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

Optimal. Leaf size=300 \[ \frac {7 b^{3/4} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{8 \sqrt {2} a^{11/4} d^{5/2}}-\frac {7 b^{3/4} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{8 \sqrt {2} a^{11/4} d^{5/2}}+\frac {7 b^{3/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{4 \sqrt {2} a^{11/4} d^{5/2}}-\frac {7 b^{3/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{4 \sqrt {2} a^{11/4} d^{5/2}}-\frac {7}{6 a^2 d (d x)^{3/2}}+\frac {1}{2 a d (d x)^{3/2} \left (a+b x^2\right )} \]

[Out]

-7/6/a^2/d/(d*x)^(3/2)+1/2/a/d/(d*x)^(3/2)/(b*x^2+a)+7/8*b^(3/4)*arctan(1-b^(1/4)*2^(1/2)*(d*x)^(1/2)/a^(1/4)/
d^(1/2))/a^(11/4)/d^(5/2)*2^(1/2)-7/8*b^(3/4)*arctan(1+b^(1/4)*2^(1/2)*(d*x)^(1/2)/a^(1/4)/d^(1/2))/a^(11/4)/d
^(5/2)*2^(1/2)+7/16*b^(3/4)*ln(a^(1/2)*d^(1/2)+x*b^(1/2)*d^(1/2)-a^(1/4)*b^(1/4)*2^(1/2)*(d*x)^(1/2))/a^(11/4)
/d^(5/2)*2^(1/2)-7/16*b^(3/4)*ln(a^(1/2)*d^(1/2)+x*b^(1/2)*d^(1/2)+a^(1/4)*b^(1/4)*2^(1/2)*(d*x)^(1/2))/a^(11/
4)/d^(5/2)*2^(1/2)

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Rubi [A]  time = 0.30, antiderivative size = 300, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {28, 290, 325, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac {7 b^{3/4} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{8 \sqrt {2} a^{11/4} d^{5/2}}-\frac {7 b^{3/4} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{8 \sqrt {2} a^{11/4} d^{5/2}}+\frac {7 b^{3/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{4 \sqrt {2} a^{11/4} d^{5/2}}-\frac {7 b^{3/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{4 \sqrt {2} a^{11/4} d^{5/2}}-\frac {7}{6 a^2 d (d x)^{3/2}}+\frac {1}{2 a d (d x)^{3/2} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-7/(6*a^2*d*(d*x)^(3/2)) + 1/(2*a*d*(d*x)^(3/2)*(a + b*x^2)) + (7*b^(3/4)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[d*x
])/(a^(1/4)*Sqrt[d])])/(4*Sqrt[2]*a^(11/4)*d^(5/2)) - (7*b^(3/4)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/
4)*Sqrt[d])])/(4*Sqrt[2]*a^(11/4)*d^(5/2)) + (7*b^(3/4)*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x - Sqrt[2]*a^(1
/4)*b^(1/4)*Sqrt[d*x]])/(8*Sqrt[2]*a^(11/4)*d^(5/2)) - (7*b^(3/4)*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x + Sq
rt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(8*Sqrt[2]*a^(11/4)*d^(5/2))

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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

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Mathematica [C]  time = 0.01, size = 32, normalized size = 0.11 \[ -\frac {2 x \, _2F_1\left (-\frac {3}{4},2;\frac {1}{4};-\frac {b x^2}{a}\right )}{3 a^2 (d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.09, size = 300, normalized size = 1.00 \[ -\frac {84 \, {\left (a^{2} b d^{3} x^{4} + a^{3} d^{3} x^{2}\right )} \left (-\frac {b^{3}}{a^{11} d^{10}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {d x} a^{8} b d^{7} \left (-\frac {b^{3}}{a^{11} d^{10}}\right )^{\frac {3}{4}} - \sqrt {a^{6} d^{6} \sqrt {-\frac {b^{3}}{a^{11} d^{10}}} + b^{2} d x} a^{8} d^{7} \left (-\frac {b^{3}}{a^{11} d^{10}}\right )^{\frac {3}{4}}}{b^{3}}\right ) + 21 \, {\left (a^{2} b d^{3} x^{4} + a^{3} d^{3} x^{2}\right )} \left (-\frac {b^{3}}{a^{11} d^{10}}\right )^{\frac {1}{4}} \log \left (7 \, a^{3} d^{3} \left (-\frac {b^{3}}{a^{11} d^{10}}\right )^{\frac {1}{4}} + 7 \, \sqrt {d x} b\right ) - 21 \, {\left (a^{2} b d^{3} x^{4} + a^{3} d^{3} x^{2}\right )} \left (-\frac {b^{3}}{a^{11} d^{10}}\right )^{\frac {1}{4}} \log \left (-7 \, a^{3} d^{3} \left (-\frac {b^{3}}{a^{11} d^{10}}\right )^{\frac {1}{4}} + 7 \, \sqrt {d x} b\right ) + 4 \, {\left (7 \, b x^{2} + 4 \, a\right )} \sqrt {d x}}{24 \, {\left (a^{2} b d^{3} x^{4} + a^{3} d^{3} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x)^(5/2)/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="fricas")

[Out]

