3.5.30 \(\int \frac {A+B x}{\sqrt {x} (a+c x^2)^3} \, dx\)

Optimal. Leaf size=320 \[ \frac {\left (5 \sqrt {a} B-21 A \sqrt {c}\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{64 \sqrt {2} a^{11/4} c^{3/4}}-\frac {\left (5 \sqrt {a} B-21 A \sqrt {c}\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{64 \sqrt {2} a^{11/4} c^{3/4}}-\frac {\left (5 \sqrt {a} B+21 A \sqrt {c}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{11/4} c^{3/4}}+\frac {\left (5 \sqrt {a} B+21 A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt {2} a^{11/4} c^{3/4}}+\frac {\sqrt {x} (7 A+5 B x)}{16 a^2 \left (a+c x^2\right )}+\frac {\sqrt {x} (A+B x)}{4 a \left (a+c x^2\right )^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.31, antiderivative size = 320, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {823, 827, 1168, 1162, 617, 204, 1165, 628} \begin {gather*} \frac {\left (5 \sqrt {a} B-21 A \sqrt {c}\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{64 \sqrt {2} a^{11/4} c^{3/4}}-\frac {\left (5 \sqrt {a} B-21 A \sqrt {c}\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{64 \sqrt {2} a^{11/4} c^{3/4}}-\frac {\left (5 \sqrt {a} B+21 A \sqrt {c}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{11/4} c^{3/4}}+\frac {\left (5 \sqrt {a} B+21 A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt {2} a^{11/4} c^{3/4}}+\frac {\sqrt {x} (7 A+5 B x)}{16 a^2 \left (a+c x^2\right )}+\frac {\sqrt {x} (A+B x)}{4 a \left (a+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(Sqrt[x]*(a + c*x^2)^3),x]

[Out]

