\(\int \frac {1}{\sqrt {a+b x^2} (c+d x^2)^{3/2}} \, dx\) [162]

Optimal result

Integrand size = 23, antiderivative size = 194 \[ \int \frac {1}{\sqrt {a+b x^2} \left (c+d x^2\right )^{3/2}} \, dx=-\frac {\sqrt {d} \sqrt {a+b x^2} E\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{\sqrt {c} (b c-a d) \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}+\frac {b \sqrt {c} \sqrt {a+b x^2} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right ),1-\frac {b c}{a d}\right )}{a \sqrt {d} (b c-a d) \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}} \] Output:

-d^(1/2)*(b*x^2+a)^(1/2)*EllipticE(d^(1/2)*x/c^(1/2)/(1+d*x^2/c)^(1/2),(1- 
b*c/a/d)^(1/2))/c^(1/2)/(-a*d+b*c)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(d*x^2+ 
c)^(1/2)+b*c^(1/2)*(b*x^2+a)^(1/2)*InverseJacobiAM(arctan(d^(1/2)*x/c^(1/2 
)),(1-b*c/a/d)^(1/2))/a/d^(1/2)/(-a*d+b*c)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2) 
/(d*x^2+c)^(1/2)
 

Mathematica [A] (verified)

Time = 1.78 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.58 \[ \int \frac {1}{\sqrt {a+b x^2} \left (c+d x^2\right )^{3/2}} \, dx=\frac {-d x \left (a+b x^2\right )+\frac {b c \sqrt {1+\frac {b x^2}{a}} \sqrt {1+\frac {d x^2}{c}} E\left (\arcsin \left (\sqrt {-\frac {b}{a}} x\right )|\frac {a d}{b c}\right )}{\sqrt {-\frac {b}{a}}}}{c (b c-a d) \sqrt {a+b x^2} \sqrt {c+d x^2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.35, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {316, 27, 324, 320, 388, 313}

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.

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

\(\Big \downarrow \) 316

\(\displaystyle \frac {\int \frac {b \sqrt {d x^2+c}}{\sqrt {b x^2+a}}dx}{c (b c-a d)}-\frac {d x \sqrt {a+b x^2}}{c \sqrt {c+d x^2} (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b \int \frac {\sqrt {d x^2+c}}{\sqrt {b x^2+a}}dx}{c (b c-a d)}-\frac {d x \sqrt {a+b x^2}}{c \sqrt {c+d x^2} (b c-a d)}\)

\(\Big \downarrow \) 324

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

\(\Big \downarrow \) 320

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

\(\Big \downarrow \) 388

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

\(\Big \downarrow \) 313

\(\displaystyle \frac {b \left (\frac {c^{3/2} \sqrt {a+b x^2} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right ),1-\frac {b c}{a d}\right )}{a \sqrt {d} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+d \left (\frac {x \sqrt {a+b x^2}}{b \sqrt {c+d x^2}}-\frac {\sqrt {c} \sqrt {a+b x^2} E\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{b \sqrt {d} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}\right )\right )}{c (b c-a d)}-\frac {d x \sqrt {a+b x^2}}{c \sqrt {c+d x^2} (b c-a d)}\)

Input:

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

Output:

-((d*x*Sqrt[a + b*x^2])/(c*(b*c - a*d)*Sqrt[c + d*x^2])) + (b*(d*((x*Sqrt[ 
a + b*x^2])/(b*Sqrt[c + d*x^2]) - (Sqrt[c]*Sqrt[a + b*x^2]*EllipticE[ArcTa 
n[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(b*Sqrt[d]*Sqrt[(c*(a + b*x^2))/ 
(a*(c + d*x^2))]*Sqrt[c + d*x^2])) + (c^(3/2)*Sqrt[a + b*x^2]*EllipticF[Ar 
cTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(a*Sqrt[d]*Sqrt[(c*(a + b*x^2 
))/(a*(c + d*x^2))]*Sqrt[c + d*x^2])))/(c*(b*c - a*d))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 313

