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

Optimal. Leaf size=273 \[ -\frac {d x \sqrt {a+b x^2}}{a (b c-a d) \sqrt {c+d x^2}}+\frac {b x \sqrt {c+d x^2}}{a (b c-a d) \sqrt {a+b x^2}}+\frac {\sqrt {c} \sqrt {d} \sqrt {a+b x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{a (b c-a d) \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}-\frac {\sqrt {c} \sqrt {d} \sqrt {a+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{a (b c-a d) \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}} \]

[Out]

-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)*EllipticF(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)

________________________________________________________________________________________

Rubi [A]
time = 0.10, antiderivative size = 273, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {425, 21, 433, 429, 506, 422} \begin {gather*} -\frac {\sqrt {c} \sqrt {d} \sqrt {a+b x^2} F\left (\text {ArcTan}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{a \sqrt {c+d x^2} (b c-a d) \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {\sqrt {c} \sqrt {d} \sqrt {a+b x^2} E\left (\text {ArcTan}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{a \sqrt {c+d x^2} (b c-a d) \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {b x \sqrt {c+d x^2}}{a \sqrt {a+b x^2} (b c-a d)}-\frac {d x \sqrt {a+b x^2}}{a \sqrt {c+d x^2} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 422

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]/(c*Rt[d/c, 2]*Sq
rt[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 425

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

Rule 429

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

Rule 433

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

Rule 506

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] - Dist[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]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}} \, dx &=\frac {b x \sqrt {c+d x^2}}{a (b c-a d) \sqrt {a+b x^2}}-\frac {\int \frac {a d+b d x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{a (b c-a d)}\\ &=\frac {b x \sqrt {c+d x^2}}{a (b c-a d) \sqrt {a+b x^2}}-\frac {d \int \frac {\sqrt {a+b x^2}}{\sqrt {c+d x^2}} \, dx}{a (b c-a d)}\\ &=\frac {b x \sqrt {c+d x^2}}{a (b c-a d) \sqrt {a+b x^2}}-\frac {d \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{b c-a d}-\frac {(b d) \int \frac {x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{a (b c-a d)}\\ &=-\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} F\left (\tan ^{-1}\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 {(c d) \int \frac {\sqrt {a+b x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{a (b c-a d)}\\ &=-\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 (\tan ^{-1}\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} F\left (\tan ^{-1}\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}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 2.07, size = 112, normalized size = 0.41 \begin {gather*} \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 (\sin ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.07, size = 248, normalized size = 0.91

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b x^{2}\right )^{\frac {3}{2}} \sqrt {c + d x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (b\,x^2+a\right )}^{3/2}\,\sqrt {d\,x^2+c}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________