Integrand size = 22, antiderivative size = 150 \[ \int \frac {c+d x^2}{x^4 \left (a+b x^2\right )^{7/4}} \, dx=-\frac {c}{3 a x^3 \left (a+b x^2\right )^{3/4}}-\frac {3 b c-2 a d}{3 a^2 x \left (a+b x^2\right )^{3/4}}+\frac {5 (3 b c-2 a d) \sqrt [4]{a+b x^2}}{6 a^3 x}+\frac {5 \sqrt {b} (3 b c-2 a d) \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{6 a^{5/2} \left (a+b x^2\right )^{3/4}} \] Output:
-1/3*c/a/x^3/(b*x^2+a)^(3/4)-1/3*(-2*a*d+3*b*c)/a^2/x/(b*x^2+a)^(3/4)+5/6* (-2*a*d+3*b*c)*(b*x^2+a)^(1/4)/a^3/x+5/6*b^(1/2)*(-2*a*d+3*b*c)*(1+b*x^2/a )^(3/4)*InverseJacobiAM(1/2*arctan(b^(1/2)*x/a^(1/2)),2^(1/2))/a^(5/2)/(b* x^2+a)^(3/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.49 \[ \int \frac {c+d x^2}{x^4 \left (a+b x^2\right )^{7/4}} \, dx=\frac {-2 a c+3 (3 b c-2 a d) x^2 \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {7}{4},\frac {1}{2},-\frac {b x^2}{a}\right )}{6 a^2 x^3 \left (a+b x^2\right )^{3/4}} \] Input:
Integrate[(c + d*x^2)/(x^4*(a + b*x^2)^(7/4)),x]
Output:
(-2*a*c + 3*(3*b*c - 2*a*d)*x^2*(1 + (b*x^2)/a)^(3/4)*Hypergeometric2F1[-1 /2, 7/4, 1/2, -((b*x^2)/a)])/(6*a^2*x^3*(a + b*x^2)^(3/4))
Time = 0.24 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {359, 253, 264, 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.
\(\displaystyle \int \frac {c+d x^2}{x^4 \left (a+b x^2\right )^{7/4}} \, dx\) |
\(\Big \downarrow \) 359 |
\(\displaystyle -\frac {(3 b c-2 a d) \int \frac {1}{x^2 \left (b x^2+a\right )^{7/4}}dx}{2 a}-\frac {c}{3 a x^3 \left (a+b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 253 |
\(\displaystyle -\frac {(3 b c-2 a d) \left (\frac {5 \int \frac {1}{x^2 \left (b x^2+a\right )^{3/4}}dx}{3 a}+\frac {2}{3 a x \left (a+b x^2\right )^{3/4}}\right )}{2 a}-\frac {c}{3 a x^3 \left (a+b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle -\frac {(3 b c-2 a d) \left (\frac {5 \left (-\frac {b \int \frac {1}{\left (b x^2+a\right )^{3/4}}dx}{2 a}-\frac {\sqrt [4]{a+b x^2}}{a x}\right )}{3 a}+\frac {2}{3 a x \left (a+b x^2\right )^{3/4}}\right )}{2 a}-\frac {c}{3 a x^3 \left (a+b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 231 |
\(\displaystyle -\frac {(3 b c-2 a d) \left (\frac {5 \left (-\frac {b \left (\frac {b x^2}{a}+1\right )^{3/4} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{3/4}}dx}{2 a \left (a+b x^2\right )^{3/4}}-\frac {\sqrt [4]{a+b x^2}}{a x}\right )}{3 a}+\frac {2}{3 a x \left (a+b x^2\right )^{3/4}}\right )}{2 a}-\frac {c}{3 a x^3 \left (a+b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 229 |
\(\displaystyle -\frac {(3 b c-2 a d) \left (\frac {5 \left (-\frac {\sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{\sqrt {a} \left (a+b x^2\right )^{3/4}}-\frac {\sqrt [4]{a+b x^2}}{a x}\right )}{3 a}+\frac {2}{3 a x \left (a+b x^2\right )^{3/4}}\right )}{2 a}-\frac {c}{3 a x^3 \left (a+b x^2\right )^{3/4}}\) |
Input:
Int[(c + d*x^2)/(x^4*(a + b*x^2)^(7/4)),x]
