3.208 \(\int \frac {x^{21/2} (A+B x^2)}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=322 \[ -\frac {7 (11 b B-3 A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} \sqrt [4]{b} c^{15/4}}+\frac {7 (11 b B-3 A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} \sqrt [4]{b} c^{15/4}}+\frac {7 (11 b B-3 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} \sqrt [4]{b} c^{15/4}}-\frac {7 (11 b B-3 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt {2} \sqrt [4]{b} c^{15/4}}+\frac {7 x^{3/2} (11 b B-3 A c)}{48 b c^3}-\frac {x^{7/2} (11 b B-3 A c)}{16 b c^2 \left (b+c x^2\right )}-\frac {x^{11/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2} \]

[Out]

7/48*(-3*A*c+11*B*b)*x^(3/2)/b/c^3-1/4*(-A*c+B*b)*x^(11/2)/b/c/(c*x^2+b)^2-1/16*(-3*A*c+11*B*b)*x^(7/2)/b/c^2/
(c*x^2+b)+7/64*(-3*A*c+11*B*b)*arctan(1-c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(1/4)/c^(15/4)*2^(1/2)-7/64*(-3*A*c
+11*B*b)*arctan(1+c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(1/4)/c^(15/4)*2^(1/2)-7/128*(-3*A*c+11*B*b)*ln(b^(1/2)+x
*c^(1/2)-b^(1/4)*c^(1/4)*2^(1/2)*x^(1/2))/b^(1/4)/c^(15/4)*2^(1/2)+7/128*(-3*A*c+11*B*b)*ln(b^(1/2)+x*c^(1/2)+
b^(1/4)*c^(1/4)*2^(1/2)*x^(1/2))/b^(1/4)/c^(15/4)*2^(1/2)

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Rubi [A]  time = 0.25, antiderivative size = 322, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {1584, 457, 288, 321, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac {x^{7/2} (11 b B-3 A c)}{16 b c^2 \left (b+c x^2\right )}+\frac {7 x^{3/2} (11 b B-3 A c)}{48 b c^3}-\frac {7 (11 b B-3 A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} \sqrt [4]{b} c^{15/4}}+\frac {7 (11 b B-3 A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} \sqrt [4]{b} c^{15/4}}+\frac {7 (11 b B-3 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} \sqrt [4]{b} c^{15/4}}-\frac {7 (11 b B-3 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt {2} \sqrt [4]{b} c^{15/4}}-\frac {x^{11/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^(21/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

(7*(11*b*B - 3*A*c)*x^(3/2))/(48*b*c^3) - ((b*B - A*c)*x^(11/2))/(4*b*c*(b + c*x^2)^2) - ((11*b*B - 3*A*c)*x^(
7/2))/(16*b*c^2*(b + c*x^2)) + (7*(11*b*B - 3*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*
b^(1/4)*c^(15/4)) - (7*(11*b*B - 3*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(1/4)*c^(
15/4)) - (7*(11*b*B - 3*A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(1/4)*c
^(15/4)) + (7*(11*b*B - 3*A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(1/4)
*c^(15/4))

