3.198 \(\int \frac {x^{13/2} (A+B x^2)}{(b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=289 \[ -\frac {(7 b B-3 A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} \sqrt [4]{b} c^{11/4}}+\frac {(7 b B-3 A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} \sqrt [4]{b} c^{11/4}}+\frac {(7 b B-3 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} \sqrt [4]{b} c^{11/4}}-\frac {(7 b B-3 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} \sqrt [4]{b} c^{11/4}}+\frac {x^{3/2} (7 b B-3 A c)}{6 b c^2}-\frac {x^{7/2} (b B-A c)}{2 b c \left (b+c x^2\right )} \]

[Out]

1/6*(-3*A*c+7*B*b)*x^(3/2)/b/c^2-1/2*(-A*c+B*b)*x^(7/2)/b/c/(c*x^2+b)+1/8*(-3*A*c+7*B*b)*arctan(1-c^(1/4)*2^(1
/2)*x^(1/2)/b^(1/4))/b^(1/4)/c^(11/4)*2^(1/2)-1/8*(-3*A*c+7*B*b)*arctan(1+c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(
1/4)/c^(11/4)*2^(1/2)-1/16*(-3*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^(1/4)/c^(11/
4)*2^(1/2)+1/16*(-3*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^(1/4)/c^(11/4)*2^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 289, 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, 321, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac {x^{3/2} (7 b B-3 A c)}{6 b c^2}-\frac {(7 b B-3 A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} \sqrt [4]{b} c^{11/4}}+\frac {(7 b B-3 A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} \sqrt [4]{b} c^{11/4}}+\frac {(7 b B-3 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} \sqrt [4]{b} c^{11/4}}-\frac {(7 b B-3 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} \sqrt [4]{b} c^{11/4}}-\frac {x^{7/2} (b B-A c)}{2 b c \left (b+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(x^(13/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

((7*b*B - 3*A*c)*x^(3/2))/(6*b*c^2) - ((b*B - A*c)*x^(7/2))/(2*b*c*(b + c*x^2)) + ((7*b*B - 3*A*c)*ArcTan[1 -
(Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*b^(1/4)*c^(11/4)) - ((7*b*B - 3*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4
)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*b^(1/4)*c^(11/4)) - ((7*b*B - 3*A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqr
t[x] + Sqrt[c]*x])/(8*Sqrt[2]*b^(1/4)*c^(11/4)) + ((7*b*B - 3*A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[
x] + Sqrt[c]*x])/(8*Sqrt[2]*b^(1/4)*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 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^{13/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac {x^{5/2} \left (A+B x^2\right )}{\left (b+c x^2\right )^2} \, dx\\ &=-\frac {(b B-A c) x^{7/2}}{2 b c \left (b+c x^2\right )}+\frac {\left (\frac {7 b B}{2}-\frac {3 A c}{2}\right ) \int \frac {x^{5/2}}{b+c x^2} \, dx}{2 b c}\\ &=\frac {(7 b B-3 A c) x^{3/2}}{6 b c^2}-\frac {(b B-A c) x^{7/2}}{2 b c \left (b+c x^2\right )}-\frac {(7 b B-3 A c) \int \frac {\sqrt {x}}{b+c x^2} \, dx}{4 c^2}\\ &=\frac {(7 b B-3 A c) x^{3/2}}{6 b c^2}-\frac {(b B-A c) x^{7/2}}{2 b c \left (b+c x^2\right )}-\frac {(7 b B-3 A c) \operatorname {Subst}\left (\int \frac {x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{2 c^2}\\ &=\frac {(7 b B-3 A c) x^{3/2}}{6 b c^2}-\frac {(b B-A c) x^{7/2}}{2 b c \left (b+c x^2\right )}+\frac {(7 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 )}{4 c^{5/2}}-\frac {(7 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 )}{4 c^{5/2}}\\ &=\frac {(7 b B-3 A c) x^{3/2}}{6 b c^2}-\frac {(b B-A c) x^{7/2}}{2 b c \left (b+c x^2\right )}-\frac {(7 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 )}{8 c^3}-\frac {(7 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 )}{8 c^3}-\frac {(7 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 )}{8 \sqrt {2} \sqrt [4]{b} c^{11/4}}-\frac {(7 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 )}{8 \sqrt {2} \sqrt [4]{b} c^{11/4}}\\ &=\frac {(7 b B-3 A c) x^{3/2}}{6 b c^2}-\frac {(b B-A c) x^{7/2}}{2 b c \left (b+c x^2\right )}-\frac {(7 b B-3 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} \sqrt [4]{b} c^{11/4}}+\frac {(7 b B-3 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} \sqrt [4]{b} c^{11/4}}-\frac {(7 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 )}{4 \sqrt {2} \sqrt [4]{b} c^{11/4}}+\frac {(7 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 )}{4 \sqrt {2} \sqrt [4]{b} c^{11/4}}\\ &=\frac {(7 b B-3 A c) x^{3/2}}{6 b c^2}-\frac {(b B-A c) x^{7/2}}{2 b c \left (b+c x^2\right )}+\frac {(7 b B-3 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} \sqrt [4]{b} c^{11/4}}-\frac {(7 b B-3 A c) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} \sqrt [4]{b} c^{11/4}}-\frac {(7 b B-3 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} \sqrt [4]{b} c^{11/4}}+\frac {(7 b B-3 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} \sqrt [4]{b} c^{11/4}}\\ \end {align*}

