3.199 \(\int \frac {x^{11/2} (A+B x^2)}{(b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=289 \[ \frac {(5 b B-A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{3/4} c^{9/4}}-\frac {(5 b B-A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{3/4} c^{9/4}}+\frac {(5 b B-A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{3/4} c^{9/4}}-\frac {(5 b B-A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} b^{3/4} c^{9/4}}+\frac {\sqrt {x} (5 b B-A c)}{2 b c^2}-\frac {x^{5/2} (b B-A c)}{2 b c \left (b+c x^2\right )} \]

[Out]

-1/2*(-A*c+B*b)*x^(5/2)/b/c/(c*x^2+b)+1/8*(-A*c+5*B*b)*arctan(1-c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(3/4)/c^(9/
4)*2^(1/2)-1/8*(-A*c+5*B*b)*arctan(1+c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(3/4)/c^(9/4)*2^(1/2)+1/16*(-A*c+5*B*b
)*ln(b^(1/2)+x*c^(1/2)-b^(1/4)*c^(1/4)*2^(1/2)*x^(1/2))/b^(3/4)/c^(9/4)*2^(1/2)-1/16*(-A*c+5*B*b)*ln(b^(1/2)+x
*c^(1/2)+b^(1/4)*c^(1/4)*2^(1/2)*x^(1/2))/b^(3/4)/c^(9/4)*2^(1/2)+1/2*(-A*c+5*B*b)*x^(1/2)/b/c^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, 211, 1165, 628, 1162, 617, 204} \[ \frac {(5 b B-A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{3/4} c^{9/4}}-\frac {(5 b B-A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{3/4} c^{9/4}}+\frac {(5 b B-A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{3/4} c^{9/4}}-\frac {(5 b B-A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} b^{3/4} c^{9/4}}+\frac {\sqrt {x} (5 b B-A c)}{2 b c^2}-\frac {x^{5/2} (b B-A c)}{2 b c \left (b+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(x^(11/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

((5*b*B - A*c)*Sqrt[x])/(2*b*c^2) - ((b*B - A*c)*x^(5/2))/(2*b*c*(b + c*x^2)) + ((5*b*B - A*c)*ArcTan[1 - (Sqr
t[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*b^(3/4)*c^(9/4)) - ((5*b*B - A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[
x])/b^(1/4)])/(4*Sqrt[2]*b^(3/4)*c^(9/4)) + ((5*b*B - A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqr
t[c]*x])/(8*Sqrt[2]*b^(3/4)*c^(9/4)) - ((5*b*B - A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*
x])/(8*Sqrt[2]*b^(3/4)*c^(9/4))

