Integrand size = 22, antiderivative size = 252 \[ \int \frac {\left (c+d x^4\right )^3}{x^6 \left (a+b x^4\right )^2} \, dx=-\frac {c^3}{5 a^2 x^5}+\frac {c^2 (2 b c-3 a d)}{a^3 x}+\frac {(b c-a d)^3 x^3}{4 a^3 b \left (a+b x^4\right )}-\frac {3 (b c-a d)^2 (3 b c+a d) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{13/4} b^{7/4}}+\frac {3 (b c-a d)^2 (3 b c+a d) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{13/4} b^{7/4}}-\frac {3 (b c-a d)^2 (3 b c+a d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x}{\sqrt {a}+\sqrt {b} x^2}\right )}{8 \sqrt {2} a^{13/4} b^{7/4}} \] Output:
-1/5*c^3/a^2/x^5+c^2*(-3*a*d+2*b*c)/a^3/x+1/4*(-a*d+b*c)^3*x^3/a^3/b/(b*x^ 4+a)+3/16*(-a*d+b*c)^2*(a*d+3*b*c)*arctan(-1+2^(1/2)*b^(1/4)*x/a^(1/4))*2^ (1/2)/a^(13/4)/b^(7/4)+3/16*(-a*d+b*c)^2*(a*d+3*b*c)*arctan(1+2^(1/2)*b^(1 /4)*x/a^(1/4))*2^(1/2)/a^(13/4)/b^(7/4)-3/16*(-a*d+b*c)^2*(a*d+3*b*c)*arct anh(2^(1/2)*a^(1/4)*b^(1/4)*x/(a^(1/2)+b^(1/2)*x^2))*2^(1/2)/a^(13/4)/b^(7 /4)
Time = 0.19 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.21 \[ \int \frac {\left (c+d x^4\right )^3}{x^6 \left (a+b x^4\right )^2} \, dx=\frac {-\frac {32 a^{5/4} c^3}{x^5}-\frac {160 \sqrt [4]{a} c^2 (-2 b c+3 a d)}{x}-\frac {40 \sqrt [4]{a} (-b c+a d)^3 x^3}{b \left (a+b x^4\right )}-\frac {30 \sqrt {2} (b c-a d)^2 (3 b c+a d) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{b^{7/4}}+\frac {30 \sqrt {2} (b c-a d)^2 (3 b c+a d) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{b^{7/4}}+\frac {15 \sqrt {2} (b c-a d)^2 (3 b c+a d) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{b^{7/4}}-\frac {15 \sqrt {2} (b c-a d)^2 (3 b c+a d) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{b^{7/4}}}{160 a^{13/4}} \] Input:
Integrate[(c + d*x^4)^3/(x^6*(a + b*x^4)^2),x]
Output:
((-32*a^(5/4)*c^3)/x^5 - (160*a^(1/4)*c^2*(-2*b*c + 3*a*d))/x - (40*a^(1/4 )*(-(b*c) + a*d)^3*x^3)/(b*(a + b*x^4)) - (30*Sqrt[2]*(b*c - a*d)^2*(3*b*c + a*d)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/b^(7/4) + (30*Sqrt[2]*(b* c - a*d)^2*(3*b*c + a*d)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/b^(7/4) + (15*Sqrt[2]*(b*c - a*d)^2*(3*b*c + a*d)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^ (1/4)*x + Sqrt[b]*x^2])/b^(7/4) - (15*Sqrt[2]*(b*c - a*d)^2*(3*b*c + a*d)* Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/b^(7/4))/(160*a^(1 3/4))
Time = 0.92 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.43, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {968, 25, 1040, 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.
