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

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

[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)
)

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Rubi [A]  time = 0.285814, antiderivative size = 331, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.45, Rules used = {821, 823, 827, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{\left (3 \sqrt{a} B-5 A \sqrt{c}\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{64 \sqrt{2} a^{9/4} c^{5/4}}+\frac{\left (3 \sqrt{a} B-5 A \sqrt{c}\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{64 \sqrt{2} a^{9/4} c^{5/4}}-\frac{\left (3 \sqrt{a} B+5 A \sqrt{c}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{9/4} c^{5/4}}+\frac{\left (3 \sqrt{a} B+5 A \sqrt{c}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt{2} a^{9/4} c^{5/4}}+\frac{\sqrt{x} (a B+5 A c x)}{16 a^2 c \left (a+c x^2\right )}-\frac{\sqrt{x} (a B-A c x)}{4 a c \left (a+c x^2\right )^2} \]

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 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 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{\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*}

Mathematica [A]  time = 0.193418, size = 356, normalized size = 1.08 \[ \frac{\frac{32 a^2 A x^{3/2}}{\left (a+c x^2\right )^2}-\frac{3 \sqrt{2} a^{5/4} B \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{c^{5/4}}+\frac{3 \sqrt{2} a^{5/4} B \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{c^{5/4}}-\frac{6 \sqrt{2} a^{5/4} B \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )}{c^{5/4}}+\frac{6 \sqrt{2} a^{5/4} B \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{c^{5/4}}+\frac{32 a^2 B x^{5/2}}{\left (a+c x^2\right )^2}-\frac{20 (-a)^{3/4} A \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-a}}\right )}{c^{3/4}}+\frac{20 (-a)^{3/4} A \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-a}}\right )}{c^{3/4}}+\frac{40 a A x^{3/2}}{a+c x^2}+\frac{24 a B x^{5/2}}{a+c x^2}-\frac{24 a B \sqrt{x}}{c}}{128 a^3} \]

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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Maple [A]  time = 0.016, size = 335, normalized size = 1. \begin{align*} 2\,{\frac{1}{ \left ( c{x}^{2}+a \right ) ^{2}} \left ({\frac{5\,Ac{x}^{7/2}}{32\,{a}^{2}}}+1/32\,{\frac{B{x}^{5/2}}{a}}+{\frac{9\,A{x}^{3/2}}{32\,a}}-{\frac{3\,B\sqrt{x}}{32\,c}} \right ) }+{\frac{3\,B\sqrt{2}}{64\,{a}^{2}c}\sqrt [4]{{\frac{a}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}+1 \right ) }+{\frac{3\,B\sqrt{2}}{64\,{a}^{2}c}\sqrt [4]{{\frac{a}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}-1 \right ) }+{\frac{3\,B\sqrt{2}}{128\,{a}^{2}c}\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{5\,A\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\,A\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\,A\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(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*x^(1/2)/c)/(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)+3/64/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)*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)))+5/128/a^2/c*A/(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/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 [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(x^(1/2)*(B*x+A)/(c*x^2+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.82098, size = 2237, normalized size = 6.76 \begin{align*} \text{result too large to display} \end{align*}

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

[Out]

Timed out

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Giac [A]  time = 1.22303, size = 414, normalized size = 1.25 \begin{align*} \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{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} + \frac{\sqrt{2}{\left (3 \, \left (a c^{3}\right )^{\frac{1}{4}} B a c^{3} + 5 \, \left (a c^{3}\right )^{\frac{3}{4}} A c^{2}\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^{5}} + \frac{\sqrt{2}{\left (3 \, \left (a c^{3}\right )^{\frac{1}{4}} B a c^{3} - 5 \, \left (a c^{3}\right )^{\frac{3}{4}} A c^{2}\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^{5}} \end{align*}

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) - 1/128*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) + 1/64*sqrt(2)*(3*(a*c^3)^(1/4)*B*a*c^3 + 5*(a*c^3)^(3/4)*A*c^2)*arctan(1/2*sqrt(2)*(sq
rt(2)*(a/c)^(1/4) + 2*sqrt(x))/(a/c)^(1/4))/(a^3*c^5) + 1/128*sqrt(2)*(3*(a*c^3)^(1/4)*B*a*c^3 - 5*(a*c^3)^(3/
4)*A*c^2)*log(sqrt(2)*sqrt(x)*(a/c)^(1/4) + x + sqrt(a/c))/(a^3*c^5)