3.5.20 \(\int \frac {x^{5/2} (A+B x)}{(a+c x^2)^2} \, dx\)

Optimal. Leaf size=304 \[ \frac {\left (5 \sqrt {a} B+3 A \sqrt {c}\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{8 \sqrt {2} \sqrt [4]{a} c^{9/4}}-\frac {\left (5 \sqrt {a} B+3 A \sqrt {c}\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{8 \sqrt {2} \sqrt [4]{a} c^{9/4}}+\frac {\left (5 \sqrt {a} B-3 A \sqrt {c}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} c^{9/4}}-\frac {\left (5 \sqrt {a} B-3 A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} \sqrt [4]{a} c^{9/4}}-\frac {x^{3/2} (A+B x)}{2 c \left (a+c x^2\right )}+\frac {5 B \sqrt {x}}{2 c^2} \]

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Rubi [A]  time = 0.31, antiderivative size = 304, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {819, 825, 827, 1168, 1162, 617, 204, 1165, 628} \begin {gather*} \frac {\left (5 \sqrt {a} B+3 A \sqrt {c}\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{8 \sqrt {2} \sqrt [4]{a} c^{9/4}}-\frac {\left (5 \sqrt {a} B+3 A \sqrt {c}\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{8 \sqrt {2} \sqrt [4]{a} c^{9/4}}+\frac {\left (5 \sqrt {a} B-3 A \sqrt {c}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} c^{9/4}}-\frac {\left (5 \sqrt {a} B-3 A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} \sqrt [4]{a} c^{9/4}}-\frac {x^{3/2} (A+B x)}{2 c \left (a+c x^2\right )}+\frac {5 B \sqrt {x}}{2 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 204

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

Rule 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 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 825

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(g*(d + e*x)^m)/
(c*m), x] + Dist[1/c, Int[((d + e*x)^(m - 1)*Simp[c*d*f - a*e*g + (g*c*d + c*e*f)*x, x])/(a + c*x^2), x], x] /
; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && FractionQ[m] && GtQ[m, 0]

Rule 827

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2, Subst[Int[(e*f
- d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 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 1168

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[-(a*c)]

