Integrand size = 22, antiderivative size = 92 \[ \int \frac {c+d x^4}{x^3 \left (a+b x^4\right )^{5/4}} \, dx=-\frac {c}{2 a x^2 \sqrt [4]{a+b x^4}}-\frac {(3 b c-2 a d) \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{2 a^{3/2} \sqrt {b} \sqrt [4]{a+b x^4}} \] Output:
-1/2*c/a/x^2/(b*x^4+a)^(1/4)-1/2*(-2*a*d+3*b*c)*(1+b*x^4/a)^(1/4)*Elliptic E(sin(1/2*arctan(b^(1/2)*x^2/a^(1/2))),2^(1/2))/a^(3/2)/b^(1/2)/(b*x^4+a)^ (1/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.93 \[ \int \frac {c+d x^4}{x^3 \left (a+b x^4\right )^{5/4}} \, dx=\frac {-2 a c-6 b c x^4+4 a d x^4+(3 b c-2 a d) x^4 \sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^4}{a}\right )}{4 a^2 x^2 \sqrt [4]{a+b x^4}} \] Input:
Integrate[(c + d*x^4)/(x^3*(a + b*x^4)^(5/4)),x]
Output:
(-2*a*c - 6*b*c*x^4 + 4*a*d*x^4 + (3*b*c - 2*a*d)*x^4*(1 + (b*x^4)/a)^(1/4 )*Hypergeometric2F1[1/4, 1/2, 3/2, -((b*x^4)/a)])/(4*a^2*x^2*(a + b*x^4)^( 1/4))
Time = 0.35 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {955, 807, 213, 212}
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.
\(\displaystyle \int \frac {c+d x^4}{x^3 \left (a+b x^4\right )^{5/4}} \, dx\) |
\(\Big \downarrow \) 955 |
\(\displaystyle -\frac {(3 b c-2 a d) \int \frac {x}{\left (b x^4+a\right )^{5/4}}dx}{2 a}-\frac {c}{2 a x^2 \sqrt [4]{a+b x^4}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle -\frac {(3 b c-2 a d) \int \frac {1}{\left (b x^4+a\right )^{5/4}}dx^2}{4 a}-\frac {c}{2 a x^2 \sqrt [4]{a+b x^4}}\) |
\(\Big \downarrow \) 213 |
\(\displaystyle -\frac {\sqrt [4]{\frac {b x^4}{a}+1} (3 b c-2 a d) \int \frac {1}{\left (\frac {b x^4}{a}+1\right )^{5/4}}dx^2}{4 a^2 \sqrt [4]{a+b x^4}}-\frac {c}{2 a x^2 \sqrt [4]{a+b x^4}}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle -\frac {\sqrt [4]{\frac {b x^4}{a}+1} (3 b c-2 a d) E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{2 a^{3/2} \sqrt {b} \sqrt [4]{a+b x^4}}-\frac {c}{2 a x^2 \sqrt [4]{a+b x^4}}\) |
Input:
Int[(c + d*x^4)/(x^3*(a + b*x^4)^(5/4)),x]
Output:
-1/2*c/(a*x^2*(a + b*x^4)^(1/4)) - ((3*b*c - 2*a*d)*(1 + (b*x^4)/a)^(1/4)* EllipticE[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(2*a^(3/2)*Sqrt[b]*(a + b*x ^4)^(1/4))
Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) )*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a , 0] && PosQ[b/a]
Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(1/4)/( a*(a + b*x^2)^(1/4)) Int[1/(1 + b*(x^2/a))^(5/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a] && PosQ[b/a]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[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]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)) Int[(e *x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b* c - a*d, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) && !ILtQ[p, -1]
\[\int \frac {d \,x^{4}+c}{x^{3} \left (b \,x^{4}+a \right )^{\frac {5}{4}}}d x\]
Input:
int((d*x^4+c)/x^3/(b*x^4+a)^(5/4),x)
Output:
int((d*x^4+c)/x^3/(b*x^4+a)^(5/4),x)
\[ \int \frac {c+d x^4}{x^3 \left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{3}} \,d x } \] Input:
integrate((d*x^4+c)/x^3/(b*x^4+a)^(5/4),x, algorithm="fricas")
Output:
integral((b*x^4 + a)^(3/4)*(d*x^4 + c)/(b^2*x^11 + 2*a*b*x^7 + a^2*x^3), x )
Result contains complex when optimal does not.
Time = 17.81 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.66 \[ \int \frac {c+d x^4}{x^3 \left (a+b x^4\right )^{5/4}} \, dx=- \frac {c {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} x^{2}} + \frac {d x^{2} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}}} \] Input:
integrate((d*x**4+c)/x**3/(b*x**4+a)**(5/4),x)
Output:
-c*hyper((-1/2, 5/4), (1/2,), b*x**4*exp_polar(I*pi)/a)/(2*a**(5/4)*x**2) + d*x**2*hyper((1/2, 5/4), (3/2,), b*x**4*exp_polar(I*pi)/a)/(2*a**(5/4))
\[ \int \frac {c+d x^4}{x^3 \left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{3}} \,d x } \] Input:
integrate((d*x^4+c)/x^3/(b*x^4+a)^(5/4),x, algorithm="maxima")
Output:
integrate((d*x^4 + c)/((b*x^4 + a)^(5/4)*x^3), x)
\[ \int \frac {c+d x^4}{x^3 \left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{3}} \,d x } \] Input:
integrate((d*x^4+c)/x^3/(b*x^4+a)^(5/4),x, algorithm="giac")
Output:
integrate((d*x^4 + c)/((b*x^4 + a)^(5/4)*x^3), x)
Timed out. \[ \int \frac {c+d x^4}{x^3 \left (a+b x^4\right )^{5/4}} \, dx=\int \frac {d\,x^4+c}{x^3\,{\left (b\,x^4+a\right )}^{5/4}} \,d x \] Input:
int((c + d*x^4)/(x^3*(a + b*x^4)^(5/4)),x)
Output:
int((c + d*x^4)/(x^3*(a + b*x^4)^(5/4)), x)
\[ \int \frac {c+d x^4}{x^3 \left (a+b x^4\right )^{5/4}} \, dx=\left (\int \frac {x}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a +\left (b \,x^{4}+a \right )^{\frac {1}{4}} b \,x^{4}}d x \right ) d +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,x^{3}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b \,x^{7}}d x \right ) c \] Input:
int((d*x^4+c)/x^3/(b*x^4+a)^(5/4),x)
Output:
int(x/((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**3 + (a + b*x**4)**(1/4)*b*x**7),x)*c