3.3.13 \(\int \frac {x^{11/2} (A+B x^2)}{(b x^2+c x^4)^3} \, dx\)

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

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Rubi [A]  time = 0.23, antiderivative size = 293, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {1584, 457, 290, 329, 211, 1165, 628, 1162, 617, 204} \begin {gather*} -\frac {3 (7 A c+b B) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{11/4} c^{5/4}}+\frac {3 (7 A c+b B) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{11/4} c^{5/4}}-\frac {3 (7 A c+b B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{11/4} c^{5/4}}+\frac {3 (7 A c+b B) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt {2} b^{11/4} c^{5/4}}+\frac {\sqrt {x} (7 A c+b B)}{16 b^2 c \left (b+c x^2\right )}-\frac {\sqrt {x} (b B-A c)}{4 b c \left (b+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^(11/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

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

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 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && 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 )^3} \, dx &=\int \frac {A+B x^2}{\sqrt {x} \left (b+c x^2\right )^3} \, dx\\ &=-\frac {(b B-A c) \sqrt {x}}{4 b c \left (b+c x^2\right )^2}+\frac {\left (\frac {b B}{2}+\frac {7 A c}{2}\right ) \int \frac {1}{\sqrt {x} \left (b+c x^2\right )^2} \, dx}{4 b c}\\ &=-\frac {(b B-A c) \sqrt {x}}{4 b c \left (b+c x^2\right )^2}+\frac {(b B+7 A c) \sqrt {x}}{16 b^2 c \left (b+c x^2\right )}+\frac {(3 (b B+7 A c)) \int \frac {1}{\sqrt {x} \left (b+c x^2\right )} \, dx}{32 b^2 c}\\ &=-\frac {(b B-A c) \sqrt {x}}{4 b c \left (b+c x^2\right )^2}+\frac {(b B+7 A c) \sqrt {x}}{16 b^2 c \left (b+c x^2\right )}+\frac {(3 (b B+7 A c)) \operatorname {Subst}\left (\int \frac {1}{b+c x^4} \, dx,x,\sqrt {x}\right )}{16 b^2 c}\\ &=-\frac {(b B-A c) \sqrt {x}}{4 b c \left (b+c x^2\right )^2}+\frac {(b B+7 A c) \sqrt {x}}{16 b^2 c \left (b+c x^2\right )}+\frac {(3 (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 )}{32 b^{5/2} c}+\frac {(3 (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 )}{32 b^{5/2} c}\\ &=-\frac {(b B-A c) \sqrt {x}}{4 b c \left (b+c x^2\right )^2}+\frac {(b B+7 A c) \sqrt {x}}{16 b^2 c \left (b+c x^2\right )}+\frac {(3 (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 )}{64 b^{5/2} c^{3/2}}+\frac {(3 (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 )}{64 b^{5/2} c^{3/2}}-\frac {(3 (b B+7 A c)) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{b}}{\sqrt [4]{c}}+2 x}{-\frac {\sqrt {b}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt {x}\right )}{64 \sqrt {2} b^{11/4} c^{5/4}}-\frac {(3 (b B+7 A c)) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{b}}{\sqrt [4]{c}}-2 x}{-\frac {\sqrt {b}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt {x}\right )}{64 \sqrt {2} b^{11/4} c^{5/4}}\\ &=-\frac {(b B-A c) \sqrt {x}}{4 b c \left (b+c x^2\right )^2}+\frac {(b B+7 A c) \sqrt {x}}{16 b^2 c \left (b+c x^2\right )}-\frac {3 (b B+7 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{11/4} c^{5/4}}+\frac {3 (b B+7 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{11/4} c^{5/4}}+\frac {(3 (b B+7 A c)) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{11/4} c^{5/4}}-\frac {(3 (b B+7 A c)) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{11/4} c^{5/4}}\\ &=-\frac {(b B-A c) \sqrt {x}}{4 b c \left (b+c x^2\right )^2}+\frac {(b B+7 A c) \sqrt {x}}{16 b^2 c \left (b+c x^2\right )}-\frac {3 (b B+7 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{11/4} c^{5/4}}+\frac {3 (b B+7 A c) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{11/4} c^{5/4}}-\frac {3 (b B+7 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{11/4} c^{5/4}}+\frac {3 (b B+7 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{11/4} c^{5/4}}\\ \end {align*}

