Integrand size = 15, antiderivative size = 167 \[ \int \frac {c+d x}{a+b x^4} \, dx=\frac {d \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b}}-\frac {c \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{b}}+\frac {c \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{b}}+\frac {c \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x}{\sqrt {a}+\sqrt {b} x^2}\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{b}} \] Output:
1/2*d*arctan(b^(1/2)*x^2/a^(1/2))/a^(1/2)/b^(1/2)+1/4*c*arctan(-1+2^(1/2)* b^(1/4)*x/a^(1/4))*2^(1/2)/a^(3/4)/b^(1/4)+1/4*c*arctan(1+2^(1/2)*b^(1/4)* x/a^(1/4))*2^(1/2)/a^(3/4)/b^(1/4)+1/4*c*arctanh(2^(1/2)*a^(1/4)*b^(1/4)*x /(a^(1/2)+b^(1/2)*x^2))*2^(1/2)/a^(3/4)/b^(1/4)
Time = 0.06 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.10 \[ \int \frac {c+d x}{a+b x^4} \, dx=\frac {-2 \left (\sqrt {2} \sqrt [4]{b} c+2 \sqrt [4]{a} d\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )+2 \left (\sqrt {2} \sqrt [4]{b} c-2 \sqrt [4]{a} d\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )+\sqrt {2} \sqrt [4]{b} c \left (-\log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )+\log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )\right )}{8 a^{3/4} \sqrt {b}} \] Input:
Integrate[(c + d*x)/(a + b*x^4),x]
Output:
(-2*(Sqrt[2]*b^(1/4)*c + 2*a^(1/4)*d)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/ 4)] + 2*(Sqrt[2]*b^(1/4)*c - 2*a^(1/4)*d)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a ^(1/4)] + Sqrt[2]*b^(1/4)*c*(-Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sq rt[b]*x^2] + Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2]))/(8*a ^(3/4)*Sqrt[b])
Time = 0.58 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.31, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {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.
\(\displaystyle \int \frac {c+d x}{a+b x^4} \, dx\) |
\(\Big \downarrow \) 2415 |
\(\displaystyle \int \left (\frac {c}{a+b x^4}+\frac {d x}{a+b x^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {c \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{b}}+\frac {c \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{b}}-\frac {c \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{4 \sqrt {2} a^{3/4} \sqrt [4]{b}}+\frac {c \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{4 \sqrt {2} a^{3/4} \sqrt [4]{b}}+\frac {d \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b}}\) |
Input:
Int[(c + d*x)/(a + b*x^4),x]
Output:
(d*ArcTan[(Sqrt[b]*x^2)/Sqrt[a]])/(2*Sqrt[a]*Sqrt[b]) - (c*ArcTan[1 - (Sqr t[2]*b^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a^(3/4)*b^(1/4)) + (c*ArcTan[1 + (Sqr t[2]*b^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a^(3/4)*b^(1/4)) - (c*Log[Sqrt[a] - S qrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(4*Sqrt[2]*a^(3/4)*b^(1/4)) + (c* 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))
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 = 0.13 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.19
method | result | size |
risch | \(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{4}+a \right )}{\sum }\frac {\left (d \textit {\_R} +c \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}}{4 b}\) | \(32\) |
default | \(\frac {c \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}{x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{8 a}+\frac {d \arctan \left (x^{2} \sqrt {\frac {b}{a}}\right )}{2 \sqrt {a b}}\) | \(124\) |
Input:
int((d*x+c)/(b*x^4+a),x,method=_RETURNVERBOSE)
Output:
1/4/b*sum((_R*d+c)/_R^3*ln(x-_R),_R=RootOf(_Z^4*b+a))
Result contains complex when optimal does not.
