Integrand size = 22, antiderivative size = 150 \[ \int \frac {c+d x^4}{x^7 \sqrt [4]{a+b x^4}} \, dx=-\frac {b (b c-2 a d) x^2}{4 a^2 \sqrt [4]{a+b x^4}}-\frac {c \left (a+b x^4\right )^{3/4}}{6 a x^6}+\frac {(b c-2 a d) \left (a+b x^4\right )^{3/4}}{4 a^2 x^2}+\frac {\sqrt {b} (b c-2 a d) \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{4 a^{3/2} \sqrt [4]{a+b x^4}} \] Output:
-1/4*b*(-2*a*d+b*c)*x^2/a^2/(b*x^4+a)^(1/4)-1/6*c*(b*x^4+a)^(3/4)/a/x^6+1/ 4*(-2*a*d+b*c)*(b*x^4+a)^(3/4)/a^2/x^2+1/4*b^(1/2)*(-2*a*d+b*c)*(1+b*x^4/a )^(1/4)*EllipticE(sin(1/2*arctan(b^(1/2)*x^2/a^(1/2))),2^(1/2))/a^(3/2)/(b *x^4+a)^(1/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.52 \[ \int \frac {c+d x^4}{x^7 \sqrt [4]{a+b x^4}} \, dx=\frac {-2 c \left (a+b x^4\right )+3 (b c-2 a d) x^4 \sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {1}{2},-\frac {b x^4}{a}\right )}{12 a x^6 \sqrt [4]{a+b x^4}} \] Input:
Integrate[(c + d*x^4)/(x^7*(a + b*x^4)^(1/4)),x]
Output:
(-2*c*(a + b*x^4) + 3*(b*c - 2*a*d)*x^4*(1 + (b*x^4)/a)^(1/4)*Hypergeometr ic2F1[-1/2, 1/4, 1/2, -((b*x^4)/a)])/(12*a*x^6*(a + b*x^4)^(1/4))
Time = 0.41 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {955, 807, 264, 227, 225, 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.
\(\displaystyle \int \frac {c+d x^4}{x^7 \sqrt [4]{a+b x^4}} \, dx\) |
\(\Big \downarrow \) 955 |
\(\displaystyle -\frac {(b c-2 a d) \int \frac {1}{x^3 \sqrt [4]{b x^4+a}}dx}{2 a}-\frac {c \left (a+b x^4\right )^{3/4}}{6 a x^6}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle -\frac {(b c-2 a d) \int \frac {1}{x^4 \sqrt [4]{b x^4+a}}dx^2}{4 a}-\frac {c \left (a+b x^4\right )^{3/4}}{6 a x^6}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle -\frac {(b c-2 a d) \left (\frac {b \int \frac {1}{\sqrt [4]{b x^4+a}}dx^2}{2 a}-\frac {\left (a+b x^4\right )^{3/4}}{a x^2}\right )}{4 a}-\frac {c \left (a+b x^4\right )^{3/4}}{6 a x^6}\) |
\(\Big \downarrow \) 227 |
\(\displaystyle -\frac {(b c-2 a d) \left (\frac {b \sqrt [4]{\frac {b x^4}{a}+1} \int \frac {1}{\sqrt [4]{\frac {b x^4}{a}+1}}dx^2}{2 a \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{a x^2}\right )}{4 a}-\frac {c \left (a+b x^4\right )^{3/4}}{6 a x^6}\) |
\(\Big \downarrow \) 225 |
\(\displaystyle -\frac {(b c-2 a d) \left (\frac {b \sqrt [4]{\frac {b x^4}{a}+1} \left (\frac {2 x^2}{\sqrt [4]{\frac {b x^4}{a}+1}}-\int \frac {1}{\left (\frac {b x^4}{a}+1\right )^{5/4}}dx^2\right )}{2 a \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{a x^2}\right )}{4 a}-\frac {c \left (a+b x^4\right )^{3/4}}{6 a x^6}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle -\frac {(b c-2 a d) \left (\frac {b \sqrt [4]{\frac {b x^4}{a}+1} \left (\frac {2 x^2}{\sqrt [4]{\frac {b x^4}{a}+1}}-\frac {2 \sqrt {a} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b}}\right )}{2 a \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{a x^2}\right )}{4 a}-\frac {c \left (a+b x^4\right )^{3/4}}{6 a x^6}\) |
