\(\int \frac {1}{(a+b x^2)^{3/4} (c+d x^2)} \, dx\) [1531]

Optimal result

Integrand size = 21, antiderivative size = 152 \[ \int \frac {1}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )} \, dx=\frac {\sqrt [4]{a} \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{(b c-a d) x}+\frac {\sqrt [4]{a} \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{(b c-a d) x} \] Output:

a^(1/4)*(-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)/(-a*d+b*c)/x+a^(1/4)*(-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)/(-a*d+b*c)/x
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )} \, dx=\frac {\sqrt [4]{a} \sqrt {-\frac {b x^2}{a}} \left (\operatorname {EllipticPi}\left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )+\operatorname {EllipticPi}\left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )\right )}{(b c-a d) x} \] Input:

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

Output:

(a^(1/4)*Sqrt[-((b*x^2)/a)]*(EllipticPi[-((Sqrt[a]*Sqrt[d])/Sqrt[-(b*c) + 
a*d]), ArcSin[(a + b*x^2)^(1/4)/a^(1/4)], -1] + EllipticPi[(Sqrt[a]*Sqrt[d 
])/Sqrt[-(b*c) + a*d], ArcSin[(a + b*x^2)^(1/4)/a^(1/4)], -1]))/((b*c - a* 
d)*x)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {312, 118, 25, 925, 1542}

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.

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

\(\Big \downarrow \) 312

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

\(\Big \downarrow \) 118

\(\displaystyle -\frac {2 \sqrt {-\frac {b x^2}{a}} \int -\frac {1}{\sqrt {1-\frac {x^8}{a}} \left (d x^8+b c-a d\right )}d\sqrt [4]{b x^2+a}}{x}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 \sqrt {-\frac {b x^2}{a}} \int \frac {1}{\sqrt {1-\frac {x^8}{a}} \left (d x^8+b c-a d\right )}d\sqrt [4]{b x^2+a}}{x}\)

\(\Big \downarrow \) 925

\(\displaystyle -\frac {2 \sqrt {-\frac {b x^2}{a}} \left (-\frac {\int \frac {1}{\left (1-\frac {\sqrt {d} x^4}{\sqrt {a d-b c}}\right ) \sqrt {1-\frac {x^8}{a}}}d\sqrt [4]{b x^2+a}}{2 (b c-a d)}-\frac {\int \frac {1}{\left (\frac {\sqrt {d} x^4}{\sqrt {a d-b c}}+1\right ) \sqrt {1-\frac {x^8}{a}}}d\sqrt [4]{b x^2+a}}{2 (b c-a d)}\right )}{x}\)

\(\Big \downarrow \) 1542

\(\displaystyle -\frac {2 \sqrt {-\frac {b x^2}{a}} \left (-\frac {\sqrt [4]{a} \operatorname {EllipticPi}\left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}},\arcsin \left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right ),-1\right )}{2 (b c-a d)}-\frac {\sqrt [4]{a} \operatorname {EllipticPi}\left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}},\arcsin \left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right ),-1\right )}{2 (b c-a d)}\right )}{x}\)

Input:

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

Output:

(-2*Sqrt[-((b*x^2)/a)]*(-1/2*(a^(1/4)*EllipticPi[-((Sqrt[a]*Sqrt[d])/Sqrt[ 
-(b*c) + a*d]), ArcSin[(a + b*x^2)^(1/4)/a^(1/4)], -1])/(b*c - a*d) - (a^( 
1/4)*EllipticPi[(Sqrt[a]*Sqrt[d])/Sqrt[-(b*c) + a*d], ArcSin[(a + b*x^2)^( 
1/4)/a^(1/4)], -1])/(2*(b*c - a*d))))/x
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 118

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^( 
3/4)), x_] :> Simp[-4   Subst[Int[1/((b*e - a*f - b*x^4)*Sqrt[c - d*(e/f) + 
 d*(x^4/f)]), x], x, (e + f*x)^(1/4)], x] /; FreeQ[{a, b, c, d, e, f}, x] & 
& GtQ[-f/(d*e - c*f), 0]
 

rule 312

Int[1/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Sim 
p[Sqrt[(-b)*(x^2/a)]/(2*x)   Subst[Int[1/(Sqrt[(-b)*(x/a)]*(a + b*x)^(3/4)* 
(c + d*x)), x], x, x^2], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
 

rule 925

Int[1/(Sqrt[(a_) + (b_.)*(x_)^4]*((c_) + (d_.)*(x_)^4)), x_Symbol] :> Simp[ 
1/(2*c)   Int[1/(Sqrt[a + b*x^4]*(1 - Rt[-d/c, 2]*x^2)), x], x] + Simp[1/(2 
*c)   Int[1/(Sqrt[a + b*x^4]*(1 + Rt[-d/c, 2]*x^2)), x], x] /; FreeQ[{a, b, 
 c, d}, x] && NeQ[b*c - a*d, 0]
 

rule 1542

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[ 
{q = Rt[-c/a, 4]}, Simp[(1/(d*Sqrt[a]*q))*EllipticPi[-e/(d*q^2), ArcSin[q*x 
], -1], x]] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && GtQ[a, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(1/(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)
 

Giac [F]

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

integrate(1/(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)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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