3.360 \(\int \frac {1}{x^5 (a+\frac {b}{c+d x^2})^{3/2}} \, dx\)

Optimal. Leaf size=212 \[ -\frac {a b d^2}{(a c+b)^3 \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}-\frac {3 b d^2 (b-4 a c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{\sqrt {a c+b}}\right )}{8 \sqrt {c} (a c+b)^{7/2}}-\frac {d (3 b-4 a c) \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{8 x^2 (a c+b)^3}-\frac {\left (c+d x^2\right )^2 \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 x^4 (a c+b)^2} \]

[Out]

-3/8*b*(-4*a*c+b)*d^2*arctanh(c^(1/2)*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/(a*c+b)^(1/2))/(a*c+b)^(7/2)/c^(1/2)-a
*b*d^2/(a*c+b)^3/((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)-1/8*(-4*a*c+3*b)*d*(d*x^2+c)*((a*d*x^2+a*c+b)/(d*x^2+c))^(1
/2)/(a*c+b)^3/x^2-1/4*(d*x^2+c)^2*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/(a*c+b)^2/x^4

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Rubi [A]  time = 0.58, antiderivative size = 246, normalized size of antiderivative = 1.16, number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6722, 1975, 446, 96, 94, 93, 208} \[ \frac {3 b d^2 (b-4 a c)}{8 c (a c+b)^3 \sqrt {a+\frac {b}{c+d x^2}}}-\frac {3 b d^2 (b-4 a c) \sqrt {a \left (c+d x^2\right )+b} \tanh ^{-1}\left (\frac {\sqrt {a c+b} \sqrt {c+d x^2}}{\sqrt {c} \sqrt {a \left (c+d x^2\right )+b}}\right )}{8 \sqrt {c} (a c+b)^{7/2} \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}-\frac {d (b-4 a c) \left (c+d x^2\right )}{8 c x^2 (a c+b)^2 \sqrt {a+\frac {b}{c+d x^2}}}-\frac {\left (c+d x^2\right )^2}{4 c x^4 (a c+b) \sqrt {a+\frac {b}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(a + b/(c + d*x^2))^(3/2)),x]

[Out]

(3*b*(b - 4*a*c)*d^2)/(8*c*(b + a*c)^3*Sqrt[a + b/(c + d*x^2)]) - ((b - 4*a*c)*d*(c + d*x^2))/(8*c*(b + a*c)^2
*x^2*Sqrt[a + b/(c + d*x^2)]) - (c + d*x^2)^2/(4*c*(b + a*c)*x^4*Sqrt[a + b/(c + d*x^2)]) - (3*b*(b - 4*a*c)*d
^2*Sqrt[b + a*(c + d*x^2)]*ArcTanh[(Sqrt[b + a*c]*Sqrt[c + d*x^2])/(Sqrt[c]*Sqrt[b + a*(c + d*x^2)])])/(8*Sqrt
[c]*(b + a*c)^(7/2)*Sqrt[c + d*x^2]*Sqrt[a + b/(c + d*x^2)])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 94

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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]

