Integrand size = 40, antiderivative size = 516 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5+i x^6}{\left (a+b x^4\right )^4} \, dx=\frac {x \left (b c-a g+(b d-a h) x+(b e-a i) x^2+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}+\frac {x \left (7 (11 b c+a g)+12 (5 b d+a h) x+15 (3 b e+a i) x^2\right )}{384 a^3 b \left (a+b x^4\right )}-\frac {8 a f-x \left (11 b c+a g+2 (5 b d+a h) x+3 (3 b e+a i) x^2\right )}{96 a^2 b \left (a+b x^4\right )^2}+\frac {(5 b d+a h) \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{32 a^{7/2} b^{3/2}}-\frac {\left (7 \sqrt {b} (11 b c+a g)+5 \sqrt {a} (3 b e+a i)\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{256 \sqrt {2} a^{15/4} b^{7/4}}+\frac {\left (7 \sqrt {b} (11 b c+a g)+5 \sqrt {a} (3 b e+a i)\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{256 \sqrt {2} a^{15/4} b^{7/4}}-\frac {\left (7 \sqrt {b} (11 b c+a g)-5 \sqrt {a} (3 b e+a i)\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{512 \sqrt {2} a^{15/4} b^{7/4}}+\frac {\left (7 \sqrt {b} (11 b c+a g)-5 \sqrt {a} (3 b e+a i)\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{512 \sqrt {2} a^{15/4} b^{7/4}} \]
1/12*x*(b*c-a*g+(-a*h+b*d)*x+(-a*i+b*e)*x^2+b*f*x^3)/a/b/(b*x^4+a)^3+1/384 *x*(7*a*g+77*b*c+12*(a*h+5*b*d)*x+15*(a*i+3*b*e)*x^2)/a^3/b/(b*x^4+a)+1/96 *(-8*a*f+x*(11*b*c+a*g+2*(a*h+5*b*d)*x+3*(a*i+3*b*e)*x^2))/a^2/b/(b*x^4+a) ^2+1/32*(a*h+5*b*d)*arctan(x^2*b^(1/2)/a^(1/2))/a^(7/2)/b^(3/2)-1/1024*ln( -a^(1/4)*b^(1/4)*x*2^(1/2)+a^(1/2)+x^2*b^(1/2))*(-5*(a*i+3*b*e)*a^(1/2)+7* (a*g+11*b*c)*b^(1/2))/a^(15/4)/b^(7/4)*2^(1/2)+1/1024*ln(a^(1/4)*b^(1/4)*x *2^(1/2)+a^(1/2)+x^2*b^(1/2))*(-5*(a*i+3*b*e)*a^(1/2)+7*(a*g+11*b*c)*b^(1/ 2))/a^(15/4)/b^(7/4)*2^(1/2)+1/512*arctan(-1+b^(1/4)*x*2^(1/2)/a^(1/4))*(5 *(a*i+3*b*e)*a^(1/2)+7*(a*g+11*b*c)*b^(1/2))/a^(15/4)/b^(7/4)*2^(1/2)+1/51 2*arctan(1+b^(1/4)*x*2^(1/2)/a^(1/4))*(5*(a*i+3*b*e)*a^(1/2)+7*(a*g+11*b*c )*b^(1/2))/a^(15/4)/b^(7/4)*2^(1/2)
Time = 0.72 (sec) , antiderivative size = 530, normalized size of antiderivative = 1.03 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5+i x^6}{\left (a+b x^4\right )^4} \, dx=\frac {\frac {32 a^{7/4} b^{3/4} x (11 b c+a g+b x (10 d+9 e x)+a x (2 h+3 i x))}{\left (a+b x^4\right )^2}+\frac {8 a^{3/4} b^{3/4} x (77 b c+7 a g+15 b x (4 d+3 e x)+3 a x (4 h+5 i x))}{a+b x^4}-\frac {256 a^{11/4} b^{3/4} (-b x (c+x (d+e x))+a (f+x (g+x (h+i x))))}{\left (a+b x^4\right )^3}-6 \left (77 \sqrt {2} b^{3/2} c+80 \sqrt [4]{a} b^{5/4} d+15 \sqrt {2} \sqrt {a} b e+7 \sqrt {2} a \sqrt {b} g+16 a^{5/4} \sqrt [4]{b} h+5 \sqrt {2} a^{3/2} i\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )+6 \left (77 \sqrt {2} b^{3/2} c-80 \sqrt [4]{a} b^{5/4} d+15 \sqrt {2} \sqrt {a} b e+7 \sqrt {2} a \sqrt {b} g-16 a^{5/4} \sqrt [4]{b} h+5 \sqrt {2} a^{3/2} i\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )+3 \sqrt {2} \left (-77 b^{3/2} c+15 \sqrt {a} b e-7 a \sqrt {b} g+5 a^{3/2} i\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )+3 \sqrt {2} \left (77 b^{3/2} c-15 \sqrt {a} b e+7 a \sqrt {b} g-5 a^{3/2} i\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{3072 a^{15/4} b^{7/4}} \]
((32*a^(7/4)*b^(3/4)*x*(11*b*c + a*g + b*x*(10*d + 9*e*x) + a*x*(2*h + 3*i *x)))/(a + b*x^4)^2 + (8*a^(3/4)*b^(3/4)*x*(77*b*c + 7*a*g + 15*b*x*(4*d + 3*e*x) + 3*a*x*(4*h + 5*i*x)))/(a + b*x^4) - (256*a^(11/4)*b^(3/4)*(-(b*x *(c + x*(d + e*x))) + a*(f + x*(g + x*(h + i*x)))))/(a + b*x^4)^3 - 6*(77* Sqrt[2]*b^(3/2)*c + 80*a^(1/4)*b^(5/4)*d + 15*Sqrt[2]*Sqrt[a]*b*e + 7*Sqrt [2]*a*Sqrt[b]*g + 16*a^(5/4)*b^(1/4)*h + 5*Sqrt[2]*a^(3/2)*i)*ArcTan[1 - ( Sqrt[2]*b^(1/4)*x)/a^(1/4)] + 6*(77*Sqrt[2]*b^(3/2)*c - 80*a^(1/4)*b^(5/4) *d + 15*Sqrt[2]*Sqrt[a]*b*e + 7*Sqrt[2]*a*Sqrt[b]*g - 16*a^(5/4)*b^(1/4)*h + 5*Sqrt[2]*a^(3/2)*i)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)] + 3*Sqrt[2 ]*(-77*b^(3/2)*c + 15*Sqrt[a]*b*e - 7*a*Sqrt[b]*g + 5*a^(3/2)*i)*Log[Sqrt[ a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2] + 3*Sqrt[2]*(77*b^(3/2)*c - 15*Sqrt[a]*b*e + 7*a*Sqrt[b]*g - 5*a^(3/2)*i)*Log[Sqrt[a] + Sqrt[2]*a^(1/4 )*b^(1/4)*x + Sqrt[b]*x^2])/(3072*a^(15/4)*b^(7/4))
Time = 1.04 (sec) , antiderivative size = 544, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2397, 25, 2393, 25, 2394, 27, 2415, 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.
| \(\displaystyle \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5+i x^6}{\left (a+b x^4\right )^4} \, dx\) |
| \(\Big \downarrow \) 2397 |
| \(\displaystyle \frac {x \left (x (b d-a h)+x^2 (b e-a i)-a g+b c+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}-\frac {\int -\frac {8 b^2 f x^3+3 b (3 b e+a i) x^2+2 b (5 b d+a h) x+b (11 b c+a g)}{\left (b x^4+a\right )^3}dx}{12 a b^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {8 b^2 f x^3+3 b (3 b e+a i) x^2+2 b (5 b d+a h) x+b (11 b c+a g)}{\left (b x^4+a\right )^3}dx}{12 a b^2}+\frac {x \left (x (b d-a h)+x^2 (b e-a i)-a g+b c+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}\) |
| \(\Big \downarrow \) 2393 |
| \(\displaystyle \frac {-\frac {\int -\frac {15 b (3 b e+a i) x^2+12 b (5 b d+a h) x+7 b (11 b c+a g)}{\left (b x^4+a\right )^2}dx}{8 a}-\frac {8 a b f-x \left (b (a g+11 b c)+2 b x (a h+5 b d)+3 b x^2 (a i+3 b e)\right )}{8 a \left (a+b x^4\right )^2}}{12 a b^2}+\frac {x \left (x (b d-a h)+x^2 (b e-a i)-a g+b c+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {15 b (3 b e+a i) x^2+12 b (5 b d+a h) x+7 b (11 b c+a g)}{\left (b x^4+a\right )^2}dx}{8 a}-\frac {8 a b f-x \left (b (a g+11 b c)+2 b x (a h+5 b d)+3 b x^2 (a i+3 b e)\right )}{8 a \left (a+b x^4\right )^2}}{12 a b^2}+\frac {x \left (x (b d-a h)+x^2 (b e-a i)-a g+b c+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}\) |