| \(\displaystyle \int \frac {\left (c+d x^4\right )^3}{x^6 \left (a+b x^4\right )^2} \, dx\) |
| \(\Big \downarrow \) 968 |
| \(\displaystyle \frac {\left (c+d x^4\right )^2 (b c-a d)}{4 a b x^5 \left (a+b x^4\right )}-\frac {\int -\frac {\left (d x^4+c\right ) \left (d (b c+3 a d) x^4+c (9 b c-5 a d)\right )}{x^6 \left (b x^4+a\right )}dx}{4 a b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\left (d x^4+c\right ) \left (d (b c+3 a d) x^4+c (9 b c-5 a d)\right )}{x^6 \left (b x^4+a\right )}dx}{4 a b}+\frac {\left (c+d x^4\right )^2 (b c-a d)}{4 a b x^5 \left (a+b x^4\right )}\) |
| \(\Big \downarrow \) 1040 |
| \(\displaystyle \frac {\int \left (-\frac {(5 a d-9 b c) c^2}{a x^6}-\frac {\left (9 b^2 c^2-15 a b d c+2 a^2 d^2\right ) c}{a^2 x^2}+\frac {3 (a d-b c)^2 (3 b c+a d) x^2}{a^2 \left (b x^4+a\right )}\right )dx}{4 a b}+\frac {\left (c+d x^4\right )^2 (b c-a d)}{4 a b x^5 \left (a+b x^4\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {3 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right ) (b c-a d)^2 (a d+3 b c)}{2 \sqrt {2} a^{9/4} b^{3/4}}+\frac {3 \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right ) (b c-a d)^2 (a d+3 b c)}{2 \sqrt {2} a^{9/4} b^{3/4}}+\frac {3 (b c-a d)^2 (a d+3 b c) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{4 \sqrt {2} a^{9/4} b^{3/4}}-\frac {3 (b c-a d)^2 (a d+3 b c) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{4 \sqrt {2} a^{9/4} b^{3/4}}+\frac {c \left (2 a^2 d^2-15 a b c d+9 b^2 c^2\right )}{a^2 x}-\frac {c^2 (9 b c-5 a d)}{5 a x^5}}{4 a b}+\frac {\left (c+d x^4\right )^2 (b c-a d)}{4 a b x^5 \left (a+b x^4\right )}\) |
Input:
Int[(c + d*x^4)^3/(x^6*(a + b*x^4)^2),x]
Output:
((b*c - a*d)*(c + d*x^4)^2)/(4*a*b*x^5*(a + b*x^4)) + (-1/5*(c^2*(9*b*c - 5*a*d))/(a*x^5) + (c*(9*b^2*c^2 - 15*a*b*c*d + 2*a^2*d^2))/(a^2*x) - (3*(b *c - a*d)^2*(3*b*c + a*d)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(2*Sqrt [2]*a^(9/4)*b^(3/4)) + (3*(b*c - a*d)^2*(3*b*c + a*d)*ArcTan[1 + (Sqrt[2]* b^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a^(9/4)*b^(3/4)) + (3*(b*c - a*d)^2*(3*b*c + a*d)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(4*Sqrt[2] *a^(9/4)*b^(3/4)) - (3*(b*c - a*d)^2*(3*b*c + a*d)*Log[Sqrt[a] + Sqrt[2]*a ^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(4*Sqrt[2]*a^(9/4)*b^(3/4)))/(4*a*b)
