3.4.11 \(\int \frac {1}{x^3 (\frac {e (a+b x^2)}{c+d x^2})^{3/2}} \, dx\) [311]

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

[Out]

3/2*(-a*d+b*c)*arctanh(c^(1/2)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/a^(1/2)/e^(1/2))*c^(1/2)/a^(5/2)/e^(3/2)-3/2*(-a*
d+b*c)/a^2/e/(e*(b*x^2+a)/(d*x^2+c))^(1/2)+1/2*(-a*d+b*c)/a/(a*e-c*e*(b*x^2+a)/(d*x^2+c))/(e*(b*x^2+a)/(d*x^2+
c))^(1/2)

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Rubi [A]
time = 0.14, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1981, 1980, 296, 331, 214} \begin {gather*} \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 )} \end {gather*}

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 214

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

Rule 296

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] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

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 1980

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)))/((c_) + (d_.)*(x_)))^(p_), x_Symbol] :> With[{q = Denominator[p]}
, Dist[q*e*(b*c - a*d), Subst[Int[x^(q*(p + 1) - 1)*(((-a)*e + c*x^q)^m/(b*e - d*x^q)^(m + 2)), x], x, (e*((a
+ b*x)/(c + d*x)))^(1/q)], x]] /; FreeQ[{a, b, c, d, e, m}, x] && FractionQ[p] && IntegerQ[m]

Rule 1981

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

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) \text {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)) \text {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)) \text {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*}

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Mathematica [A]
time = 2.12, size = 148, normalized size = 0.87 \begin {gather*} \frac {-\sqrt {a} \sqrt {c+d x^2} \left (3 b c x^2+a \left (c-2 d x^2\right )\right )+3 \sqrt {c} (b c-a d) x^2 \sqrt {a+b x^2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{5/2} e x^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(640\) vs. \(2(146)=292\).
time = 0.12, size = 641, normalized size = 3.77

method	result	size
risch	\(-\frac {c \left (b \,x^{2}+a \right )}{2 a^{2} x^{2} e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}}+\frac {\left (-\frac {3 c \ln \left (\frac {2 a c e +\left (a d e +b c e \right ) x^{2}+2 \sqrt {a c e}\, \sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}}{x^{2}}\right ) d}{4 a \sqrt {a c e}}+\frac {3 c^{2} \ln \left (\frac {2 a c e +\left (a d e +b c e \right ) x^{2}+2 \sqrt {a c e}\, \sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}}{x^{2}}\right ) b}{4 a^{2} \sqrt {a c e}}+\frac {x^{2} d^{3}}{\left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}-\frac {2 x^{2} b c \,d^{2}}{a \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}+\frac {x^{2} b^{2} c^{2} d}{a^{2} \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}+\frac {c \,d^{2}}{\left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}-\frac {2 b \,c^{2} d}{a \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}+\frac {b^{2} c^{3}}{a^{2} \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}\right ) \sqrt {\left (d \,x^{2}+c \right ) e \left (b \,x^{2}+a \right )}}{e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right )}\)	\(526\)
default	\(-\frac {\left (-2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b^{2} d \,x^{6}+3 \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}+2 a c}{x^{2}}\right ) a^{2} b c d \,x^{4}-3 \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}+2 a c}{x^{2}}\right ) a \,b^{2} c^{2} x^{4}-4 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a b d \,x^{4}-2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b^{2} c \,x^{4}+3 \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}+2 a c}{x^{2}}\right ) a^{3} c d \,x^{2}-3 \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}+2 a c}{x^{2}}\right ) a^{2} b \,c^{2} x^{2}+2 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\, b \,x^{2}-2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a^{2} d \,x^{2}-2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a b c \,x^{2}-4 \sqrt {a c}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, a^{2} d \,x^{2}+4 \sqrt {a c}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, a b c \,x^{2}+2 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\, a \right ) \left (b \,x^{2}+a \right )}{4 \sqrt {a c}\, x^{2} a^{3} \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \left (d \,x^{2}+c \right ) \left (\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}\right )^{\frac {3}{2}}}\)	\(641\)