| \(\Big \downarrow \) 2394 |
| \(\displaystyle \frac {\frac {\frac {x \left (7 b (a g+11 b c)+12 b x (a h+5 b d)+15 b x^2 (a i+3 b e)\right )}{4 a \left (a+b x^4\right )}-\frac {\int -\frac {3 \left (5 b (3 b e+a i) x^2+8 b (5 b d+a h) x+7 b (11 b c+a g)\right )}{b x^4+a}dx}{4 a}}{8 a}-\frac {8 a b f-x \left (b (a g+11 b c)+2 b x (a h+5 b d)+3 b x^2 (a i+3 b e)\right )}{8 a \left (a+b x^4\right )^2}}{12 a b^2}+\frac {x \left (x (b d-a h)+x^2 (b e-a i)-a g+b c+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {3 \int \frac {5 b (3 b e+a i) x^2+8 b (5 b d+a h) x+7 b (11 b c+a g)}{b x^4+a}dx}{4 a}+\frac {x \left (7 b (a g+11 b c)+12 b x (a h+5 b d)+15 b x^2 (a i+3 b e)\right )}{4 a \left (a+b x^4\right )}}{8 a}-\frac {8 a b f-x \left (b (a g+11 b c)+2 b x (a h+5 b d)+3 b x^2 (a i+3 b e)\right )}{8 a \left (a+b x^4\right )^2}}{12 a b^2}+\frac {x \left (x (b d-a h)+x^2 (b e-a i)-a g+b c+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}\) |
| \(\Big \downarrow \) 2415 |
| \(\displaystyle \frac {\frac {\frac {3 \int \left (\frac {8 b (5 b d+a h) x}{b x^4+a}+\frac {5 b (3 b e+a i) x^2+7 b (11 b c+a g)}{b x^4+a}\right )dx}{4 a}+\frac {x \left (7 b (a g+11 b c)+12 b x (a h+5 b d)+15 b x^2 (a i+3 b e)\right )}{4 a \left (a+b x^4\right )}}{8 a}-\frac {8 a b f-x \left (b (a g+11 b c)+2 b x (a h+5 b d)+3 b x^2 (a i+3 b e)\right )}{8 a \left (a+b x^4\right )^2}}{12 a b^2}+\frac {x \left (x (b d-a h)+x^2 (b e-a i)-a g+b c+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\frac {3 \left (-\frac {\sqrt [4]{b} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (7 \sqrt {b} (a g+11 b c)+5 \sqrt {a} (a i+3 b e)\right )}{2 \sqrt {2} a^{3/4}}+\frac {\sqrt [4]{b} \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right ) \left (7 \sqrt {b} (a g+11 b c)+5 \sqrt {a} (a i+3 b e)\right )}{2 \sqrt {2} a^{3/4}}-\frac {\sqrt [4]{b} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right ) \left (7 \sqrt {b} (a g+11 b c)-5 \sqrt {a} (a i+3 b e)\right )}{4 \sqrt {2} a^{3/4}}+\frac {\sqrt [4]{b} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right ) \left (7 \sqrt {b} (a g+11 b c)-5 \sqrt {a} (a i+3 b e)\right )}{4 \sqrt {2} a^{3/4}}+\frac {4 \sqrt {b} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ) (a h+5 b d)}{\sqrt {a}}\right )}{4 a}+\frac {x \left (7 b (a g+11 b c)+12 b x (a h+5 b d)+15 b x^2 (a i+3 b e)\right )}{4 a \left (a+b x^4\right )}}{8 a}-\frac {8 a b f-x \left (b (a g+11 b c)+2 b x (a h+5 b d)+3 b x^2 (a i+3 b e)\right )}{8 a \left (a+b x^4\right )^2}}{12 a b^2}+\frac {x \left (x (b d-a h)+x^2 (b e-a i)-a g+b c+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}\) |
