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

Optimal. Leaf size=170 \[ \frac{3 \sqrt{c} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{a} \sqrt{e}}\right )}{2 a^{5/2} e^{3/2}}-\frac{3 (b c-a d)}{2 a^2 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{b c-a d}{2 a \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (a e-\frac{c e \left (a+b x^2\right )}{c+d x^2}\right )} \]

[Out]

(-3*(b*c - a*d))/(2*a^2*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]) + (b*c - a*d)/(2*a*Sqrt[(e*(a + b*x^2))/(c + d*x^
2)]*(a*e - (c*e*(a + b*x^2))/(c + d*x^2))) + (3*Sqrt[c]*(b*c - a*d)*ArcTanh[(Sqrt[c]*Sqrt[(e*(a + b*x^2))/(c +
d*x^2)])/(Sqrt[a]*Sqrt[e])])/(2*a^(5/2)*e^(3/2))

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Rubi [A]  time = 0.111482, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {1960, 290, 325, 208} \[ \frac{3 \sqrt{c} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{a} \sqrt{e}}\right )}{2 a^{5/2} e^{3/2}}-\frac{3 (b c-a d)}{2 a^2 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{b c-a d}{2 a \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (a e-\frac{c e \left (a+b x^2\right )}{c+d x^2}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(b*c - a*d))/(2*a^2*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]) + (b*c - a*d)/(2*a*Sqrt[(e*(a + b*x^2))/(c + d*x^
2)]*(a*e - (c*e*(a + b*x^2))/(c + d*x^2))) + (3*Sqrt[c]*(b*c - a*d)*ArcTanh[(Sqrt[c]*Sqrt[(e*(a + b*x^2))/(c +
d*x^2)])/(Sqrt[a]*Sqrt[e])])/(2*a^(5/2)*e^(3/2))

Rule 1960

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> With[{q = Den
ominator[p]}, Dist[(q*e*(b*c - a*d))/n, Subst[Int[(x^(q*(p + 1) - 1)*(-(a*e) + c*x^q)^(Simplify[(m + 1)/n] - 1
))/(b*e - d*x^q)^(Simplify[(m + 1)/n] + 1), x], x, ((e*(a + b*x^n))/(c + d*x^n))^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, m, n}, x] && FractionQ[p] && IntegerQ[Simplify[(m + 1)/n]]

Rule 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{1}{x^3 \left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx &=((b c-a d) e) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (-a e+c x^2\right )^2} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )\\ &=\frac{b c-a d}{2 a \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (a e-\frac{c e \left (a+b x^2\right )}{c+d x^2}\right )}-\frac{(3 (b c-a d)) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (-a e+c x^2\right )} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )}{2 a}\\ &=-\frac{3 (b c-a d)}{2 a^2 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{b c-a d}{2 a \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (a e-\frac{c e \left (a+b x^2\right )}{c+d x^2}\right )}-\frac{(3 c (b c-a d)) \operatorname{Subst}\left (\int \frac{1}{-a e+c x^2} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )}{2 a^2 e}\\ &=-\frac{3 (b c-a d)}{2 a^2 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{b c-a d}{2 a \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (a e-\frac{c e \left (a+b x^2\right )}{c+d x^2}\right )}+\frac{3 \sqrt{c} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{a} \sqrt{e}}\right )}{2 a^{5/2} e^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0848137, size = 148, normalized size = 0.87 \[ \frac{3 \sqrt{c} x^2 \sqrt{a+b x^2} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )-\sqrt{a} \sqrt{c+d x^2} \left (a \left (c-2 d x^2\right )+3 b c x^2\right )}{2 a^{5/2} e x^2 \sqrt{c+d x^2} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[a]*Sqrt[c + d*x^2]*(3*b*c*x^2 + a*(c - 2*d*x^2))) + 3*Sqrt[c]*(b*c - a*d)*x^2*Sqrt[a + b*x^2]*ArcTanh[
(Sqrt[c]*Sqrt[a + b*x^2])/(Sqrt[a]*Sqrt[c + d*x^2])])/(2*a^(5/2)*e*x^2*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*Sqrt[
c + d*x^2])