Input:
Int[(c + d*x^4)/(x^7*(a + b*x^4)^(1/4)),x]
Output:
-1/6*(c*(a + b*x^4)^(3/4))/(a*x^6) - ((b*c - 2*a*d)*(-((a + b*x^4)^(3/4)/( a*x^2)) + (b*(1 + (b*x^4)/a)^(1/4)*((2*x^2)/(1 + (b*x^4)/a)^(1/4) - (2*Sqr t[a]*EllipticE[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/Sqrt[b]))/(2*a*(a + b* x^4)^(1/4))))/(4*a)
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)^(-1/4), x_Symbol] :> Simp[2*(x/(a + b*x^2)^(1/4)) , x] - Simp[a Int[1/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b}, x] && GtQ[ a, 0] && PosQ[b/a]
Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(1/4)/( a + b*x^2)^(1/4) Int[1/(1 + b*(x^2/a))^(1/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
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]
\[\int \frac {d \,x^{4}+c}{x^{7} \left (b \,x^{4}+a \right )^{\frac {1}{4}}}d x\]
Input:
int((d*x^4+c)/x^7/(b*x^4+a)^(1/4),x)
Output:
int((d*x^4+c)/x^7/(b*x^4+a)^(1/4),x)
\[ \int \frac {c+d x^4}{x^7 \sqrt [4]{a+b x^4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} x^{7}} \,d x } \] Input:
integrate((d*x^4+c)/x^7/(b*x^4+a)^(1/4),x, algorithm="fricas")
Output:
integral((b*x^4 + a)^(3/4)*(d*x^4 + c)/(b*x^11 + a*x^7), x)
Result contains complex when optimal does not.
Time = 4.19 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.44 \[ \int \frac {c+d x^4}{x^7 \sqrt [4]{a+b x^4}} \, dx=- \frac {c {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {1}{4} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{6 \sqrt [4]{a} x^{6}} - \frac {d {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{2 \sqrt [4]{a} x^{2}} \] Input:
integrate((d*x**4+c)/x**7/(b*x**4+a)**(1/4),x)
Output:
-c*hyper((-3/2, 1/4), (-1/2,), b*x**4*exp_polar(I*pi)/a)/(6*a**(1/4)*x**6) - d*hyper((-1/2, 1/4), (1/2,), b*x**4*exp_polar(I*pi)/a)/(2*a**(1/4)*x**2 )
\[ \int \frac {c+d x^4}{x^7 \sqrt [4]{a+b x^4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} x^{7}} \,d x } \] Input:
integrate((d*x^4+c)/x^7/(b*x^4+a)^(1/4),x, algorithm="maxima")
Output:
integrate((d*x^4 + c)/((b*x^4 + a)^(1/4)*x^7), x)
\[ \int \frac {c+d x^4}{x^7 \sqrt [4]{a+b x^4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} x^{7}} \,d x } \] Input:
integrate((d*x^4+c)/x^7/(b*x^4+a)^(1/4),x, algorithm="giac")
Output:
integrate((d*x^4 + c)/((b*x^4 + a)^(1/4)*x^7), x)
Timed out. \[ \int \frac {c+d x^4}{x^7 \sqrt [4]{a+b x^4}} \, dx=\int \frac {d\,x^4+c}{x^7\,{\left (b\,x^4+a\right )}^{1/4}} \,d x \] Input:
int((c + d*x^4)/(x^7*(a + b*x^4)^(1/4)),x)
Output:
int((c + d*x^4)/(x^7*(a + b*x^4)^(1/4)), x)
\[ \int \frac {c+d x^4}{x^7 \sqrt [4]{a+b x^4}} \, dx=\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} x^{7}}d x \right ) c +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} x^{3}}d x \right ) d \] Input:
int((d*x^4+c)/x^7/(b*x^4+a)^(1/4),x)
Output:
int(1/((a + b*x**4)**(1/4)*x**7),x)*c + int(1/((a + b*x**4)**(1/4)*x**3),x )*d