(Sqrt[x]*(A + B*x))/(4*a*(a + c*x^2)^2) + (Sqrt[x]*(7*A + 5*B*x))/(16*a^2*(a + c*x^2)) - ((5*Sqrt[a]*B + 21*A*
Sqrt[c])*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(11/4)*c^(3/4)) + ((5*Sqrt[a]*B + 21*A*S
qrt[c])*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(11/4)*c^(3/4)) + ((5*Sqrt[a]*B - 21*A*Sq
rt[c])*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*a^(11/4)*c^(3/4)) - ((5*Sqrt[a]
*B - 21*A*Sqrt[c])*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*a^(11/4)*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 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 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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 {A+B x}{\sqrt {x} \left (a+c x^2\right )^3} \, dx &=\frac {\sqrt {x} (A+B x)}{4 a \left (a+c x^2\right )^2}-\frac {\int \frac {-\frac {7}{2} a A c-\frac {5}{2} a B c x}{\sqrt {x} \left (a+c x^2\right )^2} \, dx}{4 a^2 c}\\ &=\frac {\sqrt {x} (A+B x)}{4 a \left (a+c x^2\right )^2}+\frac {\sqrt {x} (7 A+5 B x)}{16 a^2 \left (a+c x^2\right )}+\frac {\int \frac {\frac {21}{4} a^2 A c^2+\frac {5}{4} a^2 B c^2 x}{\sqrt {x} \left (a+c x^2\right )} \, dx}{8 a^4 c^2}\\ &=\frac {\sqrt {x} (A+B x)}{4 a \left (a+c x^2\right )^2}+\frac {\sqrt {x} (7 A+5 B x)}{16 a^2 \left (a+c x^2\right )}+\frac {\operatorname {Subst}\left (\int \frac {\frac {21}{4} a^2 A c^2+\frac {5}{4} a^2 B c^2 x^2}{a+c x^4} \, dx,x,\sqrt {x}\right )}{4 a^4 c^2}\\ &=\frac {\sqrt {x} (A+B x)}{4 a \left (a+c x^2\right )^2}+\frac {\sqrt {x} (7 A+5 B x)}{16 a^2 \left (a+c x^2\right )}-\frac {\left (5 \sqrt {a} B-21 A \sqrt {c}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a} \sqrt {c}-c x^2}{a+c x^4} \, dx,x,\sqrt {x}\right )}{32 a^{5/2} c}+\frac {\left (5 \sqrt {a} B+21 A \sqrt {c}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a} \sqrt {c}+c x^2}{a+c x^4} \, dx,x,\sqrt {x}\right )}{32 a^{5/2} c}\\ &=\frac {\sqrt {x} (A+B x)}{4 a \left (a+c x^2\right )^2}+\frac {\sqrt {x} (7 A+5 B x)}{16 a^2 \left (a+c x^2\right )}+\frac {\left (5 \sqrt {a} B+21 A \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 )}{64 a^{5/2} c}+\frac {\left (5 \sqrt {a} B+21 A \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 )}{64 a^{5/2} c}+\frac {\left (5 \sqrt {a} B-21 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 )}{64 \sqrt {2} a^{11/4} c^{3/4}}+\frac {\left (5 \sqrt {a} B-21 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 )}{64 \sqrt {2} a^{11/4} c^{3/4}}\\ &=\frac {\sqrt {x} (A+B x)}{4 a \left (a+c x^2\right )^2}+\frac {\sqrt {x} (7 A+5 B x)}{16 a^2 \left (a+c x^2\right )}+\frac {\left (5 \sqrt {a} B-21 A \sqrt {c}\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} a^{11/4} c^{3/4}}-\frac {\left (5 \sqrt {a} B-21 A \sqrt {c}\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} a^{11/4} c^{3/4}}+\frac {\left (5 \sqrt {a} B+21 A \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 )}{32 \sqrt {2} a^{11/4} c^{3/4}}-\frac {\left (5 \sqrt {a} B+21 A \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 )}{32 \sqrt {2} a^{11/4} c^{3/4}}\\ &=\frac {\sqrt {x} (A+B x)}{4 a \left (a+c x^2\right )^2}+\frac {\sqrt {x} (7 A+5 B x)}{16 a^2 \left (a+c x^2\right )}-\frac {\left (5 \sqrt {a} B+21 A \sqrt {c}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{11/4} c^{3/4}}+\frac {\left (5 \sqrt {a} B+21 A \sqrt {c}\right ) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{11/4} c^{3/4}}+\frac {\left (5 \sqrt {a} B-21 A \sqrt {c}\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} a^{11/4} c^{3/4}}-\frac {\left (5 \sqrt {a} B-21 A \sqrt {c}\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} a^{11/4} c^{3/4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.17, size = 344, normalized size = 1.08 \begin {gather*} \frac {\frac {32 a^2 A \sqrt {x}}{\left (a+c x^2\right )^2}+\frac {32 a^2 B x^{3/2}}{\left (a+c x^2\right )^2}+\frac {56 a A \sqrt {x}}{a+c x^2}-\frac {21 \sqrt {2} \sqrt [4]{a} A \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{\sqrt [4]{c}}+\frac {21 \sqrt {2} \sqrt [4]{a} A \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{\sqrt [4]{c}}-\frac {42 \sqrt {2} \sqrt [4]{a} A \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt [4]{c}}+\frac {42 \sqrt {2} \sqrt [4]{a} A \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt [4]{c}}-\frac {20 (-a)^{3/4} B \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-a}}\right )}{c^{3/4}}+\frac {20 (-a)^{3/4} B \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-a}}\right )}{c^{3/4}}+\frac {40 a B x^{3/2}}{a+c x^2}}{128 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(Sqrt[x]*(a + c*x^2)^3),x]

[Out]

