3.4.36 \(\int \frac {(a+\frac {b}{c+d x^2})^{3/2}}{x^5} \, dx\) [336]

Optimal. Leaf size=205 \[ \frac {b d^2 \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{c^3}+\frac {(9 b+4 a c) d \left (c+d x^2\right ) \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{8 c^3 x^2}-\frac {(b+a c) \left (c+d x^2\right )^2 \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{4 c^3 x^4}-\frac {3 b (5 b+4 a c) d^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {b+a c}}\right )}{8 c^{7/2} \sqrt {b+a c}} \]

[Out]

-3/8*b*(4*a*c+5*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))/c^(7/2)/(a*c+b)^(1/2)+
b*d^2*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/c^3+1/8*(4*a*c+9*b)*d*(d*x^2+c)*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/c^3/
x^2-1/4*(a*c+b)*(d*x^2+c)^2*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/c^3/x^4

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Rubi [A]
time = 0.28, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1985, 1981, 1980, 466, 1171, 396, 214} \begin {gather*} -\frac {3 b d^2 (4 a c+5 b) \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 c^{7/2} \sqrt {a c+b}}+\frac {b d^2 \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{c^3}+\frac {d (4 a c+9 b) \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{8 c^3 x^2}-\frac {(a c+b) \left (c+d x^2\right )^2 \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 c^3 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b/(c + d*x^2))^(3/2)/x^5,x]

[Out]

(b*d^2*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/c^3 + ((9*b + 4*a*c)*d*(c + d*x^2)*Sqrt[(b + a*c + a*d*x^2)/(c +
d*x^2)])/(8*c^3*x^2) - ((b + a*c)*(c + d*x^2)^2*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/(4*c^3*x^4) - (3*b*(5*
b + 4*a*c)*d^2*ArcTanh[(Sqrt[c]*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/Sqrt[b + a*c]])/(8*c^(7/2)*Sqrt[b + a*c
])

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 396

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

Rule 466

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x
*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1171

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1
)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] &&
NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

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]]

Rule 1985

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

Rubi steps

\begin {align*} \int \frac {\left (a+\frac {b}{c+d x^2}\right )^{3/2}}{x^5} \, dx &=\frac {\left (\sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}\right ) \int \frac {\left (b+a \left (c+d x^2\right )\right )^{3/2}}{x^5 \left (c+d x^2\right )^{3/2}} \, dx}{\sqrt {b+a \left (c+d x^2\right )}}\\ &=\frac {\left (\sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}\right ) \int \frac {\left (b+a c+a d x^2\right )^{3/2}}{x^5 \left (c+d x^2\right )^{3/2}} \, dx}{\sqrt {b+a \left (c+d x^2\right )}}\\ &=\frac {\left (\sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}\right ) \text {Subst}\left (\int \frac {(b+a c+a d x)^{3/2}}{x^3 (c+d x)^{3/2}} \, dx,x,x^2\right )}{2 \sqrt {b+a \left (c+d x^2\right )}}\\ &=-\frac {\sqrt {a+\frac {b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )^2}{4 c (b+a c) x^4}-\frac {\left ((5 b+4 a c) d \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}\right ) \text {Subst}\left (\int \frac {(b+a c+a d x)^{3/2}}{x^2 (c+d x)^{3/2}} \, dx,x,x^2\right )}{8 c (b+a c) \sqrt {b+a \left (c+d x^2\right )}}\\ &=\frac {(5 b+4 a c) d \sqrt {a+\frac {b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )}{8 c^2 (b+a c) x^2}-\frac {\sqrt {a+\frac {b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )^2}{4 c (b+a c) x^4}+\frac {\left (3 b (5 b+4 a c) d^2 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}\right ) \text {Subst}\left (\int \frac {\sqrt {b+a c+a d x}}{x (c+d x)^{3/2}} \, dx,x,x^2\right )}{16 c^2 (b+a c) \sqrt {b+a \left (c+d x^2\right )}}\\ &=\frac {3 b (5 b+4 a c) d^2 \sqrt {a+\frac {b}{c+d x^2}}}{8 c^3 (b+a c)}+\frac {(5 b+4 a c) d \sqrt {a+\frac {b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )}{8 c^2 (b+a c) x^2}-\frac {\sqrt {a+\frac {b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )^2}{4 c (b+a c) x^4}+\frac {\left (3 b (5 b+4 a c) d^2 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {c+d x} \sqrt {b+a c+a d x}} \, dx,x,x^2\right )}{16 c^3 \sqrt {b+a \left (c+d x^2\right )}}\\ &=\frac {3 b (5 b+4 a c) d^2 \sqrt {a+\frac {b}{c+d x^2}}}{8 c^3 (b+a c)}+\frac {(5 b+4 a c) d \sqrt {a+\frac {b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )}{8 c^2 (b+a c) x^2}-\frac {\sqrt {a+\frac {b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )^2}{4 c (b+a c) x^4}+\frac {\left (3 b (5 b+4 a c) d^2 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}\right ) \text {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 c^3 \sqrt {b+a \left (c+d x^2\right )}}\\ &=\frac {3 b (5 b+4 a c) d^2 \sqrt {a+\frac {b}{c+d x^2}}}{8 c^3 (b+a c)}+\frac {(5 b+4 a c) d \sqrt {a+\frac {b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )}{8 c^2 (b+a c) x^2}-\frac {\sqrt {a+\frac {b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )^2}{4 c (b+a c) x^4}-\frac {3 b (5 b+4 a c) d^2 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}} \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 c^{7/2} \sqrt {b+a c} \sqrt {b+a \left (c+d x^2\right )}}\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 153, normalized size = 0.75 \begin {gather*} \frac {\sqrt {\frac {b+a c+a d x^2}{c+d x^2}} \left (-2 b c^2-2 a c^3+5 b c d x^2+15 b d^2 x^4+2 a c d^2 x^4\right )}{8 c^3 x^4}+\frac {3 b (5 b+4 a c) d^2 \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {-b-a c}}\right )}{8 c^{7/2} \sqrt {-b-a c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(-2*b*c^2 - 2*a*c^3 + 5*b*c*d*x^2 + 15*b*d^2*x^4 + 2*a*c*d^2*x^4))/(8*c
^3*x^4) + (3*b*(5*b + 4*a*c)*d^2*ArcTan[(Sqrt[c]*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/Sqrt[-b - a*c]])/(8*c^
(7/2)*Sqrt[-b - a*c])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1652\) vs. \(2(183)=366\).
time = 0.15, size = 1653, normalized size = 8.06

