Integrand size = 22, antiderivative size = 73 \[ \int \frac {c+d x^4}{x \left (a+b x^4\right )^{3/4}} \, dx=\frac {d \sqrt [4]{a+b x^4}}{b}-\frac {c \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}-\frac {c \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}} \] Output:
d*(b*x^4+a)^(1/4)/b-1/2*c*arctan((b*x^4+a)^(1/4)/a^(1/4))/a^(3/4)-1/2*c*ar ctanh((b*x^4+a)^(1/4)/a^(1/4))/a^(3/4)
Time = 0.09 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00 \[ \int \frac {c+d x^4}{x \left (a+b x^4\right )^{3/4}} \, dx=\frac {d \sqrt [4]{a+b x^4}}{b}-\frac {c \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}-\frac {c \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}} \] Input:
Integrate[(c + d*x^4)/(x*(a + b*x^4)^(3/4)),x]
Output:
(d*(a + b*x^4)^(1/4))/b - (c*ArcTan[(a + b*x^4)^(1/4)/a^(1/4)])/(2*a^(3/4) ) - (c*ArcTanh[(a + b*x^4)^(1/4)/a^(1/4)])/(2*a^(3/4))
Time = 0.34 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.16, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {948, 90, 73, 756, 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.
\(\displaystyle \int \frac {c+d x^4}{x \left (a+b x^4\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{4} \int \frac {d x^4+c}{x^4 \left (b x^4+a\right )^{3/4}}dx^4\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{4} \left (c \int \frac {1}{x^4 \left (b x^4+a\right )^{3/4}}dx^4+\frac {4 d \sqrt [4]{a+b x^4}}{b}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{4} \left (\frac {4 c \int \frac {1}{\frac {x^{16}}{b}-\frac {a}{b}}d\sqrt [4]{b x^4+a}}{b}+\frac {4 d \sqrt [4]{a+b x^4}}{b}\right )\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {1}{4} \left (\frac {4 c \left (-\frac {b \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{b x^4+a}}{2 \sqrt {a}}-\frac {b \int \frac {1}{x^8+\sqrt {a}}d\sqrt [4]{b x^4+a}}{2 \sqrt {a}}\right )}{b}+\frac {4 d \sqrt [4]{a+b x^4}}{b}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{4} \left (\frac {4 c \left (-\frac {b \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{b x^4+a}}{2 \sqrt {a}}-\frac {b \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}\right )}{b}+\frac {4 d \sqrt [4]{a+b x^4}}{b}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} \left (\frac {4 c \left (-\frac {b \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}-\frac {b \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}\right )}{b}+\frac {4 d \sqrt [4]{a+b x^4}}{b}\right )\) |
Input:
Int[(c + d*x^4)/(x*(a + b*x^4)^(3/4)),x]
Output:
((4*d*(a + b*x^4)^(1/4))/b + (4*c*(-1/2*(b*ArcTan[(a + b*x^4)^(1/4)/a^(1/4 )])/a^(3/4) - (b*ArcTanh[(a + b*x^4)^(1/4)/a^(1/4)])/(2*a^(3/4))))/b)/4
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[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) 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 = 0.11 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.08
method | result | size |
pseudoelliptic | \(\frac {4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} d \,a^{\frac {3}{4}}-\ln \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}}\right ) c b -2 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right ) c b}{4 b \,a^{\frac {3}{4}}}\) | \(79\) |
Input:
int((d*x^4+c)/x/(b*x^4+a)^(3/4),x,method=_RETURNVERBOSE)
Output:
1/4*(4*(b*x^4+a)^(1/4)*d*a^(3/4)-ln(((b*x^4+a)^(1/4)+a^(1/4))/((b*x^4+a)^( 1/4)-a^(1/4)))*c*b-2*arctan((b*x^4+a)^(1/4)/a^(1/4))*c*b)/b/a^(3/4)
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.25 \[ \int \frac {c+d x^4}{x \left (a+b x^4\right )^{3/4}} \, dx=-\frac {b \left (\frac {c^{4}}{a^{3}}\right )^{\frac {1}{4}} \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} c + a \left (\frac {c^{4}}{a^{3}}\right )^{\frac {1}{4}}\right ) + i \, b \left (\frac {c^{4}}{a^{3}}\right )^{\frac {1}{4}} \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} c + i \, a \left (\frac {c^{4}}{a^{3}}\right )^{\frac {1}{4}}\right ) - i \, b \left (\frac {c^{4}}{a^{3}}\right )^{\frac {1}{4}} \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} c - i \, a \left (\frac {c^{4}}{a^{3}}\right )^{\frac {1}{4}}\right ) - b \left (\frac {c^{4}}{a^{3}}\right )^{\frac {1}{4}} \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} c - a \left (\frac {c^{4}}{a^{3}}\right )^{\frac {1}{4}}\right ) - 4 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} d}{4 \, b} \] Input:
integrate((d*x^4+c)/x/(b*x^4+a)^(3/4),x, algorithm="fricas")
Output:
-1/4*(b*(c^4/a^3)^(1/4)*log((b*x^4 + a)^(1/4)*c + a*(c^4/a^3)^(1/4)) + I*b *(c^4/a^3)^(1/4)*log((b*x^4 + a)^(1/4)*c + I*a*(c^4/a^3)^(1/4)) - I*b*(c^4 /a^3)^(1/4)*log((b*x^4 + a)^(1/4)*c - I*a*(c^4/a^3)^(1/4)) - b*(c^4/a^3)^( 1/4)*log((b*x^4 + a)^(1/4)*c - a*(c^4/a^3)^(1/4)) - 4*(b*x^4 + a)^(1/4)*d) /b
Time = 29.88 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.86 \[ \int \frac {c+d x^4}{x \left (a+b x^4\right )^{3/4}} \, dx=d \left (\begin {cases} \frac {x^{4}}{4 a^{\frac {3}{4}}} & \text {for}\: b = 0 \\\frac {\sqrt [4]{a + b x^{4}}}{b} & \text {otherwise} \end {cases}\right ) - \frac {c \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 b^{\frac {3}{4}} x^{3} \Gamma \left (\frac {7}{4}\right )} \] Input:
integrate((d*x**4+c)/x/(b*x**4+a)**(3/4),x)
Output:
d*Piecewise((x**4/(4*a**(3/4)), Eq(b, 0)), ((a + b*x**4)**(1/4)/b, True)) - c*gamma(3/4)*hyper((3/4, 3/4), (7/4,), a*exp_polar(I*pi)/(b*x**4))/(4*b* *(3/4)*x**3*gamma(7/4))
Time = 0.11 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.03 \[ \int \frac {c+d x^4}{x \left (a+b x^4\right )^{3/4}} \, dx=-\frac {1}{4} \, c {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {3}{4}}} - \frac {\log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {3}{4}}}\right )} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} d}{b} \] Input:
integrate((d*x^4+c)/x/(b*x^4+a)^(3/4),x, algorithm="maxima")
Output:
-1/4*c*(2*arctan((b*x^4 + a)^(1/4)/a^(1/4))/a^(3/4) - log(((b*x^4 + a)^(1/ 4) - a^(1/4))/((b*x^4 + a)^(1/4) + a^(1/4)))/a^(3/4)) + (b*x^4 + a)^(1/4)* d/b
Leaf count of result is larger than twice the leaf count of optimal. 195 vs. \(2 (55) = 110\).
Time = 0.13 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.67 \[ \int \frac {c+d x^4}{x \left (a+b x^4\right )^{3/4}} \, dx=\frac {\sqrt {2} c \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{4 \, \left (-a\right )^{\frac {3}{4}}} + \frac {\sqrt {2} c \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{4 \, \left (-a\right )^{\frac {3}{4}}} + \frac {\sqrt {2} c \log \left (\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{8 \, \left (-a\right )^{\frac {3}{4}}} + \frac {\sqrt {2} \left (-a\right )^{\frac {1}{4}} c \log \left (-\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{8 \, a} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} d}{b} \] Input:
integrate((d*x^4+c)/x/(b*x^4+a)^(3/4),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)^(3/4) + 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)^(3/4) + 1/8*sqrt(2)*c*log(s qrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/(-a)^(3/ 4) + 1/8*sqrt(2)*(-a)^(1/4)*c*log(-sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/a + (b*x^4 + a)^(1/4)*d/b
Time = 3.46 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.75 \[ \int \frac {c+d x^4}{x \left (a+b x^4\right )^{3/4}} \, dx=\frac {d\,{\left (b\,x^4+a\right )}^{1/4}}{b}-\frac {c\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{2\,a^{3/4}}-\frac {c\,\mathrm {atanh}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{2\,a^{3/4}} \] Input:
int((c + d*x^4)/(x*(a + b*x^4)^(3/4)),x)
Output:
(d*(a + b*x^4)^(1/4))/b - (c*atan((a + b*x^4)^(1/4)/a^(1/4)))/(2*a^(3/4)) - (c*atanh((a + b*x^4)^(1/4)/a^(1/4)))/(2*a^(3/4))
\[ \int \frac {c+d x^4}{x \left (a+b x^4\right )^{3/4}} \, dx=\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} d +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} x}d x \right ) b c}{b} \] Input:
int((d*x^4+c)/x/(b*x^4+a)^(3/4),x)
Output:
((a + b*x**4)**(1/4)*d + int(1/((a + b*x**4)**(3/4)*x),x)*b*c)/b