3.7.98 \(\int \frac {(d x)^{15/2}}{(a^2+2 a b x^2+b^2 x^4)^2} \, dx\) [698]

Optimal. Leaf size=350 \[ \frac {195 d^7 \sqrt {d x}}{64 b^4}-\frac {d (d x)^{13/2}}{6 b \left (a+b x^2\right )^3}-\frac {13 d^3 (d x)^{9/2}}{48 b^2 \left (a+b x^2\right )^2}-\frac {39 d^5 (d x)^{5/2}}{64 b^3 \left (a+b x^2\right )}+\frac {195 \sqrt [4]{a} d^{15/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} b^{17/4}}-\frac {195 \sqrt [4]{a} d^{15/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} b^{17/4}}+\frac {195 \sqrt [4]{a} d^{15/2} \log \left (\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}\right )}{256 \sqrt {2} b^{17/4}}-\frac {195 \sqrt [4]{a} d^{15/2} \log \left (\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}\right )}{256 \sqrt {2} b^{17/4}} \]

[Out]

-1/6*d*(d*x)^(13/2)/b/(b*x^2+a)^3-13/48*d^3*(d*x)^(9/2)/b^2/(b*x^2+a)^2-39/64*d^5*(d*x)^(5/2)/b^3/(b*x^2+a)+19
5/256*a^(1/4)*d^(15/2)*arctan(1-b^(1/4)*2^(1/2)*(d*x)^(1/2)/a^(1/4)/d^(1/2))/b^(17/4)*2^(1/2)-195/256*a^(1/4)*
d^(15/2)*arctan(1+b^(1/4)*2^(1/2)*(d*x)^(1/2)/a^(1/4)/d^(1/2))/b^(17/4)*2^(1/2)+195/512*a^(1/4)*d^(15/2)*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))/b^(17/4)*2^(1/2)-195/512*a^(1/4)*d^(15/2)
*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))/b^(17/4)*2^(1/2)+195/64*d^7*(d*x)^(
1/2)/b^4

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Rubi [A]
time = 0.25, antiderivative size = 350, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {28, 294, 327, 335, 217, 1179, 642, 1176, 631, 210} \begin {gather*} \frac {195 \sqrt [4]{a} d^{15/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} b^{17/4}}-\frac {195 \sqrt [4]{a} d^{15/2} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{128 \sqrt {2} b^{17/4}}+\frac {195 \sqrt [4]{a} d^{15/2} \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} b^{17/4}}-\frac {195 \sqrt [4]{a} d^{15/2} \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} b^{17/4}}-\frac {39 d^5 (d x)^{5/2}}{64 b^3 \left (a+b x^2\right )}-\frac {13 d^3 (d x)^{9/2}}{48 b^2 \left (a+b x^2\right )^2}-\frac {d (d x)^{13/2}}{6 b \left (a+b x^2\right )^3}+\frac {195 d^7 \sqrt {d x}}{64 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*x)^(15/2)/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

(195*d^7*Sqrt[d*x])/(64*b^4) - (d*(d*x)^(13/2))/(6*b*(a + b*x^2)^3) - (13*d^3*(d*x)^(9/2))/(48*b^2*(a + b*x^2)
^2) - (39*d^5*(d*x)^(5/2))/(64*b^3*(a + b*x^2)) + (195*a^(1/4)*d^(15/2)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[d*x])
/(a^(1/4)*Sqrt[d])])/(128*Sqrt[2]*b^(17/4)) - (195*a^(1/4)*d^(15/2)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^
(1/4)*Sqrt[d])])/(128*Sqrt[2]*b^(17/4)) + (195*a^(1/4)*d^(15/2)*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x - Sqrt
[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(256*Sqrt[2]*b^(17/4)) - (195*a^(1/4)*d^(15/2)*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*S
qrt[d]*x + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(256*Sqrt[2]*b^(17/4))

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 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 217

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 294

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 327

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*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 335

