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

Optimal. Leaf size=255 \[ -\frac {\sqrt [4]{b} (b B-A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} c^{9/4}}+\frac {\sqrt [4]{b} (b B-A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} c^{9/4}}-\frac {\sqrt [4]{b} (b B-A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{9/4}}+\frac {\sqrt [4]{b} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} c^{9/4}}-\frac {2 \sqrt {x} (b B-A c)}{c^2}+\frac {2 B x^{5/2}}{5 c} \]

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Rubi [A]  time = 0.21, antiderivative size = 255, 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, 459, 321, 329, 211, 1165, 628, 1162, 617, 204} \begin {gather*} -\frac {2 \sqrt {x} (b B-A c)}{c^2}-\frac {\sqrt [4]{b} (b B-A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} c^{9/4}}+\frac {\sqrt [4]{b} (b B-A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} c^{9/4}}-\frac {\sqrt [4]{b} (b B-A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{9/4}}+\frac {\sqrt [4]{b} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} c^{9/4}}+\frac {2 B x^{5/2}}{5 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*B - A*c)*Sqrt[x])/c^2 + (2*B*x^(5/2))/(5*c) - (b^(1/4)*(b*B - A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])
/b^(1/4)])/(Sqrt[2]*c^(9/4)) + (b^(1/4)*(b*B - A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*c^
(9/4)) - (b^(1/4)*(b*B - A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(2*Sqrt[2]*c^(9/4))
+ (b^(1/4)*(b*B - A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(2*Sqrt[2]*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 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

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 )}{b x^2+c x^4} \, dx &=\int \frac {x^{3/2} \left (A+B x^2\right )}{b+c x^2} \, dx\\ &=\frac {2 B x^{5/2}}{5 c}-\frac {\left (2 \left (\frac {5 b B}{2}-\frac {5 A c}{2}\right )\right ) \int \frac {x^{3/2}}{b+c x^2} \, dx}{5 c}\\ &=-\frac {2 (b B-A c) \sqrt {x}}{c^2}+\frac {2 B x^{5/2}}{5 c}+\frac {(b (b B-A c)) \int \frac {1}{\sqrt {x} \left (b+c x^2\right )} \, dx}{c^2}\\ &=-\frac {2 (b B-A c) \sqrt {x}}{c^2}+\frac {2 B x^{5/2}}{5 c}+\frac {(2 b (b B-A c)) \operatorname {Subst}\left (\int \frac {1}{b+c x^4} \, dx,x,\sqrt {x}\right )}{c^2}\\ &=-\frac {2 (b B-A c) \sqrt {x}}{c^2}+\frac {2 B x^{5/2}}{5 c}+\frac {\left (\sqrt {b} (b B-A c)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {b}-\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{c^2}+\frac {\left (\sqrt {b} (b B-A c)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {b}+\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{c^2}\\ &=-\frac {2 (b B-A c) \sqrt {x}}{c^2}+\frac {2 B x^{5/2}}{5 c}+\frac {\left (\sqrt {b} (b B-A c)\right ) \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 )}{2 c^{5/2}}+\frac {\left (\sqrt {b} (b B-A c)\right ) \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 )}{2 c^{5/2}}-\frac {\left (\sqrt [4]{b} (b B-A c)\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{b}}{\sqrt [4]{c}}+2 x}{-\frac {\sqrt {b}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt {x}\right )}{2 \sqrt {2} c^{9/4}}-\frac {\left (\sqrt [4]{b} (b B-A c)\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{b}}{\sqrt [4]{c}}-2 x}{-\frac {\sqrt {b}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt {x}\right )}{2 \sqrt {2} c^{9/4}}\\ &=-\frac {2 (b B-A c) \sqrt {x}}{c^2}+\frac {2 B x^{5/2}}{5 c}-\frac {\sqrt [4]{b} (b B-A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} c^{9/4}}+\frac {\sqrt [4]{b} (b B-A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} c^{9/4}}+\frac {\left (\sqrt [4]{b} (b B-A c)\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{9/4}}-\frac {\left (\sqrt [4]{b} (b B-A c)\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{9/4}}\\ &=-\frac {2 (b B-A c) \sqrt {x}}{c^2}+\frac {2 B x^{5/2}}{5 c}-\frac {\sqrt [4]{b} (b B-A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{9/4}}+\frac {\sqrt [4]{b} (b B-A c) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{9/4}}-\frac {\sqrt [4]{b} (b B-A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} c^{9/4}}+\frac {\sqrt [4]{b} (b B-A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} c^{9/4}}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 208, normalized size = 0.82 \begin {gather*} \frac {-40 \sqrt {x} (b B-A c)+\frac {5 \sqrt {2} \sqrt [4]{b} (A c-b B) \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 )\right )}{\sqrt [4]{c}}-\frac {10 \sqrt {2} \sqrt [4]{b} (b B-A c) \left (\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )-\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )\right )}{\sqrt [4]{c}}+8 B c x^{5/2}}{20 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-40*(b*B - A*c)*Sqrt[x] + 8*B*c*x^(5/2) - (10*Sqrt[2]*b^(1/4)*(b*B - A*c)*(ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x
])/b^(1/4)] - ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)]))/c^(1/4) + (5*Sqrt[2]*b^(1/4)*(-(b*B) + A*c)*(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]))/c^(1/4))/(20*c^2)