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[(-(c*b - a*d))*(e*x)^(m + 1)*(a + b*x^n)^(p + 1) *((c + d*x^n)^(q - 1)/(a*b*e*n*(p + 1))), x] + Simp[1/(a*b*n*(p + 1)) Int [(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 2)*Simp[c*(c*b*n*(p + 1) + (c *b - a*d)*(m + 1)) + d*(c*b*n*(p + 1) + (c*b - a*d)*(m + n*(q - 1) + 1))*x^ n, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[ n, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _))^(q_.)*((e_) + (f_.)*(x_)^(n_))^(r_.), x_Symbol] :> Int[ExpandIntegrand[ (g*x)^m*(a + b*x^n)^p*(c + d*x^n)^q*(e + f*x^n)^r, x], x] /; FreeQ[{a, b, c , d, e, f, g, m, n}, x] && IGtQ[p, -2] && IGtQ[q, 0] && IGtQ[r, 0]
Time = 0.11 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.90
| method | result | size |
| default | \(\frac {-\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) x^{3}}{4 b \left (b \,x^{4}+a \right )}+\frac {3 \left (a^{3} d^{3}+a^{2} b c \,d^{2}-5 a \,b^{2} c^{2} d +3 b^{3} c^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}{x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{32 b^{2} \left (\frac {a}{b}\right )^{\frac {1}{4}}}}{a^{3}}-\frac {c^{3}}{5 a^{2} x^{5}}-\frac {c^{2} \left (3 a d -2 c b \right )}{a^{3} x}\) | \(227\) |
| risch | \(\frac {-\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+15 a \,b^{2} c^{2} d -9 b^{3} c^{3}\right ) x^{8}}{4 a^{3} b}-\frac {3 c^{2} \left (5 a d -3 c b \right ) x^{4}}{5 a^{2}}-\frac {c^{3}}{5 a}}{x^{5} \left (b \,x^{4}+a \right )}+\frac {3 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{13} b^{7} \textit {\_Z}^{4}+a^{12} d^{12}+4 a^{11} b c \,d^{11}-14 a^{10} b^{2} c^{2} d^{10}-44 a^{9} b^{3} c^{3} d^{9}+127 a^{8} b^{4} c^{4} d^{8}+136 a^{7} b^{5} c^{5} d^{7}-644 a^{6} b^{6} c^{6} d^{6}+328 a^{5} b^{7} c^{7} d^{5}+1039 a^{4} b^{8} c^{8} d^{4}-1932 a^{3} b^{9} c^{9} d^{3}+1458 a^{2} b^{10} c^{10} d^{2}-540 a \,b^{11} c^{11} d +81 b^{12} c^{12}\right )}{\sum }\textit {\_R} \ln \left (\left (5 \textit {\_R}^{4} a^{13} b^{7}+4 a^{12} d^{12}+16 a^{11} b c \,d^{11}-56 a^{10} b^{2} c^{2} d^{10}-176 a^{9} b^{3} c^{3} d^{9}+508 a^{8} b^{4} c^{4} d^{8}+544 a^{7} b^{5} c^{5} d^{7}-2576 a^{6} b^{6} c^{6} d^{6}+1312 a^{5} b^{7} c^{7} d^{5}+4156 a^{4} b^{8} c^{8} d^{4}-7728 a^{3} b^{9} c^{9} d^{3}+5832 a^{2} b^{10} c^{10} d^{2}-2160 a \,b^{11} c^{11} d +324 b^{12} c^{12}\right ) x +\left (-a^{13} b^{5} d^{3}-a^{12} b^{6} c \,d^{2}+5 a^{11} b^{7} c^{2} d -3 b^{8} c^{3} a^{10}\right ) \textit {\_R}^{3}\right )\right )}{16}\) | \(500\) |
Input:
int((d*x^4+c)^3/x^6/(b*x^4+a)^2,x,method=_RETURNVERBOSE)
Output:
1/a^3*(-1/4*(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)/b*x^3/(b*x^4+a)+ 3/32*(a^3*d^3+a^2*b*c*d^2-5*a*b^2*c^2*d+3*b^3*c^3)/b^2/(a/b)^(1/4)*2^(1/2) *(ln((x^2-(a/b)^(1/4)*x*2^(1/2)+(a/b)^(1/2))/(x^2+(a/b)^(1/4)*x*2^(1/2)+(a /b)^(1/2)))+2*arctan(2^(1/2)/(a/b)^(1/4)*x+1)+2*arctan(2^(1/2)/(a/b)^(1/4) *x-1)))-1/5*c^3/a^2/x^5-c^2*(3*a*d-2*b*c)/a^3/x
Result contains complex when optimal does not.
