3.6.19 \(\int \frac {(d x)^{17/2}}{(a^2+2 a b x^2+b^2 x^4)^2} \, dx\)

Optimal. Leaf size=350 \[ -\frac {385 a^{3/4} d^{17/2} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{256 \sqrt {2} b^{19/4}}+\frac {385 a^{3/4} d^{17/2} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{256 \sqrt {2} b^{19/4}}+\frac {385 a^{3/4} d^{17/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} b^{19/4}}-\frac {385 a^{3/4} d^{17/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{128 \sqrt {2} b^{19/4}}-\frac {55 d^5 (d x)^{7/2}}{64 b^3 \left (a+b x^2\right )}-\frac {5 d^3 (d x)^{11/2}}{16 b^2 \left (a+b x^2\right )^2}-\frac {d (d x)^{15/2}}{6 b \left (a+b x^2\right )^3}+\frac {385 d^7 (d x)^{3/2}}{192 b^4} \]

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Rubi [A]  time = 0.39, 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, 288, 321, 329, 297, 1162, 617, 204, 1165, 628} \begin {gather*} -\frac {385 a^{3/4} d^{17/2} \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^{19/4}}+\frac {385 a^{3/4} d^{17/2} \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^{19/4}}+\frac {385 a^{3/4} d^{17/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} b^{19/4}}-\frac {385 a^{3/4} d^{17/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{128 \sqrt {2} b^{19/4}}-\frac {55 d^5 (d x)^{7/2}}{64 b^3 \left (a+b x^2\right )}-\frac {5 d^3 (d x)^{11/2}}{16 b^2 \left (a+b x^2\right )^2}-\frac {d (d x)^{15/2}}{6 b \left (a+b x^2\right )^3}+\frac {385 d^7 (d x)^{3/2}}{192 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*x)^(17/2)/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

(385*d^7*(d*x)^(3/2))/(192*b^4) - (d*(d*x)^(15/2))/(6*b*(a + b*x^2)^3) - (5*d^3*(d*x)^(11/2))/(16*b^2*(a + b*x
^2)^2) - (55*d^5*(d*x)^(7/2))/(64*b^3*(a + b*x^2)) + (385*a^(3/4)*d^(17/2)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[d*
x])/(a^(1/4)*Sqrt[d])])/(128*Sqrt[2]*b^(19/4)) - (385*a^(3/4)*d^(17/2)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[d*x])/
(a^(1/4)*Sqrt[d])])/(128*Sqrt[2]*b^(19/4)) - (385*a^(3/4)*d^(17/2)*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x - S
qrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(256*Sqrt[2]*b^(19/4)) + (385*a^(3/4)*d^(17/2)*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^(19/4))

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 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 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 321