(x*(b*c - a*g + (b*d - a*h)*x + (b*e - a*i)*x^2 + b*f*x^3))/(12*a*b*(a + b *x^4)^3) + (-1/8*(8*a*b*f - x*(b*(11*b*c + a*g) + 2*b*(5*b*d + a*h)*x + 3* b*(3*b*e + a*i)*x^2))/(a*(a + b*x^4)^2) + ((x*(7*b*(11*b*c + a*g) + 12*b*( 5*b*d + a*h)*x + 15*b*(3*b*e + a*i)*x^2))/(4*a*(a + b*x^4)) + (3*((4*Sqrt[ b]*(5*b*d + a*h)*ArcTan[(Sqrt[b]*x^2)/Sqrt[a]])/Sqrt[a] - (b^(1/4)*(7*Sqrt [b]*(11*b*c + a*g) + 5*Sqrt[a]*(3*b*e + a*i))*ArcTan[1 - (Sqrt[2]*b^(1/4)* x)/a^(1/4)])/(2*Sqrt[2]*a^(3/4)) + (b^(1/4)*(7*Sqrt[b]*(11*b*c + a*g) + 5* Sqrt[a]*(3*b*e + a*i))*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(2*Sqrt[2] *a^(3/4)) - (b^(1/4)*(7*Sqrt[b]*(11*b*c + a*g) - 5*Sqrt[a]*(3*b*e + a*i))* Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(4*Sqrt[2]*a^(3/4) ) + (b^(1/4)*(7*Sqrt[b]*(11*b*c + a*g) - 5*Sqrt[a]*(3*b*e + a*i))*Log[Sqrt [a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(4*Sqrt[2]*a^(3/4))))/(4*a ))/(8*a))/(12*a*b^2)
3.3.8.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], i}, Simp[(a*Coeff[Pq, x, q] - b*x*ExpandToSum[Pq - Coeff[Pq, x, q]*x^q , x])*((a + b*x^n)^(p + 1)/(a*b*n*(p + 1))), x] + Simp[1/(a*n*(p + 1)) In t[Sum[(n*(p + 1) + i + 1)*Coeff[Pq, x, i]*x^i, {i, 0, q - 1}]*(a + b*x^n)^( p + 1), x], x] /; q == n - 1] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n , 0] && LtQ[p, -1]
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(-x)*Pq*((a + b *x^n)^(p + 1)/(a*n*(p + 1))), x] + Simp[1/(a*n*(p + 1)) Int[ExpandToSum[n *(p + 1)*Pq + D[x*Pq, x], x]*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x ] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1] && LtQ[Expon[Pq, x], n - 1]
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, Module[{Q = PolynomialQuotient[b^(Floor[(q - 1)/n] + 1)*Pq, a + b*x^n, x], R = PolynomialRemainder[b^(Floor[(q - 1)/n] + 1)*Pq, a + b*x^n, x]}, S imp[(-x)*R*((a + b*x^n)^(p + 1)/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1))), x] + Simp[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)) Int[(a + b*x^n)^(p + 1)* ExpandToSum[a*n*(p + 1)*Q + n*(p + 1)*R + D[x*R, x], x], x], x]] /; GeQ[q, n]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1]
Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = Sum[x^ii*((Coeff [Pq, x, ii] + Coeff[Pq, x, n/2 + ii]*x^(n/2))/(a + b*x^n)), {ii, 0, n/2 - 1 }]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] && Expon[Pq, x] < n
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.55 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.46
| method | result | size |
