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

Optimal. Leaf size=356 \[ \frac {x (a c+2 b) \left (a c+a d x^2+b\right )}{a^2 (a c+b) \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}-\frac {\sqrt {c} (a c+2 b) \left (a c+a d x^2+b\right ) E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{a^2 \sqrt {d} (a c+b) \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}} \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}}+\frac {c^{3/2} \left (a c+a d x^2+b\right ) F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{a \sqrt {d} (a c+b) \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}} \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}}-\frac {b x}{a (a c+b) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.31, antiderivative size = 411, normalized size of antiderivative = 1.15, number of steps used = 7, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {6722, 1974, 413, 531, 418, 492, 411} \[ \frac {x (a c+2 b) \sqrt {a c+a d x^2+b} \sqrt {a \left (c+d x^2\right )+b}}{a^2 (a c+b) \left (c+d x^2\right ) \sqrt {a+\frac {b}{c+d x^2}}}-\frac {\sqrt {c} (a c+2 b) \sqrt {a c+a d x^2+b} \sqrt {a \left (c+d x^2\right )+b} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{a^2 \sqrt {d} (a c+b) \left (c+d x^2\right ) \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}} \sqrt {a+\frac {b}{c+d x^2}}}+\frac {c^{3/2} \sqrt {a c+a d x^2+b} \sqrt {a \left (c+d x^2\right )+b} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{a \sqrt {d} (a c+b) \left (c+d x^2\right ) \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}} \sqrt {a+\frac {b}{c+d x^2}}}-\frac {b x \sqrt {a \left (c+d x^2\right )+b}}{a (a c+b) \sqrt {a c+a d x^2+b} \sqrt {a+\frac {b}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/(c + d*x^2))^(-3/2),x]

[Out]

-((b*x*Sqrt[b + a*(c + d*x^2)])/(a*(b + a*c)*Sqrt[b + a*c + a*d*x^2]*Sqrt[a + b/(c + d*x^2)])) + ((2*b + a*c)*
x*Sqrt[b + a*c + a*d*x^2]*Sqrt[b + a*(c + d*x^2)])/(a^2*(b + a*c)*(c + d*x^2)*Sqrt[a + b/(c + d*x^2)]) - (Sqrt
[c]*(2*b + a*c)*Sqrt[b + a*c + a*d*x^2]*Sqrt[b + a*(c + d*x^2)]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], b/(b +
a*c)])/(a^2*(b + a*c)*Sqrt[d]*(c + d*x^2)*Sqrt[(c*(b + a*c + a*d*x^2))/((b + a*c)*(c + d*x^2))]*Sqrt[a + b/(c
+ d*x^2)]) + (c^(3/2)*Sqrt[b + a*c + a*d*x^2]*Sqrt[b + a*(c + d*x^2)]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], b
/(b + a*c)])/(a*(b + a*c)*Sqrt[d]*(c + d*x^2)*Sqrt[(c*(b + a*c + a*d*x^2))/((b + a*c)*(c + d*x^2))]*Sqrt[a + b
/(c + d*x^2)])

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 413

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

Rule 531

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Dist[
e, Int[(a + b*x^n)^p*(c + d*x^n)^q, x], x] + Dist[f, Int[x^n*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, n, p, q}, x]

Rule 1974

Int[(u_)^(p_.)*(v_)^(q_.), x_Symbol] :> Int[ExpandToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{p, q}, x] &&
BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] &&  !BinomialMatchQ[{u, v}, x]

Rule 6722

Int[(u_.)*((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[(a + b*v^n)^FracPart[p]/(v^(n*FracPart[p])*(b + a/
v^n)^FracPart[p]), Int[u*v^(n*p)*(b + a/v^n)^p, x], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[p] && ILtQ[n, 0] &
& BinomialQ[v, x] &&  !LinearQ[v, x]

Rubi steps

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

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Mathematica [C]  time = 0.52, size = 241, normalized size = 0.68 \[ -\frac {\sqrt {\frac {a d}{a c+b}} \sqrt {\frac {a c+a d x^2+b}{c+d x^2}} \left (b x \left (c+d x^2\right ) \sqrt {\frac {a d}{a c+b}}-i b c \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {a c+a d x^2+b}{a c+b}} F\left (i \sinh ^{-1}\left (\sqrt {\frac {a d}{b+a c}} x\right )|\frac {b}{a c}+1\right )+i c (a c+2 b) \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {a c+a d x^2+b}{a c+b}} E\left (i \sinh ^{-1}\left (\sqrt {\frac {a d}{b+a c}} x\right )|\frac {b}{a c}+1\right )\right )}{a^2 d \left (a \left (c+d x^2\right )+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (d^{2} x^{4} + 2 \, c d x^{2} + c^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{a^{2} d^{2} x^{4} + a^{2} c^{2} + 2 \, {\left (a^{2} c + a b\right )} d x^{2} + 2 \, a b c + b^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 466, normalized size = 1.31 \[ -\frac {\left (\sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {-\frac {a d}{a c +b}}\, b d \,x^{3}-\sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a \,c^{2} \EllipticE \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )+\sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {-\frac {a d}{a c +b}}\, b c x -2 \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b c \EllipticE \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )+\sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b c \EllipticF \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )\right ) \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}{\sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {-\frac {a d}{a c +b}}\, \left (a c +b \right ) \left (a d \,x^{2}+a c +b \right ) a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b/(c + d*x^2))^(3/2),x)

[Out]

int(1/(a + b/(c + d*x^2))^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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