\(\int \frac {1}{x^2 (a+b x^2)^{3/4} (c+d x^2)} \, dx\) [1532]

Optimal result

Integrand size = 24, antiderivative size = 243 \[ \int \frac {1}{x^2 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )} \, dx=-\frac {\sqrt [4]{a+b x^2}}{a c x}-\frac {\sqrt {b} \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{\sqrt {a} c \left (a+b x^2\right )^{3/4}}-\frac {\sqrt [4]{a} d \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{c (b c-a d) x}-\frac {\sqrt [4]{a} d \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{c (b c-a d) x} \] Output:

-(b*x^2+a)^(1/4)/a/c/x-b^(1/2)*(1+b*x^2/a)^(3/4)*InverseJacobiAM(1/2*arcta 
n(b^(1/2)*x/a^(1/2)),2^(1/2))/a^(1/2)/c/(b*x^2+a)^(3/4)-a^(1/4)*d*(-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/(-a*d+b*c)/x-a^(1/4)*d*(-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/(-a*d+b*c)/x
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 10.27 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.40 \[ \int \frac {1}{x^2 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )} \, dx=\frac {-b d x^4 \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+\frac {6 c \left (-3 a c \left (2 a c+3 b c x^2+4 a d x^2+2 b d x^4\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+x^2 \left (a+b x^2\right ) \left (c+d x^2\right ) \left (4 a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},2,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+3 b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {7}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )\right )}{\left (c+d x^2\right ) \left (6 a c \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )-x^2 \left (4 a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},2,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+3 b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {7}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )\right )}}{6 a c^2 x \left (a+b x^2\right )^{3/4}} \] Input:

Integrate[1/(x^2*(a + b*x^2)^(3/4)*(c + d*x^2)),x]
 

Output:

(-(b*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)]) + (6*c*(-3*a*c*(2*a*c + 3*b*c*x^2 + 4*a*d*x^2 + 2*b*d*x^4)* 
AppellF1[1/2, 3/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/c)] + x^2*(a + b*x^2)*( 
c + 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)])))/((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*Appell 
F1[3/2, 7/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c)]))))/(6*a*c^2*x*(a + b*x^2 
)^(3/4))
 

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 0.19 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.26, 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.

\(\displaystyle \int \frac {1}{x^2 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )} \, dx\)

\(\Big \downarrow \) 395

\(\displaystyle \frac {\left (\frac {b x^2}{a}+1\right )^{3/4} \int \frac {1}{x^2 \left (\frac {b x^2}{a}+1\right )^{3/4} \left (d x^2+c\right )}dx}{\left (a+b x^2\right )^{3/4}}\)

\(\Big \downarrow \) 394

\(\displaystyle -\frac {\left (\frac {b x^2}{a}+1\right )^{3/4} \operatorname {AppellF1}\left (-\frac {1}{2},\frac {3}{4},1,\frac {1}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{c x \left (a+b x^2\right )^{3/4}}\)

Input:

Int[1/(x^2*(a + b*x^2)^(3/4)*(c + d*x^2)),x]
 

Output:

-(((1 + (b*x^2)/a)^(3/4)*AppellF1[-1/2, 3/4, 1, 1/2, -((b*x^2)/a), -((d*x^ 
2)/c)])/(c*x*(a + b*x^2)^(3/4)))
 

Defintions of rubi rules used

rule 394

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])
 

rule 395

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])
 

Maple [F]

\[\int \frac {1}{x^{2} \left (b \,x^{2}+a \right )^{\frac {3}{4}} \left (x^{2} d +c \right )}d x\]

Input:

int(1/x^2/(b*x^2+a)^(3/4)/(d*x^2+c),x)
 

Output:

int(1/x^2/(b*x^2+a)^(3/4)/(d*x^2+c),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{x^2 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )} \, dx=\text {Timed out} \] Input:

integrate(1/x^2/(b*x^2+a)^(3/4)/(d*x^2+c),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {1}{x^2 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )} \, dx=\int \frac {1}{x^{2} \left (a + b x^{2}\right )^{\frac {3}{4}} \left (c + d x^{2}\right )}\, dx \] Input:

integrate(1/x**2/(b*x**2+a)**(3/4)/(d*x**2+c),x)
 

Output:

Integral(1/(x**2*(a + b*x**2)**(3/4)*(c + d*x**2)), x)
 

Maxima [F]

\[ \int \frac {1}{x^2 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} {\left (d x^{2} + c\right )} x^{2}} \,d x } \] Input:

integrate(1/x^2/(b*x^2+a)^(3/4)/(d*x^2+c),x, algorithm="maxima")
 

Output:

integrate(1/((b*x^2 + a)^(3/4)*(d*x^2 + c)*x^2), x)
 

Giac [F]

\[ \int \frac {1}{x^2 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} {\left (d x^{2} + c\right )} x^{2}} \,d x } \] Input:

integrate(1/x^2/(b*x^2+a)^(3/4)/(d*x^2+c),x, algorithm="giac")
 

Output:

integrate(1/((b*x^2 + a)^(3/4)*(d*x^2 + c)*x^2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^2 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )} \, dx=\int \frac {1}{x^2\,{\left (b\,x^2+a\right )}^{3/4}\,\left (d\,x^2+c\right )} \,d x \] Input:

int(1/(x^2*(a + b*x^2)^(3/4)*(c + d*x^2)),x)
 

Output:

int(1/(x^2*(a + b*x^2)^(3/4)*(c + d*x^2)), x)
 

Reduce [F]

\[ \int \frac {1}{x^2 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )} \, dx=\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} c \,x^{2}+\left (b \,x^{2}+a \right )^{\frac {3}{4}} d \,x^{4}}d x \] Input:

int(1/x^2/(b*x^2+a)^(3/4)/(d*x^2+c),x)
 

Output:

int(1/((a + b*x**2)**(3/4)*c*x**2 + (a + b*x**2)**(3/4)*d*x**4),x)