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 631

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 642

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 1176

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 1179

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)^{15/2}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx &=b^4 \int \frac {(d x)^{15/2}}{\left (a b+b^2 x^2\right )^4} \, dx\\ &=-\frac {d (d x)^{13/2}}{6 b \left (a+b x^2\right )^3}+\frac {1}{12} \left (13 b^2 d^2\right ) \int \frac {(d x)^{11/2}}{\left (a b+b^2 x^2\right )^3} \, dx\\ &=-\frac {d (d x)^{13/2}}{6 b \left (a+b x^2\right )^3}-\frac {13 d^3 (d x)^{9/2}}{48 b^2 \left (a+b x^2\right )^2}+\frac {1}{32} \left (39 d^4\right ) \int \frac {(d x)^{7/2}}{\left (a b+b^2 x^2\right )^2} \, dx\\ &=-\frac {d (d x)^{13/2}}{6 b \left (a+b x^2\right )^3}-\frac {13 d^3 (d x)^{9/2}}{48 b^2 \left (a+b x^2\right )^2}-\frac {39 d^5 (d x)^{5/2}}{64 b^3 \left (a+b x^2\right )}+\frac {\left (195 d^6\right ) \int \frac {(d x)^{3/2}}{a b+b^2 x^2} \, dx}{128 b^2}\\ &=\frac {195 d^7 \sqrt {d x}}{64 b^4}-\frac {d (d x)^{13/2}}{6 b \left (a+b x^2\right )^3}-\frac {13 d^3 (d x)^{9/2}}{48 b^2 \left (a+b x^2\right )^2}-\frac {39 d^5 (d x)^{5/2}}{64 b^3 \left (a+b x^2\right )}-\frac {\left (195 a d^8\right ) \int \frac {1}{\sqrt {d x} \left (a b+b^2 x^2\right )} \, dx}{128 b^3}\\ &=\frac {195 d^7 \sqrt {d x}}{64 b^4}-\frac {d (d x)^{13/2}}{6 b \left (a+b x^2\right )^3}-\frac {13 d^3 (d x)^{9/2}}{48 b^2 \left (a+b x^2\right )^2}-\frac {39 d^5 (d x)^{5/2}}{64 b^3 \left (a+b x^2\right )}-\frac {\left (195 a d^7\right ) \text {Subst}\left (\int \frac {1}{a b+\frac {b^2 x^4}{d^2}} \, dx,x,\sqrt {d x}\right )}{64 b^3}\\ &=\frac {195 d^7 \sqrt {d x}}{64 b^4}-\frac {d (d x)^{13/2}}{6 b \left (a+b x^2\right )^3}-\frac {13 d^3 (d x)^{9/2}}{48 b^2 \left (a+b x^2\right )^2}-\frac {39 d^5 (d x)^{5/2}}{64 b^3 \left (a+b x^2\right )}-\frac {\left (195 \sqrt {a} d^6\right ) \text {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 b^3}-\frac {\left (195 \sqrt {a} d^6\right ) \text {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 b^3}\\ &=\frac {195 d^7 \sqrt {d x}}{64 b^4}-\frac {d (d x)^{13/2}}{6 b \left (a+b x^2\right )^3}-\frac {13 d^3 (d x)^{9/2}}{48 b^2 \left (a+b x^2\right )^2}-\frac {39 d^5 (d x)^{5/2}}{64 b^3 \left (a+b x^2\right )}+\frac {\left (195 \sqrt [4]{a} d^{15/2}\right ) \text {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} b^{17/4}}+\frac {\left (195 \sqrt [4]{a} d^{15/2}\right ) \text {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} b^{17/4}}-\frac {\left (195 \sqrt {a} d^8\right ) \text {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 b^{9/2}}-\frac {\left (195 \sqrt {a} d^8\right ) \text {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 b^{9/2}}\\ &=\frac {195 d^7 \sqrt {d x}}{64 b^4}-\frac {d (d x)^{13/2}}{6 b \left (a+b x^2\right )^3}-\frac {13 d^3 (d x)^{9/2}}{48 b^2 \left (a+b x^2\right )^2}-\frac {39 d^5 (d x)^{5/2}}{64 b^3 \left (a+b x^2\right )}+\frac {195 \sqrt [4]{a} d^{15/2} \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} b^{17/4}}-\frac {195 \sqrt [4]{a} d^{15/2} \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} b^{17/4}}-\frac {\left (195 \sqrt [4]{a} d^{15/2}\right ) \text {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} b^{17/4}}+\frac {\left (195 \sqrt [4]{a} d^{15/2}\right ) \text {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} b^{17/4}}\\ &=\frac {195 d^7 \sqrt {d x}}{64 b^4}-\frac {d (d x)^{13/2}}{6 b \left (a+b x^2\right )^3}-\frac {13 d^3 (d x)^{9/2}}{48 b^2 \left (a+b x^2\right )^2}-\frac {39 d^5 (d x)^{5/2}}{64 b^3 \left (a+b x^2\right )}+\frac {195 \sqrt [4]{a} d^{15/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} b^{17/4}}-\frac {195 \sqrt [4]{a} d^{15/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} b^{17/4}}+\frac {195 \sqrt [4]{a} d^{15/2} \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} b^{17/4}}-\frac {195 \sqrt [4]{a} d^{15/2} \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} b^{17/4}}\\ \end {align*}

