3.702 \(\int \frac {(d x)^{7/2}}{(a^2+2 a b x^2+b^2 x^4)^2} \, dx\)

Optimal. Leaf size=336 \[ -\frac {5 d^{7/2} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{256 \sqrt {2} a^{7/4} b^{9/4}}+\frac {5 d^{7/2} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{256 \sqrt {2} a^{7/4} b^{9/4}}-\frac {5 d^{7/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} a^{7/4} b^{9/4}}+\frac {5 d^{7/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{128 \sqrt {2} a^{7/4} b^{9/4}}+\frac {5 d^3 \sqrt {d x}}{192 a b^2 \left (a+b x^2\right )}-\frac {5 d^3 \sqrt {d x}}{48 b^2 \left (a+b x^2\right )^2}-\frac {d (d x)^{5/2}}{6 b \left (a+b x^2\right )^3} \]

[Out]

-1/6*d*(d*x)^(5/2)/b/(b*x^2+a)^3-5/256*d^(7/2)*arctan(1-b^(1/4)*2^(1/2)*(d*x)^(1/2)/a^(1/4)/d^(1/2))/a^(7/4)/b
^(9/4)*2^(1/2)+5/256*d^(7/2)*arctan(1+b^(1/4)*2^(1/2)*(d*x)^(1/2)/a^(1/4)/d^(1/2))/a^(7/4)/b^(9/4)*2^(1/2)-5/5
12*d^(7/2)*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^(7/4)/b^(9/4)*2^(1/2)+5
/512*d^(7/2)*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^(7/4)/b^(9/4)*2^(1/2)
-5/48*d^3*(d*x)^(1/2)/b^2/(b*x^2+a)^2+5/192*d^3*(d*x)^(1/2)/a/b^2/(b*x^2+a)

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Rubi [A]  time = 0.34, antiderivative size = 336, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {28, 288, 290, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac {5 d^{7/2} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{256 \sqrt {2} a^{7/4} b^{9/4}}+\frac {5 d^{7/2} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{256 \sqrt {2} a^{7/4} b^{9/4}}-\frac {5 d^{7/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} a^{7/4} b^{9/4}}+\frac {5 d^{7/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{128 \sqrt {2} a^{7/4} b^{9/4}}+\frac {5 d^3 \sqrt {d x}}{192 a b^2 \left (a+b x^2\right )}-\frac {5 d^3 \sqrt {d x}}{48 b^2 \left (a+b x^2\right )^2}-\frac {d (d x)^{5/2}}{6 b \left (a+b x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^(7/2)/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

-(d*(d*x)^(5/2))/(6*b*(a + b*x^2)^3) - (5*d^3*Sqrt[d*x])/(48*b^2*(a + b*x^2)^2) + (5*d^3*Sqrt[d*x])/(192*a*b^2
*(a + b*x^2)) - (5*d^(7/2)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(128*Sqrt[2]*a^(7/4)*b^(
9/4)) + (5*d^(7/2)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(128*Sqrt[2]*a^(7/4)*b^(9/4)) -
(5*d^(7/2)*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(256*Sqrt[2]*a^(7/4)*
b^(9/4)) + (5*d^(7/2)*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(256*Sqrt[
2]*a^(7/4)*b^(9/4))