((32*a^2*A*Sqrt[x])/(a + c*x^2)^2 + (32*a^2*B*x^(3/2))/(a + c*x^2)^2 + (56*a*A*Sqrt[x])/(a + c*x^2) + (40*a*B*
x^(3/2))/(a + c*x^2) - (42*Sqrt[2]*a^(1/4)*A*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/c^(1/4) + (42*Sqrt
[2]*a^(1/4)*A*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/c^(1/4) - (20*(-a)^(3/4)*B*ArcTan[(c^(1/4)*Sqrt[x
])/(-a)^(1/4)])/c^(3/4) + (20*(-a)^(3/4)*B*ArcTanh[(c^(1/4)*Sqrt[x])/(-a)^(1/4)])/c^(3/4) - (21*Sqrt[2]*a^(1/4
)*A*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/c^(1/4) + (21*Sqrt[2]*a^(1/4)*A*Log[Sqrt[a] +
Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/c^(1/4))/(128*a^3)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.58, size = 199, normalized size = 0.62 \begin {gather*} -\frac {\left (5 \sqrt {a} B+21 A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}}\right )}{32 \sqrt {2} a^{11/4} c^{3/4}}-\frac {\left (5 \sqrt {a} B-21 A \sqrt {c}\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}}{\sqrt {a}+\sqrt {c} x}\right )}{32 \sqrt {2} a^{11/4} c^{3/4}}+\frac {11 a A \sqrt {x}+9 a B x^{3/2}+7 A c x^{5/2}+5 B c x^{7/2}}{16 a^2 \left (a+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/(Sqrt[x]*(a + c*x^2)^3),x]

[Out]

(11*a*A*Sqrt[x] + 9*a*B*x^(3/2) + 7*A*c*x^(5/2) + 5*B*c*x^(7/2))/(16*a^2*(a + c*x^2)^2) - ((5*Sqrt[a]*B + 21*A
*Sqrt[c])*ArcTan[(Sqrt[a] - Sqrt[c]*x)/(Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x])])/(32*Sqrt[2]*a^(11/4)*c^(3/4)) - ((5
*Sqrt[a]*B - 21*A*Sqrt[c])*ArcTanh[(Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[c]*x)])/(32*Sqrt[2]*a^(11
/4)*c^(3/4))

________________________________________________________________________________________

fricas [B]  time = 0.47, size = 981, normalized size = 3.07 \begin {gather*} \frac {{\left (a^{2} c^{2} x^{4} + 2 \, a^{3} c x^{2} + a^{4}\right )} \sqrt {-\frac {a^{5} c \sqrt {-\frac {625 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 194481 \, A^{4} c^{2}}{a^{11} c^{3}}} + 210 \, A B}{a^{5} c}} \log \left (-{\left (625 \, B^{4} a^{2} - 194481 \, A^{4} c^{2}\right )} \sqrt {x} + {\left (5 \, B a^{9} c^{2} \sqrt {-\frac {625 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 194481 \, A^{4} c^{2}}{a^{11} c^{3}}} - 525 \, A B^{2} a^{4} c + 9261 \, A^{3} a^{3} c^{2}\right )} \sqrt {-\frac {a^{5} c \sqrt {-\frac {625 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 194481 \, A^{4} c^{2}}{a^{11} c^{3}}} + 210 \, A B}{a^{5} c}}\right ) - {\left (a^{2} c^{2} x^{4} + 2 \, a^{3} c x^{2} + a^{4}\right )} \sqrt {-\frac {a^{5} c \sqrt {-\frac {625 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 194481 \, A^{4} c^{2}}{a^{11} c^{3}}} + 210 \, A B}{a^{5} c}} \log \left (-{\left (625 \, B^{4} a^{2} - 194481 \, A^{4} c^{2}\right )} \sqrt {x} - {\left (5 \, B a^{9} c^{2} \sqrt {-\frac {625 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 194481 \, A^{4} c^{2}}{a^{11} c^{3}}} - 525 \, A B^{2} a^{4} c + 9261 \, A^{3} a^{3} c^{2}\right )} \sqrt {-\frac {a^{5} c \sqrt {-\frac {625 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 194481 \, A^{4} c^{2}}{a^{11} c^{3}}} + 210 \, A B}{a^{5} c}}\right ) - {\left (a^{2} c^{2} x^{4} + 2 \, a^{3} c x^{2} + a^{4}\right )} \sqrt {\frac {a^{5} c \sqrt {-\frac {625 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 194481 \, A^{4} c^{2}}{a^{11} c^{3}}} - 210 \, A B}{a^{5} c}} \log \left (-{\left (625 \, B^{4} a^{2} - 194481 \, A^{4} c^{2}\right )} \sqrt {x} + {\left (5 \, B a^{9} c^{2} \sqrt {-\frac {625 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 194481 \, A^{4} c^{2}}{a^{11} c^{3}}} + 525 \, A B^{2} a^{4} c - 9261 \, A^{3} a^{3} c^{2}\right )} \sqrt {\frac {a^{5} c \sqrt {-\frac {625 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 194481 \, A^{4} c^{2}}{a^{11} c^{3}}} - 210 \, A B}{a^{5} c}}\right ) + {\left (a^{2} c^{2} x^{4} + 2 \, a^{3} c x^{2} + a^{4}\right )} \sqrt {\frac {a^{5} c \sqrt {-\frac {625 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 194481 \, A^{4} c^{2}}{a^{11} c^{3}}} - 210 \, A B}{a^{5} c}} \log \left (-{\left (625 \, B^{4} a^{2} - 194481 \, A^{4} c^{2}\right )} \sqrt {x} - {\left (5 \, B a^{9} c^{2} \sqrt {-\frac {625 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 194481 \, A^{4} c^{2}}{a^{11} c^{3}}} + 525 \, A B^{2} a^{4} c - 9261 \, A^{3} a^{3} c^{2}\right )} \sqrt {\frac {a^{5} c \sqrt {-\frac {625 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 194481 \, A^{4} c^{2}}{a^{11} c^{3}}} - 210 \, A B}{a^{5} c}}\right ) + 4 \, {\left (5 \, B c x^{3} + 7 \, A c x^{2} + 9 \, B a x + 11 \, A a\right )} \sqrt {x}}{64 \, {\left (a^{2} c^{2} x^{4} + 2 \, a^{3} c x^{2} + a^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(1/2)/(c*x^2+a)^3,x, algorithm="fricas")

