3.210 \(\int \frac {x^{17/2} (A+B x^2)}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=293 \[ \frac {3 (A c+7 b B) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{5/4} c^{11/4}}-\frac {3 (A c+7 b B) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{5/4} c^{11/4}}-\frac {3 (A c+7 b B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{5/4} c^{11/4}}+\frac {3 (A c+7 b B) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt {2} b^{5/4} c^{11/4}}-\frac {x^{3/2} (A c+7 b B)}{16 b c^2 \left (b+c x^2\right )}-\frac {x^{7/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2} \]

[Out]

-1/4*(-A*c+B*b)*x^(7/2)/b/c/(c*x^2+b)^2-1/16*(A*c+7*B*b)*x^(3/2)/b/c^2/(c*x^2+b)-3/64*(A*c+7*B*b)*arctan(1-c^(
1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(5/4)/c^(11/4)*2^(1/2)+3/64*(A*c+7*B*b)*arctan(1+c^(1/4)*2^(1/2)*x^(1/2)/b^(1/
4))/b^(5/4)/c^(11/4)*2^(1/2)+3/128*(A*c+7*B*b)*ln(b^(1/2)+x*c^(1/2)-b^(1/4)*c^(1/4)*2^(1/2)*x^(1/2))/b^(5/4)/c
^(11/4)*2^(1/2)-3/128*(A*c+7*B*b)*ln(b^(1/2)+x*c^(1/2)+b^(1/4)*c^(1/4)*2^(1/2)*x^(1/2))/b^(5/4)/c^(11/4)*2^(1/
2)

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Rubi [A]  time = 0.22, antiderivative size = 293, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {1584, 457, 288, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac {3 (A c+7 b B) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{5/4} c^{11/4}}-\frac {3 (A c+7 b B) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{5/4} c^{11/4}}-\frac {3 (A c+7 b B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{5/4} c^{11/4}}+\frac {3 (A c+7 b B) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt {2} b^{5/4} c^{11/4}}-\frac {x^{3/2} (A c+7 b B)}{16 b c^2 \left (b+c x^2\right )}-\frac {x^{7/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^(17/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

-((b*B - A*c)*x^(7/2))/(4*b*c*(b + c*x^2)^2) - ((7*b*B + A*c)*x^(3/2))/(16*b*c^2*(b + c*x^2)) - (3*(7*b*B + A*
c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(5/4)*c^(11/4)) + (3*(7*b*B + A*c)*ArcTan[1 +
(Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(5/4)*c^(11/4)) + (3*(7*b*B + A*c)*Log[Sqrt[b] - Sqrt[2]*b^(
1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(5/4)*c^(11/4)) - (3*(7*b*B + A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1
/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(5/4)*c^(11/4))

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 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 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 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

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]