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Mathematica [A]  time = 0.38, size = 230, normalized size = 0.78 \begin {gather*} \frac {\frac {(7 A c+b B) \left (7 \left (b+c x^2\right ) \left (8 b^{3/4} \sqrt [4]{c} \sqrt {x}-3 \sqrt {2} \left (b+c x^2\right ) \left (\log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )-\log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )+2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )-2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )\right )\right )+32 b^{7/4} \sqrt [4]{c} \sqrt {x}\right )}{b^{11/4} \sqrt [4]{c}}-256 B \sqrt {x}}{896 c \left (b+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^(11/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

(-256*B*Sqrt[x] + ((b*B + 7*A*c)*(32*b^(7/4)*c^(1/4)*Sqrt[x] + 7*(b + c*x^2)*(8*b^(3/4)*c^(1/4)*Sqrt[x] - 3*Sq
rt[2]*(b + c*x^2)*(2*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)] - 2*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^
(1/4)] + Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] - Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sq
rt[x] + Sqrt[c]*x]))))/(b^(11/4)*c^(1/4)))/(896*c*(b + c*x^2)^2)

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IntegrateAlgebraic [A]  time = 0.83, size = 181, normalized size = 0.62 \begin {gather*} -\frac {3 (7 A c+b B) \tan ^{-1}\left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )}{32 \sqrt {2} b^{11/4} c^{5/4}}+\frac {3 (7 A c+b B) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{32 \sqrt {2} b^{11/4} c^{5/4}}-\frac {\sqrt {x} \left (-11 A b c-7 A c^2 x^2+3 b^2 B-b B c x^2\right )}{16 b^2 c \left (b+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^(11/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

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

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fricas [B]  time = 0.44, size = 793, normalized size = 2.71 \begin {gather*} \frac {12 \, {\left (b^{2} c^{3} x^{4} + 2 \, b^{3} c^{2} x^{2} + b^{4} c\right )} \left (-\frac {B^{4} b^{4} + 28 \, A B^{3} b^{3} c + 294 \, A^{2} B^{2} b^{2} c^{2} + 1372 \, A^{3} B b c^{3} + 2401 \, A^{4} c^{4}}{b^{11} c^{5}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {b^{6} c^{2} \sqrt {-\frac {B^{4} b^{4} + 28 \, A B^{3} b^{3} c + 294 \, A^{2} B^{2} b^{2} c^{2} + 1372 \, A^{3} B b c^{3} + 2401 \, A^{4} c^{4}}{b^{11} c^{5}}} + {\left (B^{2} b^{2} + 14 \, A B b c + 49 \, A^{2} c^{2}\right )} x} b^{8} c^{4} \left (-\frac {B^{4} b^{4} + 28 \, A B^{3} b^{3} c + 294 \, A^{2} B^{2} b^{2} c^{2} + 1372 \, A^{3} B b c^{3} + 2401 \, A^{4} c^{4}}{b^{11} c^{5}}\right )^{\frac {3}{4}} - {\left (B b^{9} c^{4} + 7 \, A b^{8} c^{5}\right )} \sqrt {x} \left (-\frac {B^{4} b^{4} + 28 \, A B^{3} b^{3} c + 294 \, A^{2} B^{2} b^{2} c^{2} + 1372 \, A^{3} B b c^{3} + 2401 \, A^{4} c^{4}}{b^{11} c^{5}}\right )^{\frac {3}{4}}}{B^{4} b^{4} + 28 \, A B^{3} b^{3} c + 294 \, A^{2} B^{2} b^{2} c^{2} + 1372 \, A^{3} B b c^{3} + 2401 \, A^{4} c^{4}}\right ) + 3 \, {\left (b^{2} c^{3} x^{4} + 2 \, b^{3} c^{2} x^{2} + b^{4} c\right )} \left (-\frac {B^{4} b^{4} + 28 \, A B^{3} b^{3} c + 294 \, A^{2} B^{2} b^{2} c^{2} + 1372 \, A^{3} B b c^{3} + 2401 \, A^{4} c^{4}}{b^{11} c^{5}}\right )^{\frac {1}{4}} \log \left (3 \, b^{3} c \left (-\frac {B^{4} b^{4} + 28 \, A B^{3} b^{3} c + 294 \, A^{2} B^{2} b^{2} c^{2} + 1372 \, A^{3} B b c^{3} + 2401 \, A^{4} c^{4}}{b^{11} c^{5}}\right )^{\frac {1}{4}} + 3 \, {\left (B b + 7 \, A c\right )} \sqrt {x}\right ) - 3 \, {\left (b^{2} c^{3} x^{4} + 2 \, b^{3} c^{2} x^{2} + b^{4} c\right )} \left (-\frac {B^{4} b^{4} + 28 \, A B^{3} b^{3} c + 294 \, A^{2} B^{2} b^{2} c^{2} + 1372 \, A^{3} B b c^{3} + 2401 \, A^{4} c^{4}}{b^{11} c^{5}}\right )^{\frac {1}{4}} \log \left (-3 \, b^{3} c \left (-\frac {B^{4} b^{4} + 28 \, A B^{3} b^{3} c + 294 \, A^{2} B^{2} b^{2} c^{2} + 1372 \, A^{3} B b c^{3} + 2401 \, A^{4} c^{4}}{b^{11} c^{5}}\right )^{\frac {1}{4}} + 3 \, {\left (B b + 7 \, A c\right )} \sqrt {x}\right ) - 4 \, {\left (3 \, B b^{2} - 11 \, A b c - {\left (B b c + 7 \, A c^{2}\right )} x^{2}\right )} \sqrt {x}}{64 \, {\left (b^{2} c^{3} x^{4} + 2 \, b^{3} c^{2} x^{2} + b^{4} c\right )}} \end {gather*}