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Mathematica [A]
time = 0.47, size = 175, normalized size = 0.50 \begin {gather*} \frac {d^7 \sqrt {d x} \left (\frac {4 \sqrt [4]{b} \sqrt {x} \left (585 a^3+1638 a^2 b x^2+1469 a b^2 x^4+384 b^3 x^6\right )}{\left (a+b x^2\right )^3}+585 \sqrt {2} \sqrt [4]{a} \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )-585 \sqrt {2} \sqrt [4]{a} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )\right )}{768 b^{17/4} \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^(15/2)/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

(d^7*Sqrt[d*x]*((4*b^(1/4)*Sqrt[x]*(585*a^3 + 1638*a^2*b*x^2 + 1469*a*b^2*x^4 + 384*b^3*x^6))/(a + b*x^2)^3 +
585*Sqrt[2]*a^(1/4)*ArcTan[(Sqrt[a] - Sqrt[b]*x)/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])] - 585*Sqrt[2]*a^(1/4)*ArcT
anh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[b]*x)]))/(768*b^(17/4)*Sqrt[x])

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Maple [A]
time = 0.09, size = 220, normalized size = 0.63

method	result	size
derivativedivides	\(2 d^{7} \left (\frac {\sqrt {d x}}{b^{4}}-\frac {a \,d^{2} \left (\frac {-\frac {317 b^{2} \left (d x \right )^{\frac {9}{2}}}{384}-\frac {81 a b \,d^{2} \left (d x \right )^{\frac {5}{2}}}{64}-\frac {67 a^{2} d^{4} \sqrt {d x}}{128}}{\left (d^{2} x^{2} b +a \,d^{2}\right )^{3}}+\frac {195 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{1024 a \,d^{2}}\right )}{b^{4}}\right )\)	\(220\)
default	\(2 d^{7} \left (\frac {\sqrt {d x}}{b^{4}}-\frac {a \,d^{2} \left (\frac {-\frac {317 b^{2} \left (d x \right )^{\frac {9}{2}}}{384}-\frac {81 a b \,d^{2} \left (d x \right )^{\frac {5}{2}}}{64}-\frac {67 a^{2} d^{4} \sqrt {d x}}{128}}{\left (d^{2} x^{2} b +a \,d^{2}\right )^{3}}+\frac {195 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{1024 a \,d^{2}}\right )}{b^{4}}\right )\)	\(220\)
risch	\(\frac {2 x \,d^{8}}{b^{4} \sqrt {d x}}+\left (\frac {317 a d \left (d x \right )^{\frac {9}{2}}}{192 b^{2} \left (d^{2} x^{2} b +a \,d^{2}\right )^{3}}+\frac {81 a^{2} d^{3} \left (d x \right )^{\frac {5}{2}}}{32 b^{3} \left (d^{2} x^{2} b +a \,d^{2}\right )^{3}}+\frac {67 a^{3} d^{5} \sqrt {d x}}{64 b^{4} \left (d^{2} x^{2} b +a \,d^{2}\right )^{3}}-\frac {195 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )}{512 b^{4} d}-\frac {195 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )}{256 b^{4} d}-\frac {195 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )}{256 b^{4} d}\right ) d^{8}\)	\(291\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(15/2)/(b^2*x^4+2*a*b*x^2+a^2)^2,x,method=_RETURNVERBOSE)

[Out]

2*d^7*(1/b^4*(d*x)^(1/2)-1/b^4*a*d^2*((-317/384*b^2*(d*x)^(9/2)-81/64*a*b*d^2*(d*x)^(5/2)-67/128*a^2*d^4*(d*x)
^(1/2))/(b*d^2*x^2+a*d^2)^3+195/1024*(a*d^2/b)^(1/4)/a/d^2*2^(1/2)*(ln((d*x+(a*d^2/b)^(1/4)*(d*x)^(1/2)*2^(1/2
)+(a*d^2/b)^(1/2))/(d*x-(a*d^2/b)^(1/4)*(d*x)^(1/2)*2^(1/2)+(a*d^2/b)^(1/2)))+2*arctan(2^(1/2)/(a*d^2/b)^(1/4)
*(d*x)^(1/2)+1)+2*arctan(2^(1/2)/(a*d^2/b)^(1/4)*(d*x)^(1/2)-1))))

