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

Optimal. Leaf size=242 \[ -\frac{2 b \sqrt{c} \sqrt{d} \sqrt{a+b x^2} \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right ),1-\frac{b c}{a d}\right )}{a \sqrt{c+d x^2} (b c-a d)^2 \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac{b x}{a \sqrt{a+b x^2} \sqrt{c+d x^2} (b c-a d)}+\frac{\sqrt{d} \sqrt{a+b x^2} (a d+b c) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{a \sqrt{c} \sqrt{c+d x^2} (b c-a d)^2 \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}} \]

[Out]

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

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Rubi [A]  time = 0.11983, antiderivative size = 242, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {414, 525, 418, 411} \[ \frac{b x}{a \sqrt{a+b x^2} \sqrt{c+d x^2} (b c-a d)}-\frac{2 b \sqrt{c} \sqrt{d} \sqrt{a+b x^2} F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{a \sqrt{c+d x^2} (b c-a d)^2 \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac{\sqrt{d} \sqrt{a+b x^2} (a d+b c) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{a \sqrt{c} \sqrt{c+d x^2} (b c-a d)^2 \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 414

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]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 525

Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)^(3/2)), x_Symbol] :> Dist[(b*e - a*
f)/(b*c - a*d), Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[Sqrt[a + b
*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[b/a] && PosQ[d/c]

Rule 418

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

Rule 411

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

Rubi steps

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

Mathematica [C]  time = 0.673031, size = 224, normalized size = 0.93 \[ \frac{\sqrt{\frac{b}{a}} \left (i b c \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} (a d-b c) \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{b}{a}}\right ),\frac{a d}{b c}\right )+x \sqrt{\frac{b}{a}} \left (a^2 d^2+a b d^2 x^2+b^2 c \left (c+d x^2\right )\right )+i b c \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} (a d+b c) E\left (i \sinh ^{-1}\left (\sqrt{\frac{b}{a}} x\right )|\frac{a d}{b c}\right )\right )}{b c \sqrt{a+b x^2} \sqrt{c+d x^2} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.024, size = 354, normalized size = 1.5 \begin{align*}{\frac{1}{ac \left ( ad-bc \right ) ^{2} \left ( bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac \right ) } \left ( \sqrt{-{\frac{b}{a}}}{x}^{3}ab{d}^{2}+\sqrt{-{\frac{b}{a}}}{x}^{3}{b}^{2}cd-\sqrt{{\frac{b{x}^{2}+a}{a}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticF} \left ( x\sqrt{-{\frac{b}{a}}},\sqrt{{\frac{ad}{bc}}} \right ) abcd+\sqrt{{\frac{b{x}^{2}+a}{a}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticF} \left ( x\sqrt{-{\frac{b}{a}}},\sqrt{{\frac{ad}{bc}}} \right ){b}^{2}{c}^{2}-\sqrt{{\frac{b{x}^{2}+a}{a}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticE} \left ( x\sqrt{-{\frac{b}{a}}},\sqrt{{\frac{ad}{bc}}} \right ) abcd-\sqrt{{\frac{b{x}^{2}+a}{a}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticE} \left ( x\sqrt{-{\frac{b}{a}}},\sqrt{{\frac{ad}{bc}}} \right ){b}^{2}{c}^{2}+x{a}^{2}{d}^{2}\sqrt{-{\frac{b}{a}}}+x{b}^{2}{c}^{2}\sqrt{-{\frac{b}{a}}} \right ) \sqrt{d{x}^{2}+c}\sqrt{b{x}^{2}+a}{\frac{1}{\sqrt{-{\frac{b}{a}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}^{\frac{3}{2}}{\left (d x^{2} + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b x^{2} + a} \sqrt{d x^{2} + c}}{b^{2} d^{2} x^{8} + 2 \,{\left (b^{2} c d + a b d^{2}\right )} x^{6} +{\left (b^{2} c^{2} + 4 \, a b c d + a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} + 2 \,{\left (a b c^{2} + a^{2} c d\right )} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*x^2 + a)*sqrt(d*x^2 + c)/(b^2*d^2*x^8 + 2*(b^2*c*d + a*b*d^2)*x^6 + (b^2*c^2 + 4*a*b*c*d + a^2
*d^2)*x^4 + a^2*c^2 + 2*(a*b*c^2 + a^2*c*d)*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}^{\frac{3}{2}}{\left (d x^{2} + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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