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 288

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 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 {(d x)^{7/2}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx &=b^4 \int \frac {(d x)^{7/2}}{\left (a b+b^2 x^2\right )^4} \, dx\\ &=-\frac {d (d x)^{5/2}}{6 b \left (a+b x^2\right )^3}+\frac {1}{12} \left (5 b^2 d^2\right ) \int \frac {(d x)^{3/2}}{\left (a b+b^2 x^2\right )^3} \, dx\\ &=-\frac {d (d x)^{5/2}}{6 b \left (a+b x^2\right )^3}-\frac {5 d^3 \sqrt {d x}}{48 b^2 \left (a+b x^2\right )^2}+\frac {1}{96} \left (5 d^4\right ) \int \frac {1}{\sqrt {d x} \left (a b+b^2 x^2\right )^2} \, dx\\ &=-\frac {d (d x)^{5/2}}{6 b \left (a+b x^2\right )^3}-\frac {5 d^3 \sqrt {d x}}{48 b^2 \left (a+b x^2\right )^2}+\frac {5 d^3 \sqrt {d x}}{192 a b^2 \left (a+b x^2\right )}+\frac {\left (5 d^4\right ) \int \frac {1}{\sqrt {d x} \left (a b+b^2 x^2\right )} \, dx}{128 a b}\\ &=-\frac {d (d x)^{5/2}}{6 b \left (a+b x^2\right )^3}-\frac {5 d^3 \sqrt {d x}}{48 b^2 \left (a+b x^2\right )^2}+\frac {5 d^3 \sqrt {d x}}{192 a b^2 \left (a+b x^2\right )}+\frac {\left (5 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{a b+\frac {b^2 x^4}{d^2}} \, dx,x,\sqrt {d x}\right )}{64 a b}\\ &=-\frac {d (d x)^{5/2}}{6 b \left (a+b x^2\right )^3}-\frac {5 d^3 \sqrt {d x}}{48 b^2 \left (a+b x^2\right )^2}+\frac {5 d^3 \sqrt {d x}}{192 a b^2 \left (a+b x^2\right )}+\frac {\left (5 d^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 )}{128 a^{3/2} b}+\frac {\left (5 d^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 )}{128 a^{3/2} b}\\ &=-\frac {d (d x)^{5/2}}{6 b \left (a+b x^2\right )^3}-\frac {5 d^3 \sqrt {d x}}{48 b^2 \left (a+b x^2\right )^2}+\frac {5 d^3 \sqrt {d x}}{192 a b^2 \left (a+b x^2\right )}-\frac {\left (5 d^{7/2}\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 )}{256 \sqrt {2} a^{7/4} b^{9/4}}-\frac {\left (5 d^{7/2}\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 )}{256 \sqrt {2} a^{7/4} b^{9/4}}+\frac {\left (5 d^4\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 )}{256 a^{3/2} b^{5/2}}+\frac {\left (5 d^4\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 )}{256 a^{3/2} b^{5/2}}\\ &=-\frac {d (d x)^{5/2}}{6 b \left (a+b x^2\right )^3}-\frac {5 d^3 \sqrt {d x}}{48 b^2 \left (a+b x^2\right )^2}+\frac {5 d^3 \sqrt {d x}}{192 a b^2 \left (a+b x^2\right )}-\frac {5 d^{7/2} \log \left (\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}\right )}{256 \sqrt {2} a^{7/4} b^{9/4}}+\frac {5 d^{7/2} \log \left (\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}\right )}{256 \sqrt {2} a^{7/4} b^{9/4}}+\frac {\left (5 d^{7/2}\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 )}{128 \sqrt {2} a^{7/4} b^{9/4}}-\frac {\left (5 d^{7/2}\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 )}{128 \sqrt {2} a^{7/4} b^{9/4}}\\ &=-\frac {d (d x)^{5/2}}{6 b \left (a+b x^2\right )^3}-\frac {5 d^3 \sqrt {d x}}{48 b^2 \left (a+b x^2\right )^2}+\frac {5 d^3 \sqrt {d x}}{192 a b^2 \left (a+b x^2\right )}-\frac {5 d^{7/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} a^{7/4} b^{9/4}}+\frac {5 d^{7/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} a^{7/4} b^{9/4}}-\frac {5 d^{7/2} \log \left (\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}\right )}{256 \sqrt {2} a^{7/4} b^{9/4}}+\frac {5 d^{7/2} \log \left (\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}\right )}{256 \sqrt {2} a^{7/4} b^{9/4}}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 279, normalized size = 0.83 \[ \frac {d^3 \sqrt {d x} \left (-\frac {105 \sqrt {2} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{a^{7/4} \sqrt {x}}+\frac {105 \sqrt {2} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{a^{7/4} \sqrt {x}}-\frac {210 \sqrt {2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{a^{7/4} \sqrt {x}}+\frac {210 \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{a^{7/4} \sqrt {x}}+\frac {280 \sqrt [4]{b}}{a^2+a b x^2}-\frac {3072 b^{5/4} x^2}{\left (a+b x^2\right )^3}+\frac {160 \sqrt [4]{b}}{\left (a+b x^2\right )^2}-\frac {1280 a \sqrt [4]{b}}{\left (a+b x^2\right )^3}\right )}{10752 b^{9/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^(7/2)/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

