3.308 \(\int \frac {1}{\sqrt {x} (a+b x^2)^3} \, dx\)

Optimal. Leaf size=239 \[ -\frac {21 \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{11/4} \sqrt [4]{b}}+\frac {21 \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{11/4} \sqrt [4]{b}}-\frac {21 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{11/4} \sqrt [4]{b}}+\frac {21 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt {2} a^{11/4} \sqrt [4]{b}}+\frac {7 \sqrt {x}}{16 a^2 \left (a+b x^2\right )}+\frac {\sqrt {x}}{4 a \left (a+b x^2\right )^2} \]

[Out]

-21/64*arctan(1-b^(1/4)*2^(1/2)*x^(1/2)/a^(1/4))/a^(11/4)/b^(1/4)*2^(1/2)+21/64*arctan(1+b^(1/4)*2^(1/2)*x^(1/
2)/a^(1/4))/a^(11/4)/b^(1/4)*2^(1/2)-21/128*ln(a^(1/2)+x*b^(1/2)-a^(1/4)*b^(1/4)*2^(1/2)*x^(1/2))/a^(11/4)/b^(
1/4)*2^(1/2)+21/128*ln(a^(1/2)+x*b^(1/2)+a^(1/4)*b^(1/4)*2^(1/2)*x^(1/2))/a^(11/4)/b^(1/4)*2^(1/2)+1/4*x^(1/2)
/a/(b*x^2+a)^2+7/16*x^(1/2)/a^2/(b*x^2+a)

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Rubi [A]  time = 0.17, 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 = {290, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac {7 \sqrt {x}}{16 a^2 \left (a+b x^2\right )}-\frac {21 \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{11/4} \sqrt [4]{b}}+\frac {21 \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{11/4} \sqrt [4]{b}}-\frac {21 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{11/4} \sqrt [4]{b}}+\frac {21 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt {2} a^{11/4} \sqrt [4]{b}}+\frac {\sqrt {x}}{4 a \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[x]*(a + b*x^2)^3),x]

[Out]

Sqrt[x]/(4*a*(a + b*x^2)^2) + (7*Sqrt[x])/(16*a^2*(a + b*x^2)) - (21*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1
/4)])/(32*Sqrt[2]*a^(11/4)*b^(1/4)) + (21*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(11/4)*
b^(1/4)) - (21*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(11/4)*b^(1/4)) + (21
*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(11/4)*b^(1/4))

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 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}{\sqrt {x} \left (a+b x^2\right )^3} \, dx &=\frac {\sqrt {x}}{4 a \left (a+b x^2\right )^2}+\frac {7 \int \frac {1}{\sqrt {x} \left (a+b x^2\right )^2} \, dx}{8 a}\\ &=\frac {\sqrt {x}}{4 a \left (a+b x^2\right )^2}+\frac {7 \sqrt {x}}{16 a^2 \left (a+b x^2\right )}+\frac {21 \int \frac {1}{\sqrt {x} \left (a+b x^2\right )} \, dx}{32 a^2}\\ &=\frac {\sqrt {x}}{4 a \left (a+b x^2\right )^2}+\frac {7 \sqrt {x}}{16 a^2 \left (a+b x^2\right )}+\frac {21 \operatorname {Subst}\left (\int \frac {1}{a+b x^4} \, dx,x,\sqrt {x}\right )}{16 a^2}\\ &=\frac {\sqrt {x}}{4 a \left (a+b x^2\right )^2}+\frac {7 \sqrt {x}}{16 a^2 \left (a+b x^2\right )}+\frac {21 \operatorname {Subst}\left (\int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{32 a^{5/2}}+\frac {21 \operatorname {Subst}\left (\int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{32 a^{5/2}}\\ &=\frac {\sqrt {x}}{4 a \left (a+b x^2\right )^2}+\frac {7 \sqrt {x}}{16 a^2 \left (a+b x^2\right )}+\frac {21 \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^{5/2} \sqrt {b}}+\frac {21 \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^{5/2} \sqrt {b}}-\frac {21 \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^{11/4} \sqrt [4]{b}}-\frac {21 \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^{11/4} \sqrt [4]{b}}\\ &=\frac {\sqrt {x}}{4 a \left (a+b x^2\right )^2}+\frac {7 \sqrt {x}}{16 a^2 \left (a+b x^2\right )}-\frac {21 \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{11/4} \sqrt [4]{b}}+\frac {21 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{11/4} \sqrt [4]{b}}+\frac {21 \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^{11/4} \sqrt [4]{b}}-\frac {21 \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^{11/4} \sqrt [4]{b}}\\ &=\frac {\sqrt {x}}{4 a \left (a+b x^2\right )^2}+\frac {7 \sqrt {x}}{16 a^2 \left (a+b x^2\right )}-\frac {21 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{11/4} \sqrt [4]{b}}+\frac {21 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{11/4} \sqrt [4]{b}}-\frac {21 \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{11/4} \sqrt [4]{b}}+\frac {21 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{11/4} \sqrt [4]{b}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 220, normalized size = 0.92 \[ \frac {\frac {32 a^{7/4} \sqrt {x}}{\left (a+b x^2\right )^2}+\frac {56 a^{3/4} \sqrt {x}}{a+b x^2}-\frac {21 \sqrt {2} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{\sqrt [4]{b}}+\frac {21 \sqrt {2} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{\sqrt [4]{b}}-\frac {42 \sqrt {2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt [4]{b}}+\frac {42 \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt [4]{b}}}{128 a^{11/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((32*a^(7/4)*Sqrt[x])/(a + b*x^2)^2 + (56*a^(3/4)*Sqrt[x])/(a + b*x^2) - (42*Sqrt[2]*ArcTan[1 - (Sqrt[2]*b^(1/
4)*Sqrt[x])/a^(1/4)])/b^(1/4) + (42*Sqrt[2]*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/b^(1/4) - (21*Sqrt[
2]*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/b^(1/4) + (21*Sqrt[2]*Log[Sqrt[a] + Sqrt[2]*a^(
1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/b^(1/4))/(128*a^(11/4))

