-b c^4\right ) \arctan \left (\frac {\sqrt {b} (c+d x)^2}{\sqrt {a}}\right )}{\sqrt {a} \left (a+b c^4\right )^2}-\frac {1}{d x \left (a+b c^4\right )}-\frac {4 b c^3 \log (-d x)}{\left (a+b c^4\right )^2}+\frac {b c^3 \log \left (a+b (c+d x)^4\right )}{\left (a+b c^4\right )^2}\right )\) |
d*(-(1/((a + b*c^4)*d*x)) - (Sqrt[b]*c*(a - b*c^4)*ArcTan[(Sqrt[b]*(c + d* x)^2)/Sqrt[a]])/(Sqrt[a]*(a + b*c^4)^2) + (b^(1/4)*(Sqrt[a]*(a - 3*b*c^4) + Sqrt[b]*c^2*(3*a - b*c^4))*ArcTan[1 - (Sqrt[2]*b^(1/4)*(c + d*x))/a^(1/4 )])/(2*Sqrt[2]*a^(3/4)*(a + b*c^4)^2) - (b^(1/4)*(Sqrt[a]*(a - 3*b*c^4) + Sqrt[b]*c^2*(3*a - b*c^4))*ArcTan[1 + (Sqrt[2]*b^(1/4)*(c + d*x))/a^(1/4)] )/(2*Sqrt[2]*a^(3/4)*(a + b*c^4)^2) - (4*b*c^3*Log[-(d*x)])/(a + b*c^4)^2 - (b^(1/4)*(Sqrt[a]*(a - 3*b*c^4) - Sqrt[b]*c^2*(3*a - b*c^4))*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*(c + d*x) + Sqrt[b]*(c + d*x)^2])/(4*Sqrt[2]*a^ (3/4)*(a + b*c^4)^2) + (b^(1/4)*(Sqrt[a]*(a - 3*b*c^4) - Sqrt[b]*c^2*(3*a - b*c^4))*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*(c + d*x) + Sqrt[b]*(c + d *x)^2])/(4*Sqrt[2]*a^(3/4)*(a + b*c^4)^2) + (b*c^3*Log[a + b*(c + d*x)^4]) /(a + b*c^4)^2)
3.2.15.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coeff icient[v, x, 0], d = Coefficient[v, x, 1]}, Simp[1/d^(m + 1) Subst[Int[Si mplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.14 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.37
| method | result | size |
| default | \(-\frac {1}{\left (b \,c^{4}+a \right ) x}-\frac {4 b \,c^{3} d \ln \left (x \right )}{\left (b \,c^{4}+a \right )^{2}}-\frac {d \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,d^{4} \textit {\_Z}^{4}+4 b c \,d^{3} \textit {\_Z}^{3}+6 b \,c^{2} d^{2} \textit {\_Z}^{2}+4 b \,c^{3} d \textit {\_Z} +b \,c^{4}+a \right )}{\sum }\frac {\left (-4 b \,d^{3} c^{3} \textit {\_R}^{3}+d^{2} \left (-15 b \,c^{4}+a \right ) \textit {\_R}^{2}+4 c d \left (-5 b \,c^{4}+a \right ) \textit {\_R} -10 b \,c^{6}+6 a \,c^{2}\right ) \ln \left (x -\textit {\_R} \right )}{d^{3} \textit {\_R}^{3}+3 c \,d^{2} \textit {\_R}^{2}+3 c^{2} d \textit {\_R} +c^{3}}\right )}{4 \left (b \,c^{4}+a \right )^{2}}\) | \(184\) |
| risch | \(-\frac {1}{\left (b \,c^{4}+a \right ) x}-\frac {4 b \,c^{3} d \ln \left (x \right )}{b^{2} c^{8}+2 a b \,c^{4}+a^{2}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (a^{3} b^{2} c^{8}+2 a^{4} b \,c^{4}+a^{5}\right ) \textit {\_Z}^{4}-16 a^{3} b \,c^{3} d \,\textit {\_Z}^{3}+20 a^{2} b \,c^{2} d^{2} \textit {\_Z}^{2}-8 a b c \,d^{3} \textit {\_Z} +b \,d^{4}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (-3 a^{2} b^{3} c^{12} d -a^{3} b^{2} c^{8} d +7 a^{4} b \,c^{4} d +5 d \,a^{5}\right ) \textit {\_R}^{4}+\left (4 a \,b^{3} c^{11} d^{2}-40 a^{2} b^{2} c^{7} d^{2}-44 a^{3} b \,c^{3} d^{2}\right ) \textit {\_R}^{3}+\left (-b^{3} c^{10} d^{3}+62 a \,b^{2} c^{6} d^{3}+79 a^{2} b \,c^{2} d^{3}\right ) \textit {\_R}^{2}+\left (-16 b^{2} c^{5} d^{4}-32 b c \,d^{4} a \right ) \textit {\_R} +4 b \,d^{5}\right ) x +\left (-a^{2} b^{3} c^{13}-3 a^{3} b^{2} c^{9}-3 a^{4} b \,c^{5}-c \,a^{5}\right ) \textit {\_R}^{4}+\left (3 a \,b^{3} c^{12} d -9 a^{2} b^{2} c^{8} d -11 a^{3} b \,c^{4} d +a^{4} d \right ) \textit {\_R}^{3}+\left (-b^{3} c^{11} d^{2}+46 a \,b^{2} c^{7} d^{2}+15 a^{2} b \,c^{3} d^{2}\right ) \textit {\_R}^{2}-16 b^{2} c^{6} d^{3} \textit {\_R} \right )\right )}{4}\) | \(420\) |
-1/(b*c^4+a)/x-4*b*c^3*d*ln(x)/(b*c^4+a)^2-1/4*d/(b*c^4+a)^2*sum((-4*b*d^3 *c^3*_R^3+d^2*(-15*b*c^4+a)*_R^2+4*c*d*(-5*b*c^4+a)*_R-10*b*c^6+6*a*c^2)/( _R^3*d^3+3*_R^2*c*d^2+3*_R*c^2*d+c^3)*ln(x-_R),_R=RootOf(_Z^4*b*d^4+4*_Z^3 *b*c*d^3+6*_Z^2*b*c^2*d^2+4*_Z*b*c^3*d+b*c^4+a))
Result contains complex when optimal does not.