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Mathematica [C]  time = 0.26, size = 136, normalized size = 0.47 \[ \frac {(3 A c-6 b B) \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b}}\right )+(6 b B-3 A c) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b}}\right )+2 \sqrt [4]{-b} B c^{3/4} x^{3/2}}{3 \sqrt [4]{-b} c^{11/4}}+\frac {2 x^{3/2} (b B-A c) \, _2F_1\left (\frac {3}{4},2;\frac {7}{4};-\frac {c x^2}{b}\right )}{3 b c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(13/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

(2*(-b)^(1/4)*B*c^(3/4)*x^(3/2) + (-6*b*B + 3*A*c)*ArcTan[(c^(1/4)*Sqrt[x])/(-b)^(1/4)] + (6*b*B - 3*A*c)*ArcT
anh[(c^(1/4)*Sqrt[x])/(-b)^(1/4)])/(3*(-b)^(1/4)*c^(11/4)) + (2*(b*B - A*c)*x^(3/2)*Hypergeometric2F1[3/4, 2,
7/4, -((c*x^2)/b)])/(3*b*c^2)

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fricas [B]  time = 1.05, size = 925, normalized size = 3.20 \[ -\frac {12 \, {\left (c^{3} x^{2} + b c^{2}\right )} \left (-\frac {2401 \, B^{4} b^{4} - 4116 \, A B^{3} b^{3} c + 2646 \, A^{2} B^{2} b^{2} c^{2} - 756 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}{b c^{11}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {{\left (117649 \, B^{6} b^{6} - 302526 \, A B^{5} b^{5} c + 324135 \, A^{2} B^{4} b^{4} c^{2} - 185220 \, A^{3} B^{3} b^{3} c^{3} + 59535 \, A^{4} B^{2} b^{2} c^{4} - 10206 \, A^{5} B b c^{5} + 729 \, A^{6} c^{6}\right )} x - {\left (2401 \, B^{4} b^{5} c^{5} - 4116 \, A B^{3} b^{4} c^{6} + 2646 \, A^{2} B^{2} b^{3} c^{7} - 756 \, A^{3} B b^{2} c^{8} + 81 \, A^{4} b c^{9}\right )} \sqrt {-\frac {2401 \, B^{4} b^{4} - 4116 \, A B^{3} b^{3} c + 2646 \, A^{2} B^{2} b^{2} c^{2} - 756 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}{b c^{11}}}} c^{3} \left (-\frac {2401 \, B^{4} b^{4} - 4116 \, A B^{3} b^{3} c + 2646 \, A^{2} B^{2} b^{2} c^{2} - 756 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}{b c^{11}}\right )^{\frac {1}{4}} + {\left (343 \, B^{3} b^{3} c^{3} - 441 \, A B^{2} b^{2} c^{4} + 189 \, A^{2} B b c^{5} - 27 \, A^{3} c^{6}\right )} \sqrt {x} \left (-\frac {2401 \, B^{4} b^{4} - 4116 \, A B^{3} b^{3} c + 2646 \, A^{2} B^{2} b^{2} c^{2} - 756 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}{b c^{11}}\right )^{\frac {1}{4}}}{2401 \, B^{4} b^{4} - 4116 \, A B^{3} b^{3} c + 2646 \, A^{2} B^{2} b^{2} c^{2} - 756 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}\right ) - 3 \, {\left (c^{3} x^{2} + b c^{2}\right )} \left (-\frac {2401 \, B^{4} b^{4} - 4116 \, A B^{3} b^{3} c + 2646 \, A^{2} B^{2} b^{2} c^{2} - 756 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}{b c^{11}}\right )^{\frac {1}{4}} \log \left (b c^{8} \left (-\frac {2401 \, B^{4} b^{4} - 4116 \, A B^{3} b^{3} c + 2646 \, A^{2} B^{2} b^{2} c^{2} - 756 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}{b c^{11}}\right )^{\frac {3}{4}} - {\left (343 \, B^{3} b^{3} - 441 \, A B^{2} b^{2} c + 189 \, A^{2} B b c^{2} - 27 \, A^{3} c^{3}\right )} \sqrt {x}\right ) + 3 \, {\left (c^{3} x^{2} + b c^{2}\right )} \left (-\frac {2401 \, B^{4} b^{4} - 4116 \, A B^{3} b^{3} c + 2646 \, A^{2} B^{2} b^{2} c^{2} - 756 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}{b c^{11}}\right )^{\frac {1}{4}} \log \left (-b c^{8} \left (-\frac {2401 \, B^{4} b^{4} - 4116 \, A B^{3} b^{3} c + 2646 \, A^{2} B^{2} b^{2} c^{2} - 756 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}{b c^{11}}\right )^{\frac {3}{4}} - {\left (343 \, B^{3} b^{3} - 441 \, A B^{2} b^{2} c + 189 \, A^{2} B b c^{2} - 27 \, A^{3} c^{3}\right )} \sqrt {x}\right ) - 4 \, {\left (4 \, B c x^{3} + {\left (7 \, B b - 3 \, A c\right )} x\right )} \sqrt {x}}{24 \, {\left (c^{3} x^{2} + b c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

-1/24*(12*(c^3*x^2 + b*c^2)*(-(2401*B^4*b^4 - 4116*A*B^3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 756*A^3*B*b*c^3 + 81*A
^4*c^4)/(b*c^11))^(1/4)*arctan((sqrt((117649*B^6*b^6 - 302526*A*B^5*b^5*c + 324135*A^2*B^4*b^4*c^2 - 185220*A^
3*B^3*b^3*c^3 + 59535*A^4*B^2*b^2*c^4 - 10206*A^5*B*b*c^5 + 729*A^6*c^6)*x - (2401*B^4*b^5*c^5 - 4116*A*B^3*b^
4*c^6 + 2646*A^2*B^2*b^3*c^7 - 756*A^3*B*b^2*c^8 + 81*A^4*b*c^9)*sqrt(-(2401*B^4*b^4 - 4116*A*B^3*b^3*c + 2646
*A^2*B^2*b^2*c^2 - 756*A^3*B*b*c^3 + 81*A^4*c^4)/(b*c^11)))*c^3*(-(2401*B^4*b^4 - 4116*A*B^3*b^3*c + 2646*A^2*
B^2*b^2*c^2 - 756*A^3*B*b*c^3 + 81*A^4*c^4)/(b*c^11))^(1/4) + (343*B^3*b^3*c^3 - 441*A*B^2*b^2*c^4 + 189*A^2*B
*b*c^5 - 27*A^3*c^6)*sqrt(x)*(-(2401*B^4*b^4 - 4116*A*B^3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 756*A^3*B*b*c^3 + 81*
A^4*c^4)/(b*c^11))^(1/4))/(2401*B^4*b^4 - 4116*A*B^3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 756*A^3*B*b*c^3 + 81*A^4*c
^4)) - 3*(c^3*x^2 + b*c^2)*(-(2401*B^4*b^4 - 4116*A*B^3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 756*A^3*B*b*c^3 + 81*A^
4*c^4)/(b*c^11))^(1/4)*log(b*c^8*(-(2401*B^4*b^4 - 4116*A*B^3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 756*A^3*B*b*c^3 +
81*A^4*c^4)/(b*c^11))^(3/4) - (343*B^3*b^3 - 441*A*B^2*b^2*c + 189*A^2*B*b*c^2 - 27*A^3*c^3)*sqrt(x)) + 3*(c^
3*x^2 + b*c^2)*(-(2401*B^4*b^4 - 4116*A*B^3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 756*A^3*B*b*c^3 + 81*A^4*c^4)/(b*c^
11))^(1/4)*log(-b*c^8*(-(2401*B^4*b^4 - 4116*A*B^3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 756*A^3*B*b*c^3 + 81*A^4*c^4
)/(b*c^11))^(3/4) - (343*B^3*b^3 - 441*A*B^2*b^2*c + 189*A^2*B*b*c^2 - 27*A^3*c^3)*sqrt(x)) - 4*(4*B*c*x^3 + (
7*B*b - 3*A*c)*x)*sqrt(x))/(c^3*x^2 + b*c^2)