| risch | \(\frac {\frac {5 \left (a i +3 b e \right ) b \,x^{11}}{128 a^{3}}+\frac {\left (a h +5 b d \right ) b \,x^{10}}{32 a^{3}}+\frac {7 \left (a g +11 b c \right ) b \,x^{9}}{384 a^{3}}+\frac {7 \left (a i +3 b e \right ) x^{7}}{64 a^{2}}+\frac {\left (a h +5 b d \right ) x^{6}}{12 a^{2}}+\frac {3 \left (a g +11 b c \right ) x^{5}}{64 a^{2}}-\frac {\left (5 a i -113 b e \right ) x^{3}}{384 a b}-\frac {\left (a h -11 b d \right ) x^{2}}{32 a b}-\frac {\left (7 a g -51 b c \right ) x}{128 a b}-\frac {f}{12 b}}{\left (b \,x^{4}+a \right )^{3}}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\left (5 \left (a i +3 b e \right ) \textit {\_R}^{2}+8 \left (a h +5 b d \right ) \textit {\_R} +7 a g +77 b c \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}}{512 a^{3} b^{2}}\) | \(237\) |
| default | \(\frac {\frac {5 \left (a i +3 b e \right ) b \,x^{11}}{128 a^{3}}+\frac {\left (a h +5 b d \right ) b \,x^{10}}{32 a^{3}}+\frac {7 \left (a g +11 b c \right ) b \,x^{9}}{384 a^{3}}+\frac {7 \left (a i +3 b e \right ) x^{7}}{64 a^{2}}+\frac {\left (a h +5 b d \right ) x^{6}}{12 a^{2}}+\frac {3 \left (a g +11 b c \right ) x^{5}}{64 a^{2}}-\frac {\left (5 a i -113 b e \right ) x^{3}}{384 a b}-\frac {\left (a h -11 b d \right ) x^{2}}{32 a b}-\frac {\left (7 a g -51 b c \right ) x}{128 a b}-\frac {f}{12 b}}{\left (b \,x^{4}+a \right )^{3}}+\frac {\frac {\left (7 a g +77 b c \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}{x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{8 a}+\frac {\left (8 a h +40 b d \right ) \arctan \left (x^{2} \sqrt {\frac {b}{a}}\right )}{2 \sqrt {a b}}+\frac {\left (5 a i +15 b e \right ) \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}{x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{8 b \left (\frac {a}{b}\right )^{\frac {1}{4}}}}{128 a^{3} b}\) | \(432\) |
(5/128*(a*i+3*b*e)/a^3*b*x^11+1/32*(a*h+5*b*d)/a^3*b*x^10+7/384*(a*g+11*b* c)/a^3*b*x^9+7/64*(a*i+3*b*e)/a^2*x^7+1/12/a^2*(a*h+5*b*d)*x^6+3/64/a^2*(a *g+11*b*c)*x^5-1/384*(5*a*i-113*b*e)/a/b*x^3-1/32*(a*h-11*b*d)/a/b*x^2-1/1 28*(7*a*g-51*b*c)/a/b*x-1/12*f/b)/(b*x^4+a)^3+1/512/a^3/b^2*sum((5*(a*i+3* b*e)*_R^2+8*(a*h+5*b*d)*_R+7*a*g+77*b*c)/_R^3*ln(x-_R),_R=RootOf(_Z^4*b+a) )
Timed out. \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5+i x^6}{\left (a+b x^4\right )^4} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5+i x^6}{\left (a+b x^4\right )^4} \, dx=\text {Timed out} \]