[Out]

1/64*((a^2*c^2*x^4 + 2*a^3*c*x^2 + a^4)*sqrt(-(a^5*c*sqrt(-(625*B^4*a^2 - 22050*A^2*B^2*a*c + 194481*A^4*c^2)/
(a^11*c^3)) + 210*A*B)/(a^5*c))*log(-(625*B^4*a^2 - 194481*A^4*c^2)*sqrt(x) + (5*B*a^9*c^2*sqrt(-(625*B^4*a^2
- 22050*A^2*B^2*a*c + 194481*A^4*c^2)/(a^11*c^3)) - 525*A*B^2*a^4*c + 9261*A^3*a^3*c^2)*sqrt(-(a^5*c*sqrt(-(62
5*B^4*a^2 - 22050*A^2*B^2*a*c + 194481*A^4*c^2)/(a^11*c^3)) + 210*A*B)/(a^5*c))) - (a^2*c^2*x^4 + 2*a^3*c*x^2
+ a^4)*sqrt(-(a^5*c*sqrt(-(625*B^4*a^2 - 22050*A^2*B^2*a*c + 194481*A^4*c^2)/(a^11*c^3)) + 210*A*B)/(a^5*c))*l
og(-(625*B^4*a^2 - 194481*A^4*c^2)*sqrt(x) - (5*B*a^9*c^2*sqrt(-(625*B^4*a^2 - 22050*A^2*B^2*a*c + 194481*A^4*
c^2)/(a^11*c^3)) - 525*A*B^2*a^4*c + 9261*A^3*a^3*c^2)*sqrt(-(a^5*c*sqrt(-(625*B^4*a^2 - 22050*A^2*B^2*a*c + 1
94481*A^4*c^2)/(a^11*c^3)) + 210*A*B)/(a^5*c))) - (a^2*c^2*x^4 + 2*a^3*c*x^2 + a^4)*sqrt((a^5*c*sqrt(-(625*B^4
*a^2 - 22050*A^2*B^2*a*c + 194481*A^4*c^2)/(a^11*c^3)) - 210*A*B)/(a^5*c))*log(-(625*B^4*a^2 - 194481*A^4*c^2)
*sqrt(x) + (5*B*a^9*c^2*sqrt(-(625*B^4*a^2 - 22050*A^2*B^2*a*c + 194481*A^4*c^2)/(a^11*c^3)) + 525*A*B^2*a^4*c
- 9261*A^3*a^3*c^2)*sqrt((a^5*c*sqrt(-(625*B^4*a^2 - 22050*A^2*B^2*a*c + 194481*A^4*c^2)/(a^11*c^3)) - 210*A*
B)/(a^5*c))) + (a^2*c^2*x^4 + 2*a^3*c*x^2 + a^4)*sqrt((a^5*c*sqrt(-(625*B^4*a^2 - 22050*A^2*B^2*a*c + 194481*A
^4*c^2)/(a^11*c^3)) - 210*A*B)/(a^5*c))*log(-(625*B^4*a^2 - 194481*A^4*c^2)*sqrt(x) - (5*B*a^9*c^2*sqrt(-(625*
B^4*a^2 - 22050*A^2*B^2*a*c + 194481*A^4*c^2)/(a^11*c^3)) + 525*A*B^2*a^4*c - 9261*A^3*a^3*c^2)*sqrt((a^5*c*sq
rt(-(625*B^4*a^2 - 22050*A^2*B^2*a*c + 194481*A^4*c^2)/(a^11*c^3)) - 210*A*B)/(a^5*c))) + 4*(5*B*c*x^3 + 7*A*c
*x^2 + 9*B*a*x + 11*A*a)*sqrt(x))/(a^2*c^2*x^4 + 2*a^3*c*x^2 + a^4)

