3.293 \(\int \frac{1}{x^{3/2} (a+b x^2)} \, dx\)

Optimal. Leaf size=202 \[ -\frac{\sqrt [4]{b} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} a^{5/4}}+\frac{\sqrt [4]{b} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} a^{5/4}}+\frac{\sqrt [4]{b} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} a^{5/4}}-\frac{\sqrt [4]{b} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{\sqrt{2} a^{5/4}}-\frac{2}{a \sqrt{x}} \]

[Out]

-2/(a*Sqrt[x]) + (b^(1/4)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2]*a^(5/4)) - (b^(1/4)*ArcTan[1
+ (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2]*a^(5/4)) - (b^(1/4)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt
[x] + Sqrt[b]*x])/(2*Sqrt[2]*a^(5/4)) + (b^(1/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(
2*Sqrt[2]*a^(5/4))

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Rubi [A]  time = 0.166747, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.533, Rules used = {325, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac{\sqrt [4]{b} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} a^{5/4}}+\frac{\sqrt [4]{b} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} a^{5/4}}+\frac{\sqrt [4]{b} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} a^{5/4}}-\frac{\sqrt [4]{b} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{\sqrt{2} a^{5/4}}-\frac{2}{a \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(3/2)*(a + b*x^2)),x]

[Out]

-2/(a*Sqrt[x]) + (b^(1/4)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2]*a^(5/4)) - (b^(1/4)*ArcTan[1
+ (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2]*a^(5/4)) - (b^(1/4)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt
[x] + Sqrt[b]*x])/(2*Sqrt[2]*a^(5/4)) + (b^(1/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(
2*Sqrt[2]*a^(5/4))

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{1}{x^{3/2} \left (a+b x^2\right )} \, dx &=-\frac{2}{a \sqrt{x}}-\frac{b \int \frac{\sqrt{x}}{a+b x^2} \, dx}{a}\\ &=-\frac{2}{a \sqrt{x}}-\frac{(2 b) \operatorname{Subst}\left (\int \frac{x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{a}\\ &=-\frac{2}{a \sqrt{x}}+\frac{\sqrt{b} \operatorname{Subst}\left (\int \frac{\sqrt{a}-\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{a}-\frac{\sqrt{b} \operatorname{Subst}\left (\int \frac{\sqrt{a}+\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{a}\\ &=-\frac{2}{a \sqrt{x}}-\frac{\operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt{a}}{\sqrt{b}}-\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\sqrt{x}\right )}{2 a}-\frac{\operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt{a}}{\sqrt{b}}+\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\sqrt{x}\right )}{2 a}-\frac{\sqrt [4]{b} \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [4]{a}}{\sqrt [4]{b}}+2 x}{-\frac{\sqrt{a}}{\sqrt{b}}-\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\sqrt{x}\right )}{2 \sqrt{2} a^{5/4}}-\frac{\sqrt [4]{b} \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [4]{a}}{\sqrt [4]{b}}-2 x}{-\frac{\sqrt{a}}{\sqrt{b}}+\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\sqrt{x}\right )}{2 \sqrt{2} a^{5/4}}\\ &=-\frac{2}{a \sqrt{x}}-\frac{\sqrt [4]{b} \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{2 \sqrt{2} a^{5/4}}+\frac{\sqrt [4]{b} \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{2 \sqrt{2} a^{5/4}}-\frac{\sqrt [4]{b} \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} a^{5/4}}+\frac{\sqrt [4]{b} \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} a^{5/4}}\\ &=-\frac{2}{a \sqrt{x}}+\frac{\sqrt [4]{b} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} a^{5/4}}-\frac{\sqrt [4]{b} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} a^{5/4}}-\frac{\sqrt [4]{b} \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{2 \sqrt{2} a^{5/4}}+\frac{\sqrt [4]{b} \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{2 \sqrt{2} a^{5/4}}\\ \end{align*}

Mathematica [C]  time = 0.0053555, size = 27, normalized size = 0.13 \[ -\frac{2 \, _2F_1\left (-\frac{1}{4},1;\frac{3}{4};-\frac{b x^2}{a}\right )}{a \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(3/2)*(a + b*x^2)),x]

[Out]

(-2*Hypergeometric2F1[-1/4, 1, 3/4, -((b*x^2)/a)])/(a*Sqrt[x])

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Maple [A]  time = 0.007, size = 140, normalized size = 0.7 \begin{align*} -{\frac{\sqrt{2}}{4\,a}\ln \left ({ \left ( x-\sqrt [4]{{\frac{a}{b}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) \left ( x+\sqrt [4]{{\frac{a}{b}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-{\frac{\sqrt{2}}{2\,a}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-{\frac{\sqrt{2}}{2\,a}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-2\,{\frac{1}{a\sqrt{x}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(3/2)/(b*x^2+a),x)

[Out]

