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

Optimal. Leaf size=331 \[ -\frac {\left (3 \sqrt {a} B-5 A \sqrt {c}\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{64 \sqrt {2} a^{9/4} c^{5/4}}+\frac {\left (3 \sqrt {a} B-5 A \sqrt {c}\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{64 \sqrt {2} a^{9/4} c^{5/4}}-\frac {\left (3 \sqrt {a} B+5 A \sqrt {c}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{9/4} c^{5/4}}+\frac {\left (3 \sqrt {a} B+5 A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt {2} a^{9/4} c^{5/4}}+\frac {\sqrt {x} (a B+5 A c x)}{16 a^2 c \left (a+c x^2\right )}-\frac {\sqrt {x} (a B-A c x)}{4 a c \left (a+c x^2\right )^2} \]

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Rubi [A]  time = 0.29, antiderivative size = 331, 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 = {821, 823, 827, 1168, 1162, 617, 204, 1165, 628} \begin {gather*} -\frac {\left (3 \sqrt {a} B-5 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^{9/4} c^{5/4}}+\frac {\left (3 \sqrt {a} B-5 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^{9/4} c^{5/4}}-\frac {\left (3 \sqrt {a} B+5 A \sqrt {c}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{9/4} c^{5/4}}+\frac {\left (3 \sqrt {a} B+5 A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt {2} a^{9/4} c^{5/4}}+\frac {\sqrt {x} (a B+5 A c x)}{16 a^2 c \left (a+c x^2\right )}-\frac {\sqrt {x} (a B-A c x)}{4 a c \left (a+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[x]*(a*B - A*c*x))/(4*a*c*(a + c*x^2)^2) + (Sqrt[x]*(a*B + 5*A*c*x))/(16*a^2*c*(a + c*x^2)) - ((3*Sqrt[a
]*B + 5*A*Sqrt[c])*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(9/4)*c^(5/4)) + ((3*Sqrt[a]*B
+ 5*A*Sqrt[c])*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(9/4)*c^(5/4)) - ((3*Sqrt[a]*B -
5*A*Sqrt[c])*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*a^(9/4)*c^(5/4)) + ((3*Sq
rt[a]*B - 5*A*Sqrt[c])*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*a^(9/4)*c^(5/4)
)

Rule 204

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

Rule 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 821

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

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 {\sqrt {x} (A+B x)}{\left (a+c x^2\right )^3} \, dx &=-\frac {\sqrt {x} (a B-A c x)}{4 a c \left (a+c x^2\right )^2}+\frac {\int \frac {\frac {a B}{2}+\frac {5 A c x}{2}}{\sqrt {x} \left (a+c x^2\right )^2} \, dx}{4 a c}\\ &=-\frac {\sqrt {x} (a B-A c x)}{4 a c \left (a+c x^2\right )^2}+\frac {\sqrt {x} (a B+5 A c x)}{16 a^2 c \left (a+c x^2\right )}-\frac {\int \frac {-\frac {3}{4} a^2 B c-\frac {5}{4} a A c^2 x}{\sqrt {x} \left (a+c x^2\right )} \, dx}{8 a^3 c^2}\\ &=-\frac {\sqrt {x} (a B-A c x)}{4 a c \left (a+c x^2\right )^2}+\frac {\sqrt {x} (a B+5 A c x)}{16 a^2 c \left (a+c x^2\right )}-\frac {\operatorname {Subst}\left (\int \frac {-\frac {3}{4} a^2 B c-\frac {5}{4} a A c^2 x^2}{a+c x^4} \, dx,x,\sqrt {x}\right )}{4 a^3 c^2}\\ &=-\frac {\sqrt {x} (a B-A c x)}{4 a c \left (a+c x^2\right )^2}+\frac {\sqrt {x} (a B+5 A c x)}{16 a^2 c \left (a+c x^2\right )}+\frac {\left (3 \sqrt {a} B-5 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^2 c^{3/2}}+\frac {\left (3 \sqrt {a} B+5 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^2 c^{3/2}}\\ &=-\frac {\sqrt {x} (a B-A c x)}{4 a c \left (a+c x^2\right )^2}+\frac {\sqrt {x} (a B+5 A c x)}{16 a^2 c \left (a+c x^2\right )}+\frac {\left (3 \sqrt {a} B+5 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^2 c^{3/2}}+\frac {\left (3 \sqrt {a} B+5 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^2 c^{3/2}}-\frac {\left (3 \sqrt {a} B-5 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^{9/4} c^{5/4}}-\frac {\left (3 \sqrt {a} B-5 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^{9/4} c^{5/4}}\\ &=-\frac {\sqrt {x} (a B-A c x)}{4 a c \left (a+c x^2\right )^2}+\frac {\sqrt {x} (a B+5 A c x)}{16 a^2 c \left (a+c x^2\right )}-\frac {\left (3 \sqrt {a} B-5 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^{9/4} c^{5/4}}+\frac {\left (3 \sqrt {a} B-5 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^{9/4} c^{5/4}}+\frac {\left (3 \sqrt {a} B+5 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^{9/4} c^{5/4}}-\frac {\left (3 \sqrt {a} B+5 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^{9/4} c^{5/4}}\\ &=-\frac {\sqrt {x} (a B-A c x)}{4 a c \left (a+c x^2\right )^2}+\frac {\sqrt {x} (a B+5 A c x)}{16 a^2 c \left (a+c x^2\right )}-\frac {\left (3 \sqrt {a} B+5 A \sqrt {c}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{9/4} c^{5/4}}+\frac {\left (3 \sqrt {a} B+5 A \sqrt {c}\right ) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{9/4} c^{5/4}}-\frac {\left (3 \sqrt {a} B-5 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^{9/4} c^{5/4}}+\frac {\left (3 \sqrt {a} B-5 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^{9/4} c^{5/4}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 356, normalized size = 1.08 \begin {gather*} \frac {-\frac {3 \sqrt {2} a^{5/4} B \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{c^{5/4}}+\frac {3 \sqrt {2} a^{5/4} B \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{c^{5/4}}-\frac {6 \sqrt {2} a^{5/4} B \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{c^{5/4}}+\frac {6 \sqrt {2} a^{5/4} B \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{c^{5/4}}+\frac {32 a^2 A x^{3/2}}{\left (a+c x^2\right )^2}+\frac {32 a^2 B x^{5/2}}{\left (a+c x^2\right )^2}-\frac {20 (-a)^{3/4} A \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-a}}\right )}{c^{3/4}}+\frac {20 (-a)^{3/4} A \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-a}}\right )}{c^{3/4}}+\frac {40 a A x^{3/2}}{a+c x^2}+\frac {24 a B x^{5/2}}{a+c x^2}-\frac {24 a B \sqrt {x}}{c}}{128 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-24*a*B*Sqrt[x])/c + (32*a^2*A*x^(3/2))/(a + c*x^2)^2 + (32*a^2*B*x^(5/2))/(a + c*x^2)^2 + (40*a*A*x^(3/2))/
(a + c*x^2) + (24*a*B*x^(5/2))/(a + c*x^2) - (6*Sqrt[2]*a^(5/4)*B*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)
])/c^(5/4) + (6*Sqrt[2]*a^(5/4)*B*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/c^(5/4) - (20*(-a)^(3/4)*A*Ar
cTan[(c^(1/4)*Sqrt[x])/(-a)^(1/4)])/c^(3/4) + (20*(-a)^(3/4)*A*ArcTanh[(c^(1/4)*Sqrt[x])/(-a)^(1/4)])/c^(3/4)
- (3*Sqrt[2]*a^(5/4)*B*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/c^(5/4) + (3*Sqrt[2]*a^(5/4
)*B*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/c^(5/4))/(128*a^3)

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IntegrateAlgebraic [A]  time = 0.89, size = 207, normalized size = 0.63 \begin {gather*} -\frac {\left (3 \sqrt {a} B+5 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^{9/4} c^{5/4}}+\frac {\left (3 \sqrt {a} B-5 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^{9/4} c^{5/4}}+\frac {-3 a^2 B \sqrt {x}+9 a A c x^{3/2}+a B c x^{5/2}+5 A c^2 x^{7/2}}{16 a^2 c \left (a+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.46, size = 1022, normalized size = 3.09 \begin {gather*} -\frac {{\left (a^{2} c^{3} x^{4} + 2 \, a^{3} c^{2} x^{2} + a^{4} c\right )} \sqrt {-\frac {a^{4} c^{2} \sqrt {-\frac {81 \, B^{4} a^{2} - 450 \, A^{2} B^{2} a c + 625 \, A^{4} c^{2}}{a^{9} c^{5}}} + 30 \, A B}{a^{4} c^{2}}} \log \left (-{\left (81 \, B^{4} a^{2} - 625 \, A^{4} c^{2}\right )} \sqrt {x} + {\left (5 \, A a^{7} c^{4} \sqrt {-\frac {81 \, B^{4} a^{2} - 450 \, A^{2} B^{2} a c + 625 \, A^{4} c^{2}}{a^{9} c^{5}}} + 27 \, B^{3} a^{4} c - 75 \, A^{2} B a^{3} c^{2}\right )} \sqrt {-\frac {a^{4} c^{2} \sqrt {-\frac {81 \, B^{4} a^{2} - 450 \, A^{2} B^{2} a c + 625 \, A^{4} c^{2}}{a^{9} c^{5}}} + 30 \, A B}{a^{4} c^{2}}}\right ) - {\left (a^{2} c^{3} x^{4} + 2 \, a^{3} c^{2} x^{2} + a^{4} c\right )} \sqrt {-\frac {a^{4} c^{2} \sqrt {-\frac {81 \, B^{4} a^{2} - 450 \, A^{2} B^{2} a c + 625 \, A^{4} c^{2}}{a^{9} c^{5}}} + 30 \, A B}{a^{4} c^{2}}} \log \left (-{\left (81 \, B^{4} a^{2} - 625 \, A^{4} c^{2}\right )} \sqrt {x} - {\left (5 \, A a^{7} c^{4} \sqrt {-\frac {81 \, B^{4} a^{2} - 450 \, A^{2} B^{2} a c + 625 \, A^{4} c^{2}}{a^{9} c^{5}}} + 27 \, B^{3} a^{4} c - 75 \, A^{2} B a^{3} c^{2}\right )} \sqrt {-\frac {a^{4} c^{2} \sqrt {-\frac {81 \, B^{4} a^{2} - 450 \, A^{2} B^{2} a c + 625 \, A^{4} c^{2}}{a^{9} c^{5}}} + 30 \, A B}{a^{4} c^{2}}}\right ) - {\left (a^{2} c^{3} x^{4} + 2 \, a^{3} c^{2} x^{2} + a^{4} c\right )} \sqrt {\frac {a^{4} c^{2} \sqrt {-\frac {81 \, B^{4} a^{2} - 450 \, A^{2} B^{2} a c + 625 \, A^{4} c^{2}}{a^{9} c^{5}}} - 30 \, A B}{a^{4} c^{2}}} \log \left (-{\left (81 \, B^{4} a^{2} - 625 \, A^{4} c^{2}\right )} \sqrt {x} + {\left (5 \, A a^{7} c^{4} \sqrt {-\frac {81 \, B^{4} a^{2} - 450 \, A^{2} B^{2} a c + 625 \, A^{4} c^{2}}{a^{9} c^{5}}} - 27 \, B^{3} a^{4} c + 75 \, A^{2} B a^{3} c^{2}\right )} \sqrt {\frac {a^{4} c^{2} \sqrt {-\frac {81 \, B^{4} a^{2} - 450 \, A^{2} B^{2} a c + 625 \, A^{4} c^{2}}{a^{9} c^{5}}} - 30 \, A B}{a^{4} c^{2}}}\right ) + {\left (a^{2} c^{3} x^{4} + 2 \, a^{3} c^{2} x^{2} + a^{4} c\right )} \sqrt {\frac {a^{4} c^{2} \sqrt {-\frac {81 \, B^{4} a^{2} - 450 \, A^{2} B^{2} a c + 625 \, A^{4} c^{2}}{a^{9} c^{5}}} - 30 \, A B}{a^{4} c^{2}}} \log \left (-{\left (81 \, B^{4} a^{2} - 625 \, A^{4} c^{2}\right )} \sqrt {x} - {\left (5 \, A a^{7} c^{4} \sqrt {-\frac {81 \, B^{4} a^{2} - 450 \, A^{2} B^{2} a c + 625 \, A^{4} c^{2}}{a^{9} c^{5}}} - 27 \, B^{3} a^{4} c + 75 \, A^{2} B a^{3} c^{2}\right )} \sqrt {\frac {a^{4} c^{2} \sqrt {-\frac {81 \, B^{4} a^{2} - 450 \, A^{2} B^{2} a c + 625 \, A^{4} c^{2}}{a^{9} c^{5}}} - 30 \, A B}{a^{4} c^{2}}}\right ) - 4 \, {\left (5 \, A c^{2} x^{3} + B a c x^{2} + 9 \, A a c x - 3 \, B a^{2}\right )} \sqrt {x}}{64 \, {\left (a^{2} c^{3} x^{4} + 2 \, a^{3} c^{2} x^{2} + a^{4} c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*((a^2*c^3*x^4 + 2*a^3*c^2*x^2 + a^4*c)*sqrt(-(a^4*c^2*sqrt(-(81*B^4*a^2 - 450*A^2*B^2*a*c + 625*A^4*c^2)
/(a^9*c^5)) + 30*A*B)/(a^4*c^2))*log(-(81*B^4*a^2 - 625*A^4*c^2)*sqrt(x) + (5*A*a^7*c^4*sqrt(-(81*B^4*a^2 - 45
0*A^2*B^2*a*c + 625*A^4*c^2)/(a^9*c^5)) + 27*B^3*a^4*c - 75*A^2*B*a^3*c^2)*sqrt(-(a^4*c^2*sqrt(-(81*B^4*a^2 -
450*A^2*B^2*a*c + 625*A^4*c^2)/(a^9*c^5)) + 30*A*B)/(a^4*c^2))) - (a^2*c^3*x^4 + 2*a^3*c^2*x^2 + a^4*c)*sqrt(-
(a^4*c^2*sqrt(-(81*B^4*a^2 - 450*A^2*B^2*a*c + 625*A^4*c^2)/(a^9*c^5)) + 30*A*B)/(a^4*c^2))*log(-(81*B^4*a^2 -
625*A^4*c^2)*sqrt(x) - (5*A*a^7*c^4*sqrt(-(81*B^4*a^2 - 450*A^2*B^2*a*c + 625*A^4*c^2)/(a^9*c^5)) + 27*B^3*a^
4*c - 75*A^2*B*a^3*c^2)*sqrt(-(a^4*c^2*sqrt(-(81*B^4*a^2 - 450*A^2*B^2*a*c + 625*A^4*c^2)/(a^9*c^5)) + 30*A*B)
/(a^4*c^2))) - (a^2*c^3*x^4 + 2*a^3*c^2*x^2 + a^4*c)*sqrt((a^4*c^2*sqrt(-(81*B^4*a^2 - 450*A^2*B^2*a*c + 625*A