Rule 446

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

Rule 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, 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}{x^5 \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}}{x^5 \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}}{x^5 \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 {\sqrt {b+a \left (c+d x^2\right )} \operatorname {Subst}\left (\int \frac {(c+d x)^{3/2}}{x^3 (b+a c+a d x)^{3/2}} \, dx,x,x^2\right )}{2 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=-\frac {\left (c+d x^2\right )^2}{4 c (b+a c) x^4 \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\left ((b-4 a c) d \sqrt {b+a \left (c+d x^2\right )}\right ) \operatorname {Subst}\left (\int \frac {(c+d x)^{3/2}}{x^2 (b+a c+a d x)^{3/2}} \, dx,x,x^2\right )}{8 c (b+a c) \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=-\frac {(b-4 a c) d \left (c+d x^2\right )}{8 c (b+a c)^2 x^2 \sqrt {a+\frac {b}{c+d x^2}}}-\frac {\left (c+d x^2\right )^2}{4 c (b+a c) x^4 \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\left (3 b (b-4 a c) d^2 \sqrt {b+a \left (c+d x^2\right )}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {c+d x}}{x (b+a c+a d x)^{3/2}} \, dx,x,x^2\right )}{16 c (b+a c)^2 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=\frac {3 b (b-4 a c) d^2}{8 c (b+a c)^3 \sqrt {a+\frac {b}{c+d x^2}}}-\frac {(b-4 a c) d \left (c+d x^2\right )}{8 c (b+a c)^2 x^2 \sqrt {a+\frac {b}{c+d x^2}}}-\frac {\left (c+d x^2\right )^2}{4 c (b+a c) x^4 \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\left (3 b (b-4 a c) d^2 \sqrt {b+a \left (c+d x^2\right )}\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c+d x} \sqrt {b+a c+a d x}} \, dx,x,x^2\right )}{16 (b+a c)^3 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=\frac {3 b (b-4 a c) d^2}{8 c (b+a c)^3 \sqrt {a+\frac {b}{c+d x^2}}}-\frac {(b-4 a c) d \left (c+d x^2\right )}{8 c (b+a c)^2 x^2 \sqrt {a+\frac {b}{c+d x^2}}}-\frac {\left (c+d x^2\right )^2}{4 c (b+a c) x^4 \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\left (3 b (b-4 a c) d^2 \sqrt {b+a \left (c+d x^2\right )}\right ) \operatorname {Subst}\left (\int \frac {1}{-c-(-b-a c) x^2} \, dx,x,\frac {\sqrt {c+d x^2}}{\sqrt {b+a \left (c+d x^2\right )}}\right )}{8 (b+a c)^3 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=\frac {3 b (b-4 a c) d^2}{8 c (b+a c)^3 \sqrt {a+\frac {b}{c+d x^2}}}-\frac {(b-4 a c) d \left (c+d x^2\right )}{8 c (b+a c)^2 x^2 \sqrt {a+\frac {b}{c+d x^2}}}-\frac {\left (c+d x^2\right )^2}{4 c (b+a c) x^4 \sqrt {a+\frac {b}{c+d x^2}}}-\frac {3 b (b-4 a c) d^2 \sqrt {b+a \left (c+d x^2\right )} \tanh ^{-1}\left (\frac {\sqrt {b+a c} \sqrt {c+d x^2}}{\sqrt {c} \sqrt {b+a \left (c+d x^2\right )}}\right )}{8 \sqrt {c} (b+a c)^{7/2} \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ \end {align*}

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Mathematica [A]  time = 0.57, size = 281, normalized size = 1.33 \[ -\frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}} \left (2 \sqrt {c (a c+b)} \left (c+d x^2\right ) \left (2 a^2 c \left (c^2-d^2 x^4\right )+a b \left (4 c^2+5 c d x^2+13 d^2 x^4\right )+b^2 \left (2 c+5 d x^2\right )\right )+6 b d^2 x^4 \log (x) (4 a c-b) \sqrt {\left (c+d x^2\right ) \left (a \left (c+d x^2\right )+b\right )}+3 b d^2 x^4 (b-4 a c) \sqrt {\left (c+d x^2\right ) \left (a \left (c+d x^2\right )+b\right )} \log \left (2 \sqrt {c (a c+b)} \sqrt {\left (c+d x^2\right ) \left (a c+a d x^2+b\right )}+2 a c \left (c+d x^2\right )+b \left (2 c+d x^2\right )\right )\right )}{16 x^4 (a c+b)^3 \sqrt {c (a c+b)} \left (a \left (c+d x^2\right )+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*(a + b/(c + d*x^2))^(3/2)),x]

[Out]

-1/16*(Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(2*Sqrt[c*(b + a*c)]*(c + d*x^2)*(b^2*(2*c + 5*d*x^2) + 2*a^2*c*(
c^2 - d^2*x^4) + a*b*(4*c^2 + 5*c*d*x^2 + 13*d^2*x^4)) + 6*b*(-b + 4*a*c)*d^2*x^4*Sqrt[(c + d*x^2)*(b + a*(c +
d*x^2))]*Log[x] + 3*b*(b - 4*a*c)*d^2*x^4*Sqrt[(c + d*x^2)*(b + a*(c + d*x^2))]*Log[2*a*c*(c + d*x^2) + b*(2*
c + d*x^2) + 2*Sqrt[c*(b + a*c)]*Sqrt[(c + d*x^2)*(b + a*c + a*d*x^2)]]))/((b + a*c)^3*Sqrt[c*(b + a*c)]*x^4*(
b + a*(c + d*x^2)))

