3.4.2 \(\int \frac {1}{x^{5/2} (a+b x^2)^2} \, dx\)

Optimal. Leaf size=230 \[ \frac {7 b^{3/4} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} a^{11/4}}-\frac {7 b^{3/4} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} a^{11/4}}+\frac {7 b^{3/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{11/4}}-\frac {7 b^{3/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} a^{11/4}}-\frac {7}{6 a^2 x^{3/2}}+\frac {1}{2 a x^{3/2} \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.18, antiderivative size = 230, 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 = {290, 325, 329, 211, 1165, 628, 1162, 617, 204} \begin {gather*} \frac {7 b^{3/4} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} a^{11/4}}-\frac {7 b^{3/4} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} a^{11/4}}+\frac {7 b^{3/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{11/4}}-\frac {7 b^{3/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} a^{11/4}}-\frac {7}{6 a^2 x^{3/2}}+\frac {1}{2 a x^{3/2} \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/2)*(a + b*x^2)^2),x]

[Out]

-7/(6*a^2*x^(3/2)) + 1/(2*a*x^(3/2)*(a + b*x^2)) + (7*b^(3/4)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(
4*Sqrt[2]*a^(11/4)) - (7*b^(3/4)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(11/4)) + (7*b^(3
/4)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(11/4)) - (7*b^(3/4)*Log[Sqrt[a]
+ Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(11/4))

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 211

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 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 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 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 {1}{x^{5/2} \left (a+b x^2\right )^2} \, dx &=\frac {1}{2 a x^{3/2} \left (a+b x^2\right )}+\frac {7 \int \frac {1}{x^{5/2} \left (a+b x^2\right )} \, dx}{4 a}\\ &=-\frac {7}{6 a^2 x^{3/2}}+\frac {1}{2 a x^{3/2} \left (a+b x^2\right )}-\frac {(7 b) \int \frac {1}{\sqrt {x} \left (a+b x^2\right )} \, dx}{4 a^2}\\ &=-\frac {7}{6 a^2 x^{3/2}}+\frac {1}{2 a x^{3/2} \left (a+b x^2\right )}-\frac {(7 b) \operatorname {Subst}\left (\int \frac {1}{a+b x^4} \, dx,x,\sqrt {x}\right )}{2 a^2}\\ &=-\frac {7}{6 a^2 x^{3/2}}+\frac {1}{2 a x^{3/2} \left (a+b x^2\right )}-\frac {(7 b) \operatorname {Subst}\left (\int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{4 a^{5/2}}-\frac {(7 b) \operatorname {Subst}\left (\int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{4 a^{5/2}}\\ &=-\frac {7}{6 a^2 x^{3/2}}+\frac {1}{2 a x^{3/2} \left (a+b x^2\right )}-\frac {\left (7 \sqrt {b}\right ) \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 a^{5/2}}-\frac {\left (7 \sqrt {b}\right ) \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 a^{5/2}}+\frac {\left (7 b^{3/4}\right ) \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} a^{11/4}}+\frac {\left (7 b^{3/4}\right ) \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} a^{11/4}}\\ &=-\frac {7}{6 a^2 x^{3/2}}+\frac {1}{2 a x^{3/2} \left (a+b x^2\right )}+\frac {7 b^{3/4} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{11/4}}-\frac {7 b^{3/4} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{11/4}}-\frac {\left (7 b^{3/4}\right ) \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} a^{11/4}}+\frac {\left (7 b^{3/4}\right ) \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} a^{11/4}}\\ &=-\frac {7}{6 a^2 x^{3/2}}+\frac {1}{2 a x^{3/2} \left (a+b x^2\right )}+\frac {7 b^{3/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{11/4}}-\frac {7 b^{3/4} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{11/4}}+\frac {7 b^{3/4} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{11/4}}-\frac {7 b^{3/4} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{11/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 29, normalized size = 0.13 \begin {gather*} -\frac {2 \, _2F_1\left (-\frac {3}{4},2;\frac {1}{4};-\frac {b x^2}{a}\right )}{3 a^2 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.34, size = 149, normalized size = 0.65 \begin {gather*} \frac {7 b^{3/4} \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 )}{4 \sqrt {2} a^{11/4}}-\frac {7 b^{3/4} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{4 \sqrt {2} a^{11/4}}+\frac {-4 a-7 b x^2}{6 a^2 x^{3/2} \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a - 7*b*x^2)/(6*a^2*x^(3/2)*(a + b*x^2)) + (7*b^(3/4)*ArcTan[(a^(1/4)/(Sqrt[2]*b^(1/4)) - (b^(1/4)*x)/(Sqr
t[2]*a^(1/4)))/Sqrt[x]])/(4*Sqrt[2]*a^(11/4)) - (7*b^(3/4)*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/(Sqrt[a]
+ Sqrt[b]*x)])/(4*Sqrt[2]*a^(11/4))

