Integrand size = 22, antiderivative size = 76 \[ \int \frac {c+d x^4}{x \sqrt [4]{a+b x^4}} \, dx=\frac {d \left (a+b x^4\right )^{3/4}}{3 b}+\frac {c \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}-\frac {c \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}} \] Output:
1/3*d*(b*x^4+a)^(3/4)/b+1/2*c*arctan((b*x^4+a)^(1/4)/a^(1/4))/a^(1/4)-1/2* c*arctanh((b*x^4+a)^(1/4)/a^(1/4))/a^(1/4)
Time = 0.09 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00 \[ \int \frac {c+d x^4}{x \sqrt [4]{a+b x^4}} \, dx=\frac {d \left (a+b x^4\right )^{3/4}}{3 b}+\frac {c \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}-\frac {c \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}} \] Input:
Integrate[(c + d*x^4)/(x*(a + b*x^4)^(1/4)),x]
Output:
(d*(a + b*x^4)^(3/4))/(3*b) + (c*ArcTan[(a + b*x^4)^(1/4)/a^(1/4)])/(2*a^( 1/4)) - (c*ArcTanh[(a + b*x^4)^(1/4)/a^(1/4)])/(2*a^(1/4))
Time = 0.35 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {948, 90, 73, 25, 27, 827, 216, 219}
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.
\(\displaystyle \int \frac {c+d x^4}{x \sqrt [4]{a+b x^4}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{4} \int \frac {d x^4+c}{x^4 \sqrt [4]{b x^4+a}}dx^4\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{4} \left (c \int \frac {1}{x^4 \sqrt [4]{b x^4+a}}dx^4+\frac {4 d \left (a+b x^4\right )^{3/4}}{3 b}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{4} \left (\frac {4 c \int -\frac {b x^8}{a-x^{16}}d\sqrt [4]{b x^4+a}}{b}+\frac {4 d \left (a+b x^4\right )^{3/4}}{3 b}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} \left (\frac {4 d \left (a+b x^4\right )^{3/4}}{3 b}-\frac {4 c \int \frac {b x^8}{a-x^{16}}d\sqrt [4]{b x^4+a}}{b}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \left (\frac {4 d \left (a+b x^4\right )^{3/4}}{3 b}-4 c \int \frac {x^8}{a-x^{16}}d\sqrt [4]{b x^4+a}\right )\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {1}{4} \left (\frac {4 d \left (a+b x^4\right )^{3/4}}{3 b}-4 c \left (\frac {1}{2} \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{b x^4+a}-\frac {1}{2} \int \frac {1}{x^8+\sqrt {a}}d\sqrt [4]{b x^4+a}\right )\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{4} \left (\frac {4 d \left (a+b x^4\right )^{3/4}}{3 b}-4 c \left (\frac {1}{2} \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{b x^4+a}-\frac {\arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}\right )\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} \left (\frac {4 d \left (a+b x^4\right )^{3/4}}{3 b}-4 c \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}-\frac {\arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}\right )\right )\) |
Input:
Int[(c + d*x^4)/(x*(a + b*x^4)^(1/4)),x]
Output:
((4*d*(a + b*x^4)^(3/4))/(3*b) - 4*c*(-1/2*ArcTan[(a + b*x^4)^(1/4)/a^(1/4 )]/a^(1/4) + ArcTanh[(a + b*x^4)^(1/4)/a^(1/4)]/(2*a^(1/4))))/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Time = 0.14 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.04
method | result | size |
pseudoelliptic | \(\frac {4 \left (b \,x^{4}+a \right )^{\frac {3}{4}} d \,a^{\frac {1}{4}}+6 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right ) c b -3 \ln \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}}\right ) c b}{12 b \,a^{\frac {1}{4}}}\) | \(79\) |
Input:
int((d*x^4+c)/x/(b*x^4+a)^(1/4),x,method=_RETURNVERBOSE)
Output:
1/12*(4*(b*x^4+a)^(3/4)*d*a^(1/4)+6*arctan((b*x^4+a)^(1/4)/a^(1/4))*c*b-3* ln(((b*x^4+a)^(1/4)+a^(1/4))/((b*x^4+a)^(1/4)-a^(1/4)))*c*b)/b/a^(1/4)
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.28 \[ \int \frac {c+d x^4}{x \sqrt [4]{a+b x^4}} \, dx=-\frac {3 \, \left (\frac {c^{4}}{a}\right )^{\frac {1}{4}} b \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} c^{3} + \left (\frac {c^{4}}{a}\right )^{\frac {3}{4}} a\right ) - 3 i \, \left (\frac {c^{4}}{a}\right )^{\frac {1}{4}} b \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} c^{3} + i \, \left (\frac {c^{4}}{a}\right )^{\frac {3}{4}} a\right ) + 3 i \, \left (\frac {c^{4}}{a}\right )^{\frac {1}{4}} b \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} c^{3} - i \, \left (\frac {c^{4}}{a}\right )^{\frac {3}{4}} a\right ) - 3 \, \left (\frac {c^{4}}{a}\right )^{\frac {1}{4}} b \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} c^{3} - \left (\frac {c^{4}}{a}\right )^{\frac {3}{4}} a\right ) - 4 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} d}{12 \, b} \] Input:
integrate((d*x^4+c)/x/(b*x^4+a)^(1/4),x, algorithm="fricas")
Output:
-1/12*(3*(c^4/a)^(1/4)*b*log((b*x^4 + a)^(1/4)*c^3 + (c^4/a)^(3/4)*a) - 3* I*(c^4/a)^(1/4)*b*log((b*x^4 + a)^(1/4)*c^3 + I*(c^4/a)^(3/4)*a) + 3*I*(c^ 4/a)^(1/4)*b*log((b*x^4 + a)^(1/4)*c^3 - I*(c^4/a)^(3/4)*a) - 3*(c^4/a)^(1 /4)*b*log((b*x^4 + a)^(1/4)*c^3 - (c^4/a)^(3/4)*a) - 4*(b*x^4 + a)^(3/4)*d )/b
Time = 15.79 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.83 \[ \int \frac {c+d x^4}{x \sqrt [4]{a+b x^4}} \, dx=d \left (\begin {cases} \frac {x^{4}}{4 \sqrt [4]{a}} & \text {for}\: b = 0 \\\frac {\left (a + b x^{4}\right )^{\frac {3}{4}}}{3 b} & \text {otherwise} \end {cases}\right ) - \frac {c \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 \sqrt [4]{b} x \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate((d*x**4+c)/x/(b*x**4+a)**(1/4),x)
Output:
d*Piecewise((x**4/(4*a**(1/4)), Eq(b, 0)), ((a + b*x**4)**(3/4)/(3*b), Tru e)) - c*gamma(1/4)*hyper((1/4, 1/4), (5/4,), a*exp_polar(I*pi)/(b*x**4))/( 4*b**(1/4)*x*gamma(5/4))
Time = 0.12 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.99 \[ \int \frac {c+d x^4}{x \sqrt [4]{a+b x^4}} \, dx=\frac {1}{4} \, c {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}}\right )} + \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}} d}{3 \, b} \] Input:
integrate((d*x^4+c)/x/(b*x^4+a)^(1/4),x, algorithm="maxima")
Output:
1/4*c*(2*arctan((b*x^4 + a)^(1/4)/a^(1/4))/a^(1/4) + log(((b*x^4 + a)^(1/4 ) - a^(1/4))/((b*x^4 + a)^(1/4) + a^(1/4)))/a^(1/4)) + 1/3*(b*x^4 + a)^(3/ 4)*d/b
Leaf count of result is larger than twice the leaf count of optimal. 196 vs. \(2 (56) = 112\).
Time = 0.13 (sec) , antiderivative size = 196, normalized size of antiderivative = 2.58 \[ \int \frac {c+d x^4}{x \sqrt [4]{a+b x^4}} \, dx=\frac {\sqrt {2} c \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{4 \, \left (-a\right )^{\frac {1}{4}}} + \frac {\sqrt {2} c \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{4 \, \left (-a\right )^{\frac {1}{4}}} + \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} c \log \left (\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{8 \, a} + \frac {\sqrt {2} c \log \left (-\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{8 \, \left (-a\right )^{\frac {1}{4}}} + \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}} d}{3 \, b} \] Input:
integrate((d*x^4+c)/x/(b*x^4+a)^(1/4),x, algorithm="giac")
Output:
1/4*sqrt(2)*c*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(b*x^4 + a)^(1/4) )/(-a)^(1/4))/(-a)^(1/4) + 1/4*sqrt(2)*c*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a) ^(1/4) - 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/(-a)^(1/4) + 1/8*sqrt(2)*(-a)^(3 /4)*c*log(sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a ))/a + 1/8*sqrt(2)*c*log(-sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^ 4 + a) + sqrt(-a))/(-a)^(1/4) + 1/3*(b*x^4 + a)^(3/4)*d/b
Time = 4.06 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.74 \[ \int \frac {c+d x^4}{x \sqrt [4]{a+b x^4}} \, dx=\frac {d\,{\left (b\,x^4+a\right )}^{3/4}}{3\,b}+\frac {c\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{2\,a^{1/4}}-\frac {c\,\mathrm {atanh}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{2\,a^{1/4}} \] Input:
int((c + d*x^4)/(x*(a + b*x^4)^(1/4)),x)
Output:
(d*(a + b*x^4)^(3/4))/(3*b) + (c*atan((a + b*x^4)^(1/4)/a^(1/4)))/(2*a^(1/ 4)) - (c*atanh((a + b*x^4)^(1/4)/a^(1/4)))/(2*a^(1/4))
\[ \int \frac {c+d x^4}{x \sqrt [4]{a+b x^4}} \, dx=\left (\int \frac {x^{3}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}}d x \right ) d +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} x}d x \right ) c \] Input:
int((d*x^4+c)/x/(b*x^4+a)^(1/4),x)
Output:
int(x**3/(a + b*x**4)**(1/4),x)*d + int(1/((a + b*x**4)**(1/4)*x),x)*c