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IntegrateAlgebraic [A]  time = 0.22, size = 165, normalized size = 0.65 \begin {gather*} -\frac {\left (b^{5/4} B-A \sqrt [4]{b} c\right ) \tan ^{-1}\left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )}{\sqrt {2} c^{9/4}}+\frac {\left (b^{5/4} B-A \sqrt [4]{b} c\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{\sqrt {2} c^{9/4}}+\frac {2 \left (5 A c \sqrt {x}-5 b B \sqrt {x}+B c x^{5/2}\right )}{5 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-5*b*B*Sqrt[x] + 5*A*c*Sqrt[x] + B*c*x^(5/2)))/(5*c^2) - ((b^(5/4)*B - A*b^(1/4)*c)*ArcTan[(Sqrt[b] - Sqrt
[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])])/(Sqrt[2]*c^(9/4)) + ((b^(5/4)*B - A*b^(1/4)*c)*ArcTanh[(Sqrt[2]*b^(
1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(Sqrt[2]*c^(9/4))

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fricas [B]  time = 0.43, size = 660, normalized size = 2.59 \begin {gather*} -\frac {20 \, c^{2} \left (-\frac {B^{4} b^{5} - 4 \, A B^{3} b^{4} c + 6 \, A^{2} B^{2} b^{3} c^{2} - 4 \, A^{3} B b^{2} c^{3} + A^{4} b c^{4}}{c^{9}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {c^{4} \sqrt {-\frac {B^{4} b^{5} - 4 \, A B^{3} b^{4} c + 6 \, A^{2} B^{2} b^{3} c^{2} - 4 \, A^{3} B b^{2} c^{3} + A^{4} b c^{4}}{c^{9}}} + {\left (B^{2} b^{2} - 2 \, A B b c + A^{2} c^{2}\right )} x} c^{7} \left (-\frac {B^{4} b^{5} - 4 \, A B^{3} b^{4} c + 6 \, A^{2} B^{2} b^{3} c^{2} - 4 \, A^{3} B b^{2} c^{3} + A^{4} b c^{4}}{c^{9}}\right )^{\frac {3}{4}} + {\left (B b c^{7} - A c^{8}\right )} \sqrt {x} \left (-\frac {B^{4} b^{5} - 4 \, A B^{3} b^{4} c + 6 \, A^{2} B^{2} b^{3} c^{2} - 4 \, A^{3} B b^{2} c^{3} + A^{4} b c^{4}}{c^{9}}\right )^{\frac {3}{4}}}{B^{4} b^{5} - 4 \, A B^{3} b^{4} c + 6 \, A^{2} B^{2} b^{3} c^{2} - 4 \, A^{3} B b^{2} c^{3} + A^{4} b c^{4}}\right ) + 5 \, c^{2} \left (-\frac {B^{4} b^{5} - 4 \, A B^{3} b^{4} c + 6 \, A^{2} B^{2} b^{3} c^{2} - 4 \, A^{3} B b^{2} c^{3} + A^{4} b c^{4}}{c^{9}}\right )^{\frac {1}{4}} \log \left (c^{2} \left (-\frac {B^{4} b^{5} - 4 \, A B^{3} b^{4} c + 6 \, A^{2} B^{2} b^{3} c^{2} - 4 \, A^{3} B b^{2} c^{3} + A^{4} b c^{4}}{c^{9}}\right )^{\frac {1}{4}} - {\left (B b - A c\right )} \sqrt {x}\right ) - 5 \, c^{2} \left (-\frac {B^{4} b^{5} - 4 \, A B^{3} b^{4} c + 6 \, A^{2} B^{2} b^{3} c^{2} - 4 \, A^{3} B b^{2} c^{3} + A^{4} b c^{4}}{c^{9}}\right )^{\frac {1}{4}} \log \left (-c^{2} \left (-\frac {B^{4} b^{5} - 4 \, A B^{3} b^{4} c + 6 \, A^{2} B^{2} b^{3} c^{2} - 4 \, A^{3} B b^{2} c^{3} + A^{4} b c^{4}}{c^{9}}\right )^{\frac {1}{4}} - {\left (B b - A c\right )} \sqrt {x}\right ) - 4 \, {\left (B c x^{2} - 5 \, B b + 5 \, A c\right )} \sqrt {x}}{10 \, c^{2}} \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),x, algorithm="fricas")