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

rule 316

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) 
), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d))   Int[(a + b*x^2)^(p + 1)*(c + d*x 
^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x 
], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  ! 
( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, 
 p, q, x]
 

rule 320

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S 
imp[(Sqrt[a + b*x^2]/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*( 
c + d*x^2)))]))*EllipticF[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /; Fre 
eQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]
 

rule 324

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

rule 388

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] 
 :> Simp[x*(Sqrt[a + b*x^2]/(b*Sqrt[c + d*x^2])), x] - Simp[c/b   Int[Sqrt[ 
a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]
 

Maple [A] (verified)

Time = 6.05 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.74

method	result	size
default	\(\frac {\left (\sqrt {-\frac {b}{a}}\, b d \,x^{3}-\sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {x^{2} d +c}{c}}\, \operatorname {EllipticE}\left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right ) b c +\sqrt {-\frac {b}{a}}\, a d x \right ) \sqrt {x^{2} d +c}\, \sqrt {b \,x^{2}+a}}{c \sqrt {-\frac {b}{a}}\, \left (a d -b c \right ) \left (b d \,x^{4}+a d \,x^{2}+x^{2} b c +a c \right )}\)	\(144\)
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \left (\frac {\left (b d \,x^{2}+a d \right ) x}{c \left (a d -b c \right ) \sqrt {\left (x^{2}+\frac {c}{d}\right ) \left (b d \,x^{2}+a d \right )}}+\frac {\left (\frac {1}{c}-\frac {a d}{\left (a d -b c \right ) c}\right ) \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \operatorname {EllipticF}\left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )}{\sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+x^{2} b c +a c}}+\frac {b \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \left (\operatorname {EllipticF}\left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )-\operatorname {EllipticE}\left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )\right )}{\left (a d -b c \right ) \sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+x^{2} b c +a c}}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {x^{2} d +c}}\)	\(321\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.82 \[ \int \frac {1}{\sqrt {a+b x^2} \left (c+d x^2\right )^{3/2}} \, dx=-\frac {\sqrt {b x^{2} + a} \sqrt {d x^{2} + c} a d x - {\left (b d x^{2} + b c\right )} \sqrt {a c} \sqrt {-\frac {b}{a}} E(\arcsin \left (x \sqrt {-\frac {b}{a}}\right )\,|\,\frac {a d}{b c}) + {\left ({\left (a + b\right )} d x^{2} + {\left (a + b\right )} c\right )} \sqrt {a c} \sqrt {-\frac {b}{a}} F(\arcsin \left (x \sqrt {-\frac {b}{a}}\right )\,|\,\frac {a d}{b c})}{a b c^{3} - a^{2} c^{2} d + {\left (a b c^{2} d - a^{2} c d^{2}\right )} x^{2}} \] Input:

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

Output:

-(sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*a*d*x - (b*d*x^2 + b*c)*sqrt(a*c)*sqrt(- 
b/a)*elliptic_e(arcsin(x*sqrt(-b/a)), a*d/(b*c)) + ((a + b)*d*x^2 + (a + b 
)*c)*sqrt(a*c)*sqrt(-b/a)*elliptic_f(arcsin(x*sqrt(-b/a)), a*d/(b*c)))/(a* 
b*c^3 - a^2*c^2*d + (a*b*c^2*d - a^2*c*d^2)*x^2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\sqrt {a+b x^2} \left (c+d x^2\right )^{3/2}} \, dx=\int \frac {\sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}}{b \,d^{2} x^{6}+a \,d^{2} x^{4}+2 b c d \,x^{4}+2 a c d \,x^{2}+b \,c^{2} x^{2}+a \,c^{2}}d x \] Input:

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

Output:

int((sqrt(c + d*x**2)*sqrt(a + b*x**2))/(a*c**2 + 2*a*c*d*x**2 + a*d**2*x* 
*4 + b*c**2*x**2 + 2*b*c*d*x**4 + b*d**2*x**6),x)