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 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)^{17/2}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx &=b^4 \int \frac {(d x)^{17/2}}{\left (a b+b^2 x^2\right )^4} \, dx\\ &=-\frac {d (d x)^{15/2}}{6 b \left (a+b x^2\right )^3}+\frac {1}{4} \left (5 b^2 d^2\right ) \int \frac {(d x)^{13/2}}{\left (a b+b^2 x^2\right )^3} \, dx\\ &=-\frac {d (d x)^{15/2}}{6 b \left (a+b x^2\right )^3}-\frac {5 d^3 (d x)^{11/2}}{16 b^2 \left (a+b x^2\right )^2}+\frac {1}{32} \left (55 d^4\right ) \int \frac {(d x)^{9/2}}{\left (a b+b^2 x^2\right )^2} \, dx\\ &=-\frac {d (d x)^{15/2}}{6 b \left (a+b x^2\right )^3}-\frac {5 d^3 (d x)^{11/2}}{16 b^2 \left (a+b x^2\right )^2}-\frac {55 d^5 (d x)^{7/2}}{64 b^3 \left (a+b x^2\right )}+\frac {\left (385 d^6\right ) \int \frac {(d x)^{5/2}}{a b+b^2 x^2} \, dx}{128 b^2}\\ &=\frac {385 d^7 (d x)^{3/2}}{192 b^4}-\frac {d (d x)^{15/2}}{6 b \left (a+b x^2\right )^3}-\frac {5 d^3 (d x)^{11/2}}{16 b^2 \left (a+b x^2\right )^2}-\frac {55 d^5 (d x)^{7/2}}{64 b^3 \left (a+b x^2\right )}-\frac {\left (385 a d^8\right ) \int \frac {\sqrt {d x}}{a b+b^2 x^2} \, dx}{128 b^3}\\ &=\frac {385 d^7 (d x)^{3/2}}{192 b^4}-\frac {d (d x)^{15/2}}{6 b \left (a+b x^2\right )^3}-\frac {5 d^3 (d x)^{11/2}}{16 b^2 \left (a+b x^2\right )^2}-\frac {55 d^5 (d x)^{7/2}}{64 b^3 \left (a+b x^2\right )}-\frac {\left (385 a d^7\right ) \operatorname {Subst}\left (\int \frac {x^2}{a b+\frac {b^2 x^4}{d^2}} \, dx,x,\sqrt {d x}\right )}{64 b^3}\\ &=\frac {385 d^7 (d x)^{3/2}}{192 b^4}-\frac {d (d x)^{15/2}}{6 b \left (a+b x^2\right )^3}-\frac {5 d^3 (d x)^{11/2}}{16 b^2 \left (a+b x^2\right )^2}-\frac {55 d^5 (d x)^{7/2}}{64 b^3 \left (a+b x^2\right )}+\frac {\left (385 a d^7\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 b^{7/2}}-\frac {\left (385 a d^7\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 b^{7/2}}\\ &=\frac {385 d^7 (d x)^{3/2}}{192 b^4}-\frac {d (d x)^{15/2}}{6 b \left (a+b x^2\right )^3}-\frac {5 d^3 (d x)^{11/2}}{16 b^2 \left (a+b x^2\right )^2}-\frac {55 d^5 (d x)^{7/2}}{64 b^3 \left (a+b x^2\right )}-\frac {\left (385 a^{3/4} d^{17/2}\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} b^{19/4}}-\frac {\left (385 a^{3/4} d^{17/2}\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} b^{19/4}}-\frac {\left (385 a d^9\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 b^5}-\frac {\left (385 a d^9\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 b^5}\\ &=\frac {385 d^7 (d x)^{3/2}}{192 b^4}-\frac {d (d x)^{15/2}}{6 b \left (a+b x^2\right )^3}-\frac {5 d^3 (d x)^{11/2}}{16 b^2 \left (a+b x^2\right )^2}-\frac {55 d^5 (d x)^{7/2}}{64 b^3 \left (a+b x^2\right )}-\frac {385 a^{3/4} d^{17/2} \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^{19/4}}+\frac {385 a^{3/4} d^{17/2} \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^{19/4}}-\frac {\left (385 a^{3/4} d^{17/2}\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} b^{19/4}}+\frac {\left (385 a^{3/4} d^{17/2}\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} b^{19/4}}\\ &=\frac {385 d^7 (d x)^{3/2}}{192 b^4}-\frac {d (d x)^{15/2}}{6 b \left (a+b x^2\right )^3}-\frac {5 d^3 (d x)^{11/2}}{16 b^2 \left (a+b x^2\right )^2}-\frac {55 d^5 (d x)^{7/2}}{64 b^3 \left (a+b x^2\right )}+\frac {385 a^{3/4} d^{17/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} b^{19/4}}-\frac {385 a^{3/4} d^{17/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} b^{19/4}}-\frac {385 a^{3/4} d^{17/2} \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^{19/4}}+\frac {385 a^{3/4} d^{17/2} \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^{19/4}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 87, normalized size = 0.25 \begin {gather*} -\frac {2 d^8 x \sqrt {d x} \left (-77 a^3-99 a^2 b x^2-45 a b^2 x^4+77 \left (a+b x^2\right )^3 \, _2F_1\left (\frac {3}{4},4;\frac {7}{4};-\frac {b x^2}{a}\right )-3 b^3 x^6\right )}{9 b^4 \left (a+b x^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^(17/2)/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