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 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{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^{11/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac {x^{3/2} \left (A+B x^2\right )}{\left (b+c x^2\right )^2} \, dx\\ &=-\frac {(b B-A c) x^{5/2}}{2 b c \left (b+c x^2\right )}+\frac {\left (\frac {5 b B}{2}-\frac {A c}{2}\right ) \int \frac {x^{3/2}}{b+c x^2} \, dx}{2 b c}\\ &=\frac {(5 b B-A c) \sqrt {x}}{2 b c^2}-\frac {(b B-A c) x^{5/2}}{2 b c \left (b+c x^2\right )}-\frac {(5 b B-A c) \int \frac {1}{\sqrt {x} \left (b+c x^2\right )} \, dx}{4 c^2}\\ &=\frac {(5 b B-A c) \sqrt {x}}{2 b c^2}-\frac {(b B-A c) x^{5/2}}{2 b c \left (b+c x^2\right )}-\frac {(5 b B-A c) \operatorname {Subst}\left (\int \frac {1}{b+c x^4} \, dx,x,\sqrt {x}\right )}{2 c^2}\\ &=\frac {(5 b B-A c) \sqrt {x}}{2 b c^2}-\frac {(b B-A c) x^{5/2}}{2 b c \left (b+c x^2\right )}-\frac {(5 b B-A c) \operatorname {Subst}\left (\int \frac {\sqrt {b}-\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{4 \sqrt {b} c^2}-\frac {(5 b B-A c) \operatorname {Subst}\left (\int \frac {\sqrt {b}+\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{4 \sqrt {b} c^2}\\ &=\frac {(5 b B-A c) \sqrt {x}}{2 b c^2}-\frac {(b B-A c) x^{5/2}}{2 b c \left (b+c x^2\right )}-\frac {(5 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 )}{8 \sqrt {b} c^{5/2}}-\frac {(5 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 )}{8 \sqrt {b} c^{5/2}}+\frac {(5 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 )}{8 \sqrt {2} b^{3/4} c^{9/4}}+\frac {(5 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 )}{8 \sqrt {2} b^{3/4} c^{9/4}}\\ &=\frac {(5 b B-A c) \sqrt {x}}{2 b c^2}-\frac {(b B-A c) x^{5/2}}{2 b c \left (b+c x^2\right )}+\frac {(5 b B-A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{3/4} c^{9/4}}-\frac {(5 b B-A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{3/4} c^{9/4}}-\frac {(5 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 )}{4 \sqrt {2} b^{3/4} c^{9/4}}+\frac {(5 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 )}{4 \sqrt {2} b^{3/4} c^{9/4}}\\ &=\frac {(5 b B-A c) \sqrt {x}}{2 b c^2}-\frac {(b B-A c) x^{5/2}}{2 b c \left (b+c x^2\right )}+\frac {(5 b B-A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{3/4} c^{9/4}}-\frac {(5 b B-A c) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{3/4} c^{9/4}}+\frac {(5 b B-A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{3/4} c^{9/4}}-\frac {(5 b B-A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{3/4} c^{9/4}}\\ \end {align*}

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Mathematica [A]  time = 0.53, size = 354, normalized size = 1.22 \[ \frac {\frac {2 \sqrt {2} (5 b B-A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{b^{3/4}}-\frac {2 \sqrt {2} (5 b B-A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{b^{3/4}}-\frac {\sqrt {2} A c \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{b^{3/4}}+\frac {\sqrt {2} A c \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{b^{3/4}}-\frac {8 A c^{5/4} \sqrt {x}}{b+c x^2}+\frac {8 b B \sqrt [4]{c} \sqrt {x}}{b+c x^2}+5 \sqrt {2} \sqrt [4]{b} B \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )-5 \sqrt {2} \sqrt [4]{b} B \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )+32 B \sqrt [4]{c} \sqrt {x}}{16 c^{9/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(11/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

(32*B*c^(1/4)*Sqrt[x] + (8*b*B*c^(1/4)*Sqrt[x])/(b + c*x^2) - (8*A*c^(5/4)*Sqrt[x])/(b + c*x^2) + (2*Sqrt[2]*(
5*b*B - A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/b^(3/4) - (2*Sqrt[2]*(5*b*B - A*c)*ArcTan[1 + (Sqr
t[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/b^(3/4) + 5*Sqrt[2]*b^(1/4)*B*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] +
Sqrt[c]*x] - (Sqrt[2]*A*c*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/b^(3/4) - 5*Sqrt[2]*b^(1
/4)*B*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] + (Sqrt[2]*A*c*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*
c^(1/4)*Sqrt[x] + Sqrt[c]*x])/b^(3/4))/(16*c^(9/4))

