3.4.10 \(\int \frac {1}{x^{5/2} (a+b x^2)^3} \, dx\) [310]

Optimal. Leaf size=251 \[ -\frac {77}{48 a^3 x^{3/2}}+\frac {1}{4 a x^{3/2} \left (a+b x^2\right )^2}+\frac {11}{16 a^2 x^{3/2} \left (a+b x^2\right )}+\frac {77 b^{3/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{15/4}}-\frac {77 b^{3/4} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{15/4}}+\frac {77 b^{3/4} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{15/4}}-\frac {77 b^{3/4} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{15/4}} \]

[Out]

-77/48/a^3/x^(3/2)+1/4/a/x^(3/2)/(b*x^2+a)^2+11/16/a^2/x^(3/2)/(b*x^2+a)+77/64*b^(3/4)*arctan(1-b^(1/4)*2^(1/2
)*x^(1/2)/a^(1/4))/a^(15/4)*2^(1/2)-77/64*b^(3/4)*arctan(1+b^(1/4)*2^(1/2)*x^(1/2)/a^(1/4))/a^(15/4)*2^(1/2)+7
7/128*b^(3/4)*ln(a^(1/2)+x*b^(1/2)-a^(1/4)*b^(1/4)*2^(1/2)*x^(1/2))/a^(15/4)*2^(1/2)-77/128*b^(3/4)*ln(a^(1/2)
+x*b^(1/2)+a^(1/4)*b^(1/4)*2^(1/2)*x^(1/2))/a^(15/4)*2^(1/2)

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Rubi [A]
time = 0.13, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {296, 331, 335, 217, 1179, 642, 1176, 631, 210} \begin {gather*} \frac {77 b^{3/4} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{15/4}}-\frac {77 b^{3/4} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt {2} a^{15/4}}+\frac {77 b^{3/4} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{15/4}}-\frac {77 b^{3/4} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{15/4}}-\frac {77}{48 a^3 x^{3/2}}+\frac {11}{16 a^2 x^{3/2} \left (a+b x^2\right )}+\frac {1}{4 a x^{3/2} \left (a+b x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/2)*(a + b*x^2)^3),x]

[Out]

-77/(48*a^3*x^(3/2)) + 1/(4*a*x^(3/2)*(a + b*x^2)^2) + 11/(16*a^2*x^(3/2)*(a + b*x^2)) + (77*b^(3/4)*ArcTan[1
- (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(15/4)) - (77*b^(3/4)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])
/a^(1/4)])/(32*Sqrt[2]*a^(15/4)) + (77*b^(3/4)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64
*Sqrt[2]*a^(15/4)) - (77*b^(3/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(15
/4))

Rule 210

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

Rule 217

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

Rule 296

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] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