________________________________________________________________________________________

giac [A]  time = 0.23, size = 293, normalized size = 0.92 \begin {gather*} \frac {5 \, B c x^{\frac {7}{2}} + 7 \, A c x^{\frac {5}{2}} + 9 \, B a x^{\frac {3}{2}} + 11 \, A a \sqrt {x}}{16 \, {\left (c x^{2} + a\right )}^{2} a^{2}} + \frac {\sqrt {2} {\left (21 \, \left (a c^{3}\right )^{\frac {1}{4}} A c^{2} + 5 \, \left (a c^{3}\right )^{\frac {3}{4}} B\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 )}{64 \, a^{3} c^{3}} + \frac {\sqrt {2} {\left (21 \, \left (a c^{3}\right )^{\frac {1}{4}} A c^{2} + 5 \, \left (a c^{3}\right )^{\frac {3}{4}} B\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 )}{64 \, a^{3} c^{3}} + \frac {\sqrt {2} {\left (21 \, \left (a c^{3}\right )^{\frac {1}{4}} A c^{2} - 5 \, \left (a c^{3}\right )^{\frac {3}{4}} B\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{c}}\right )}{128 \, a^{3} c^{3}} - \frac {\sqrt {2} {\left (21 \, \left (a c^{3}\right )^{\frac {1}{4}} A c^{2} - 5 \, \left (a c^{3}\right )^{\frac {3}{4}} B\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{c}}\right )}{128 \, a^{3} c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(1/2)/(c*x^2+a)^3,x, algorithm="giac")

[Out]

1/16*(5*B*c*x^(7/2) + 7*A*c*x^(5/2) + 9*B*a*x^(3/2) + 11*A*a*sqrt(x))/((c*x^2 + a)^2*a^2) + 1/64*sqrt(2)*(21*(
a*c^3)^(1/4)*A*c^2 + 5*(a*c^3)^(3/4)*B)*arctan(1/2*sqrt(2)*(sqrt(2)*(a/c)^(1/4) + 2*sqrt(x))/(a/c)^(1/4))/(a^3
*c^3) + 1/64*sqrt(2)*(21*(a*c^3)^(1/4)*A*c^2 + 5*(a*c^3)^(3/4)*B)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a/c)^(1/4) - 2
*sqrt(x))/(a/c)^(1/4))/(a^3*c^3) + 1/128*sqrt(2)*(21*(a*c^3)^(1/4)*A*c^2 - 5*(a*c^3)^(3/4)*B)*log(sqrt(2)*sqrt
(x)*(a/c)^(1/4) + x + sqrt(a/c))/(a^3*c^3) - 1/128*sqrt(2)*(21*(a*c^3)^(1/4)*A*c^2 - 5*(a*c^3)^(3/4)*B)*log(-s
qrt(2)*sqrt(x)*(a/c)^(1/4) + x + sqrt(a/c))/(a^3*c^3)