Time = 0.31 (sec) , antiderivative size = 579, normalized size of antiderivative = 1.12 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5+i x^6}{\left (a+b x^4\right )^4} \, dx=\frac {15 \, {\left (3 \, b^{3} e + a b^{2} i\right )} x^{11} + 12 \, {\left (5 \, b^{3} d + a b^{2} h\right )} x^{10} + 7 \, {\left (11 \, b^{3} c + a b^{2} g\right )} x^{9} + 42 \, {\left (3 \, a b^{2} e + a^{2} b i\right )} x^{7} + 32 \, {\left (5 \, a b^{2} d + a^{2} b h\right )} x^{6} + 18 \, {\left (11 \, a b^{2} c + a^{2} b g\right )} x^{5} - 32 \, a^{3} f + {\left (113 \, a^{2} b e - 5 \, a^{3} i\right )} x^{3} + 12 \, {\left (11 \, a^{2} b d - a^{3} h\right )} x^{2} + 3 \, {\left (51 \, a^{2} b c - 7 \, a^{3} g\right )} x}{384 \, {\left (a^{3} b^{4} x^{12} + 3 \, a^{4} b^{3} x^{8} + 3 \, a^{5} b^{2} x^{4} + a^{6} b\right )}} + \frac {\frac {\sqrt {2} {\left (77 \, b^{\frac {3}{2}} c - 15 \, \sqrt {a} b e + 7 \, a \sqrt {b} g - 5 \, a^{\frac {3}{2}} i\right )} \log \left (\sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {3}{4}}} - \frac {\sqrt {2} {\left (77 \, b^{\frac {3}{2}} c - 15 \, \sqrt {a} b e + 7 \, a \sqrt {b} g - 5 \, a^{\frac {3}{2}} i\right )} \log \left (\sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {3}{4}}} + \frac {2 \, {\left (77 \, \sqrt {2} a^{\frac {1}{4}} b^{\frac {7}{4}} c + 15 \, \sqrt {2} a^{\frac {3}{4}} b^{\frac {5}{4}} e + 7 \, \sqrt {2} a^{\frac {5}{4}} b^{\frac {3}{4}} g + 5 \, \sqrt {2} a^{\frac {7}{4}} b^{\frac {1}{4}} i - 80 \, \sqrt {a} b^{\frac {3}{2}} d - 16 \, a^{\frac {3}{2}} \sqrt {b} h\right )} \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 )}{a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {b}} b^{\frac {3}{4}}} + \frac {2 \, {\left (77 \, \sqrt {2} a^{\frac {1}{4}} b^{\frac {7}{4}} c + 15 \, \sqrt {2} a^{\frac {3}{4}} b^{\frac {5}{4}} e + 7 \, \sqrt {2} a^{\frac {5}{4}} b^{\frac {3}{4}} g + 5 \, \sqrt {2} a^{\frac {7}{4}} b^{\frac {1}{4}} i + 80 \, \sqrt {a} b^{\frac {3}{2}} d + 16 \, a^{\frac {3}{2}} \sqrt {b} h\right )} \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 )}{a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {b}} b^{\frac {3}{4}}}}{1024 \, a^{3} b} \]
1/384*(15*(3*b^3*e + a*b^2*i)*x^11 + 12*(5*b^3*d + a*b^2*h)*x^10 + 7*(11*b ^3*c + a*b^2*g)*x^9 + 42*(3*a*b^2*e + a^2*b*i)*x^7 + 32*(5*a*b^2*d + a^2*b *h)*x^6 + 18*(11*a*b^2*c + a^2*b*g)*x^5 - 32*a^3*f + (113*a^2*b*e - 5*a^3* i)*x^3 + 12*(11*a^2*b*d - a^3*h)*x^2 + 3*(51*a^2*b*c - 7*a^3*g)*x)/(a^3*b^ 4*x^12 + 3*a^4*b^3*x^8 + 3*a^5*b^2*x^4 + a^6*b) + 1/1024*(sqrt(2)*(77*b^(3 /2)*c - 15*sqrt(a)*b*e + 7*a*sqrt(b)*g - 5*a^(3/2)*i)*log(sqrt(b)*x^2 + sq rt(2)*a^(1/4)*b^(1/4)*x + sqrt(a))/(a^(3/4)*b^(3/4)) - sqrt(2)*(77*b^(3/2) *c - 15*sqrt(a)*b*e + 7*a*sqrt(b)*g - 5*a^(3/2)*i)*log(sqrt(b)*x^2 - sqrt( 2)*a^(1/4)*b^(1/4)*x + sqrt(a))/(a^(3/4)*b^(3/4)) + 2*(77*sqrt(2)*a^(1/4)* b^(7/4)*c + 15*sqrt(2)*a^(3/4)*b^(5/4)*e + 7*sqrt(2)*a^(5/4)*b^(3/4)*g + 5 *sqrt(2)*a^(7/4)*b^(1/4)*i - 80*sqrt(a)*b^(3/2)*d - 16*a^(3/2)*sqrt(b)*h)* arctan(1/2*sqrt(2)*(2*sqrt(b)*x + sqrt(2)*a^(1/4)*b^(1/4))/sqrt(sqrt(a)*sq rt(b)))/(a^(3/4)*sqrt(sqrt(a)*sqrt(b))*b^(3/4)) + 2*(77*sqrt(2)*a^(1/4)*b^ (7/4)*c + 15*sqrt(2)*a^(3/4)*b^(5/4)*e + 7*sqrt(2)*a^(5/4)*b^(3/4)*g + 5*s qrt(2)*a^(7/4)*b^(1/4)*i + 80*sqrt(a)*b^(3/2)*d + 16*a^(3/2)*sqrt(b)*h)*ar ctan(1/2*sqrt(2)*(2*sqrt(b)*x - sqrt(2)*a^(1/4)*b^(1/4))/sqrt(sqrt(a)*sqrt (b)))/(a^(3/4)*sqrt(sqrt(a)*sqrt(b))*b^(3/4)))/(a^3*b)