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {align*} \int \frac {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac {x^{5/2} \left (A+B x^2\right )}{\left (b+c x^2\right )^3} \, dx\\ &=-\frac {(b B-A c) x^{7/2}}{4 b c \left (b+c x^2\right )^2}+\frac {\left (\frac {7 b B}{2}+\frac {A c}{2}\right ) \int \frac {x^{5/2}}{\left (b+c x^2\right )^2} \, dx}{4 b c}\\ &=-\frac {(b B-A c) x^{7/2}}{4 b c \left (b+c x^2\right )^2}-\frac {(7 b B+A c) x^{3/2}}{16 b c^2 \left (b+c x^2\right )}+\frac {(3 (7 b B+A c)) \int \frac {\sqrt {x}}{b+c x^2} \, dx}{32 b c^2}\\ &=-\frac {(b B-A c) x^{7/2}}{4 b c \left (b+c x^2\right )^2}-\frac {(7 b B+A c) x^{3/2}}{16 b c^2 \left (b+c x^2\right )}+\frac {(3 (7 b B+A c)) \operatorname {Subst}\left (\int \frac {x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{16 b c^2}\\ &=-\frac {(b B-A c) x^{7/2}}{4 b c \left (b+c x^2\right )^2}-\frac {(7 b B+A c) x^{3/2}}{16 b c^2 \left (b+c x^2\right )}-\frac {(3 (7 b B+A c)) \operatorname {Subst}\left (\int \frac {\sqrt {b}-\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{32 b c^{5/2}}+\frac {(3 (7 b B+A c)) \operatorname {Subst}\left (\int \frac {\sqrt {b}+\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{32 b c^{5/2}}\\ &=-\frac {(b B-A c) x^{7/2}}{4 b c \left (b+c x^2\right )^2}-\frac {(7 b B+A c) x^{3/2}}{16 b c^2 \left (b+c x^2\right )}+\frac {(3 (7 b B+A c)) \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {b}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt {x}\right )}{64 b c^3}+\frac {(3 (7 b B+A c)) \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {b}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt {x}\right )}{64 b c^3}+\frac {(3 (7 b B+A c)) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{b}}{\sqrt [4]{c}}+2 x}{-\frac {\sqrt {b}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt {x}\right )}{64 \sqrt {2} b^{5/4} c^{11/4}}+\frac {(3 (7 b B+A c)) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{b}}{\sqrt [4]{c}}-2 x}{-\frac {\sqrt {b}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt {x}\right )}{64 \sqrt {2} b^{5/4} c^{11/4}}\\ &=-\frac {(b B-A c) x^{7/2}}{4 b c \left (b+c x^2\right )^2}-\frac {(7 b B+A c) x^{3/2}}{16 b c^2 \left (b+c x^2\right )}+\frac {3 (7 b B+A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{5/4} c^{11/4}}-\frac {3 (7 b B+A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{5/4} c^{11/4}}+\frac {(3 (7 b B+A c)) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{5/4} c^{11/4}}-\frac {(3 (7 b B+A c)) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{5/4} c^{11/4}}\\ &=-\frac {(b B-A c) x^{7/2}}{4 b c \left (b+c x^2\right )^2}-\frac {(7 b B+A c) x^{3/2}}{16 b c^2 \left (b+c x^2\right )}-\frac {3 (7 b B+A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{5/4} c^{11/4}}+\frac {3 (7 b B+A c) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{5/4} c^{11/4}}+\frac {3 (7 b B+A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{5/4} c^{11/4}}-\frac {3 (7 b B+A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{5/4} c^{11/4}}\\ \end {align*}

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Mathematica [C]  time = 0.32, size = 137, normalized size = 0.47 \[ \frac {2 c^{3/4} x^{3/2} (A c-2 b B) \, _2F_1\left (\frac {3}{4},2;\frac {7}{4};-\frac {c x^2}{b}\right )+2 c^{3/4} x^{3/2} (b B-A c) \, _2F_1\left (\frac {3}{4},3;\frac {7}{4};-\frac {c x^2}{b}\right )+3 (-b)^{7/4} B \left (\tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b}}\right )+\tanh ^{-1}\left (\frac {b \sqrt [4]{c} \sqrt {x}}{(-b)^{5/4}}\right )\right )}{3 b^2 c^{11/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(17/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

(3*(-b)^(7/4)*B*(ArcTan[(c^(1/4)*Sqrt[x])/(-b)^(1/4)] + ArcTanh[(b*c^(1/4)*Sqrt[x])/(-b)^(5/4)]) + 2*c^(3/4)*(
-2*b*B + A*c)*x^(3/2)*Hypergeometric2F1[3/4, 2, 7/4, -((c*x^2)/b)] + 2*c^(3/4)*(b*B - A*c)*x^(3/2)*Hypergeomet
ric2F1[3/4, 3, 7/4, -((c*x^2)/b)])/(3*b^2*c^(11/4))

