3.4.34 \(\int \frac {(a+\frac {b}{c+d x^2})^{3/2}}{x} \, dx\) [334]

Optimal. Leaf size=126 \[ \frac {b \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{c}+a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {a}}\right )-\frac {(b+a c)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {b+a c}}\right )}{c^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.25, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1985, 1981, 1980, 490, 536, 214} \begin {gather*} a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{\sqrt {a}}\right )-\frac {(a c+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{\sqrt {a c+b}}\right )}{c^{3/2}}+\frac {b \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b/(c + d*x^2))^(3/2)/x,x]

[Out]

(b*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/c + a^(3/2)*ArcTanh[Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]/Sqrt[a]] -
((b + a*c)^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/Sqrt[b + a*c]])/c^(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 490

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

Rule 536

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, 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]]

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

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Mathematica [A]
time = 0.19, size = 139, normalized size = 1.10 \begin {gather*} \frac {b \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{c}-\frac {(-b-a c)^{3/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {-b-a c} \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{b+a c}\right )}{c^{3/2}}+a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(651\) vs. \(2(108)=216\).
time = 0.07, size = 652, normalized size = 5.17

method	result	size
default	\(\frac {\left (\ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) a^{2} c^{2} d^{2} x^{2}-\sqrt {a \,d^{2}}\, \sqrt {c^{2} a +b c}\, \ln \left (\frac {2 a c d \,x^{2}+b d \,x^{2}+2 c^{2} a +2 \sqrt {c^{2} a +b c}\, \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}+2 b c}{x^{2}}\right ) a c d \,x^{2}+\ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) a^{2} c^{3} d -\sqrt {a \,d^{2}}\, \sqrt {c^{2} a +b c}\, \ln \left (\frac {2 a c d \,x^{2}+b d \,x^{2}+2 c^{2} a +2 \sqrt {c^{2} a +b c}\, \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}+2 b c}{x^{2}}\right ) b d \,x^{2}-\sqrt {a \,d^{2}}\, \sqrt {c^{2} a +b c}\, \ln \left (\frac {2 a c d \,x^{2}+b d \,x^{2}+2 c^{2} a +2 \sqrt {c^{2} a +b c}\, \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}+2 b c}{x^{2}}\right ) a \,c^{2}-\sqrt {a \,d^{2}}\, \sqrt {c^{2} a +b c}\, \ln \left (\frac {2 a c d \,x^{2}+b d \,x^{2}+2 c^{2} a +2 \sqrt {c^{2} a +b c}\, \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}+2 b c}{x^{2}}\right ) b c +2 \sqrt {a \,d^{2}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, b c \right ) \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}{2 \sqrt {a \,d^{2}}\, c^{2} \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}}\)	\(652\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.53, size = 201, normalized size = 1.60 \begin {gather*} -\frac {1}{2} \, a^{\frac {3}{2}} \log \left (-\frac {\sqrt {a} - \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{\sqrt {a} + \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}\right ) + \frac {b \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{c} + \frac {{\left (a^{2} c^{2} + 2 \, a b c + b^{2}\right )} \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 )}{2 \, \sqrt {{\left (a c + b\right )} c} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.47, size = 1073, normalized size = 8.52 \begin {gather*} \left [\frac {a^{\frac {3}{2}} c \log \left (8 \, a^{2} d^{2} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (2 \, a^{2} c + a b\right )} d x^{2} + 8 \, a b c + b^{2} + 4 \, {\left (2 \, a d^{2} x^{4} + {\left (4 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + b c\right )} \sqrt {a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}\right ) + {\left (a c + b\right )} \sqrt {\frac {a c + 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^{2} + b c\right )} d^{2} x^{4} + 2 \, a c^{4} + 2 \, b c^{3} + {\left (4 \, a c^{3} + 3 \, b c^{2}\right )} d x^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} \sqrt {\frac {a c + b}{c}}}{x^{4}}\right ) + 4 \, b \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{4 \, c}, -\frac {2 \, \sqrt {-a} a c \arctan \left (\frac {{\left (2 \, a d x^{2} + 2 \, a c + b\right )} \sqrt {-a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{2 \, {\left (a^{2} d x^{2} + a^{2} c + a b\right )}}\right ) - {\left (a c + b\right )} \sqrt {\frac {a c + 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^{2} + b c\right )} d^{2} x^{4} + 2 \, a c^{4} + 2 \, b c^{3} + {\left (4 \, a c^{3} + 3 \, b c^{2}\right )} d x^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} \sqrt {\frac {a c + b}{c}}}{x^{4}}\right ) - 4 \, b \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{4 \, c}, \frac {a^{\frac {3}{2}} c \log \left (8 \, a^{2} d^{2} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (2 \, a^{2} c + a b\right )} d x^{2} + 8 \, a b c + b^{2} + 4 \, {\left (2 \, a d^{2} x^{4} + {\left (4 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + b c\right )} \sqrt {a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}\right ) + 2 \, {\left (a c + b\right )} \sqrt {-\frac {a c + b}{c}} \arctan \left (\frac {{\left ({\left (2 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + 2 \, b c\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} \sqrt {-\frac {a c + b}{c}}}{2 \, {\left (a^{2} c^{2} + {\left (a^{2} c + a b\right )} d x^{2} + 2 \, a b c + b^{2}\right )}}\right ) + 4 \, b \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{4 \, c}, -\frac {\sqrt {-a} a c \arctan \left (\frac {{\left (2 \, a d x^{2} + 2 \, a c + b\right )} \sqrt {-a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{2 \, {\left (a^{2} d x^{2} + a^{2} c + a b\right )}}\right ) - {\left (a c + b\right )} \sqrt {-\frac {a c + b}{c}} \arctan \left (\frac {{\left ({\left (2 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + 2 \, b c\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} \sqrt {-\frac {a c + b}{c}}}{2 \, {\left (a^{2} c^{2} + {\left (a^{2} c + a b\right )} d x^{2} + 2 \, a b c + b^{2}\right )}}\right ) - 2 \, b \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{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,x, algorithm="fricas")

