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

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

[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 = 0.59, antiderivative size = 260, normalized size of antiderivative = 1.27, 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} \[ \frac {3 b d^2 (4 a c+5 b) \sqrt {a+\frac {b}{c+d x^2}}}{8 c^3 (a c+b)}-\frac {3 b d^2 (4 a c+5 b) \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}} \tanh ^{-1}\left (\frac {\sqrt {a c+b} \sqrt {c+d x^2}}{\sqrt {c} \sqrt {a \left (c+d x^2\right )+b}}\right )}{8 c^{7/2} \sqrt {a c+b} \sqrt {a \left (c+d x^2\right )+b}}+\frac {d (4 a c+5 b) \sqrt {a+\frac {b}{c+d x^2}} \left (a \left (c+d x^2\right )+b\right )}{8 c^2 x^2 (a c+b)}-\frac {\sqrt {a+\frac {b}{c+d x^2}} \left (a \left (c+d x^2\right )+b\right )^2}{4 c x^4 (a c+b)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/(c + d*x^2))^(3/2)/x^5,x]

[Out]

(3*b*(5*b + 4*a*c)*d^2*Sqrt[a + b/(c + d*x^2)])/(8*c^3*(b + a*c)) + ((5*b + 4*a*c)*d*Sqrt[a + b/(c + d*x^2)]*(
b + a*(c + d*x^2)))/(8*c^2*(b + a*c)*x^2) - (Sqrt[a + b/(c + d*x^2)]*(b + a*(c + d*x^2))^2)/(4*c*(b + a*c)*x^4
) - (3*b*(5*b + 4*a*c)*d^2*Sqrt[c + d*x^2]*Sqrt[a + b/(c + d*x^2)]*ArcTanh[(Sqrt[b + a*c]*Sqrt[c + d*x^2])/(Sq
rt[c]*Sqrt[b + a*(c + d*x^2)])])/(8*c^(7/2)*Sqrt[b + a*c]*Sqrt[b + a*(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 {\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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \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 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 = 0.33, size = 190, normalized size = 0.93 \[ \frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}} \left (-\frac {2 c^2 (a c+b)}{x^4}+\frac {3 b d^2 (4 a c+5 b) \left (c+d x^2\right ) \left (2 \log (x)-\log \left (2 \sqrt {c (a c+b)} \sqrt {\left (c+d x^2\right ) \left (a c+a d x^2+b\right )}+2 a c \left (c+d x^2\right )+b \left (2 c+d x^2\right )\right )\right )}{2 \sqrt {c (a c+b)} \sqrt {\left (c+d x^2\right ) \left (a \left (c+d x^2\right )+b\right )}}+d^2 (2 a c+15 b)+\frac {5 b c d}{x^2}\right )}{8 c^3} \]

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

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fricas [A]  time = 1.52, size = 557, normalized size = 2.72 \[ \left [\frac {3 \, {\left (4 \, a b c + 5 \, b^{2}\right )} \sqrt {a c^{2} + b c} d^{2} x^{4} \log \left (\frac {{\left (8 \, a^{2} c^{2} + 8 \, a b c + b^{2}\right )} d^{2} x^{4} + 8 \, a^{2} c^{4} + 16 \, a b c^{3} + 8 \, b^{2} c^{2} + 8 \, {\left (2 \, a^{2} c^{3} + 3 \, a b c^{2} + b^{2} c\right )} d x^{2} - 4 \, {\left ({\left (2 \, a c + b\right )} d^{2} x^{4} + 2 \, a c^{3} + {\left (4 \, a c^{2} + 3 \, b c\right )} d x^{2} + 2 \, b c^{2}\right )} \sqrt {a c^{2} + b c} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{x^{4}}\right ) - 4 \, {\left (2 \, a^{2} c^{5} - {\left (2 \, a^{2} c^{3} + 17 \, a b c^{2} + 15 \, b^{2} c\right )} d^{2} x^{4} + 4 \, a b c^{4} + 2 \, b^{2} c^{3} - 5 \, {\left (a b c^{3} + b^{2} c^{2}\right )} d x^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{32 \, {\left (a c^{5} + b c^{4}\right )} x^{4}}, \frac {3 \, {\left (4 \, a b c + 5 \, b^{2}\right )} \sqrt {-a c^{2} - b c} d^{2} x^{4} \arctan \left (\frac {{\left ({\left (2 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + 2 \, b c\right )} \sqrt {-a c^{2} - b c} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{2 \, {\left (a^{2} c^{3} + 2 \, a b c^{2} + {\left (a^{2} c^{2} + a b c\right )} d x^{2} + b^{2} c\right )}}\right ) - 2 \, {\left (2 \, a^{2} c^{5} - {\left (2 \, a^{2} c^{3} + 17 \, a b c^{2} + 15 \, b^{2} c\right )} d^{2} x^{4} + 4 \, a b c^{4} + 2 \, b^{2} c^{3} - 5 \, {\left (a b c^{3} + b^{2} c^{2}\right )} d x^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{16 \, {\left (a c^{5} + b c^{4}\right )} x^{4}}\right ] \]

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

[Out]

undef

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maple [B]  time = 0.08, size = 1653, normalized size = 8.06 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)*
x^8*a^2*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^3*b*c^5*d^3-18*(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)*x^8*
a*b*d^4-39*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^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^2*
c^2*d^3-42*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^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^3*b*c^6*d^2-62*(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)*x^6*a*b*c*d^3-15*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*b^4*c^2*d^3-39*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^2*c^5*d^2-18*(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)*x^6*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^2*c^3*d^2-42*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^3*c^4*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*c^2*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*c*d^2-44*(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)*x^4*a*b*c^2*d^2-15*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^4*c^3*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*b^2*c*d^2+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)*x^
4*b*d^2-18*(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)*x^4*b^2*c*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*c^2*d+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)*x^2*b*c*d-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 = 2.42, size = 313, normalized size = 1.53 \[ \frac {b d^{2} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{c^{3}} + \frac {3 \, {\left (4 \, a b c + 5 \, b^{2}\right )} d^{2} \log \left (\frac {c \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} - \sqrt {{\left (a c + b\right )} c}}{c \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} + \sqrt {{\left (a c + b\right )} c}}\right )}{16 \, \sqrt {{\left (a c + b\right )} c} c^{3}} - \frac {{\left (4 \, a b c^{2} + 9 \, b^{2} c\right )} d^{2} \left (\frac {a d x^{2} + a c + b}{d x^{2} + c}\right )^{\frac {3}{2}} - {\left (4 \, a^{2} b c^{2} + 11 \, a b^{2} c + 7 \, b^{3}\right )} d^{2} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{8 \, {\left (a^{2} c^{5} + 2 \, a b c^{4} + b^{2} c^{3} + \frac {{\left (a d x^{2} + a c + b\right )}^{2} c^{5}}{{\left (d x^{2} + c\right )}^{2}} - \frac {2 \, {\left (a c^{5} + b c^{4}\right )} {\left (a d x^{2} + a c + b\right )}}{d x^{2} + c}\right )}} \]

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

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

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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