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fricas [B]  time = 1.09, size = 725, normalized size = 2.51 \[ \frac {4 \, {\left (c^{3} x^{2} + b c^{2}\right )} \left (-\frac {625 \, B^{4} b^{4} - 500 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 20 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{9}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {b^{2} c^{4} \sqrt {-\frac {625 \, B^{4} b^{4} - 500 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 20 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{9}}} + {\left (25 \, B^{2} b^{2} - 10 \, A B b c + A^{2} c^{2}\right )} x} b^{2} c^{7} \left (-\frac {625 \, B^{4} b^{4} - 500 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 20 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{9}}\right )^{\frac {3}{4}} + {\left (5 \, B b^{3} c^{7} - A b^{2} c^{8}\right )} \sqrt {x} \left (-\frac {625 \, B^{4} b^{4} - 500 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 20 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{9}}\right )^{\frac {3}{4}}}{625 \, B^{4} b^{4} - 500 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 20 \, A^{3} B b c^{3} + A^{4} c^{4}}\right ) + {\left (c^{3} x^{2} + b c^{2}\right )} \left (-\frac {625 \, B^{4} b^{4} - 500 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 20 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{9}}\right )^{\frac {1}{4}} \log \left (b c^{2} \left (-\frac {625 \, B^{4} b^{4} - 500 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 20 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{9}}\right )^{\frac {1}{4}} - {\left (5 \, B b - A c\right )} \sqrt {x}\right ) - {\left (c^{3} x^{2} + b c^{2}\right )} \left (-\frac {625 \, B^{4} b^{4} - 500 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 20 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{9}}\right )^{\frac {1}{4}} \log \left (-b c^{2} \left (-\frac {625 \, B^{4} b^{4} - 500 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 20 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{9}}\right )^{\frac {1}{4}} - {\left (5 \, B b - A c\right )} \sqrt {x}\right ) + 4 \, {\left (4 \, B c x^{2} + 5 \, B b - A c\right )} \sqrt {x}}{8 \, {\left (c^{3} x^{2} + b c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(11/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

1/8*(4*(c^3*x^2 + b*c^2)*(-(625*B^4*b^4 - 500*A*B^3*b^3*c + 150*A^2*B^2*b^2*c^2 - 20*A^3*B*b*c^3 + A^4*c^4)/(b
^3*c^9))^(1/4)*arctan((sqrt(b^2*c^4*sqrt(-(625*B^4*b^4 - 500*A*B^3*b^3*c + 150*A^2*B^2*b^2*c^2 - 20*A^3*B*b*c^
3 + A^4*c^4)/(b^3*c^9)) + (25*B^2*b^2 - 10*A*B*b*c + A^2*c^2)*x)*b^2*c^7*(-(625*B^4*b^4 - 500*A*B^3*b^3*c + 15
0*A^2*B^2*b^2*c^2 - 20*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^9))^(3/4) + (5*B*b^3*c^7 - A*b^2*c^8)*sqrt(x)*(-(625*B^4*
b^4 - 500*A*B^3*b^3*c + 150*A^2*B^2*b^2*c^2 - 20*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^9))^(3/4))/(625*B^4*b^4 - 500*A
*B^3*b^3*c + 150*A^2*B^2*b^2*c^2 - 20*A^3*B*b*c^3 + A^4*c^4)) + (c^3*x^2 + b*c^2)*(-(625*B^4*b^4 - 500*A*B^3*b
^3*c + 150*A^2*B^2*b^2*c^2 - 20*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^9))^(1/4)*log(b*c^2*(-(625*B^4*b^4 - 500*A*B^3*b
^3*c + 150*A^2*B^2*b^2*c^2 - 20*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^9))^(1/4) - (5*B*b - A*c)*sqrt(x)) - (c^3*x^2 +
b*c^2)*(-(625*B^4*b^4 - 500*A*B^3*b^3*c + 150*A^2*B^2*b^2*c^2 - 20*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^9))^(1/4)*log
(-b*c^2*(-(625*B^4*b^4 - 500*A*B^3*b^3*c + 150*A^2*B^2*b^2*c^2 - 20*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^9))^(1/4) -
(5*B*b - A*c)*sqrt(x)) + 4*(4*B*c*x^2 + 5*B*b - A*c)*sqrt(x))/(c^3*x^2 + b*c^2)

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giac [A]  time = 0.19, size = 283, normalized size = 0.98 \[ \frac {2 \, B \sqrt {x}}{c^{2}} - \frac {\sqrt {2} {\left (5 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, b c^{3}} - \frac {\sqrt {2} {\left (5 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, b c^{3}} - \frac {\sqrt {2} {\left (5 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, b c^{3}} + \frac {\sqrt {2} {\left (5 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, b c^{3}} + \frac {B b \sqrt {x} - A c \sqrt {x}}{2 \, {\left (c x^{2} + b\right )} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(11/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