Output:
-1/3*c/(a*x^3*(a + b*x^2)^(3/4)) - ((3*b*c - 2*a*d)*(2/(3*a*x*(a + b*x^2)^ (3/4)) + (5*(-((a + b*x^2)^(1/4)/(a*x)) - (Sqrt[b]*(1 + (b*x^2)/a)^(3/4)*E llipticF[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(Sqrt[a]*(a + b*x^2)^(3/4))))/ (3*a)))/(2*a)
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*x )^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m }, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
\[\int \frac {x^{2} d +c}{x^{4} \left (b \,x^{2}+a \right )^{\frac {7}{4}}}d x\]
Input:
int((d*x^2+c)/x^4/(b*x^2+a)^(7/4),x)
Output:
int((d*x^2+c)/x^4/(b*x^2+a)^(7/4),x)
\[ \int \frac {c+d x^2}{x^4 \left (a+b x^2\right )^{7/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {7}{4}} x^{4}} \,d x } \] Input:
integrate((d*x^2+c)/x^4/(b*x^2+a)^(7/4),x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(1/4)*(d*x^2 + c)/(b^2*x^8 + 2*a*b*x^6 + a^2*x^4), x)
Result contains complex when optimal does not.
Time = 3.87 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.42 \[ \int \frac {c+d x^2}{x^4 \left (a+b x^2\right )^{7/4}} \, dx=- \frac {c {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {7}{4} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3 a^{\frac {7}{4}} x^{3}} - \frac {d {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {7}{4} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac {7}{4}} x} \] Input:
integrate((d*x**2+c)/x**4/(b*x**2+a)**(7/4),x)
Output:
-c*hyper((-3/2, 7/4), (-1/2,), b*x**2*exp_polar(I*pi)/a)/(3*a**(7/4)*x**3) - d*hyper((-1/2, 7/4), (1/2,), b*x**2*exp_polar(I*pi)/a)/(a**(7/4)*x)
\[ \int \frac {c+d x^2}{x^4 \left (a+b x^2\right )^{7/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {7}{4}} x^{4}} \,d x } \] Input:
integrate((d*x^2+c)/x^4/(b*x^2+a)^(7/4),x, algorithm="maxima")
Output:
integrate((d*x^2 + c)/((b*x^2 + a)^(7/4)*x^4), x)
\[ \int \frac {c+d x^2}{x^4 \left (a+b x^2\right )^{7/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {7}{4}} x^{4}} \,d x } \] Input:
integrate((d*x^2+c)/x^4/(b*x^2+a)^(7/4),x, algorithm="giac")
Output:
integrate((d*x^2 + c)/((b*x^2 + a)^(7/4)*x^4), x)
Timed out. \[ \int \frac {c+d x^2}{x^4 \left (a+b x^2\right )^{7/4}} \, dx=\int \frac {d\,x^2+c}{x^4\,{\left (b\,x^2+a\right )}^{7/4}} \,d x \] Input:
int((c + d*x^2)/(x^4*(a + b*x^2)^(7/4)),x)
Output:
int((c + d*x^2)/(x^4*(a + b*x^2)^(7/4)), x)
\[ \int \frac {c+d x^2}{x^4 \left (a+b x^2\right )^{7/4}} \, dx=\left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} a \,x^{4}+\left (b \,x^{2}+a \right )^{\frac {3}{4}} b \,x^{6}}d x \right ) c +\left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} a \,x^{2}+\left (b \,x^{2}+a \right )^{\frac {3}{4}} b \,x^{4}}d x \right ) d \] Input:
int((d*x^2+c)/x^4/(b*x^2+a)^(7/4),x)
Output:
int(1/((a + b*x**2)**(3/4)*a*x**4 + (a + b*x**2)**(3/4)*b*x**6),x)*c + int (1/((a + b*x**2)**(3/4)*a*x**2 + (a + b*x**2)**(3/4)*b*x**4),x)*d