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)^3,x, algorithm="fricas")

[Out]

1/64*(12*(b^2*c^3*x^4 + 2*b^3*c^2*x^2 + b^4*c)*(-(B^4*b^4 + 28*A*B^3*b^3*c + 294*A^2*B^2*b^2*c^2 + 1372*A^3*B*
b*c^3 + 2401*A^4*c^4)/(b^11*c^5))^(1/4)*arctan((sqrt(b^6*c^2*sqrt(-(B^4*b^4 + 28*A*B^3*b^3*c + 294*A^2*B^2*b^2
*c^2 + 1372*A^3*B*b*c^3 + 2401*A^4*c^4)/(b^11*c^5)) + (B^2*b^2 + 14*A*B*b*c + 49*A^2*c^2)*x)*b^8*c^4*(-(B^4*b^
4 + 28*A*B^3*b^3*c + 294*A^2*B^2*b^2*c^2 + 1372*A^3*B*b*c^3 + 2401*A^4*c^4)/(b^11*c^5))^(3/4) - (B*b^9*c^4 + 7
*A*b^8*c^5)*sqrt(x)*(-(B^4*b^4 + 28*A*B^3*b^3*c + 294*A^2*B^2*b^2*c^2 + 1372*A^3*B*b*c^3 + 2401*A^4*c^4)/(b^11
*c^5))^(3/4))/(B^4*b^4 + 28*A*B^3*b^3*c + 294*A^2*B^2*b^2*c^2 + 1372*A^3*B*b*c^3 + 2401*A^4*c^4)) + 3*(b^2*c^3
*x^4 + 2*b^3*c^2*x^2 + b^4*c)*(-(B^4*b^4 + 28*A*B^3*b^3*c + 294*A^2*B^2*b^2*c^2 + 1372*A^3*B*b*c^3 + 2401*A^4*
c^4)/(b^11*c^5))^(1/4)*log(3*b^3*c*(-(B^4*b^4 + 28*A*B^3*b^3*c + 294*A^2*B^2*b^2*c^2 + 1372*A^3*B*b*c^3 + 2401
*A^4*c^4)/(b^11*c^5))^(1/4) + 3*(B*b + 7*A*c)*sqrt(x)) - 3*(b^2*c^3*x^4 + 2*b^3*c^2*x^2 + b^4*c)*(-(B^4*b^4 +
28*A*B^3*b^3*c + 294*A^2*B^2*b^2*c^2 + 1372*A^3*B*b*c^3 + 2401*A^4*c^4)/(b^11*c^5))^(1/4)*log(-3*b^3*c*(-(B^4*
b^4 + 28*A*B^3*b^3*c + 294*A^2*B^2*b^2*c^2 + 1372*A^3*B*b*c^3 + 2401*A^4*c^4)/(b^11*c^5))^(1/4) + 3*(B*b + 7*A
*c)*sqrt(x)) - 4*(3*B*b^2 - 11*A*b*c - (B*b*c + 7*A*c^2)*x^2)*sqrt(x))/(b^2*c^3*x^4 + 2*b^3*c^2*x^2 + b^4*c)