^4*c^2)/(a^9*c^5)) - 30*A*B)/(a^4*c^2))*log(-(81*B^4*a^2 - 625*A^4*c^2)*sqrt(x) + (5*A*a^7*c^4*sqrt(-(81*B^4*a
^2 - 450*A^2*B^2*a*c + 625*A^4*c^2)/(a^9*c^5)) - 27*B^3*a^4*c + 75*A^2*B*a^3*c^2)*sqrt((a^4*c^2*sqrt(-(81*B^4*
a^2 - 450*A^2*B^2*a*c + 625*A^4*c^2)/(a^9*c^5)) - 30*A*B)/(a^4*c^2))) + (a^2*c^3*x^4 + 2*a^3*c^2*x^2 + a^4*c)*
sqrt((a^4*c^2*sqrt(-(81*B^4*a^2 - 450*A^2*B^2*a*c + 625*A^4*c^2)/(a^9*c^5)) - 30*A*B)/(a^4*c^2))*log(-(81*B^4*
a^2 - 625*A^4*c^2)*sqrt(x) - (5*A*a^7*c^4*sqrt(-(81*B^4*a^2 - 450*A^2*B^2*a*c + 625*A^4*c^2)/(a^9*c^5)) - 27*B
^3*a^4*c + 75*A^2*B*a^3*c^2)*sqrt((a^4*c^2*sqrt(-(81*B^4*a^2 - 450*A^2*B^2*a*c + 625*A^4*c^2)/(a^9*c^5)) - 30*
A*B)/(a^4*c^2))) - 4*(5*A*c^2*x^3 + B*a*c*x^2 + 9*A*a*c*x - 3*B*a^2)*sqrt(x))/(a^2*c^3*x^4 + 2*a^3*c^2*x^2 + a
^4*c)

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giac [A]  time = 0.21, size = 297, normalized size = 0.90 \begin {gather*} \frac {\sqrt {2} {\left (3 \, \left (a c^{3}\right )^{\frac {1}{4}} B a c + 5 \, \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 )}{64 \, a^{3} c^{3}} + \frac {\sqrt {2} {\left (3 \, \left (a c^{3}\right )^{\frac {1}{4}} B a c + 5 \, \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 )}{64 \, a^{3} c^{3}} + \frac {\sqrt {2} {\left (3 \, \left (a c^{3}\right )^{\frac {1}{4}} B a c - 5 \, \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 )}{128 \, a^{3} c^{3}} - \frac {\sqrt {2} {\left (3 \, \left (a c^{3}\right )^{\frac {1}{4}} B a c - 5 \, \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 )}{128 \, a^{3} c^{3}} + \frac {5 \, A c^{2} x^{\frac {7}{2}} + B a c x^{\frac {5}{2}} + 9 \, A a c x^{\frac {3}{2}} - 3 \, B a^{2} \sqrt {x}}{16 \, {\left (c x^{2} + a\right )}^{2} a^{2} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*sqrt(2)*(3*(a*c^3)^(1/4)*B*a*c + 5*(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^3*c^3) + 1/64*sqrt(2)*(3*(a*c^3)^(1/4)*B*a*c + 5*(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^3*c^3) + 1/128*sqrt(2)*(3*(a*c^3)^(1/4)*B*a*c - 5*(a*c^3)^(3/4)*A)*l
og(sqrt(2)*sqrt(x)*(a/c)^(1/4) + x + sqrt(a/c))/(a^3*c^3) - 1/128*sqrt(2)*(3*(a*c^3)^(1/4)*B*a*c - 5*(a*c^3)^(
3/4)*A)*log(-sqrt(2)*sqrt(x)*(a/c)^(1/4) + x + sqrt(a/c))/(a^3*c^3) + 1/16*(5*A*c^2*x^(7/2) + B*a*c*x^(5/2) +
9*A*a*c*x^(3/2) - 3*B*a^2*sqrt(x))/((c*x^2 + a)^2*a^2*c)

