3.3.1 \(\int \frac {x^{7/2} (A+B x^2)}{(b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=261 \[ -\frac {(3 A c+b B) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{7/4} c^{5/4}}+\frac {(3 A c+b B) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{7/4} c^{5/4}}-\frac {(3 A c+b B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{7/4} c^{5/4}}+\frac {(3 A c+b B) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} b^{7/4} c^{5/4}}-\frac {\sqrt {x} (b B-A c)}{2 b c \left (b+c x^2\right )} \]

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Rubi [A]  time = 0.20, antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {1584, 457, 329, 211, 1165, 628, 1162, 617, 204} \begin {gather*} -\frac {(3 A c+b B) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{7/4} c^{5/4}}+\frac {(3 A c+b B) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{7/4} c^{5/4}}-\frac {(3 A c+b B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{7/4} c^{5/4}}+\frac {(3 A c+b B) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} b^{7/4} c^{5/4}}-\frac {\sqrt {x} (b B-A c)}{2 b c \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^(7/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

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

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Mathematica [A]  time = 0.37, size = 203, normalized size = 0.78 \begin {gather*} \frac {\frac {(3 A c+b B) \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 )}{b^{7/4} \sqrt [4]{c}}-32 B \sqrt {x}}{48 c \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^(7/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

(-32*B*Sqrt[x] + ((b*B + 3*A*c)*(8*b^(3/4)*c^(1/4)*Sqrt[x] - 3*Sqrt[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)*Sqrt[x] + Sqrt[c]*x])))/(b^(7/4)*c^(1/4)))/(48*
c*(b + c*x^2))

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IntegrateAlgebraic [A]  time = 0.63, size = 160, normalized size = 0.61 \begin {gather*} -\frac {(3 A c+b B) \tan ^{-1}\left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )}{4 \sqrt {2} b^{7/4} c^{5/4}}+\frac {(3 A c+b B) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{4 \sqrt {2} b^{7/4} c^{5/4}}+\frac {\sqrt {x} (A c-b B)}{2 b c \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^(7/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

((-(b*B) + A*c)*Sqrt[x])/(2*b*c*(b + c*x^2)) - ((b*B + 3*A*c)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^
(1/4)*Sqrt[x])])/(4*Sqrt[2]*b^(7/4)*c^(5/4)) + ((b*B + 3*A*c)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[
b] + Sqrt[c]*x)])/(4*Sqrt[2]*b^(7/4)*c^(5/4))

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fricas [B]  time = 0.44, size = 717, normalized size = 2.75 \begin {gather*} \frac {4 \, {\left (b c^{2} x^{2} + b^{2} c\right )} \left (-\frac {B^{4} b^{4} + 12 \, A B^{3} b^{3} c + 54 \, A^{2} B^{2} b^{2} c^{2} + 108 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}{b^{7} c^{5}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {b^{4} c^{2} \sqrt {-\frac {B^{4} b^{4} + 12 \, A B^{3} b^{3} c + 54 \, A^{2} B^{2} b^{2} c^{2} + 108 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}{b^{7} c^{5}}} + {\left (B^{2} b^{2} + 6 \, A B b c + 9 \, A^{2} c^{2}\right )} x} b^{5} c^{4} \left (-\frac {B^{4} b^{4} + 12 \, A B^{3} b^{3} c + 54 \, A^{2} B^{2} b^{2} c^{2} + 108 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}{b^{7} c^{5}}\right )^{\frac {3}{4}} - {\left (B b^{6} c^{4} + 3 \, A b^{5} c^{5}\right )} \sqrt {x} \left (-\frac {B^{4} b^{4} + 12 \, A B^{3} b^{3} c + 54 \, A^{2} B^{2} b^{2} c^{2} + 108 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}{b^{7} c^{5}}\right )^{\frac {3}{4}}}{B^{4} b^{4} + 12 \, A B^{3} b^{3} c + 54 \, A^{2} B^{2} b^{2} c^{2} + 108 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}\right ) + {\left (b c^{2} x^{2} + b^{2} c\right )} \left (-\frac {B^{4} b^{4} + 12 \, A B^{3} b^{3} c + 54 \, A^{2} B^{2} b^{2} c^{2} + 108 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}{b^{7} c^{5}}\right )^{\frac {1}{4}} \log \left (b^{2} c \left (-\frac {B^{4} b^{4} + 12 \, A B^{3} b^{3} c + 54 \, A^{2} B^{2} b^{2} c^{2} + 108 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}{b^{7} c^{5}}\right )^{\frac {1}{4}} + {\left (B b + 3 \, A c\right )} \sqrt {x}\right ) - {\left (b c^{2} x^{2} + b^{2} c\right )} \left (-\frac {B^{4} b^{4} + 12 \, A B^{3} b^{3} c + 54 \, A^{2} B^{2} b^{2} c^{2} + 108 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}{b^{7} c^{5}}\right )^{\frac {1}{4}} \log \left (-b^{2} c \left (-\frac {B^{4} b^{4} + 12 \, A B^{3} b^{3} c + 54 \, A^{2} B^{2} b^{2} c^{2} + 108 \, A^{3} B b c^{3} + 81 \, A^{4} c^{4}}{b^{7} c^{5}}\right )^{\frac {1}{4}} + {\left (B b + 3 \, A c\right )} \sqrt {x}\right ) - 4 \, {\left (B b - A c\right )} \sqrt {x}}{8 \, {\left (b c^{2} x^{2} + b^{2} c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