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fricas [A]  time = 0.91, size = 241, normalized size = 1.01 \[ \frac {84 \, {\left (a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}\right )} \left (-\frac {1}{a^{11} b}\right )^{\frac {1}{4}} \arctan \left (\sqrt {a^{6} \sqrt {-\frac {1}{a^{11} b}} + x} a^{8} b \left (-\frac {1}{a^{11} b}\right )^{\frac {3}{4}} - a^{8} b \sqrt {x} \left (-\frac {1}{a^{11} b}\right )^{\frac {3}{4}}\right ) + 21 \, {\left (a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}\right )} \left (-\frac {1}{a^{11} b}\right )^{\frac {1}{4}} \log \left (a^{3} \left (-\frac {1}{a^{11} b}\right )^{\frac {1}{4}} + \sqrt {x}\right ) - 21 \, {\left (a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}\right )} \left (-\frac {1}{a^{11} b}\right )^{\frac {1}{4}} \log \left (-a^{3} \left (-\frac {1}{a^{11} b}\right )^{\frac {1}{4}} + \sqrt {x}\right ) + 4 \, {\left (7 \, b x^{2} + 11 \, a\right )} \sqrt {x}}{64 \, {\left (a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(84*(a^2*b^2*x^4 + 2*a^3*b*x^2 + a^4)*(-1/(a^11*b))^(1/4)*arctan(sqrt(a^6*sqrt(-1/(a^11*b)) + x)*a^8*b*(-
1/(a^11*b))^(3/4) - a^8*b*sqrt(x)*(-1/(a^11*b))^(3/4)) + 21*(a^2*b^2*x^4 + 2*a^3*b*x^2 + a^4)*(-1/(a^11*b))^(1
/4)*log(a^3*(-1/(a^11*b))^(1/4) + sqrt(x)) - 21*(a^2*b^2*x^4 + 2*a^3*b*x^2 + a^4)*(-1/(a^11*b))^(1/4)*log(-a^3
*(-1/(a^11*b))^(1/4) + sqrt(x)) + 4*(7*b*x^2 + 11*a)*sqrt(x))/(a^2*b^2*x^4 + 2*a^3*b*x^2 + a^4)

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giac [A]  time = 0.65, size = 209, normalized size = 0.87 \[ \frac {21 \, \sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{64 \, a^{3} b} + \frac {21 \, \sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{64 \, a^{3} b} + \frac {21 \, \sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{128 \, a^{3} b} - \frac {21 \, \sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{128 \, a^{3} b} + \frac {7 \, b x^{\frac {5}{2}} + 11 \, a \sqrt {x}}{16 \, {\left (b x^{2} + a\right )}^{2} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

21/64*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^3*b) + 21/64*
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^3*b) + 21/128*sqrt
(2)*(a*b^3)^(1/4)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^3*b) - 21/128*sqrt(2)*(a*b^3)^(1/4)*log(
-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^3*b) + 1/16*(7*b*x^(5/2) + 11*a*sqrt(x))/((b*x^2 + a)^2*a^2)

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maple [A]  time = 0.01, size = 166, normalized size = 0.69 \[ \frac {\sqrt {x}}{4 \left (b \,x^{2}+a \right )^{2} a}+\frac {7 \sqrt {x}}{16 \left (b \,x^{2}+a \right ) a^{2}}+\frac {21 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{64 a^{3}}+\frac {21 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{64 a^{3}}+\frac {21 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}\right )}{128 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^2+a)^3/x^(1/2),x)

[Out]

1/4*x^(1/2)/a/(b*x^2+a)^2+7/16*x^(1/2)/a^2/(b*x^2+a)+21/128/a^3*(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)))+21/64/a^3*(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)
/(a/b)^(1/4)*x^(1/2)+1)+21/64/a^3*(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 = 3.02, size = 217, normalized size = 0.91 \[ \frac {7 \, b x^{\frac {5}{2}} + 11 \, a \sqrt {x}}{16 \, {\left (a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}\right )}} + \frac {21 \, {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {\sqrt {2} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}} - \frac {\sqrt {2} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}}\right )}}{128 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(7*b*x^(5/2) + 11*a*sqrt(x))/(a^2*b^2*x^4 + 2*a^3*b*x^2 + a^4) + 21/128*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sq
rt(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)*
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))) + 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)*lo
g(-sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a))/(a^(3/4)*b^(1/4)))/a^2

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mupad [B]  time = 4.67, size = 86, normalized size = 0.36 \[ \frac {\frac {11\,\sqrt {x}}{16\,a}+\frac {7\,b\,x^{5/2}}{16\,a^2}}{a^2+2\,a\,b\,x^2+b^2\,x^4}-\frac {21\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )}{32\,{\left (-a\right )}^{11/4}\,b^{1/4}}-\frac {21\,\mathrm {atanh}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )}{32\,{\left (-a\right )}^{11/4}\,b^{1/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(1/2)*(a + b*x^2)^3),x)

[Out]

((11*x^(1/2))/(16*a) + (7*b*x^(5/2))/(16*a^2))/(a^2 + b^2*x^4 + 2*a*b*x^2) - (21*atan((b^(1/4)*x^(1/2))/(-a)^(
1/4)))/(32*(-a)^(11/4)*b^(1/4)) - (21*atanh((b^(1/4)*x^(1/2))/(-a)^(1/4)))/(32*(-a)^(11/4)*b^(1/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**2+a)**3/x**(1/2),x)

[Out]

Timed out

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