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fricas [B]  time = 1.26, size = 961, normalized size = 4.53 \[ \left [\frac {3 \, {\left ({\left (4 \, a^{2} b c - a b^{2}\right )} d^{3} x^{6} + {\left (4 \, a^{2} b c^{2} + 3 \, a b^{2} c - b^{3}\right )} d^{2} x^{4}\right )} \sqrt {a c^{2} + b c} \log \left (\frac {{\left (8 \, a^{2} c^{2} + 8 \, a b c + b^{2}\right )} d^{2} x^{4} + 8 \, a^{2} c^{4} + 16 \, a b c^{3} + 8 \, b^{2} c^{2} + 8 \, {\left (2 \, a^{2} c^{3} + 3 \, a b c^{2} + b^{2} c\right )} d x^{2} + 4 \, {\left ({\left (2 \, a c + b\right )} d^{2} x^{4} + 2 \, a c^{3} + {\left (4 \, a c^{2} + 3 \, b c\right )} d x^{2} + 2 \, b c^{2}\right )} \sqrt {a c^{2} + b c} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{x^{4}}\right ) + 4 \, {\left ({\left (2 \, a^{3} c^{3} - 11 \, a^{2} b c^{2} - 13 \, a b^{2} c\right )} d^{3} x^{6} - 2 \, a^{3} c^{6} - 6 \, a^{2} b c^{5} - 6 \, a b^{2} c^{4} + {\left (2 \, a^{3} c^{4} - 16 \, a^{2} b c^{3} - 23 \, a b^{2} c^{2} - 5 \, b^{3} c\right )} d^{2} x^{4} - 2 \, b^{3} c^{3} - {\left (2 \, a^{3} c^{5} + 11 \, a^{2} b c^{4} + 16 \, a b^{2} c^{3} + 7 \, b^{3} c^{2}\right )} d x^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{32 \, {\left ({\left (a^{5} c^{5} + 4 \, a^{4} b c^{4} + 6 \, a^{3} b^{2} c^{3} + 4 \, a^{2} b^{3} c^{2} + a b^{4} c\right )} d x^{6} + {\left (a^{5} c^{6} + 5 \, a^{4} b c^{5} + 10 \, a^{3} b^{2} c^{4} + 10 \, a^{2} b^{3} c^{3} + 5 \, a b^{4} c^{2} + b^{5} c\right )} x^{4}\right )}}, -\frac {3 \, {\left ({\left (4 \, a^{2} b c - a b^{2}\right )} d^{3} x^{6} + {\left (4 \, a^{2} b c^{2} + 3 \, a b^{2} c - b^{3}\right )} d^{2} x^{4}\right )} \sqrt {-a c^{2} - b c} \arctan \left (\frac {{\left ({\left (2 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + 2 \, b c\right )} \sqrt {-a c^{2} - b c} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{2 \, {\left (a^{2} c^{3} + 2 \, a b c^{2} + {\left (a^{2} c^{2} + a b c\right )} d x^{2} + b^{2} c\right )}}\right ) - 2 \, {\left ({\left (2 \, a^{3} c^{3} - 11 \, a^{2} b c^{2} - 13 \, a b^{2} c\right )} d^{3} x^{6} - 2 \, a^{3} c^{6} - 6 \, a^{2} b c^{5} - 6 \, a b^{2} c^{4} + {\left (2 \, a^{3} c^{4} - 16 \, a^{2} b c^{3} - 23 \, a b^{2} c^{2} - 5 \, b^{3} c\right )} d^{2} x^{4} - 2 \, b^{3} c^{3} - {\left (2 \, a^{3} c^{5} + 11 \, a^{2} b c^{4} + 16 \, a b^{2} c^{3} + 7 \, b^{3} c^{2}\right )} d x^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{16 \, {\left ({\left (a^{5} c^{5} + 4 \, a^{4} b c^{4} + 6 \, a^{3} b^{2} c^{3} + 4 \, a^{2} b^{3} c^{2} + a b^{4} c\right )} d x^{6} + {\left (a^{5} c^{6} + 5 \, a^{4} b c^{5} + 10 \, a^{3} b^{2} c^{4} + 10 \, a^{2} b^{3} c^{3} + 5 \, a b^{4} c^{2} + b^{5} c\right )} x^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(a+b/(d*x^2+c))^(3/2),x, algorithm="fricas")

