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

Optimal. Leaf size=310 \[ \frac {b^{3/4} (11 b B-7 A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} c^{15/4}}-\frac {b^{3/4} (11 b B-7 A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} c^{15/4}}-\frac {b^{3/4} (11 b B-7 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{15/4}}+\frac {b^{3/4} (11 b B-7 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} c^{15/4}}-\frac {x^{3/2} (11 b B-7 A c)}{6 c^3}+\frac {x^{7/2} (11 b B-7 A c)}{14 b c^2}-\frac {x^{11/2} (b B-A c)}{2 b c \left (b+c x^2\right )} \]

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Rubi [A]  time = 0.24, antiderivative size = 310, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {1584, 457, 321, 329, 297, 1162, 617, 204, 1165, 628} \begin {gather*} \frac {b^{3/4} (11 b B-7 A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} c^{15/4}}-\frac {b^{3/4} (11 b B-7 A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} c^{15/4}}-\frac {b^{3/4} (11 b B-7 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{15/4}}+\frac {b^{3/4} (11 b B-7 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} c^{15/4}}+\frac {x^{7/2} (11 b B-7 A c)}{14 b c^2}-\frac {x^{3/2} (11 b B-7 A c)}{6 c^3}-\frac {x^{11/2} (b B-A c)}{2 b c \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((11*b*B - 7*A*c)*x^(3/2))/(6*c^3) + ((11*b*B - 7*A*c)*x^(7/2))/(14*b*c^2) - ((b*B - A*c)*x^(11/2))/(2*b*c*(b
+ c*x^2)) - (b^(3/4)*(11*b*B - 7*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*c^(15/4)) + (
b^(3/4)*(11*b*B - 7*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*c^(15/4)) + (b^(3/4)*(11*b*
B - 7*A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*c^(15/4)) - (b^(3/4)*(11*b*B
- 7*A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*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 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^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac {x^{9/2} \left (A+B x^2\right )}{\left (b+c x^2\right )^2} \, dx\\ &=-\frac {(b B-A c) x^{11/2}}{2 b c \left (b+c x^2\right )}+\frac {\left (\frac {11 b B}{2}-\frac {7 A c}{2}\right ) \int \frac {x^{9/2}}{b+c x^2} \, dx}{2 b c}\\ &=\frac {(11 b B-7 A c) x^{7/2}}{14 b c^2}-\frac {(b B-A c) x^{11/2}}{2 b c \left (b+c x^2\right )}-\frac {(11 b B-7 A c) \int \frac {x^{5/2}}{b+c x^2} \, dx}{4 c^2}\\ &=-\frac {(11 b B-7 A c) x^{3/2}}{6 c^3}+\frac {(11 b B-7 A c) x^{7/2}}{14 b c^2}-\frac {(b B-A c) x^{11/2}}{2 b c \left (b+c x^2\right )}+\frac {(b (11 b B-7 A c)) \int \frac {\sqrt {x}}{b+c x^2} \, dx}{4 c^3}\\ &=-\frac {(11 b B-7 A c) x^{3/2}}{6 c^3}+\frac {(11 b B-7 A c) x^{7/2}}{14 b c^2}-\frac {(b B-A c) x^{11/2}}{2 b c \left (b+c x^2\right )}+\frac {(b (11 b B-7 A c)) \operatorname {Subst}\left (\int \frac {x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{2 c^3}\\ &=-\frac {(11 b B-7 A c) x^{3/2}}{6 c^3}+\frac {(11 b B-7 A c) x^{7/2}}{14 b c^2}-\frac {(b B-A c) x^{11/2}}{2 b c \left (b+c x^2\right )}-\frac {(b (11 b B-7 A c)) \operatorname {Subst}\left (\int \frac {\sqrt {b}-\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{4 c^{7/2}}+\frac {(b (11 b B-7 A c)) \operatorname {Subst}\left (\int \frac {\sqrt {b}+\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{4 c^{7/2}}\\ &=-\frac {(11 b B-7 A c) x^{3/2}}{6 c^3}+\frac {(11 b B-7 A c) x^{7/2}}{14 b c^2}-\frac {(b B-A c) x^{11/2}}{2 b c \left (b+c x^2\right )}+\frac {(b (11 b B-7 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 )}{8 c^4}+\frac {(b (11 b B-7 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 )}{8 c^4}+\frac {\left (b^{3/4} (11 b B-7 A c)\right ) \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 )}{8 \sqrt {2} c^{15/4}}+\frac {\left (b^{3/4} (11 b B-7 A c)\right ) \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 )}{8 \sqrt {2} c^{15/4}}\\ &=-\frac {(11 b B-7 A c) x^{3/2}}{6 c^3}+\frac {(11 b B-7 A c) x^{7/2}}{14 b c^2}-\frac {(b B-A c) x^{11/2}}{2 b c \left (b+c x^2\right )}+\frac {b^{3/4} (11 b B-7 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} c^{15/4}}-\frac {b^{3/4} (11 b B-7 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} c^{15/4}}+\frac {\left (b^{3/4} (11 b B-7 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{15/4}}-\frac {\left (b^{3/4} (11 b B-7 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{15/4}}\\ &=-\frac {(11 b B-7 A c) x^{3/2}}{6 c^3}+\frac {(11 b B-7 A c) x^{7/2}}{14 b c^2}-\frac {(b B-A c) x^{11/2}}{2 b c \left (b+c x^2\right )}-\frac {b^{3/4} (11 b B-7 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{15/4}}+\frac {b^{3/4} (11 b B-7 A c) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{15/4}}+\frac {b^{3/4} (11 b B-7 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} c^{15/4}}-\frac {b^{3/4} (11 b B-7 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} c^{15/4}}\\ \end {align*}

