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

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

[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))

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Rubi [A]  time = 0.305759, antiderivative size = 320, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {823, 827, 1168, 1162, 617, 204, 1165, 628} \[ \frac{\left (5 \sqrt{a} B-21 A \sqrt{c}\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{64 \sqrt{2} a^{11/4} c^{3/4}}-\frac{\left (5 \sqrt{a} B-21 A \sqrt{c}\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{64 \sqrt{2} a^{11/4} c^{3/4}}-\frac{\left (5 \sqrt{a} B+21 A \sqrt{c}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{11/4} c^{3/4}}+\frac{\left (5 \sqrt{a} B+21 A \sqrt{c}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt{2} a^{11/4} c^{3/4}}+\frac{\sqrt{x} (7 A+5 B x)}{16 a^2 \left (a+c x^2\right )}+\frac{\sqrt{x} (A+B x)}{4 a \left (a+c x^2\right )^2} \]

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 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 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)]

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 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 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 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 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]

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 = 0.188193, size = 344, normalized size = 1.08 \[ \frac{\frac{32 a^2 A \sqrt{x}}{\left (a+c x^2\right )^2}+\frac{32 a^2 B x^{3/2}}{\left (a+c x^2\right )^2}+\frac{56 a A \sqrt{x}}{a+c x^2}-\frac{21 \sqrt{2} \sqrt [4]{a} A \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{\sqrt [4]{c}}+\frac{21 \sqrt{2} \sqrt [4]{a} A \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{\sqrt [4]{c}}-\frac{42 \sqrt{2} \sqrt [4]{a} A \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt [4]{c}}+\frac{42 \sqrt{2} \sqrt [4]{a} A \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{\sqrt [4]{c}}-\frac{20 (-a)^{3/4} B \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-a}}\right )}{c^{3/4}}+\frac{20 (-a)^{3/4} B \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-a}}\right )}{c^{3/4}}+\frac{40 a B x^{3/2}}{a+c x^2}}{128 a^3} \]

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)

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Maple [A]  time = 0.009, size = 349, normalized size = 1.1 \begin{align*}{\frac{A}{4\,a \left ( c{x}^{2}+a \right ) ^{2}}\sqrt{x}}+{\frac{7\,A}{16\,{a}^{2} \left ( c{x}^{2}+a \right ) }\sqrt{x}}+{\frac{21\,A\sqrt{2}}{128\,{a}^{3}}\sqrt [4]{{\frac{a}{c}}}\ln \left ({ \left ( x+\sqrt [4]{{\frac{a}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{c}}} \right ) \left ( x-\sqrt [4]{{\frac{a}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{c}}} \right ) ^{-1}} \right ) }+{\frac{21\,A\sqrt{2}}{64\,{a}^{3}}\sqrt [4]{{\frac{a}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}+1 \right ) }+{\frac{21\,A\sqrt{2}}{64\,{a}^{3}}\sqrt [4]{{\frac{a}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}-1 \right ) }+{\frac{B}{4\,a \left ( c{x}^{2}+a \right ) ^{2}}{x}^{{\frac{3}{2}}}}+{\frac{5\,B}{16\,{a}^{2} \left ( c{x}^{2}+a \right ) }{x}^{{\frac{3}{2}}}}+{\frac{5\,B\sqrt{2}}{128\,{a}^{2}c}\ln \left ({ \left ( x-\sqrt [4]{{\frac{a}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{c}}} \right ) \left ( x+\sqrt [4]{{\frac{a}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{c}}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}+{\frac{5\,B\sqrt{2}}{64\,{a}^{2}c}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}+{\frac{5\,B\sqrt{2}}{64\,{a}^{2}c}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}} \end{align*}

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)*x^
(1/2)*2^(1/2)+(a/c)^(1/2))/(x-(a/c)^(1/4)*x^(1/2)*2^(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)*x^(1/2)*2
^(1/2)+(a/c)^(1/2))/(x+(a/c)^(1/4)*x^(1/2)*2^(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)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [B]  time = 1.85141, size = 2321, normalized size = 7.25 \begin{align*} \text{result too large to display} \end{align*}

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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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

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Giac [A]  time = 1.17315, size = 396, normalized size = 1.24 \begin{align*} \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{align*}

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)