[Out]

-1/10*(20*c^2*(-(B^4*b^5 - 4*A*B^3*b^4*c + 6*A^2*B^2*b^3*c^2 - 4*A^3*B*b^2*c^3 + A^4*b*c^4)/c^9)^(1/4)*arctan(
(sqrt(c^4*sqrt(-(B^4*b^5 - 4*A*B^3*b^4*c + 6*A^2*B^2*b^3*c^2 - 4*A^3*B*b^2*c^3 + A^4*b*c^4)/c^9) + (B^2*b^2 -
2*A*B*b*c + A^2*c^2)*x)*c^7*(-(B^4*b^5 - 4*A*B^3*b^4*c + 6*A^2*B^2*b^3*c^2 - 4*A^3*B*b^2*c^3 + A^4*b*c^4)/c^9)
^(3/4) + (B*b*c^7 - A*c^8)*sqrt(x)*(-(B^4*b^5 - 4*A*B^3*b^4*c + 6*A^2*B^2*b^3*c^2 - 4*A^3*B*b^2*c^3 + A^4*b*c^
4)/c^9)^(3/4))/(B^4*b^5 - 4*A*B^3*b^4*c + 6*A^2*B^2*b^3*c^2 - 4*A^3*B*b^2*c^3 + A^4*b*c^4)) + 5*c^2*(-(B^4*b^5
- 4*A*B^3*b^4*c + 6*A^2*B^2*b^3*c^2 - 4*A^3*B*b^2*c^3 + A^4*b*c^4)/c^9)^(1/4)*log(c^2*(-(B^4*b^5 - 4*A*B^3*b^
4*c + 6*A^2*B^2*b^3*c^2 - 4*A^3*B*b^2*c^3 + A^4*b*c^4)/c^9)^(1/4) - (B*b - A*c)*sqrt(x)) - 5*c^2*(-(B^4*b^5 -
4*A*B^3*b^4*c + 6*A^2*B^2*b^3*c^2 - 4*A^3*B*b^2*c^3 + A^4*b*c^4)/c^9)^(1/4)*log(-c^2*(-(B^4*b^5 - 4*A*B^3*b^4*
c + 6*A^2*B^2*b^3*c^2 - 4*A^3*B*b^2*c^3 + A^4*b*c^4)/c^9)^(1/4) - (B*b - A*c)*sqrt(x)) - 4*(B*c*x^2 - 5*B*b +
5*A*c)*sqrt(x))/c^2

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giac [A]  time = 0.20, size = 263, normalized size = 1.03 \begin {gather*} \frac {\sqrt {2} {\left (\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 )}{2 \, c^{3}} + \frac {\sqrt {2} {\left (\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 )}{2 \, c^{3}} + \frac {\sqrt {2} {\left (\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 )}{4 \, c^{3}} - \frac {\sqrt {2} {\left (\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 )}{4 \, c^{3}} + \frac {2 \, {\left (B c^{4} x^{\frac {5}{2}} - 5 \, B b c^{3} \sqrt {x} + 5 \, A c^{4} \sqrt {x}\right )}}{5 \, c^{5}} \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),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 299, normalized size = 1.17 \begin {gather*} \frac {2 B \,x^{\frac {5}{2}}}{5 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 )}{2 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 )}{2 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 )}{4 c}+\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{2 c^{2}}+\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{2 c^{2}}+\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \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 )}{4 c^{2}}+\frac {2 A \sqrt {x}}{c}-\frac {2 B b \sqrt {x}}{c^{2}} \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),x)

