Integrand size = 22, antiderivative size = 83 \[ \int \frac {c+d x^4}{x \left (a+b x^4\right )^{5/4}} \, dx=\frac {b c-a d}{a b \sqrt [4]{a+b x^4}}+\frac {c \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{5/4}}-\frac {c \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{5/4}} \] Output:
(-a*d+b*c)/a/b/(b*x^4+a)^(1/4)+1/2*c*arctan((b*x^4+a)^(1/4)/a^(1/4))/a^(5/ 4)-1/2*c*arctanh((b*x^4+a)^(1/4)/a^(1/4))/a^(5/4)
Time = 0.13 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96 \[ \int \frac {c+d x^4}{x \left (a+b x^4\right )^{5/4}} \, dx=\frac {\frac {2 \sqrt [4]{a} (b c-a d)}{b \sqrt [4]{a+b x^4}}+c \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )-c \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{5/4}} \] Input:
Integrate[(c + d*x^4)/(x*(a + b*x^4)^(5/4)),x]
Output:
((2*a^(1/4)*(b*c - a*d))/(b*(a + b*x^4)^(1/4)) + c*ArcTan[(a + b*x^4)^(1/4 )/a^(1/4)] - c*ArcTanh[(a + b*x^4)^(1/4)/a^(1/4)])/(2*a^(5/4))
Time = 0.36 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {948, 87, 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.
\(\displaystyle \int \frac {c+d x^4}{x \left (a+b x^4\right )^{5/4}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{4} \int \frac {d x^4+c}{x^4 \left (b x^4+a\right )^{5/4}}dx^4\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {1}{4} \left (\frac {c \int \frac {1}{x^4 \sqrt [4]{b x^4+a}}dx^4}{a}+\frac {4 (b c-a d)}{a b \sqrt [4]{a+b x^4}}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{4} \left (\frac {4 c \int -\frac {b x^8}{a-x^{16}}d\sqrt [4]{b x^4+a}}{a b}+\frac {4 (b c-a d)}{a b \sqrt [4]{a+b x^4}}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} \left (\frac {4 (b c-a d)}{a b \sqrt [4]{a+b x^4}}-\frac {4 c \int \frac {b x^8}{a-x^{16}}d\sqrt [4]{b x^4+a}}{a b}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \left (\frac {4 (b c-a d)}{a b \sqrt [4]{a+b x^4}}-\frac {4 c \int \frac {x^8}{a-x^{16}}d\sqrt [4]{b x^4+a}}{a}\right )\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {1}{4} \left (\frac {4 (b c-a d)}{a b \sqrt [4]{a+b x^4}}-\frac {4 c \left (\frac {1}{2} \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{b x^4+a}-\frac {1}{2} \int \frac {1}{x^8+\sqrt {a}}d\sqrt [4]{b x^4+a}\right )}{a}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{4} \left (\frac {4 (b c-a d)}{a b \sqrt [4]{a+b x^4}}-\frac {4 c \left (\frac {1}{2} \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{b x^4+a}-\frac {\arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}\right )}{a}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} \left (\frac {4 (b c-a d)}{a b \sqrt [4]{a+b x^4}}-\frac {4 c \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}-\frac {\arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}\right )}{a}\right )\) |
Input:
Int[(c + d*x^4)/(x*(a + b*x^4)^(5/4)),x]
Output:
((4*(b*c - a*d))/(a*b*(a + b*x^4)^(1/4)) - (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))))/a )/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*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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 = 88, normalized size of antiderivative = 1.06
method | result | size |
pseudoelliptic | \(\frac {\frac {\arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right ) b c}{2 a^{\frac {5}{4}}}-\frac {\ln \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}}\right ) b c}{4 a^{\frac {5}{4}}}-\frac {a d -c b}{a \left (b \,x^{4}+a \right )^{\frac {1}{4}}}}{b}\) | \(88\) |
Input:
int((d*x^4+c)/x/(b*x^4+a)^(5/4),x,method=_RETURNVERBOSE)
Output:
(1/2*arctan((b*x^4+a)^(1/4)/a^(1/4))*b/a^(5/4)*c-1/4*ln(((b*x^4+a)^(1/4)+a ^(1/4))/((b*x^4+a)^(1/4)-a^(1/4)))*b/a^(5/4)*c-(a*d-b*c)/a/(b*x^4+a)^(1/4) )/b
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 256, normalized size of antiderivative = 3.08 \[ \int \frac {c+d x^4}{x \left (a+b x^4\right )^{5/4}} \, dx=-\frac {{\left (a b^{2} x^{4} + a^{2} b\right )} \left (\frac {c^{4}}{a^{5}}\right )^{\frac {1}{4}} \log \left (a^{4} \left (\frac {c^{4}}{a^{5}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} c^{3}\right ) - {\left (i \, a b^{2} x^{4} + i \, a^{2} b\right )} \left (\frac {c^{4}}{a^{5}}\right )^{\frac {1}{4}} \log \left (i \, a^{4} \left (\frac {c^{4}}{a^{5}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} c^{3}\right ) - {\left (-i \, a b^{2} x^{4} - i \, a^{2} b\right )} \left (\frac {c^{4}}{a^{5}}\right )^{\frac {1}{4}} \log \left (-i \, a^{4} \left (\frac {c^{4}}{a^{5}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} c^{3}\right ) - {\left (a b^{2} x^{4} + a^{2} b\right )} \left (\frac {c^{4}}{a^{5}}\right )^{\frac {1}{4}} \log \left (-a^{4} \left (\frac {c^{4}}{a^{5}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} c^{3}\right ) - 4 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} {\left (b c - a d\right )}}{4 \, {\left (a b^{2} x^{4} + a^{2} b\right )}} \] Input:
integrate((d*x^4+c)/x/(b*x^4+a)^(5/4),x, algorithm="fricas")
Output:
-1/4*((a*b^2*x^4 + a^2*b)*(c^4/a^5)^(1/4)*log(a^4*(c^4/a^5)^(3/4) + (b*x^4 + a)^(1/4)*c^3) - (I*a*b^2*x^4 + I*a^2*b)*(c^4/a^5)^(1/4)*log(I*a^4*(c^4/ a^5)^(3/4) + (b*x^4 + a)^(1/4)*c^3) - (-I*a*b^2*x^4 - I*a^2*b)*(c^4/a^5)^( 1/4)*log(-I*a^4*(c^4/a^5)^(3/4) + (b*x^4 + a)^(1/4)*c^3) - (a*b^2*x^4 + a^ 2*b)*(c^4/a^5)^(1/4)*log(-a^4*(c^4/a^5)^(3/4) + (b*x^4 + a)^(1/4)*c^3) - 4 *(b*x^4 + a)^(3/4)*(b*c - a*d))/(a*b^2*x^4 + a^2*b)
Time = 33.58 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.80 \[ \int \frac {c+d x^4}{x \left (a+b x^4\right )^{5/4}} \, dx=d \left (\begin {cases} - \frac {1}{b \sqrt [4]{a + b x^{4}}} & \text {for}\: b \neq 0 \\\frac {x^{4}}{4 a^{\frac {5}{4}}} & \text {otherwise} \end {cases}\right ) - \frac {c \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 b^{\frac {5}{4}} x^{5} \Gamma \left (\frac {9}{4}\right )} \] Input:
integrate((d*x**4+c)/x/(b*x**4+a)**(5/4),x)
Output:
d*Piecewise((-1/(b*(a + b*x**4)**(1/4)), Ne(b, 0)), (x**4/(4*a**(5/4)), Tr ue)) - c*gamma(5/4)*hyper((5/4, 5/4), (9/4,), a*exp_polar(I*pi)/(b*x**4))/ (4*b**(5/4)*x**5*gamma(9/4))
Time = 0.11 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.13 \[ \int \frac {c+d x^4}{x \left (a+b x^4\right )^{5/4}} \, dx=\frac {1}{4} \, c {\left (\frac {\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}}}{a} + \frac {4}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} a}\right )} - \frac {d}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} b} \] Input:
integrate((d*x^4+c)/x/(b*x^4+a)^(5/4),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))/a + 4/((b*x^4 + a)^( 1/4)*a)) - d/((b*x^4 + a)^(1/4)*b)
Leaf count of result is larger than twice the leaf count of optimal. 214 vs. \(2 (65) = 130\).
Time = 0.13 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.58 \[ \int \frac {c+d x^4}{x \left (a+b x^4\right )^{5/4}} \, dx=\frac {\sqrt {2} c \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{4 \, \left (-a\right )^{\frac {1}{4}} a} + \frac {\sqrt {2} c \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{4 \, \left (-a\right )^{\frac {1}{4}} a} + \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} c \log \left (\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{8 \, a^{2}} + \frac {\sqrt {2} c \log \left (-\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{8 \, \left (-a\right )^{\frac {1}{4}} a} + \frac {b c - a d}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} a b} \] Input:
integrate((d*x^4+c)/x/(b*x^4+a)^(5/4),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)*a) + 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)*a) + 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^2 + 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)*a) + (b*c - a*d)/((b*x^4 + a)^(1/ 4)*a*b)
Time = 3.93 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.84 \[ \int \frac {c+d x^4}{x \left (a+b x^4\right )^{5/4}} \, dx=\frac {c}{a\,{\left (b\,x^4+a\right )}^{1/4}}-\frac {d}{b\,{\left (b\,x^4+a\right )}^{1/4}}+\frac {c\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{2\,a^{5/4}}-\frac {c\,\mathrm {atanh}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{2\,a^{5/4}} \] Input:
int((c + d*x^4)/(x*(a + b*x^4)^(5/4)),x)
Output:
c/(a*(a + b*x^4)^(1/4)) - d/(b*(a + b*x^4)^(1/4)) + (c*atan((a + b*x^4)^(1 /4)/a^(1/4)))/(2*a^(5/4)) - (c*atanh((a + b*x^4)^(1/4)/a^(1/4)))/(2*a^(5/4 ))
\[ \int \frac {c+d x^4}{x \left (a+b x^4\right )^{5/4}} \, dx=\left (\int \frac {x^{3}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a +\left (b \,x^{4}+a \right )^{\frac {1}{4}} b \,x^{4}}d x \right ) d +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a x +\left (b \,x^{4}+a \right )^{\frac {1}{4}} b \,x^{5}}d x \right ) c \] Input:
int((d*x^4+c)/x/(b*x^4+a)^(5/4),x)
Output:
int(x**3/((a + b*x**4)**(1/4)*a + (a + b*x**4)**(1/4)*b*x**4),x)*d + int(1 /((a + b*x**4)**(1/4)*a*x + (a + b*x**4)**(1/4)*b*x**5),x)*c