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maple [A]  time = 0.07, size = 335, normalized size = 1.01 \begin {gather*} \frac {5 \sqrt {2}\, A \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}\, A \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}\, 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 \left (\frac {a}{c}\right )^{\frac {1}{4}} a^{2} c}+\frac {3 \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 )}{64 a^{2} c}+\frac {3 \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 )}{64 a^{2} c}+\frac {3 \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 )}{128 a^{2} c}+\frac {\frac {5 A c \,x^{\frac {7}{2}}}{16 a^{2}}+\frac {B \,x^{\frac {5}{2}}}{16 a}+\frac {9 A \,x^{\frac {3}{2}}}{16 a}-\frac {3 B \sqrt {x}}{16 c}}{\left (c \,x^{2}+a \right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(5/32*A*c/a^2*x^(7/2)+1/32*B/a*x^(5/2)+9/32*A/a*x^(3/2)-3/32*B/c*x^(1/2))/(c*x^2+a)^2+3/64/a^2/c*B*(a/c)^(1/
4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)-1)+3/128/a^2/c*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)))+3/64/a^2/c*B*(a/c)^(1/4)*2^(1/2)*arctan(2^(1/
2)/(a/c)^(1/4)*x^(1/2)+1)+5/128/a^2/c*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)))+5/64/a^2/c*A/(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)+
1)+5/64/a^2/c*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 = 1.29, size = 311, normalized size = 0.94 \begin {gather*} \frac {5 \, A c^{2} x^{\frac {7}{2}} + B a c x^{\frac {5}{2}} + 9 \, A a c x^{\frac {3}{2}} - 3 \, B a^{2} \sqrt {x}}{16 \, {\left (a^{2} c^{3} x^{4} + 2 \, a^{3} c^{2} x^{2} + a^{4} c\right )}} + \frac {\frac {2 \, \sqrt {2} {\left (3 \, B a \sqrt {c} + 5 \, 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 (3 \, B a \sqrt {c} + 5 \, 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 (3 \, B a \sqrt {c} - 5 \, 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 (3 \, B a \sqrt {c} - 5 \, 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}}}}{128 \, a^{2} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.29, size = 690, normalized size = 2.08 \begin {gather*} \frac {\frac {9\,A\,x^{3/2}}{16\,a}+\frac {B\,x^{5/2}}{16\,a}-\frac {3\,B\,\sqrt {x}}{16\,c}+\frac {5\,A\,c\,x^{7/2}}{16\,a^2}}{a^2+2\,a\,c\,x^2+c^2\,x^4}-2\,\mathrm {atanh}\left (\frac {9\,B^2\,c\,\sqrt {x}\,\sqrt {\frac {25\,A^2\,\sqrt {-a^9\,c^5}}{4096\,a^9\,c^4}-\frac {15\,A\,B}{2048\,a^4\,c^2}-\frac {9\,B^2\,\sqrt {-a^9\,c^5}}{4096\,a^8\,c^5}}}{32\,\left (\frac {45\,A\,B^2}{2048\,a^2}-\frac {125\,A^3\,c}{2048\,a^3}+\frac {27\,B^3\,\sqrt {-a^9\,c^5}}{2048\,a^6\,c^3}-\frac {75\,A^2\,B\,\sqrt {-a^9\,c^5}}{2048\,a^7\,c^2}\right )}-\frac {25\,A^2\,c^2\,\sqrt {x}\,\sqrt {\frac {25\,A^2\,\sqrt {-a^9\,c^5}}{4096\,a^9\,c^4}-\frac {15\,A\,B}{2048\,a^4\,c^2}-\frac {9\,B^2\,\sqrt {-a^9\,c^5}}{4096\,a^8\,c^5}}}{32\,\left (\frac {45\,A\,B^2}{2048\,a}-\frac {125\,A^3\,c}{2048\,a^2}+\frac {27\,B^3\,\sqrt {-a^9\,c^5}}{2048\,a^5\,c^3}-\frac {75\,A^2\,B\,\sqrt {-a^9\,c^5}}{2048\,a^6\,c^2}\right )}\right )\,\sqrt {-\frac {9\,B^2\,a\,\sqrt {-a^9\,c^5}-25\,A^2\,c\,\sqrt {-a^9\,c^5}+30\,A\,B\,a^5\,c^3}{4096\,a^9\,c^5}}-2\,\mathrm {atanh}\left (\frac {9\,B^2\,c\,\sqrt {x}\,\sqrt {\frac {9\,B^2\,\sqrt {-a^9\,c^5}}{4096\,a^8\,c^5}-\frac {25\,A^2\,\sqrt {-a^9\,c^5}}{4096\,a^9\,c^4}-\frac {15\,A\,B}{2048\,a^4\,c^2}}}{32\,\left (\frac {45\,A\,B^2}{2048\,a^2}-\frac {125\,A^3\,c}{2048\,a^3}-\frac {27\,B^3\,\sqrt {-a^9\,c^5}}{2048\,a^6\,c^3}+\frac {75\,A^2\,B\,\sqrt {-a^9\,c^5}}{2048\,a^7\,c^2}\right )}-\frac {25\,A^2\,c^2\,\sqrt {x}\,\sqrt {\frac {9\,B^2\,\sqrt {-a^9\,c^5}}{4096\,a^8\,c^5}-\frac {25\,A^2\,\sqrt {-a^9\,c^5}}{4096\,a^9\,c^4}-\frac {15\,A\,B}{2048\,a^4\,c^2}}}{32\,\left (\frac {45\,A\,B^2}{2048\,a}-\frac {125\,A^3\,c}{2048\,a^2}-\frac {27\,B^3\,\sqrt {-a^9\,c^5}}{2048\,a^5\,c^3}+\frac {75\,A^2\,B\,\sqrt {-a^9\,c^5}}{2048\,a^6\,c^2}\right )}\right )\,\sqrt {-\frac {25\,A^2\,c\,\sqrt {-a^9\,c^5}-9\,B^2\,a\,\sqrt {-a^9\,c^5}+30\,A\,B\,a^5\,c^3}{4096\,a^9\,c^5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((9*A*x^(3/2))/(16*a) + (B*x^(5/2))/(16*a) - (3*B*x^(1/2))/(16*c) + (5*A*c*x^(7/2))/(16*a^2))/(a^2 + c^2*x^4 +
2*a*c*x^2) - 2*atanh((9*B^2*c*x^(1/2)*((25*A^2*(-a^9*c^5)^(1/2))/(4096*a^9*c^4) - (15*A*B)/(2048*a^4*c^2) - (
9*B^2*(-a^9*c^5)^(1/2))/(4096*a^8*c^5))^(1/2))/(32*((45*A*B^2)/(2048*a^2) - (125*A^3*c)/(2048*a^3) + (27*B^3*(
-a^9*c^5)^(1/2))/(2048*a^6*c^3) - (75*A^2*B*(-a^9*c^5)^(1/2))/(2048*a^7*c^2))) - (25*A^2*c^2*x^(1/2)*((25*A^2*
(-a^9*c^5)^(1/2))/(4096*a^9*c^4) - (15*A*B)/(2048*a^4*c^2) - (9*B^2*(-a^9*c^5)^(1/2))/(4096*a^8*c^5))^(1/2))/(
32*((45*A*B^2)/(2048*a) - (125*A^3*c)/(2048*a^2) + (27*B^3*(-a^9*c^5)^(1/2))/(2048*a^5*c^3) - (75*A^2*B*(-a^9*
c^5)^(1/2))/(2048*a^6*c^2))))*(-(9*B^2*a*(-a^9*c^5)^(1/2) - 25*A^2*c*(-a^9*c^5)^(1/2) + 30*A*B*a^5*c^3)/(4096*
a^9*c^5))^(1/2) - 2*atanh((9*B^2*c*x^(1/2)*((9*B^2*(-a^9*c^5)^(1/2))/(4096*a^8*c^5) - (25*A^2*(-a^9*c^5)^(1/2)
)/(4096*a^9*c^4) - (15*A*B)/(2048*a^4*c^2))^(1/2))/(32*((45*A*B^2)/(2048*a^2) - (125*A^3*c)/(2048*a^3) - (27*B
^3*(-a^9*c^5)^(1/2))/(2048*a^6*c^3) + (75*A^2*B*(-a^9*c^5)^(1/2))/(2048*a^7*c^2))) - (25*A^2*c^2*x^(1/2)*((9*B
^2*(-a^9*c^5)^(1/2))/(4096*a^8*c^5) - (25*A^2*(-a^9*c^5)^(1/2))/(4096*a^9*c^4) - (15*A*B)/(2048*a^4*c^2))^(1/2
))/(32*((45*A*B^2)/(2048*a) - (125*A^3*c)/(2048*a^2) - (27*B^3*(-a^9*c^5)^(1/2))/(2048*a^5*c^3) + (75*A^2*B*(-
a^9*c^5)^(1/2))/(2048*a^6*c^2))))*(-(25*A^2*c*(-a^9*c^5)^(1/2) - 9*B^2*a*(-a^9*c^5)^(1/2) + 30*A*B*a^5*c^3)/(4
096*a^9*c^5))^(1/2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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