\(\int \frac {c+d x}{a-b x^4} \, dx\) [115]

Optimal result

Integrand size = 16, antiderivative size = 87 \[ \int \frac {c+d x}{a-b x^4} \, dx=\frac {c \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt [4]{b}}+\frac {c \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt [4]{b}}+\frac {d \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b}} \]

[Out]

1/2*c*arctan(b^(1/4)*x/a^(1/4))/a^(3/4)/b^(1/4)+1/2*c*arctanh(b^(1/4)*x/a^(1/4))/a^(3/4)/b^(1/4)+1/2*d*arctanh
(x^2*b^(1/2)/a^(1/2))/a^(1/2)/b^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {1890, 218, 214, 211, 281} \[ \int \frac {c+d x}{a-b x^4} \, dx=\frac {c \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt [4]{b}}+\frac {c \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt [4]{b}}+\frac {d \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b}} \]

[In]

Int[(c + d*x)/(a - b*x^4),x]

[Out]

(c*ArcTan[(b^(1/4)*x)/a^(1/4)])/(2*a^(3/4)*b^(1/4)) + (c*ArcTanh[(b^(1/4)*x)/a^(1/4)])/(2*a^(3/4)*b^(1/4)) + (
d*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a]])/(2*Sqrt[a]*Sqrt[b])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 1890

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

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {c}{a-b x^4}+\frac {d x}{a-b x^4}\right ) \, dx \\ & = c \int \frac {1}{a-b x^4} \, dx+d \int \frac {x}{a-b x^4} \, dx \\ & = \frac {c \int \frac {1}{\sqrt {a}-\sqrt {b} x^2} \, dx}{2 \sqrt {a}}+\frac {c \int \frac {1}{\sqrt {a}+\sqrt {b} x^2} \, dx}{2 \sqrt {a}}+\frac {1}{2} d \text {Subst}\left (\int \frac {1}{a-b x^2} \, dx,x,x^2\right ) \\ & = \frac {c \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt [4]{b}}+\frac {c \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt [4]{b}}+\frac {d \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.54 \[ \int \frac {c+d x}{a-b x^4} \, dx=\frac {2 \sqrt [4]{b} c \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )-\left (\sqrt [4]{b} c+\sqrt [4]{a} d\right ) \log \left (\sqrt [4]{a}-\sqrt [4]{b} x\right )+\sqrt [4]{b} c \log \left (\sqrt [4]{a}+\sqrt [4]{b} x\right )-\sqrt [4]{a} d \log \left (\sqrt [4]{a}+\sqrt [4]{b} x\right )+\sqrt [4]{a} d \log \left (\sqrt {a}+\sqrt {b} x^2\right )}{4 a^{3/4} \sqrt {b}} \]

[In]

Integrate[(c + d*x)/(a - b*x^4),x]

[Out]

(2*b^(1/4)*c*ArcTan[(b^(1/4)*x)/a^(1/4)] - (b^(1/4)*c + a^(1/4)*d)*Log[a^(1/4) - b^(1/4)*x] + b^(1/4)*c*Log[a^
(1/4) + b^(1/4)*x] - a^(1/4)*d*Log[a^(1/4) + b^(1/4)*x] + a^(1/4)*d*Log[Sqrt[a] + Sqrt[b]*x^2])/(4*a^(3/4)*Sqr
t[b])

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.47 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.39

method	result	size
risch	\(-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4} b -a \right )}{\sum }\frac {\left (\textit {\_R} d +c \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}}{4 b}\)	\(34\)
default	\(\frac {c \left (\frac {a}{b}\right )^{\frac {1}{4}} \left (\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )\right )}{4 a}+\frac {d \ln \left (\frac {a +x^{2} \sqrt {a b}}{a -x^{2} \sqrt {a b}}\right )}{4 \sqrt {a b}}\)	\(87\)

[In]

int((d*x+c)/(-b*x^4+a),x,method=_RETURNVERBOSE)

[Out]

-1/4/b*sum((_R*d+c)/_R^3*ln(x-_R),_R=RootOf(_Z^4*b-a))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.80 (sec) , antiderivative size = 39057, normalized size of antiderivative = 448.93 \[ \int \frac {c+d x}{a-b x^4} \, dx=\text {Too large to display} \]

[In]

integrate((d*x+c)/(-b*x^4+a),x, algorithm="fricas")

[Out]

Too large to include

Sympy [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.45 \[ \int \frac {c+d x}{a-b x^4} \, dx=- \operatorname {RootSum} {\left (256 t^{4} a^{3} b^{2} - 32 t^{2} a^{2} b d^{2} - 16 t a b c^{2} d + a d^{4} - b c^{4}, \left ( t \mapsto t \log {\left (x + \frac {- 128 t^{3} a^{3} b d^{2} + 16 t^{2} a^{2} b c^{2} d + 8 t a^{2} d^{4} - 4 t a b c^{4} + 5 a c^{2} d^{3}}{4 a c d^{4} + b c^{5}} \right )} \right )\right )} \]

[In]

integrate((d*x+c)/(-b*x**4+a),x)

[Out]

-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))))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 126 vs. \(2 (57) = 114\).