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giac [A]  time = 0.20, size = 293, normalized size = 1.00 \begin {gather*} \frac {3 \, \sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b + 7 \, \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 )}{64 \, b^{3} c^{2}} + \frac {3 \, \sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b + 7 \, \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 )}{64 \, b^{3} c^{2}} + \frac {3 \, \sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b + 7 \, \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 )}{128 \, b^{3} c^{2}} - \frac {3 \, \sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b + 7 \, \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 )}{128 \, b^{3} c^{2}} + \frac {B b c x^{\frac {5}{2}} + 7 \, A c^{2} x^{\frac {5}{2}} - 3 \, B b^{2} \sqrt {x} + 11 \, A b c \sqrt {x}}{16 \, {\left (c x^{2} + b\right )}^{2} b^{2} c} \end {gather*}

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)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 325, normalized size = 1.11 \begin {gather*} \frac {21 \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 )}{64 b^{3}}+\frac {21 \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 )}{64 b^{3}}+\frac {21 \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 )}{128 b^{3}}+\frac {3 \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 )}{64 b^{2} c}+\frac {3 \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 )}{64 b^{2} c}+\frac {3 \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 )}{128 b^{2} c}+\frac {\frac {\left (7 A c +b B \right ) x^{\frac {5}{2}}}{16 b^{2}}+\frac {\left (11 A c -3 b B \right ) \sqrt {x}}{16 b c}}{\left (c \,x^{2}+b \right )^{2}} \end {gather*}

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)^3,x)

[Out]

2*(1/32*(7*A*c+B*b)/b^2*x^(5/2)+1/32*(11*A*c-3*B*b)/b/c*x^(1/2))/(c*x^2+b)^2+21/64/b^3*(b/c)^(1/4)*2^(1/2)*A*a
rctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)+21/128/b^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/b^3*(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)
*x^(1/2)+1)+3/64/b^2/c*(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)+3/128/b^2/c*(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)))+3/64/b^2/c*
(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 = 3.05, size = 276, normalized size = 0.94 \begin {gather*} \frac {{\left (B b c + 7 \, A c^{2}\right )} x^{\frac {5}{2}} - {\left (3 \, B b^{2} - 11 \, A b c\right )} \sqrt {x}}{16 \, {\left (b^{2} c^{3} x^{4} + 2 \, b^{3} c^{2} x^{2} + b^{4} c\right )}} + \frac {3 \, {\left (\frac {2 \, \sqrt {2} {\left (B b + 7 \, 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 (B b + 7 \, 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 (B b + 7 \, 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 (B b + 7 \, 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}}}\right )}}{128 \, b^{2} c} \end {gather*}