[Out]

[1/32*(3*((4*a^2*b*c - a*b^2)*d^3*x^6 + (4*a^2*b*c^2 + 3*a*b^2*c - b^3)*d^2*x^4)*sqrt(a*c^2 + b*c)*log(((8*a^2
*c^2 + 8*a*b*c + b^2)*d^2*x^4 + 8*a^2*c^4 + 16*a*b*c^3 + 8*b^2*c^2 + 8*(2*a^2*c^3 + 3*a*b*c^2 + b^2*c)*d*x^2 +
4*((2*a*c + b)*d^2*x^4 + 2*a*c^3 + (4*a*c^2 + 3*b*c)*d*x^2 + 2*b*c^2)*sqrt(a*c^2 + b*c)*sqrt((a*d*x^2 + a*c +
b)/(d*x^2 + c)))/x^4) + 4*((2*a^3*c^3 - 11*a^2*b*c^2 - 13*a*b^2*c)*d^3*x^6 - 2*a^3*c^6 - 6*a^2*b*c^5 - 6*a*b^
2*c^4 + (2*a^3*c^4 - 16*a^2*b*c^3 - 23*a*b^2*c^2 - 5*b^3*c)*d^2*x^4 - 2*b^3*c^3 - (2*a^3*c^5 + 11*a^2*b*c^4 +
16*a*b^2*c^3 + 7*b^3*c^2)*d*x^2)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/((a^5*c^5 + 4*a^4*b*c^4 + 6*a^3*b^2*c^
3 + 4*a^2*b^3*c^2 + a*b^4*c)*d*x^6 + (a^5*c^6 + 5*a^4*b*c^5 + 10*a^3*b^2*c^4 + 10*a^2*b^3*c^3 + 5*a*b^4*c^2 +
b^5*c)*x^4), -1/16*(3*((4*a^2*b*c - a*b^2)*d^3*x^6 + (4*a^2*b*c^2 + 3*a*b^2*c - b^3)*d^2*x^4)*sqrt(-a*c^2 - b*
c)*arctan(1/2*((2*a*c + b)*d*x^2 + 2*a*c^2 + 2*b*c)*sqrt(-a*c^2 - b*c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))/(
a^2*c^3 + 2*a*b*c^2 + (a^2*c^2 + a*b*c)*d*x^2 + b^2*c)) - 2*((2*a^3*c^3 - 11*a^2*b*c^2 - 13*a*b^2*c)*d^3*x^6 -
2*a^3*c^6 - 6*a^2*b*c^5 - 6*a*b^2*c^4 + (2*a^3*c^4 - 16*a^2*b*c^3 - 23*a*b^2*c^2 - 5*b^3*c)*d^2*x^4 - 2*b^3*c
^3 - (2*a^3*c^5 + 11*a^2*b*c^4 + 16*a*b^2*c^3 + 7*b^3*c^2)*d*x^2)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/((a^5
*c^5 + 4*a^4*b*c^4 + 6*a^3*b^2*c^3 + 4*a^2*b^3*c^2 + a*b^4*c)*d*x^6 + (a^5*c^6 + 5*a^4*b*c^5 + 10*a^3*b^2*c^4
+ 10*a^2*b^3*c^3 + 5*a*b^4*c^2 + b^5*c)*x^4)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {undef} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(a+b/(d*x^2+c))^(3/2),x, algorithm="giac")

