3.202 \(\int \frac {x^{5/2} (A+B x^2)}{(b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=284 \[ \frac {(b B-5 A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{9/4} c^{3/4}}-\frac {(b B-5 A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{9/4} c^{3/4}}-\frac {(b B-5 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{9/4} c^{3/4}}+\frac {(b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} b^{9/4} c^{3/4}}+\frac {b B-5 A c}{2 b^2 c \sqrt {x}}-\frac {b B-A c}{2 b c \sqrt {x} \left (b+c x^2\right )} \]

[Out]

-1/8*(-5*A*c+B*b)*arctan(1-c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(9/4)/c^(3/4)*2^(1/2)+1/8*(-5*A*c+B*b)*arctan(1+
c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(9/4)/c^(3/4)*2^(1/2)+1/16*(-5*A*c+B*b)*ln(b^(1/2)+x*c^(1/2)-b^(1/4)*c^(1/4
)*2^(1/2)*x^(1/2))/b^(9/4)/c^(3/4)*2^(1/2)-1/16*(-5*A*c+B*b)*ln(b^(1/2)+x*c^(1/2)+b^(1/4)*c^(1/4)*2^(1/2)*x^(1
/2))/b^(9/4)/c^(3/4)*2^(1/2)+1/2*(-5*A*c+B*b)/b^2/c/x^(1/2)+1/2*(A*c-B*b)/b/c/(c*x^2+b)/x^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 284, 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, 325, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac {(b B-5 A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{9/4} c^{3/4}}-\frac {(b B-5 A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{9/4} c^{3/4}}-\frac {(b B-5 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{9/4} c^{3/4}}+\frac {(b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} b^{9/4} c^{3/4}}+\frac {b B-5 A c}{2 b^2 c \sqrt {x}}-\frac {b B-A c}{2 b c \sqrt {x} \left (b+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(x^(5/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

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

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -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^{5/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac {A+B x^2}{x^{3/2} \left (b+c x^2\right )^2} \, dx\\ &=-\frac {b B-A c}{2 b c \sqrt {x} \left (b+c x^2\right )}+\frac {\left (-\frac {b B}{2}+\frac {5 A c}{2}\right ) \int \frac {1}{x^{3/2} \left (b+c x^2\right )} \, dx}{2 b c}\\ &=\frac {b B-5 A c}{2 b^2 c \sqrt {x}}-\frac {b B-A c}{2 b c \sqrt {x} \left (b+c x^2\right )}+\frac {(b B-5 A c) \int \frac {\sqrt {x}}{b+c x^2} \, dx}{4 b^2}\\ &=\frac {b B-5 A c}{2 b^2 c \sqrt {x}}-\frac {b B-A c}{2 b c \sqrt {x} \left (b+c x^2\right )}+\frac {(b B-5 A c) \operatorname {Subst}\left (\int \frac {x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{2 b^2}\\ &=\frac {b B-5 A c}{2 b^2 c \sqrt {x}}-\frac {b B-A c}{2 b c \sqrt {x} \left (b+c x^2\right )}-\frac {(b B-5 A c) \operatorname {Subst}\left (\int \frac {\sqrt {b}-\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{4 b^2 \sqrt {c}}+\frac {(b B-5 A c) \operatorname {Subst}\left (\int \frac {\sqrt {b}+\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{4 b^2 \sqrt {c}}\\ &=\frac {b B-5 A c}{2 b^2 c \sqrt {x}}-\frac {b B-A c}{2 b c \sqrt {x} \left (b+c x^2\right )}+\frac {(b B-5 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^2 c}+\frac {(b B-5 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^2 c}+\frac {(b B-5 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^{9/4} c^{3/4}}+\frac {(b B-5 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^{9/4} c^{3/4}}\\ &=\frac {b B-5 A c}{2 b^2 c \sqrt {x}}-\frac {b B-A c}{2 b c \sqrt {x} \left (b+c x^2\right )}+\frac {(b B-5 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{9/4} c^{3/4}}-\frac {(b B-5 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{9/4} c^{3/4}}+\frac {(b B-5 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^{9/4} c^{3/4}}-\frac {(b B-5 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^{9/4} c^{3/4}}\\ &=\frac {b B-5 A c}{2 b^2 c \sqrt {x}}-\frac {b B-A c}{2 b c \sqrt {x} \left (b+c x^2\right )}-\frac {(b B-5 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{9/4} c^{3/4}}+\frac {(b B-5 A c) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{9/4} c^{3/4}}+\frac {(b B-5 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{9/4} c^{3/4}}-\frac {(b B-5 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{9/4} c^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.32, size = 117, normalized size = 0.41 \[ \frac {2 x^{3/2} (b B-A c) \, _2F_1\left (\frac {3}{4},2;\frac {7}{4};-\frac {c x^2}{b}\right )+3 A \left ((-b)^{3/4} \sqrt [4]{c} \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b}}\right )-(-b)^{3/4} \sqrt [4]{c} \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b}}\right )-\frac {2 b}{\sqrt {x}}\right )}{3 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(5/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

