3.297 \(\int \frac{x^{5/2}}{(a+b x^2)^2} \, dx\)

Optimal. Leaf size=218 \[ \frac{3 \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} \sqrt [4]{a} b^{7/4}}-\frac{3 \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} \sqrt [4]{a} b^{7/4}}-\frac{3 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} \sqrt [4]{a} b^{7/4}}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt{2} \sqrt [4]{a} b^{7/4}}-\frac{x^{3/2}}{2 b \left (a+b x^2\right )} \]

[Out]

-x^(3/2)/(2*b*(a + b*x^2)) - (3*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(1/4)*b^(7/4)) + (
3*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(1/4)*b^(7/4)) + (3*Log[Sqrt[a] - Sqrt[2]*a^(1/4
)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(1/4)*b^(7/4)) - (3*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x]
+ Sqrt[b]*x])/(8*Sqrt[2]*a^(1/4)*b^(7/4))

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Rubi [A]  time = 0.148484, antiderivative size = 218, 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 = {288, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac{3 \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} \sqrt [4]{a} b^{7/4}}-\frac{3 \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} \sqrt [4]{a} b^{7/4}}-\frac{3 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} \sqrt [4]{a} b^{7/4}}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt{2} \sqrt [4]{a} b^{7/4}}-\frac{x^{3/2}}{2 b \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/(a + b*x^2)^2,x]

[Out]

-x^(3/2)/(2*b*(a + b*x^2)) - (3*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(1/4)*b^(7/4)) + (
3*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(1/4)*b^(7/4)) + (3*Log[Sqrt[a] - Sqrt[2]*a^(1/4
)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(1/4)*b^(7/4)) - (3*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x]
+ Sqrt[b]*x])/(8*Sqrt[2]*a^(1/4)*b^(7/4))

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& 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{x^{5/2}}{\left (a+b x^2\right )^2} \, dx &=-\frac{x^{3/2}}{2 b \left (a+b x^2\right )}+\frac{3 \int \frac{\sqrt{x}}{a+b x^2} \, dx}{4 b}\\ &=-\frac{x^{3/2}}{2 b \left (a+b x^2\right )}+\frac{3 \operatorname{Subst}\left (\int \frac{x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{2 b}\\ &=-\frac{x^{3/2}}{2 b \left (a+b x^2\right )}-\frac{3 \operatorname{Subst}\left (\int \frac{\sqrt{a}-\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{4 b^{3/2}}+\frac{3 \operatorname{Subst}\left (\int \frac{\sqrt{a}+\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{4 b^{3/2}}\\ &=-\frac{x^{3/2}}{2 b \left (a+b x^2\right )}+\frac{3 \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 )}{8 b^2}+\frac{3 \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 )}{8 b^2}+\frac{3 \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 )}{8 \sqrt{2} \sqrt [4]{a} b^{7/4}}+\frac{3 \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 )}{8 \sqrt{2} \sqrt [4]{a} b^{7/4}}\\ &=-\frac{x^{3/2}}{2 b \left (a+b x^2\right )}+\frac{3 \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{8 \sqrt{2} \sqrt [4]{a} b^{7/4}}-\frac{3 \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{8 \sqrt{2} \sqrt [4]{a} b^{7/4}}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} \sqrt [4]{a} b^{7/4}}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} \sqrt [4]{a} b^{7/4}}\\ &=-\frac{x^{3/2}}{2 b \left (a+b x^2\right )}-\frac{3 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} \sqrt [4]{a} b^{7/4}}+\frac{3 \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} \sqrt [4]{a} b^{7/4}}+\frac{3 \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{8 \sqrt{2} \sqrt [4]{a} b^{7/4}}-\frac{3 \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{8 \sqrt{2} \sqrt [4]{a} b^{7/4}}\\ \end{align*}