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)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.39, size = 780, normalized size = 2.66 \begin {gather*} \frac {\frac {x^{5/2}\,\left (7\,A\,c+B\,b\right )}{16\,b^2}+\frac {\sqrt {x}\,\left (11\,A\,c-3\,B\,b\right )}{16\,b\,c}}{b^2+2\,b\,c\,x^2+c^2\,x^4}-\frac {\mathrm {atan}\left (\frac {\frac {\left (7\,A\,c+B\,b\right )\,\left (\frac {9\,\sqrt {x}\,\left (49\,A^2\,c^3+14\,A\,B\,b\,c^2+B^2\,b^2\,c\right )}{64\,b^4}-\frac {9\,\left (7\,A\,c+B\,b\right )\,\left (7\,A\,c^3+B\,b\,c^2\right )}{64\,{\left (-b\right )}^{15/4}\,c^{5/4}}\right )\,3{}\mathrm {i}}{64\,{\left (-b\right )}^{11/4}\,c^{5/4}}+\frac {\left (7\,A\,c+B\,b\right )\,\left (\frac {9\,\sqrt {x}\,\left (49\,A^2\,c^3+14\,A\,B\,b\,c^2+B^2\,b^2\,c\right )}{64\,b^4}+\frac {9\,\left (7\,A\,c+B\,b\right )\,\left (7\,A\,c^3+B\,b\,c^2\right )}{64\,{\left (-b\right )}^{15/4}\,c^{5/4}}\right )\,3{}\mathrm {i}}{64\,{\left (-b\right )}^{11/4}\,c^{5/4}}}{\frac {3\,\left (7\,A\,c+B\,b\right )\,\left (\frac {9\,\sqrt {x}\,\left (49\,A^2\,c^3+14\,A\,B\,b\,c^2+B^2\,b^2\,c\right )}{64\,b^4}-\frac {9\,\left (7\,A\,c+B\,b\right )\,\left (7\,A\,c^3+B\,b\,c^2\right )}{64\,{\left (-b\right )}^{15/4}\,c^{5/4}}\right )}{64\,{\left (-b\right )}^{11/4}\,c^{5/4}}-\frac {3\,\left (7\,A\,c+B\,b\right )\,\left (\frac {9\,\sqrt {x}\,\left (49\,A^2\,c^3+14\,A\,B\,b\,c^2+B^2\,b^2\,c\right )}{64\,b^4}+\frac {9\,\left (7\,A\,c+B\,b\right )\,\left (7\,A\,c^3+B\,b\,c^2\right )}{64\,{\left (-b\right )}^{15/4}\,c^{5/4}}\right )}{64\,{\left (-b\right )}^{11/4}\,c^{5/4}}}\right )\,\left (7\,A\,c+B\,b\right )\,3{}\mathrm {i}}{32\,{\left (-b\right )}^{11/4}\,c^{5/4}}-\frac {3\,\mathrm {atan}\left (\frac {\frac {3\,\left (7\,A\,c+B\,b\right )\,\left (\frac {9\,\sqrt {x}\,\left (49\,A^2\,c^3+14\,A\,B\,b\,c^2+B^2\,b^2\,c\right )}{64\,b^4}-\frac {\left (7\,A\,c+B\,b\right )\,\left (7\,A\,c^3+B\,b\,c^2\right )\,9{}\mathrm {i}}{64\,{\left (-b\right )}^{15/4}\,c^{5/4}}\right )}{64\,{\left (-b\right )}^{11/4}\,c^{5/4}}+\frac {3\,\left (7\,A\,c+B\,b\right )\,\left (\frac {9\,\sqrt {x}\,\left (49\,A^2\,c^3+14\,A\,B\,b\,c^2+B^2\,b^2\,c\right )}{64\,b^4}+\frac {\left (7\,A\,c+B\,b\right )\,\left (7\,A\,c^3+B\,b\,c^2\right )\,9{}\mathrm {i}}{64\,{\left (-b\right )}^{15/4}\,c^{5/4}}\right )}{64\,{\left (-b\right )}^{11/4}\,c^{5/4}}}{\frac {\left (7\,A\,c+B\,b\right )\,\left (\frac {9\,\sqrt {x}\,\left (49\,A^2\,c^3+14\,A\,B\,b\,c^2+B^2\,b^2\,c\right )}{64\,b^4}-\frac {\left (7\,A\,c+B\,b\right )\,\left (7\,A\,c^3+B\,b\,c^2\right )\,9{}\mathrm {i}}{64\,{\left (-b\right )}^{15/4}\,c^{5/4}}\right )\,3{}\mathrm {i}}{64\,{\left (-b\right )}^{11/4}\,c^{5/4}}-\frac {\left (7\,A\,c+B\,b\right )\,\left (\frac {9\,\sqrt {x}\,\left (49\,A^2\,c^3+14\,A\,B\,b\,c^2+B^2\,b^2\,c\right )}{64\,b^4}+\frac {\left (7\,A\,c+B\,b\right )\,\left (7\,A\,c^3+B\,b\,c^2\right )\,9{}\mathrm {i}}{64\,{\left (-b\right )}^{15/4}\,c^{5/4}}\right )\,3{}\mathrm {i}}{64\,{\left (-b\right )}^{11/4}\,c^{5/4}}}\right )\,\left (7\,A\,c+B\,b\right )}{32\,{\left (-b\right )}^{11/4}\,c^{5/4}} \end {gather*}