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 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 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^{21/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac {x^{9/2} \left (A+B x^2\right )}{\left (b+c x^2\right )^3} \, dx\\ &=-\frac {(b B-A c) x^{11/2}}{4 b c \left (b+c x^2\right )^2}+\frac {\left (\frac {11 b B}{2}-\frac {3 A c}{2}\right ) \int \frac {x^{9/2}}{\left (b+c x^2\right )^2} \, dx}{4 b c}\\ &=-\frac {(b B-A c) x^{11/2}}{4 b c \left (b+c x^2\right )^2}-\frac {(11 b B-3 A c) x^{7/2}}{16 b c^2 \left (b+c x^2\right )}+\frac {(7 (11 b B-3 A c)) \int \frac {x^{5/2}}{b+c x^2} \, dx}{32 b c^2}\\ &=\frac {7 (11 b B-3 A c) x^{3/2}}{48 b c^3}-\frac {(b B-A c) x^{11/2}}{4 b c \left (b+c x^2\right )^2}-\frac {(11 b B-3 A c) x^{7/2}}{16 b c^2 \left (b+c x^2\right )}-\frac {(7 (11 b B-3 A c)) \int \frac {\sqrt {x}}{b+c x^2} \, dx}{32 c^3}\\ &=\frac {7 (11 b B-3 A c) x^{3/2}}{48 b c^3}-\frac {(b B-A c) x^{11/2}}{4 b c \left (b+c x^2\right )^2}-\frac {(11 b B-3 A c) x^{7/2}}{16 b c^2 \left (b+c x^2\right )}-\frac {(7 (11 b B-3 A c)) \operatorname {Subst}\left (\int \frac {x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{16 c^3}\\ &=\frac {7 (11 b B-3 A c) x^{3/2}}{48 b c^3}-\frac {(b B-A c) x^{11/2}}{4 b c \left (b+c x^2\right )^2}-\frac {(11 b B-3 A c) x^{7/2}}{16 b c^2 \left (b+c x^2\right )}+\frac {(7 (11 b B-3 A c)) \operatorname {Subst}\left (\int \frac {\sqrt {b}-\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{32 c^{7/2}}-\frac {(7 (11 b B-3 A c)) \operatorname {Subst}\left (\int \frac {\sqrt {b}+\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{32 c^{7/2}}\\ &=\frac {7 (11 b B-3 A c) x^{3/2}}{48 b c^3}-\frac {(b B-A c) x^{11/2}}{4 b c \left (b+c x^2\right )^2}-\frac {(11 b B-3 A c) x^{7/2}}{16 b c^2 \left (b+c x^2\right )}-\frac {(7 (11 b B-3 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 c^4}-\frac {(7 (11 b B-3 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 c^4}-\frac {(7 (11 b B-3 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} \sqrt [4]{b} c^{15/4}}-\frac {(7 (11 b B-3 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} \sqrt [4]{b} c^{15/4}}\\ &=\frac {7 (11 b B-3 A c) x^{3/2}}{48 b c^3}-\frac {(b B-A c) x^{11/2}}{4 b c \left (b+c x^2\right )^2}-\frac {(11 b B-3 A c) x^{7/2}}{16 b c^2 \left (b+c x^2\right )}-\frac {7 (11 b B-3 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} \sqrt [4]{b} c^{15/4}}+\frac {7 (11 b B-3 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} \sqrt [4]{b} c^{15/4}}-\frac {(7 (11 b B-3 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} \sqrt [4]{b} c^{15/4}}+\frac {(7 (11 b B-3 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} \sqrt [4]{b} c^{15/4}}\\ &=\frac {7 (11 b B-3 A c) x^{3/2}}{48 b c^3}-\frac {(b B-A c) x^{11/2}}{4 b c \left (b+c x^2\right )^2}-\frac {(11 b B-3 A c) x^{7/2}}{16 b c^2 \left (b+c x^2\right )}+\frac {7 (11 b B-3 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} \sqrt [4]{b} c^{15/4}}-\frac {7 (11 b B-3 A c) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} \sqrt [4]{b} c^{15/4}}-\frac {7 (11 b B-3 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} \sqrt [4]{b} c^{15/4}}+\frac {7 (11 b B-3 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} \sqrt [4]{b} c^{15/4}}\\ \end {align*}

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Mathematica [C]  time = 0.47, size = 176, normalized size = 0.55 \[ \frac {\frac {2 c^{3/4} x^{3/2} (3 b B-2 A c) \, _2F_1\left (\frac {3}{4},2;\frac {7}{4};-\frac {c x^2}{b}\right )}{b}+\frac {2 c^{3/4} x^{3/2} (A c-b B) \, _2F_1\left (\frac {3}{4},3;\frac {7}{4};-\frac {c x^2}{b}\right )}{b}+\frac {(3 A c-9 b B) \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b}}\right )}{\sqrt [4]{-b}}+\frac {(9 b B-3 A c) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b}}\right )}{\sqrt [4]{-b}}+2 B c^{3/4} x^{3/2}}{3 c^{15/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(21/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