Mathematica [C]  time = 0.0124453, size = 43, normalized size = 0.2 \[ \frac{2 x^{3/2} \left (\frac{\, _2F_1\left (\frac{3}{4},2;\frac{7}{4};-\frac{b x^2}{a}\right )}{a}-\frac{1}{a+b x^2}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(a + b*x^2)^2,x]

[Out]

(2*x^(3/2)*(-(a + b*x^2)^(-1) + Hypergeometric2F1[3/4, 2, 7/4, -((b*x^2)/a)]/a))/b

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Maple [A]  time = 0.009, size = 149, normalized size = 0.7 \begin{align*} -{\frac{1}{2\,b \left ( b{x}^{2}+a \right ) }{x}^{{\frac{3}{2}}}}+{\frac{3\,\sqrt{2}}{16\,{b}^{2}}\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{3\,\sqrt{2}}{8\,{b}^{2}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+{\frac{3\,\sqrt{2}}{8\,{b}^{2}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/(b*x^2+a)^2,x)

[Out]

-1/2*x^(3/2)/b/(b*x^2+a)+3/16/b^2/(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)))+3/8/b^2/(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/
2)+1)+3/8/b^2/(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)-1)

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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^(5/2)/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.37307, size = 447, normalized size = 2.05 \begin{align*} -\frac{12 \,{\left (b^{2} x^{2} + a b\right )} \left (-\frac{1}{a b^{7}}\right )^{\frac{1}{4}} \arctan \left (\sqrt{-a b^{3} \sqrt{-\frac{1}{a b^{7}}} + x} b^{2} \left (-\frac{1}{a b^{7}}\right )^{\frac{1}{4}} - b^{2} \sqrt{x} \left (-\frac{1}{a b^{7}}\right )^{\frac{1}{4}}\right ) - 3 \,{\left (b^{2} x^{2} + a b\right )} \left (-\frac{1}{a b^{7}}\right )^{\frac{1}{4}} \log \left (a b^{5} \left (-\frac{1}{a b^{7}}\right )^{\frac{3}{4}} + \sqrt{x}\right ) + 3 \,{\left (b^{2} x^{2} + a b\right )} \left (-\frac{1}{a b^{7}}\right )^{\frac{1}{4}} \log \left (-a b^{5} \left (-\frac{1}{a b^{7}}\right )^{\frac{3}{4}} + \sqrt{x}\right ) + 4 \, x^{\frac{3}{2}}}{8 \,{\left (b^{2} x^{2} + a b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

-1/8*(12*(b^2*x^2 + a*b)*(-1/(a*b^7))^(1/4)*arctan(sqrt(-a*b^3*sqrt(-1/(a*b^7)) + x)*b^2*(-1/(a*b^7))^(1/4) -
b^2*sqrt(x)*(-1/(a*b^7))^(1/4)) - 3*(b^2*x^2 + a*b)*(-1/(a*b^7))^(1/4)*log(a*b^5*(-1/(a*b^7))^(3/4) + sqrt(x))
+ 3*(b^2*x^2 + a*b)*(-1/(a*b^7))^(1/4)*log(-a*b^5*(-1/(a*b^7))^(3/4) + sqrt(x)) + 4*x^(3/2))/(b^2*x^2 + a*b)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)/(b*x**2+a)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.75973, size = 269, normalized size = 1.23 \begin{align*} -\frac{x^{\frac{3}{2}}}{2 \,{\left (b x^{2} + a\right )} b} + \frac{3 \, \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 )}{8 \, a b^{4}} + \frac{3 \, \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 )}{8 \, a b^{4}} - \frac{3 \, \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 )}{16 \, a b^{4}} + \frac{3 \, \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 )}{16 \, a b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(b*x^2+a)^2,x, algorithm="giac")

[Out]

-1/2*x^(3/2)/((b*x^2 + a)*b) + 3/8*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*b^4) + 3/8*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*b^4) - 3/16*sqrt(2)*(a*b^3)^(3/4)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a*b^4) + 3/16*sqr
t(2)*(a*b^3)^(3/4)*log(-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a*b^4)