(-2*d^8*x*Sqrt[d*x]*(-77*a^3 - 99*a^2*b*x^2 - 45*a*b^2*x^4 - 3*b^3*x^6 + 77*(a + b*x^2)^3*Hypergeometric2F1[3/
4, 4, 7/4, -((b*x^2)/a)]))/(9*b^4*(a + b*x^2)^3)

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IntegrateAlgebraic [A]  time = 0.89, size = 212, normalized size = 0.61 \begin {gather*} \frac {385 a^{3/4} d^{17/2} \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 )}{128 \sqrt {2} b^{19/4}}+\frac {385 a^{3/4} d^{17/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}}{\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x}\right )}{128 \sqrt {2} b^{19/4}}+\frac {d^8 \sqrt {d x} \left (385 a^3 x+990 a^2 b x^3+765 a b^2 x^5+128 b^3 x^7\right )}{192 b^4 \left (a+b x^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d*x)^(17/2)/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

(d^8*Sqrt[d*x]*(385*a^3*x + 990*a^2*b*x^3 + 765*a*b^2*x^5 + 128*b^3*x^7))/(192*b^4*(a + b*x^2)^3) + (385*a^(3/
4)*d^(17/2)*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]])/(
128*Sqrt[2]*b^(19/4)) + (385*a^(3/4)*d^(17/2)*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x])/(Sqrt[a]*Sqrt[d] + S
qrt[b]*Sqrt[d]*x)])/(128*Sqrt[2]*b^(19/4))

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fricas [A]  time = 2.46, size = 399, normalized size = 1.14 \begin {gather*} \frac {4620 \, \left (-\frac {a^{3} d^{34}}{b^{19}}\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^{3} d^{34}}{b^{19}}\right )^{\frac {1}{4}} \sqrt {d x} a^{2} b^{5} d^{25} - \sqrt {a^{4} d^{51} x - \sqrt {-\frac {a^{3} d^{34}}{b^{19}}} a^{3} b^{9} d^{34}} \left (-\frac {a^{3} d^{34}}{b^{19}}\right )^{\frac {1}{4}} b^{5}}{a^{3} d^{34}}\right ) - 1155 \, \left (-\frac {a^{3} d^{34}}{b^{19}}\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 (57066625 \, \sqrt {d x} a^{2} d^{25} + 57066625 \, \left (-\frac {a^{3} d^{34}}{b^{19}}\right )^{\frac {3}{4}} b^{14}\right ) + 1155 \, \left (-\frac {a^{3} d^{34}}{b^{19}}\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 (57066625 \, \sqrt {d x} a^{2} d^{25} - 57066625 \, \left (-\frac {a^{3} d^{34}}{b^{19}}\right )^{\frac {3}{4}} b^{14}\right ) + 4 \, {\left (128 \, b^{3} d^{8} x^{7} + 765 \, a b^{2} d^{8} x^{5} + 990 \, a^{2} b d^{8} x^{3} + 385 \, a^{3} d^{8} x\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)^(17/2)/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="fricas")

[Out]