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

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fricas [B]  time = 1.16, size = 920, normalized size = 3.24 \[ \frac {4 \, {\left (b^{2} c x^{3} + b^{3} x\right )} \left (-\frac {B^{4} b^{4} - 20 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 500 \, A^{3} B b c^{3} + 625 \, A^{4} c^{4}}{b^{9} c^{3}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {{\left (B^{6} b^{6} - 30 \, A B^{5} b^{5} c + 375 \, A^{2} B^{4} b^{4} c^{2} - 2500 \, A^{3} B^{3} b^{3} c^{3} + 9375 \, A^{4} B^{2} b^{2} c^{4} - 18750 \, A^{5} B b c^{5} + 15625 \, A^{6} c^{6}\right )} x - {\left (B^{4} b^{9} c - 20 \, A B^{3} b^{8} c^{2} + 150 \, A^{2} B^{2} b^{7} c^{3} - 500 \, A^{3} B b^{6} c^{4} + 625 \, A^{4} b^{5} c^{5}\right )} \sqrt {-\frac {B^{4} b^{4} - 20 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 500 \, A^{3} B b c^{3} + 625 \, A^{4} c^{4}}{b^{9} c^{3}}}} b^{2} c \left (-\frac {B^{4} b^{4} - 20 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 500 \, A^{3} B b c^{3} + 625 \, A^{4} c^{4}}{b^{9} c^{3}}\right )^{\frac {1}{4}} + {\left (B^{3} b^{5} c - 15 \, A B^{2} b^{4} c^{2} + 75 \, A^{2} B b^{3} c^{3} - 125 \, A^{3} b^{2} c^{4}\right )} \sqrt {x} \left (-\frac {B^{4} b^{4} - 20 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 500 \, A^{3} B b c^{3} + 625 \, A^{4} c^{4}}{b^{9} c^{3}}\right )^{\frac {1}{4}}}{B^{4} b^{4} - 20 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 500 \, A^{3} B b c^{3} + 625 \, A^{4} c^{4}}\right ) - {\left (b^{2} c x^{3} + b^{3} x\right )} \left (-\frac {B^{4} b^{4} - 20 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 500 \, A^{3} B b c^{3} + 625 \, A^{4} c^{4}}{b^{9} c^{3}}\right )^{\frac {1}{4}} \log \left (b^{7} c^{2} \left (-\frac {B^{4} b^{4} - 20 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 500 \, A^{3} B b c^{3} + 625 \, A^{4} c^{4}}{b^{9} c^{3}}\right )^{\frac {3}{4}} - {\left (B^{3} b^{3} - 15 \, A B^{2} b^{2} c + 75 \, A^{2} B b c^{2} - 125 \, A^{3} c^{3}\right )} \sqrt {x}\right ) + {\left (b^{2} c x^{3} + b^{3} x\right )} \left (-\frac {B^{4} b^{4} - 20 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 500 \, A^{3} B b c^{3} + 625 \, A^{4} c^{4}}{b^{9} c^{3}}\right )^{\frac {1}{4}} \log \left (-b^{7} c^{2} \left (-\frac {B^{4} b^{4} - 20 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 500 \, A^{3} B b c^{3} + 625 \, A^{4} c^{4}}{b^{9} c^{3}}\right )^{\frac {3}{4}} - {\left (B^{3} b^{3} - 15 \, A B^{2} b^{2} c + 75 \, A^{2} B b c^{2} - 125 \, A^{3} c^{3}\right )} \sqrt {x}\right ) + 4 \, {\left ({\left (B b - 5 \, A c\right )} x^{2} - 4 \, A b\right )} \sqrt {x}}{8 \, {\left (b^{2} c x^{3} + b^{3} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