2*B*sqrt(x)/c^2 - 1/8*sqrt(2)*(5*(b*c^3)^(1/4)*B*b - (b*c^3)^(1/4)*A*c)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4
) + 2*sqrt(x))/(b/c)^(1/4))/(b*c^3) - 1/8*sqrt(2)*(5*(b*c^3)^(1/4)*B*b - (b*c^3)^(1/4)*A*c)*arctan(-1/2*sqrt(2
)*(sqrt(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/(b*c^3) - 1/16*sqrt(2)*(5*(b*c^3)^(1/4)*B*b - (b*c^3)^(1/4)*A
*c)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b*c^3) + 1/16*sqrt(2)*(5*(b*c^3)^(1/4)*B*b - (b*c^3)^(1/
4)*A*c)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b*c^3) + 1/2*(B*b*sqrt(x) - A*c*sqrt(x))/((c*x^2 +
b)*c^2)

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maple [A]  time = 0.06, size = 323, normalized size = 1.12 \[ -\frac {A \sqrt {x}}{2 \left (c \,x^{2}+b \right ) c}+\frac {B b \sqrt {x}}{2 \left (c \,x^{2}+b \right ) c^{2}}+\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 b c}+\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 b c}+\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{16 b c}-\frac {5 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 c^{2}}-\frac {5 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 c^{2}}-\frac {5 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{16 c^{2}}+\frac {2 B \sqrt {x}}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(11/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x)

[Out]

2*B/c^2*x^(1/2)-1/2/c*x^(1/2)/(c*x^2+b)*A+1/2/c^2*x^(1/2)/(c*x^2+b)*b*B+1/8/c*(b/c)^(1/4)/b*2^(1/2)*A*arctan(2
^(1/2)/(b/c)^(1/4)*x^(1/2)-1)+1/16/c*(b/c)^(1/4)/b*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)))+1/8/c*(b/c)^(1/4)/b*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)
-5/8/c^2*(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)-5/16/c^2*(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)))-5/8/c^2*(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 = 250, normalized size = 0.87 \[ \frac {{\left (B b - A c\right )} \sqrt {x}}{2 \, {\left (c^{3} x^{2} + b c^{2}\right )}} + \frac {2 \, B \sqrt {x}}{c^{2}} - \frac {\frac {2 \, \sqrt {2} {\left (5 \, B b - A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} {\left (5 \, B b - A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} {\left (5 \, B b - A c\right )} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (5 \, B b - A c\right )} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}}}{16 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(11/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

1/2*(B*b - A*c)*sqrt(x)/(c^3*x^2 + b*c^2) + 2*B*sqrt(x)/c^2 - 1/16*(2*sqrt(2)*(5*B*b - A*c)*arctan(1/2*sqrt(2)
*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c)))/(sqrt(b)*sqrt(sqrt(b)*sqrt(c))) + 2*sqrt
(2)*(5*B*b - A*c)*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) - 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c)))/(sq
rt(b)*sqrt(sqrt(b)*sqrt(c))) + sqrt(2)*(5*B*b - A*c)*log(sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b)
)/(b^(3/4)*c^(1/4)) - sqrt(2)*(5*B*b - A*c)*log(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(3/
4)*c^(1/4)))/c^2