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 335

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 631

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 642

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 1176

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 1179

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 {1}{x^{5/2} \left (a+b x^2\right )^3} \, dx &=\frac {1}{4 a x^{3/2} \left (a+b x^2\right )^2}+\frac {11 \int \frac {1}{x^{5/2} \left (a+b x^2\right )^2} \, dx}{8 a}\\ &=\frac {1}{4 a x^{3/2} \left (a+b x^2\right )^2}+\frac {11}{16 a^2 x^{3/2} \left (a+b x^2\right )}+\frac {77 \int \frac {1}{x^{5/2} \left (a+b x^2\right )} \, dx}{32 a^2}\\ &=-\frac {77}{48 a^3 x^{3/2}}+\frac {1}{4 a x^{3/2} \left (a+b x^2\right )^2}+\frac {11}{16 a^2 x^{3/2} \left (a+b x^2\right )}-\frac {(77 b) \int \frac {1}{\sqrt {x} \left (a+b x^2\right )} \, dx}{32 a^3}\\ &=-\frac {77}{48 a^3 x^{3/2}}+\frac {1}{4 a x^{3/2} \left (a+b x^2\right )^2}+\frac {11}{16 a^2 x^{3/2} \left (a+b x^2\right )}-\frac {(77 b) \text {Subst}\left (\int \frac {1}{a+b x^4} \, dx,x,\sqrt {x}\right )}{16 a^3}\\ &=-\frac {77}{48 a^3 x^{3/2}}+\frac {1}{4 a x^{3/2} \left (a+b x^2\right )^2}+\frac {11}{16 a^2 x^{3/2} \left (a+b x^2\right )}-\frac {(77 b) \text {Subst}\left (\int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{32 a^{7/2}}-\frac {(77 b) \text {Subst}\left (\int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{32 a^{7/2}}\\ &=-\frac {77}{48 a^3 x^{3/2}}+\frac {1}{4 a x^{3/2} \left (a+b x^2\right )^2}+\frac {11}{16 a^2 x^{3/2} \left (a+b x^2\right )}-\frac {\left (77 \sqrt {b}\right ) \text {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^{7/2}}-\frac {\left (77 \sqrt {b}\right ) \text {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^{7/2}}+\frac {\left (77 b^{3/4}\right ) \text {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^{15/4}}+\frac {\left (77 b^{3/4}\right ) \text {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^{15/4}}\\ &=-\frac {77}{48 a^3 x^{3/2}}+\frac {1}{4 a x^{3/2} \left (a+b x^2\right )^2}+\frac {11}{16 a^2 x^{3/2} \left (a+b x^2\right )}+\frac {77 b^{3/4} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{15/4}}-\frac {77 b^{3/4} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{15/4}}-\frac {\left (77 b^{3/4}\right ) \text {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^{15/4}}+\frac {\left (77 b^{3/4}\right ) \text {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^{15/4}}\\ &=-\frac {77}{48 a^3 x^{3/2}}+\frac {1}{4 a x^{3/2} \left (a+b x^2\right )^2}+\frac {11}{16 a^2 x^{3/2} \left (a+b x^2\right )}+\frac {77 b^{3/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{15/4}}-\frac {77 b^{3/4} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{15/4}}+\frac {77 b^{3/4} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{15/4}}-\frac {77 b^{3/4} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{15/4}}\\ \end {align*}

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Mathematica [A]
time = 0.36, size = 149, normalized size = 0.59 \begin {gather*} \frac {-\frac {4 a^{3/4} \left (32 a^2+121 a b x^2+77 b^2 x^4\right )}{x^{3/2} \left (a+b x^2\right )^2}+231 \sqrt {2} b^{3/4} \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )-231 \sqrt {2} b^{3/4} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{192 a^{15/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/2)*(a + b*x^2)^3),x]

[Out]

((-4*a^(3/4)*(32*a^2 + 121*a*b*x^2 + 77*b^2*x^4))/(x^(3/2)*(a + b*x^2)^2) + 231*Sqrt[2]*b^(3/4)*ArcTan[(Sqrt[a
] - Sqrt[b]*x)/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])] - 231*Sqrt[2]*b^(3/4)*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[
x])/(Sqrt[a] + Sqrt[b]*x)])/(192*a^(15/4))

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Maple [A]
time = 0.11, size = 145, normalized size = 0.58

method	result	size
derivativedivides	\(-\frac {2 b \left (\frac {\frac {15 b \,x^{\frac {5}{2}}}{32}+\frac {19 a \sqrt {x}}{32}}{\left (b \,x^{2}+a \right )^{2}}+\frac {77 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{256 a}\right )}{a^{3}}-\frac {2}{3 a^{3} x^{\frac {3}{2}}}\)	\(145\)
default	\(-\frac {2 b \left (\frac {\frac {15 b \,x^{\frac {5}{2}}}{32}+\frac {19 a \sqrt {x}}{32}}{\left (b \,x^{2}+a \right )^{2}}+\frac {77 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{256 a}\right )}{a^{3}}-\frac {2}{3 a^{3} x^{\frac {3}{2}}}\)	\(145\)
risch	\(-\frac {2}{3 a^{3} x^{\frac {3}{2}}}-\frac {15 b^{2} x^{\frac {5}{2}}}{16 a^{3} \left (b \,x^{2}+a \right )^{2}}-\frac {19 b \sqrt {x}}{16 a^{2} \left (b \,x^{2}+a \right )^{2}}-\frac {77 b \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )}{128 a^{4}}-\frac {77 b \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{64 a^{4}}-\frac {77 b \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{64 a^{4}}\)	\(181\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(5/2)/(b*x^2+a)^3,x,method=_RETURNVERBOSE)