Time = 78.19 (sec) , antiderivative size = 1128605, normalized size of antiderivative = 2275.41 \[ \int \frac {1}{x^2 \left (a+b (c+d x)^4\right )} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {1}{x^2 \left (a+b (c+d x)^4\right )} \, dx=\text {Timed out} \]
\[ \int \frac {1}{x^2 \left (a+b (c+d x)^4\right )} \, dx=\int { \frac {1}{{\left ({\left (d x + c\right )}^{4} b + a\right )} x^{2}} \,d x } \]
-4*b*c^3*d*log(x)/(b^2*c^8 + 2*a*b*c^4 + a^2) + b*d^2*integrate((4*b*c^3*d ^3*x^3 + 10*b*c^6 + (15*b*c^4 - a)*d^2*x^2 - 6*a*c^2 + 4*(5*b*c^5 - a*c)*d *x)/(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^2 + 4*b*c^3*d*x + b*c^4 + a ), x)/(b^2*c^8 + 2*a*b*c^4 + a^2) - 1/((b*c^4 + a)*x)
\[ \int \frac {1}{x^2 \left (a+b (c+d x)^4\right )} \, dx=\int { \frac {1}{{\left ({\left (d x + c\right )}^{4} b + a\right )} x^{2}} \,d x } \]
Time = 9.53 (sec) , antiderivative size = 2440, normalized size of antiderivative = 4.92 \[ \int \frac {1}{x^2 \left (a+b (c+d x)^4\right )} \, dx=\text {Too large to display} \]
symsum(log(-(4*root(256*a^3*b^2*c^8*z^4 + 512*a^4*b*c^4*z^4 + 256*a^5*z^4 - 1024*a^3*b*c^3*d*z^3 + 320*a^2*b*c^2*d^2*z^2 - 32*a*b*c*d^3*z + b*d^4, z , k)^2*b^7*c^11*d^17 - 16*root(256*a^3*b^2*c^8*z^4 + 512*a^4*b*c^4*z^4 + 2 56*a^5*z^4 - 1024*a^3*b*c^3*d*z^3 + 320*a^2*b*c^2*d^2*z^2 - 32*a*b*c*d^3*z + b*d^4, z, k)^3*a^4*b^4*d^16 - b^5*d^20*x + 16*root(256*a^3*b^2*c^8*z^4 + 512*a^4*b*c^4*z^4 + 256*a^5*z^4 - 1024*a^3*b*c^3*d*z^3 + 320*a^2*b*c^2*d ^2*z^2 - 32*a*b*c*d^3*z + b*d^4, z, k)*b^6*c^6*d^18 - 60*root(256*a^3*b^2* c^8*z^4 + 512*a^4*b*c^4*z^4 + 256*a^5*z^4 - 1024*a^3*b*c^3*d*z^3 + 320*a^2 *b*c^2*d^2*z^2 - 32*a*b*c*d^3*z + b*d^4, z, k)^2*a^2*b^5*c^3*d^17 + 176*ro ot(256*a^3*b^2*c^8*z^4 + 512*a^4*b*c^4*z^4 + 256*a^5*z^4 - 1024*a^3*b*c^3* d*z^3 + 320*a^2*b*c^2*d^2*z^2 - 32*a*b*c*d^3*z + b*d^4, z, k)^3*a^3*b^5*c^ 4*d^16 + 192*root(256*a^3*b^2*c^8*z^4 + 512*a^4*b*c^4*z^4 + 256*a^5*z^4 - 1024*a^3*b*c^3*d*z^3 + 320*a^2*b*c^2*d^2*z^2 - 32*a*b*c*d^3*z + b*d^4, z, k)^4*a^4*b^5*c^5*d^15 + 144*root(256*a^3*b^2*c^8*z^4 + 512*a^4*b*c^4*z^4 + 256*a^5*z^4 - 1024*a^3*b*c^3*d*z^3 + 320*a^2*b*c^2*d^2*z^2 - 32*a*b*c*d^3 *z + b*d^4, z, k)^3*a^2*b^6*c^8*d^16 + 192*root(256*a^3*b^2*c^8*z^4 + 512* a^4*b*c^4*z^4 + 256*a^5*z^4 - 1024*a^3*b*c^3*d*z^3 + 320*a^2*b*c^2*d^2*z^2 - 32*a*b*c*d^3*z + b*d^4, z, k)^4*a^3*b^6*c^9*d^15 + 64*root(256*a^3*b^2* c^8*z^4 + 512*a^4*b*c^4*z^4 + 256*a^5*z^4 - 1024*a^3*b*c^3*d*z^3 + 320*a^2 *b*c^2*d^2*z^2 - 32*a*b*c*d^3*z + b*d^4, z, k)^4*a^2*b^7*c^13*d^15 + 16...