3.4.4 \(\int \frac {x^{7/2}}{(a+b x^2)^3} \, dx\) [304]

Optimal. Leaf size=239 \[ -\frac {x^{5/2}}{4 b \left (a+b x^2\right )^2}-\frac {5 \sqrt {x}}{16 b^2 \left (a+b x^2\right )}-\frac {5 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{3/4} b^{9/4}}+\frac {5 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{3/4} b^{9/4}}-\frac {5 \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{3/4} b^{9/4}}+\frac {5 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{3/4} b^{9/4}} \]

[Out]

-1/4*x^(5/2)/b/(b*x^2+a)^2-5/64*arctan(1-b^(1/4)*2^(1/2)*x^(1/2)/a^(1/4))/a^(3/4)/b^(9/4)*2^(1/2)+5/64*arctan(
1+b^(1/4)*2^(1/2)*x^(1/2)/a^(1/4))/a^(3/4)/b^(9/4)*2^(1/2)-5/128*ln(a^(1/2)+x*b^(1/2)-a^(1/4)*b^(1/4)*2^(1/2)*
x^(1/2))/a^(3/4)/b^(9/4)*2^(1/2)+5/128*ln(a^(1/2)+x*b^(1/2)+a^(1/4)*b^(1/4)*2^(1/2)*x^(1/2))/a^(3/4)/b^(9/4)*2
^(1/2)-5/16*x^(1/2)/b^2/(b*x^2+a)

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Rubi [A]
time = 0.12, antiderivative size = 239, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {294, 335, 217, 1179, 642, 1176, 631, 210} \begin {gather*} -\frac {5 \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{3/4} b^{9/4}}+\frac {5 \text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt {2} a^{3/4} b^{9/4}}-\frac {5 \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{3/4} b^{9/4}}+\frac {5 \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{3/4} b^{9/4}}-\frac {5 \sqrt {x}}{16 b^2 \left (a+b x^2\right )}-\frac {x^{5/2}}{4 b \left (a+b x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(7/2)/(a + b*x^2)^3,x]

[Out]

-1/4*x^(5/2)/(b*(a + b*x^2)^2) - (5*Sqrt[x])/(16*b^2*(a + b*x^2)) - (5*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^
(1/4)])/(32*Sqrt[2]*a^(3/4)*b^(9/4)) + (5*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(3/4)*b
^(9/4)) - (5*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(3/4)*b^(9/4)) + (5*Log
[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(3/4)*b^(9/4))

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 294

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 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 {x^{7/2}}{\left (a+b x^2\right )^3} \, dx &=-\frac {x^{5/2}}{4 b \left (a+b x^2\right )^2}+\frac {5 \int \frac {x^{3/2}}{\left (a+b x^2\right )^2} \, dx}{8 b}\\ &=-\frac {x^{5/2}}{4 b \left (a+b x^2\right )^2}-\frac {5 \sqrt {x}}{16 b^2 \left (a+b x^2\right )}+\frac {5 \int \frac {1}{\sqrt {x} \left (a+b x^2\right )} \, dx}{32 b^2}\\ &=-\frac {x^{5/2}}{4 b \left (a+b x^2\right )^2}-\frac {5 \sqrt {x}}{16 b^2 \left (a+b x^2\right )}+\frac {5 \text {Subst}\left (\int \frac {1}{a+b x^4} \, dx,x,\sqrt {x}\right )}{16 b^2}\\ &=-\frac {x^{5/2}}{4 b \left (a+b x^2\right )^2}-\frac {5 \sqrt {x}}{16 b^2 \left (a+b x^2\right )}+\frac {5 \text {Subst}\left (\int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{32 \sqrt {a} b^2}+\frac {5 \text {Subst}\left (\int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{32 \sqrt {a} b^2}\\ &=-\frac {x^{5/2}}{4 b \left (a+b x^2\right )^2}-\frac {5 \sqrt {x}}{16 b^2 \left (a+b x^2\right )}+\frac {5 \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 \sqrt {a} b^{5/2}}+\frac {5 \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 \sqrt {a} b^{5/2}}-\frac {5 \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^{3/4} b^{9/4}}-\frac {5 \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^{3/4} b^{9/4}}\\ &=-\frac {x^{5/2}}{4 b \left (a+b x^2\right )^2}-\frac {5 \sqrt {x}}{16 b^2 \left (a+b x^2\right )}-\frac {5 \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{3/4} b^{9/4}}+\frac {5 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{3/4} b^{9/4}}+\frac {5 \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^{3/4} b^{9/4}}-\frac {5 \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^{3/4} b^{9/4}}\\ &=-\frac {x^{5/2}}{4 b \left (a+b x^2\right )^2}-\frac {5 \sqrt {x}}{16 b^2 \left (a+b x^2\right )}-\frac {5 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{3/4} b^{9/4}}+\frac {5 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{3/4} b^{9/4}}-\frac {5 \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{3/4} b^{9/4}}+\frac {5 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{3/4} b^{9/4}}\\ \end {align*}