Time = 0.27 (sec) , antiderivative size = 603, normalized size of antiderivative = 1.17 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5+i x^6}{\left (a+b x^4\right )^4} \, dx=\frac {\sqrt {2} {\left (40 \, \sqrt {2} \sqrt {a b} b^{3} d + 8 \, \sqrt {2} \sqrt {a b} a b^{2} h + 77 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{3} c + 7 \, \left (a b^{3}\right )^{\frac {1}{4}} a b^{2} g + 15 \, \left (a b^{3}\right )^{\frac {3}{4}} b e + 5 \, \left (a b^{3}\right )^{\frac {3}{4}} a i\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 )}{512 \, a^{4} b^{4}} + \frac {\sqrt {2} {\left (40 \, \sqrt {2} \sqrt {a b} b^{3} d + 8 \, \sqrt {2} \sqrt {a b} a b^{2} h + 77 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{3} c + 7 \, \left (a b^{3}\right )^{\frac {1}{4}} a b^{2} g + 15 \, \left (a b^{3}\right )^{\frac {3}{4}} b e + 5 \, \left (a b^{3}\right )^{\frac {3}{4}} a i\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 )}{512 \, a^{4} b^{4}} + \frac {\sqrt {2} {\left (77 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{3} c + 7 \, \left (a b^{3}\right )^{\frac {1}{4}} a b^{2} g - 15 \, \left (a b^{3}\right )^{\frac {3}{4}} b e - 5 \, \left (a b^{3}\right )^{\frac {3}{4}} a i\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{1024 \, a^{4} b^{4}} - \frac {\sqrt {2} {\left (77 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{3} c + 7 \, \left (a b^{3}\right )^{\frac {1}{4}} a b^{2} g - 15 \, \left (a b^{3}\right )^{\frac {3}{4}} b e - 5 \, \left (a b^{3}\right )^{\frac {3}{4}} a i\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{1024 \, a^{4} b^{4}} + \frac {45 \, b^{3} e x^{11} + 15 \, a b^{2} i x^{11} + 60 \, b^{3} d x^{10} + 12 \, a b^{2} h x^{10} + 77 \, b^{3} c x^{9} + 7 \, a b^{2} g x^{9} + 126 \, a b^{2} e x^{7} + 42 \, a^{2} b i x^{7} + 160 \, a b^{2} d x^{6} + 32 \, a^{2} b h x^{6} + 198 \, a b^{2} c x^{5} + 18 \, a^{2} b g x^{5} + 113 \, a^{2} b e x^{3} - 5 \, a^{3} i x^{3} + 132 \, a^{2} b d x^{2} - 12 \, a^{3} h x^{2} + 153 \, a^{2} b c x - 21 \, a^{3} g x - 32 \, a^{3} f}{384 \, {\left (b x^{4} + a\right )}^{3} a^{3} b} \]