(d^3*Sqrt[d*x]*((-1280*a*b^(1/4))/(a + b*x^2)^3 - (3072*b^(5/4)*x^2)/(a + b*x^2)^3 + (160*b^(1/4))/(a + b*x^2)
^2 + (280*b^(1/4))/(a^2 + a*b*x^2) - (210*Sqrt[2]*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(a^(7/4)*Sqrt
[x]) + (210*Sqrt[2]*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(a^(7/4)*Sqrt[x]) - (105*Sqrt[2]*Log[Sqrt[a
] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(a^(7/4)*Sqrt[x]) + (105*Sqrt[2]*Log[Sqrt[a] + Sqrt[2]*a^(1/
4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(a^(7/4)*Sqrt[x])))/(10752*b^(9/4))

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fricas [A]  time = 0.94, size = 389, normalized size = 1.16 \[ \frac {60 \, {\left (a b^{5} x^{6} + 3 \, a^{2} b^{4} x^{4} + 3 \, a^{3} b^{3} x^{2} + a^{4} b^{2}\right )} \left (-\frac {d^{14}}{a^{7} b^{9}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {d x} a^{5} b^{7} d^{3} \left (-\frac {d^{14}}{a^{7} b^{9}}\right )^{\frac {3}{4}} - \sqrt {a^{4} b^{4} \sqrt {-\frac {d^{14}}{a^{7} b^{9}}} + d^{7} x} a^{5} b^{7} \left (-\frac {d^{14}}{a^{7} b^{9}}\right )^{\frac {3}{4}}}{d^{14}}\right ) + 15 \, {\left (a b^{5} x^{6} + 3 \, a^{2} b^{4} x^{4} + 3 \, a^{3} b^{3} x^{2} + a^{4} b^{2}\right )} \left (-\frac {d^{14}}{a^{7} b^{9}}\right )^{\frac {1}{4}} \log \left (5 \, a^{2} b^{2} \left (-\frac {d^{14}}{a^{7} b^{9}}\right )^{\frac {1}{4}} + 5 \, \sqrt {d x} d^{3}\right ) - 15 \, {\left (a b^{5} x^{6} + 3 \, a^{2} b^{4} x^{4} + 3 \, a^{3} b^{3} x^{2} + a^{4} b^{2}\right )} \left (-\frac {d^{14}}{a^{7} b^{9}}\right )^{\frac {1}{4}} \log \left (-5 \, a^{2} b^{2} \left (-\frac {d^{14}}{a^{7} b^{9}}\right )^{\frac {1}{4}} + 5 \, \sqrt {d x} d^{3}\right ) + 4 \, {\left (5 \, b^{2} d^{3} x^{4} - 42 \, a b d^{3} x^{2} - 15 \, a^{2} d^{3}\right )} \sqrt {d x}}{768 \, {\left (a b^{5} x^{6} + 3 \, a^{2} b^{4} x^{4} + 3 \, a^{3} b^{3} x^{2} + a^{4} b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(7/2)/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="fricas")

[Out]