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fricas [A]  time = 0.53, size = 228, normalized size = 0.99 \begin {gather*} -\frac {84 \, {\left (a^{2} b x^{4} + a^{3} x^{2}\right )} \left (-\frac {b^{3}}{a^{11}}\right )^{\frac {1}{4}} \arctan \left (-\frac {a^{8} b \sqrt {x} \left (-\frac {b^{3}}{a^{11}}\right )^{\frac {3}{4}} - \sqrt {a^{6} \sqrt {-\frac {b^{3}}{a^{11}}} + b^{2} x} a^{8} \left (-\frac {b^{3}}{a^{11}}\right )^{\frac {3}{4}}}{b^{3}}\right ) + 21 \, {\left (a^{2} b x^{4} + a^{3} x^{2}\right )} \left (-\frac {b^{3}}{a^{11}}\right )^{\frac {1}{4}} \log \left (7 \, a^{3} \left (-\frac {b^{3}}{a^{11}}\right )^{\frac {1}{4}} + 7 \, b \sqrt {x}\right ) - 21 \, {\left (a^{2} b x^{4} + a^{3} x^{2}\right )} \left (-\frac {b^{3}}{a^{11}}\right )^{\frac {1}{4}} \log \left (-7 \, a^{3} \left (-\frac {b^{3}}{a^{11}}\right )^{\frac {1}{4}} + 7 \, b \sqrt {x}\right ) + 4 \, {\left (7 \, b x^{2} + 4 \, a\right )} \sqrt {x}}{24 \, {\left (a^{2} b x^{4} + a^{3} 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)^2,x, algorithm="fricas")

[Out]

-1/24*(84*(a^2*b*x^4 + a^3*x^2)*(-b^3/a^11)^(1/4)*arctan(-(a^8*b*sqrt(x)*(-b^3/a^11)^(3/4) - sqrt(a^6*sqrt(-b^
3/a^11) + b^2*x)*a^8*(-b^3/a^11)^(3/4))/b^3) + 21*(a^2*b*x^4 + a^3*x^2)*(-b^3/a^11)^(1/4)*log(7*a^3*(-b^3/a^11
)^(1/4) + 7*b*sqrt(x)) - 21*(a^2*b*x^4 + a^3*x^2)*(-b^3/a^11)^(1/4)*log(-7*a^3*(-b^3/a^11)^(1/4) + 7*b*sqrt(x)
) + 4*(7*b*x^2 + 4*a)*sqrt(x))/(a^2*b*x^4 + a^3*x^2)

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giac [A]  time = 0.64, size = 196, normalized size = 0.85 \begin {gather*} -\frac {7 \, \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 )}{8 \, a^{3}} - \frac {7 \, \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 )}{8 \, a^{3}} - \frac {7 \, \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 )}{16 \, a^{3}} + \frac {7 \, \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 )}{16 \, a^{3}} - \frac {b \sqrt {x}}{2 \, {\left (b x^{2} + a\right )} a^{2}} - \frac {2}{3 \, a^{2} 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)^2,x, algorithm="giac")

[Out]

-7/8*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^3 - 7/8*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^3 - 7/16*sqrt(2)*(a*b^3)^(
1/4)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/a^3 + 7/16*sqrt(2)*(a*b^3)^(1/4)*log(-sqrt(2)*sqrt(x)*(a
/b)^(1/4) + x + sqrt(a/b))/a^3 - 1/2*b*sqrt(x)/((b*x^2 + a)*a^2) - 2/3/(a^2*x^(3/2))

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maple [A]  time = 0.02, size = 161, normalized size = 0.70 \begin {gather*} -\frac {b \sqrt {x}}{2 \left (b \,x^{2}+a \right ) a^{2}}-\frac {7 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{8 a^{3}}-\frac {7 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{8 a^{3}}-\frac {7 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, b \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 )}{16 a^{3}}-\frac {2}{3 a^{2} x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(5/2)/(b*x^2+a)^2,x)

[Out]

-2/3/a^2/x^(3/2)-1/2/a^2*b*x^(1/2)/(b*x^2+a)-7/16/a^3*b*(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)))-7/8/a^3*b*(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(
1/4)*x^(1/2)+1)-7/8/a^3*b*(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 = 3.06, size = 209, normalized size = 0.91 \begin {gather*} -\frac {7 \, b x^{2} + 4 \, a}{6 \, {\left (a^{2} b x^{\frac {7}{2}} + a^{3} x^{\frac {3}{2}}\right )}} - \frac {7 \, {\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 )}}{16 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(7*b*x^2 + 4*a)/(a^2*b*x^(7/2) + a^3*x^(3/2)) - 7/16*(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(sqrt(a)*sqrt(b))) + 2*sqrt(2)*b*arctan(-1/2*sqr
t(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))) + sq
rt(2)*b^(3/4)*log(sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(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^2

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mupad [B]  time = 4.68, size = 77, normalized size = 0.33 \begin {gather*} \frac {7\,{\left (-b\right )}^{3/4}\,\mathrm {atan}\left (\frac {{\left (-b\right )}^{1/4}\,\sqrt {x}}{a^{1/4}}\right )}{4\,a^{11/4}}-\frac {\frac {2}{3\,a}+\frac {7\,b\,x^2}{6\,a^2}}{a\,x^{3/2}+b\,x^{7/2}}+\frac {7\,{\left (-b\right )}^{3/4}\,\mathrm {atanh}\left (\frac {{\left (-b\right )}^{1/4}\,\sqrt {x}}{a^{1/4}}\right )}{4\,a^{11/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(5/2)*(a + b*x^2)^2),x)

[Out]

(7*(-b)^(3/4)*atan(((-b)^(1/4)*x^(1/2))/a^(1/4)))/(4*a^(11/4)) - (2/(3*a) + (7*b*x^2)/(6*a^2))/(a*x^(3/2) + b*
x^(7/2)) + (7*(-b)^(3/4)*atanh(((-b)^(1/4)*x^(1/2))/a^(1/4)))/(4*a^(11/4))

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

[Out]

Timed out

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