1/8*(4*(b*c^2*x^2 + b^2*c)*(-(B^4*b^4 + 12*A*B^3*b^3*c + 54*A^2*B^2*b^2*c^2 + 108*A^3*B*b*c^3 + 81*A^4*c^4)/(b
^7*c^5))^(1/4)*arctan((sqrt(b^4*c^2*sqrt(-(B^4*b^4 + 12*A*B^3*b^3*c + 54*A^2*B^2*b^2*c^2 + 108*A^3*B*b*c^3 + 8
1*A^4*c^4)/(b^7*c^5)) + (B^2*b^2 + 6*A*B*b*c + 9*A^2*c^2)*x)*b^5*c^4*(-(B^4*b^4 + 12*A*B^3*b^3*c + 54*A^2*B^2*
b^2*c^2 + 108*A^3*B*b*c^3 + 81*A^4*c^4)/(b^7*c^5))^(3/4) - (B*b^6*c^4 + 3*A*b^5*c^5)*sqrt(x)*(-(B^4*b^4 + 12*A
*B^3*b^3*c + 54*A^2*B^2*b^2*c^2 + 108*A^3*B*b*c^3 + 81*A^4*c^4)/(b^7*c^5))^(3/4))/(B^4*b^4 + 12*A*B^3*b^3*c +
54*A^2*B^2*b^2*c^2 + 108*A^3*B*b*c^3 + 81*A^4*c^4)) + (b*c^2*x^2 + b^2*c)*(-(B^4*b^4 + 12*A*B^3*b^3*c + 54*A^2
*B^2*b^2*c^2 + 108*A^3*B*b*c^3 + 81*A^4*c^4)/(b^7*c^5))^(1/4)*log(b^2*c*(-(B^4*b^4 + 12*A*B^3*b^3*c + 54*A^2*B
^2*b^2*c^2 + 108*A^3*B*b*c^3 + 81*A^4*c^4)/(b^7*c^5))^(1/4) + (B*b + 3*A*c)*sqrt(x)) - (b*c^2*x^2 + b^2*c)*(-(
B^4*b^4 + 12*A*B^3*b^3*c + 54*A^2*B^2*b^2*c^2 + 108*A^3*B*b*c^3 + 81*A^4*c^4)/(b^7*c^5))^(1/4)*log(-b^2*c*(-(B
^4*b^4 + 12*A*B^3*b^3*c + 54*A^2*B^2*b^2*c^2 + 108*A^3*B*b*c^3 + 81*A^4*c^4)/(b^7*c^5))^(1/4) + (B*b + 3*A*c)*
sqrt(x)) - 4*(B*b - A*c)*sqrt(x))/(b*c^2*x^2 + b^2*c)

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giac [A]  time = 0.21, size = 273, normalized size = 1.05 \begin {gather*} \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b + 3 \, \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^{2} c^{2}} + \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b + 3 \, \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^{2} c^{2}} + \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b + 3 \, \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^{2} c^{2}} - \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b + 3 \, \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^{2} c^{2}} - \frac {B b \sqrt {x} - A c \sqrt {x}}{2 \, {\left (c x^{2} + b\right )} b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 305, normalized size = 1.17 \begin {gather*} \frac {3 \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^{2}}+\frac {3 \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^{2}}+\frac {3 \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^{2}}+\frac {\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 b c}+\frac {\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 b c}+\frac {\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 b c}+\frac {\left (A c -b B \right ) \sqrt {x}}{2 \left (c \,x^{2}+b \right ) b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x)

[Out]