[Out]

2/5*B*x^(5/2)/c+2/c*A*x^(1/2)-2/c^2*b*B*x^(1/2)-1/2/c*(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)
+1)-1/2/c*(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)-1/4/c*(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)))+1/2/c^2*(b/c)^(1/4)*2^(1/2)*B*
arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)*b+1/2/c^2*(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)*b+
1/4/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)))*b

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maxima [A]  time = 2.99, size = 235, normalized size = 0.92 \begin {gather*} \frac {{\left (\frac {2 \, \sqrt {2} {\left (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 (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 (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 (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}}}\right )} b}{4 \, c^{2}} + \frac {2 \, {\left (B c x^{\frac {5}{2}} - 5 \, {\left (B b - A c\right )} \sqrt {x}\right )}}{5 \, c^{2}} \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),x, algorithm="maxima")

[Out]

1/4*(2*sqrt(2)*(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)*(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))) + sqrt(2)*(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)*(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)))*b/c^2 + 2/5*(B*c*x^(5/2) - 5*(B*b - A*c)*sqrt(x))/c^2

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mupad [B]  time = 0.27, size = 789, normalized size = 3.09 \begin {gather*} \sqrt {x}\,\left (\frac {2\,A}{c}-\frac {2\,B\,b}{c^2}\right )+\frac {2\,B\,x^{5/2}}{5\,c}-\frac {{\left (-b\right )}^{1/4}\,\mathrm {atan}\left (\frac {\frac {{\left (-b\right )}^{1/4}\,\left (A\,c-B\,b\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,b^2\,c^2-2\,A\,B\,b^3\,c+B^2\,b^4\right )}{c}-\frac {{\left (-b\right )}^{1/4}\,\left (32\,A\,b^2\,c^2-32\,B\,b^3\,c\right )\,\left (A\,c-B\,b\right )}{2\,c^{9/4}}\right )\,1{}\mathrm {i}}{2\,c^{9/4}}+\frac {{\left (-b\right )}^{1/4}\,\left (A\,c-B\,b\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,b^2\,c^2-2\,A\,B\,b^3\,c+B^2\,b^4\right )}{c}+\frac {{\left (-b\right )}^{1/4}\,\left (32\,A\,b^2\,c^2-32\,B\,b^3\,c\right )\,\left (A\,c-B\,b\right )}{2\,c^{9/4}}\right )\,1{}\mathrm {i}}{2\,c^{9/4}}}{\frac {{\left (-b\right )}^{1/4}\,\left (A\,c-B\,b\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,b^2\,c^2-2\,A\,B\,b^3\,c+B^2\,b^4\right )}{c}-\frac {{\left (-b\right )}^{1/4}\,\left (32\,A\,b^2\,c^2-32\,B\,b^3\,c\right )\,\left (A\,c-B\,b\right )}{2\,c^{9/4}}\right )}{2\,c^{9/4}}-\frac {{\left (-b\right )}^{1/4}\,\left (A\,c-B\,b\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,b^2\,c^2-2\,A\,B\,b^3\,c+B^2\,b^4\right )}{c}+\frac {{\left (-b\right )}^{1/4}\,\left (32\,A\,b^2\,c^2-32\,B\,b^3\,c\right )\,\left (A\,c-B\,b\right )}{2\,c^{9/4}}\right )}{2\,c^{9/4}}}\right )\,\left (A\,c-B\,b\right )\,1{}\mathrm {i}}{c^{9/4}}-\frac {{\left (-b\right )}^{1/4}\,\mathrm {atan}\left (\frac {\frac {{\left (-b\right )}^{1/4}\,\left (A\,c-B\,b\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,b^2\,c^2-2\,A\,B\,b^3\,c+B^2\,b^4\right )}{c}-\frac {{\left (-b\right )}^{1/4}\,\left (32\,A\,b^2\,c^2-32\,B\,b^3\,c\right )\,\left (A\,c-B\,b\right )\,1{}\mathrm {i}}{2\,c^{9/4}}\right )}{2\,c^{9/4}}+\frac {{\left (-b\right )}^{1/4}\,\left (A\,c-B\,b\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,b^2\,c^2-2\,A\,B\,b^3\,c+B^2\,b^4\right )}{c}+\frac {{\left (-b\right )}^{1/4}\,\left (32\,A\,b^2\,c^2-32\,B\,b^3\,c\right )\,\left (A\,c-B\,b\right )\,1{}\mathrm {i}}{2\,c^{9/4}}\right )}{2\,c^{9/4}}}{\frac {{\left (-b\right )}^{1/4}\,\left (A\,c-B\,b\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,b^2\,c^2-2\,A\,B\,b^3\,c+B^2\,b^4\right )}{c}-\frac {{\left (-b\right )}^{1/4}\,\left (32\,A\,b^2\,c^2-32\,B\,b^3\,c\right )\,\left (A\,c-B\,b\right )\,1{}\mathrm {i}}{2\,c^{9/4}}\right )\,1{}\mathrm {i}}{2\,c^{9/4}}-\frac {{\left (-b\right )}^{1/4}\,\left (A\,c-B\,b\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,b^2\,c^2-2\,A\,B\,b^3\,c+B^2\,b^4\right )}{c}+\frac {{\left (-b\right )}^{1/4}\,\left (32\,A\,b^2\,c^2-32\,B\,b^3\,c\right )\,\left (A\,c-B\,b\right )\,1{}\mathrm {i}}{2\,c^{9/4}}\right )\,1{}\mathrm {i}}{2\,c^{9/4}}}\right )\,\left (A\,c-B\,b\right )}{c^{9/4}} \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),x)