1/768*(60*(a*b^5*x^6 + 3*a^2*b^4*x^4 + 3*a^3*b^3*x^2 + a^4*b^2)*(-d^14/(a^7*b^9))^(1/4)*arctan(-(sqrt(d*x)*a^5
*b^7*d^3*(-d^14/(a^7*b^9))^(3/4) - sqrt(a^4*b^4*sqrt(-d^14/(a^7*b^9)) + d^7*x)*a^5*b^7*(-d^14/(a^7*b^9))^(3/4)
)/d^14) + 15*(a*b^5*x^6 + 3*a^2*b^4*x^4 + 3*a^3*b^3*x^2 + a^4*b^2)*(-d^14/(a^7*b^9))^(1/4)*log(5*a^2*b^2*(-d^1
4/(a^7*b^9))^(1/4) + 5*sqrt(d*x)*d^3) - 15*(a*b^5*x^6 + 3*a^2*b^4*x^4 + 3*a^3*b^3*x^2 + a^4*b^2)*(-d^14/(a^7*b
^9))^(1/4)*log(-5*a^2*b^2*(-d^14/(a^7*b^9))^(1/4) + 5*sqrt(d*x)*d^3) + 4*(5*b^2*d^3*x^4 - 42*a*b*d^3*x^2 - 15*
a^2*d^3)*sqrt(d*x))/(a*b^5*x^6 + 3*a^2*b^4*x^4 + 3*a^3*b^3*x^2 + a^4*b^2)

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giac [A]  time = 0.24, size = 304, normalized size = 0.90 \[ \frac {1}{1536} \, d^{3} {\left (\frac {30 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a^{2} b^{3}} + \frac {30 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a^{2} b^{3}} + \frac {15 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} \log \left (d x + \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{a^{2} b^{3}} - \frac {15 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} \log \left (d x - \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{a^{2} b^{3}} + \frac {8 \, {\left (5 \, \sqrt {d x} b^{2} d^{6} x^{4} - 42 \, \sqrt {d x} a b d^{6} x^{2} - 15 \, \sqrt {d x} a^{2} d^{6}\right )}}{{\left (b d^{2} x^{2} + a d^{2}\right )}^{3} a b^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(7/2)/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="giac")

[Out]

1/1536*d^3*(30*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*b^3) + 30*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*b^3) + 15*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*b^3) - 15*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*b^3) + 8*(5*sqrt(d*x)*b^2*d^6*x^4 - 42*sqrt(d*x)*a*b*d^6*x^2 - 15*sqrt(d*x)*a^2*d^6)/((b*d^2*x^2 + a*d^2)^
3*a*b^2))

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maple [A]  time = 0.02, size = 277, normalized size = 0.82 \[ -\frac {5 \sqrt {d x}\, a \,d^{9}}{64 \left (b \,d^{2} x^{2}+d^{2} a \right )^{3} b^{2}}-\frac {7 \left (d x \right )^{\frac {5}{2}} d^{7}}{32 \left (b \,d^{2} x^{2}+d^{2} a \right )^{3} b}+\frac {5 \left (d x \right )^{\frac {9}{2}} d^{5}}{192 \left (b \,d^{2} x^{2}+d^{2} a \right )^{3} a}+\frac {5 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, d^{3} \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )}{256 a^{2} b^{2}}+\frac {5 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, d^{3} \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )}{256 a^{2} b^{2}}+\frac {5 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, d^{3} \ln \left (\frac {d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )}{512 a^{2} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(7/2)/(b^2*x^4+2*a*b*x^2+a^2)^2,x)

[Out]

5/192*d^5/(b*d^2*x^2+a*d^2)^3/a*(d*x)^(9/2)-7/32*d^7/(b*d^2*x^2+a*d^2)^3/b*(d*x)^(5/2)-5/64*d^9/(b*d^2*x^2+a*d
^2)^3/b^2*a*(d*x)^(1/2)+5/512*d^3/a^2/b^2*(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)))+5/256*d^3/a^2/b^2*(a/b*d^2)^(1/4)*
2^(1/2)*arctan(2^(1/2)/(a/b*d^2)^(1/4)*(d*x)^(1/2)+1)+5/256*d^3/a^2/b^2*(a/b*d^2)^(1/4)*2^(1/2)*arctan(2^(1/2)
/(a/b*d^2)^(1/4)*(d*x)^(1/2)-1)