[Out]

-2/a^3*b*((15/32*b*x^(5/2)+19/32*a*x^(1/2))/(b*x^2+a)^2+77/256*(a/b)^(1/4)/a*2^(1/2)*(ln((x+(a/b)^(1/4)*x^(1/2
)*2^(1/2)+(a/b)^(1/2))/(x-(a/b)^(1/4)*x^(1/2)*2^(1/2)+(a/b)^(1/2)))+2*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)+2*
arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)))-2/3/a^3/x^(3/2)

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Maxima [A]
time = 0.49, size = 231, normalized size = 0.92 \begin {gather*} -\frac {77 \, b^{2} x^{4} + 121 \, a b x^{2} + 32 \, a^{2}}{48 \, {\left (a^{3} b^{2} x^{\frac {11}{2}} + 2 \, a^{4} b x^{\frac {7}{2}} + a^{5} x^{\frac {3}{2}}\right )}} - \frac {77 \, {\left (\frac {2 \, \sqrt {2} b \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 {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {2 \, \sqrt {2} b \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 {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {\sqrt {2} b^{\frac {3}{4}} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {3}{4}}} - \frac {\sqrt {2} b^{\frac {3}{4}} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {3}{4}}}\right )}}{128 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

-1/48*(77*b^2*x^4 + 121*a*b*x^2 + 32*a^2)/(a^3*b^2*x^(11/2) + 2*a^4*b*x^(7/2) + a^5*x^(3/2)) - 77/128*(2*sqrt(
2)*b*arctan(1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) + 2*sqrt(b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(a)*sqrt(sqr
t(a)*sqrt(b))) + 2*sqrt(2)*b*arctan(-1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) - 2*sqrt(b)*sqrt(x))/sqrt(sqrt(a)*sq
rt(b)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b))) + sqrt(2)*b^(3/4)*log(sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sq
rt(a))/a^(3/4) - sqrt(2)*b^(3/4)*log(-sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a))/a^(3/4))/a^3

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Fricas [A]
time = 1.71, size = 283, normalized size = 1.13 \begin {gather*} -\frac {924 \, {\left (a^{3} b^{2} x^{6} + 2 \, a^{4} b x^{4} + a^{5} x^{2}\right )} \left (-\frac {b^{3}}{a^{15}}\right )^{\frac {1}{4}} \arctan \left (-\frac {a^{11} b \sqrt {x} \left (-\frac {b^{3}}{a^{15}}\right )^{\frac {3}{4}} - \sqrt {a^{8} \sqrt {-\frac {b^{3}}{a^{15}}} + b^{2} x} a^{11} \left (-\frac {b^{3}}{a^{15}}\right )^{\frac {3}{4}}}{b^{3}}\right ) + 231 \, {\left (a^{3} b^{2} x^{6} + 2 \, a^{4} b x^{4} + a^{5} x^{2}\right )} \left (-\frac {b^{3}}{a^{15}}\right )^{\frac {1}{4}} \log \left (77 \, a^{4} \left (-\frac {b^{3}}{a^{15}}\right )^{\frac {1}{4}} + 77 \, b \sqrt {x}\right ) - 231 \, {\left (a^{3} b^{2} x^{6} + 2 \, a^{4} b x^{4} + a^{5} x^{2}\right )} \left (-\frac {b^{3}}{a^{15}}\right )^{\frac {1}{4}} \log \left (-77 \, a^{4} \left (-\frac {b^{3}}{a^{15}}\right )^{\frac {1}{4}} + 77 \, b \sqrt {x}\right ) + 4 \, {\left (77 \, b^{2} x^{4} + 121 \, a b x^{2} + 32 \, a^{2}\right )} \sqrt {x}}{192 \, {\left (a^{3} b^{2} x^{6} + 2 \, a^{4} b x^{4} + a^{5} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