[Out]

undef

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maple [B]  time = 0.08, size = 1947, normalized size = 9.18 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(a+b/(d*x^2+c))^(3/2),x)

[Out]

-1/16*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)*(d*x^2+c)*(-33*ln((2*a*c*d*x^2+b*d*x^2+2*a*c^2+2*b*c+2*(a*c^2+b*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))/x^2)*x^4*a^3*b^2*c^5*d^2+16*((d*x^2+c)*(a*d*x^2+a*c+b))^(1
/2)*(a*c^2+b*c)^(3/2)*x^4*a^2*b*c^2*d^2+16*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*c^2+b*c)^(3/2)*x
^4*a^2*b*c^2*d^2+16*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*(a*c^2+b*c)^(3/2)*x^4*a*b^2*c*d^2-10*(a*d^2*x^4+2*a*c*d*
x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*c^2+b*c)^(3/2)*x^4*a*b^2*c*d^2-2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(3/
2)*(a*c^2+b*c)^(3/2)*x^2*a*b*c*d+2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*c^2+b*c)^(3/2)*x^6*a^2*b
*c*d^3+3*ln((2*a*c*d*x^2+b*d*x^2+2*a*c^2+2*b*c+2*(a*c^2+b*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))/x^2)*x^4*b^5*c^2*d^2-6*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*c^2+b*c)^(3/2)*x^4*b^3*d^2+6*(
a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(3/2)*(a*c^2+b*c)^(3/2)*x^2*b^2*d+8*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*
c^2+b*c)^(3/2)*(a*c^2+b*c)^(3/2)*a*b*c^2-3*ln((2*a*c*d*x^2+b*d*x^2+2*a*c^2+2*b*c+2*(a*c^2+b*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))/x^2)*x^4*a*b^4*c^3*d^2+6*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(3/
2)*(a*c^2+b*c)^(3/2)*x^4*a*b*d^2-8*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(3/2)*(a*c^2+b*c)^(3/2)*x^2*a^2*c
^2*d+12*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*c^2+b*c)^(3/2)*x^8*a^3*c*d^4-12*ln((2*a*c*d*x^2+b*d
*x^2+2*a*c^2+2*b*c+2*(a*c^2+b*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))/x^2)*x^6*a^4*b*c^5*d^3
-6*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*c^2+b*c)^(3/2)*x^8*a^2*b*d^4-21*ln((2*a*c*d*x^2+b*d*x^2+
2*a*c^2+2*b*c+2*(a*c^2+b*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))/x^2)*x^6*a^3*b^2*c^4*d^3+32
*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*c^2+b*c)^(3/2)*x^6*a^3*c^2*d^3-6*ln((2*a*c*d*x^2+b*d*x^2+2
*a*c^2+2*b*c+2*(a*c^2+b*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))/x^2)*x^6*a^2*b^3*c^3*d^3-12*
ln((2*a*c*d*x^2+b*d*x^2+2*a*c^2+2*b*c+2*(a*c^2+b*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))/x^2
)*x^4*a^4*b*c^6*d^2+3*ln((2*a*c*d*x^2+b*d*x^2+2*a*c^2+2*b*c+2*(a*c^2+b*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))/x^2)*x^6*a*b^4*c^2*d^3+4*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(3/2)*(a*c^2+b*c)^(3/2)*
a^2*c^3+4*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(3/2)*(a*c^2+b*c)^(3/2)*b^2*c-12*(a*d^2*x^4+2*a*c*d*x^2+b*
d*x^2+a*c^2+b*c)^(1/2)*(a*c^2+b*c)^(3/2)*x^6*a*b^2*d^3+20*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*c
^2+b*c)^(3/2)*x^4*a^3*c^3*d^2-27*ln((2*a*c*d*x^2+b*d*x^2+2*a*c^2+2*b*c+2*(a*c^2+b*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))/x^2)*x^4*a^2*b^3*c^4*d^2-12*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(3/2)*(a*c
^2+b*c)^(3/2)*x^4*a^2*c*d^2)/c/((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)/(a*c+b)^4/x^4/(a*c^2+b*c)^(3/2)/(a*d*x^2+a*c+
b)