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,method=_RETURNVERBOSE)

[Out]

-1/4*(-2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*b^2*d*x^6+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)*a^2*b*c*d*x^4-3*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)*a*b^2*c^2*x^4-4*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*a*b*d*x^4-2*(b*d*
x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*b^2*c*x^4+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)*a^3*c*d*x^2-3*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)*a^2*b*c^2*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*(a*c)^(1/2)*b*x^2-2*(b*d*x^4+a*d*x^2+b*c*x^2+
a*c)^(1/2)*(a*c)^(1/2)*a^2*d*x^2-2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*a*b*c*x^2-4*(a*c)^(1/2)*((d
*x^2+c)*(b*x^2+a))^(1/2)*a^2*d*x^2+4*(a*c)^(1/2)*((d*x^2+c)*(b*x^2+a))^(1/2)*a*b*c*x^2+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)/(a*c)^(1/2)/x^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 [A]
time = 0.50, size = 178, normalized size = 1.05 \begin {gather*} -\frac {1}{4} \, {\left (\frac {3 \, {\left (b c - a d\right )} c \log \left (\frac {c \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} - \sqrt {a c}}{c \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} + \sqrt {a c}}\right )}{\sqrt {a c} a^{2}} - \frac {2 \, {\left (2 \, a b c - 2 \, a^{2} d - \frac {3 \, {\left (b c^{2} - a c d\right )} {\left (b x^{2} + a\right )}}{d x^{2} + c}\right )}}{a^{2} c \left (\frac {b x^{2} + a}{d x^{2} + c}\right )^{\frac {3}{2}} - a^{3} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}}\right )} e^{\left (-\frac {3}{2}\right )} \end {gather*}

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]

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

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Fricas [A]
time = 1.03, size = 433, normalized size = 2.55 \begin {gather*} \left [-\frac {{\left (3 \, {\left ({\left (b^{2} c - a b d\right )} x^{4} + {\left (a b c - a^{2} d\right )} x^{2}\right )} \sqrt {\frac {c}{a}} \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 x^{2} + a}{d x^{2} + c}} \sqrt {\frac {c}{a}}}{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 x^{2} + a}{d x^{2} + c}}\right )} e^{\left (-\frac {3}{2}\right )}}{8 \, {\left (a^{2} b x^{4} + a^{3} x^{2}\right )}}, -\frac {{\left (3 \, {\left ({\left (b^{2} c - a b d\right )} x^{4} + {\left (a b c - a^{2} d\right )} x^{2}\right )} \sqrt {-\frac {c}{a}} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} \sqrt {-\frac {c}{a}}}{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 x^{2} + a}{d x^{2} + c}}\right )} e^{\left (-\frac {3}{2}\right )}}{4 \, {\left (a^{2} b x^{4} + a^{3} x^{2}\right )}}\right ] \end {gather*}

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)*x^4 + (a*b*c - a^2*d)*x^2)*sqrt(c/a)*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*x^2 + a)/(d*x^2 + c))*sqrt(c/a))/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*
x^2 + a)/(d*x^2 + c)))*e^(-3/2)/(a^2*b*x^4 + a^3*x^2), -1/4*(3*((b^2*c - a*b*d)*x^4 + (a*b*c - a^2*d)*x^2)*sqr
t(-c/a)*arctan(1/2*((b*c + a*d)*x^2 + 2*a*c)*sqrt((b*x^2 + a)/(d*x^2 + c))*sqrt(-c/a)/(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*x^2 + a)/(d*x^2 + c)))*e^(-3/2)/(a^2*b*x^4 + a^
3*x^2)]

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

[Out]

Timed out

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^3\,{\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/(x^3*((e*(a + b*x^2))/(c + d*x^2))^(3/2)),x)

[Out]

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

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