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Mathematica [A]
time = 0.34, size = 138, normalized size = 0.58 \begin {gather*} \frac {-\frac {4 \sqrt [4]{b} \sqrt {x} \left (5 a+9 b x^2\right )}{\left (a+b x^2\right )^2}-\frac {5 \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )}{a^{3/4}}+\frac {5 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{a^{3/4}}}{64 b^{9/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(7/2)/(a + b*x^2)^3,x]

[Out]

((-4*b^(1/4)*Sqrt[x]*(5*a + 9*b*x^2))/(a + b*x^2)^2 - (5*Sqrt[2]*ArcTan[(Sqrt[a] - Sqrt[b]*x)/(Sqrt[2]*a^(1/4)
*b^(1/4)*Sqrt[x])])/a^(3/4) + (5*Sqrt[2]*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[b]*x)])/a^(
3/4))/(64*b^(9/4))

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Maple [A]
time = 0.06, size = 139, normalized size = 0.58

method	result	size
derivativedivides	\(\frac {-\frac {9 x^{\frac {5}{2}}}{16 b}-\frac {5 a \sqrt {x}}{16 b^{2}}}{\left (b \,x^{2}+a \right )^{2}}+\frac {5 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{128 b^{2} a}\)	\(139\)
default	\(\frac {-\frac {9 x^{\frac {5}{2}}}{16 b}-\frac {5 a \sqrt {x}}{16 b^{2}}}{\left (b \,x^{2}+a \right )^{2}}+\frac {5 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{128 b^{2} a}\)	\(139\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)/(b*x^2+a)^3,x,method=_RETURNVERBOSE)

[Out]

2*(-9/32*x^(5/2)/b-5/32*a*x^(1/2)/b^2)/(b*x^2+a)^2+5/128/b^2*(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*ar
ctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1))

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Maxima [A]
time = 0.50, size = 218, normalized size = 0.91 \begin {gather*} -\frac {9 \, b x^{\frac {5}{2}} + 5 \, a \sqrt {x}}{16 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{2} + a^{2} b^{2}\right )}} + \frac {5 \, {\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 {a} \sqrt {\sqrt {a} \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 {a} \sqrt {\sqrt {a} \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 {3}{4}} b^{\frac {1}{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 {3}{4}} b^{\frac {1}{4}}}\right )}}{128 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

-1/16*(9*b*x^(5/2) + 5*a*sqrt(x))/(b^4*x^4 + 2*a*b^3*x^2 + a^2*b^2) + 5/128*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqr
t(2)*a^(1/4)*b^(1/4) + 2*sqrt(b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b))) + 2*sqrt(2)*a
rctan(-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(sqrt(a)*
sqrt(b))) + sqrt(2)*log(sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a))/(a^(3/4)*b^(1/4)) - sqrt(2)*log
(-sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a))/(a^(3/4)*b^(1/4)))/b^2