Rubi steps

\begin {align*} \int \frac {x^{5/2} (A+B x)}{\left (a+c x^2\right )^2} \, dx &=-\frac {x^{3/2} (A+B x)}{2 c \left (a+c x^2\right )}+\frac {\int \frac {\sqrt {x} \left (\frac {3 a A}{2}+\frac {5 a B x}{2}\right )}{a+c x^2} \, dx}{2 a c}\\ &=\frac {5 B \sqrt {x}}{2 c^2}-\frac {x^{3/2} (A+B x)}{2 c \left (a+c x^2\right )}+\frac {\int \frac {-\frac {5 a^2 B}{2}+\frac {3}{2} a A c x}{\sqrt {x} \left (a+c x^2\right )} \, dx}{2 a c^2}\\ &=\frac {5 B \sqrt {x}}{2 c^2}-\frac {x^{3/2} (A+B x)}{2 c \left (a+c x^2\right )}+\frac {\operatorname {Subst}\left (\int \frac {-\frac {5 a^2 B}{2}+\frac {3}{2} a A c x^2}{a+c x^4} \, dx,x,\sqrt {x}\right )}{a c^2}\\ &=\frac {5 B \sqrt {x}}{2 c^2}-\frac {x^{3/2} (A+B x)}{2 c \left (a+c x^2\right )}+\frac {\left (3 A-\frac {5 \sqrt {a} B}{\sqrt {c}}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a} \sqrt {c}+c x^2}{a+c x^4} \, dx,x,\sqrt {x}\right )}{4 c^2}-\frac {\left (3 A+\frac {5 \sqrt {a} B}{\sqrt {c}}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a} \sqrt {c}-c x^2}{a+c x^4} \, dx,x,\sqrt {x}\right )}{4 c^2}\\ &=\frac {5 B \sqrt {x}}{2 c^2}-\frac {x^{3/2} (A+B x)}{2 c \left (a+c x^2\right )}+\frac {\left (5 \sqrt {a} B+3 A \sqrt {c}\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{c}}+2 x}{-\frac {\sqrt {a}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt {x}\right )}{8 \sqrt {2} \sqrt [4]{a} c^{9/4}}+\frac {\left (5 \sqrt {a} B+3 A \sqrt {c}\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{c}}-2 x}{-\frac {\sqrt {a}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt {x}\right )}{8 \sqrt {2} \sqrt [4]{a} c^{9/4}}+\frac {\left (3 A-\frac {5 \sqrt {a} B}{\sqrt {c}}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {a}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt {x}\right )}{8 c^2}+\frac {\left (3 A-\frac {5 \sqrt {a} B}{\sqrt {c}}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {a}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt {x}\right )}{8 c^2}\\ &=\frac {5 B \sqrt {x}}{2 c^2}-\frac {x^{3/2} (A+B x)}{2 c \left (a+c x^2\right )}+\frac {\left (5 \sqrt {a} B+3 A \sqrt {c}\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} \sqrt [4]{a} c^{9/4}}-\frac {\left (5 \sqrt {a} B+3 A \sqrt {c}\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} \sqrt [4]{a} c^{9/4}}+\frac {\left (3 A-\frac {5 \sqrt {a} B}{\sqrt {c}}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} c^{7/4}}-\frac {\left (3 A-\frac {5 \sqrt {a} B}{\sqrt {c}}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} c^{7/4}}\\ &=\frac {5 B \sqrt {x}}{2 c^2}-\frac {x^{3/2} (A+B x)}{2 c \left (a+c x^2\right )}-\frac {\left (3 A-\frac {5 \sqrt {a} B}{\sqrt {c}}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} c^{7/4}}+\frac {\left (3 A-\frac {5 \sqrt {a} B}{\sqrt {c}}\right ) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} c^{7/4}}+\frac {\left (5 \sqrt {a} B+3 A \sqrt {c}\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} \sqrt [4]{a} c^{9/4}}-\frac {\left (5 \sqrt {a} B+3 A \sqrt {c}\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} \sqrt [4]{a} c^{9/4}}\\ \end {align*}

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Mathematica [A]  time = 0.52, size = 344, normalized size = 1.13 \begin {gather*} \frac {1}{16} \left (\frac {8 A x^{7/2}}{a^2+a c x^2}+\frac {8 B x^{9/2}}{a^2+a c x^2}+\frac {12 A \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-a}}\right )}{\sqrt [4]{-a} c^{7/4}}-\frac {12 A \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-a}}\right )}{\sqrt [4]{-a} c^{7/4}}-\frac {8 A x^{3/2}}{a c}+\frac {5 \sqrt {2} \sqrt [4]{a} B \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{c^{9/4}}-\frac {5 \sqrt {2} \sqrt [4]{a} B \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{c^{9/4}}+\frac {10 \sqrt {2} \sqrt [4]{a} B \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{c^{9/4}}-\frac {10 \sqrt {2} \sqrt [4]{a} B \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{c^{9/4}}-\frac {8 B x^{5/2}}{a c}+\frac {40 B \sqrt {x}}{c^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((40*B*Sqrt[x])/c^2 - (8*A*x^(3/2))/(a*c) - (8*B*x^(5/2))/(a*c) + (8*A*x^(7/2))/(a^2 + a*c*x^2) + (8*B*x^(9/2)
)/(a^2 + a*c*x^2) + (10*Sqrt[2]*a^(1/4)*B*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/c^(9/4) - (10*Sqrt[2]
*a^(1/4)*B*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/c^(9/4) + (12*A*ArcTan[(c^(1/4)*Sqrt[x])/(-a)^(1/4)]
)/((-a)^(1/4)*c^(7/4)) - (12*A*ArcTanh[(c^(1/4)*Sqrt[x])/(-a)^(1/4)])/((-a)^(1/4)*c^(7/4)) + (5*Sqrt[2]*a^(1/4
)*B*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/c^(9/4) - (5*Sqrt[2]*a^(1/4)*B*Log[Sqrt[a] + S
qrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/c^(9/4))/16