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maxima [B]  time = 2.35, size = 450, normalized size = 2.12 \[ -\frac {3 \, {\left (4 \, a b c - b^{2}\right )} d^{2} \log \left (\frac {c \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} - \sqrt {{\left (a c + b\right )} c}}{c \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} + \sqrt {{\left (a c + b\right )} c}}\right )}{16 \, {\left (a^{3} c^{3} + 3 \, a^{2} b c^{2} + 3 \, a b^{2} c + b^{3}\right )} \sqrt {{\left (a c + b\right )} c}} - \frac {8 \, {\left (a^{3} b c^{2} + 2 \, a^{2} b^{2} c + a b^{3}\right )} d^{2} + \frac {3 \, {\left (4 \, a b c^{2} - b^{2} c\right )} {\left (a d x^{2} + a c + b\right )}^{2} d^{2}}{{\left (d x^{2} + c\right )}^{2}} - \frac {5 \, {\left (4 \, a^{2} b c^{2} + 3 \, a b^{2} c - b^{3}\right )} {\left (a d x^{2} + a c + b\right )} d^{2}}{d x^{2} + c}}{8 \, {\left ({\left (a^{3} c^{5} + 3 \, a^{2} b c^{4} + 3 \, a b^{2} c^{3} + b^{3} c^{2}\right )} \left (\frac {a d x^{2} + a c + b}{d x^{2} + c}\right )^{\frac {5}{2}} - 2 \, {\left (a^{4} c^{5} + 4 \, a^{3} b c^{4} + 6 \, a^{2} b^{2} c^{3} + 4 \, a b^{3} c^{2} + b^{4} c\right )} \left (\frac {a d x^{2} + a c + b}{d x^{2} + c}\right )^{\frac {3}{2}} + {\left (a^{5} c^{5} + 5 \, a^{4} b c^{4} + 10 \, a^{3} b^{2} c^{3} + 10 \, a^{2} b^{3} c^{2} + 5 \, a b^{4} c + b^{5}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(a+b/(d*x^2+c))^(3/2),x, algorithm="maxima")

[Out]

-3/16*(4*a*b*c - b^2)*d^2*log((c*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)) - sqrt((a*c + b)*c))/(c*sqrt((a*d*x^2 +
a*c + b)/(d*x^2 + c)) + sqrt((a*c + b)*c)))/((a^3*c^3 + 3*a^2*b*c^2 + 3*a*b^2*c + b^3)*sqrt((a*c + b)*c)) - 1
/8*(8*(a^3*b*c^2 + 2*a^2*b^2*c + a*b^3)*d^2 + 3*(4*a*b*c^2 - b^2*c)*(a*d*x^2 + a*c + b)^2*d^2/(d*x^2 + c)^2 -
5*(4*a^2*b*c^2 + 3*a*b^2*c - b^3)*(a*d*x^2 + a*c + b)*d^2/(d*x^2 + c))/((a^3*c^5 + 3*a^2*b*c^4 + 3*a*b^2*c^3 +
b^3*c^2)*((a*d*x^2 + a*c + b)/(d*x^2 + c))^(5/2) - 2*(a^4*c^5 + 4*a^3*b*c^4 + 6*a^2*b^2*c^3 + 4*a*b^3*c^2 + b
^4*c)*((a*d*x^2 + a*c + b)/(d*x^2 + c))^(3/2) + (a^5*c^5 + 5*a^4*b*c^4 + 10*a^3*b^2*c^3 + 10*a^2*b^3*c^2 + 5*a
*b^4*c + b^5)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{x^5\,{\left (a+\frac {b}{d\,x^2+c}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5*(a + b/(c + d*x^2))^(3/2)),x)

[Out]

int(1/(x^5*(a + b/(c + d*x^2))^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{5} \left (\frac {a c + a d x^{2} + b}{c + d x^{2}}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(a+b/(d*x**2+c))**(3/2),x)

[Out]

Integral(1/(x**5*((a*c + a*d*x**2 + b)/(c + d*x**2))**(3/2)), x)

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