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maxima [A]  time = 2.96, size = 332, normalized size = 0.99 \[ \frac {\frac {8 \, {\left (5 \, \left (d x\right )^{\frac {9}{2}} b^{2} d^{6} - 42 \, \left (d x\right )^{\frac {5}{2}} a b d^{8} - 15 \, \sqrt {d x} a^{2} d^{10}\right )}}{a b^{5} d^{6} x^{6} + 3 \, a^{2} b^{4} d^{6} x^{4} + 3 \, a^{3} b^{3} d^{6} x^{2} + a^{4} b^{2} d^{6}} + \frac {15 \, {\left (\frac {\sqrt {2} d^{6} \log \left (\sqrt {b} d x + \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {3}{4}} b^{\frac {1}{4}}} - \frac {\sqrt {2} d^{6} \log \left (\sqrt {b} d x - \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {3}{4}} b^{\frac {1}{4}}} + \frac {2 \, \sqrt {2} d^{5} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {a}} + \frac {2 \, \sqrt {2} d^{5} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {a}}\right )}}{a b^{2}}}{1536 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(7/2)/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="maxima")

[Out]

1/1536*(8*(5*(d*x)^(9/2)*b^2*d^6 - 42*(d*x)^(5/2)*a*b*d^8 - 15*sqrt(d*x)*a^2*d^10)/(a*b^5*d^6*x^6 + 3*a^2*b^4*
d^6*x^4 + 3*a^3*b^3*d^6*x^2 + a^4*b^2*d^6) + 15*(sqrt(2)*d^6*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)*b^(1/4)) - sqrt(2)*d^6*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)*b^(1/4)) + 2*sqrt(2)*d^5*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)) + 2*sqrt(2)*d^5*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)))/(a*b^2))/d

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mupad [B]  time = 4.26, size = 150, normalized size = 0.45 \[ \frac {5\,d^{7/2}\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {d\,x}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )}{128\,{\left (-a\right )}^{7/4}\,b^{9/4}}-\frac {\frac {7\,d^7\,{\left (d\,x\right )}^{5/2}}{32\,b}-\frac {5\,d^5\,{\left (d\,x\right )}^{9/2}}{192\,a}+\frac {5\,a\,d^9\,\sqrt {d\,x}}{64\,b^2}}{a^3\,d^6+3\,a^2\,b\,d^6\,x^2+3\,a\,b^2\,d^6\,x^4+b^3\,d^6\,x^6}+\frac {5\,d^{7/2}\,\mathrm {atanh}\left (\frac {b^{1/4}\,\sqrt {d\,x}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )}{128\,{\left (-a\right )}^{7/4}\,b^{9/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(7/2)/(a^2 + b^2*x^4 + 2*a*b*x^2)^2,x)

[Out]

(5*d^(7/2)*atan((b^(1/4)*(d*x)^(1/2))/((-a)^(1/4)*d^(1/2))))/(128*(-a)^(7/4)*b^(9/4)) - ((7*d^7*(d*x)^(5/2))/(
32*b) - (5*d^5*(d*x)^(9/2))/(192*a) + (5*a*d^9*(d*x)^(1/2))/(64*b^2))/(a^3*d^6 + b^3*d^6*x^6 + 3*a^2*b*d^6*x^2
+ 3*a*b^2*d^6*x^4) + (5*d^(7/2)*atanh((b^(1/4)*(d*x)^(1/2))/((-a)^(1/4)*d^(1/2))))/(128*(-a)^(7/4)*b^(9/4))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x\right )^{\frac {7}{2}}}{\left (a + b x^{2}\right )^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**(7/2)/(b**2*x**4+2*a*b*x**2+a**2)**2,x)

[Out]

Integral((d*x)**(7/2)/(a + b*x**2)**4, x)

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