Time = 1.03 (sec) , antiderivative size = 41851, normalized size of antiderivative = 250.60 \[ \int \frac {c+d x}{a+b x^4} \, dx=\text {Too large to display} \] Input:
integrate((d*x+c)/(b*x^4+a),x, algorithm="fricas")
Output:
Too large to include
Time = 0.58 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.74 \[ \int \frac {c+d x}{a+b x^4} \, dx=\operatorname {RootSum} {\left (256 t^{4} a^{3} b^{2} + 32 t^{2} a^{2} b d^{2} - 16 t a b c^{2} d + a d^{4} + b c^{4}, \left ( t \mapsto t \log {\left (x + \frac {- 128 t^{3} a^{3} b d^{2} - 16 t^{2} a^{2} b c^{2} d - 8 t a^{2} d^{4} - 4 t a b c^{4} + 5 a c^{2} d^{3}}{4 a c d^{4} - b c^{5}} \right )} \right )\right )} \] Input:
integrate((d*x+c)/(b*x**4+a),x)
Output:
RootSum(256*_t**4*a**3*b**2 + 32*_t**2*a**2*b*d**2 - 16*_t*a*b*c**2*d + a* d**4 + b*c**4, Lambda(_t, _t*log(x + (-128*_t**3*a**3*b*d**2 - 16*_t**2*a* *2*b*c**2*d - 8*_t*a**2*d**4 - 4*_t*a*b*c**4 + 5*a*c**2*d**3)/(4*a*c*d**4 - b*c**5))))
Time = 0.11 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.24 \[ \int \frac {c+d x}{a+b x^4} \, dx=\frac {\sqrt {2} c \log \left (\sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{8 \, a^{\frac {3}{4}} b^{\frac {1}{4}}} - \frac {\sqrt {2} c \log \left (\sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{8 \, a^{\frac {3}{4}} b^{\frac {1}{4}}} + \frac {{\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} c - 2 \, \sqrt {a} d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{4 \, a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {b}} b^{\frac {1}{4}}} + \frac {{\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} c + 2 \, \sqrt {a} d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{4 \, a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {b}} b^{\frac {1}{4}}} \] Input:
integrate((d*x+c)/(b*x^4+a),x, algorithm="maxima")
Output:
1/8*sqrt(2)*c*log(sqrt(b)*x^2 + sqrt(2)*a^(1/4)*b^(1/4)*x + sqrt(a))/(a^(3 /4)*b^(1/4)) - 1/8*sqrt(2)*c*log(sqrt(b)*x^2 - sqrt(2)*a^(1/4)*b^(1/4)*x + sqrt(a))/(a^(3/4)*b^(1/4)) + 1/4*(sqrt(2)*a^(1/4)*b^(1/4)*c - 2*sqrt(a)*d )*arctan(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^(1/4)) + 1/4*(sqrt(2)*a^(1/4)*b ^(1/4)*c + 2*sqrt(a)*d)*arctan(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^(1/4))
Time = 0.13 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.28 \[ \int \frac {c+d x}{a+b x^4} \, dx=\frac {\sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} c \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{8 \, a b} - \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} c \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{8 \, a b} - \frac {\sqrt {2} {\left (\sqrt {2} \sqrt {a b} b d - \left (a b^{3}\right )^{\frac {1}{4}} b c\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{4 \, a b^{2}} - \frac {\sqrt {2} {\left (\sqrt {2} \sqrt {a b} b d - \left (a b^{3}\right )^{\frac {1}{4}} b c\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{4 \, a b^{2}} \] Input:
integrate((d*x+c)/(b*x^4+a),x, algorithm="giac")
Output:
1/8*sqrt(2)*(a*b^3)^(1/4)*c*log(x^2 + sqrt(2)*x*(a/b)^(1/4) + sqrt(a/b))/( a*b) - 1/8*sqrt(2)*(a*b^3)^(1/4)*c*log(x^2 - sqrt(2)*x*(a/b)^(1/4) + sqrt( a/b))/(a*b) - 1/4*sqrt(2)*(sqrt(2)*sqrt(a*b)*b*d - (a*b^3)^(1/4)*b*c)*arct an(1/2*sqrt(2)*(2*x + sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/(a*b^2) - 1/4*sqrt (2)*(sqrt(2)*sqrt(a*b)*b*d - (a*b^3)^(1/4)*b*c)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/(a*b^2)
Time = 6.40 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.96 \[ \int \frac {c+d x}{a+b x^4} \, dx=\left \{\begin {array}{cl} -\frac {2\,c+3\,d\,x}{6\,b\,x^3} & \text {\ if\ \ }a=0\\ \frac {\mathrm {atan}\left (\frac {\sqrt {2}\,b^{1/4}\,x}{a^{1/4}}-1\right )\,\left (2\,a^{1/4}\,d+\sqrt {2}\,b^{1/4}\,c\right )}{4\,a^{3/4}\,\sqrt {b}}-\frac {\mathrm {atan}\left (\frac {\sqrt {2}\,b^{1/4}\,x}{a^{1/4}}+1\right )\,\left (4\,a^{1/4}\,d-2\,\sqrt {2}\,b^{1/4}\,c\right )}{8\,a^{3/4}\,\sqrt {b}}+\frac {\sqrt {2}\,c\,\ln \left (\frac {\sqrt {a}+\sqrt {b}\,x^2+\sqrt {2}\,a^{1/4}\,b^{1/4}\,x}{\sqrt {a}+\sqrt {b}\,x^2-\sqrt {2}\,a^{1/4}\,b^{1/4}\,x}\right )}{8\,a^{3/4}\,b^{1/4}} & \text {\ if\ \ }a\neq 0 \end {array}\right . \] Input:
int((c + d*x)/(a + b*x^4),x)
Output:
piecewise(a == 0, -(2*c + 3*d*x)/(6*b*x^3), a ~= 0, (atan((2^(1/2)*b^(1/4) *x)/a^(1/4) - 1)*(2*a^(1/4)*d + 2^(1/2)*b^(1/4)*c))/(4*a^(3/4)*b^(1/2)) - (atan((2^(1/2)*b^(1/4)*x)/a^(1/4) + 1)*(4*a^(1/4)*d - 2*2^(1/2)*b^(1/4)*c) )/(8*a^(3/4)*b^(1/2)) + (2^(1/2)*c*log((a^(1/2) + b^(1/2)*x^2 + 2^(1/2)*a^ (1/4)*b^(1/4)*x)/(a^(1/2) + b^(1/2)*x^2 - 2^(1/2)*a^(1/4)*b^(1/4)*x)))/(8* a^(3/4)*b^(1/4)))
Time = 0.16 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.28 \[ \int \frac {c+d x}{a+b x^4} \, dx=\frac {\sqrt {b}\, \left (-2 b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 \sqrt {b}\, x}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) c -4 \sqrt {a}\, \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 \sqrt {b}\, x}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) d +2 b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 \sqrt {b}\, x}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) c -4 \sqrt {a}\, \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 \sqrt {b}\, x}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) d -b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, \mathrm {log}\left (-b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}+\sqrt {b}\, x^{2}\right ) c +b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, \mathrm {log}\left (b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}+\sqrt {b}\, x^{2}\right ) c \right )}{8 a b} \] Input:
int((d*x+c)/(b*x^4+a),x)
Output:
(sqrt(b)*( - 2*b**(1/4)*a**(1/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(b)*x)/(b**(1/4)*a**(1/4)*sqrt(2)))*c - 4*sqrt(a)*atan((b**(1/4)*a* *(1/4)*sqrt(2) - 2*sqrt(b)*x)/(b**(1/4)*a**(1/4)*sqrt(2)))*d + 2*b**(1/4)* a**(1/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(b)*x)/(b**(1/4)* a**(1/4)*sqrt(2)))*c - 4*sqrt(a)*atan((b**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt( b)*x)/(b**(1/4)*a**(1/4)*sqrt(2)))*d - b**(1/4)*a**(1/4)*sqrt(2)*log( - b* *(1/4)*a**(1/4)*sqrt(2)*x + sqrt(a) + sqrt(b)*x**2)*c + b**(1/4)*a**(1/4)* sqrt(2)*log(b**(1/4)*a**(1/4)*sqrt(2)*x + sqrt(a) + sqrt(b)*x**2)*c))/(8*a *b)