Integrand size = 24, antiderivative size = 256 \[ \int \frac {x^4}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )} \, dx=\frac {2 x \sqrt [4]{a+b x^2}}{3 b d}-\frac {2 \sqrt {a} (3 b c+2 a d) \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{3 b^{3/2} d^2 \left (a+b x^2\right )^{3/4}}+\frac {\sqrt [4]{a} c^2 \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{d^2 (b c-a d) x}+\frac {\sqrt [4]{a} c^2 \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{d^2 (b c-a d) x} \] Output:
2/3*x*(b*x^2+a)^(1/4)/b/d-2/3*a^(1/2)*(2*a*d+3*b*c)*(1+b*x^2/a)^(3/4)*Inve rseJacobiAM(1/2*arctan(b^(1/2)*x/a^(1/2)),2^(1/2))/b^(3/2)/d^2/(b*x^2+a)^( 3/4)+a^(1/4)*c^2*(-b*x^2/a)^(1/2)*EllipticPi((b*x^2+a)^(1/4)/a^(1/4),-a^(1 /2)*d^(1/2)/(a*d-b*c)^(1/2),I)/d^2/(-a*d+b*c)/x+a^(1/4)*c^2*(-b*x^2/a)^(1/ 2)*EllipticPi((b*x^2+a)^(1/4)/a^(1/4),a^(1/2)*d^(1/2)/(a*d-b*c)^(1/2),I)/d ^2/(-a*d+b*c)/x
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 9.66 (sec) , antiderivative size = 250, normalized size of antiderivative = 0.98 \[ \int \frac {x^4}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )} \, dx=\frac {x \left (-\frac {(3 b c+2 a d) x^2 \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{b c}+6 \left (\frac {a}{b}+x^2+\frac {6 a^2 c^2 \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{b \left (c+d x^2\right ) \left (-6 a c \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+x^2 \left (4 a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},2,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+3 b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {7}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )\right )}\right )\right )}{9 d \left (a+b x^2\right )^{3/4}} \] Input:
Integrate[x^4/((a + b*x^2)^(3/4)*(c + d*x^2)),x]
Output:
(x*(-(((3*b*c + 2*a*d)*x^2*(1 + (b*x^2)/a)^(3/4)*AppellF1[3/2, 3/4, 1, 5/2 , -((b*x^2)/a), -((d*x^2)/c)])/(b*c)) + 6*(a/b + x^2 + (6*a^2*c^2*AppellF1 [1/2, 3/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/c)])/(b*(c + d*x^2)*(-6*a*c*App ellF1[1/2, 3/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/c)] + x^2*(4*a*d*AppellF1[ 3/2, 3/4, 2, 5/2, -((b*x^2)/a), -((d*x^2)/c)] + 3*b*c*AppellF1[3/2, 7/4, 1 , 5/2, -((b*x^2)/a), -((d*x^2)/c)]))))))/(9*d*(a + b*x^2)^(3/4))
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 0.20 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.25, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {395, 394}
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.
\(\displaystyle \int \frac {x^4}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )} \, dx\) |
\(\Big \downarrow \) 395 |
\(\displaystyle \frac {\left (\frac {b x^2}{a}+1\right )^{3/4} \int \frac {x^4}{\left (\frac {b x^2}{a}+1\right )^{3/4} \left (d x^2+c\right )}dx}{\left (a+b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 394 |
\(\displaystyle \frac {x^5 \left (\frac {b x^2}{a}+1\right )^{3/4} \operatorname {AppellF1}\left (\frac {5}{2},\frac {3}{4},1,\frac {7}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{5 c \left (a+b x^2\right )^{3/4}}\) |
Input:
Int[x^4/((a + b*x^2)^(3/4)*(c + d*x^2)),x]
Output:
(x^5*(1 + (b*x^2)/a)^(3/4)*AppellF1[5/2, 3/4, 1, 7/2, -((b*x^2)/a), -((d*x ^2)/c)])/(5*c*(a + b*x^2)^(3/4))
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 , -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^ FracPart[p]) Int[(e*x)^m*(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ [{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && !(IntegerQ[p] || GtQ[a, 0])
\[\int \frac {x^{4}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} \left (x^{2} d +c \right )}d x\]
Input:
int(x^4/(b*x^2+a)^(3/4)/(d*x^2+c),x)
Output:
int(x^4/(b*x^2+a)^(3/4)/(d*x^2+c),x)
Timed out. \[ \int \frac {x^4}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )} \, dx=\text {Timed out} \] Input:
integrate(x^4/(b*x^2+a)^(3/4)/(d*x^2+c),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {x^4}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )} \, dx=\int \frac {x^{4}}{\left (a + b x^{2}\right )^{\frac {3}{4}} \left (c + d x^{2}\right )}\, dx \] Input:
integrate(x**4/(b*x**2+a)**(3/4)/(d*x**2+c),x)
Output:
Integral(x**4/((a + b*x**2)**(3/4)*(c + d*x**2)), x)
\[ \int \frac {x^4}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )} \, dx=\int { \frac {x^{4}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} {\left (d x^{2} + c\right )}} \,d x } \] Input:
integrate(x^4/(b*x^2+a)^(3/4)/(d*x^2+c),x, algorithm="maxima")
Output:
integrate(x^4/((b*x^2 + a)^(3/4)*(d*x^2 + c)), x)
\[ \int \frac {x^4}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )} \, dx=\int { \frac {x^{4}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} {\left (d x^{2} + c\right )}} \,d x } \] Input:
integrate(x^4/(b*x^2+a)^(3/4)/(d*x^2+c),x, algorithm="giac")
Output:
integrate(x^4/((b*x^2 + a)^(3/4)*(d*x^2 + c)), x)
Timed out. \[ \int \frac {x^4}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )} \, dx=\int \frac {x^4}{{\left (b\,x^2+a\right )}^{3/4}\,\left (d\,x^2+c\right )} \,d x \] Input:
int(x^4/((a + b*x^2)^(3/4)*(c + d*x^2)),x)
Output:
int(x^4/((a + b*x^2)^(3/4)*(c + d*x^2)), x)
\[ \int \frac {x^4}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )} \, dx=\int \frac {x^{4}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} c +\left (b \,x^{2}+a \right )^{\frac {3}{4}} d \,x^{2}}d x \] Input:
int(x^4/(b*x^2+a)^(3/4)/(d*x^2+c),x)
Output:
int(x**4/((a + b*x**2)**(3/4)*c + (a + b*x**2)**(3/4)*d*x**2),x)