3.4.15 \(\int \frac {1}{(\frac {e (a+b x^2)}{c+d x^2})^{3/2}} \, dx\) [315]

Optimal. Leaf size=327 \[ \frac {(b c-a d) x}{a b e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}-\frac {d (b c-2 a d) x \left (a+b x^2\right )}{a b^2 e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}+\frac {\sqrt {c} \sqrt {d} (b c-2 a d) \left (a+b x^2\right ) E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{a b^2 e \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}+\frac {c^{3/2} \sqrt {d} \left (a+b x^2\right ) F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{a b e \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 327, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1986, 424, 545, 429, 506, 422} \begin {gather*} \frac {\sqrt {c} \sqrt {d} \left (a+b x^2\right ) (b c-2 a d) E\left (\text {ArcTan}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{a b^2 e \left (c+d x^2\right ) \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}+\frac {c^{3/2} \sqrt {d} \left (a+b x^2\right ) F\left (\text {ArcTan}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{a b e \left (c+d x^2\right ) \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}-\frac {d x \left (a+b x^2\right ) (b c-2 a d)}{a b^2 e \left (c+d x^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}+\frac {x (b c-a d)}{a b e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)*x)/(a*b*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]) - (d*(b*c - 2*a*d)*x*(a + b*x^2))/(a*b^2*e*Sqrt[(e*(
a + b*x^2))/(c + d*x^2)]*(c + d*x^2)) + (Sqrt[c]*Sqrt[d]*(b*c - 2*a*d)*(a + b*x^2)*EllipticE[ArcTan[(Sqrt[d]*x
)/Sqrt[c]], 1 - (b*c)/(a*d)])/(a*b^2*e*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]
*(c + d*x^2)) + (c^(3/2)*Sqrt[d]*(a + b*x^2)*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(a*b*e*S
qrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2))

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 424

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 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 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]

Rule 545

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 1986

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

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 4.00, size = 203, normalized size = 0.62 \begin {gather*} \frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (-i c (-b c+2 a d) \sqrt {1+\frac {b x^2}{a}} \sqrt {1+\frac {d x^2}{c}} E\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )+(b c-a d) \left (\sqrt {\frac {b}{a}} x \left (c+d x^2\right )-i c \sqrt {1+\frac {b x^2}{a}} \sqrt {1+\frac {d x^2}{c}} F\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )\right )\right )}{a^2 \left (\frac {b}{a}\right )^{3/2} e^2 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*((-I)*c*(-(b*c) + 2*a*d)*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticE[
I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)] + (b*c - a*d)*(Sqrt[b/a]*x*(c + d*x^2) - I*c*Sqrt[1 + (b*x^2)/a]*Sqrt[1 +
(d*x^2)/c]*EllipticF[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)])))/(a^2*(b/a)^(3/2)*e^2*(a + b*x^2))

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Maple [A]
time = 0.04, size = 514, normalized size = 1.57

method	result	size
default	\(-\frac {\left (b \,x^{2}+a \right ) \left (\sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {-\frac {b}{a}}\, a \,d^{2} x^{3}-\sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {-\frac {b}{a}}\, b c d \,x^{3}+\sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, \EllipticF \left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right ) \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, a c d -\sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, \EllipticF \left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right ) \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, b \,c^{2}-2 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, \EllipticE \left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right ) \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, a c d +\sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, \EllipticE \left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right ) \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, b \,c^{2}+\sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {-\frac {b}{a}}\, a c d x -\sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {-\frac {b}{a}}\, b \,c^{2} x \right )}{b \left (\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}\right )^{\frac {3}{2}} \left (d \,x^{2}+c \right )^{2} a \sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}}\)	\(514\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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