________________________________________________________________________________________

maple [A]  time = 0.06, size = 349, normalized size = 1.09 \begin {gather*} \frac {B \,x^{\frac {3}{2}}}{4 \left (c \,x^{2}+a \right )^{2} a}+\frac {A \sqrt {x}}{4 \left (c \,x^{2}+a \right )^{2} a}+\frac {5 B \,x^{\frac {3}{2}}}{16 \left (c \,x^{2}+a \right ) a^{2}}+\frac {7 A \sqrt {x}}{16 \left (c \,x^{2}+a \right ) a^{2}}+\frac {21 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{64 a^{3}}+\frac {21 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{64 a^{3}}+\frac {21 \left (\frac {a}{c}\right )^{\frac {1}{4}} \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 )}{128 a^{3}}+\frac {5 \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{64 \left (\frac {a}{c}\right )^{\frac {1}{4}} a^{2} c}+\frac {5 \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{64 \left (\frac {a}{c}\right )^{\frac {1}{4}} a^{2} c}+\frac {5 \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 )}{128 \left (\frac {a}{c}\right )^{\frac {1}{4}} a^{2} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(1/2)/(c*x^2+a)^3,x)

[Out]

1/4*A*x^(1/2)/a/(c*x^2+a)^2+7/16*A/a^2*x^(1/2)/(c*x^2+a)+21/128*A/a^3*(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)))+21/64*A/a^3*(a/c)^(1/4)*2^(1/2)*arctan
(2^(1/2)/(a/c)^(1/4)*x^(1/2)+1)+21/64*A/a^3*(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)-1)+1/4*B*x^
(3/2)/a/(c*x^2+a)^2+5/16*B/a^2*x^(3/2)/(c*x^2+a)+5/128*B/a^2/c/(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)))+5/64*B/a^2/c/(a/c)^(1/4)*2^(1/2)*arctan(2^(1/
2)/(a/c)^(1/4)*x^(1/2)+1)+5/64*B/a^2/c/(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)-1)

________________________________________________________________________________________

maxima [A]  time = 1.50, size = 291, normalized size = 0.91 \begin {gather*} \frac {5 \, B c x^{\frac {7}{2}} + 7 \, A c x^{\frac {5}{2}} + 9 \, B a x^{\frac {3}{2}} + 11 \, A a \sqrt {x}}{16 \, {\left (a^{2} c^{2} x^{4} + 2 \, a^{3} c x^{2} + a^{4}\right )}} + \frac {\frac {2 \, \sqrt {2} {\left (5 \, B \sqrt {a} + 21 \, A \sqrt {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 \sqrt {a} + 21 \, A \sqrt {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 \sqrt {a} - 21 \, A \sqrt {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 \sqrt {a} - 21 \, A \sqrt {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}}}}{128 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(1/2)/(c*x^2+a)^3,x, algorithm="maxima")

[Out]

1/16*(5*B*c*x^(7/2) + 7*A*c*x^(5/2) + 9*B*a*x^(3/2) + 11*A*a*sqrt(x))/(a^2*c^2*x^4 + 2*a^3*c*x^2 + a^4) + 1/12
8*(2*sqrt(2)*(5*B*sqrt(a) + 21*A*sqrt(c))*arctan(1/2*sqrt(2)*(sqrt(2)*a^(1/4)*c^(1/4) + 2*sqrt(c)*sqrt(x))/sqr
t(sqrt(a)*sqrt(c)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(c))*sqrt(c)) + 2*sqrt(2)*(5*B*sqrt(a) + 21*A*sqrt(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*sqrt(a) - 21*A*sqrt(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*sqrt(a) - 21*A*sqrt(c))*log(-sqrt(2)*a^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt
(a))/(a^(3/4)*c^(3/4)))/a^2

