3.4.5 \(\int \frac {x^{5/2}}{(a+b x^2)^3} \, dx\)

Optimal. Leaf size=242 \[ \frac {3 \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{5/4} b^{7/4}}-\frac {3 \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{5/4} b^{7/4}}-\frac {3 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{5/4} b^{7/4}}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt {2} a^{5/4} b^{7/4}}+\frac {3 x^{3/2}}{16 a b \left (a+b x^2\right )}-\frac {x^{3/2}}{4 b \left (a+b x^2\right )^2} \]

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Rubi [A]  time = 0.17, antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {288, 290, 329, 297, 1162, 617, 204, 1165, 628} \begin {gather*} \frac {3 \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{5/4} b^{7/4}}-\frac {3 \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{5/4} b^{7/4}}-\frac {3 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{5/4} b^{7/4}}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt {2} a^{5/4} b^{7/4}}+\frac {3 x^{3/2}}{16 a b \left (a+b x^2\right )}-\frac {x^{3/2}}{4 b \left (a+b x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/(a + b*x^2)^3,x]

[Out]

-x^(3/2)/(4*b*(a + b*x^2)^2) + (3*x^(3/2))/(16*a*b*(a + b*x^2)) - (3*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1
/4)])/(32*Sqrt[2]*a^(5/4)*b^(7/4)) + (3*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(5/4)*b^(
7/4)) + (3*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(5/4)*b^(7/4)) - (3*Log[S
qrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(5/4)*b^(7/4))

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 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 290

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

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 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]

Rubi steps

\begin {align*} \int \frac {x^{5/2}}{\left (a+b x^2\right )^3} \, dx &=-\frac {x^{3/2}}{4 b \left (a+b x^2\right )^2}+\frac {3 \int \frac {\sqrt {x}}{\left (a+b x^2\right )^2} \, dx}{8 b}\\ &=-\frac {x^{3/2}}{4 b \left (a+b x^2\right )^2}+\frac {3 x^{3/2}}{16 a b \left (a+b x^2\right )}+\frac {3 \int \frac {\sqrt {x}}{a+b x^2} \, dx}{32 a b}\\ &=-\frac {x^{3/2}}{4 b \left (a+b x^2\right )^2}+\frac {3 x^{3/2}}{16 a 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 )}{16 a b}\\ &=-\frac {x^{3/2}}{4 b \left (a+b x^2\right )^2}+\frac {3 x^{3/2}}{16 a 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 )}{32 a 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 )}{32 a b^{3/2}}\\ &=-\frac {x^{3/2}}{4 b \left (a+b x^2\right )^2}+\frac {3 x^{3/2}}{16 a 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 )}{64 a 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 )}{64 a 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 )}{64 \sqrt {2} a^{5/4} 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 )}{64 \sqrt {2} a^{5/4} b^{7/4}}\\ &=-\frac {x^{3/2}}{4 b \left (a+b x^2\right )^2}+\frac {3 x^{3/2}}{16 a 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 )}{64 \sqrt {2} a^{5/4} b^{7/4}}-\frac {3 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{5/4} 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 )}{32 \sqrt {2} a^{5/4} 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 )}{32 \sqrt {2} a^{5/4} b^{7/4}}\\ &=-\frac {x^{3/2}}{4 b \left (a+b x^2\right )^2}+\frac {3 x^{3/2}}{16 a 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 )}{32 \sqrt {2} a^{5/4} b^{7/4}}+\frac {3 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{5/4} b^{7/4}}+\frac {3 \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{5/4} b^{7/4}}-\frac {3 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{5/4} b^{7/4}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 45, normalized size = 0.19 \begin {gather*} \frac {2 x^{3/2} \left (\frac {\, _2F_1\left (\frac {3}{4},3;\frac {7}{4};-\frac {b x^2}{a}\right )}{a^2}-\frac {1}{\left (a+b x^2\right )^2}\right )}{5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(a + b*x^2)^3,x]

[Out]

(2*x^(3/2)*(-(a + b*x^2)^(-2) + Hypergeometric2F1[3/4, 3, 7/4, -((b*x^2)/a)]/a^2))/(5*b)