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IntegrateAlgebraic [A]  time = 0.81, size = 190, normalized size = 0.62 \begin {gather*} \frac {\left (5 \sqrt {a} B-3 A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}}\right )}{4 \sqrt {2} \sqrt [4]{a} c^{9/4}}-\frac {\left (5 \sqrt {a} B+3 A \sqrt {c}\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}}{\sqrt {a}+\sqrt {c} x}\right )}{4 \sqrt {2} \sqrt [4]{a} c^{9/4}}+\frac {5 a B \sqrt {x}-A c x^{3/2}+4 B c x^{5/2}}{2 c^2 \left (a+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^(5/2)*(A + B*x))/(a + c*x^2)^2,x]

[Out]

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

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fricas [B]  time = 0.44, size = 884, normalized size = 2.91 \begin {gather*} \frac {{\left (c^{3} x^{2} + a c^{2}\right )} \sqrt {\frac {c^{4} \sqrt {-\frac {625 \, B^{4} a^{2} - 450 \, A^{2} B^{2} a c + 81 \, A^{4} c^{2}}{a c^{9}}} + 30 \, A B}{c^{4}}} \log \left (-{\left (625 \, B^{4} a^{2} - 81 \, A^{4} c^{2}\right )} \sqrt {x} + {\left (3 \, A a c^{7} \sqrt {-\frac {625 \, B^{4} a^{2} - 450 \, A^{2} B^{2} a c + 81 \, A^{4} c^{2}}{a c^{9}}} + 125 \, B^{3} a^{2} c^{2} - 45 \, A^{2} B a c^{3}\right )} \sqrt {\frac {c^{4} \sqrt {-\frac {625 \, B^{4} a^{2} - 450 \, A^{2} B^{2} a c + 81 \, A^{4} c^{2}}{a c^{9}}} + 30 \, A B}{c^{4}}}\right ) - {\left (c^{3} x^{2} + a c^{2}\right )} \sqrt {\frac {c^{4} \sqrt {-\frac {625 \, B^{4} a^{2} - 450 \, A^{2} B^{2} a c + 81 \, A^{4} c^{2}}{a c^{9}}} + 30 \, A B}{c^{4}}} \log \left (-{\left (625 \, B^{4} a^{2} - 81 \, A^{4} c^{2}\right )} \sqrt {x} - {\left (3 \, A a c^{7} \sqrt {-\frac {625 \, B^{4} a^{2} - 450 \, A^{2} B^{2} a c + 81 \, A^{4} c^{2}}{a c^{9}}} + 125 \, B^{3} a^{2} c^{2} - 45 \, A^{2} B a c^{3}\right )} \sqrt {\frac {c^{4} \sqrt {-\frac {625 \, B^{4} a^{2} - 450 \, A^{2} B^{2} a c + 81 \, A^{4} c^{2}}{a c^{9}}} + 30 \, A B}{c^{4}}}\right ) - {\left (c^{3} x^{2} + a c^{2}\right )} \sqrt {-\frac {c^{4} \sqrt {-\frac {625 \, B^{4} a^{2} - 450 \, A^{2} B^{2} a c + 81 \, A^{4} c^{2}}{a c^{9}}} - 30 \, A B}{c^{4}}} \log \left (-{\left (625 \, B^{4} a^{2} - 81 \, A^{4} c^{2}\right )} \sqrt {x} + {\left (3 \, A a c^{7} \sqrt {-\frac {625 \, B^{4} a^{2} - 450 \, A^{2} B^{2} a c + 81 \, A^{4} c^{2}}{a c^{9}}} - 125 \, B^{3} a^{2} c^{2} + 45 \, A^{2} B a c^{3}\right )} \sqrt {-\frac {c^{4} \sqrt {-\frac {625 \, B^{4} a^{2} - 450 \, A^{2} B^{2} a c + 81 \, A^{4} c^{2}}{a c^{9}}} - 30 \, A B}{c^{4}}}\right ) + {\left (c^{3} x^{2} + a c^{2}\right )} \sqrt {-\frac {c^{4} \sqrt {-\frac {625 \, B^{4} a^{2} - 450 \, A^{2} B^{2} a c + 81 \, A^{4} c^{2}}{a c^{9}}} - 30 \, A B}{c^{4}}} \log \left (-{\left (625 \, B^{4} a^{2} - 81 \, A^{4} c^{2}\right )} \sqrt {x} - {\left (3 \, A a c^{7} \sqrt {-\frac {625 \, B^{4} a^{2} - 450 \, A^{2} B^{2} a c + 81 \, A^{4} c^{2}}{a c^{9}}} - 125 \, B^{3} a^{2} c^{2} + 45 \, A^{2} B a c^{3}\right )} \sqrt {-\frac {c^{4} \sqrt {-\frac {625 \, B^{4} a^{2} - 450 \, A^{2} B^{2} a c + 81 \, A^{4} c^{2}}{a c^{9}}} - 30 \, A B}{c^{4}}}\right ) + 4 \, {\left (4 \, B c x^{2} - A c x + 5 \, B a\right )} \sqrt {x}}{8 \, {\left (c^{3} x^{2} + a c^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((c^3*x^2 + a*c^2)*sqrt((c^4*sqrt(-(625*B^4*a^2 - 450*A^2*B^2*a*c + 81*A^4*c^2)/(a*c^9)) + 30*A*B)/c^4)*lo
g(-(625*B^4*a^2 - 81*A^4*c^2)*sqrt(x) + (3*A*a*c^7*sqrt(-(625*B^4*a^2 - 450*A^2*B^2*a*c + 81*A^4*c^2)/(a*c^9))
+ 125*B^3*a^2*c^2 - 45*A^2*B*a*c^3)*sqrt((c^4*sqrt(-(625*B^4*a^2 - 450*A^2*B^2*a*c + 81*A^4*c^2)/(a*c^9)) + 3
0*A*B)/c^4)) - (c^3*x^2 + a*c^2)*sqrt((c^4*sqrt(-(625*B^4*a^2 - 450*A^2*B^2*a*c + 81*A^4*c^2)/(a*c^9)) + 30*A*
B)/c^4)*log(-(625*B^4*a^2 - 81*A^4*c^2)*sqrt(x) - (3*A*a*c^7*sqrt(-(625*B^4*a^2 - 450*A^2*B^2*a*c + 81*A^4*c^2
)/(a*c^9)) + 125*B^3*a^2*c^2 - 45*A^2*B*a*c^3)*sqrt((c^4*sqrt(-(625*B^4*a^2 - 450*A^2*B^2*a*c + 81*A^4*c^2)/(a
*c^9)) + 30*A*B)/c^4)) - (c^3*x^2 + a*c^2)*sqrt(-(c^4*sqrt(-(625*B^4*a^2 - 450*A^2*B^2*a*c + 81*A^4*c^2)/(a*c^
9)) - 30*A*B)/c^4)*log(-(625*B^4*a^2 - 81*A^4*c^2)*sqrt(x) + (3*A*a*c^7*sqrt(-(625*B^4*a^2 - 450*A^2*B^2*a*c +
81*A^4*c^2)/(a*c^9)) - 125*B^3*a^2*c^2 + 45*A^2*B*a*c^3)*sqrt(-(c^4*sqrt(-(625*B^4*a^2 - 450*A^2*B^2*a*c + 81
*A^4*c^2)/(a*c^9)) - 30*A*B)/c^4)) + (c^3*x^2 + a*c^2)*sqrt(-(c^4*sqrt(-(625*B^4*a^2 - 450*A^2*B^2*a*c + 81*A^
4*c^2)/(a*c^9)) - 30*A*B)/c^4)*log(-(625*B^4*a^2 - 81*A^4*c^2)*sqrt(x) - (3*A*a*c^7*sqrt(-(625*B^4*a^2 - 450*A
^2*B^2*a*c + 81*A^4*c^2)/(a*c^9)) - 125*B^3*a^2*c^2 + 45*A^2*B*a*c^3)*sqrt(-(c^4*sqrt(-(625*B^4*a^2 - 450*A^2*
B^2*a*c + 81*A^4*c^2)/(a*c^9)) - 30*A*B)/c^4)) + 4*(4*B*c*x^2 - A*c*x + 5*B*a)*sqrt(x))/(c^3*x^2 + a*c^2)