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fricas [B]  time = 0.94, size = 990, normalized size = 3.38 \[ -\frac {12 \, {\left (b c^{4} x^{4} + 2 \, b^{2} c^{3} x^{2} + b^{3} c^{2}\right )} \left (-\frac {2401 \, B^{4} b^{4} + 1372 \, A B^{3} b^{3} c + 294 \, A^{2} B^{2} b^{2} c^{2} + 28 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{5} c^{11}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {{\left (117649 \, B^{6} b^{6} + 100842 \, A B^{5} b^{5} c + 36015 \, A^{2} B^{4} b^{4} c^{2} + 6860 \, A^{3} B^{3} b^{3} c^{3} + 735 \, A^{4} B^{2} b^{2} c^{4} + 42 \, A^{5} B b c^{5} + A^{6} c^{6}\right )} x - {\left (2401 \, B^{4} b^{7} c^{5} + 1372 \, A B^{3} b^{6} c^{6} + 294 \, A^{2} B^{2} b^{5} c^{7} + 28 \, A^{3} B b^{4} c^{8} + A^{4} b^{3} c^{9}\right )} \sqrt {-\frac {2401 \, B^{4} b^{4} + 1372 \, A B^{3} b^{3} c + 294 \, A^{2} B^{2} b^{2} c^{2} + 28 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{5} c^{11}}}} b c^{3} \left (-\frac {2401 \, B^{4} b^{4} + 1372 \, A B^{3} b^{3} c + 294 \, A^{2} B^{2} b^{2} c^{2} + 28 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{5} c^{11}}\right )^{\frac {1}{4}} - {\left (343 \, B^{3} b^{4} c^{3} + 147 \, A B^{2} b^{3} c^{4} + 21 \, A^{2} B b^{2} c^{5} + A^{3} b c^{6}\right )} \sqrt {x} \left (-\frac {2401 \, B^{4} b^{4} + 1372 \, A B^{3} b^{3} c + 294 \, A^{2} B^{2} b^{2} c^{2} + 28 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{5} c^{11}}\right )^{\frac {1}{4}}}{2401 \, B^{4} b^{4} + 1372 \, A B^{3} b^{3} c + 294 \, A^{2} B^{2} b^{2} c^{2} + 28 \, A^{3} B b c^{3} + A^{4} c^{4}}\right ) - 3 \, {\left (b c^{4} x^{4} + 2 \, b^{2} c^{3} x^{2} + b^{3} c^{2}\right )} \left (-\frac {2401 \, B^{4} b^{4} + 1372 \, A B^{3} b^{3} c + 294 \, A^{2} B^{2} b^{2} c^{2} + 28 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{5} c^{11}}\right )^{\frac {1}{4}} \log \left (27 \, b^{4} c^{8} \left (-\frac {2401 \, B^{4} b^{4} + 1372 \, A B^{3} b^{3} c + 294 \, A^{2} B^{2} b^{2} c^{2} + 28 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{5} c^{11}}\right )^{\frac {3}{4}} + 27 \, {\left (343 \, B^{3} b^{3} + 147 \, A B^{2} b^{2} c + 21 \, A^{2} B b c^{2} + A^{3} c^{3}\right )} \sqrt {x}\right ) + 3 \, {\left (b c^{4} x^{4} + 2 \, b^{2} c^{3} x^{2} + b^{3} c^{2}\right )} \left (-\frac {2401 \, B^{4} b^{4} + 1372 \, A B^{3} b^{3} c + 294 \, A^{2} B^{2} b^{2} c^{2} + 28 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{5} c^{11}}\right )^{\frac {1}{4}} \log \left (-27 \, b^{4} c^{8} \left (-\frac {2401 \, B^{4} b^{4} + 1372 \, A B^{3} b^{3} c + 294 \, A^{2} B^{2} b^{2} c^{2} + 28 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{5} c^{11}}\right )^{\frac {3}{4}} + 27 \, {\left (343 \, B^{3} b^{3} + 147 \, A B^{2} b^{2} c + 21 \, A^{2} B b c^{2} + A^{3} c^{3}\right )} \sqrt {x}\right ) + 4 \, {\left ({\left (11 \, B b c - 3 \, A c^{2}\right )} x^{3} + {\left (7 \, B b^{2} + A b c\right )} x\right )} \sqrt {x}}{64 \, {\left (b c^{4} x^{4} + 2 \, b^{2} c^{3} x^{2} + b^{3} c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(17/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

