3.6.77 \(\int \frac {1}{(d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx\)

Optimal. Leaf size=414 \[ -\frac {2 \left (a+b x^2\right )}{3 a d (d x)^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {b^{3/4} \left (a+b x^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{2 \sqrt {2} a^{7/4} d^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b^{3/4} \left (a+b x^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{2 \sqrt {2} a^{7/4} d^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {b^{3/4} \left (a+b x^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{\sqrt {2} a^{7/4} d^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b^{3/4} \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{\sqrt {2} a^{7/4} d^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

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Rubi [A]  time = 0.28, antiderivative size = 414, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1112, 325, 329, 211, 1165, 628, 1162, 617, 204} \begin {gather*} \frac {b^{3/4} \left (a+b x^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{2 \sqrt {2} a^{7/4} d^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b^{3/4} \left (a+b x^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{2 \sqrt {2} a^{7/4} d^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {b^{3/4} \left (a+b x^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{\sqrt {2} a^{7/4} d^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b^{3/4} \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{\sqrt {2} a^{7/4} d^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {2 \left (a+b x^2\right )}{3 a d (d x)^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d*x)^(5/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]),x]

[Out]

(-2*(a + b*x^2))/(3*a*d*(d*x)^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (b^(3/4)*(a + b*x^2)*ArcTan[1 - (Sqrt[2
]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(Sqrt[2]*a^(7/4)*d^(5/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (b^(3/4)*
(a + b*x^2)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(Sqrt[2]*a^(7/4)*d^(5/2)*Sqrt[a^2 + 2*a
*b*x^2 + b^2*x^4]) + (b^(3/4)*(a + b*x^2)*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x - Sqrt[2]*a^(1/4)*b^(1/4)*Sq
rt[d*x]])/(2*Sqrt[2]*a^(7/4)*d^(5/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (b^(3/4)*(a + b*x^2)*Log[Sqrt[a]*Sqrt[
d] + Sqrt[b]*Sqrt[d]*x + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(2*Sqrt[2]*a^(7/4)*d^(5/2)*Sqrt[a^2 + 2*a*b*x^2 +
b^2*x^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 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 1112

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

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

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Mathematica [C]  time = 0.01, size = 52, normalized size = 0.13 \begin {gather*} -\frac {2 x \left (a+b x^2\right ) \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};-\frac {b x^2}{a}\right )}{3 a (d x)^{5/2} \sqrt {\left (a+b x^2\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d*x)^(5/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]),x]

[Out]

(-2*x*(a + b*x^2)*Hypergeometric2F1[-3/4, 1, 1/4, -((b*x^2)/a)])/(3*a*(d*x)^(5/2)*Sqrt[(a + b*x^2)^2])

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IntegrateAlgebraic [A]  time = 44.93, size = 204, normalized size = 0.49 \begin {gather*} \frac {\left (a d^2+b d^2 x^2\right ) \left (\frac {b^{3/4} \tan ^{-1}\left (\frac {\frac {\sqrt [4]{a} \sqrt {d}}{\sqrt {2} \sqrt [4]{b}}-\frac {\sqrt [4]{b} \sqrt {d} x}{\sqrt {2} \sqrt [4]{a}}}{\sqrt {d x}}\right )}{\sqrt {2} a^{7/4} d^{5/2}}-\frac {b^{3/4} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d} \sqrt {d x}}{\sqrt {a} d+\sqrt {b} d x}\right )}{\sqrt {2} a^{7/4} d^{5/2}}-\frac {2}{3 a d (d x)^{3/2}}\right )}{d^2 \sqrt {\frac {\left (a d^2+b d^2 x^2\right )^2}{d^4}}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((d*x)^(5/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]),x]

[Out]

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

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fricas [A]  time = 0.96, size = 227, normalized size = 0.55 \begin {gather*} -\frac {12 \, a d^{3} x^{2} \left (-\frac {b^{3}}{a^{7} d^{10}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {d x} a^{5} b d^{7} \left (-\frac {b^{3}}{a^{7} d^{10}}\right )^{\frac {3}{4}} - \sqrt {a^{4} d^{6} \sqrt {-\frac {b^{3}}{a^{7} d^{10}}} + b^{2} d x} a^{5} d^{7} \left (-\frac {b^{3}}{a^{7} d^{10}}\right )^{\frac {3}{4}}}{b^{3}}\right ) + 3 \, a d^{3} x^{2} \left (-\frac {b^{3}}{a^{7} d^{10}}\right )^{\frac {1}{4}} \log \left (a^{2} d^{3} \left (-\frac {b^{3}}{a^{7} d^{10}}\right )^{\frac {1}{4}} + \sqrt {d x} b\right ) - 3 \, a d^{3} x^{2} \left (-\frac {b^{3}}{a^{7} d^{10}}\right )^{\frac {1}{4}} \log \left (-a^{2} d^{3} \left (-\frac {b^{3}}{a^{7} d^{10}}\right )^{\frac {1}{4}} + \sqrt {d x} b\right ) + 4 \, \sqrt {d x}}{6 \, a d^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(12*a*d^3*x^2*(-b^3/(a^7*d^10))^(1/4)*arctan(-(sqrt(d*x)*a^5*b*d^7*(-b^3/(a^7*d^10))^(3/4) - sqrt(a^4*d^6
*sqrt(-b^3/(a^7*d^10)) + b^2*d*x)*a^5*d^7*(-b^3/(a^7*d^10))^(3/4))/b^3) + 3*a*d^3*x^2*(-b^3/(a^7*d^10))^(1/4)*
log(a^2*d^3*(-b^3/(a^7*d^10))^(1/4) + sqrt(d*x)*b) - 3*a*d^3*x^2*(-b^3/(a^7*d^10))^(1/4)*log(-a^2*d^3*(-b^3/(a
^7*d^10))^(1/4) + sqrt(d*x)*b) + 4*sqrt(d*x))/(a*d^3*x^2)