(2*B*c^(3/4)*x^(3/2) + ((-9*b*B + 3*A*c)*ArcTan[(c^(1/4)*Sqrt[x])/(-b)^(1/4)])/(-b)^(1/4) + ((9*b*B - 3*A*c)*A
rcTanh[(c^(1/4)*Sqrt[x])/(-b)^(1/4)])/(-b)^(1/4) + (2*c^(3/4)*(3*b*B - 2*A*c)*x^(3/2)*Hypergeometric2F1[3/4, 2
, 7/4, -((c*x^2)/b)])/b + (2*c^(3/4)*(-(b*B) + A*c)*x^(3/2)*Hypergeometric2F1[3/4, 3, 7/4, -((c*x^2)/b)])/b)/(
3*c^(15/4))

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fricas [B]  time = 0.78, size = 993, normalized size = 3.08 \[ -\frac {84 \, {\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\right )} \left (-\frac {14641 \, B^{4} b^{4} - 15972 \, A B^{3} b^{3} c + 6534 \, A^{2} B^{2} b^{2} c^{2} - 1188 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}{b c^{15}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {{\left (1771561 \, B^{6} b^{6} - 2898918 \, A B^{5} b^{5} c + 1976535 \, A^{2} B^{4} b^{4} c^{2} - 718740 \, A^{3} B^{3} b^{3} c^{3} + 147015 \, A^{4} B^{2} b^{2} c^{4} - 16038 \, A^{5} B b c^{5} + 729 \, A^{6} c^{6}\right )} x - {\left (14641 \, B^{4} b^{5} c^{7} - 15972 \, A B^{3} b^{4} c^{8} + 6534 \, A^{2} B^{2} b^{3} c^{9} - 1188 \, A^{3} B b^{2} c^{10} + 81 \, A^{4} b c^{11}\right )} \sqrt {-\frac {14641 \, B^{4} b^{4} - 15972 \, A B^{3} b^{3} c + 6534 \, A^{2} B^{2} b^{2} c^{2} - 1188 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}{b c^{15}}}} c^{4} \left (-\frac {14641 \, B^{4} b^{4} - 15972 \, A B^{3} b^{3} c + 6534 \, A^{2} B^{2} b^{2} c^{2} - 1188 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}{b c^{15}}\right )^{\frac {1}{4}} + {\left (1331 \, B^{3} b^{3} c^{4} - 1089 \, A B^{2} b^{2} c^{5} + 297 \, A^{2} B b c^{6} - 27 \, A^{3} c^{7}\right )} \sqrt {x} \left (-\frac {14641 \, B^{4} b^{4} - 15972 \, A B^{3} b^{3} c + 6534 \, A^{2} B^{2} b^{2} c^{2} - 1188 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}{b c^{15}}\right )^{\frac {1}{4}}}{14641 \, B^{4} b^{4} - 15972 \, A B^{3} b^{3} c + 6534 \, A^{2} B^{2} b^{2} c^{2} - 1188 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}\right ) - 21 \, {\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\right )} \left (-\frac {14641 \, B^{4} b^{4} - 15972 \, A B^{3} b^{3} c + 6534 \, A^{2} B^{2} b^{2} c^{2} - 1188 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}{b c^{15}}\right )^{\frac {1}{4}} \log \left (343 \, b c^{11} \left (-\frac {14641 \, B^{4} b^{4} - 15972 \, A B^{3} b^{3} c + 6534 \, A^{2} B^{2} b^{2} c^{2} - 1188 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}{b c^{15}}\right )^{\frac {3}{4}} - 343 \, {\left (1331 \, B^{3} b^{3} - 1089 \, A B^{2} b^{2} c + 297 \, A^{2} B b c^{2} - 27 \, A^{3} c^{3}\right )} \sqrt {x}\right ) + 21 \, {\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\right )} \left (-\frac {14641 \, B^{4} b^{4} - 15972 \, A B^{3} b^{3} c + 6534 \, A^{2} B^{2} b^{2} c^{2} - 1188 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}{b c^{15}}\right )^{\frac {1}{4}} \log \left (-343 \, b c^{11} \left (-\frac {14641 \, B^{4} b^{4} - 15972 \, A B^{3} b^{3} c + 6534 \, A^{2} B^{2} b^{2} c^{2} - 1188 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}{b c^{15}}\right )^{\frac {3}{4}} - 343 \, {\left (1331 \, B^{3} b^{3} - 1089 \, A B^{2} b^{2} c + 297 \, A^{2} B b c^{2} - 27 \, A^{3} c^{3}\right )} \sqrt {x}\right ) - 4 \, {\left (32 \, B c^{2} x^{5} + 11 \, {\left (11 \, B b c - 3 \, A c^{2}\right )} x^{3} + 7 \, {\left (11 \, B b^{2} - 3 \, A b c\right )} x\right )} \sqrt {x}}{192 \, {\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(21/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