1/2*(A*c-B*b)/b/c*x^(1/2)/(c*x^2+b)+3/16/b^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/b^2*(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1
/2)+1)+3/8/b^2*(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)+1/16/b/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)))+1/8/b/c*(b/c)^(1/4)*2^
(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+1/8/b/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 = 3.11, size = 241, normalized size = 0.92 \begin {gather*} -\frac {{\left (B b - A c\right )} \sqrt {x}}{2 \, {\left (b c^{2} x^{2} + b^{2} c\right )}} + \frac {\frac {2 \, \sqrt {2} {\left (B b + 3 \, 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 + 3 \, 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 + 3 \, 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 + 3 \, 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 \, b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

-1/2*(B*b - A*c)*sqrt(x)/(b*c^2*x^2 + b^2*c) + 1/16*(2*sqrt(2)*(B*b + 3*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 + 3*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 + 3*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 + 3*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
*c)

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mupad [B]  time = 0.33, size = 750, normalized size = 2.87 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\frac {\left (3\,A\,c+B\,b\right )\,\left (\frac {\sqrt {x}\,\left (9\,A^2\,c^3+6\,A\,B\,b\,c^2+B^2\,b^2\,c\right )}{b^2}-\frac {\left (3\,A\,c+B\,b\right )\,\left (24\,A\,c^3+8\,B\,b\,c^2\right )\,1{}\mathrm {i}}{8\,{\left (-b\right )}^{7/4}\,c^{5/4}}\right )}{8\,{\left (-b\right )}^{7/4}\,c^{5/4}}+\frac {\left (3\,A\,c+B\,b\right )\,\left (\frac {\sqrt {x}\,\left (9\,A^2\,c^3+6\,A\,B\,b\,c^2+B^2\,b^2\,c\right )}{b^2}+\frac {\left (3\,A\,c+B\,b\right )\,\left (24\,A\,c^3+8\,B\,b\,c^2\right )\,1{}\mathrm {i}}{8\,{\left (-b\right )}^{7/4}\,c^{5/4}}\right )}{8\,{\left (-b\right )}^{7/4}\,c^{5/4}}}{\frac {\left (3\,A\,c+B\,b\right )\,\left (\frac {\sqrt {x}\,\left (9\,A^2\,c^3+6\,A\,B\,b\,c^2+B^2\,b^2\,c\right )}{b^2}-\frac {\left (3\,A\,c+B\,b\right )\,\left (24\,A\,c^3+8\,B\,b\,c^2\right )\,1{}\mathrm {i}}{8\,{\left (-b\right )}^{7/4}\,c^{5/4}}\right )\,1{}\mathrm {i}}{8\,{\left (-b\right )}^{7/4}\,c^{5/4}}-\frac {\left (3\,A\,c+B\,b\right )\,\left (\frac {\sqrt {x}\,\left (9\,A^2\,c^3+6\,A\,B\,b\,c^2+B^2\,b^2\,c\right )}{b^2}+\frac {\left (3\,A\,c+B\,b\right )\,\left (24\,A\,c^3+8\,B\,b\,c^2\right )\,1{}\mathrm {i}}{8\,{\left (-b\right )}^{7/4}\,c^{5/4}}\right )\,1{}\mathrm {i}}{8\,{\left (-b\right )}^{7/4}\,c^{5/4}}}\right )\,\left (3\,A\,c+B\,b\right )}{4\,{\left (-b\right )}^{7/4}\,c^{5/4}}+\frac {\sqrt {x}\,\left (A\,c-B\,b\right )}{2\,b\,c\,\left (c\,x^2+b\right )}+\frac {\mathrm {atan}\left (\frac {\frac {\left (3\,A\,c+B\,b\right )\,\left (\frac {\sqrt {x}\,\left (9\,A^2\,c^3+6\,A\,B\,b\,c^2+B^2\,b^2\,c\right )}{b^2}-\frac {\left (3\,A\,c+B\,b\right )\,\left (24\,A\,c^3+8\,B\,b\,c^2\right )}{8\,{\left (-b\right )}^{7/4}\,c^{5/4}}\right )\,1{}\mathrm {i}}{8\,{\left (-b\right )}^{7/4}\,c^{5/4}}+\frac {\left (3\,A\,c+B\,b\right )\,\left (\frac {\sqrt {x}\,\left (9\,A^2\,c^3+6\,A\,B\,b\,c^2+B^2\,b^2\,c\right )}{b^2}+\frac {\left (3\,A\,c+B\,b\right )\,\left (24\,A\,c^3+8\,B\,b\,c^2\right )}{8\,{\left (-b\right )}^{7/4}\,c^{5/4}}\right )\,1{}\mathrm {i}}{8\,{\left (-b\right )}^{7/4}\,c^{5/4}}}{\frac {\left (3\,A\,c+B\,b\right )\,\left (\frac {\sqrt {x}\,\left (9\,A^2\,c^3+6\,A\,B\,b\,c^2+B^2\,b^2\,c\right )}{b^2}-\frac {\left (3\,A\,c+B\,b\right )\,\left (24\,A\,c^3+8\,B\,b\,c^2\right )}{8\,{\left (-b\right )}^{7/4}\,c^{5/4}}\right )}{8\,{\left (-b\right )}^{7/4}\,c^{5/4}}-\frac {\left (3\,A\,c+B\,b\right )\,\left (\frac {\sqrt {x}\,\left (9\,A^2\,c^3+6\,A\,B\,b\,c^2+B^2\,b^2\,c\right )}{b^2}+\frac {\left (3\,A\,c+B\,b\right )\,\left (24\,A\,c^3+8\,B\,b\,c^2\right )}{8\,{\left (-b\right )}^{7/4}\,c^{5/4}}\right )}{8\,{\left (-b\right )}^{7/4}\,c^{5/4}}}\right )\,\left (3\,A\,c+B\,b\right )\,1{}\mathrm {i}}{4\,{\left (-b\right )}^{7/4}\,c^{5/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(7/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x)