method	result	size
risch	\(-\frac {\left (d \,x^{2}+c \right ) \left (-2 a c d \,x^{2}-7 b d \,x^{2}+2 c^{2} a +2 b c \right ) \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}{8 c^{3} x^{4}}+\frac {\left (-\frac {3 d^{2} b \ln \left (\frac {2 c^{2} a +2 b c +\left (2 a c d +b d \right ) x^{2}+2 \sqrt {c^{2} a +b c}\, \sqrt {c^{2} a +b c +\left (2 a c d +b d \right ) x^{2}+a \,d^{2} x^{4}}}{x^{2}}\right ) a}{4 c^{2} \sqrt {c^{2} a +b c}}-\frac {15 d^{2} b^{2} \ln \left (\frac {2 c^{2} a +2 b c +\left (2 a c d +b d \right ) x^{2}+2 \sqrt {c^{2} a +b c}\, \sqrt {c^{2} a +b c +\left (2 a c d +b d \right ) x^{2}+a \,d^{2} x^{4}}}{x^{2}}\right )}{16 c^{3} \sqrt {c^{2} a +b c}}+\frac {d^{3} b \,x^{2} a}{c^{3} \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}}+\frac {d^{2} b a}{c^{2} \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}}+\frac {d^{2} b^{2}}{c^{3} \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}}\right ) \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}}{a d \,x^{2}+a c +b}\)	\(446\)
default	\(\text {Expression too large to display}\)	\(1653\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)*(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)*
a^2*c*d^4*x^8+12*ln((2*a*c*d*x^2+b*d*x^2+2*c^2*a+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)+2*b*c)/x^2)*a^3*b*c^5*d^3*x^6+18*(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)*a*b*
d^4*x^8+39*ln((2*a*c*d*x^2+b*d*x^2+2*c^2*a+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)
+2*b*c)/x^2)*a^2*b^2*c^4*d^3*x^6+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)*a^2*c^2*
d^3*x^6+12*ln((2*a*c*d*x^2+b*d*x^2+2*c^2*a+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)
+2*b*c)/x^2)*a^3*b*c^6*d^2*x^4+42*ln((2*a*c*d*x^2+b*d*x^2+2*c^2*a+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)+2*b*c)/x^2)*a*b^3*c^3*d^3*x^6+62*(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)*a*b*c*d^3*x^6+39*ln((2*a*c*d*x^2+b*d*x^2+2*c^2*a+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)+2*b*c)/x^2)*a^2*b^2*c^5*d^2*x^4+15*ln((2*a*c*d*x^2+b*d*x^2+2*c^2*a+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)+2*b*c)/x^2)*b^4*c^2*d^3*x^6+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)*a^2*c^3*d^2*x^4+18*(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)*b^2*d^3*x^6+42*ln((2*a*c*d*x^2+b*d*x^2+2*c^2*a+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)+2*b*c)/x^2)*a*b^3*c^4*d^2*x^4-12*(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)
*a*c*d^2*x^4+44*(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)*a*b*c^2*d^2*x^4-16*(a*c^2+b*
c)^(3/2)*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*a*b*c^2*d^2*x^4+15*ln((2*a*c*d*x^2+b*d*x^2+2*c^2*a+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)+2*b*c)/x^2)*b^4*c^3*d^2*x^4-18*(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*d^2*x^4+18*(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)*b^2*c*d^2*x^4-16*(a*c^2+b*c)^(3/2)*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*b^2*c*d^2*x^4-8*(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*c^2*d*x^2-14*(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*c*d*x^2+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*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*c^2)/(a*c^2+b*c)^(3/2)/x^4/(a*c+b)/c^4/((d*x^2
+c)*(a*d*x^2+a*c+b))^(1/2)

