Integrand size = 22, antiderivative size = 120 \[ \int \frac {x^5 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=-\frac {(b c-a d) x^2}{3 b^2 \left (a+b x^4\right )^{3/4}}+\frac {d x^2 \sqrt [4]{a+b x^4}}{3 b^2}+\frac {2 \sqrt {a} (b c-2 a d) \left (1+\frac {b x^4}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{3 b^{5/2} \left (a+b x^4\right )^{3/4}} \] Output:
-1/3*(-a*d+b*c)*x^2/b^2/(b*x^4+a)^(3/4)+1/3*d*x^2*(b*x^4+a)^(1/4)/b^2+2/3* a^(1/2)*(-2*a*d+b*c)*(1+b*x^4/a)^(3/4)*InverseJacobiAM(1/2*arctan(b^(1/2)* x^2/a^(1/2)),2^(1/2))/b^(5/2)/(b*x^4+a)^(3/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.06 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.65 \[ \int \frac {x^5 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\frac {x^2 \left (-b c+2 a d+b d x^4+(b c-2 a d) \left (1+\frac {b x^4}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},-\frac {b x^4}{a}\right )\right )}{3 b^2 \left (a+b x^4\right )^{3/4}} \] Input:
Integrate[(x^5*(c + d*x^4))/(a + b*x^4)^(7/4),x]
Output:
(x^2*(-(b*c) + 2*a*d + b*d*x^4 + (b*c - 2*a*d)*(1 + (b*x^4)/a)^(3/4)*Hyper geometric2F1[1/2, 3/4, 3/2, -((b*x^4)/a)]))/(3*b^2*(a + b*x^4)^(3/4))
Time = 0.41 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {957, 807, 262, 231, 229}
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.
\(\displaystyle \int \frac {x^5 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx\) |
\(\Big \downarrow \) 957 |
\(\displaystyle \frac {x^6 (b c-a d)}{3 a b \left (a+b x^4\right )^{3/4}}-\frac {(b c-2 a d) \int \frac {x^5}{\left (b x^4+a\right )^{3/4}}dx}{a b}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {x^6 (b c-a d)}{3 a b \left (a+b x^4\right )^{3/4}}-\frac {(b c-2 a d) \int \frac {x^4}{\left (b x^4+a\right )^{3/4}}dx^2}{2 a b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {x^6 (b c-a d)}{3 a b \left (a+b x^4\right )^{3/4}}-\frac {(b c-2 a d) \left (\frac {2 x^2 \sqrt [4]{a+b x^4}}{3 b}-\frac {2 a \int \frac {1}{\left (b x^4+a\right )^{3/4}}dx^2}{3 b}\right )}{2 a b}\) |
\(\Big \downarrow \) 231 |
\(\displaystyle \frac {x^6 (b c-a d)}{3 a b \left (a+b x^4\right )^{3/4}}-\frac {(b c-2 a d) \left (\frac {2 x^2 \sqrt [4]{a+b x^4}}{3 b}-\frac {2 a \left (\frac {b x^4}{a}+1\right )^{3/4} \int \frac {1}{\left (\frac {b x^4}{a}+1\right )^{3/4}}dx^2}{3 b \left (a+b x^4\right )^{3/4}}\right )}{2 a b}\) |
\(\Big \downarrow \) 229 |
\(\displaystyle \frac {x^6 (b c-a d)}{3 a b \left (a+b x^4\right )^{3/4}}-\frac {(b c-2 a d) \left (\frac {2 x^2 \sqrt [4]{a+b x^4}}{3 b}-\frac {4 a^{3/2} \left (\frac {b x^4}{a}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{3 b^{3/2} \left (a+b x^4\right )^{3/4}}\right )}{2 a b}\) |
Input:
Int[(x^5*(c + d*x^4))/(a + b*x^4)^(7/4),x]
Output:
((b*c - a*d)*x^6)/(3*a*b*(a + b*x^4)^(3/4)) - ((b*c - 2*a*d)*((2*x^2*(a + b*x^4)^(1/4))/(3*b) - (4*a^(3/2)*(1 + (b*x^4)/a)^(3/4)*EllipticF[ArcTan[(S qrt[b]*x^2)/Sqrt[a]]/2, 2])/(3*b^(3/2)*(a + b*x^4)^(3/4))))/(2*a*b)
Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) )*EllipticF[(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)^(-3/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(3/4)/( a + b*x^2)^(3/4) Int[1/(1 + b*(x^2/a))^(3/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && 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[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a *b*e*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* (p + 1)) Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N eQ[p, -5/4]) || !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 , m, (-n)*(p + 1)]))
\[\int \frac {x^{5} \left (d \,x^{4}+c \right )}{\left (b \,x^{4}+a \right )^{\frac {7}{4}}}d x\]
Input:
int(x^5*(d*x^4+c)/(b*x^4+a)^(7/4),x)
Output:
int(x^5*(d*x^4+c)/(b*x^4+a)^(7/4),x)
\[ \int \frac {x^5 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{5}}{{\left (b x^{4} + a\right )}^{\frac {7}{4}}} \,d x } \] Input:
integrate(x^5*(d*x^4+c)/(b*x^4+a)^(7/4),x, algorithm="fricas")
Output:
integral((d*x^9 + c*x^5)*(b*x^4 + a)^(1/4)/(b^2*x^8 + 2*a*b*x^4 + a^2), x)
Result contains complex when optimal does not.
Time = 9.57 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.50 \[ \int \frac {x^5 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\frac {c x^{6} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {7}{4} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{6 a^{\frac {7}{4}}} + \frac {d x^{10} {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{10 a^{\frac {7}{4}}} \] Input:
integrate(x**5*(d*x**4+c)/(b*x**4+a)**(7/4),x)
Output:
c*x**6*hyper((3/2, 7/4), (5/2,), b*x**4*exp_polar(I*pi)/a)/(6*a**(7/4)) + d*x**10*hyper((7/4, 5/2), (7/2,), b*x**4*exp_polar(I*pi)/a)/(10*a**(7/4))
\[ \int \frac {x^5 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{5}}{{\left (b x^{4} + a\right )}^{\frac {7}{4}}} \,d x } \] Input:
integrate(x^5*(d*x^4+c)/(b*x^4+a)^(7/4),x, algorithm="maxima")
Output:
integrate((d*x^4 + c)*x^5/(b*x^4 + a)^(7/4), x)
\[ \int \frac {x^5 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{5}}{{\left (b x^{4} + a\right )}^{\frac {7}{4}}} \,d x } \] Input:
integrate(x^5*(d*x^4+c)/(b*x^4+a)^(7/4),x, algorithm="giac")
Output:
integrate((d*x^4 + c)*x^5/(b*x^4 + a)^(7/4), x)
Timed out. \[ \int \frac {x^5 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\int \frac {x^5\,\left (d\,x^4+c\right )}{{\left (b\,x^4+a\right )}^{7/4}} \,d x \] Input:
int((x^5*(c + d*x^4))/(a + b*x^4)^(7/4),x)
Output:
int((x^5*(c + d*x^4))/(a + b*x^4)^(7/4), x)
\[ \int \frac {x^5 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\left (\int \frac {x^{9}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} a +\left (b \,x^{4}+a \right )^{\frac {3}{4}} b \,x^{4}}d x \right ) d +\left (\int \frac {x^{5}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} a +\left (b \,x^{4}+a \right )^{\frac {3}{4}} b \,x^{4}}d x \right ) c \] Input:
int(x^5*(d*x^4+c)/(b*x^4+a)^(7/4),x)
Output:
int(x**9/((a + b*x**4)**(3/4)*a + (a + b*x**4)**(3/4)*b*x**4),x)*d + int(x **5/((a + b*x**4)**(3/4)*a + (a + b*x**4)**(3/4)*b*x**4),x)*c