1/8*(4*(b^2*c*x^3 + b^3*x)*(-(B^4*b^4 - 20*A*B^3*b^3*c + 150*A^2*B^2*b^2*c^2 - 500*A^3*B*b*c^3 + 625*A^4*c^4)/
(b^9*c^3))^(1/4)*arctan((sqrt((B^6*b^6 - 30*A*B^5*b^5*c + 375*A^2*B^4*b^4*c^2 - 2500*A^3*B^3*b^3*c^3 + 9375*A^
4*B^2*b^2*c^4 - 18750*A^5*B*b*c^5 + 15625*A^6*c^6)*x - (B^4*b^9*c - 20*A*B^3*b^8*c^2 + 150*A^2*B^2*b^7*c^3 - 5
00*A^3*B*b^6*c^4 + 625*A^4*b^5*c^5)*sqrt(-(B^4*b^4 - 20*A*B^3*b^3*c + 150*A^2*B^2*b^2*c^2 - 500*A^3*B*b*c^3 +
625*A^4*c^4)/(b^9*c^3)))*b^2*c*(-(B^4*b^4 - 20*A*B^3*b^3*c + 150*A^2*B^2*b^2*c^2 - 500*A^3*B*b*c^3 + 625*A^4*c
^4)/(b^9*c^3))^(1/4) + (B^3*b^5*c - 15*A*B^2*b^4*c^2 + 75*A^2*B*b^3*c^3 - 125*A^3*b^2*c^4)*sqrt(x)*(-(B^4*b^4
- 20*A*B^3*b^3*c + 150*A^2*B^2*b^2*c^2 - 500*A^3*B*b*c^3 + 625*A^4*c^4)/(b^9*c^3))^(1/4))/(B^4*b^4 - 20*A*B^3*
b^3*c + 150*A^2*B^2*b^2*c^2 - 500*A^3*B*b*c^3 + 625*A^4*c^4)) - (b^2*c*x^3 + b^3*x)*(-(B^4*b^4 - 20*A*B^3*b^3*
c + 150*A^2*B^2*b^2*c^2 - 500*A^3*B*b*c^3 + 625*A^4*c^4)/(b^9*c^3))^(1/4)*log(b^7*c^2*(-(B^4*b^4 - 20*A*B^3*b^
3*c + 150*A^2*B^2*b^2*c^2 - 500*A^3*B*b*c^3 + 625*A^4*c^4)/(b^9*c^3))^(3/4) - (B^3*b^3 - 15*A*B^2*b^2*c + 75*A
^2*B*b*c^2 - 125*A^3*c^3)*sqrt(x)) + (b^2*c*x^3 + b^3*x)*(-(B^4*b^4 - 20*A*B^3*b^3*c + 150*A^2*B^2*b^2*c^2 - 5
00*A^3*B*b*c^3 + 625*A^4*c^4)/(b^9*c^3))^(1/4)*log(-b^7*c^2*(-(B^4*b^4 - 20*A*B^3*b^3*c + 150*A^2*B^2*b^2*c^2
- 500*A^3*B*b*c^3 + 625*A^4*c^4)/(b^9*c^3))^(3/4) - (B^3*b^3 - 15*A*B^2*b^2*c + 75*A^2*B*b*c^2 - 125*A^3*c^3)*
sqrt(x)) + 4*((B*b - 5*A*c)*x^2 - 4*A*b)*sqrt(x))/(b^2*c*x^3 + b^3*x)

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giac [A]  time = 0.21, size = 278, normalized size = 0.98 \[ \frac {B b x^{2} - 5 \, A c x^{2} - 4 \, A b}{2 \, {\left (c x^{\frac {5}{2}} + b \sqrt {x}\right )} b^{2}} + \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {3}{4}} B b - 5 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, b^{3} c^{3}} + \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {3}{4}} B b - 5 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, b^{3} c^{3}} - \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {3}{4}} B b - 5 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, b^{3} c^{3}} + \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {3}{4}} B b - 5 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, b^{3} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 323, normalized size = 1.14 \[ -\frac {A c \,x^{\frac {3}{2}}}{2 \left (c \,x^{2}+b \right ) b^{2}}+\frac {B \,x^{\frac {3}{2}}}{2 \left (c \,x^{2}+b \right ) b}-\frac {5 \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {b}{c}\right )^{\frac {1}{4}} b^{2}}-\frac {5 \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {b}{c}\right )^{\frac {1}{4}} b^{2}}-\frac {5 \sqrt {2}\, A \ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{16 \left (\frac {b}{c}\right )^{\frac {1}{4}} b^{2}}+\frac {\sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {b}{c}\right )^{\frac {1}{4}} b c}+\frac {\sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {b}{c}\right )^{\frac {1}{4}} b c}+\frac {\sqrt {2}\, B \ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{16 \left (\frac {b}{c}\right )^{\frac {1}{4}} b c}-\frac {2 A}{b^{2} \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x)

[Out]

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

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maxima [A]  time = 3.04, size = 222, normalized size = 0.78 \[ \frac {{\left (B b - 5 \, A c\right )} x^{2} - 4 \, A b}{2 \, {\left (b^{2} c x^{\frac {5}{2}} + b^{3} \sqrt {x}\right )}} + \frac {{\left (B b - 5 \, A c\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{16 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.23, size = 104, normalized size = 0.37 \[ \frac {\mathrm {atanh}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )\,\left (5\,A\,c-B\,b\right )}{4\,{\left (-b\right )}^{9/4}\,c^{3/4}}-\frac {\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )\,\left (5\,A\,c-B\,b\right )}{4\,{\left (-b\right )}^{9/4}\,c^{3/4}}-\frac {\frac {2\,A}{b}+\frac {x^2\,\left (5\,A\,c-B\,b\right )}{2\,b^2}}{b\,\sqrt {x}+c\,x^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(5/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x)

[Out]

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

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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**(5/2)*(B*x**2+A)/(c*x**4+b*x**2)**2,x)

[Out]

Timed out

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