1/768*(4620*(-a^3*d^34/b^19)^(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^3*d^34/b^19)
^(1/4)*sqrt(d*x)*a^2*b^5*d^25 - sqrt(a^4*d^51*x - sqrt(-a^3*d^34/b^19)*a^3*b^9*d^34)*(-a^3*d^34/b^19)^(1/4)*b^
5)/(a^3*d^34)) - 1155*(-a^3*d^34/b^19)^(1/4)*(b^7*x^6 + 3*a*b^6*x^4 + 3*a^2*b^5*x^2 + a^3*b^4)*log(57066625*sq
rt(d*x)*a^2*d^25 + 57066625*(-a^3*d^34/b^19)^(3/4)*b^14) + 1155*(-a^3*d^34/b^19)^(1/4)*(b^7*x^6 + 3*a*b^6*x^4
+ 3*a^2*b^5*x^2 + a^3*b^4)*log(57066625*sqrt(d*x)*a^2*d^25 - 57066625*(-a^3*d^34/b^19)^(3/4)*b^14) + 4*(128*b^
3*d^8*x^7 + 765*a*b^2*d^8*x^5 + 990*a^2*b*d^8*x^3 + 385*a^3*d^8*x)*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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giac [A]  time = 0.24, size = 316, normalized size = 0.90 \begin {gather*} \frac {1}{1536} \, d^{8} {\left (\frac {1024 \, \sqrt {d x} x}{b^{4}} - \frac {2310 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{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^{7} d} - \frac {2310 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{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^{7} d} + \frac {1155 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{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^{7} d} - \frac {1155 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{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^{7} d} + \frac {8 \, {\left (381 \, \sqrt {d x} a b^{2} d^{6} x^{5} + 606 \, \sqrt {d x} a^{2} b d^{6} x^{3} + 257 \, \sqrt {d x} a^{3} d^{6} x\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)^(17/2)/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="giac")

[Out]

1/1536*d^8*(1024*sqrt(d*x)*x/b^4 - 2310*sqrt(2)*(a*b^3*d^2)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(a*d^2/b)^(1/4)
+ 2*sqrt(d*x))/(a*d^2/b)^(1/4))/(b^7*d) - 2310*sqrt(2)*(a*b^3*d^2)^(3/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a*d^2/b
)^(1/4) - 2*sqrt(d*x))/(a*d^2/b)^(1/4))/(b^7*d) + 1155*sqrt(2)*(a*b^3*d^2)^(3/4)*log(d*x + sqrt(2)*(a*d^2/b)^(
1/4)*sqrt(d*x) + sqrt(a*d^2/b))/(b^7*d) - 1155*sqrt(2)*(a*b^3*d^2)^(3/4)*log(d*x - sqrt(2)*(a*d^2/b)^(1/4)*sqr
t(d*x) + sqrt(a*d^2/b))/(b^7*d) + 8*(381*sqrt(d*x)*a*b^2*d^6*x^5 + 606*sqrt(d*x)*a^2*b*d^6*x^3 + 257*sqrt(d*x)
*a^3*d^6*x)/((b*d^2*x^2 + a*d^2)^3*b^4))

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maple [A]  time = 0.02, size = 290, normalized size = 0.83 \begin {gather*} \frac {257 \left (d x \right )^{\frac {3}{2}} a^{3} d^{13}}{192 \left (b \,d^{2} x^{2}+d^{2} a \right )^{3} b^{4}}+\frac {101 \left (d x \right )^{\frac {7}{2}} a^{2} d^{11}}{32 \left (b \,d^{2} x^{2}+d^{2} a \right )^{3} b^{3}}+\frac {127 \left (d x \right )^{\frac {11}{2}} a \,d^{9}}{64 \left (b \,d^{2} x^{2}+d^{2} a \right )^{3} b^{2}}-\frac {385 \sqrt {2}\, a \,d^{9} \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )}{256 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} b^{5}}-\frac {385 \sqrt {2}\, a \,d^{9} \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )}{256 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} b^{5}}-\frac {385 \sqrt {2}\, a \,d^{9} \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 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} b^{5}}+\frac {2 \left (d x \right )^{\frac {3}{2}} d^{7}}{3 b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(17/2)/(b^2*x^4+2*a*b*x^2+a^2)^2,x)

[Out]

