Integrand size = 24, antiderivative size = 348 \[ \int \frac {1}{x^2 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^2} \, dx=-\frac {(2 b c-3 a d) \sqrt [4]{a+b x^2}}{2 a c^2 (b c-a d) x}-\frac {d \sqrt [4]{a+b x^2}}{2 c (b c-a d) x \left (c+d x^2\right )}-\frac {\sqrt {b} (2 b c-3 a d) \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{2 \sqrt {a} c^2 (b c-a d) \left (a+b x^2\right )^{3/4}}-\frac {3 \sqrt [4]{a} d (3 b c-2 a d) \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{4 c^2 (b c-a d)^2 x}-\frac {3 \sqrt [4]{a} d (3 b c-2 a d) \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{4 c^2 (b c-a d)^2 x} \] Output:
-1/2*(-3*a*d+2*b*c)*(b*x^2+a)^(1/4)/a/c^2/(-a*d+b*c)/x-1/2*d*(b*x^2+a)^(1/ 4)/c/(-a*d+b*c)/x/(d*x^2+c)-1/2*b^(1/2)*(-3*a*d+2*b*c)*(1+b*x^2/a)^(3/4)*I nverseJacobiAM(1/2*arctan(b^(1/2)*x/a^(1/2)),2^(1/2))/a^(1/2)/c^2/(-a*d+b* c)/(b*x^2+a)^(3/4)-3/4*a^(1/4)*d*(-2*a*d+3*b*c)*(-b*x^2/a)^(1/2)*EllipticP i((b*x^2+a)^(1/4)/a^(1/4),-a^(1/2)*d^(1/2)/(a*d-b*c)^(1/2),I)/c^2/(-a*d+b* c)^2/x-3/4*a^(1/4)*d*(-2*a*d+3*b*c)*(-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)/c^2/(-a*d+b*c)^2/x
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 10.58 (sec) , antiderivative size = 418, normalized size of antiderivative = 1.20 \[ \int \frac {1}{x^2 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^2} \, dx=\frac {-\frac {b d (-2 b c+3 a d) x^4 \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{-b c+a d}+\frac {6 c \left (-6 a c \left (-b^2 c x^2 \left (3 c+2 d x^2\right )+2 a^2 d \left (c+3 d x^2\right )+a b \left (-2 c^2-3 c d x^2+3 d^2 x^4\right )\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+x^2 \left (a+b x^2\right ) \left (-2 b c \left (c+d x^2\right )+a d \left (2 c+3 d x^2\right )\right ) \left (4 a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},2,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+3 b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {7}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )\right )}{(b c-a d) \left (c+d x^2\right ) \left (-6 a c \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+x^2 \left (4 a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},2,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+3 b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {7}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )\right )}}{12 a c^3 x \left (a+b x^2\right )^{3/4}} \] Input:
Integrate[1/(x^2*(a + b*x^2)^(3/4)*(c + d*x^2)^2),x]
Output:
(-((b*d*(-2*b*c + 3*a*d)*x^4*(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) + a*d)) + (6*c*(-6*a*c*(-(b^2*c*x ^2*(3*c + 2*d*x^2)) + 2*a^2*d*(c + 3*d*x^2) + a*b*(-2*c^2 - 3*c*d*x^2 + 3* d^2*x^4))*AppellF1[1/2, 3/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/c)] + x^2*(a + b*x^2)*(-2*b*c*(c + d*x^2) + a*d*(2*c + 3*d*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)])))/((b*c - a*d)*(c + d*x^2)*(-6*a*c*AppellF1[ 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)]))))/(12*a*c^3*x*(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 = 62, normalized size of antiderivative = 0.18, 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.
\(\displaystyle \int \frac {1}{x^2 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 395 |
\(\displaystyle \frac {\left (\frac {b x^2}{a}+1\right )^{3/4} \int \frac {1}{x^2 \left (\frac {b x^2}{a}+1\right )^{3/4} \left (d x^2+c\right )^2}dx}{\left (a+b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 394 |
\(\displaystyle -\frac {\left (\frac {b x^2}{a}+1\right )^{3/4} \operatorname {AppellF1}\left (-\frac {1}{2},\frac {3}{4},2,\frac {1}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{c^2 x \left (a+b x^2\right )^{3/4}}\) |
Input:
Int[1/(x^2*(a + b*x^2)^(3/4)*(c + d*x^2)^2),x]
Output:
-(((1 + (b*x^2)/a)^(3/4)*AppellF1[-1/2, 3/4, 2, 1/2, -((b*x^2)/a), -((d*x^ 2)/c)])/(c^2*x*(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])
\[\int \frac {1}{x^{2} \left (b \,x^{2}+a \right )^{\frac {3}{4}} \left (x^{2} d +c \right )^{2}}d x\]
Input:
int(1/x^2/(b*x^2+a)^(3/4)/(d*x^2+c)^2,x)
Output:
int(1/x^2/(b*x^2+a)^(3/4)/(d*x^2+c)^2,x)
Timed out. \[ \int \frac {1}{x^2 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^2} \, dx=\text {Timed out} \] Input:
integrate(1/x^2/(b*x^2+a)^(3/4)/(d*x^2+c)^2,x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {1}{x^2 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^2} \, dx=\int \frac {1}{x^{2} \left (a + b x^{2}\right )^{\frac {3}{4}} \left (c + d x^{2}\right )^{2}}\, dx \] Input:
integrate(1/x**2/(b*x**2+a)**(3/4)/(d*x**2+c)**2,x)
Output:
Integral(1/(x**2*(a + b*x**2)**(3/4)*(c + d*x**2)**2), x)
\[ \int \frac {1}{x^2 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^2} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} {\left (d x^{2} + c\right )}^{2} x^{2}} \,d x } \] Input:
integrate(1/x^2/(b*x^2+a)^(3/4)/(d*x^2+c)^2,x, algorithm="maxima")
Output:
integrate(1/((b*x^2 + a)^(3/4)*(d*x^2 + c)^2*x^2), x)
\[ \int \frac {1}{x^2 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^2} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} {\left (d x^{2} + c\right )}^{2} x^{2}} \,d x } \] Input:
integrate(1/x^2/(b*x^2+a)^(3/4)/(d*x^2+c)^2,x, algorithm="giac")
Output:
integrate(1/((b*x^2 + a)^(3/4)*(d*x^2 + c)^2*x^2), x)
Timed out. \[ \int \frac {1}{x^2 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^2} \, dx=\int \frac {1}{x^2\,{\left (b\,x^2+a\right )}^{3/4}\,{\left (d\,x^2+c\right )}^2} \,d x \] Input:
int(1/(x^2*(a + b*x^2)^(3/4)*(c + d*x^2)^2),x)
Output:
int(1/(x^2*(a + b*x^2)^(3/4)*(c + d*x^2)^2), x)
\[ \int \frac {1}{x^2 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^2} \, dx=\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} c^{2} x^{2}+2 \left (b \,x^{2}+a \right )^{\frac {3}{4}} c d \,x^{4}+\left (b \,x^{2}+a \right )^{\frac {3}{4}} d^{2} x^{6}}d x \] Input:
int(1/x^2/(b*x^2+a)^(3/4)/(d*x^2+c)^2,x)
Output:
int(1/((a + b*x**2)**(3/4)*c**2*x**2 + 2*(a + b*x**2)**(3/4)*c*d*x**4 + (a + b*x**2)**(3/4)*d**2*x**6),x)