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giac [A]  time = 0.21, size = 283, normalized size = 0.98 \[ \frac {2 \, B x^{\frac {3}{2}}}{3 \, c^{2}} + \frac {B b x^{\frac {3}{2}} - A c x^{\frac {3}{2}}}{2 \, {\left (c x^{2} + b\right )} c^{2}} - \frac {\sqrt {2} {\left (7 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 3 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, b c^{5}} - \frac {\sqrt {2} {\left (7 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 3 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, b c^{5}} + \frac {\sqrt {2} {\left (7 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 3 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, b c^{5}} - \frac {\sqrt {2} {\left (7 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 3 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, b c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

2/3*B*x^(3/2)/c^2 + 1/2*(B*b*x^(3/2) - A*c*x^(3/2))/((c*x^2 + b)*c^2) - 1/8*sqrt(2)*(7*(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^5) - 1/8*sqrt(2)*(7*(
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^5) + 1/16*sqrt(2)*(7*(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^5) - 1/16*sqrt(2)*(7*(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 + s
qrt(b/c))/(b*c^5)

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maple [A]  time = 0.06, size = 317, normalized size = 1.10 \[ -\frac {A \,x^{\frac {3}{2}}}{2 \left (c \,x^{2}+b \right ) c}+\frac {B b \,x^{\frac {3}{2}}}{2 \left (c \,x^{2}+b \right ) c^{2}}+\frac {2 B \,x^{\frac {3}{2}}}{3 c^{2}}+\frac {3 \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{2}}+\frac {3 \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{2}}+\frac {3 \sqrt {2}\, A \ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{16 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{2}}-\frac {7 \sqrt {2}\, B b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{3}}-\frac {7 \sqrt {2}\, B b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{3}}-\frac {7 \sqrt {2}\, B b \ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{16 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(13/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x)

[Out]

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

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maxima [A]  time = 3.31, size = 223, normalized size = 0.77 \[ \frac {{\left (B b - A c\right )} x^{\frac {3}{2}}}{2 \, {\left (c^{3} x^{2} + b c^{2}\right )}} + \frac {2 \, B x^{\frac {3}{2}}}{3 \, c^{2}} - \frac {{\left (7 \, B b - 3 \, A c\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{16 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.25, size = 106, normalized size = 0.37 \[ \frac {2\,B\,x^{3/2}}{3\,c^2}-\frac {x^{3/2}\,\left (\frac {A\,c}{2}-\frac {B\,b}{2}\right )}{c^3\,x^2+b\,c^2}+\frac {\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )\,\left (3\,A\,c-7\,B\,b\right )}{4\,{\left (-b\right )}^{1/4}\,c^{11/4}}+\frac {\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}\,1{}\mathrm {i}}{{\left (-b\right )}^{1/4}}\right )\,\left (3\,A\,c-7\,B\,b\right )\,1{}\mathrm {i}}{4\,{\left (-b\right )}^{1/4}\,c^{11/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(13/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x)

[Out]

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

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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**(13/2)*(B*x**2+A)/(c*x**4+b*x**2)**2,x)

[Out]

Timed out

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