Integrand size = 22, antiderivative size = 143 \[ \int \frac {x^4 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {2 a (b c-a d) x}{5 b^3 \left (a+b x^2\right )^{5/4}}+\frac {2 (5 b c-12 a d) x}{5 b^3 \sqrt [4]{a+b x^2}}+\frac {2 d x \left (a+b x^2\right )^{3/4}}{5 b^3}-\frac {24 \sqrt {a} (b c-2 a d) \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 b^{7/2} \sqrt [4]{a+b x^2}} \] Output:
2/5*a*(-a*d+b*c)*x/b^3/(b*x^2+a)^(5/4)+2/5*(-12*a*d+5*b*c)*x/b^3/(b*x^2+a) ^(1/4)+2/5*d*x*(b*x^2+a)^(3/4)/b^3-24/5*a^(1/2)*(-2*a*d+b*c)*(1+b*x^2/a)^( 1/4)*EllipticE(sin(1/2*arctan(b^(1/2)*x/a^(1/2))),2^(1/2))/b^(7/2)/(b*x^2+ a)^(1/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.07 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.73 \[ \int \frac {x^4 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {2 x \left (12 a^2 d+b^2 x^2 \left (-7 c+d x^2\right )+a b \left (-6 c+14 d x^2\right )+6 (b c-2 a d) \left (a+b x^2\right ) \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{5 b^3 \left (a+b x^2\right )^{5/4}} \] Input:
Integrate[(x^4*(c + d*x^2))/(a + b*x^2)^(9/4),x]
Output:
(2*x*(12*a^2*d + b^2*x^2*(-7*c + d*x^2) + a*b*(-6*c + 14*d*x^2) + 6*(b*c - 2*a*d)*(a + b*x^2)*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, -((b*x^2)/a)]))/(5*b^3*(a + b*x^2)^(5/4))
Time = 0.24 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {362, 250, 250, 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.
\(\displaystyle \int \frac {x^4 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx\) |
\(\Big \downarrow \) 362 |
\(\displaystyle \frac {2 x^5 (b c-a d)}{5 a b \left (a+b x^2\right )^{5/4}}-\frac {(b c-2 a d) \int \frac {x^4}{\left (b x^2+a\right )^{5/4}}dx}{a b}\) |
\(\Big \downarrow \) 250 |
\(\displaystyle \frac {2 x^5 (b c-a d)}{5 a b \left (a+b x^2\right )^{5/4}}-\frac {(b c-2 a d) \left (\frac {2 x^3}{5 b \sqrt [4]{a+b x^2}}-\frac {6 a \int \frac {x^2}{\left (b x^2+a\right )^{5/4}}dx}{5 b}\right )}{a b}\) |
\(\Big \downarrow \) 250 |
\(\displaystyle \frac {2 x^5 (b c-a d)}{5 a b \left (a+b x^2\right )^{5/4}}-\frac {(b c-2 a d) \left (\frac {2 x^3}{5 b \sqrt [4]{a+b x^2}}-\frac {6 a \left (\frac {2 x}{b \sqrt [4]{a+b x^2}}-\frac {2 a \int \frac {1}{\left (b x^2+a\right )^{5/4}}dx}{b}\right )}{5 b}\right )}{a b}\) |
\(\Big \downarrow \) 213 |
\(\displaystyle \frac {2 x^5 (b c-a d)}{5 a b \left (a+b x^2\right )^{5/4}}-\frac {(b c-2 a d) \left (\frac {2 x^3}{5 b \sqrt [4]{a+b x^2}}-\frac {6 a \left (\frac {2 x}{b \sqrt [4]{a+b x^2}}-\frac {2 \sqrt [4]{\frac {b x^2}{a}+1} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{5/4}}dx}{b \sqrt [4]{a+b x^2}}\right )}{5 b}\right )}{a b}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle \frac {2 x^5 (b c-a d)}{5 a b \left (a+b x^2\right )^{5/4}}-\frac {(b c-2 a d) \left (\frac {2 x^3}{5 b \sqrt [4]{a+b x^2}}-\frac {6 a \left (\frac {2 x}{b \sqrt [4]{a+b x^2}}-\frac {4 \sqrt {a} \sqrt [4]{\frac {b x^2}{a}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{b^{3/2} \sqrt [4]{a+b x^2}}\right )}{5 b}\right )}{a b}\) |
Input:
Int[(x^4*(c + d*x^2))/(a + b*x^2)^(9/4),x]
Output:
(2*(b*c - a*d)*x^5)/(5*a*b*(a + b*x^2)^(5/4)) - ((b*c - 2*a*d)*((2*x^3)/(5 *b*(a + b*x^2)^(1/4)) - (6*a*((2*x)/(b*(a + b*x^2)^(1/4)) - (4*Sqrt[a]*(1 + (b*x^2)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(b^(3/2)*( a + b*x^2)^(1/4))))/(5*b)))/(a*b)