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mupad [B]  time = 0.21, size = 744, normalized size = 2.57 \[ \frac {2\,B\,\sqrt {x}}{c^2}-\frac {\sqrt {x}\,\left (\frac {A\,c}{2}-\frac {B\,b}{2}\right )}{c^3\,x^2+b\,c^2}+\frac {\mathrm {atan}\left (\frac {\frac {\left (A\,c-5\,B\,b\right )\,\left (\frac {\sqrt {x}\,\left (A^2\,c^2-10\,A\,B\,b\,c+25\,B^2\,b^2\right )}{c}-\frac {\left (A\,c-5\,B\,b\right )\,\left (8\,A\,b\,c^2-40\,B\,b^2\,c\right )}{8\,{\left (-b\right )}^{3/4}\,c^{9/4}}\right )\,1{}\mathrm {i}}{8\,{\left (-b\right )}^{3/4}\,c^{9/4}}+\frac {\left (A\,c-5\,B\,b\right )\,\left (\frac {\sqrt {x}\,\left (A^2\,c^2-10\,A\,B\,b\,c+25\,B^2\,b^2\right )}{c}+\frac {\left (A\,c-5\,B\,b\right )\,\left (8\,A\,b\,c^2-40\,B\,b^2\,c\right )}{8\,{\left (-b\right )}^{3/4}\,c^{9/4}}\right )\,1{}\mathrm {i}}{8\,{\left (-b\right )}^{3/4}\,c^{9/4}}}{\frac {\left (A\,c-5\,B\,b\right )\,\left (\frac {\sqrt {x}\,\left (A^2\,c^2-10\,A\,B\,b\,c+25\,B^2\,b^2\right )}{c}-\frac {\left (A\,c-5\,B\,b\right )\,\left (8\,A\,b\,c^2-40\,B\,b^2\,c\right )}{8\,{\left (-b\right )}^{3/4}\,c^{9/4}}\right )}{8\,{\left (-b\right )}^{3/4}\,c^{9/4}}-\frac {\left (A\,c-5\,B\,b\right )\,\left (\frac {\sqrt {x}\,\left (A^2\,c^2-10\,A\,B\,b\,c+25\,B^2\,b^2\right )}{c}+\frac {\left (A\,c-5\,B\,b\right )\,\left (8\,A\,b\,c^2-40\,B\,b^2\,c\right )}{8\,{\left (-b\right )}^{3/4}\,c^{9/4}}\right )}{8\,{\left (-b\right )}^{3/4}\,c^{9/4}}}\right )\,\left (A\,c-5\,B\,b\right )\,1{}\mathrm {i}}{4\,{\left (-b\right )}^{3/4}\,c^{9/4}}+\frac {\mathrm {atan}\left (\frac {\frac {\left (A\,c-5\,B\,b\right )\,\left (\frac {\sqrt {x}\,\left (A^2\,c^2-10\,A\,B\,b\,c+25\,B^2\,b^2\right )}{c}-\frac {\left (A\,c-5\,B\,b\right )\,\left (8\,A\,b\,c^2-40\,B\,b^2\,c\right )\,1{}\mathrm {i}}{8\,{\left (-b\right )}^{3/4}\,c^{9/4}}\right )}{8\,{\left (-b\right )}^{3/4}\,c^{9/4}}+\frac {\left (A\,c-5\,B\,b\right )\,\left (\frac {\sqrt {x}\,\left (A^2\,c^2-10\,A\,B\,b\,c+25\,B^2\,b^2\right )}{c}+\frac {\left (A\,c-5\,B\,b\right )\,\left (8\,A\,b\,c^2-40\,B\,b^2\,c\right )\,1{}\mathrm {i}}{8\,{\left (-b\right )}^{3/4}\,c^{9/4}}\right )}{8\,{\left (-b\right )}^{3/4}\,c^{9/4}}}{\frac {\left (A\,c-5\,B\,b\right )\,\left (\frac {\sqrt {x}\,\left (A^2\,c^2-10\,A\,B\,b\,c+25\,B^2\,b^2\right )}{c}-\frac {\left (A\,c-5\,B\,b\right )\,\left (8\,A\,b\,c^2-40\,B\,b^2\,c\right )\,1{}\mathrm {i}}{8\,{\left (-b\right )}^{3/4}\,c^{9/4}}\right )\,1{}\mathrm {i}}{8\,{\left (-b\right )}^{3/4}\,c^{9/4}}-\frac {\left (A\,c-5\,B\,b\right )\,\left (\frac {\sqrt {x}\,\left (A^2\,c^2-10\,A\,B\,b\,c+25\,B^2\,b^2\right )}{c}+\frac {\left (A\,c-5\,B\,b\right )\,\left (8\,A\,b\,c^2-40\,B\,b^2\,c\right )\,1{}\mathrm {i}}{8\,{\left (-b\right )}^{3/4}\,c^{9/4}}\right )\,1{}\mathrm {i}}{8\,{\left (-b\right )}^{3/4}\,c^{9/4}}}\right )\,\left (A\,c-5\,B\,b\right )}{4\,{\left (-b\right )}^{3/4}\,c^{9/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(11/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x)

