\(\int \frac {x}{(a+\frac {b}{c+d x^2})^{3/2}} \, dx\) [357]

Optimal result

Integrand size = 19, antiderivative size = 100 \[ \int \frac {x}{\left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=\frac {3 b}{2 a^2 d \sqrt {a+\frac {b}{c+d x^2}}}+\frac {c+d x^2}{2 a d \sqrt {a+\frac {b}{c+d x^2}}}-\frac {3 b \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{c+d x^2}}}{\sqrt {a}}\right )}{2 a^{5/2} d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {1605, 248, 44, 53, 65, 214} \[ \int \frac {x}{\left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=-\frac {3 b \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{c+d x^2}}}{\sqrt {a}}\right )}{2 a^{5/2} d}+\frac {3 b}{2 a^2 d \sqrt {a+\frac {b}{c+d x^2}}}+\frac {c+d x^2}{2 a d \sqrt {a+\frac {b}{c+d x^2}}} \]

[In]

Int[x/(a + b/(c + d*x^2))^(3/2),x]

[Out]

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

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 248

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^2, x], x, 1/x] /; FreeQ[{a, b, p},
x] && ILtQ[n, 0]

Rule 1605

Int[((a_.) + (b_.)*(Pq_)^(n_.))^(p_.)*(Qr_), x_Symbol] :> With[{q = Expon[Pq, x], r = Expon[Qr, x]}, Dist[Coef
f[Qr, x, r]/(q*Coeff[Pq, x, q]), Subst[Int[(a + b*x^n)^p, x], x, Pq], x] /; EqQ[r, q - 1] && EqQ[Coeff[Qr, x,
r]*D[Pq, x], q*Coeff[Pq, x, q]*Qr]] /; FreeQ[{a, b, n, p}, x] && PolyQ[Pq, x] && PolyQ[Qr, x]

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.14 \[ \int \frac {x}{\left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=\frac {\left (c+d x^2\right ) \sqrt {\frac {b+a \left (c+d x^2\right )}{c+d x^2}} \left (3 b+a \left (c+d x^2\right )\right )}{2 a^2 d \left (b+a \left (c+d x^2\right )\right )}-\frac {3 b \text {arctanh}\left (\frac {\sqrt {\frac {b+a \left (c+d x^2\right )}{c+d x^2}}}{\sqrt {a}}\right )}{2 a^{5/2} d} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(216\) vs. \(2(84)=168\).

Time = 1.14 (sec) , antiderivative size = 217, normalized size of antiderivative = 2.17

method	result	size
risch	\(\frac {a d \,x^{2}+a c +b}{2 d \,a^{2} \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}-\frac {b \left (\frac {3 \ln \left (\frac {a c d +\frac {1}{2} b d +a \,d^{2} x^{2}}{\sqrt {a \,d^{2}}}+\sqrt {a \,c^{2}+b c +\left (2 a c d +b d \right ) x^{2}+a \,d^{2} x^{4}}\right )}{2 \sqrt {a \,d^{2}}}-\frac {2 \left (d \,x^{2}+c \right )}{d \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}}\right ) \sqrt {\left (a d \,x^{2}+a c +b \right ) \left (d \,x^{2}+c \right )}}{2 a^{2} \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right )}\)	\(217\)
derivativedivides	\(-\frac {\sqrt {\frac {a \left (d \,x^{2}+c \right )+b}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right ) \left (-6 \sqrt {\left (a \left (d \,x^{2}+c \right )+b \right ) \left (d \,x^{2}+c \right )}\, a^{\frac {5}{2}} \left (d \,x^{2}+c \right )^{2}+3 \ln \left (\frac {2 \sqrt {\left (a \left (d \,x^{2}+c \right )+b \right ) \left (d \,x^{2}+c \right )}\, \sqrt {a}+2 a \left (d \,x^{2}+c \right )+b}{2 \sqrt {a}}\right ) a^{2} b \left (d \,x^{2}+c \right )^{2}+4 a^{\frac {3}{2}} {\left (\left (a \left (d \,x^{2}+c \right )+b \right ) \left (d \,x^{2}+c \right )\right )}^{\frac {3}{2}}-12 \sqrt {\left (a \left (d \,x^{2}+c \right )+b \right ) \left (d \,x^{2}+c \right )}\, a^{\frac {3}{2}} b \left (d \,x^{2}+c \right )+6 \ln \left (\frac {2 \sqrt {\left (a \left (d \,x^{2}+c \right )+b \right ) \left (d \,x^{2}+c \right )}\, \sqrt {a}+2 a \left (d \,x^{2}+c \right )+b}{2 \sqrt {a}}\right ) a \,b^{2} \left (d \,x^{2}+c \right )-6 \sqrt {\left (a \left (d \,x^{2}+c \right )+b \right ) \left (d \,x^{2}+c \right )}\, \sqrt {a}\, b^{2}+3 \ln \left (\frac {2 \sqrt {\left (a \left (d \,x^{2}+c \right )+b \right ) \left (d \,x^{2}+c \right )}\, \sqrt {a}+2 a \left (d \,x^{2}+c \right )+b}{2 \sqrt {a}}\right ) b^{3}\right )}{4 d \,a^{\frac {5}{2}} \sqrt {\left (a \left (d \,x^{2}+c \right )+b \right ) \left (d \,x^{2}+c \right )}\, {\left (a \left (d \,x^{2}+c \right )+b \right )}^{2}}\)	\(363\)
default	\(\frac {\sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right ) \left (-3 \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) a b \,d^{2} x^{2}+2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, a d \,x^{2}-3 \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) a b c d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, a c -3 \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) b^{2} d +4 \sqrt {\left (a d \,x^{2}+a c +b \right ) \left (d \,x^{2}+c \right )}\, \sqrt {a \,d^{2}}\, b +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, b \right )}{4 a^{2} d \sqrt {\left (a d \,x^{2}+a c +b \right ) \left (d \,x^{2}+c \right )}\, \sqrt {a \,d^{2}}\, \left (a d \,x^{2}+a c +b \right )}\)	\(478\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (84) = 168\).

