\(\int \frac {1}{a+b (c+d x)^4} \, dx\) [113]

Optimal result

Integrand size = 13, antiderivative size = 221 \[ \int \frac {1}{a+b (c+d x)^4} \, dx=-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{b} d}+\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{b} d}-\frac {\log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} (c+d x)+\sqrt {b} (c+d x)^2\right )}{4 \sqrt {2} a^{3/4} \sqrt [4]{b} d}+\frac {\log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} (c+d x)+\sqrt {b} (c+d x)^2\right )}{4 \sqrt {2} a^{3/4} \sqrt [4]{b} d} \]

[Out]

1/4*arctan(-1+b^(1/4)*(d*x+c)*2^(1/2)/a^(1/4))/a^(3/4)/b^(1/4)/d*2^(1/2)+1/4*arctan(1+b^(1/4)*(d*x+c)*2^(1/2)/
a^(1/4))/a^(3/4)/b^(1/4)/d*2^(1/2)-1/8*ln(-a^(1/4)*b^(1/4)*(d*x+c)*2^(1/2)+a^(1/2)+(d*x+c)^2*b^(1/2))/a^(3/4)/
b^(1/4)/d*2^(1/2)+1/8*ln(a^(1/4)*b^(1/4)*(d*x+c)*2^(1/2)+a^(1/2)+(d*x+c)^2*b^(1/2))/a^(3/4)/b^(1/4)/d*2^(1/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {253, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {1}{a+b (c+d x)^4} \, dx=-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{b} d}+\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}+1\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{b} d}-\frac {\log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} (c+d x)+\sqrt {a}+\sqrt {b} (c+d x)^2\right )}{4 \sqrt {2} a^{3/4} \sqrt [4]{b} d}+\frac {\log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} (c+d x)+\sqrt {a}+\sqrt {b} (c+d x)^2\right )}{4 \sqrt {2} a^{3/4} \sqrt [4]{b} d} \]

[In]

Int[(a + b*(c + d*x)^4)^(-1),x]

[Out]

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

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 253

Int[((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[1/Coefficient[v, x, 1], Subst[Int[(a + b*x^n)^p, x], x,
v], x] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && NeQ[v, x]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.73 \[ \int \frac {1}{a+b (c+d x)^4} \, dx=\frac {-2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )+2 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )-\log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} (c+d x)+\sqrt {b} (c+d x)^2\right )+\log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} (c+d x)+\sqrt {b} (c+d x)^2\right )}{4 \sqrt {2} a^{3/4} \sqrt [4]{b} d} \]

[In]

Integrate[(a + b*(c + d*x)^4)^(-1),x]

[Out]

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

Maple [C] (verified)

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

Time = 0.06 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.43

method	result	size
default	\(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,d^{4} \textit {\_Z}^{4}+4 b c \,d^{3} \textit {\_Z}^{3}+6 b \,c^{2} d^{2} \textit {\_Z}^{2}+4 b \,c^{3} d \textit {\_Z} +b \,c^{4}+a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{d^{3} \textit {\_R}^{3}+3 c \,d^{2} \textit {\_R}^{2}+3 c^{2} d \textit {\_R} +c^{3}}}{4 b d}\)	\(94\)
risch	\(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,d^{4} \textit {\_Z}^{4}+4 b c \,d^{3} \textit {\_Z}^{3}+6 b \,c^{2} d^{2} \textit {\_Z}^{2}+4 b \,c^{3} d \textit {\_Z} +b \,c^{4}+a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{d^{3} \textit {\_R}^{3}+3 c \,d^{2} \textit {\_R}^{2}+3 c^{2} d \textit {\_R} +c^{3}}}{4 b d}\)	\(94\)

[In]

int(1/(a+b*(d*x+c)^4),x,method=_RETURNVERBOSE)

[Out]