-1/192*(84*(c^5*x^4 + 2*b*c^4*x^2 + b^2*c^3)*(-(14641*B^4*b^4 - 15972*A*B^3*b^3*c + 6534*A^2*B^2*b^2*c^2 - 118
8*A^3*B*b*c^3 + 81*A^4*c^4)/(b*c^15))^(1/4)*arctan((sqrt((1771561*B^6*b^6 - 2898918*A*B^5*b^5*c + 1976535*A^2*
B^4*b^4*c^2 - 718740*A^3*B^3*b^3*c^3 + 147015*A^4*B^2*b^2*c^4 - 16038*A^5*B*b*c^5 + 729*A^6*c^6)*x - (14641*B^
4*b^5*c^7 - 15972*A*B^3*b^4*c^8 + 6534*A^2*B^2*b^3*c^9 - 1188*A^3*B*b^2*c^10 + 81*A^4*b*c^11)*sqrt(-(14641*B^4
*b^4 - 15972*A*B^3*b^3*c + 6534*A^2*B^2*b^2*c^2 - 1188*A^3*B*b*c^3 + 81*A^4*c^4)/(b*c^15)))*c^4*(-(14641*B^4*b
^4 - 15972*A*B^3*b^3*c + 6534*A^2*B^2*b^2*c^2 - 1188*A^3*B*b*c^3 + 81*A^4*c^4)/(b*c^15))^(1/4) + (1331*B^3*b^3
*c^4 - 1089*A*B^2*b^2*c^5 + 297*A^2*B*b*c^6 - 27*A^3*c^7)*sqrt(x)*(-(14641*B^4*b^4 - 15972*A*B^3*b^3*c + 6534*
A^2*B^2*b^2*c^2 - 1188*A^3*B*b*c^3 + 81*A^4*c^4)/(b*c^15))^(1/4))/(14641*B^4*b^4 - 15972*A*B^3*b^3*c + 6534*A^
2*B^2*b^2*c^2 - 1188*A^3*B*b*c^3 + 81*A^4*c^4)) - 21*(c^5*x^4 + 2*b*c^4*x^2 + b^2*c^3)*(-(14641*B^4*b^4 - 1597
2*A*B^3*b^3*c + 6534*A^2*B^2*b^2*c^2 - 1188*A^3*B*b*c^3 + 81*A^4*c^4)/(b*c^15))^(1/4)*log(343*b*c^11*(-(14641*
B^4*b^4 - 15972*A*B^3*b^3*c + 6534*A^2*B^2*b^2*c^2 - 1188*A^3*B*b*c^3 + 81*A^4*c^4)/(b*c^15))^(3/4) - 343*(133
1*B^3*b^3 - 1089*A*B^2*b^2*c + 297*A^2*B*b*c^2 - 27*A^3*c^3)*sqrt(x)) + 21*(c^5*x^4 + 2*b*c^4*x^2 + b^2*c^3)*(
-(14641*B^4*b^4 - 15972*A*B^3*b^3*c + 6534*A^2*B^2*b^2*c^2 - 1188*A^3*B*b*c^3 + 81*A^4*c^4)/(b*c^15))^(1/4)*lo
g(-343*b*c^11*(-(14641*B^4*b^4 - 15972*A*B^3*b^3*c + 6534*A^2*B^2*b^2*c^2 - 1188*A^3*B*b*c^3 + 81*A^4*c^4)/(b*
c^15))^(3/4) - 343*(1331*B^3*b^3 - 1089*A*B^2*b^2*c + 297*A^2*B*b*c^2 - 27*A^3*c^3)*sqrt(x)) - 4*(32*B*c^2*x^5
+ 11*(11*B*b*c - 3*A*c^2)*x^3 + 7*(11*B*b^2 - 3*A*b*c)*x)*sqrt(x))/(c^5*x^4 + 2*b*c^4*x^2 + b^2*c^3)