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Maple [B]  time = 0.016, size = 641, normalized size = 3.8 \begin{align*} -{\frac{b{x}^{2}+a}{4\,{a}^{3}{x}^{2} \left ( d{x}^{2}+c \right ) } \left ( -2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{ac}{x}^{6}{b}^{2}d+3\,\ln \left ({\frac{ad{x}^{2}+bc{x}^{2}+2\,\sqrt{ac}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}+2\,ac}{{x}^{2}}} \right ){x}^{4}{a}^{2}bcd-3\,\ln \left ({\frac{ad{x}^{2}+bc{x}^{2}+2\,\sqrt{ac}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}+2\,ac}{{x}^{2}}} \right ){x}^{4}a{b}^{2}{c}^{2}-4\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{ac}{x}^{4}abd-2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{ac}{x}^{4}{b}^{2}c+3\,\ln \left ({\frac{ad{x}^{2}+bc{x}^{2}+2\,\sqrt{ac}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}+2\,ac}{{x}^{2}}} \right ){x}^{2}{a}^{3}cd-3\,\ln \left ({\frac{ad{x}^{2}+bc{x}^{2}+2\,\sqrt{ac}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}+2\,ac}{{x}^{2}}} \right ){x}^{2}{a}^{2}b{c}^{2}+2\, \left ( bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac \right ) ^{3/2}\sqrt{ac}{x}^{2}b-2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{ac}{x}^{2}{a}^{2}d-2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{ac}{x}^{2}abc-4\,\sqrt{ac}\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }{x}^{2}{a}^{2}d+4\,\sqrt{ac}\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }{x}^{2}abc+2\, \left ( bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac \right ) ^{3/2}\sqrt{ac}a \right ){\frac{1}{\sqrt{ac}}}{\frac{1}{\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }}} \left ({\frac{e \left ( b{x}^{2}+a \right ) }{d{x}^{2}+c}} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(-2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*x^6*b^2*d+3*ln((a*d*x^2+b*c*x^2+2*(a*c)^(1/2)*(b*d*x^
4+a*d*x^2+b*c*x^2+a*c)^(1/2)+2*a*c)/x^2)*x^4*a^2*b*c*d-3*ln((a*d*x^2+b*c*x^2+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*
c*x^2+a*c)^(1/2)+2*a*c)/x^2)*x^4*a*b^2*c^2-4*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*x^4*a*b*d-2*(b*d*
x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*x^4*b^2*c+3*ln((a*d*x^2+b*c*x^2+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*
x^2+a*c)^(1/2)+2*a*c)/x^2)*x^2*a^3*c*d-3*ln((a*d*x^2+b*c*x^2+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)
+2*a*c)/x^2)*x^2*a^2*b*c^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*(a*c)^(1/2)*x^2*b-2*(b*d*x^4+a*d*x^2+b*c*x^2+
a*c)^(1/2)*(a*c)^(1/2)*x^2*a^2*d-2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*x^2*a*b*c-4*(a*c)^(1/2)*((d
*x^2+c)*(b*x^2+a))^(1/2)*x^2*a^2*d+4*(a*c)^(1/2)*((d*x^2+c)*(b*x^2+a))^(1/2)*x^2*a*b*c+2*(b*d*x^4+a*d*x^2+b*c*
x^2+a*c)^(3/2)*(a*c)^(1/2)*a)*(b*x^2+a)/x^2/(a*c)^(1/2)/a^3/((d*x^2+c)*(b*x^2+a))^(1/2)/(d*x^2+c)/(e*(b*x^2+a)
/(d*x^2+c))^(3/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 12.3692, size = 957, normalized size = 5.63 \begin{align*} \left [-\frac{3 \,{\left ({\left (b^{2} c - a b d\right )} e x^{4} +{\left (a b c - a^{2} d\right )} e x^{2}\right )} \sqrt{\frac{c}{a e}} \log \left (\frac{{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{4} + 8 \, a^{2} c^{2} + 8 \,{\left (a b c^{2} + a^{2} c d\right )} x^{2} - 4 \,{\left ({\left (a b c d + a^{2} d^{2}\right )} x^{4} + 2 \, a^{2} c^{2} +{\left (a b c^{2} + 3 \, a^{2} c d\right )} x^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}} \sqrt{\frac{c}{a e}}}{x^{4}}\right ) + 4 \,{\left ({\left (3 \, b c d - 2 \, a d^{2}\right )} x^{4} + a c^{2} +{\left (3 \, b c^{2} - a c d\right )} x^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{8 \,{\left (a^{2} b e^{2} x^{4} + a^{3} e^{2} x^{2}\right )}}, -\frac{3 \,{\left ({\left (b^{2} c - a b d\right )} e x^{4} +{\left (a b c - a^{2} d\right )} e x^{2}\right )} \sqrt{-\frac{c}{a e}} \arctan \left (\frac{{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}} \sqrt{-\frac{c}{a e}}}{2 \,{\left (b c x^{2} + a c\right )}}\right ) + 2 \,{\left ({\left (3 \, b c d - 2 \, a d^{2}\right )} x^{4} + a c^{2} +{\left (3 \, b c^{2} - a c d\right )} x^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{4 \,{\left (a^{2} b e^{2} x^{4} + a^{3} e^{2} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(3*((b^2*c - a*b*d)*e*x^4 + (a*b*c - a^2*d)*e*x^2)*sqrt(c/(a*e))*log(((b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^
4 + 8*a^2*c^2 + 8*(a*b*c^2 + a^2*c*d)*x^2 - 4*((a*b*c*d + a^2*d^2)*x^4 + 2*a^2*c^2 + (a*b*c^2 + 3*a^2*c*d)*x^2
)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))*sqrt(c/(a*e)))/x^4) + 4*((3*b*c*d - 2*a*d^2)*x^4 + a*c^2 + (3*b*c^2 - a*c*
d)*x^2)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(a^2*b*e^2*x^4 + a^3*e^2*x^2), -1/4*(3*((b^2*c - a*b*d)*e*x^4 + (a*
b*c - a^2*d)*e*x^2)*sqrt(-c/(a*e))*arctan(1/2*((b*c + a*d)*x^2 + 2*a*c)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))*sqrt
(-c/(a*e))/(b*c*x^2 + a*c)) + 2*((3*b*c*d - 2*a*d^2)*x^4 + a*c^2 + (3*b*c^2 - a*c*d)*x^2)*sqrt((b*e*x^2 + a*e)
/(d*x^2 + c)))/(a^2*b*e^2*x^4 + a^3*e^2*x^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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