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Mathematica [C]  time = 0.30, size = 154, normalized size = 0.50 \begin {gather*} -\frac {(-b)^{3/4} (3 b B-2 A c) \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b}}\right )}{c^{15/4}}+\frac {(-b)^{3/4} (3 b B-2 A c) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b}}\right )}{c^{15/4}}+\frac {2 x^{3/2} (A c-b B) \, _2F_1\left (\frac {3}{4},2;\frac {7}{4};-\frac {c x^2}{b}\right )}{3 c^3}+\frac {2 x^{3/2} (A c-2 b B)}{3 c^3}+\frac {2 B x^{7/2}}{7 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.75, size = 362, normalized size = 1.17 \begin {gather*} \frac {\sqrt {2} A b^{3/4} \tan ^{-1}\left (\frac {\frac {\sqrt [4]{b}}{\sqrt {2} \sqrt [4]{c}}-\frac {\sqrt [4]{c} x}{\sqrt {2} \sqrt [4]{b}}}{\sqrt {x}}\right )}{c^{11/4}}-\frac {A b^{3/4} \tan ^{-1}\left (\frac {\frac {\sqrt [4]{b}}{\sqrt {2} \sqrt [4]{c}}-\frac {\sqrt [4]{c} x}{\sqrt {2} \sqrt [4]{b}}}{\sqrt {x}}\right )}{4 \sqrt {2} c^{11/4}}+\frac {7 A b^{3/4} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{4 \sqrt {2} c^{11/4}}+\frac {x^{3/2} \left (49 A b c-77 b^2 B\right )+x^{7/2} \left (28 A c^2-44 b B c\right )+12 B c^2 x^{11/2}}{42 c^3 \left (b+c x^2\right )}-\frac {11 b^{7/4} B \tan ^{-1}\left (\frac {\frac {\sqrt [4]{b}}{\sqrt {2} \sqrt [4]{c}}-\frac {\sqrt [4]{c} x}{\sqrt {2} \sqrt [4]{b}}}{\sqrt {x}}\right )}{4 \sqrt {2} c^{15/4}}-\frac {11 b^{7/4} B \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{4 \sqrt {2} c^{15/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-77*b^2*B + 49*A*b*c)*x^(3/2) + (-44*b*B*c + 28*A*c^2)*x^(7/2) + 12*B*c^2*x^(11/2))/(42*c^3*(b + c*x^2)) - (
11*b^(7/4)*B*ArcTan[(b^(1/4)/(Sqrt[2]*c^(1/4)) - (c^(1/4)*x)/(Sqrt[2]*b^(1/4)))/Sqrt[x]])/(4*Sqrt[2]*c^(15/4))
- (A*b^(3/4)*ArcTan[(b^(1/4)/(Sqrt[2]*c^(1/4)) - (c^(1/4)*x)/(Sqrt[2]*b^(1/4)))/Sqrt[x]])/(4*Sqrt[2]*c^(11/4)
) + (Sqrt[2]*A*b^(3/4)*ArcTan[(b^(1/4)/(Sqrt[2]*c^(1/4)) - (c^(1/4)*x)/(Sqrt[2]*b^(1/4)))/Sqrt[x]])/c^(11/4) -
(11*b^(7/4)*B*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(4*Sqrt[2]*c^(15/4)) + (7*A*b
^(3/4)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(4*Sqrt[2]*c^(11/4))