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Maxima [A]
time = 0.50, size = 343, normalized size = 0.98 \begin {gather*} \frac {\frac {3072 \, \sqrt {d x} d^{8}}{b^{4}} + \frac {8 \, {\left (317 \, \left (d x\right )^{\frac {9}{2}} a b^{2} d^{10} + 486 \, \left (d x\right )^{\frac {5}{2}} a^{2} b d^{12} + 201 \, \sqrt {d x} a^{3} d^{14}\right )}}{b^{7} d^{6} x^{6} + 3 \, a b^{6} d^{6} x^{4} + 3 \, a^{2} b^{5} d^{6} x^{2} + a^{3} b^{4} d^{6}} - \frac {585 \, {\left (\frac {\sqrt {2} d^{10} \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^{10} \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^{9} \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^{9} \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^{4}}}{1536 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(15/2)/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="maxima")

[Out]

1/1536*(3072*sqrt(d*x)*d^8/b^4 + 8*(317*(d*x)^(9/2)*a*b^2*d^10 + 486*(d*x)^(5/2)*a^2*b*d^12 + 201*sqrt(d*x)*a^
3*d^14)/(b^7*d^6*x^6 + 3*a*b^6*d^6*x^4 + 3*a^2*b^5*d^6*x^2 + a^3*b^4*d^6) - 585*(sqrt(2)*d^10*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^10*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^9*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)*sq
rt(a)) + 2*sqrt(2)*d^9*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^4)/d

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Fricas [A]
time = 0.37, size = 363, normalized size = 1.04 \begin {gather*} -\frac {2340 \, \left (-\frac {a d^{30}}{b^{17}}\right )^{\frac {1}{4}} {\left (b^{7} x^{6} + 3 \, a b^{6} x^{4} + 3 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )} \arctan \left (-\frac {\left (-\frac {a d^{30}}{b^{17}}\right )^{\frac {3}{4}} \sqrt {d x} b^{13} d^{7} - \sqrt {d^{15} x + \sqrt {-\frac {a d^{30}}{b^{17}}} b^{8}} \left (-\frac {a d^{30}}{b^{17}}\right )^{\frac {3}{4}} b^{13}}{a d^{30}}\right ) + 585 \, \left (-\frac {a d^{30}}{b^{17}}\right )^{\frac {1}{4}} {\left (b^{7} x^{6} + 3 \, a b^{6} x^{4} + 3 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )} \log \left (195 \, \sqrt {d x} d^{7} + 195 \, \left (-\frac {a d^{30}}{b^{17}}\right )^{\frac {1}{4}} b^{4}\right ) - 585 \, \left (-\frac {a d^{30}}{b^{17}}\right )^{\frac {1}{4}} {\left (b^{7} x^{6} + 3 \, a b^{6} x^{4} + 3 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )} \log \left (195 \, \sqrt {d x} d^{7} - 195 \, \left (-\frac {a d^{30}}{b^{17}}\right )^{\frac {1}{4}} b^{4}\right ) - 4 \, {\left (384 \, b^{3} d^{7} x^{6} + 1469 \, a b^{2} d^{7} x^{4} + 1638 \, a^{2} b d^{7} x^{2} + 585 \, a^{3} d^{7}\right )} \sqrt {d x}}{768 \, {\left (b^{7} x^{6} + 3 \, a b^{6} x^{4} + 3 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(15/2)/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="fricas")

[Out]