Time = 0.12 (sec) , antiderivative size = 2096, normalized size of antiderivative = 8.32 \[ \int \frac {\left (c+d x^4\right )^3}{x^6 \left (a+b x^4\right )^2} \, dx=\text {Too large to display} \] Input:
integrate((d*x^4+c)^3/x^6/(b*x^4+a)^2,x, algorithm="fricas")
Output:
1/80*(20*(9*b^3*c^3 - 15*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*x^8 - 16*a ^2*b*c^3 + 48*(3*a*b^2*c^3 - 5*a^2*b*c^2*d)*x^4 + 15*(a^3*b^2*x^9 + a^4*b* x^5)*(-(81*b^12*c^12 - 540*a*b^11*c^11*d + 1458*a^2*b^10*c^10*d^2 - 1932*a ^3*b^9*c^9*d^3 + 1039*a^4*b^8*c^8*d^4 + 328*a^5*b^7*c^7*d^5 - 644*a^6*b^6* c^6*d^6 + 136*a^7*b^5*c^5*d^7 + 127*a^8*b^4*c^4*d^8 - 44*a^9*b^3*c^3*d^9 - 14*a^10*b^2*c^2*d^10 + 4*a^11*b*c*d^11 + a^12*d^12)/(a^13*b^7))^(1/4)*log (27*a^10*b^5*(-(81*b^12*c^12 - 540*a*b^11*c^11*d + 1458*a^2*b^10*c^10*d^2 - 1932*a^3*b^9*c^9*d^3 + 1039*a^4*b^8*c^8*d^4 + 328*a^5*b^7*c^7*d^5 - 644* a^6*b^6*c^6*d^6 + 136*a^7*b^5*c^5*d^7 + 127*a^8*b^4*c^4*d^8 - 44*a^9*b^3*c ^3*d^9 - 14*a^10*b^2*c^2*d^10 + 4*a^11*b*c*d^11 + a^12*d^12)/(a^13*b^7))^( 3/4) + 27*(27*b^9*c^9 - 135*a*b^8*c^8*d + 252*a^2*b^7*c^7*d^2 - 188*a^3*b^ 6*c^6*d^3 - 6*a^4*b^5*c^5*d^4 + 78*a^5*b^4*c^4*d^5 - 20*a^6*b^3*c^3*d^6 - 12*a^7*b^2*c^2*d^7 + 3*a^8*b*c*d^8 + a^9*d^9)*x) - 15*(I*a^3*b^2*x^9 + I*a ^4*b*x^5)*(-(81*b^12*c^12 - 540*a*b^11*c^11*d + 1458*a^2*b^10*c^10*d^2 - 1 932*a^3*b^9*c^9*d^3 + 1039*a^4*b^8*c^8*d^4 + 328*a^5*b^7*c^7*d^5 - 644*a^6 *b^6*c^6*d^6 + 136*a^7*b^5*c^5*d^7 + 127*a^8*b^4*c^4*d^8 - 44*a^9*b^3*c^3* d^9 - 14*a^10*b^2*c^2*d^10 + 4*a^11*b*c*d^11 + a^12*d^12)/(a^13*b^7))^(1/4 )*log(27*I*a^10*b^5*(-(81*b^12*c^12 - 540*a*b^11*c^11*d + 1458*a^2*b^10*c^ 10*d^2 - 1932*a^3*b^9*c^9*d^3 + 1039*a^4*b^8*c^8*d^4 + 328*a^5*b^7*c^7*d^5 - 644*a^6*b^6*c^6*d^6 + 136*a^7*b^5*c^5*d^7 + 127*a^8*b^4*c^4*d^8 - 44...
Time = 175.52 (sec) , antiderivative size = 440, normalized size of antiderivative = 1.75 \[ \int \frac {\left (c+d x^4\right )^3}{x^6 \left (a+b x^4\right )^2} \, dx=\operatorname {RootSum} {\left (65536 t^{4} a^{13} b^{7} + 81 a^{12} d^{12} + 324 a^{11} b c d^{11} - 1134 a^{10} b^{2} c^{2} d^{10} - 3564 a^{9} b^{3} c^{3} d^{9} + 10287 a^{8} b^{4} c^{4} d^{8} + 11016 a^{7} b^{5} c^{5} d^{7} - 52164 a^{6} b^{6} c^{6} d^{6} + 26568 a^{5} b^{7} c^{7} d^{5} + 84159 a^{4} b^{8} c^{8} d^{4} - 156492 a^{3} b^{9} c^{9} d^{3} + 118098 a^{2} b^{10} c^{10} d^{2} - 43740 a b^{11} c^{11} d + 6561 b^{12} c^{12}, \left ( t \mapsto t \log {\left (\frac {4096 t^{3} a^{10} b^{5}}{27 a^{9} d^{9} + 81 a^{8} b c d^{8} - 324 a^{7} b^{2} c^{2} d^{7} - 540 a^{6} b^{3} c^{3} d^{6} + 2106 a^{5} b^{4} c^{4} d^{5} - 162 a^{4} b^{5} c^{5} d^{4} - 5076 a^{3} b^{6} c^{6} d^{3} + 6804 a^{2} b^{7} c^{7} d^{2} - 3645 a b^{8} c^{8} d + 729 b^{9} c^{9}} + x \right )} \right )\right )} + \frac {- 4 a^{2} b c^{3} + x^{8} \left (- 5 a^{3} d^{3} + 15 a^{2} b c d^{2} - 75 a b^{2} c^{2} d + 45 b^{3} c^{3}\right ) + x^{4} \left (- 60 a^{2} b c^{2} d + 36 a b^{2} c^{3}\right )}{20 a^{4} b x^{5} + 20 a^{3} b^{2} x^{9}} \] Input:
integrate((d*x**4+c)**3/x**6/(b*x**4+a)**2,x)
Output:
RootSum(65536*_t**4*a**13*b**7 + 81*a**12*d**12 + 324*a**11*b*c*d**11 - 11 34*a**10*b**2*c**2*d**10 - 3564*a**9*b**3*c**3*d**9 + 10287*a**8*b**4*c**4 *d**8 + 11016*a**7*b**5*c**5*d**7 - 52164*a**6*b**6*c**6*d**6 + 26568*a**5 *b**7*c**7*d**5 + 84159*a**4*b**8*c**8*d**4 - 156492*a**3*b**9*c**9*d**3 + 118098*a**2*b**10*c**10*d**2 - 43740*a*b**11*c**11*d + 6561*b**12*c**12, Lambda(_t, _t*log(4096*_t**3*a**10*b**5/(27*a**9*d**9 + 81*a**8*b*c*d**8 - 324*a**7*b**2*c**2*d**7 - 540*a**6*b**3*c**3*d**6 + 2106*a**5*b**4*c**4*d **5 - 162*a**4*b**5*c**5*d**4 - 5076*a**3*b**6*c**6*d**3 + 6804*a**2*b**7* c**7*d**2 - 3645*a*b**8*c**8*d + 729*b**9*c**9) + x))) + (-4*a**2*b*c**3 + x**8*(-5*a**3*d**3 + 15*a**2*b*c*d**2 - 75*a*b**2*c**2*d + 45*b**3*c**3) + x**4*(-60*a**2*b*c**2*d + 36*a*b**2*c**3))/(20*a**4*b*x**5 + 20*a**3*b** 2*x**9)
Time = 0.11 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.24 \[ \int \frac {\left (c+d x^4\right )^3}{x^6 \left (a+b x^4\right )^2} \, dx=\frac {5 \, {\left (9 \, b^{3} c^{3} - 15 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} x^{8} - 4 \, a^{2} b c^{3} + 12 \, {\left (3 \, a b^{2} c^{3} - 5 \, a^{2} b c^{2} d\right )} x^{4}}{20 \, {\left (a^{3} b^{2} x^{9} + a^{4} b x^{5}\right )}} + \frac {3 \, {\left (3 \, b^{3} c^{3} - 5 \, a b^{2} c^{2} d + a^{2} b c d^{2} + a^{3} d^{3}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{32 \, a^{3} b} \] Input:
integrate((d*x^4+c)^3/x^6/(b*x^4+a)^2,x, algorithm="maxima")
Output:
1/20*(5*(9*b^3*c^3 - 15*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*x^8 - 4*a^2 *b*c^3 + 12*(3*a*b^2*c^3 - 5*a^2*b*c^2*d)*x^4)/(a^3*b^2*x^9 + a^4*b*x^5) + 3/32*(3*b^3*c^3 - 5*a*b^2*c^2*d + a^2*b*c*d^2 + a^3*d^3)*(2*sqrt(2)*arcta n(1/2*sqrt(2)*(2*sqrt(b)*x + sqrt(2)*a^(1/4)*b^(1/4))/sqrt(sqrt(a)*sqrt(b) ))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) + 2*sqrt(2)*arctan(1/2*sqrt(2)*(2*sqrt( b)*x - sqrt(2)*a^(1/4)*b^(1/4))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(sqrt(a)*sqrt( b))*sqrt(b)) - sqrt(2)*log(sqrt(b)*x^2 + sqrt(2)*a^(1/4)*b^(1/4)*x + sqrt( a))/(a^(1/4)*b^(3/4)) + sqrt(2)*log(sqrt(b)*x^2 - sqrt(2)*a^(1/4)*b^(1/4)* x + sqrt(a))/(a^(1/4)*b^(3/4)))/(a^3*b)
Leaf count of result is larger than twice the leaf count of optimal. 502 vs. \(2 (201) = 402\).