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fricas [B]  time = 0.45, size = 989, normalized size = 3.19 \begin {gather*} \frac {84 \, {\left (c^{4} x^{2} + b c^{3}\right )} \left (-\frac {14641 \, B^{4} b^{7} - 37268 \, A B^{3} b^{6} c + 35574 \, A^{2} B^{2} b^{5} c^{2} - 15092 \, A^{3} B b^{4} c^{3} + 2401 \, A^{4} b^{3} c^{4}}{c^{15}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {{\left (1771561 \, B^{6} b^{10} - 6764142 \, A B^{5} b^{9} c + 10761135 \, A^{2} B^{4} b^{8} c^{2} - 9130660 \, A^{3} B^{3} b^{7} c^{3} + 4357815 \, A^{4} B^{2} b^{6} c^{4} - 1109262 \, A^{5} B b^{5} c^{5} + 117649 \, A^{6} b^{4} c^{6}\right )} x - {\left (14641 \, B^{4} b^{7} c^{7} - 37268 \, A B^{3} b^{6} c^{8} + 35574 \, A^{2} B^{2} b^{5} c^{9} - 15092 \, A^{3} B b^{4} c^{10} + 2401 \, A^{4} b^{3} c^{11}\right )} \sqrt {-\frac {14641 \, B^{4} b^{7} - 37268 \, A B^{3} b^{6} c + 35574 \, A^{2} B^{2} b^{5} c^{2} - 15092 \, A^{3} B b^{4} c^{3} + 2401 \, A^{4} b^{3} c^{4}}{c^{15}}}} c^{4} \left (-\frac {14641 \, B^{4} b^{7} - 37268 \, A B^{3} b^{6} c + 35574 \, A^{2} B^{2} b^{5} c^{2} - 15092 \, A^{3} B b^{4} c^{3} + 2401 \, A^{4} b^{3} c^{4}}{c^{15}}\right )^{\frac {1}{4}} + {\left (1331 \, B^{3} b^{5} c^{4} - 2541 \, A B^{2} b^{4} c^{5} + 1617 \, A^{2} B b^{3} c^{6} - 343 \, A^{3} b^{2} c^{7}\right )} \sqrt {x} \left (-\frac {14641 \, B^{4} b^{7} - 37268 \, A B^{3} b^{6} c + 35574 \, A^{2} B^{2} b^{5} c^{2} - 15092 \, A^{3} B b^{4} c^{3} + 2401 \, A^{4} b^{3} c^{4}}{c^{15}}\right )^{\frac {1}{4}}}{14641 \, B^{4} b^{7} - 37268 \, A B^{3} b^{6} c + 35574 \, A^{2} B^{2} b^{5} c^{2} - 15092 \, A^{3} B b^{4} c^{3} + 2401 \, A^{4} b^{3} c^{4}}\right ) - 21 \, {\left (c^{4} x^{2} + b c^{3}\right )} \left (-\frac {14641 \, B^{4} b^{7} - 37268 \, A B^{3} b^{6} c + 35574 \, A^{2} B^{2} b^{5} c^{2} - 15092 \, A^{3} B b^{4} c^{3} + 2401 \, A^{4} b^{3} c^{4}}{c^{15}}\right )^{\frac {1}{4}} \log \left (c^{11} \left (-\frac {14641 \, B^{4} b^{7} - 37268 \, A B^{3} b^{6} c + 35574 \, A^{2} B^{2} b^{5} c^{2} - 15092 \, A^{3} B b^{4} c^{3} + 2401 \, A^{4} b^{3} c^{4}}{c^{15}}\right )^{\frac {3}{4}} - {\left (1331 \, B^{3} b^{5} - 2541 \, A B^{2} b^{4} c + 1617 \, A^{2} B b^{3} c^{2} - 343 \, A^{3} b^{2} c^{3}\right )} \sqrt {x}\right ) + 21 \, {\left (c^{4} x^{2} + b c^{3}\right )} \left (-\frac {14641 \, B^{4} b^{7} - 37268 \, A B^{3} b^{6} c + 35574 \, A^{2} B^{2} b^{5} c^{2} - 15092 \, A^{3} B b^{4} c^{3} + 2401 \, A^{4} b^{3} c^{4}}{c^{15}}\right )^{\frac {1}{4}} \log \left (-c^{11} \left (-\frac {14641 \, B^{4} b^{7} - 37268 \, A B^{3} b^{6} c + 35574 \, A^{2} B^{2} b^{5} c^{2} - 15092 \, A^{3} B b^{4} c^{3} + 2401 \, A^{4} b^{3} c^{4}}{c^{15}}\right )^{\frac {3}{4}} - {\left (1331 \, B^{3} b^{5} - 2541 \, A B^{2} b^{4} c + 1617 \, A^{2} B b^{3} c^{2} - 343 \, A^{3} b^{2} c^{3}\right )} \sqrt {x}\right ) + 4 \, {\left (12 \, B c^{2} x^{5} - 4 \, {\left (11 \, B b c - 7 \, A c^{2}\right )} x^{3} - 7 \, {\left (11 \, B b^{2} - 7 \, A b c\right )} x\right )} \sqrt {x}}{168 \, {\left (c^{4} x^{2} + b c^{3}\right )}} \end {gather*}