________________________________________________________________________________________

mupad [B]  time = 1.28, size = 687, normalized size = 2.15 \begin {gather*} 2\,\mathrm {atanh}\left (\frac {441\,A^2\,c^3\,\sqrt {x}\,\sqrt {\frac {25\,B^2\,\sqrt {-a^{11}\,c^3}}{4096\,a^{10}\,c^3}-\frac {441\,A^2\,\sqrt {-a^{11}\,c^3}}{4096\,a^{11}\,c^2}-\frac {105\,A\,B}{2048\,a^5\,c}}}{32\,\left (\frac {125\,B^3\,c}{2048\,a}+\frac {525\,A\,B^2\,\sqrt {-a^{11}\,c^3}}{2048\,a^7}-\frac {9261\,A^3\,c\,\sqrt {-a^{11}\,c^3}}{2048\,a^8}-\frac {2205\,A^2\,B\,c^2}{2048\,a^2}\right )}-\frac {25\,B^2\,c^2\,\sqrt {x}\,\sqrt {\frac {25\,B^2\,\sqrt {-a^{11}\,c^3}}{4096\,a^{10}\,c^3}-\frac {441\,A^2\,\sqrt {-a^{11}\,c^3}}{4096\,a^{11}\,c^2}-\frac {105\,A\,B}{2048\,a^5\,c}}}{32\,\left (\frac {125\,B^3\,c}{2048\,a^2}+\frac {525\,A\,B^2\,\sqrt {-a^{11}\,c^3}}{2048\,a^8}-\frac {9261\,A^3\,c\,\sqrt {-a^{11}\,c^3}}{2048\,a^9}-\frac {2205\,A^2\,B\,c^2}{2048\,a^3}\right )}\right )\,\sqrt {-\frac {441\,A^2\,c\,\sqrt {-a^{11}\,c^3}-25\,B^2\,a\,\sqrt {-a^{11}\,c^3}+210\,A\,B\,a^6\,c^2}{4096\,a^{11}\,c^3}}+2\,\mathrm {atanh}\left (\frac {441\,A^2\,c^3\,\sqrt {x}\,\sqrt {\frac {441\,A^2\,\sqrt {-a^{11}\,c^3}}{4096\,a^{11}\,c^2}-\frac {105\,A\,B}{2048\,a^5\,c}-\frac {25\,B^2\,\sqrt {-a^{11}\,c^3}}{4096\,a^{10}\,c^3}}}{32\,\left (\frac {125\,B^3\,c}{2048\,a}-\frac {525\,A\,B^2\,\sqrt {-a^{11}\,c^3}}{2048\,a^7}+\frac {9261\,A^3\,c\,\sqrt {-a^{11}\,c^3}}{2048\,a^8}-\frac {2205\,A^2\,B\,c^2}{2048\,a^2}\right )}-\frac {25\,B^2\,c^2\,\sqrt {x}\,\sqrt {\frac {441\,A^2\,\sqrt {-a^{11}\,c^3}}{4096\,a^{11}\,c^2}-\frac {105\,A\,B}{2048\,a^5\,c}-\frac {25\,B^2\,\sqrt {-a^{11}\,c^3}}{4096\,a^{10}\,c^3}}}{32\,\left (\frac {125\,B^3\,c}{2048\,a^2}-\frac {525\,A\,B^2\,\sqrt {-a^{11}\,c^3}}{2048\,a^8}+\frac {9261\,A^3\,c\,\sqrt {-a^{11}\,c^3}}{2048\,a^9}-\frac {2205\,A^2\,B\,c^2}{2048\,a^3}\right )}\right )\,\sqrt {-\frac {25\,B^2\,a\,\sqrt {-a^{11}\,c^3}-441\,A^2\,c\,\sqrt {-a^{11}\,c^3}+210\,A\,B\,a^6\,c^2}{4096\,a^{11}\,c^3}}+\frac {\frac {11\,A\,\sqrt {x}}{16\,a}+\frac {9\,B\,x^{3/2}}{16\,a}+\frac {7\,A\,c\,x^{5/2}}{16\,a^2}+\frac {5\,B\,c\,x^{7/2}}{16\,a^2}}{a^2+2\,a\,c\,x^2+c^2\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(1/2)*(a + c*x^2)^3),x)