Time = 0.13 (sec) , antiderivative size = 502, normalized size of antiderivative = 1.99 \[ \int \frac {\left (c+d x^4\right )^3}{x^6 \left (a+b x^4\right )^2} \, dx=\frac {b^{3} c^{3} x^{3} - 3 \, a b^{2} c^{2} d x^{3} + 3 \, a^{2} b c d^{2} x^{3} - a^{3} d^{3} x^{3}}{4 \, {\left (b x^{4} + a\right )} a^{3} b} + \frac {3 \, \sqrt {2} {\left (3 \, \left (a b^{3}\right )^{\frac {3}{4}} b^{3} c^{3} - 5 \, \left (a b^{3}\right )^{\frac {3}{4}} a b^{2} c^{2} d + \left (a b^{3}\right )^{\frac {3}{4}} a^{2} b c d^{2} + \left (a b^{3}\right )^{\frac {3}{4}} a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{16 \, a^{4} b^{4}} + \frac {3 \, \sqrt {2} {\left (3 \, \left (a b^{3}\right )^{\frac {3}{4}} b^{3} c^{3} - 5 \, \left (a b^{3}\right )^{\frac {3}{4}} a b^{2} c^{2} d + \left (a b^{3}\right )^{\frac {3}{4}} a^{2} b c d^{2} + \left (a b^{3}\right )^{\frac {3}{4}} a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{16 \, a^{4} b^{4}} - \frac {3 \, \sqrt {2} {\left (3 \, \left (a b^{3}\right )^{\frac {3}{4}} b^{3} c^{3} - 5 \, \left (a b^{3}\right )^{\frac {3}{4}} a b^{2} c^{2} d + \left (a b^{3}\right )^{\frac {3}{4}} a^{2} b c d^{2} + \left (a b^{3}\right )^{\frac {3}{4}} a^{3} d^{3}\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{32 \, a^{4} b^{4}} + \frac {3 \, \sqrt {2} {\left (3 \, \left (a b^{3}\right )^{\frac {3}{4}} b^{3} c^{3} - 5 \, \left (a b^{3}\right )^{\frac {3}{4}} a b^{2} c^{2} d + \left (a b^{3}\right )^{\frac {3}{4}} a^{2} b c d^{2} + \left (a b^{3}\right )^{\frac {3}{4}} a^{3} d^{3}\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{32 \, a^{4} b^{4}} + \frac {10 \, b c^{3} x^{4} - 15 \, a c^{2} d x^{4} - a c^{3}}{5 \, a^{3} x^{5}} \] Input:
integrate((d*x^4+c)^3/x^6/(b*x^4+a)^2,x, algorithm="giac")
Output:
1/4*(b^3*c^3*x^3 - 3*a*b^2*c^2*d*x^3 + 3*a^2*b*c*d^2*x^3 - a^3*d^3*x^3)/(( b*x^4 + a)*a^3*b) + 3/16*sqrt(2)*(3*(a*b^3)^(3/4)*b^3*c^3 - 5*(a*b^3)^(3/4 )*a*b^2*c^2*d + (a*b^3)^(3/4)*a^2*b*c*d^2 + (a*b^3)^(3/4)*a^3*d^3)*arctan( 1/2*sqrt(2)*(2*x + sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/(a^4*b^4) + 3/16*sqrt (2)*(3*(a*b^3)^(3/4)*b^3*c^3 - 5*(a*b^3)^(3/4)*a*b^2*c^2*d + (a*b^3)^(3/4) *a^2*b*c*d^2 + (a*b^3)^(3/4)*a^3*d^3)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(a /b)^(1/4))/(a/b)^(1/4))/(a^4*b^4) - 3/32*sqrt(2)*(3*(a*b^3)^(3/4)*b^3*c^3 - 5*(a*b^3)^(3/4)*a*b^2*c^2*d + (a*b^3)^(3/4)*a^2*b*c*d^2 + (a*b^3)^(3/4)* a^3*d^3)*log(x^2 + sqrt(2)*x*(a/b)^(1/4) + sqrt(a/b))/(a^4*b^4) + 3/32*sqr t(2)*(3*(a*b^3)^(3/4)*b^3*c^3 - 5*(a*b^3)^(3/4)*a*b^2*c^2*d + (a*b^3)^(3/4 )*a^2*b*c*d^2 + (a*b^3)^(3/4)*a^3*d^3)*log(x^2 - sqrt(2)*x*(a/b)^(1/4) + s qrt(a/b))/(a^4*b^4) + 1/5*(10*b*c^3*x^4 - 15*a*c^2*d*x^4 - a*c^3)/(a^3*x^5 )