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)^2,x, algorithm="fricas")

[Out]

1/168*(84*(c^4*x^2 + b*c^3)*(-(14641*B^4*b^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3*B*b^4*c^3
+ 2401*A^4*b^3*c^4)/c^15)^(1/4)*arctan((sqrt((1771561*B^6*b^10 - 6764142*A*B^5*b^9*c + 10761135*A^2*B^4*b^8*c
^2 - 9130660*A^3*B^3*b^7*c^3 + 4357815*A^4*B^2*b^6*c^4 - 1109262*A^5*B*b^5*c^5 + 117649*A^6*b^4*c^6)*x - (1464
1*B^4*b^7*c^7 - 37268*A*B^3*b^6*c^8 + 35574*A^2*B^2*b^5*c^9 - 15092*A^3*B*b^4*c^10 + 2401*A^4*b^3*c^11)*sqrt(-
(14641*B^4*b^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3*B*b^4*c^3 + 2401*A^4*b^3*c^4)/c^15))*c^
4*(-(14641*B^4*b^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3*B*b^4*c^3 + 2401*A^4*b^3*c^4)/c^15)
^(1/4) + (1331*B^3*b^5*c^4 - 2541*A*B^2*b^4*c^5 + 1617*A^2*B*b^3*c^6 - 343*A^3*b^2*c^7)*sqrt(x)*(-(14641*B^4*b
^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3*B*b^4*c^3 + 2401*A^4*b^3*c^4)/c^15)^(1/4))/(14641*B
^4*b^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3*B*b^4*c^3 + 2401*A^4*b^3*c^4)) - 21*(c^4*x^2 +
b*c^3)*(-(14641*B^4*b^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3*B*b^4*c^3 + 2401*A^4*b^3*c^4)/
c^15)^(1/4)*log(c^11*(-(14641*B^4*b^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3*B*b^4*c^3 + 2401
*A^4*b^3*c^4)/c^15)^(3/4) - (1331*B^3*b^5 - 2541*A*B^2*b^4*c + 1617*A^2*B*b^3*c^2 - 343*A^3*b^2*c^3)*sqrt(x))
+ 21*(c^4*x^2 + b*c^3)*(-(14641*B^4*b^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3*B*b^4*c^3 + 24
01*A^4*b^3*c^4)/c^15)^(1/4)*log(-c^11*(-(14641*B^4*b^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3
*B*b^4*c^3 + 2401*A^4*b^3*c^4)/c^15)^(3/4) - (1331*B^3*b^5 - 2541*A*B^2*b^4*c + 1617*A^2*B*b^3*c^2 - 343*A^3*b
^2*c^3)*sqrt(x)) + 4*(12*B*c^2*x^5 - 4*(11*B*b*c - 7*A*c^2)*x^3 - 7*(11*B*b^2 - 7*A*b*c)*x)*sqrt(x))/(c^4*x^2
+ b*c^3)

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giac [A]  time = 0.27, size = 299, normalized size = 0.96 \begin {gather*} -\frac {B b^{2} x^{\frac {3}{2}} - A b c x^{\frac {3}{2}}}{2 \, {\left (c x^{2} + b\right )} c^{3}} + \frac {\sqrt {2} {\left (11 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 7 \, \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 )}{8 \, c^{6}} + \frac {\sqrt {2} {\left (11 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 7 \, \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 )}{8 \, c^{6}} - \frac {\sqrt {2} {\left (11 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 7 \, \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 )}{16 \, c^{6}} + \frac {\sqrt {2} {\left (11 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 7 \, \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 )}{16 \, c^{6}} + \frac {2 \, {\left (3 \, B c^{12} x^{\frac {7}{2}} - 14 \, B b c^{11} x^{\frac {3}{2}} + 7 \, A c^{12} x^{\frac {3}{2}}\right )}}{21 \, c^{14}} \end {gather*}

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)^2,x, algorithm="giac")

[Out]