[Out]

[1/4*(a^(3/2)*c*log(8*a^2*d^2*x^4 + 8*a^2*c^2 + 8*(2*a^2*c + a*b)*d*x^2 + 8*a*b*c + b^2 + 4*(2*a*d^2*x^4 + (4*
a*c + b)*d*x^2 + 2*a*c^2 + b*c)*sqrt(a)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))) + (a*c + b)*sqrt((a*c + b)/c)*l
og(((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^2 + b*c)*d^2*x^4 + 2*a*c^4 + 2*b*c^3 + (4*a*c^3 + 3*b*c^2)*d*x^2)*sqrt((a*d*x^2 + a*c + b
)/(d*x^2 + c))*sqrt((a*c + b)/c))/x^4) + 4*b*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/c, -1/4*(2*sqrt(-a)*a*c*ar
ctan(1/2*(2*a*d*x^2 + 2*a*c + b)*sqrt(-a)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))/(a^2*d*x^2 + a^2*c + a*b)) - (
a*c + b)*sqrt((a*c + 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^2 + b*c)*d^2*x^4 + 2*a*c^4 + 2*b*c^3 + (4*a*c^3 + 3*b*c^2)*d*
x^2)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))*sqrt((a*c + b)/c))/x^4) - 4*b*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))
)/c, 1/4*(a^(3/2)*c*log(8*a^2*d^2*x^4 + 8*a^2*c^2 + 8*(2*a^2*c + a*b)*d*x^2 + 8*a*b*c + b^2 + 4*(2*a*d^2*x^4 +
(4*a*c + b)*d*x^2 + 2*a*c^2 + b*c)*sqrt(a)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))) + 2*(a*c + b)*sqrt(-(a*c +
b)/c)*arctan(1/2*((2*a*c + b)*d*x^2 + 2*a*c^2 + 2*b*c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))*sqrt(-(a*c + b)/c
)/(a^2*c^2 + (a^2*c + a*b)*d*x^2 + 2*a*b*c + b^2)) + 4*b*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/c, -1/2*(sqrt(
-a)*a*c*arctan(1/2*(2*a*d*x^2 + 2*a*c + b)*sqrt(-a)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))/(a^2*d*x^2 + a^2*c +
a*b)) - (a*c + b)*sqrt(-(a*c + b)/c)*arctan(1/2*((2*a*c + b)*d*x^2 + 2*a*c^2 + 2*b*c)*sqrt((a*d*x^2 + a*c + b
)/(d*x^2 + c))*sqrt(-(a*c + b)/c)/(a^2*c^2 + (a^2*c + a*b)*d*x^2 + 2*a*b*c + b^2)) - 2*b*sqrt((a*d*x^2 + a*c +
b)/(d*x^2 + c)))/c]

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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}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((a+b/(d*x^2+c))^(3/2)/x,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 {{\left (a+\frac {b}{d\,x^2+c}\right )}^{3/2}}{x} \,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,x)

[Out]

int((a + b/(c + d*x^2))^(3/2)/x, x)

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