[Out]

x^(1/2)*((2*A)/c - (2*B*b)/c^2) + (2*B*x^(5/2))/(5*c) - ((-b)^(1/4)*atan((((-b)^(1/4)*(A*c - B*b)*((16*x^(1/2)
*(B^2*b^4 + A^2*b^2*c^2 - 2*A*B*b^3*c))/c - ((-b)^(1/4)*(32*A*b^2*c^2 - 32*B*b^3*c)*(A*c - B*b))/(2*c^(9/4)))*
1i)/(2*c^(9/4)) + ((-b)^(1/4)*(A*c - B*b)*((16*x^(1/2)*(B^2*b^4 + A^2*b^2*c^2 - 2*A*B*b^3*c))/c + ((-b)^(1/4)*
(32*A*b^2*c^2 - 32*B*b^3*c)*(A*c - B*b))/(2*c^(9/4)))*1i)/(2*c^(9/4)))/(((-b)^(1/4)*(A*c - B*b)*((16*x^(1/2)*(
B^2*b^4 + A^2*b^2*c^2 - 2*A*B*b^3*c))/c - ((-b)^(1/4)*(32*A*b^2*c^2 - 32*B*b^3*c)*(A*c - B*b))/(2*c^(9/4))))/(
2*c^(9/4)) - ((-b)^(1/4)*(A*c - B*b)*((16*x^(1/2)*(B^2*b^4 + A^2*b^2*c^2 - 2*A*B*b^3*c))/c + ((-b)^(1/4)*(32*A
*b^2*c^2 - 32*B*b^3*c)*(A*c - B*b))/(2*c^(9/4))))/(2*c^(9/4))))*(A*c - B*b)*1i)/c^(9/4) - ((-b)^(1/4)*atan((((
-b)^(1/4)*(A*c - B*b)*((16*x^(1/2)*(B^2*b^4 + A^2*b^2*c^2 - 2*A*B*b^3*c))/c - ((-b)^(1/4)*(32*A*b^2*c^2 - 32*B
*b^3*c)*(A*c - B*b)*1i)/(2*c^(9/4))))/(2*c^(9/4)) + ((-b)^(1/4)*(A*c - B*b)*((16*x^(1/2)*(B^2*b^4 + A^2*b^2*c^
2 - 2*A*B*b^3*c))/c + ((-b)^(1/4)*(32*A*b^2*c^2 - 32*B*b^3*c)*(A*c - B*b)*1i)/(2*c^(9/4))))/(2*c^(9/4)))/(((-b
)^(1/4)*(A*c - B*b)*((16*x^(1/2)*(B^2*b^4 + A^2*b^2*c^2 - 2*A*B*b^3*c))/c - ((-b)^(1/4)*(32*A*b^2*c^2 - 32*B*b
^3*c)*(A*c - B*b)*1i)/(2*c^(9/4)))*1i)/(2*c^(9/4)) - ((-b)^(1/4)*(A*c - B*b)*((16*x^(1/2)*(B^2*b^4 + A^2*b^2*c
^2 - 2*A*B*b^3*c))/c + ((-b)^(1/4)*(32*A*b^2*c^2 - 32*B*b^3*c)*(A*c - B*b)*1i)/(2*c^(9/4)))*1i)/(2*c^(9/4))))*
(A*c - B*b))/c^(9/4)