Time = 0.34 (sec) , antiderivative size = 395, normalized size of antiderivative = 3.95 \[ \int \frac {x}{\left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (a b d x^{2} + a b c + b^{2}\right )} \sqrt {a} \log \left (8 \, a^{2} d^{2} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (2 \, a^{2} c + a b\right )} d x^{2} + 8 \, a b c + b^{2} - 4 \, {\left (2 \, a d^{2} x^{4} + {\left (4 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + b c\right )} \sqrt {a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}\right ) + 4 \, {\left (a^{2} d^{2} x^{4} + a^{2} c^{2} + {\left (2 \, a^{2} c + 3 \, a b\right )} d x^{2} + 3 \, a b c\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{8 \, {\left (a^{4} d^{2} x^{2} + {\left (a^{4} c + a^{3} b\right )} d\right )}}, \frac {3 \, {\left (a b d x^{2} + a b c + b^{2}\right )} \sqrt {-a} \arctan \left (\frac {{\left (2 \, a d x^{2} + 2 \, a c + b\right )} \sqrt {-a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{2 \, {\left (a^{2} d x^{2} + a^{2} c + a b\right )}}\right ) + 2 \, {\left (a^{2} d^{2} x^{4} + a^{2} c^{2} + {\left (2 \, a^{2} c + 3 \, a b\right )} d x^{2} + 3 \, a b c\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{4 \, {\left (a^{4} d^{2} x^{2} + {\left (a^{4} c + a^{3} b\right )} d\right )}}\right ] \]

[In]

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

[Out]

[1/8*(3*(a*b*d*x^2 + a*b*c + b^2)*sqrt(a)*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))) + 4*(
a^2*d^2*x^4 + a^2*c^2 + (2*a^2*c + 3*a*b)*d*x^2 + 3*a*b*c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(a^4*d^2*x^2
+ (a^4*c + a^3*b)*d), 1/4*(3*(a*b*d*x^2 + a*b*c + b^2)*sqrt(-a)*arctan(1/2*(2*a*d*x^2 + 2*a*c + b)*sqrt(-a)*s
qrt((a*d*x^2 + a*c + b)/(d*x^2 + c))/(a^2*d*x^2 + a^2*c + a*b)) + 2*(a^2*d^2*x^4 + a^2*c^2 + (2*a^2*c + 3*a*b)
*d*x^2 + 3*a*b*c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(a^4*d^2*x^2 + (a^4*c + a^3*b)*d)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.61 \[ \int \frac {x}{\left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=\frac {2 \, a b - \frac {3 \, {\left (a d x^{2} + a c + b\right )} b}{d x^{2} + c}}{2 \, {\left (a^{3} d \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} - a^{2} d \left (\frac {a d x^{2} + a c + b}{d x^{2} + c}\right )^{\frac {3}{2}}\right )}} + \frac {3 \, b \log \left (-\frac {\sqrt {a} - \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{\sqrt {a} + \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}\right )}{4 \, a^{\frac {5}{2}} d} \]

[In]

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

[Out]

1/2*(2*a*b - 3*(a*d*x^2 + a*c + b)*b/(d*x^2 + c))/(a^3*d*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)) - a^2*d*((a*d*x
^2 + a*c + b)/(d*x^2 + c))^(3/2)) + 3/4*b*log(-(sqrt(a) - sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(sqrt(a) + sq
rt((a*d*x^2 + a*c + b)/(d*x^2 + c))))/(a^(5/2)*d)

Giac [F(-2)]

Exception generated. \[ \int \frac {x}{\left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x/(a+b/(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:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [B] (verification not implemented)

Time = 18.32 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.61 \[ \int \frac {x}{\left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=\frac {{\left (\frac {a\,\left (d\,x^2+c\right )}{b}+1\right )}^{3/2}\,\left (d\,x^2+c\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{2},\frac {5}{2};\ \frac {7}{2};\ -\frac {a\,\left (d\,x^2+c\right )}{b}\right )}{5\,d\,{\left (a+\frac {b}{d\,x^2+c}\right )}^{3/2}} \]

[In]

int(x/(a + b/(c + d*x^2))^(3/2),x)

[Out]

(((a*(c + d*x^2))/b + 1)^(3/2)*(c + d*x^2)*hypergeom([3/2, 5/2], 7/2, -(a*(c + d*x^2))/b))/(5*d*(a + b/(c + d*
x^2))^(3/2))