-1/768*(2340*(-a*d^30/b^17)^(1/4)*(b^7*x^6 + 3*a*b^6*x^4 + 3*a^2*b^5*x^2 + a^3*b^4)*arctan(-((-a*d^30/b^17)^(3
/4)*sqrt(d*x)*b^13*d^7 - sqrt(d^15*x + sqrt(-a*d^30/b^17)*b^8)*(-a*d^30/b^17)^(3/4)*b^13)/(a*d^30)) + 585*(-a*
d^30/b^17)^(1/4)*(b^7*x^6 + 3*a*b^6*x^4 + 3*a^2*b^5*x^2 + a^3*b^4)*log(195*sqrt(d*x)*d^7 + 195*(-a*d^30/b^17)^
(1/4)*b^4) - 585*(-a*d^30/b^17)^(1/4)*(b^7*x^6 + 3*a*b^6*x^4 + 3*a^2*b^5*x^2 + a^3*b^4)*log(195*sqrt(d*x)*d^7
- 195*(-a*d^30/b^17)^(1/4)*b^4) - 4*(384*b^3*d^7*x^6 + 1469*a*b^2*d^7*x^4 + 1638*a^2*b*d^7*x^2 + 585*a^3*d^7)*
sqrt(d*x))/(b^7*x^6 + 3*a*b^6*x^4 + 3*a^2*b^5*x^2 + a^3*b^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d x\right )^{\frac {15}{2}}}{\left (a + b x^{2}\right )^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**(15/2)/(b**2*x**4+2*a*b*x**2+a**2)**2,x)

[Out]

Integral((d*x)**(15/2)/(a + b*x**2)**4, x)

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Giac [A]
time = 3.17, size = 302, normalized size = 0.86 \begin {gather*} -\frac {1}{1536} \, d^{7} {\left (\frac {1170 \, \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 )}{b^{5}} + \frac {1170 \, \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 )}{b^{5}} + \frac {585 \, \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 )}{b^{5}} - \frac {585 \, \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 )}{b^{5}} - \frac {3072 \, \sqrt {d x}}{b^{4}} - \frac {8 \, {\left (317 \, \sqrt {d x} a b^{2} d^{6} x^{4} + 486 \, \sqrt {d x} a^{2} b d^{6} x^{2} + 201 \, \sqrt {d x} a^{3} d^{6}\right )}}{{\left (b d^{2} x^{2} + a d^{2}\right )}^{3} b^{4}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(15/2)/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="giac")

[Out]

-1/1536*d^7*(1170*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))/b^5 + 1170*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))/b^5 + 585*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))
/b^5 - 585*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))/b^5 - 3072*s
qrt(d*x)/b^4 - 8*(317*sqrt(d*x)*a*b^2*d^6*x^4 + 486*sqrt(d*x)*a^2*b*d^6*x^2 + 201*sqrt(d*x)*a^3*d^6)/((b*d^2*x
^2 + a*d^2)^3*b^4))

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Mupad [B]
time = 4.30, size = 171, normalized size = 0.49 \begin {gather*} \frac {\frac {67\,a^3\,d^{13}\,\sqrt {d\,x}}{64}+\frac {81\,a^2\,b\,d^{11}\,{\left (d\,x\right )}^{5/2}}{32}+\frac {317\,a\,b^2\,d^9\,{\left (d\,x\right )}^{9/2}}{192}}{a^3\,b^4\,d^6+3\,a^2\,b^5\,d^6\,x^2+3\,a\,b^6\,d^6\,x^4+b^7\,d^6\,x^6}+\frac {2\,d^7\,\sqrt {d\,x}}{b^4}-\frac {195\,{\left (-a\right )}^{1/4}\,d^{15/2}\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {d\,x}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )}{128\,b^{17/4}}+\frac {{\left (-a\right )}^{1/4}\,d^{15/2}\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {d\,x}\,1{}\mathrm {i}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )\,195{}\mathrm {i}}{128\,b^{17/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(15/2)/(a^2 + b^2*x^4 + 2*a*b*x^2)^2,x)

[Out]

((67*a^3*d^13*(d*x)^(1/2))/64 + (81*a^2*b*d^11*(d*x)^(5/2))/32 + (317*a*b^2*d^9*(d*x)^(9/2))/192)/(a^3*b^4*d^6
+ b^7*d^6*x^6 + 3*a*b^6*d^6*x^4 + 3*a^2*b^5*d^6*x^2) + (2*d^7*(d*x)^(1/2))/b^4 - (195*(-a)^(1/4)*d^(15/2)*ata
n((b^(1/4)*(d*x)^(1/2))/((-a)^(1/4)*d^(1/2))))/(128*b^(17/4)) + ((-a)^(1/4)*d^(15/2)*atan((b^(1/4)*(d*x)^(1/2)
*1i)/((-a)^(1/4)*d^(1/2)))*195i)/(128*b^(17/4))

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