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giac [A]  time = 0.23, size = 304, normalized size = 0.94 \[ \frac {2 \, B x^{\frac {3}{2}}}{3 \, c^{3}} + \frac {19 \, B b c x^{\frac {7}{2}} - 11 \, A c^{2} x^{\frac {7}{2}} + 15 \, B b^{2} x^{\frac {3}{2}} - 7 \, A b c x^{\frac {3}{2}}}{16 \, {\left (c x^{2} + b\right )}^{2} c^{3}} - \frac {7 \, \sqrt {2} {\left (11 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 3 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b c^{6}} - \frac {7 \, \sqrt {2} {\left (11 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 3 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b c^{6}} + \frac {7 \, \sqrt {2} {\left (11 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 3 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b c^{6}} - \frac {7 \, \sqrt {2} {\left (11 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 3 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b c^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(21/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

2/3*B*x^(3/2)/c^3 + 1/16*(19*B*b*c*x^(7/2) - 11*A*c^2*x^(7/2) + 15*B*b^2*x^(3/2) - 7*A*b*c*x^(3/2))/((c*x^2 +
b)^2*c^3) - 7/64*sqrt(2)*(11*(b*c^3)^(3/4)*B*b - 3*(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*c^6) - 7/64*sqrt(2)*(11*(b*c^3)^(3/4)*B*b - 3*(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*c^6) + 7/128*sqrt(2)*(11*(b*c^3)^(3/4)*B*b - 3*(b*c^3)^(
3/4)*A*c)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b*c^6) - 7/128*sqrt(2)*(11*(b*c^3)^(3/4)*B*b - 3*(
b*c^3)^(3/4)*A*c)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b*c^6)

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maple [A]  time = 0.07, size = 357, normalized size = 1.11 \[ -\frac {11 A \,x^{\frac {7}{2}}}{16 \left (c \,x^{2}+b \right )^{2} c}+\frac {19 B b \,x^{\frac {7}{2}}}{16 \left (c \,x^{2}+b \right )^{2} c^{2}}-\frac {7 A b \,x^{\frac {3}{2}}}{16 \left (c \,x^{2}+b \right )^{2} c^{2}}+\frac {15 B \,b^{2} x^{\frac {3}{2}}}{16 \left (c \,x^{2}+b \right )^{2} c^{3}}+\frac {2 B \,x^{\frac {3}{2}}}{3 c^{3}}+\frac {21 \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{64 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{3}}+\frac {21 \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{64 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{3}}+\frac {21 \sqrt {2}\, A \ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{128 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{3}}-\frac {77 \sqrt {2}\, B b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{64 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{4}}-\frac {77 \sqrt {2}\, B b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{64 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{4}}-\frac {77 \sqrt {2}\, B b \ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{128 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(21/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x)

