Integrand size = 26, antiderivative size = 88 \[ \int \frac {x^4}{\sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \, dx=\frac {x^5 \sqrt [4]{1+\frac {b x^2}{a}} \sqrt [4]{1+\frac {d x^2}{c}} \operatorname {AppellF1}\left (\frac {5}{2},\frac {1}{4},\frac {1}{4},\frac {7}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{5 \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \] Output:
1/5*x^5*(1+b*x^2/a)^(1/4)*(1+d*x^2/c)^(1/4)*AppellF1(5/2,1/4,1/4,7/2,-b*x^ 2/a,-d*x^2/c)/(b*x^2+a)^(1/4)/(d*x^2+c)^(1/4)
Leaf count is larger than twice the leaf count of optimal. \(271\) vs. \(2(88)=176\).
Time = 2.69 (sec) , antiderivative size = 271, normalized size of antiderivative = 3.08 \[ \int \frac {x^4}{\sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \, dx=\frac {x \left (6 \left (a+b x^2\right ) \left (c+d x^2\right )-5 (b c+a d) x^2 \sqrt [4]{1+\frac {b x^2}{a}} \sqrt [4]{1+\frac {d x^2}{c}} \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},\frac {1}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+\frac {36 a^2 c^2 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},\frac {1}{4},\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{-6 a c \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},\frac {1}{4},\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+x^2 \left (a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},\frac {5}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{4},\frac {1}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )}\right )}{24 b d \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \] Input:
Integrate[x^4/((a + b*x^2)^(1/4)*(c + d*x^2)^(1/4)),x]
Output:
(x*(6*(a + b*x^2)*(c + d*x^2) - 5*(b*c + a*d)*x^2*(1 + (b*x^2)/a)^(1/4)*(1 + (d*x^2)/c)^(1/4)*AppellF1[3/2, 1/4, 1/4, 5/2, -((b*x^2)/a), -((d*x^2)/c )] + (36*a^2*c^2*AppellF1[1/2, 1/4, 1/4, 3/2, -((b*x^2)/a), -((d*x^2)/c)]) /(-6*a*c*AppellF1[1/2, 1/4, 1/4, 3/2, -((b*x^2)/a), -((d*x^2)/c)] + x^2*(a *d*AppellF1[3/2, 1/4, 5/4, 5/2, -((b*x^2)/a), -((d*x^2)/c)] + b*c*AppellF1 [3/2, 5/4, 1/4, 5/2, -((b*x^2)/a), -((d*x^2)/c)]))))/(24*b*d*(a + b*x^2)^( 1/4)*(c + d*x^2)^(1/4))
Time = 0.24 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {395, 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.
\(\displaystyle \int \frac {x^4}{\sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \, dx\) |
\(\Big \downarrow \) 395 |
\(\displaystyle \frac {\sqrt [4]{\frac {b x^2}{a}+1} \int \frac {x^4}{\sqrt [4]{\frac {b x^2}{a}+1} \sqrt [4]{d x^2+c}}dx}{\sqrt [4]{a+b x^2}}\) |
\(\Big \downarrow \) 395 |
\(\displaystyle \frac {\sqrt [4]{\frac {b x^2}{a}+1} \sqrt [4]{\frac {d x^2}{c}+1} \int \frac {x^4}{\sqrt [4]{\frac {b x^2}{a}+1} \sqrt [4]{\frac {d x^2}{c}+1}}dx}{\sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}}\) |
\(\Big \downarrow \) 394 |
\(\displaystyle \frac {x^5 \sqrt [4]{\frac {b x^2}{a}+1} \sqrt [4]{\frac {d x^2}{c}+1} \operatorname {AppellF1}\left (\frac {5}{2},\frac {1}{4},\frac {1}{4},\frac {7}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{5 \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}}\) |
Input:
Int[x^4/((a + b*x^2)^(1/4)*(c + d*x^2)^(1/4)),x]
Output:
(x^5*(1 + (b*x^2)/a)^(1/4)*(1 + (d*x^2)/c)^(1/4)*AppellF1[5/2, 1/4, 1/4, 7 /2, -((b*x^2)/a), -((d*x^2)/c)])/(5*(a + b*x^2)^(1/4)*(c + d*x^2)^(1/4))
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])
\[\int \frac {x^{4}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (x^{2} d +c \right )^{\frac {1}{4}}}d x\]
Input:
int(x^4/(b*x^2+a)^(1/4)/(d*x^2+c)^(1/4),x)
Output:
int(x^4/(b*x^2+a)^(1/4)/(d*x^2+c)^(1/4),x)
Timed out. \[ \int \frac {x^4}{\sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \, dx=\text {Timed out} \] Input:
integrate(x^4/(b*x^2+a)^(1/4)/(d*x^2+c)^(1/4),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {x^4}{\sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \, dx=\int \frac {x^{4}}{\sqrt [4]{a + b x^{2}} \sqrt [4]{c + d x^{2}}}\, dx \] Input:
integrate(x**4/(b*x**2+a)**(1/4)/(d*x**2+c)**(1/4),x)
Output:
Integral(x**4/((a + b*x**2)**(1/4)*(c + d*x**2)**(1/4)), x)
\[ \int \frac {x^4}{\sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \, dx=\int { \frac {x^{4}}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (d x^{2} + c\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate(x^4/(b*x^2+a)^(1/4)/(d*x^2+c)^(1/4),x, algorithm="maxima")
Output:
integrate(x^4/((b*x^2 + a)^(1/4)*(d*x^2 + c)^(1/4)), x)
\[ \int \frac {x^4}{\sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \, dx=\int { \frac {x^{4}}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (d x^{2} + c\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate(x^4/(b*x^2+a)^(1/4)/(d*x^2+c)^(1/4),x, algorithm="giac")
Output:
integrate(x^4/((b*x^2 + a)^(1/4)*(d*x^2 + c)^(1/4)), x)
Timed out. \[ \int \frac {x^4}{\sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \, dx=\int \frac {x^4}{{\left (b\,x^2+a\right )}^{1/4}\,{\left (d\,x^2+c\right )}^{1/4}} \,d x \] Input:
int(x^4/((a + b*x^2)^(1/4)*(c + d*x^2)^(1/4)),x)
Output:
int(x^4/((a + b*x^2)^(1/4)*(c + d*x^2)^(1/4)), x)
\[ \int \frac {x^4}{\sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \, dx=\int \frac {x^{4}}{\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}}}d x \] Input:
int(x^4/(b*x^2+a)^(1/4)/(d*x^2+c)^(1/4),x)
Output:
int(x**4/((c + d*x**2)**(1/4)*(a + b*x**2)**(1/4)),x)