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giac [A]  time = 0.24, size = 256, normalized size = 0.62 \begin {gather*} -\frac {1}{12} \, {\left (\frac {6 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a^{2} d^{3}} + \frac {6 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a^{2} d^{3}} + \frac {3 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} \log \left (d x + \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{a^{2} d^{3}} - \frac {3 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} \log \left (d x - \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{a^{2} d^{3}} + \frac {8}{\sqrt {d x} a d^{2} x}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(6*sqrt(2)*(a*b^3*d^2)^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(a*d^2/b)^(1/4) + 2*sqrt(d*x))/(a*d^2/b)^(1/4))
/(a^2*d^3) + 6*sqrt(2)*(a*b^3*d^2)^(1/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a*d^2/b)^(1/4) - 2*sqrt(d*x))/(a*d^2/b)
^(1/4))/(a^2*d^3) + 3*sqrt(2)*(a*b^3*d^2)^(1/4)*log(d*x + sqrt(2)*(a*d^2/b)^(1/4)*sqrt(d*x) + sqrt(a*d^2/b))/(
a^2*d^3) - 3*sqrt(2)*(a*b^3*d^2)^(1/4)*log(d*x - sqrt(2)*(a*d^2/b)^(1/4)*sqrt(d*x) + sqrt(a*d^2/b))/(a^2*d^3)
+ 8/(sqrt(d*x)*a*d^2*x))*sgn(b*x^2 + a)

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maple [A]  time = 0.01, size = 239, normalized size = 0.58 \begin {gather*} -\frac {\left (b \,x^{2}+a \right ) \left (8 a \,d^{2}+6 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (d x \right )^{\frac {3}{2}} b \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}-\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )+6 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (d x \right )^{\frac {3}{2}} b \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}+\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )+3 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (d x \right )^{\frac {3}{2}} b \ln \left (\frac {d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )\right )}{12 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (d x \right )^{\frac {3}{2}} a^{2} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(b*x^2+a)/d^3*(3*b*(a/b*d^2)^(1/4)*2^(1/2)*(d*x)^(3/2)*ln((d*x+(a/b*d^2)^(1/4)*(d*x)^(1/2)*2^(1/2)+(a/b*
d^2)^(1/2))/(d*x-(a/b*d^2)^(1/4)*(d*x)^(1/2)*2^(1/2)+(a/b*d^2)^(1/2)))+6*b*(a/b*d^2)^(1/4)*2^(1/2)*(d*x)^(3/2)
*arctan((2^(1/2)*(d*x)^(1/2)+(a/b*d^2)^(1/4))/(a/b*d^2)^(1/4))+6*b*(a/b*d^2)^(1/4)*2^(1/2)*(d*x)^(3/2)*arctan(
(2^(1/2)*(d*x)^(1/2)-(a/b*d^2)^(1/4))/(a/b*d^2)^(1/4))+8*d^2*a)/((b*x^2+a)^2)^(1/2)/a^2/(d*x)^(3/2)

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maxima [A]  time = 3.08, size = 242, normalized size = 0.58 \begin {gather*} -\frac {\frac {3 \, {\left (\frac {\sqrt {2} b^{\frac {3}{4}} \log \left (\sqrt {b} d x + \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {3}{4}}} - \frac {\sqrt {2} b^{\frac {3}{4}} \log \left (\sqrt {b} d x - \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {3}{4}}} + \frac {2 \, \sqrt {2} b \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {a} d} + \frac {2 \, \sqrt {2} b \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {a} d}\right )}}{a} + \frac {8}{\left (d x\right )^{\frac {3}{2}} a}}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*(sqrt(2)*b^(3/4)*log(sqrt(b)*d*x + sqrt(2)*(a*d^2)^(1/4)*sqrt(d*x)*b^(1/4) + sqrt(a)*d)/(a*d^2)^(3/4)
- sqrt(2)*b^(3/4)*log(sqrt(b)*d*x - sqrt(2)*(a*d^2)^(1/4)*sqrt(d*x)*b^(1/4) + sqrt(a)*d)/(a*d^2)^(3/4) + 2*sq
rt(2)*b*arctan(1/2*sqrt(2)*(sqrt(2)*(a*d^2)^(1/4)*b^(1/4) + 2*sqrt(d*x)*sqrt(b))/sqrt(sqrt(a)*sqrt(b)*d))/(sqr
t(sqrt(a)*sqrt(b)*d)*sqrt(a)*d) + 2*sqrt(2)*b*arctan(-1/2*sqrt(2)*(sqrt(2)*(a*d^2)^(1/4)*b^(1/4) - 2*sqrt(d*x)
*sqrt(b))/sqrt(sqrt(a)*sqrt(b)*d))/(sqrt(sqrt(a)*sqrt(b)*d)*sqrt(a)*d))/a + 8/((d*x)^(3/2)*a))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (d\,x\right )}^{5/2}\,\sqrt {{\left (b\,x^2+a\right )}^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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