-1/24*(84*(a^2*b*d^3*x^4 + a^3*d^3*x^2)*(-b^3/(a^11*d^10))^(1/4)*arctan(-(sqrt(d*x)*a^8*b*d^7*(-b^3/(a^11*d^10
))^(3/4) - sqrt(a^6*d^6*sqrt(-b^3/(a^11*d^10)) + b^2*d*x)*a^8*d^7*(-b^3/(a^11*d^10))^(3/4))/b^3) + 21*(a^2*b*d
^3*x^4 + a^3*d^3*x^2)*(-b^3/(a^11*d^10))^(1/4)*log(7*a^3*d^3*(-b^3/(a^11*d^10))^(1/4) + 7*sqrt(d*x)*b) - 21*(a
^2*b*d^3*x^4 + a^3*d^3*x^2)*(-b^3/(a^11*d^10))^(1/4)*log(-7*a^3*d^3*(-b^3/(a^11*d^10))^(1/4) + 7*sqrt(d*x)*b)
+ 4*(7*b*x^2 + 4*a)*sqrt(d*x))/(a^2*b*d^3*x^4 + a^3*d^3*x^2)

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giac [A]  time = 0.19, size = 276, normalized size = 0.92 \[ -\frac {\sqrt {d x} b}{2 \, {\left (b d^{2} x^{2} + a d^{2}\right )} a^{2} d} - \frac {7 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{8 \, a^{3} d^{3}} - \frac {7 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{8 \, a^{3} d^{3}} - \frac {7 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} \log \left (d x + \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{16 \, a^{3} d^{3}} + \frac {7 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} \log \left (d x - \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{16 \, a^{3} d^{3}} - \frac {2}{3 \, \sqrt {d x} a^{2} d^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x)^(5/2)/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 226, normalized size = 0.75 \[ -\frac {\sqrt {d x}\, b}{2 \left (b \,d^{2} x^{2}+d^{2} a \right ) a^{2} d}-\frac {2}{3 \left (d x \right )^{\frac {3}{2}} a^{2} d}-\frac {7 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, b \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )}{8 a^{3} d^{3}}-\frac {7 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, b \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )}{8 a^{3} d^{3}}-\frac {7 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, b \ln \left (\frac {d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )}{16 a^{3} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*x)^(5/2)/(b^2*x^4+2*a*b*x^2+a^2),x)

[Out]

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

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maxima [A]  time = 3.08, size = 275, normalized size = 0.92 \[ -\frac {\frac {8 \, {\left (7 \, b d^{2} x^{2} + 4 \, a d^{2}\right )}}{\left (d x\right )^{\frac {7}{2}} a^{2} b + \left (d x\right )^{\frac {3}{2}} a^{3} d^{2}} + \frac {21 \, {\left (\frac {\sqrt {2} b^{\frac {3}{4}} \log \left (\sqrt {b} d x + \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {3}{4}}} - \frac {\sqrt {2} b^{\frac {3}{4}} \log \left (\sqrt {b} d x - \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {3}{4}}} + \frac {2 \, \sqrt {2} b \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {a} d} + \frac {2 \, \sqrt {2} b \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {a} d}\right )}}{a^{2}}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x)^(5/2)/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="maxima")

[Out]

-1/48*(8*(7*b*d^2*x^2 + 4*a*d^2)/((d*x)^(7/2)*a^2*b + (d*x)^(3/2)*a^3*d^2) + 21*(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 - sq
rt(2)*(a*d^2)^(1/4)*sqrt(d*x)*b^(1/4) + sqrt(a)*d)/(a*d^2)^(3/4) + 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) + 2*sqr
t(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))/a^2)/d

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mupad [B]  time = 4.40, size = 102, normalized size = 0.34 \[ \frac {7\,{\left (-b\right )}^{3/4}\,\mathrm {atan}\left (\frac {{\left (-b\right )}^{1/4}\,\sqrt {d\,x}}{a^{1/4}\,\sqrt {d}}\right )}{4\,a^{11/4}\,d^{5/2}}-\frac {\frac {2\,d}{3\,a}+\frac {7\,b\,d\,x^2}{6\,a^2}}{b\,{\left (d\,x\right )}^{7/2}+a\,d^2\,{\left (d\,x\right )}^{3/2}}+\frac {7\,{\left (-b\right )}^{3/4}\,\mathrm {atanh}\left (\frac {{\left (-b\right )}^{1/4}\,\sqrt {d\,x}}{a^{1/4}\,\sqrt {d}}\right )}{4\,a^{11/4}\,d^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((d*x)^(5/2)*(a^2 + b^2*x^4 + 2*a*b*x^2)),x)

[Out]

(7*(-b)^(3/4)*atan(((-b)^(1/4)*(d*x)^(1/2))/(a^(1/4)*d^(1/2))))/(4*a^(11/4)*d^(5/2)) - ((2*d)/(3*a) + (7*b*d*x
^2)/(6*a^2))/(b*(d*x)^(7/2) + a*d^2*(d*x)^(3/2)) + (7*(-b)^(3/4)*atanh(((-b)^(1/4)*(d*x)^(1/2))/(a^(1/4)*d^(1/
2))))/(4*a^(11/4)*d^(5/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (d x\right )^{\frac {5}{2}} \left (a + b x^{2}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x)**(5/2)/(b**2*x**4+2*a*b*x**2+a**2),x)

[Out]

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

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