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[((c_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Simp[2*c*(( c*x)^(m - 1)/(b*(2*m - 3)*(a + b*x^2)^(1/4))), x] - Simp[2*a*c^2*((m - 1)/( b*(2*m - 3))) Int[(c*x)^(m - 2)/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b, c}, x] && PosQ[b/a] && IntegerQ[2*m] && GtQ[m, 3/2]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*b*e *(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(2*a*b*(p + 1)) I nt[(e*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && N eQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) || !RationalQ[m] || (ILtQ[p + 1/2, 0] && LeQ[-1, m, -2*(p + 1)]))
\[\int \frac {x^{4} \left (x^{2} d +c \right )}{\left (b \,x^{2}+a \right )^{\frac {9}{4}}}d x\]
Input:
int(x^4*(d*x^2+c)/(b*x^2+a)^(9/4),x)
Output:
int(x^4*(d*x^2+c)/(b*x^2+a)^(9/4),x)
\[ \int \frac {x^4 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} x^{4}}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}} \,d x } \] Input:
integrate(x^4*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="fricas")
Output:
integral((d*x^6 + c*x^4)*(b*x^2 + a)^(3/4)/(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2* b*x^2 + a^3), x)
Result contains complex when optimal does not.
Time = 4.72 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.42 \[ \int \frac {x^4 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {c x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {9}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5 a^{\frac {9}{4}}} + \frac {d x^{7} {{}_{2}F_{1}\left (\begin {matrix} \frac {9}{4}, \frac {7}{2} \\ \frac {9}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{7 a^{\frac {9}{4}}} \] Input:
integrate(x**4*(d*x**2+c)/(b*x**2+a)**(9/4),x)
Output:
c*x**5*hyper((9/4, 5/2), (7/2,), b*x**2*exp_polar(I*pi)/a)/(5*a**(9/4)) + d*x**7*hyper((9/4, 7/2), (9/2,), b*x**2*exp_polar(I*pi)/a)/(7*a**(9/4))
\[ \int \frac {x^4 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} x^{4}}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}} \,d x } \] Input:
integrate(x^4*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="maxima")
Output:
integrate((d*x^2 + c)*x^4/(b*x^2 + a)^(9/4), x)
\[ \int \frac {x^4 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} x^{4}}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}} \,d x } \] Input:
integrate(x^4*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="giac")
Output:
integrate((d*x^2 + c)*x^4/(b*x^2 + a)^(9/4), x)
Timed out. \[ \int \frac {x^4 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\int \frac {x^4\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{9/4}} \,d x \] Input:
int((x^4*(c + d*x^2))/(a + b*x^2)^(9/4),x)
Output:
int((x^4*(c + d*x^2))/(a + b*x^2)^(9/4), x)
\[ \int \frac {x^4 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\left (\int \frac {x^{6}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2}+2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a b \,x^{2}+\left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{2} x^{4}}d x \right ) d +\left (\int \frac {x^{4}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2}+2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a b \,x^{2}+\left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{2} x^{4}}d x \right ) c \] Input:
int(x^4*(d*x^2+c)/(b*x^2+a)^(9/4),x)
Output:
int(x**6/((a + b*x**2)**(1/4)*a**2 + 2*(a + b*x**2)**(1/4)*a*b*x**2 + (a + b*x**2)**(1/4)*b**2*x**4),x)*d + int(x**4/((a + b*x**2)**(1/4)*a**2 + 2*( a + b*x**2)**(1/4)*a*b*x**2 + (a + b*x**2)**(1/4)*b**2*x**4),x)*c