-1/4/a/(1/b*a)^(1/4)*2^(1/2)*ln((x-(1/b*a)^(1/4)*x^(1/2)*2^(1/2)+(1/b*a)^(1/2))/(x+(1/b*a)^(1/4)*x^(1/2)*2^(1/
2)+(1/b*a)^(1/2)))-1/2/a/(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)+1)-1/2/a/(1/b*a)^(1/4)*2^(
1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)-1)-2/a/x^(1/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.3265, size = 343, normalized size = 1.7 \begin{align*} \frac{4 \, a x \left (-\frac{b}{a^{5}}\right )^{\frac{1}{4}} \arctan \left (-\frac{a b \sqrt{x} \left (-\frac{b}{a^{5}}\right )^{\frac{1}{4}} - \sqrt{-a^{3} b \sqrt{-\frac{b}{a^{5}}} + b^{2} x} a \left (-\frac{b}{a^{5}}\right )^{\frac{1}{4}}}{b}\right ) - a x \left (-\frac{b}{a^{5}}\right )^{\frac{1}{4}} \log \left (a^{4} \left (-\frac{b}{a^{5}}\right )^{\frac{3}{4}} + b \sqrt{x}\right ) + a x \left (-\frac{b}{a^{5}}\right )^{\frac{1}{4}} \log \left (-a^{4} \left (-\frac{b}{a^{5}}\right )^{\frac{3}{4}} + b \sqrt{x}\right ) - 4 \, \sqrt{x}}{2 \, a x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(3/2)/(b*x^2+a),x, algorithm="fricas")

[Out]

1/2*(4*a*x*(-b/a^5)^(1/4)*arctan(-(a*b*sqrt(x)*(-b/a^5)^(1/4) - sqrt(-a^3*b*sqrt(-b/a^5) + b^2*x)*a*(-b/a^5)^(
1/4))/b) - a*x*(-b/a^5)^(1/4)*log(a^4*(-b/a^5)^(3/4) + b*sqrt(x)) + a*x*(-b/a^5)^(1/4)*log(-a^4*(-b/a^5)^(3/4)
+ b*sqrt(x)) - 4*sqrt(x))/(a*x)

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Sympy [A]  time = 17.5696, size = 180, normalized size = 0.89 \begin{align*} \begin{cases} \frac{\tilde{\infty }}{x^{\frac{5}{2}}} & \text{for}\: a = 0 \wedge b = 0 \\- \frac{2}{5 b x^{\frac{5}{2}}} & \text{for}\: a = 0 \\- \frac{2}{a \sqrt{x}} & \text{for}\: b = 0 \\- \frac{2}{a \sqrt{x}} + \frac{\left (-1\right )^{\frac{3}{4}} b^{3} \left (\frac{1}{b}\right )^{\frac{11}{4}} \log{\left (- \sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac{1}{b}} + \sqrt{x} \right )}}{2 a^{\frac{5}{4}}} - \frac{\left (-1\right )^{\frac{3}{4}} b^{3} \left (\frac{1}{b}\right )^{\frac{11}{4}} \log{\left (\sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac{1}{b}} + \sqrt{x} \right )}}{2 a^{\frac{5}{4}}} - \frac{\left (-1\right )^{\frac{3}{4}} b^{3} \left (\frac{1}{b}\right )^{\frac{11}{4}} \operatorname{atan}{\left (\frac{\left (-1\right )^{\frac{3}{4}} \sqrt{x}}{\sqrt [4]{a} \sqrt [4]{\frac{1}{b}}} \right )}}{a^{\frac{5}{4}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(3/2)/(b*x**2+a),x)

[Out]

Piecewise((zoo/x**(5/2), Eq(a, 0) & Eq(b, 0)), (-2/(5*b*x**(5/2)), Eq(a, 0)), (-2/(a*sqrt(x)), Eq(b, 0)), (-2/
(a*sqrt(x)) + (-1)**(3/4)*b**3*(1/b)**(11/4)*log(-(-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x))/(2*a**(5/4)) -
(-1)**(3/4)*b**3*(1/b)**(11/4)*log((-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x))/(2*a**(5/4)) - (-1)**(3/4)*b**
3*(1/b)**(11/4)*atan((-1)**(3/4)*sqrt(x)/(a**(1/4)*(1/b)**(1/4)))/a**(5/4), True))

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Giac [A]  time = 3.06703, size = 257, normalized size = 1.27 \begin{align*} -\frac{2}{a \sqrt{x}} - \frac{\sqrt{2} \left (a b^{3}\right )^{\frac{3}{4}} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{a}{b}\right )^{\frac{1}{4}} + 2 \, \sqrt{x}\right )}}{2 \, \left (\frac{a}{b}\right )^{\frac{1}{4}}}\right )}{2 \, a^{2} b^{2}} - \frac{\sqrt{2} \left (a b^{3}\right )^{\frac{3}{4}} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{a}{b}\right )^{\frac{1}{4}} - 2 \, \sqrt{x}\right )}}{2 \, \left (\frac{a}{b}\right )^{\frac{1}{4}}}\right )}{2 \, a^{2} b^{2}} + \frac{\sqrt{2} \left (a b^{3}\right )^{\frac{3}{4}} \log \left (\sqrt{2} \sqrt{x} \left (\frac{a}{b}\right )^{\frac{1}{4}} + x + \sqrt{\frac{a}{b}}\right )}{4 \, a^{2} b^{2}} - \frac{\sqrt{2} \left (a b^{3}\right )^{\frac{3}{4}} \log \left (-\sqrt{2} \sqrt{x} \left (\frac{a}{b}\right )^{\frac{1}{4}} + x + \sqrt{\frac{a}{b}}\right )}{4 \, a^{2} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(3/2)/(b*x^2+a),x, algorithm="giac")

[Out]

-2/(a*sqrt(x)) - 1/2*sqrt(2)*(a*b^3)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) + 2*sqrt(x))/(a/b)^(1/4))/(
a^2*b^2) - 1/2*sqrt(2)*(a*b^3)^(3/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) - 2*sqrt(x))/(a/b)^(1/4))/(a^2*b
^2) + 1/4*sqrt(2)*(a*b^3)^(3/4)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^2*b^2) - 1/4*sqrt(2)*(a*b^
3)^(3/4)*log(-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^2*b^2)