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

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

[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)

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Rubi [A]  time = 0.14, 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 = {414, 21, 422, 418, 492, 411} \[ \frac {b x \sqrt {c+d x^2}}{a \sqrt {a+b x^2} (b c-a d)}-\frac {d x \sqrt {a+b x^2}}{a \sqrt {c+d x^2} (b c-a d)}-\frac {\sqrt {c} \sqrt {d} \sqrt {a+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{a \sqrt {c+d x^2} (b c-a d) \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {\sqrt {c} \sqrt {d} \sqrt {a+b x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{a \sqrt {c+d x^2} (b c-a d) \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^2)^(3/2)*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 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]

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 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 422

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 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[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*}

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Mathematica [A]  time = 0.25, size = 112, normalized size = 0.41 \[ \frac {\frac {a d \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} E\left (\sin ^{-1}\left (\sqrt {-\frac {d}{c}} x\right )|\frac {b c}{a d}\right )}{\sqrt {-\frac {d}{c}}}-b x \left (c+d x^2\right )}{a \sqrt {a+b x^2} \sqrt {c+d x^2} (a d-b c)} \]

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])

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fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{b^{2} d x^{6} + {\left (b^{2} c + 2 \, a b d\right )} x^{4} + a^{2} c + {\left (2 \, a b c + a^{2} d\right )} x^{2}}, x\right ) \]

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]

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

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

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)

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maple [A]  time = 0.04, size = 248, normalized size = 0.91 \[ \frac {\left (-\sqrt {-\frac {b}{a}}\, b d \,x^{3}+\sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, a d \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-\sqrt {-\frac {b}{a}}\, b c x +\sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, b c \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-\sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, b c \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )\right ) \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}}{\sqrt {-\frac {b}{a}}\, \left (a d -b c \right ) \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right ) a} \]

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)

[Out]

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

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

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)

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

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)

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

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)

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