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giac [A]  time = 0.20, size = 283, normalized size = 0.93 \begin {gather*} \frac {2 \, B \sqrt {x}}{c^{2}} - \frac {A c x^{\frac {3}{2}} - B a \sqrt {x}}{2 \, {\left (c x^{2} + a\right )} c^{2}} - \frac {\sqrt {2} {\left (5 \, \left (a c^{3}\right )^{\frac {1}{4}} B a c - 3 \, \left (a c^{3}\right )^{\frac {3}{4}} A\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{8 \, a c^{4}} - \frac {\sqrt {2} {\left (5 \, \left (a c^{3}\right )^{\frac {1}{4}} B a c - 3 \, \left (a c^{3}\right )^{\frac {3}{4}} A\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{8 \, a c^{4}} - \frac {\sqrt {2} {\left (5 \, \left (a c^{3}\right )^{\frac {1}{4}} B a c + 3 \, \left (a c^{3}\right )^{\frac {3}{4}} A\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{c}}\right )}{16 \, a c^{4}} + \frac {\sqrt {2} {\left (5 \, \left (a c^{3}\right )^{\frac {1}{4}} B a c + 3 \, \left (a c^{3}\right )^{\frac {3}{4}} A\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{c}}\right )}{16 \, a c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 314, normalized size = 1.03 \begin {gather*} -\frac {A \,x^{\frac {3}{2}}}{2 \left (c \,x^{2}+a \right ) c}+\frac {B a \sqrt {x}}{2 \left (c \,x^{2}+a \right ) c^{2}}+\frac {3 \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {a}{c}\right )^{\frac {1}{4}} c^{2}}+\frac {3 \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {a}{c}\right )^{\frac {1}{4}} c^{2}}+\frac {3 \sqrt {2}\, A \ln \left (\frac {x -\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{c}}}{x +\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{c}}}\right )}{16 \left (\frac {a}{c}\right )^{\frac {1}{4}} c^{2}}-\frac {5 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{8 c^{2}}-\frac {5 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{8 c^{2}}-\frac {5 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \ln \left (\frac {x +\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{c}}}{x -\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{c}}}\right )}{16 c^{2}}+\frac {2 B \sqrt {x}}{c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.36, size = 283, normalized size = 0.93 \begin {gather*} -\frac {A c x^{\frac {3}{2}} - B a \sqrt {x}}{2 \, {\left (c^{3} x^{2} + a c^{2}\right )}} + \frac {2 \, B \sqrt {x}}{c^{2}} - \frac {\frac {2 \, \sqrt {2} {\left (5 \, B a \sqrt {c} - 3 \, A \sqrt {a} c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} {\left (5 \, B a \sqrt {c} - 3 \, A \sqrt {a} c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {c}} \sqrt {c}} + \frac {\sqrt {2} {\left (5 \, B a \sqrt {c} + 3 \, A \sqrt {a} c\right )} \log \left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {3}{4}}} - \frac {\sqrt {2} {\left (5 \, B a \sqrt {c} + 3 \, A \sqrt {a} c\right )} \log \left (-\sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {3}{4}}}}{16 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.27, size = 617, normalized size = 2.03 \begin {gather*} \frac {\frac {B\,a\,\sqrt {x}}{2}-\frac {A\,c\,x^{3/2}}{2}}{c^3\,x^2+a\,c^2}+\frac {2\,B\,\sqrt {x}}{c^2}-\mathrm {atan}\left (\frac {B^2\,a^2\,\sqrt {x}\,\sqrt {\frac {15\,A\,B}{32\,c^4}-\frac {25\,B^2\,\sqrt {-a\,c^9}}{64\,c^9}+\frac {9\,A^2\,\sqrt {-a\,c^9}}{64\,a\,c^8}}\,50{}\mathrm {i}}{\frac {27\,A^3\,a}{4\,c}+\frac {125\,B^3\,a^2\,\sqrt {-a\,c^9}}{4\,c^7}-\frac {75\,A\,B^2\,a^2}{4\,c^2}-\frac {45\,A^2\,B\,a\,\sqrt {-a\,c^9}}{4\,c^6}}-\frac {A^2\,a\,\sqrt {x}\,\sqrt {\frac {15\,A\,B}{32\,c^4}-\frac {25\,B^2\,\sqrt {-a\,c^9}}{64\,c^9}+\frac {9\,A^2\,\sqrt {-a\,c^9}}{64\,a\,c^8}}\,18{}\mathrm {i}}{\frac {27\,A^3\,a}{4\,c^2}+\frac {125\,B^3\,a^2\,\sqrt {-a\,c^9}}{4\,c^8}-\frac {75\,A\,B^2\,a^2}{4\,c^3}-\frac {45\,A^2\,B\,a\,\sqrt {-a\,c^9}}{4\,c^7}}\right )\,\sqrt {\frac {9\,A^2\,c\,\sqrt {-a\,c^9}-25\,B^2\,a\,\sqrt {-a\,c^9}+30\,A\,B\,a\,c^5}{64\,a\,c^9}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {B^2\,a^2\,\sqrt {x}\,\sqrt {\frac {15\,A\,B}{32\,c^4}+\frac {25\,B^2\,\sqrt {-a\,c^9}}{64\,c^9}-\frac {9\,A^2\,\sqrt {-a\,c^9}}{64\,a\,c^8}}\,50{}\mathrm {i}}{\frac {27\,A^3\,a}{4\,c}-\frac {125\,B^3\,a^2\,\sqrt {-a\,c^9}}{4\,c^7}-\frac {75\,A\,B^2\,a^2}{4\,c^2}+\frac {45\,A^2\,B\,a\,\sqrt {-a\,c^9}}{4\,c^6}}-\frac {A^2\,a\,\sqrt {x}\,\sqrt {\frac {15\,A\,B}{32\,c^4}+\frac {25\,B^2\,\sqrt {-a\,c^9}}{64\,c^9}-\frac {9\,A^2\,\sqrt {-a\,c^9}}{64\,a\,c^8}}\,18{}\mathrm {i}}{\frac {27\,A^3\,a}{4\,c^2}-\frac {125\,B^3\,a^2\,\sqrt {-a\,c^9}}{4\,c^8}-\frac {75\,A\,B^2\,a^2}{4\,c^3}+\frac {45\,A^2\,B\,a\,\sqrt {-a\,c^9}}{4\,c^7}}\right )\,\sqrt {\frac {25\,B^2\,a\,\sqrt {-a\,c^9}-9\,A^2\,c\,\sqrt {-a\,c^9}+30\,A\,B\,a\,c^5}{64\,a\,c^9}}\,2{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((B*a*x^(1/2))/2 - (A*c*x^(3/2))/2)/(a*c^2 + c^3*x^2) - atan((B^2*a^2*x^(1/2)*((15*A*B)/(32*c^4) + (25*B^2*(-a
*c^9)^(1/2))/(64*c^9) - (9*A^2*(-a*c^9)^(1/2))/(64*a*c^8))^(1/2)*50i)/((27*A^3*a)/(4*c) - (125*B^3*a^2*(-a*c^9
)^(1/2))/(4*c^7) - (75*A*B^2*a^2)/(4*c^2) + (45*A^2*B*a*(-a*c^9)^(1/2))/(4*c^6)) - (A^2*a*x^(1/2)*((15*A*B)/(3
2*c^4) + (25*B^2*(-a*c^9)^(1/2))/(64*c^9) - (9*A^2*(-a*c^9)^(1/2))/(64*a*c^8))^(1/2)*18i)/((27*A^3*a)/(4*c^2)
- (125*B^3*a^2*(-a*c^9)^(1/2))/(4*c^8) - (75*A*B^2*a^2)/(4*c^3) + (45*A^2*B*a*(-a*c^9)^(1/2))/(4*c^7)))*((25*B
^2*a*(-a*c^9)^(1/2) - 9*A^2*c*(-a*c^9)^(1/2) + 30*A*B*a*c^5)/(64*a*c^9))^(1/2)*2i - atan((B^2*a^2*x^(1/2)*((15
*A*B)/(32*c^4) - (25*B^2*(-a*c^9)^(1/2))/(64*c^9) + (9*A^2*(-a*c^9)^(1/2))/(64*a*c^8))^(1/2)*50i)/((27*A^3*a)/
(4*c) + (125*B^3*a^2*(-a*c^9)^(1/2))/(4*c^7) - (75*A*B^2*a^2)/(4*c^2) - (45*A^2*B*a*(-a*c^9)^(1/2))/(4*c^6)) -
(A^2*a*x^(1/2)*((15*A*B)/(32*c^4) - (25*B^2*(-a*c^9)^(1/2))/(64*c^9) + (9*A^2*(-a*c^9)^(1/2))/(64*a*c^8))^(1/
2)*18i)/((27*A^3*a)/(4*c^2) + (125*B^3*a^2*(-a*c^9)^(1/2))/(4*c^8) - (75*A*B^2*a^2)/(4*c^3) - (45*A^2*B*a*(-a*
c^9)^(1/2))/(4*c^7)))*((9*A^2*c*(-a*c^9)^(1/2) - 25*B^2*a*(-a*c^9)^(1/2) + 30*A*B*a*c^5)/(64*a*c^9))^(1/2)*2i
+ (2*B*x^(1/2))/c^2