-1/64*(12*(b*c^4*x^4 + 2*b^2*c^3*x^2 + b^3*c^2)*(-(2401*B^4*b^4 + 1372*A*B^3*b^3*c + 294*A^2*B^2*b^2*c^2 + 28*
A^3*B*b*c^3 + A^4*c^4)/(b^5*c^11))^(1/4)*arctan((sqrt((117649*B^6*b^6 + 100842*A*B^5*b^5*c + 36015*A^2*B^4*b^4
*c^2 + 6860*A^3*B^3*b^3*c^3 + 735*A^4*B^2*b^2*c^4 + 42*A^5*B*b*c^5 + A^6*c^6)*x - (2401*B^4*b^7*c^5 + 1372*A*B
^3*b^6*c^6 + 294*A^2*B^2*b^5*c^7 + 28*A^3*B*b^4*c^8 + A^4*b^3*c^9)*sqrt(-(2401*B^4*b^4 + 1372*A*B^3*b^3*c + 29
4*A^2*B^2*b^2*c^2 + 28*A^3*B*b*c^3 + A^4*c^4)/(b^5*c^11)))*b*c^3*(-(2401*B^4*b^4 + 1372*A*B^3*b^3*c + 294*A^2*
B^2*b^2*c^2 + 28*A^3*B*b*c^3 + A^4*c^4)/(b^5*c^11))^(1/4) - (343*B^3*b^4*c^3 + 147*A*B^2*b^3*c^4 + 21*A^2*B*b^
2*c^5 + A^3*b*c^6)*sqrt(x)*(-(2401*B^4*b^4 + 1372*A*B^3*b^3*c + 294*A^2*B^2*b^2*c^2 + 28*A^3*B*b*c^3 + A^4*c^4
)/(b^5*c^11))^(1/4))/(2401*B^4*b^4 + 1372*A*B^3*b^3*c + 294*A^2*B^2*b^2*c^2 + 28*A^3*B*b*c^3 + A^4*c^4)) - 3*(
b*c^4*x^4 + 2*b^2*c^3*x^2 + b^3*c^2)*(-(2401*B^4*b^4 + 1372*A*B^3*b^3*c + 294*A^2*B^2*b^2*c^2 + 28*A^3*B*b*c^3
+ A^4*c^4)/(b^5*c^11))^(1/4)*log(27*b^4*c^8*(-(2401*B^4*b^4 + 1372*A*B^3*b^3*c + 294*A^2*B^2*b^2*c^2 + 28*A^3
*B*b*c^3 + A^4*c^4)/(b^5*c^11))^(3/4) + 27*(343*B^3*b^3 + 147*A*B^2*b^2*c + 21*A^2*B*b*c^2 + A^3*c^3)*sqrt(x))
+ 3*(b*c^4*x^4 + 2*b^2*c^3*x^2 + b^3*c^2)*(-(2401*B^4*b^4 + 1372*A*B^3*b^3*c + 294*A^2*B^2*b^2*c^2 + 28*A^3*B
*b*c^3 + A^4*c^4)/(b^5*c^11))^(1/4)*log(-27*b^4*c^8*(-(2401*B^4*b^4 + 1372*A*B^3*b^3*c + 294*A^2*B^2*b^2*c^2 +
28*A^3*B*b*c^3 + A^4*c^4)/(b^5*c^11))^(3/4) + 27*(343*B^3*b^3 + 147*A*B^2*b^2*c + 21*A^2*B*b*c^2 + A^3*c^3)*s
qrt(x)) + 4*((11*B*b*c - 3*A*c^2)*x^3 + (7*B*b^2 + A*b*c)*x)*sqrt(x))/(b*c^4*x^4 + 2*b^2*c^3*x^2 + b^3*c^2)

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giac [A]  time = 0.23, size = 293, normalized size = 1.00 \[ -\frac {11 \, B b c x^{\frac {7}{2}} - 3 \, A c^{2} x^{\frac {7}{2}} + 7 \, B b^{2} x^{\frac {3}{2}} + A b c x^{\frac {3}{2}}}{16 \, {\left (c x^{2} + b\right )}^{2} b c^{2}} + \frac {3 \, \sqrt {2} {\left (7 \, \left (b c^{3}\right )^{\frac {3}{4}} B b + \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b^{2} c^{5}} + \frac {3 \, \sqrt {2} {\left (7 \, \left (b c^{3}\right )^{\frac {3}{4}} B b + \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b^{2} c^{5}} - \frac {3 \, \sqrt {2} {\left (7 \, \left (b c^{3}\right )^{\frac {3}{4}} B b + \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{2} c^{5}} + \frac {3 \, \sqrt {2} {\left (7 \, \left (b c^{3}\right )^{\frac {3}{4}} B b + \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{2} c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(17/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

-1/16*(11*B*b*c*x^(7/2) - 3*A*c^2*x^(7/2) + 7*B*b^2*x^(3/2) + A*b*c*x^(3/2))/((c*x^2 + b)^2*b*c^2) + 3/64*sqrt
(2)*(7*(b*c^3)^(3/4)*B*b + (b*c^3)^(3/4)*A*c)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/(b/c)^(1/4)
)/(b^2*c^5) + 3/64*sqrt(2)*(7*(b*c^3)^(3/4)*B*b + (b*c^3)^(3/4)*A*c)*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4)
- 2*sqrt(x))/(b/c)^(1/4))/(b^2*c^5) - 3/128*sqrt(2)*(7*(b*c^3)^(3/4)*B*b + (b*c^3)^(3/4)*A*c)*log(sqrt(2)*sqrt
(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^2*c^5) + 3/128*sqrt(2)*(7*(b*c^3)^(3/4)*B*b + (b*c^3)^(3/4)*A*c)*log(-sqrt
(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^2*c^5)