[Out]

(2*B*x^(1/2))/c^2 - (x^(1/2)*((A*c)/2 - (B*b)/2))/(b*c^2 + c^3*x^2) + (atan((((A*c - 5*B*b)*((x^(1/2)*(A^2*c^2
+ 25*B^2*b^2 - 10*A*B*b*c))/c - ((A*c - 5*B*b)*(8*A*b*c^2 - 40*B*b^2*c))/(8*(-b)^(3/4)*c^(9/4)))*1i)/(8*(-b)^
(3/4)*c^(9/4)) + ((A*c - 5*B*b)*((x^(1/2)*(A^2*c^2 + 25*B^2*b^2 - 10*A*B*b*c))/c + ((A*c - 5*B*b)*(8*A*b*c^2 -
40*B*b^2*c))/(8*(-b)^(3/4)*c^(9/4)))*1i)/(8*(-b)^(3/4)*c^(9/4)))/(((A*c - 5*B*b)*((x^(1/2)*(A^2*c^2 + 25*B^2*
b^2 - 10*A*B*b*c))/c - ((A*c - 5*B*b)*(8*A*b*c^2 - 40*B*b^2*c))/(8*(-b)^(3/4)*c^(9/4))))/(8*(-b)^(3/4)*c^(9/4)
) - ((A*c - 5*B*b)*((x^(1/2)*(A^2*c^2 + 25*B^2*b^2 - 10*A*B*b*c))/c + ((A*c - 5*B*b)*(8*A*b*c^2 - 40*B*b^2*c))
/(8*(-b)^(3/4)*c^(9/4))))/(8*(-b)^(3/4)*c^(9/4))))*(A*c - 5*B*b)*1i)/(4*(-b)^(3/4)*c^(9/4)) + (atan((((A*c - 5
*B*b)*((x^(1/2)*(A^2*c^2 + 25*B^2*b^2 - 10*A*B*b*c))/c - ((A*c - 5*B*b)*(8*A*b*c^2 - 40*B*b^2*c)*1i)/(8*(-b)^(
3/4)*c^(9/4))))/(8*(-b)^(3/4)*c^(9/4)) + ((A*c - 5*B*b)*((x^(1/2)*(A^2*c^2 + 25*B^2*b^2 - 10*A*B*b*c))/c + ((A
*c - 5*B*b)*(8*A*b*c^2 - 40*B*b^2*c)*1i)/(8*(-b)^(3/4)*c^(9/4))))/(8*(-b)^(3/4)*c^(9/4)))/(((A*c - 5*B*b)*((x^
(1/2)*(A^2*c^2 + 25*B^2*b^2 - 10*A*B*b*c))/c - ((A*c - 5*B*b)*(8*A*b*c^2 - 40*B*b^2*c)*1i)/(8*(-b)^(3/4)*c^(9/
4)))*1i)/(8*(-b)^(3/4)*c^(9/4)) - ((A*c - 5*B*b)*((x^(1/2)*(A^2*c^2 + 25*B^2*b^2 - 10*A*B*b*c))/c + ((A*c - 5*
B*b)*(8*A*b*c^2 - 40*B*b^2*c)*1i)/(8*(-b)^(3/4)*c^(9/4)))*1i)/(8*(-b)^(3/4)*c^(9/4))))*(A*c - 5*B*b))/(4*(-b)^
(3/4)*c^(9/4))

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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**(11/2)*(B*x**2+A)/(c*x**4+b*x**2)**2,x)

[Out]

Timed out

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