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sympy [A]  time = 170.57, size = 1374, normalized size = 4.52

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*(B*x+A)/(c*x**2+a)**2,x)

[Out]

Piecewise((zoo*(-2*A/sqrt(x) + 2*B*sqrt(x)), Eq(a, 0) & Eq(c, 0)), ((-2*A/sqrt(x) + 2*B*sqrt(x))/c**2, Eq(a, 0
)), ((2*A*x**(7/2)/7 + 2*B*x**(9/2)/9)/a**2, Eq(c, 0)), (-4*(-1)**(1/4)*A*a**(1/4)*c*x**(3/2)*(1/c)**(1/4)/(8*
(-1)**(1/4)*a**(5/4)*c**2*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(1/4)*c**3*x**2*(1/c)**(1/4)) + 3*A*a*log(-(-1)**(1/
4)*a**(1/4)*(1/c)**(1/4) + sqrt(x))/(8*(-1)**(1/4)*a**(5/4)*c**2*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(1/4)*c**3*x*
*2*(1/c)**(1/4)) - 3*A*a*log((-1)**(1/4)*a**(1/4)*(1/c)**(1/4) + sqrt(x))/(8*(-1)**(1/4)*a**(5/4)*c**2*(1/c)**
(1/4) + 8*(-1)**(1/4)*a**(1/4)*c**3*x**2*(1/c)**(1/4)) - 6*A*a*atan((-1)**(3/4)*sqrt(x)/(a**(1/4)*(1/c)**(1/4)
))/(8*(-1)**(1/4)*a**(5/4)*c**2*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(1/4)*c**3*x**2*(1/c)**(1/4)) + 3*A*c*x**2*log
(-(-1)**(1/4)*a**(1/4)*(1/c)**(1/4) + sqrt(x))/(8*(-1)**(1/4)*a**(5/4)*c**2*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(1
/4)*c**3*x**2*(1/c)**(1/4)) - 3*A*c*x**2*log((-1)**(1/4)*a**(1/4)*(1/c)**(1/4) + sqrt(x))/(8*(-1)**(1/4)*a**(5
/4)*c**2*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(1/4)*c**3*x**2*(1/c)**(1/4)) - 6*A*c*x**2*atan((-1)**(3/4)*sqrt(x)/(
a**(1/4)*(1/c)**(1/4)))/(8*(-1)**(1/4)*a**(5/4)*c**2*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(1/4)*c**3*x**2*(1/c)**(1
/4)) + 20*(-1)**(1/4)*B*a**(5/4)*sqrt(x)*(1/c)**(1/4)/(8*(-1)**(1/4)*a**(5/4)*c**2*(1/c)**(1/4) + 8*(-1)**(1/4
)*a**(1/4)*c**3*x**2*(1/c)**(1/4)) + 16*(-1)**(1/4)*B*a**(1/4)*c*x**(5/2)*(1/c)**(1/4)/(8*(-1)**(1/4)*a**(5/4)
*c**2*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(1/4)*c**3*x**2*(1/c)**(1/4)) + 5*I*B*a**(3/2)*sqrt(1/c)*log(-(-1)**(1/4
)*a**(1/4)*(1/c)**(1/4) + sqrt(x))/(8*(-1)**(1/4)*a**(5/4)*c**2*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(1/4)*c**3*x**
2*(1/c)**(1/4)) - 5*I*B*a**(3/2)*sqrt(1/c)*log((-1)**(1/4)*a**(1/4)*(1/c)**(1/4) + sqrt(x))/(8*(-1)**(1/4)*a**
(5/4)*c**2*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(1/4)*c**3*x**2*(1/c)**(1/4)) + 10*I*B*a**(3/2)*sqrt(1/c)*atan((-1)
**(3/4)*sqrt(x)/(a**(1/4)*(1/c)**(1/4)))/(8*(-1)**(1/4)*a**(5/4)*c**2*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(1/4)*c*
*3*x**2*(1/c)**(1/4)) + 5*I*B*sqrt(a)*c*x**2*sqrt(1/c)*log(-(-1)**(1/4)*a**(1/4)*(1/c)**(1/4) + sqrt(x))/(8*(-
1)**(1/4)*a**(5/4)*c**2*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(1/4)*c**3*x**2*(1/c)**(1/4)) - 5*I*B*sqrt(a)*c*x**2*s
qrt(1/c)*log((-1)**(1/4)*a**(1/4)*(1/c)**(1/4) + sqrt(x))/(8*(-1)**(1/4)*a**(5/4)*c**2*(1/c)**(1/4) + 8*(-1)**
(1/4)*a**(1/4)*c**3*x**2*(1/c)**(1/4)) + 10*I*B*sqrt(a)*c*x**2*sqrt(1/c)*atan((-1)**(3/4)*sqrt(x)/(a**(1/4)*(1
/c)**(1/4)))/(8*(-1)**(1/4)*a**(5/4)*c**2*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(1/4)*c**3*x**2*(1/c)**(1/4)), True)
)

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