-1/192*(924*(a^3*b^2*x^6 + 2*a^4*b*x^4 + a^5*x^2)*(-b^3/a^15)^(1/4)*arctan(-(a^11*b*sqrt(x)*(-b^3/a^15)^(3/4)
- sqrt(a^8*sqrt(-b^3/a^15) + b^2*x)*a^11*(-b^3/a^15)^(3/4))/b^3) + 231*(a^3*b^2*x^6 + 2*a^4*b*x^4 + a^5*x^2)*(
-b^3/a^15)^(1/4)*log(77*a^4*(-b^3/a^15)^(1/4) + 77*b*sqrt(x)) - 231*(a^3*b^2*x^6 + 2*a^4*b*x^4 + a^5*x^2)*(-b^
3/a^15)^(1/4)*log(-77*a^4*(-b^3/a^15)^(1/4) + 77*b*sqrt(x)) + 4*(77*b^2*x^4 + 121*a*b*x^2 + 32*a^2)*sqrt(x))/(
a^3*b^2*x^6 + 2*a^4*b*x^4 + a^5*x^2)

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Sympy [F(-1)] Timed out
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(1/x**(5/2)/(b*x**2+a)**3,x)

[Out]

Timed out

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Giac [A]
time = 0.82, size = 208, normalized size = 0.83 \begin {gather*} -\frac {77 \, \sqrt {2} \left (a b^{3}\right )^{\frac {1}{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^{4}} - \frac {77 \, \sqrt {2} \left (a b^{3}\right )^{\frac {1}{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^{4}} - \frac {77 \, \sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{128 \, a^{4}} + \frac {77 \, \sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{128 \, a^{4}} - \frac {15 \, b^{2} x^{\frac {5}{2}} + 19 \, a b \sqrt {x}}{16 \, {\left (b x^{2} + a\right )}^{2} a^{3}} - \frac {2}{3 \, a^{3} x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(b*x^2+a)^3,x, algorithm="giac")

[Out]

-77/64*sqrt(2)*(a*b^3)^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) + 2*sqrt(x))/(a/b)^(1/4))/a^4 - 77/64*sqr
t(2)*(a*b^3)^(1/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) - 2*sqrt(x))/(a/b)^(1/4))/a^4 - 77/128*sqrt(2)*(a*
b^3)^(1/4)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/a^4 + 77/128*sqrt(2)*(a*b^3)^(1/4)*log(-sqrt(2)*sq
rt(x)*(a/b)^(1/4) + x + sqrt(a/b))/a^4 - 1/16*(15*b^2*x^(5/2) + 19*a*b*sqrt(x))/((b*x^2 + a)^2*a^3) - 2/3/(a^3
*x^(3/2))

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Mupad [B]
time = 4.66, size = 99, normalized size = 0.39 \begin {gather*} \frac {77\,{\left (-b\right )}^{3/4}\,\mathrm {atan}\left (\frac {{\left (-b\right )}^{1/4}\,\sqrt {x}}{a^{1/4}}\right )}{32\,a^{15/4}}-\frac {\frac {2}{3\,a}+\frac {121\,b\,x^2}{48\,a^2}+\frac {77\,b^2\,x^4}{48\,a^3}}{a^2\,x^{3/2}+b^2\,x^{11/2}+2\,a\,b\,x^{7/2}}+\frac {77\,{\left (-b\right )}^{3/4}\,\mathrm {atanh}\left (\frac {{\left (-b\right )}^{1/4}\,\sqrt {x}}{a^{1/4}}\right )}{32\,a^{15/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(5/2)*(a + b*x^2)^3),x)

[Out]

(77*(-b)^(3/4)*atan(((-b)^(1/4)*x^(1/2))/a^(1/4)))/(32*a^(15/4)) - (2/(3*a) + (121*b*x^2)/(48*a^2) + (77*b^2*x
^4)/(48*a^3))/(a^2*x^(3/2) + b^2*x^(11/2) + 2*a*b*x^(7/2)) + (77*(-b)^(3/4)*atanh(((-b)^(1/4)*x^(1/2))/a^(1/4)
))/(32*a^(15/4))

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