1/4/b/d*sum(1/(_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))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.69 \[ \int \frac {1}{a+b (c+d x)^4} \, dx=\frac {1}{4} \, \left (-\frac {1}{a^{3} b d^{4}}\right )^{\frac {1}{4}} \log \left (a d \left (-\frac {1}{a^{3} b d^{4}}\right )^{\frac {1}{4}} + d x + c\right ) + \frac {1}{4} i \, \left (-\frac {1}{a^{3} b d^{4}}\right )^{\frac {1}{4}} \log \left (i \, a d \left (-\frac {1}{a^{3} b d^{4}}\right )^{\frac {1}{4}} + d x + c\right ) - \frac {1}{4} i \, \left (-\frac {1}{a^{3} b d^{4}}\right )^{\frac {1}{4}} \log \left (-i \, a d \left (-\frac {1}{a^{3} b d^{4}}\right )^{\frac {1}{4}} + d x + c\right ) - \frac {1}{4} \, \left (-\frac {1}{a^{3} b d^{4}}\right )^{\frac {1}{4}} \log \left (-a d \left (-\frac {1}{a^{3} b d^{4}}\right )^{\frac {1}{4}} + d x + c\right ) \]

[In]

integrate(1/(a+b*(d*x+c)^4),x, algorithm="fricas")

[Out]

1/4*(-1/(a^3*b*d^4))^(1/4)*log(a*d*(-1/(a^3*b*d^4))^(1/4) + d*x + c) + 1/4*I*(-1/(a^3*b*d^4))^(1/4)*log(I*a*d*
(-1/(a^3*b*d^4))^(1/4) + d*x + c) - 1/4*I*(-1/(a^3*b*d^4))^(1/4)*log(-I*a*d*(-1/(a^3*b*d^4))^(1/4) + d*x + c)
- 1/4*(-1/(a^3*b*d^4))^(1/4)*log(-a*d*(-1/(a^3*b*d^4))^(1/4) + d*x + c)

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.12 \[ \int \frac {1}{a+b (c+d x)^4} \, dx=\frac {\operatorname {RootSum} {\left (256 t^{4} a^{3} b + 1, \left ( t \mapsto t \log {\left (x + \frac {4 t a + c}{d} \right )} \right )\right )}}{d} \]

[In]

integrate(1/(a+b*(d*x+c)**4),x)

[Out]

RootSum(256*_t**4*a**3*b + 1, Lambda(_t, _t*log(x + (4*_t*a + c)/d)))/d

Maxima [F]

\[ \int \frac {1}{a+b (c+d x)^4} \, dx=\int { \frac {1}{{\left (d x + c\right )}^{4} b + a} \,d x } \]

[In]

integrate(1/(a+b*(d*x+c)^4),x, algorithm="maxima")

[Out]

integrate(1/((d*x + c)^4*b + a), x)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.47 \[ \int \frac {1}{a+b (c+d x)^4} \, dx=-\frac {1}{2} \, \left (-\frac {1}{a^{3} b d^{4}}\right )^{\frac {1}{4}} \arctan \left (-\frac {b d x + b c}{\left (-a b^{3}\right )^{\frac {1}{4}}}\right ) + \frac {1}{4} \, \left (-\frac {1}{a^{3} b d^{4}}\right )^{\frac {1}{4}} \log \left ({\left | b d x + b c + \left (-a b^{3}\right )^{\frac {1}{4}} \right |}\right ) - \frac {1}{4} \, \left (-\frac {1}{a^{3} b d^{4}}\right )^{\frac {1}{4}} \log \left ({\left | -b d x - b c + \left (-a b^{3}\right )^{\frac {1}{4}} \right |}\right ) \]

[In]

integrate(1/(a+b*(d*x+c)^4),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.27 \[ \int \frac {1}{a+b (c+d x)^4} \, dx=-\frac {\mathrm {atan}\left (\frac {b^{1/4}\,c}{{\left (-a\right )}^{1/4}}+\frac {b^{1/4}\,d\,x}{{\left (-a\right )}^{1/4}}\right )+\mathrm {atanh}\left (\frac {b^{1/4}\,c}{{\left (-a\right )}^{1/4}}+\frac {b^{1/4}\,d\,x}{{\left (-a\right )}^{1/4}}\right )}{2\,{\left (-a\right )}^{3/4}\,b^{1/4}\,d} \]

[In]

int(1/(a + b*(c + d*x)^4),x)

[Out]

-(atan((b^(1/4)*c)/(-a)^(1/4) + (b^(1/4)*d*x)/(-a)^(1/4)) + atanh((b^(1/4)*c)/(-a)^(1/4) + (b^(1/4)*d*x)/(-a)^
(1/4)))/(2*(-a)^(3/4)*b^(1/4)*d)