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IntegrateAlgebraic [A]  time = 0.45, size = 154, normalized size = 0.64 \begin {gather*} -\frac {3 \tan ^{-1}\left (\frac {\frac {\sqrt [4]{a}}{\sqrt {2} \sqrt [4]{b}}-\frac {\sqrt [4]{b} x}{\sqrt {2} \sqrt [4]{a}}}{\sqrt {x}}\right )}{32 \sqrt {2} a^{5/4} b^{7/4}}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{32 \sqrt {2} a^{5/4} b^{7/4}}+\frac {3 b x^{7/2}-a x^{3/2}}{16 a b \left (a+b x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(5/2)/(a + b*x^2)^3,x]

[Out]

(-(a*x^(3/2)) + 3*b*x^(7/2))/(16*a*b*(a + b*x^2)^2) - (3*ArcTan[(a^(1/4)/(Sqrt[2]*b^(1/4)) - (b^(1/4)*x)/(Sqrt
[2]*a^(1/4)))/Sqrt[x]])/(32*Sqrt[2]*a^(5/4)*b^(7/4)) - (3*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/(Sqrt[a] +
Sqrt[b]*x)])/(32*Sqrt[2]*a^(5/4)*b^(7/4))

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fricas [A]  time = 0.68, size = 260, normalized size = 1.07 \begin {gather*} -\frac {12 \, {\left (a b^{3} x^{4} + 2 \, a^{2} b^{2} x^{2} + a^{3} b\right )} \left (-\frac {1}{a^{5} b^{7}}\right )^{\frac {1}{4}} \arctan \left (\sqrt {-a^{3} b^{3} \sqrt {-\frac {1}{a^{5} b^{7}}} + x} a b^{2} \left (-\frac {1}{a^{5} b^{7}}\right )^{\frac {1}{4}} - a b^{2} \sqrt {x} \left (-\frac {1}{a^{5} b^{7}}\right )^{\frac {1}{4}}\right ) - 3 \, {\left (a b^{3} x^{4} + 2 \, a^{2} b^{2} x^{2} + a^{3} b\right )} \left (-\frac {1}{a^{5} b^{7}}\right )^{\frac {1}{4}} \log \left (a^{4} b^{5} \left (-\frac {1}{a^{5} b^{7}}\right )^{\frac {3}{4}} + \sqrt {x}\right ) + 3 \, {\left (a b^{3} x^{4} + 2 \, a^{2} b^{2} x^{2} + a^{3} b\right )} \left (-\frac {1}{a^{5} b^{7}}\right )^{\frac {1}{4}} \log \left (-a^{4} b^{5} \left (-\frac {1}{a^{5} b^{7}}\right )^{\frac {3}{4}} + \sqrt {x}\right ) - 4 \, {\left (3 \, b x^{3} - a x\right )} \sqrt {x}}{64 \, {\left (a b^{3} x^{4} + 2 \, a^{2} b^{2} x^{2} + a^{3} b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

-1/64*(12*(a*b^3*x^4 + 2*a^2*b^2*x^2 + a^3*b)*(-1/(a^5*b^7))^(1/4)*arctan(sqrt(-a^3*b^3*sqrt(-1/(a^5*b^7)) + x
)*a*b^2*(-1/(a^5*b^7))^(1/4) - a*b^2*sqrt(x)*(-1/(a^5*b^7))^(1/4)) - 3*(a*b^3*x^4 + 2*a^2*b^2*x^2 + a^3*b)*(-1
/(a^5*b^7))^(1/4)*log(a^4*b^5*(-1/(a^5*b^7))^(3/4) + sqrt(x)) + 3*(a*b^3*x^4 + 2*a^2*b^2*x^2 + a^3*b)*(-1/(a^5
*b^7))^(1/4)*log(-a^4*b^5*(-1/(a^5*b^7))^(3/4) + sqrt(x)) - 4*(3*b*x^3 - a*x)*sqrt(x))/(a*b^3*x^4 + 2*a^2*b^2*
x^2 + a^3*b)

