3.3.4 \(\int \frac {1}{(a+b x^2)^{3/2} \sqrt {c+d x^2}} \, dx\) [204]

3.3.4.1 Optimal result

Integrand size = 23, antiderivative size = 273 \[ \int \frac {1}{\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}} \, dx=-\frac {d x \sqrt {a+b x^2}}{a (b c-a d) \sqrt {c+d x^2}}+\frac {b x \sqrt {c+d x^2}}{a (b c-a d) \sqrt {a+b x^2}}+\frac {\sqrt {c} \sqrt {d} \sqrt {a+b x^2} E\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{a (b c-a d) \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}-\frac {\sqrt {c} \sqrt {d} \sqrt {a+b x^2} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right ),1-\frac {b c}{a d}\right )}{a (b c-a d) \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}} \]

output

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

3.3.4.2 Mathematica [A] (verified)

Time = 1.82 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.41 \[ \int \frac {1}{\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}} \, dx=\frac {-b x \left (c+d x^2\right )+\frac {a d \sqrt {1+\frac {b x^2}{a}} \sqrt {1+\frac {d x^2}{c}} E\left (\arcsin \left (\sqrt {-\frac {d}{c}} x\right )|\frac {b c}{a d}\right )}{\sqrt {-\frac {d}{c}}}}{a (-b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}} \]

input

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

output

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

3.3.4.3 Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 258, normalized size of antiderivative = 0.95, 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.

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 324

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

\(\Big \downarrow \) 320

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

\(\Big \downarrow \) 388

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

\(\Big \downarrow \) 313

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

input

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

output

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

3.3.4.3.1 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]
 

3.3.4.4 Maple [A] (verified)

Time = 4.39 (sec) , antiderivative size = 248, normalized size of antiderivative = 0.91

method	result	size
default	\(\frac {\left (-\sqrt {-\frac {b}{a}}\, b d \,x^{3}+a \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, F\left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right ) d -\sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, F\left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right ) b c +\sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, E\left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right ) b c -\sqrt {-\frac {b}{a}}\, b c x \right ) \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}}{a \sqrt {-\frac {b}{a}}\, \left (a d -b c \right ) \left (b d \,x^{4}+a d \,x^{2}+c b \,x^{2}+a c \right )}\)	\(248\)
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \left (-\frac {\left (b d \,x^{2}+b c \right ) x}{a \left (a d -b c \right ) \sqrt {\left (x^{2}+\frac {a}{b}\right ) \left (b d \,x^{2}+b c \right )}}+\frac {\left (\frac {1}{a}+\frac {b c}{a \left (a d -b c \right )}\right ) \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, F\left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )}{\sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+c b \,x^{2}+a c}}-\frac {b c \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \left (F\left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )-E\left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )\right )}{\left (a d -b c \right ) a \sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+c b \,x^{2}+a c}}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}\)	\(326\)

input

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

output

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

3.3.4.5 Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.68 \[ \int \frac {1}{\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}} \, dx=\frac {\sqrt {b x^{2} + a} \sqrt {d x^{2} + c} a b^{2} c x - {\left (b^{3} c x^{2} + a b^{2} c\right )} \sqrt {a c} \sqrt {-\frac {b}{a}} E(\arcsin \left (x \sqrt {-\frac {b}{a}}\right )\,|\,\frac {a d}{b c}) + {\left (a b^{2} c + a^{3} d + {\left (b^{3} c + a^{2} b d\right )} x^{2}\right )} \sqrt {a c} \sqrt {-\frac {b}{a}} F(\arcsin \left (x \sqrt {-\frac {b}{a}}\right )\,|\,\frac {a d}{b c})}{a^{3} b^{2} c^{2} - a^{4} b c d + {\left (a^{2} b^{3} c^{2} - a^{3} b^{2} c d\right )} x^{2}} \]

input

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

output

(sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*a*b^2*c*x - (b^3*c*x^2 + a*b^2*c)*sqrt(a* 
c)*sqrt(-b/a)*elliptic_e(arcsin(x*sqrt(-b/a)), a*d/(b*c)) + (a*b^2*c + a^3 
*d + (b^3*c + a^2*b*d)*x^2)*sqrt(a*c)*sqrt(-b/a)*elliptic_f(arcsin(x*sqrt( 
-b/a)), a*d/(b*c)))/(a^3*b^2*c^2 - a^4*b*c*d + (a^2*b^3*c^2 - a^3*b^2*c*d) 
*x^2)
 

3.3.4.6 Sympy [F]

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

input

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

output

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

3.3.4.7 Maxima [F]

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

input

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

output

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

3.3.4.8 Giac [F]

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

input

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

output

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

3.3.4.9 Mupad [F(-1)]

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

input

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

output

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