[Out]

2/3*B*x^(3/2)/c^3-11/16/c/(c*x^2+b)^2*A*x^(7/2)+19/16/c^2/(c*x^2+b)^2*B*x^(7/2)*b-7/16/c^2/(c*x^2+b)^2*x^(3/2)
*A*b+15/16/c^3/(c*x^2+b)^2*x^(3/2)*B*b^2+21/128/c^3/(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)*A*arctan(2^(1/2)/(b/c)^(
1/4)*x^(1/2)+1)+21/64/c^3/(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)-77/128/c^4/(b/c)^(1/4)*2
^(1/2)*b*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)))-77/64/c
^4/(b/c)^(1/4)*2^(1/2)*b*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)-77/64/c^4/(b/c)^(1/4)*2^(1/2)*b*B*arctan(2^(1
/2)/(b/c)^(1/4)*x^(1/2)-1)

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maxima [A]  time = 3.10, size = 256, normalized size = 0.80 \[ \frac {{\left (19 \, B b c - 11 \, A c^{2}\right )} x^{\frac {7}{2}} + {\left (15 \, B b^{2} - 7 \, A b c\right )} x^{\frac {3}{2}}}{16 \, {\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\right )}} + \frac {2 \, B x^{\frac {3}{2}}}{3 \, c^{3}} - \frac {7 \, {\left (11 \, B b - 3 \, A c\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{128 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(21/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

1/16*((19*B*b*c - 11*A*c^2)*x^(7/2) + (15*B*b^2 - 7*A*b*c)*x^(3/2))/(c^5*x^4 + 2*b*c^4*x^2 + b^2*c^3) + 2/3*B*
x^(3/2)/c^3 - 7/128*(11*B*b - 3*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) + sqrt(c)*x + sqrt(b))/(b^(1/4)*c^(3/4)) + 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)))/c^3

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mupad [B]  time = 0.26, size = 138, normalized size = 0.43 \[ \frac {x^{3/2}\,\left (\frac {15\,B\,b^2}{16}-\frac {7\,A\,b\,c}{16}\right )-x^{7/2}\,\left (\frac {11\,A\,c^2}{16}-\frac {19\,B\,b\,c}{16}\right )}{b^2\,c^3+2\,b\,c^4\,x^2+c^5\,x^4}+\frac {2\,B\,x^{3/2}}{3\,c^3}+\frac {7\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )\,\left (3\,A\,c-11\,B\,b\right )}{32\,{\left (-b\right )}^{1/4}\,c^{15/4}}+\frac {\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}\,1{}\mathrm {i}}{{\left (-b\right )}^{1/4}}\right )\,\left (3\,A\,c-11\,B\,b\right )\,7{}\mathrm {i}}{32\,{\left (-b\right )}^{1/4}\,c^{15/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(21/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x)

[Out]

(x^(3/2)*((15*B*b^2)/16 - (7*A*b*c)/16) - x^(7/2)*((11*A*c^2)/16 - (19*B*b*c)/16))/(b^2*c^3 + c^5*x^4 + 2*b*c^
4*x^2) + (2*B*x^(3/2))/(3*c^3) + (7*atan((c^(1/4)*x^(1/2))/(-b)^(1/4))*(3*A*c - 11*B*b))/(32*(-b)^(1/4)*c^(15/
4)) + (atan((c^(1/4)*x^(1/2)*1i)/(-b)^(1/4))*(3*A*c - 11*B*b)*7i)/(32*(-b)^(1/4)*c^(15/4))

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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**(21/2)*(B*x**2+A)/(c*x**4+b*x**2)**3,x)

[Out]

Timed out

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