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

Optimal. Leaf size=100 \[ -\frac {3 b \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{c+d x^2}}}{\sqrt {a}}\right )}{2 a^{5/2} d}+\frac {3 b}{2 a^2 d \sqrt {a+\frac {b}{c+d x^2}}}+\frac {c+d x^2}{2 a d \sqrt {a+\frac {b}{c+d x^2}}} \]

[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)

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Rubi [A]  time = 0.07, antiderivative size = 104, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1591, 242, 51, 63, 208} \[ \frac {3 \left (c+d x^2\right ) \sqrt {a+\frac {b}{c+d x^2}}}{2 a^2 d}-\frac {3 b \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{c+d x^2}}}{\sqrt {a}}\right )}{2 a^{5/2} d}-\frac {c+d x^2}{a d \sqrt {a+\frac {b}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b/(c + d*x^2))^(3/2),x]

[Out]

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

Rule 51

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 63

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 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 242

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 1591

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*} \int \frac {x}{\left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx,x,c+d x^2\right )}{2 d}\\ &=-\frac {\operatorname {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}{a d \sqrt {a+\frac {b}{c+d x^2}}}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,\frac {1}{c+d x^2}\right )}{2 a d}\\ &=-\frac {c+d x^2}{a d \sqrt {a+\frac {b}{c+d x^2}}}+\frac {3 \left (c+d x^2\right ) \sqrt {a+\frac {b}{c+d x^2}}}{2 a^2 d}+\frac {(3 b) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{c+d x^2}\right )}{4 a^2 d}\\ &=-\frac {c+d x^2}{a d \sqrt {a+\frac {b}{c+d x^2}}}+\frac {3 \left (c+d x^2\right ) \sqrt {a+\frac {b}{c+d x^2}}}{2 a^2 d}+\frac {3 \operatorname {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 {c+d x^2}{a d \sqrt {a+\frac {b}{c+d x^2}}}+\frac {3 \left (c+d x^2\right ) \sqrt {a+\frac {b}{c+d x^2}}}{2 a^2 d}-\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*}

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Mathematica [C]  time = 0.06, size = 50, normalized size = 0.50 \[ \frac {b \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};\frac {a+\frac {b}{d x^2+c}}{a}\right )}{a^2 d \sqrt {a+\frac {b}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Hypergeometric2F1[-1/2, 2, 1/2, (a + b/(c + d*x^2))/a])/(a^2*d*Sqrt[a + b/(c + d*x^2)])

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fricas [B]  time = 0.87, size = 395, normalized size = 3.95 \[ \left [\frac {3 \, {\left (a b d x^{2} + a b c + b^{2}\right )} \sqrt {a} \log \left (8 \, a^{2} d^{2} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (2 \, a^{2} c + a b\right )} d x^{2} + 8 \, a b c + b^{2} - 4 \, {\left (2 \, a d^{2} x^{4} + {\left (4 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + b c\right )} \sqrt {a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}\right ) + 4 \, {\left (a^{2} d^{2} x^{4} + a^{2} c^{2} + {\left (2 \, a^{2} c + 3 \, a b\right )} d x^{2} + 3 \, a b c\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{8 \, {\left (a^{4} d^{2} x^{2} + {\left (a^{4} c + a^{3} b\right )} d\right )}}, \frac {3 \, {\left (a b d x^{2} + a b c + b^{2}\right )} \sqrt {-a} \arctan \left (\frac {{\left (2 \, a d x^{2} + 2 \, a c + b\right )} \sqrt {-a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{2 \, {\left (a^{2} d x^{2} + a^{2} c + a b\right )}}\right ) + 2 \, {\left (a^{2} d^{2} x^{4} + a^{2} c^{2} + {\left (2 \, a^{2} c + 3 \, a b\right )} d x^{2} + 3 \, a b c\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{4 \, {\left (a^{4} d^{2} x^{2} + {\left (a^{4} c + a^{3} b\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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giac [B]  time = 1.55, size = 449, normalized size = 4.49 \[ \frac {b \log \left ({\left | -2 \, a^{\frac {7}{2}} c^{3} d - 6 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} a^{3} c^{2} {\left | d \right |} - 6 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )}^{2} a^{\frac {5}{2}} c d - 5 \, a^{\frac {5}{2}} b c^{2} d - 2 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )}^{3} a^{2} {\left | d \right |} - 10 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} a^{2} b c {\left | d \right |} - 5 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )}^{2} a^{\frac {3}{2}} b d - 4 \, a^{\frac {3}{2}} b^{2} c d - 4 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} a b^{2} {\left | d \right |} - \sqrt {a} b^{3} d \right |}\right )}{4 \, a^{\frac {5}{2}} {\left | d \right |} \mathrm {sgn}\left (d x^{2} + c\right )} + \frac {\sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}}{2 \, a^{2} d \mathrm {sgn}\left (d x^{2} + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*log(abs(-2*a^(7/2)*c^3*d - 6*(sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c))*a
^3*c^2*abs(d) - 6*(sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c))^2*a^(5/2)*c*d - 5*
a^(5/2)*b*c^2*d - 2*(sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c))^3*a^2*abs(d) - 1
0*(sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c))*a^2*b*c*abs(d) - 5*(sqrt(a*d^2)*x^
2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c))^2*a^(3/2)*b*d - 4*a^(3/2)*b^2*c*d - 4*(sqrt(a*d^2)*
x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c))*a*b^2*abs(d) - sqrt(a)*b^3*d))/(a^(5/2)*abs(d)*sg
n(d*x^2 + c)) + 1/2*sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c)/(a^2*d*sgn(d*x^2 + c))

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maple [B]  time = 0.05, size = 478, normalized size = 4.78 \[ \frac {\sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right ) \left (-3 a b \,d^{2} x^{2} \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +b d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}}{2 \sqrt {a \,d^{2}}}\right )-3 a b c d \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +b d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}}{2 \sqrt {a \,d^{2}}}\right )+2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, a d \,x^{2}-3 b^{2} d \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +b d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}}{2 \sqrt {a \,d^{2}}}\right )+2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, a c +4 \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \sqrt {a \,d^{2}}\, b +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, b \right )}{4 \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \sqrt {a \,d^{2}}\, \left (a d \,x^{2}+a c +b \right ) a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.22, size = 161, normalized size = 1.61 \[ \frac {2 \, a b - \frac {3 \, {\left (a d x^{2} + a c + b\right )} b}{d x^{2} + c}}{2 \, {\left (a^{3} d \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} - a^{2} d \left (\frac {a d x^{2} + a c + b}{d x^{2} + c}\right )^{\frac {3}{2}}\right )}} + \frac {3 \, b \log \left (-\frac {\sqrt {a} - \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{\sqrt {a} + \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}\right )}{4 \, a^{\frac {5}{2}} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 3.93, size = 61, normalized size = 0.61 \[ \frac {{\left (\frac {a\,\left (d\,x^2+c\right )}{b}+1\right )}^{3/2}\,\left (d\,x^2+c\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{2},\frac {5}{2};\ \frac {7}{2};\ -\frac {a\,\left (d\,x^2+c\right )}{b}\right )}{5\,d\,{\left (a+\frac {b}{d\,x^2+c}\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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)

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