[Out]

2*atanh((441*A^2*c^3*x^(1/2)*((25*B^2*(-a^11*c^3)^(1/2))/(4096*a^10*c^3) - (441*A^2*(-a^11*c^3)^(1/2))/(4096*a
^11*c^2) - (105*A*B)/(2048*a^5*c))^(1/2))/(32*((125*B^3*c)/(2048*a) + (525*A*B^2*(-a^11*c^3)^(1/2))/(2048*a^7)
- (9261*A^3*c*(-a^11*c^3)^(1/2))/(2048*a^8) - (2205*A^2*B*c^2)/(2048*a^2))) - (25*B^2*c^2*x^(1/2)*((25*B^2*(-
a^11*c^3)^(1/2))/(4096*a^10*c^3) - (441*A^2*(-a^11*c^3)^(1/2))/(4096*a^11*c^2) - (105*A*B)/(2048*a^5*c))^(1/2)
)/(32*((125*B^3*c)/(2048*a^2) + (525*A*B^2*(-a^11*c^3)^(1/2))/(2048*a^8) - (9261*A^3*c*(-a^11*c^3)^(1/2))/(204
8*a^9) - (2205*A^2*B*c^2)/(2048*a^3))))*(-(441*A^2*c*(-a^11*c^3)^(1/2) - 25*B^2*a*(-a^11*c^3)^(1/2) + 210*A*B*
a^6*c^2)/(4096*a^11*c^3))^(1/2) + 2*atanh((441*A^2*c^3*x^(1/2)*((441*A^2*(-a^11*c^3)^(1/2))/(4096*a^11*c^2) -
(105*A*B)/(2048*a^5*c) - (25*B^2*(-a^11*c^3)^(1/2))/(4096*a^10*c^3))^(1/2))/(32*((125*B^3*c)/(2048*a) - (525*A
*B^2*(-a^11*c^3)^(1/2))/(2048*a^7) + (9261*A^3*c*(-a^11*c^3)^(1/2))/(2048*a^8) - (2205*A^2*B*c^2)/(2048*a^2)))
- (25*B^2*c^2*x^(1/2)*((441*A^2*(-a^11*c^3)^(1/2))/(4096*a^11*c^2) - (105*A*B)/(2048*a^5*c) - (25*B^2*(-a^11*
c^3)^(1/2))/(4096*a^10*c^3))^(1/2))/(32*((125*B^3*c)/(2048*a^2) - (525*A*B^2*(-a^11*c^3)^(1/2))/(2048*a^8) + (
9261*A^3*c*(-a^11*c^3)^(1/2))/(2048*a^9) - (2205*A^2*B*c^2)/(2048*a^3))))*(-(25*B^2*a*(-a^11*c^3)^(1/2) - 441*
A^2*c*(-a^11*c^3)^(1/2) + 210*A*B*a^6*c^2)/(4096*a^11*c^3))^(1/2) + ((11*A*x^(1/2))/(16*a) + (9*B*x^(3/2))/(16
*a) + (7*A*c*x^(5/2))/(16*a^2) + (5*B*c*x^(7/2))/(16*a^2))/(a^2 + c^2*x^4 + 2*a*c*x^2)

________________________________________________________________________________________

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((B*x+A)/x**(1/2)/(c*x**2+a)**3,x)

[Out]

Timed out

________________________________________________________________________________________