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maple [A]  time = 0.06, size = 325, normalized size = 1.11 \[ \frac {3 \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{64 \left (\frac {b}{c}\right )^{\frac {1}{4}} b \,c^{2}}+\frac {3 \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{64 \left (\frac {b}{c}\right )^{\frac {1}{4}} b \,c^{2}}+\frac {3 \sqrt {2}\, A \ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{128 \left (\frac {b}{c}\right )^{\frac {1}{4}} b \,c^{2}}+\frac {21 \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{64 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{3}}+\frac {21 \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{64 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{3}}+\frac {21 \sqrt {2}\, B \ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{128 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{3}}+\frac {\frac {\left (3 A c -11 b B \right ) x^{\frac {7}{2}}}{16 b c}-\frac {\left (A c +7 b B \right ) x^{\frac {3}{2}}}{16 c^{2}}}{\left (c \,x^{2}+b \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(17/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x)

[Out]

2*(1/32*(3*A*c-11*B*b)/b/c*x^(7/2)-1/32*(A*c+7*B*b)/c^2*x^(3/2))/(c*x^2+b)^2+3/64/c^2/b/(b/c)^(1/4)*2^(1/2)*A*
arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+3/64/c^2/b/(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)+3
/128/c^2/b/(b/c)^(1/4)*2^(1/2)*A*ln((x-(b/c)^(1/4)*2^(1/2)*x^(1/2)+(b/c)^(1/2))/(x+(b/c)^(1/4)*2^(1/2)*x^(1/2)
+(b/c)^(1/2)))+21/64/c^3/(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+21/64/c^3/(b/c)^(1/4)*2^(
1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)+21/128/c^3/(b/c)^(1/4)*2^(1/2)*B*ln((x-(b/c)^(1/4)*2^(1/2)*x^(1/2
)+(b/c)^(1/2))/(x+(b/c)^(1/4)*2^(1/2)*x^(1/2)+(b/c)^(1/2)))

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maxima [A]  time = 3.11, size = 251, normalized size = 0.86 \[ -\frac {{\left (11 \, B b c - 3 \, A c^{2}\right )} x^{\frac {7}{2}} + {\left (7 \, B b^{2} + A b c\right )} x^{\frac {3}{2}}}{16 \, {\left (b c^{4} x^{4} + 2 \, b^{2} c^{3} x^{2} + b^{3} c^{2}\right )}} + \frac {3 \, {\left (7 \, B b + A c\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{128 \, b c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(17/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

-1/16*((11*B*b*c - 3*A*c^2)*x^(7/2) + (7*B*b^2 + A*b*c)*x^(3/2))/(b*c^4*x^4 + 2*b^2*c^3*x^2 + b^3*c^2) + 3/128
*(7*B*b + A*c)*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c
)))/(sqrt(sqrt(b)*sqrt(c))*sqrt(c)) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) - 2*sqrt(c)*sqrt(
x))/sqrt(sqrt(b)*sqrt(c)))/(sqrt(sqrt(b)*sqrt(c))*sqrt(c)) - sqrt(2)*log(sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqr
t(c)*x + sqrt(b))/(b^(1/4)*c^(3/4)) + sqrt(2)*log(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(
1/4)*c^(3/4)))/(b*c^2)

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mupad [B]  time = 0.17, size = 122, normalized size = 0.42 \[ \frac {3\,\mathrm {atanh}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )\,\left (A\,c+7\,B\,b\right )}{32\,{\left (-b\right )}^{5/4}\,c^{11/4}}-\frac {3\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )\,\left (A\,c+7\,B\,b\right )}{32\,{\left (-b\right )}^{5/4}\,c^{11/4}}-\frac {\frac {x^{3/2}\,\left (A\,c+7\,B\,b\right )}{16\,c^2}-\frac {x^{7/2}\,\left (3\,A\,c-11\,B\,b\right )}{16\,b\,c}}{b^2+2\,b\,c\,x^2+c^2\,x^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(17/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x)

[Out]

(3*atanh((c^(1/4)*x^(1/2))/(-b)^(1/4))*(A*c + 7*B*b))/(32*(-b)^(5/4)*c^(11/4)) - (3*atan((c^(1/4)*x^(1/2))/(-b
)^(1/4))*(A*c + 7*B*b))/(32*(-b)^(5/4)*c^(11/4)) - ((x^(3/2)*(A*c + 7*B*b))/(16*c^2) - (x^(7/2)*(3*A*c - 11*B*
b))/(16*b*c))/(b^2 + c^2*x^4 + 2*b*c*x^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(17/2)*(B*x**2+A)/(c*x**4+b*x**2)**3,x)

[Out]

Timed out

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