Time = 3.48 (sec) , antiderivative size = 652, normalized size of antiderivative = 2.59 \[ \int \frac {\left (c+d x^4\right )^3}{x^6 \left (a+b x^4\right )^2} \, dx=\frac {3\,\mathrm {atan}\left (\frac {3\,x\,{\left (a\,d-b\,c\right )}^2\,\left (a\,d+3\,b\,c\right )\,\left (2304\,a^{16}\,b^5\,d^6+4608\,a^{15}\,b^6\,c\,d^5-20736\,a^{14}\,b^7\,c^2\,d^4-9216\,a^{13}\,b^8\,c^3\,d^3+71424\,a^{12}\,b^9\,c^4\,d^2-69120\,a^{11}\,b^{10}\,c^5\,d+20736\,a^{10}\,b^{11}\,c^6\right )}{8\,{\left (-a\right )}^{13/4}\,b^{7/4}\,\left (864\,a^{16}\,b^3\,d^9+2592\,a^{15}\,b^4\,c\,d^8-10368\,a^{14}\,b^5\,c^2\,d^7-17280\,a^{13}\,b^6\,c^3\,d^6+67392\,a^{12}\,b^7\,c^4\,d^5-5184\,a^{11}\,b^8\,c^5\,d^4-162432\,a^{10}\,b^9\,c^6\,d^3+217728\,a^9\,b^{10}\,c^7\,d^2-116640\,a^8\,b^{11}\,c^8\,d+23328\,a^7\,b^{12}\,c^9\right )}\right )\,{\left (a\,d-b\,c\right )}^2\,\left (a\,d+3\,b\,c\right )}{8\,{\left (-a\right )}^{13/4}\,b^{7/4}}-\frac {\frac {c^3}{5\,a}+\frac {x^8\,\left (a^3\,d^3-3\,a^2\,b\,c\,d^2+15\,a\,b^2\,c^2\,d-9\,b^3\,c^3\right )}{4\,a^3\,b}+\frac {3\,c^2\,x^4\,\left (5\,a\,d-3\,b\,c\right )}{5\,a^2}}{b\,x^9+a\,x^5}-\frac {3\,\mathrm {atanh}\left (\frac {3\,x\,{\left (a\,d-b\,c\right )}^2\,\left (a\,d+3\,b\,c\right )\,\left (2304\,a^{16}\,b^5\,d^6+4608\,a^{15}\,b^6\,c\,d^5-20736\,a^{14}\,b^7\,c^2\,d^4-9216\,a^{13}\,b^8\,c^3\,d^3+71424\,a^{12}\,b^9\,c^4\,d^2-69120\,a^{11}\,b^{10}\,c^5\,d+20736\,a^{10}\,b^{11}\,c^6\right )}{8\,{\left (-a\right )}^{13/4}\,b^{7/4}\,\left (864\,a^{16}\,b^3\,d^9+2592\,a^{15}\,b^4\,c\,d^8-10368\,a^{14}\,b^5\,c^2\,d^7-17280\,a^{13}\,b^6\,c^3\,d^6+67392\,a^{12}\,b^7\,c^4\,d^5-5184\,a^{11}\,b^8\,c^5\,d^4-162432\,a^{10}\,b^9\,c^6\,d^3+217728\,a^9\,b^{10}\,c^7\,d^2-116640\,a^8\,b^{11}\,c^8\,d+23328\,a^7\,b^{12}\,c^9\right )}\right )\,{\left (a\,d-b\,c\right )}^2\,\left (a\,d+3\,b\,c\right )}{8\,{\left (-a\right )}^{13/4}\,b^{7/4}} \] Input:
int((c + d*x^4)^3/(x^6*(a + b*x^4)^2),x)
Output:
(3*atan((3*x*(a*d - b*c)^2*(a*d + 3*b*c)*(20736*a^10*b^11*c^6 + 2304*a^16* b^5*d^6 - 69120*a^11*b^10*c^5*d + 4608*a^15*b^6*c*d^5 + 71424*a^12*b^9*c^4 *d^2 - 9216*a^13*b^8*c^3*d^3 - 20736*a^14*b^7*c^2*d^4))/(8*(-a)^(13/4)*b^( 7/4)*(23328*a^7*b^12*c^9 + 864*a^16*b^3*d^9 - 116640*a^8*b^11*c^8*d + 2592 *a^15*b^4*c*d^8 + 217728*a^9*b^10*c^7*d^2 - 162432*a^10*b^9*c^6*d^3 - 5184 *a^11*b^8*c^5*d^4 + 67392*a^12*b^7*c^4*d^5 - 17280*a^13*b^6*c^3*d^6 - 1036 