Time = 0.29 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.45 \[ \int \frac {c+d x}{a-b x^4} \, dx=\frac {c \arctan \left (\frac {\sqrt {b} x}{\sqrt {\sqrt {a} \sqrt {b}}}\right )}{2 \, \sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {d \log \left (\sqrt {b} x^{2} + \sqrt {a}\right )}{4 \, \sqrt {a} \sqrt {b}} - \frac {d \log \left (\sqrt {b} x^{2} - \sqrt {a}\right )}{4 \, \sqrt {a} \sqrt {b}} - \frac {c \log \left (\frac {\sqrt {b} x - \sqrt {\sqrt {a} \sqrt {b}}}{\sqrt {b} x + \sqrt {\sqrt {a} \sqrt {b}}}\right )}{4 \, \sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} \]

[In]

integrate((d*x+c)/(-b*x^4+a),x, algorithm="maxima")

[Out]

1/2*c*arctan(sqrt(b)*x/sqrt(sqrt(a)*sqrt(b)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b))) + 1/4*d*log(sqrt(b)*x^2 + sqrt(a
))/(sqrt(a)*sqrt(b)) - 1/4*d*log(sqrt(b)*x^2 - sqrt(a))/(sqrt(a)*sqrt(b)) - 1/4*c*log((sqrt(b)*x - sqrt(sqrt(a
)*sqrt(b)))/(sqrt(b)*x + sqrt(sqrt(a)*sqrt(b))))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b)))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 227 vs. \(2 (57) = 114\).

Time = 0.28 (sec) , antiderivative size = 227, normalized size of antiderivative = 2.61 \[ \int \frac {c+d x}{a-b x^4} \, dx=\frac {\sqrt {2} \left (-a b^{3}\right )^{\frac {1}{4}} c \log \left (x^{2} + \sqrt {2} x \left (-\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {-\frac {a}{b}}\right )}{8 \, a b} - \frac {\sqrt {2} \left (-a b^{3}\right )^{\frac {1}{4}} c \log \left (x^{2} - \sqrt {2} x \left (-\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {-\frac {a}{b}}\right )}{8 \, a b} - \frac {\sqrt {2} {\left (\sqrt {2} \sqrt {-a b} b d - \left (-a b^{3}\right )^{\frac {1}{4}} b c\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (-\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{4 \, a b^{2}} - \frac {\sqrt {2} {\left (\sqrt {2} \sqrt {-a b} b d - \left (-a b^{3}\right )^{\frac {1}{4}} b c\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (-\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{4 \, a b^{2}} \]

[In]

integrate((d*x+c)/(-b*x^4+a),x, algorithm="giac")

[Out]

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)*arctan(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)

Mupad [B] (verification not implemented)

Time = 9.25 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.09 \[ \int \frac {c+d x}{a-b x^4} \, dx=\left \{\begin {array}{cl} \frac {2\,c+3\,d\,x}{6\,b\,x^3} & \text {\ if\ \ }a=0\\ \frac {\mathrm {atan}\left (\frac {\sqrt {2}\,{\left (-b\right )}^{1/4}\,x}{a^{1/4}}-1\right )\,\left (2\,a^{1/4}\,d+\sqrt {2}\,{\left (-b\right )}^{1/4}\,c\right )}{4\,a^{3/4}\,\sqrt {-b}}-\frac {\mathrm {atan}\left (\frac {\sqrt {2}\,{\left (-b\right )}^{1/4}\,x}{a^{1/4}}+1\right )\,\left (4\,a^{1/4}\,d-2\,\sqrt {2}\,{\left (-b\right )}^{1/4}\,c\right )}{8\,a^{3/4}\,\sqrt {-b}}+\frac {\sqrt {2}\,c\,\ln \left (\frac {\sqrt {-b}\,x^2+\sqrt {a}+\sqrt {2}\,a^{1/4}\,{\left (-b\right )}^{1/4}\,x}{\sqrt {-b}\,x^2+\sqrt {a}-\sqrt {2}\,a^{1/4}\,{\left (-b\right )}^{1/4}\,x}\right )}{8\,a^{3/4}\,{\left (-b\right )}^{1/4}} & \text {\ if\ \ }a\neq 0 \end {array}\right . \]

[In]

int((c + d*x)/(a - b*x^4),x)

[Out]

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(((-b)^(1/2)*x^2 + a^(1/2) + 2^(1/2)*a^(1/4)*(-b)^(1/
4)*x)/((-b)^(1/2)*x^2 + a^(1/2) - 2^(1/2)*a^(1/4)*(-b)^(1/4)*x)))/(8*a^(3/4)*(-b)^(1/4)))