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giac [A]  time = 0.66, size = 212, normalized size = 0.88 \begin {gather*} \frac {3 \, b x^{\frac {7}{2}} - a x^{\frac {3}{2}}}{16 \, {\left (b x^{2} + a\right )}^{2} a 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 )}{64 \, a^{2} 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 )}{64 \, a^{2} 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 )}{128 \, a^{2} 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 )}{128 \, a^{2} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(b*x^2+a)^3,x, algorithm="giac")

[Out]

1/16*(3*b*x^(7/2) - a*x^(3/2))/((b*x^2 + a)^2*a*b) + 3/64*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^4) + 3/64*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^4) - 3/128*sqrt(2)*(a*b^3)^(3/4)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x +
sqrt(a/b))/(a^2*b^4) + 3/128*sqrt(2)*(a*b^3)^(3/4)*log(-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^2*b^4)

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maple [A]  time = 0.02, size = 169, normalized size = 0.70 \begin {gather*} \frac {3 \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{64 \left (\frac {a}{b}\right )^{\frac {1}{4}} a \,b^{2}}+\frac {3 \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{64 \left (\frac {a}{b}\right )^{\frac {1}{4}} a \,b^{2}}+\frac {3 \sqrt {2}\, \ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}\right )}{128 \left (\frac {a}{b}\right )^{\frac {1}{4}} a \,b^{2}}+\frac {\frac {3 x^{\frac {7}{2}}}{16 a}-\frac {x^{\frac {3}{2}}}{16 b}}{\left (b \,x^{2}+a \right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/(b*x^2+a)^3,x)

[Out]

2*(3/32/a*x^(7/2)-1/32/b*x^(3/2))/(b*x^2+a)^2+3/128/b^2/a/(a/b)^(1/4)*2^(1/2)*ln((x-(a/b)^(1/4)*2^(1/2)*x^(1/2
)+(a/b)^(1/2))/(x+(a/b)^(1/4)*2^(1/2)*x^(1/2)+(a/b)^(1/2)))+3/64/b^2/a/(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b
)^(1/4)*x^(1/2)+1)+3/64/b^2/a/(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)

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maxima [A]  time = 2.97, size = 222, normalized size = 0.92 \begin {gather*} \frac {3 \, b x^{\frac {7}{2}} - a x^{\frac {3}{2}}}{16 \, {\left (a b^{3} x^{4} + 2 \, a^{2} b^{2} x^{2} + a^{3} b\right )}} + \frac {3 \, {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{128 \, a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

1/16*(3*b*x^(7/2) - a*x^(3/2))/(a*b^3*x^4 + 2*a^2*b^2*x^2 + a^3*b) + 3/128*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt
(2)*a^(1/4)*b^(1/4) + 2*sqrt(b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) + 2*sqrt(2)*ar
ctan(-1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) - 2*sqrt(b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(sqrt(a)*sqrt(b))*
sqrt(b)) - sqrt(2)*log(sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a))/(a^(1/4)*b^(3/4)) + sqrt(2)*log(
-sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a))/(a^(1/4)*b^(3/4)))/(a*b)

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mupad [B]  time = 0.08, size = 85, normalized size = 0.35 \begin {gather*} \frac {\frac {3\,x^{7/2}}{16\,a}-\frac {x^{3/2}}{16\,b}}{a^2+2\,a\,b\,x^2+b^2\,x^4}-\frac {3\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )}{32\,{\left (-a\right )}^{5/4}\,b^{7/4}}+\frac {3\,\mathrm {atanh}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )}{32\,{\left (-a\right )}^{5/4}\,b^{7/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/(a + b*x^2)^3,x)

[Out]

((3*x^(7/2))/(16*a) - x^(3/2)/(16*b))/(a^2 + b^2*x^4 + 2*a*b*x^2) - (3*atan((b^(1/4)*x^(1/2))/(-a)^(1/4)))/(32
*(-a)^(5/4)*b^(7/4)) + (3*atanh((b^(1/4)*x^(1/2))/(-a)^(1/4)))/(32*(-a)^(5/4)*b^(7/4))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)/(b*x**2+a)**3,x)

[Out]

Timed out

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