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sympy [A]  time = 164.46, size = 393, normalized size = 1.54 \begin {gather*} \begin {cases} \tilde {\infty } \left (2 A \sqrt {x} + \frac {2 B x^{\frac {5}{2}}}{5}\right ) & \text {for}\: b = 0 \wedge c = 0 \\\frac {\frac {2 A x^{\frac {5}{2}}}{5} + \frac {2 B x^{\frac {9}{2}}}{9}}{b} & \text {for}\: c = 0 \\\frac {2 A \sqrt {x} + \frac {2 B x^{\frac {5}{2}}}{5}}{c} & \text {for}\: b = 0 \\\frac {\sqrt [4]{-1} A \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 c} - \frac {\sqrt [4]{-1} A \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} \log {\left (\sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 c} + \frac {\sqrt [4]{-1} A \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{b} \sqrt [4]{\frac {1}{c}}} \right )}}{c} + \frac {2 A \sqrt {x}}{c} - \frac {\sqrt [4]{-1} B b^{\frac {5}{4}} \sqrt [4]{\frac {1}{c}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 c^{2}} + \frac {\sqrt [4]{-1} B b^{\frac {5}{4}} \sqrt [4]{\frac {1}{c}} \log {\left (\sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 c^{2}} - \frac {\sqrt [4]{-1} B b^{\frac {5}{4}} \sqrt [4]{\frac {1}{c}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{b} \sqrt [4]{\frac {1}{c}}} \right )}}{c^{2}} - \frac {2 B b \sqrt {x}}{c^{2}} + \frac {2 B x^{\frac {5}{2}}}{5 c} & \text {otherwise} \end {cases} \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),x)

[Out]

Piecewise((zoo*(2*A*sqrt(x) + 2*B*x**(5/2)/5), Eq(b, 0) & Eq(c, 0)), ((2*A*x**(5/2)/5 + 2*B*x**(9/2)/9)/b, Eq(
c, 0)), ((2*A*sqrt(x) + 2*B*x**(5/2)/5)/c, Eq(b, 0)), ((-1)**(1/4)*A*b**(1/4)*(1/c)**(1/4)*log(-(-1)**(1/4)*b*
*(1/4)*(1/c)**(1/4) + sqrt(x))/(2*c) - (-1)**(1/4)*A*b**(1/4)*(1/c)**(1/4)*log((-1)**(1/4)*b**(1/4)*(1/c)**(1/
4) + sqrt(x))/(2*c) + (-1)**(1/4)*A*b**(1/4)*(1/c)**(1/4)*atan((-1)**(3/4)*sqrt(x)/(b**(1/4)*(1/c)**(1/4)))/c
+ 2*A*sqrt(x)/c - (-1)**(1/4)*B*b**(5/4)*(1/c)**(1/4)*log(-(-1)**(1/4)*b**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*c**
2) + (-1)**(1/4)*B*b**(5/4)*(1/c)**(1/4)*log((-1)**(1/4)*b**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*c**2) - (-1)**(1/
4)*B*b**(5/4)*(1/c)**(1/4)*atan((-1)**(3/4)*sqrt(x)/(b**(1/4)*(1/c)**(1/4)))/c**2 - 2*B*b*sqrt(x)/c**2 + 2*B*x
**(5/2)/(5*c), True))

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