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)^3,x)

[Out]

((x^(5/2)*(7*A*c + B*b))/(16*b^2) + (x^(1/2)*(11*A*c - 3*B*b))/(16*b*c))/(b^2 + c^2*x^4 + 2*b*c*x^2) - (atan((
((7*A*c + B*b)*((9*x^(1/2)*(49*A^2*c^3 + B^2*b^2*c + 14*A*B*b*c^2))/(64*b^4) - (9*(7*A*c + B*b)*(7*A*c^3 + B*b
*c^2))/(64*(-b)^(15/4)*c^(5/4)))*3i)/(64*(-b)^(11/4)*c^(5/4)) + ((7*A*c + B*b)*((9*x^(1/2)*(49*A^2*c^3 + B^2*b
^2*c + 14*A*B*b*c^2))/(64*b^4) + (9*(7*A*c + B*b)*(7*A*c^3 + B*b*c^2))/(64*(-b)^(15/4)*c^(5/4)))*3i)/(64*(-b)^
(11/4)*c^(5/4)))/((3*(7*A*c + B*b)*((9*x^(1/2)*(49*A^2*c^3 + B^2*b^2*c + 14*A*B*b*c^2))/(64*b^4) - (9*(7*A*c +
B*b)*(7*A*c^3 + B*b*c^2))/(64*(-b)^(15/4)*c^(5/4))))/(64*(-b)^(11/4)*c^(5/4)) - (3*(7*A*c + B*b)*((9*x^(1/2)*
(49*A^2*c^3 + B^2*b^2*c + 14*A*B*b*c^2))/(64*b^4) + (9*(7*A*c + B*b)*(7*A*c^3 + B*b*c^2))/(64*(-b)^(15/4)*c^(5
/4))))/(64*(-b)^(11/4)*c^(5/4))))*(7*A*c + B*b)*3i)/(32*(-b)^(11/4)*c^(5/4)) - (3*atan(((3*(7*A*c + B*b)*((9*x
^(1/2)*(49*A^2*c^3 + B^2*b^2*c + 14*A*B*b*c^2))/(64*b^4) - ((7*A*c + B*b)*(7*A*c^3 + B*b*c^2)*9i)/(64*(-b)^(15
/4)*c^(5/4))))/(64*(-b)^(11/4)*c^(5/4)) + (3*(7*A*c + B*b)*((9*x^(1/2)*(49*A^2*c^3 + B^2*b^2*c + 14*A*B*b*c^2)
)/(64*b^4) + ((7*A*c + B*b)*(7*A*c^3 + B*b*c^2)*9i)/(64*(-b)^(15/4)*c^(5/4))))/(64*(-b)^(11/4)*c^(5/4)))/(((7*
A*c + B*b)*((9*x^(1/2)*(49*A^2*c^3 + B^2*b^2*c + 14*A*B*b*c^2))/(64*b^4) - ((7*A*c + B*b)*(7*A*c^3 + B*b*c^2)*
9i)/(64*(-b)^(15/4)*c^(5/4)))*3i)/(64*(-b)^(11/4)*c^(5/4)) - ((7*A*c + B*b)*((9*x^(1/2)*(49*A^2*c^3 + B^2*b^2*
c + 14*A*B*b*c^2))/(64*b^4) + ((7*A*c + B*b)*(7*A*c^3 + B*b*c^2)*9i)/(64*(-b)^(15/4)*c^(5/4)))*3i)/(64*(-b)^(1
1/4)*c^(5/4))))*(7*A*c + B*b))/(32*(-b)^(11/4)*c^(5/4))

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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**(11/2)*(B*x**2+A)/(c*x**4+b*x**2)**3,x)

[Out]

Timed out

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