-1/2*(B*b^2*x^(3/2) - A*b*c*x^(3/2))/((c*x^2 + b)*c^3) + 1/8*sqrt(2)*(11*(b*c^3)^(3/4)*B*b - 7*(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))/c^6 + 1/8*sqrt(2)*(11*(b*c^3)^(3/4)*B*b
- 7*(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))/c^6 - 1/16*sqrt(2)*(
11*(b*c^3)^(3/4)*B*b - 7*(b*c^3)^(3/4)*A*c)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^6 + 1/16*sqrt(2
)*(11*(b*c^3)^(3/4)*B*b - 7*(b*c^3)^(3/4)*A*c)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^6 + 2/21*(3
*B*c^12*x^(7/2) - 14*B*b*c^11*x^(3/2) + 7*A*c^12*x^(3/2))/c^14

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maple [A]  time = 0.06, size = 348, normalized size = 1.12 \begin {gather*} \frac {2 B \,x^{\frac {7}{2}}}{7 c^{2}}+\frac {A b \,x^{\frac {3}{2}}}{2 \left (c \,x^{2}+b \right ) c^{2}}-\frac {B \,b^{2} x^{\frac {3}{2}}}{2 \left (c \,x^{2}+b \right ) c^{3}}+\frac {2 A \,x^{\frac {3}{2}}}{3 c^{2}}-\frac {4 B b \,x^{\frac {3}{2}}}{3 c^{3}}-\frac {7 \sqrt {2}\, A b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{3}}-\frac {7 \sqrt {2}\, A b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{3}}-\frac {7 \sqrt {2}\, A 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 )}{16 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{3}}+\frac {11 \sqrt {2}\, B \,b^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{4}}+\frac {11 \sqrt {2}\, B \,b^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{4}}+\frac {11 \sqrt {2}\, B \,b^{2} \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 )}{16 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{4}} \end {gather*}

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)^2,x)

[Out]

2/7/c^2*B*x^(7/2)+2/3/c^2*x^(3/2)*A-4/3/c^3*x^(3/2)*b*B+1/2*b/c^2*x^(3/2)/(c*x^2+b)*A-1/2*b^2/c^3*x^(3/2)/(c*x
^2+b)*B-7/16*b/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)))-7/8*b/c^3/(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)-7/8*b/c^3/(b/c)^(
1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)+11/16*b^2/c^4/(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)))+11/8*b^2/c^4/(b/c)^(1/4)*2^(1/2)*B*arct
an(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+11/8*b^2/c^4/(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)

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maxima [A]  time = 3.11, size = 247, normalized size = 0.80 \begin {gather*} -\frac {{\left (B b^{2} - A b c\right )} x^{\frac {3}{2}}}{2 \, {\left (c^{4} x^{2} + b c^{3}\right )}} + \frac {{\left (11 \, B b^{2} - 7 \, A b 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 )}}{16 \, c^{3}} + \frac {2 \, {\left (3 \, B c x^{\frac {7}{2}} - 7 \, {\left (2 \, B b - A c\right )} x^{\frac {3}{2}}\right )}}{21 \, c^{3}} \end {gather*}

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)^2,x, algorithm="maxima")

[Out]

-1/2*(B*b^2 - A*b*c)*x^(3/2)/(c^4*x^2 + b*c^3) + 1/16*(11*B*b^2 - 7*A*b*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)*ar
ctan(-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) + sqrt(c)*x + sqrt(b))/(b^(1/4)*c^(3/4)))/c^3 + 2/21*(3*B*c*x^(7/2) - 7*(2*B*
b - A*c)*x^(3/2))/c^3

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mupad [B]  time = 0.17, size = 127, normalized size = 0.41 \begin {gather*} x^{3/2}\,\left (\frac {2\,A}{3\,c^2}-\frac {4\,B\,b}{3\,c^3}\right )+\frac {2\,B\,x^{7/2}}{7\,c^2}-\frac {x^{3/2}\,\left (\frac {B\,b^2}{2}-\frac {A\,b\,c}{2}\right )}{c^4\,x^2+b\,c^3}+\frac {{\left (-b\right )}^{3/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )\,\left (7\,A\,c-11\,B\,b\right )}{4\,c^{15/4}}+\frac {{\left (-b\right )}^{3/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}\,1{}\mathrm {i}}{{\left (-b\right )}^{1/4}}\right )\,\left (7\,A\,c-11\,B\,b\right )\,1{}\mathrm {i}}{4\,c^{15/4}} \end {gather*}

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)^2,x)

[Out]

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

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

[Out]

Timed out

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