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Fricas [A]
time = 1.66, size = 254, normalized size = 1.06 \begin {gather*} \frac {20 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{2} + a^{2} b^{2}\right )} \left (-\frac {1}{a^{3} b^{9}}\right )^{\frac {1}{4}} \arctan \left (\sqrt {a^{2} b^{4} \sqrt {-\frac {1}{a^{3} b^{9}}} + x} a^{2} b^{7} \left (-\frac {1}{a^{3} b^{9}}\right )^{\frac {3}{4}} - a^{2} b^{7} \sqrt {x} \left (-\frac {1}{a^{3} b^{9}}\right )^{\frac {3}{4}}\right ) + 5 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{2} + a^{2} b^{2}\right )} \left (-\frac {1}{a^{3} b^{9}}\right )^{\frac {1}{4}} \log \left (a b^{2} \left (-\frac {1}{a^{3} b^{9}}\right )^{\frac {1}{4}} + \sqrt {x}\right ) - 5 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{2} + a^{2} b^{2}\right )} \left (-\frac {1}{a^{3} b^{9}}\right )^{\frac {1}{4}} \log \left (-a b^{2} \left (-\frac {1}{a^{3} b^{9}}\right )^{\frac {1}{4}} + \sqrt {x}\right ) - 4 \, {\left (9 \, b x^{2} + 5 \, a\right )} \sqrt {x}}{64 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{2} + a^{2} b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

1/64*(20*(b^4*x^4 + 2*a*b^3*x^2 + a^2*b^2)*(-1/(a^3*b^9))^(1/4)*arctan(sqrt(a^2*b^4*sqrt(-1/(a^3*b^9)) + x)*a^
2*b^7*(-1/(a^3*b^9))^(3/4) - a^2*b^7*sqrt(x)*(-1/(a^3*b^9))^(3/4)) + 5*(b^4*x^4 + 2*a*b^3*x^2 + a^2*b^2)*(-1/(
a^3*b^9))^(1/4)*log(a*b^2*(-1/(a^3*b^9))^(1/4) + sqrt(x)) - 5*(b^4*x^4 + 2*a*b^3*x^2 + a^2*b^2)*(-1/(a^3*b^9))
^(1/4)*log(-a*b^2*(-1/(a^3*b^9))^(1/4) + sqrt(x)) - 4*(9*b*x^2 + 5*a)*sqrt(x))/(b^4*x^4 + 2*a*b^3*x^2 + a^2*b^
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(x**(7/2)/(b*x**2+a)**3,x)

[Out]

Timed out

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Giac [A]
time = 1.50, size = 209, normalized size = 0.87 \begin {gather*} \frac {5 \, \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 b^{3}} + \frac {5 \, \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 b^{3}} + \frac {5 \, \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 b^{3}} - \frac {5 \, \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 b^{3}} - \frac {9 \, b x^{\frac {5}{2}} + 5 \, a \sqrt {x}}{16 \, {\left (b x^{2} + a\right )}^{2} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(b*x^2+a)^3,x, algorithm="giac")

[Out]

5/64*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*b^3) + 5/64*sq
rt(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*b^3) + 5/128*sqrt(2)
*(a*b^3)^(1/4)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a*b^3) - 5/128*sqrt(2)*(a*b^3)^(1/4)*log(-sqr
t(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a*b^3) - 1/16*(9*b*x^(5/2) + 5*a*sqrt(x))/((b*x^2 + a)^2*b^2)

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Mupad [B]
time = 4.69, size = 87, normalized size = 0.36 \begin {gather*} -\frac {\frac {9\,x^{5/2}}{16\,b}+\frac {5\,a\,\sqrt {x}}{16\,b^2}}{a^2+2\,a\,b\,x^2+b^2\,x^4}-\frac {5\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )}{32\,{\left (-a\right )}^{3/4}\,b^{9/4}}-\frac {5\,\mathrm {atanh}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )}{32\,{\left (-a\right )}^{3/4}\,b^{9/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)/(a + b*x^2)^3,x)

[Out]

- ((9*x^(5/2))/(16*b) + (5*a*x^(1/2))/(16*b^2))/(a^2 + b^2*x^4 + 2*a*b*x^2) - (5*atan((b^(1/4)*x^(1/2))/(-a)^(
1/4)))/(32*(-a)^(3/4)*b^(9/4)) - (5*atanh((b^(1/4)*x^(1/2))/(-a)^(1/4)))/(32*(-a)^(3/4)*b^(9/4))

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