1/512*sqrt(2)*(40*sqrt(2)*sqrt(a*b)*b^3*d + 8*sqrt(2)*sqrt(a*b)*a*b^2*h + 77*(a*b^3)^(1/4)*b^3*c + 7*(a*b^3)^(1/4)*a*b^2*g + 15*(a*b^3)^(3/4)*b*e + 5*(a*b^3)^(3/4)*a*i)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(a/b)^(1/4))/(a/b)^ (1/4))/(a^4*b^4) + 1/512*sqrt(2)*(40*sqrt(2)*sqrt(a*b)*b^3*d + 8*sqrt(2)*s qrt(a*b)*a*b^2*h + 77*(a*b^3)^(1/4)*b^3*c + 7*(a*b^3)^(1/4)*a*b^2*g + 15*( a*b^3)^(3/4)*b*e + 5*(a*b^3)^(3/4)*a*i)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)* (a/b)^(1/4))/(a/b)^(1/4))/(a^4*b^4) + 1/1024*sqrt(2)*(77*(a*b^3)^(1/4)*b^3 *c + 7*(a*b^3)^(1/4)*a*b^2*g - 15*(a*b^3)^(3/4)*b*e - 5*(a*b^3)^(3/4)*a*i) *log(x^2 + sqrt(2)*x*(a/b)^(1/4) + sqrt(a/b))/(a^4*b^4) - 1/1024*sqrt(2)*( 77*(a*b^3)^(1/4)*b^3*c + 7*(a*b^3)^(1/4)*a*b^2*g - 15*(a*b^3)^(3/4)*b*e - 5*(a*b^3)^(3/4)*a*i)*log(x^2 - sqrt(2)*x*(a/b)^(1/4) + sqrt(a/b))/(a^4*b^4 ) + 1/384*(45*b^3*e*x^11 + 15*a*b^2*i*x^11 + 60*b^3*d*x^10 + 12*a*b^2*h*x^ 10 + 77*b^3*c*x^9 + 7*a*b^2*g*x^9 + 126*a*b^2*e*x^7 + 42*a^2*b*i*x^7 + 160 *a*b^2*d*x^6 + 32*a^2*b*h*x^6 + 198*a*b^2*c*x^5 + 18*a^2*b*g*x^5 + 113*a^2 *b*e*x^3 - 5*a^3*i*x^3 + 132*a^2*b*d*x^2 - 12*a^3*h*x^2 + 153*a^2*b*c*x - 21*a^3*g*x - 32*a^3*f)/((b*x^4 + a)^3*a^3*b)
Time = 10.16 (sec) , antiderivative size = 2741, normalized size of antiderivative = 5.31 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5+i x^6}{\left (a+b x^4\right )^4} \, dx=\text {Too large to display} \]
((3*x^5*(11*b*c + a*g))/(64*a^2) - f/(12*b) + (x^6*(5*b*d + a*h))/(12*a^2) + (7*x^7*(3*b*e + a*i))/(64*a^2) + (7*b*x^9*(11*b*c + a*g))/(384*a^3) + ( x*(51*b*c - 7*a*g))/(128*a*b) + (b*x^10*(5*b*d + a*h))/(32*a^3) + (5*b*x^1 1*(3*b*e + a*i))/(128*a^3) + (x^2*(11*b*d - a*h))/(32*a*b) + (x^3*(113*b*e - 5*a*i))/(384*a*b))/(a^3 + b^3*x^12 + 3*a^2*b*x^4 + 3*a*b^2*x^8) + symsu m(log(- root(68719476736*a^15*b^7*z^4 + 1211105280*a^8*b^6*c*e*z^2 + 40370 1760*a^9*b^5*c*i*z^2 + 335544320*a^9*b^5*d*h*z^2 + 110100480*a^9*b^5*e*g*z ^2 + 36700160*a^10*b^4*g*i*z^2 + 838860800*a^8*b^6*d^2*z^2 + 33554432*a^10 *b^4*h^2*z^2 + 2457600*a^7*b^3*e*h*i*z - 88309760*a^5*b^5*c*d*g*z - 176619 52*a^6*b^4*c*g*h*z + 12288000*a^6*b^4*d*e*i*z - 485703680*a^4*b^6*c^2*d*z + 409600*a^8*b^2*h*i^2*z - 97140736*a^5*b^5*c^2*h*z - 802816*a^7*b^3*g^2*h *z + 3686400*a^6*b^4*e^2*h*z + 2048000*a^7*b^3*d*i^2*z - 4014080*a^6*b^4*d *g^2*z + 18432000*a^5*b^5*d*e^2*z - 89600*a^4*b^2*d*g*h*i - 985600*a^3*b^3 *c*d*h*i + 323400*a^3*b^3*c*e*g*i - 268800*a^3*b^3*d*e*g*h - 2956800*a^2*b ^4*c*d*e*h + 14700*a^4*b^2*e*g^2*i - 224000*a^3*b^3*d^2*g*i - 98560*a^4*b^ 2*c*h^2*i - 26880*a^4*b^2*e*g*h^2 + 53900*a^4*b^2*c*g*i^2 + 1778700*a^2*b^ 4*c^2*e*i - 2464000*a^2*b^4*c*d^2*i - 672000*a^2*b^4*d^2*e*g - 295680*a^3* b^3*c*e*h^2 + 485100*a^2*b^4*c*e^2*g - 8960*a^5*b*g*h^2*i - 7392000*a*b^5* c*d^2*e + 7500*a^5*b*e*i^3 + 12782924*a*b^5*c^3*g + 33750*a^4*b^2*e^2*i^2 + 614400*a^3*b^3*d^2*h^2 + 296450*a^3*b^3*c^2*i^2 + 22050*a^3*b^3*e^2*g...