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Maxima [A]
time = 0.51, size = 313, normalized size = 1.53 \begin {gather*} \frac {b d^{2} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{c^{3}} + \frac {3 \, {\left (4 \, a b c + 5 \, 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 \, \sqrt {{\left (a c + b\right )} c} c^{3}} - \frac {{\left (4 \, a b c^{2} + 9 \, b^{2} c\right )} d^{2} \left (\frac {a d x^{2} + a c + b}{d x^{2} + c}\right )^{\frac {3}{2}} - {\left (4 \, a^{2} b c^{2} + 11 \, a b^{2} c + 7 \, b^{3}\right )} d^{2} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{8 \, {\left (a^{2} c^{5} + 2 \, a b c^{4} + b^{2} c^{3} + \frac {{\left (a d x^{2} + a c + b\right )}^{2} c^{5}}{{\left (d x^{2} + c\right )}^{2}} - \frac {2 \, {\left (a c^{5} + b c^{4}\right )} {\left (a d x^{2} + a c + b\right )}}{d x^{2} + c}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*d^2*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))/c^3 + 3/16*(4*a*b*c + 5*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)))/(sqrt((a*c + b
)*c)*c^3) - 1/8*((4*a*b*c^2 + 9*b^2*c)*d^2*((a*d*x^2 + a*c + b)/(d*x^2 + c))^(3/2) - (4*a^2*b*c^2 + 11*a*b^2*c
+ 7*b^3)*d^2*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(a^2*c^5 + 2*a*b*c^4 + b^2*c^3 + (a*d*x^2 + a*c + b)^2*c^
5/(d*x^2 + c)^2 - 2*(a*c^5 + b*c^4)*(a*d*x^2 + a*c + b)/(d*x^2 + c))

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Fricas [A]
time = 0.70, size = 557, normalized size = 2.72 \begin {gather*} \left [\frac {3 \, {\left (4 \, a b c + 5 \, b^{2}\right )} \sqrt {a c^{2} + b c} d^{2} x^{4} \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 (2 \, a^{2} c^{5} - {\left (2 \, a^{2} c^{3} + 17 \, a b c^{2} + 15 \, b^{2} c\right )} d^{2} x^{4} + 4 \, a b c^{4} + 2 \, b^{2} c^{3} - 5 \, {\left (a b c^{3} + b^{2} c^{2}\right )} d x^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{32 \, {\left (a c^{5} + b c^{4}\right )} x^{4}}, \frac {3 \, {\left (4 \, a b c + 5 \, b^{2}\right )} \sqrt {-a c^{2} - b c} d^{2} x^{4} \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 (2 \, a^{2} c^{5} - {\left (2 \, a^{2} c^{3} + 17 \, a b c^{2} + 15 \, b^{2} c\right )} d^{2} x^{4} + 4 \, a b c^{4} + 2 \, b^{2} c^{3} - 5 \, {\left (a b c^{3} + b^{2} c^{2}\right )} d x^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{16 \, {\left (a c^{5} + b c^{4}\right )} x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/32*(3*(4*a*b*c + 5*b^2)*sqrt(a*c^2 + b*c)*d^2*x^4*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^2*c^5 - (2*a^2
*c^3 + 17*a*b*c^2 + 15*b^2*c)*d^2*x^4 + 4*a*b*c^4 + 2*b^2*c^3 - 5*(a*b*c^3 + b^2*c^2)*d*x^2)*sqrt((a*d*x^2 + a
*c + b)/(d*x^2 + c)))/((a*c^5 + b*c^4)*x^4), 1/16*(3*(4*a*b*c + 5*b^2)*sqrt(-a*c^2 - b*c)*d^2*x^4*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^2*c^5 - (2*a^2*c^3 + 17*a*b*c^2 + 15*b^2*c)*d^2*x^4 + 4*a*b*
c^4 + 2*b^2*c^3 - 5*(a*b*c^3 + b^2*c^2)*d*x^2)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/((a*c^5 + b*c^4)*x^4)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

undef

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/(c + d*x^2))^(3/2)/x^5,x)

[Out]

int((a + b/(c + d*x^2))^(3/2)/x^5, x)

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