2/3*d^7*(d*x)^(3/2)/b^4+127/64*d^9*a/b^2/(b*d^2*x^2+a*d^2)^3*(d*x)^(11/2)+101/32*d^11*a^2/b^3/(b*d^2*x^2+a*d^2
)^3*(d*x)^(7/2)+257/192*d^13*a^3/b^4/(b*d^2*x^2+a*d^2)^3*(d*x)^(3/2)-385/512*d^9*a/b^5/(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)))-385/256*d^9*a/b^5/(a/b*d^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b*d^2)^(1/4)*(d*x)^(1/2)+1)-385/256*d^9
*a/b^5/(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 = 3.08, size = 334, normalized size = 0.95 \begin {gather*} -\frac {\frac {1155 \, a d^{10} {\left (\frac {2 \, \sqrt {2} \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 {b}} + \frac {2 \, \sqrt {2} \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 {b}} - \frac {\sqrt {2} \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 {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \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 {1}{4}} b^{\frac {3}{4}}}\right )}}{b^{4}} - \frac {1024 \, \left (d x\right )^{\frac {3}{2}} d^{8}}{b^{4}} - \frac {8 \, {\left (381 \, \left (d x\right )^{\frac {11}{2}} a b^{2} d^{10} + 606 \, \left (d x\right )^{\frac {7}{2}} a^{2} b d^{12} + 257 \, \left (d x\right )^{\frac {3}{2}} 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}}}{1536 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(17/2)/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="maxima")

[Out]

-1/1536*(1155*a*d^10*(2*sqrt(2)*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(b)) + 2*sqrt(2)*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(b)) - sqrt(2)*log(sqrt(b)
*d*x + sqrt(2)*(a*d^2)^(1/4)*sqrt(d*x)*b^(1/4) + sqrt(a)*d)/((a*d^2)^(1/4)*b^(3/4)) + sqrt(2)*log(sqrt(b)*d*x
- sqrt(2)*(a*d^2)^(1/4)*sqrt(d*x)*b^(1/4) + sqrt(a)*d)/((a*d^2)^(1/4)*b^(3/4)))/b^4 - 1024*(d*x)^(3/2)*d^8/b^4
- 8*(381*(d*x)^(11/2)*a*b^2*d^10 + 606*(d*x)^(7/2)*a^2*b*d^12 + 257*(d*x)^(3/2)*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))/d

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mupad [B]  time = 4.33, size = 171, normalized size = 0.49 \begin {gather*} \frac {\frac {257\,a^3\,d^{13}\,{\left (d\,x\right )}^{3/2}}{192}+\frac {101\,a^2\,b\,d^{11}\,{\left (d\,x\right )}^{7/2}}{32}+\frac {127\,a\,b^2\,d^9\,{\left (d\,x\right )}^{11/2}}{64}}{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\,{\left (d\,x\right )}^{3/2}}{3\,b^4}+\frac {385\,{\left (-a\right )}^{3/4}\,d^{17/2}\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {d\,x}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )}{128\,b^{19/4}}+\frac {{\left (-a\right )}^{3/4}\,d^{17/2}\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {d\,x}\,1{}\mathrm {i}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )\,385{}\mathrm {i}}{128\,b^{19/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(17/2)/(a^2 + b^2*x^4 + 2*a*b*x^2)^2,x)

[Out]

((257*a^3*d^13*(d*x)^(3/2))/192 + (101*a^2*b*d^11*(d*x)^(7/2))/32 + (127*a*b^2*d^9*(d*x)^(11/2))/64)/(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)^(3/2))/(3*b^4) + (385*(-a)^(3/4)*d^(17
/2)*atan((b^(1/4)*(d*x)^(1/2))/((-a)^(1/4)*d^(1/2))))/(128*b^(19/4)) + ((-a)^(3/4)*d^(17/2)*atan((b^(1/4)*(d*x
)^(1/2)*1i)/((-a)^(1/4)*d^(1/2)))*385i)/(128*b^(19/4))

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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((d*x)**(17/2)/(b**2*x**4+2*a*b*x**2+a**2)**2,x)

[Out]

Timed out

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