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

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

[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)

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Rubi [A]  time = 0.0733392, 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 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]

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 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]

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*}

Mathematica [C]  time = 0.0538379, size = 50, normalized size = 0.5 \[ \frac{b \, _2F_1\left (-\frac{1}{2},2;\frac{1}{2};\frac{a+\frac{b}{d x^2+c}}{a}\right )}{a^2 d \sqrt{a+\frac{b}{c+d x^2}}} \]

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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Maple [B]  time = 0.014, size = 478, normalized size = 4.8 \begin{align*}{\frac{d{x}^{2}+c}{4\,{a}^{2}d \left ( ad{x}^{2}+ac+b \right ) }\sqrt{{\frac{ad{x}^{2}+ac+b}{d{x}^{2}+c}}} \left ( -3\,\ln \left ( 1/2\,{\frac{2\,a{d}^{2}{x}^{2}+2\,acd+2\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}\sqrt{a{d}^{2}}+bd}{\sqrt{a{d}^{2}}}} \right ){x}^{2}ab{d}^{2}+2\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}\sqrt{a{d}^{2}}{x}^{2}ad-3\,\ln \left ( 1/2\,{\frac{2\,a{d}^{2}{x}^{2}+2\,acd+2\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}\sqrt{a{d}^{2}}+bd}{\sqrt{a{d}^{2}}}} \right ) abcd+2\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}\sqrt{a{d}^{2}}ac-3\,\ln \left ( 1/2\,{\frac{2\,a{d}^{2}{x}^{2}+2\,acd+2\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}\sqrt{a{d}^{2}}+bd}{\sqrt{a{d}^{2}}}} \right ){b}^{2}d+2\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}\sqrt{a{d}^{2}}b+4\,\sqrt{a{d}^{2}}\sqrt{ \left ( d{x}^{2}+c \right ) \left ( ad{x}^{2}+ac+b \right ) }b \right ){\frac{1}{\sqrt{ \left ( d{x}^{2}+c \right ) \left ( ad{x}^{2}+ac+b \right ) }}}{\frac{1}{\sqrt{a{d}^{2}}}}} \end{align*}

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*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))*x^2*a*b*d^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)*x^2*a*d-3*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*b*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)*
a*c-3*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))*b^2*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+4*(a*d^2)^(1/2)*((d*x^2+c)*(a*
d*x^2+a*c+b))^(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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [B]  time = 2.13275, size = 871, normalized size = 8.71 \begin{align*} \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 ] \end{align*}

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

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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Giac [B]  time = 9.17871, size = 657, normalized size = 6.57 \begin{align*} \frac{b{\left | d \right |} \log \left (12 \, a^{4}{\left | d \right |}{\left | \mathrm{sgn}\left (d x^{2} + c\right ) \right |}\right )}{2 \, a^{\frac{5}{2}} d^{2} \mathrm{sgn}\left (d x^{2} + c\right )} + \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 )} \end{align*}

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]

1/2*b*abs(d)*log(12*a^4*abs(d)*abs(sgn(d*x^2 + c)))/(a^(5/2)*d^2*sgn(d*x^2 + c)) + 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) - 10*(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)*sgn(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))