[Out]

(atan((((3*A*c + B*b)*((x^(1/2)*(9*A^2*c^3 + B^2*b^2*c + 6*A*B*b*c^2))/b^2 - ((3*A*c + B*b)*(24*A*c^3 + 8*B*b*
c^2))/(8*(-b)^(7/4)*c^(5/4)))*1i)/(8*(-b)^(7/4)*c^(5/4)) + ((3*A*c + B*b)*((x^(1/2)*(9*A^2*c^3 + B^2*b^2*c + 6
*A*B*b*c^2))/b^2 + ((3*A*c + B*b)*(24*A*c^3 + 8*B*b*c^2))/(8*(-b)^(7/4)*c^(5/4)))*1i)/(8*(-b)^(7/4)*c^(5/4)))/
(((3*A*c + B*b)*((x^(1/2)*(9*A^2*c^3 + B^2*b^2*c + 6*A*B*b*c^2))/b^2 - ((3*A*c + B*b)*(24*A*c^3 + 8*B*b*c^2))/
(8*(-b)^(7/4)*c^(5/4))))/(8*(-b)^(7/4)*c^(5/4)) - ((3*A*c + B*b)*((x^(1/2)*(9*A^2*c^3 + B^2*b^2*c + 6*A*B*b*c^
2))/b^2 + ((3*A*c + B*b)*(24*A*c^3 + 8*B*b*c^2))/(8*(-b)^(7/4)*c^(5/4))))/(8*(-b)^(7/4)*c^(5/4))))*(3*A*c + B*
b)*1i)/(4*(-b)^(7/4)*c^(5/4)) + (atan((((3*A*c + B*b)*((x^(1/2)*(9*A^2*c^3 + B^2*b^2*c + 6*A*B*b*c^2))/b^2 - (
(3*A*c + B*b)*(24*A*c^3 + 8*B*b*c^2)*1i)/(8*(-b)^(7/4)*c^(5/4))))/(8*(-b)^(7/4)*c^(5/4)) + ((3*A*c + B*b)*((x^
(1/2)*(9*A^2*c^3 + B^2*b^2*c + 6*A*B*b*c^2))/b^2 + ((3*A*c + B*b)*(24*A*c^3 + 8*B*b*c^2)*1i)/(8*(-b)^(7/4)*c^(
5/4))))/(8*(-b)^(7/4)*c^(5/4)))/(((3*A*c + B*b)*((x^(1/2)*(9*A^2*c^3 + B^2*b^2*c + 6*A*B*b*c^2))/b^2 - ((3*A*c
+ B*b)*(24*A*c^3 + 8*B*b*c^2)*1i)/(8*(-b)^(7/4)*c^(5/4)))*1i)/(8*(-b)^(7/4)*c^(5/4)) - ((3*A*c + B*b)*((x^(1/
2)*(9*A^2*c^3 + B^2*b^2*c + 6*A*B*b*c^2))/b^2 + ((3*A*c + B*b)*(24*A*c^3 + 8*B*b*c^2)*1i)/(8*(-b)^(7/4)*c^(5/4
)))*1i)/(8*(-b)^(7/4)*c^(5/4))))*(3*A*c + B*b))/(4*(-b)^(7/4)*c^(5/4)) + (x^(1/2)*(A*c - B*b))/(2*b*c*(b + c*x
^2))

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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**(7/2)*(B*x**2+A)/(c*x**4+b*x**2)**2,x)

[Out]

Timed out

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