8*a^14*b^5*c^2*d^7)))*(a*d - b*c)^2*(a*d + 3*b*c))/(8*(-a)^(13/4)*b^(7/4)) - (c^3/(5*a) + (x^8*(a^3*d^3 - 9*b^3*c^3 + 15*a*b^2*c^2*d - 3*a^2*b*c*d^2 ))/(4*a^3*b) + (3*c^2*x^4*(5*a*d - 3*b*c))/(5*a^2))/(a*x^5 + b*x^9) - (3*a tanh((3*x*(a*d - b*c)^2*(a*d + 3*b*c)*(20736*a^10*b^11*c^6 + 2304*a^16*b^5 *d^6 - 69120*a^11*b^10*c^5*d + 4608*a^15*b^6*c*d^5 + 71424*a^12*b^9*c^4*d^ 2 - 9216*a^13*b^8*c^3*d^3 - 20736*a^14*b^7*c^2*d^4))/(8*(-a)^(13/4)*b^(7/4 )*(23328*a^7*b^12*c^9 + 864*a^16*b^3*d^9 - 116640*a^8*b^11*c^8*d + 2592*a^ 15*b^4*c*d^8 + 217728*a^9*b^10*c^7*d^2 - 162432*a^10*b^9*c^6*d^3 - 5184*a^ 11*b^8*c^5*d^4 + 67392*a^12*b^7*c^4*d^5 - 17280*a^13*b^6*c^3*d^6 - 10368*a ^14*b^5*c^2*d^7)))*(a*d - b*c)^2*(a*d + 3*b*c))/(8*(-a)^(13/4)*b^(7/4))
Time = 0.22 (sec) , antiderivative size = 1403, normalized size of antiderivative = 5.57 \[ \int \frac {\left (c+d x^4\right )^3}{x^6 \left (a+b x^4\right )^2} \, dx =\text {Too large to display} \] Input:
int((d*x^4+c)^3/x^6/(b*x^4+a)^2,x)
Output:
( - 30*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt( b)*x)/(b**(1/4)*a**(1/4)*sqrt(2)))*a**4*d**3*x**5 - 30*b**(1/4)*a**(3/4)*s qrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(b)*x)/(b**(1/4)*a**(1/4)*s qrt(2)))*a**3*b*c*d**2*x**5 - 30*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)* a**(1/4)*sqrt(2) - 2*sqrt(b)*x)/(b**(1/4)*a**(1/4)*sqrt(2)))*a**3*b*d**3*x **9 + 150*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sq rt(b)*x)/(b**(1/4)*a**(1/4)*sqrt(2)))*a**2*b**2*c**2*d*x**5 - 30*b**(1/4)* a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(b)*x)/(b**(1/4)* a**(1/4)*sqrt(2)))*a**2*b**2*c*d**2*x**9 - 90*b**(1/4)*a**(3/4)*sqrt(2)*at an((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(b)*x)/(b**(1/4)*a**(1/4)*sqrt(2)))* a*b**3*c**3*x**5 + 150*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*s qrt(2) - 2*sqrt(b)*x)/(b**(1/4)*a**(1/4)*sqrt(2)))*a*b**3*c**2*d*x**9 - 90 *b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(b)*x)/ (b**(1/4)*a**(1/4)*sqrt(2)))*b**4*c**3*x**9 + 30*b**(1/4)*a**(3/4)*sqrt(2) *atan((b**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(b)*x)/(b**(1/4)*a**(1/4)*sqrt(2) ))*a**4*d**3*x**5 + 30*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*s qrt(2) + 2*sqrt(b)*x)/(b**(1/4)*a**(1/4)*sqrt(2)))*a**3*b*c*d**2*x**5 + 30 *b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(b)*x)/ (b**(1/4)*a**(1/4)*sqrt(2)))*a**3*b*d**3*x**9 - 150*b**(1/4)*a**(3/4)*sqrt (2)*atan((b**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(b)*x)/(b**(1/4)*a**(1/4)*s...