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

Optimal. Leaf size=315 \[ \frac{3 \left (\sqrt{a} B-A \sqrt{c}\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{64 \sqrt{2} a^{7/4} c^{7/4}}-\frac{3 \left (\sqrt{a} B-A \sqrt{c}\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{64 \sqrt{2} a^{7/4} c^{7/4}}-\frac{3 \left (\sqrt{a} B+A \sqrt{c}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{7/4} c^{7/4}}+\frac{3 \left (\sqrt{a} B+A \sqrt{c}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt{2} a^{7/4} c^{7/4}}-\frac{\sqrt{x} (A+B x)}{4 c \left (a+c x^2\right )^2}+\frac{\sqrt{x} (A+3 B x)}{16 a c \left (a+c x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.280052, antiderivative size = 315, 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 = {819, 823, 827, 1168, 1162, 617, 204, 1165, 628} \[ \frac{3 \left (\sqrt{a} B-A \sqrt{c}\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{64 \sqrt{2} a^{7/4} c^{7/4}}-\frac{3 \left (\sqrt{a} B-A \sqrt{c}\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{64 \sqrt{2} a^{7/4} c^{7/4}}-\frac{3 \left (\sqrt{a} B+A \sqrt{c}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{7/4} c^{7/4}}+\frac{3 \left (\sqrt{a} B+A \sqrt{c}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt{2} a^{7/4} c^{7/4}}-\frac{\sqrt{x} (A+B x)}{4 c \left (a+c x^2\right )^2}+\frac{\sqrt{x} (A+3 B x)}{16 a c \left (a+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 819

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

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

Mathematica [A]  time = 0.19621, size = 360, normalized size = 1.14 \[ \frac{-\frac{3 \sqrt{2} \sqrt [4]{a} A \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{c^{5/4}}+\frac{3 \sqrt{2} \sqrt [4]{a} A \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{c^{5/4}}-\frac{6 \sqrt{2} \sqrt [4]{a} A \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )}{c^{5/4}}+\frac{6 \sqrt{2} \sqrt [4]{a} A \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{c^{5/4}}+\frac{24 A x^{5/2}}{a+c x^2}+\frac{32 a A x^{5/2}}{\left (a+c x^2\right )^2}-\frac{12 (-a)^{3/4} B \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-a}}\right )}{c^{7/4}}+\frac{12 (-a)^{3/4} B \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-a}}\right )}{c^{7/4}}+\frac{8 B x^{7/2}}{a+c x^2}+\frac{32 a B x^{7/2}}{\left (a+c x^2\right )^2}-\frac{24 A \sqrt{x}}{c}-\frac{8 B x^{3/2}}{c}}{128 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-24*A*Sqrt[x])/c - (8*B*x^(3/2))/c + (32*a*A*x^(5/2))/(a + c*x^2)^2 + (32*a*B*x^(7/2))/(a + c*x^2)^2 + (24*A
*x^(5/2))/(a + c*x^2) + (8*B*x^(7/2))/(a + c*x^2) - (6*Sqrt[2]*a^(1/4)*A*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/
a^(1/4)])/c^(5/4) + (6*Sqrt[2]*a^(1/4)*A*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/c^(5/4) - (12*(-a)^(3/
4)*B*ArcTan[(c^(1/4)*Sqrt[x])/(-a)^(1/4)])/c^(7/4) + (12*(-a)^(3/4)*B*ArcTanh[(c^(1/4)*Sqrt[x])/(-a)^(1/4)])/c
^(7/4) - (3*Sqrt[2]*a^(1/4)*A*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/c^(5/4) + (3*Sqrt[2]
*a^(1/4)*A*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/c^(5/4))/(128*a^2)

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Maple [A]  time = 0.015, size = 334, normalized size = 1.1 \begin{align*} 2\,{\frac{1}{ \left ( c{x}^{2}+a \right ) ^{2}} \left ({\frac{3\,B{x}^{7/2}}{32\,a}}+1/32\,{\frac{A{x}^{5/2}}{a}}-1/32\,{\frac{B{x}^{3/2}}{c}}-{\frac{3\,A\sqrt{x}}{32\,c}} \right ) }+{\frac{3\,A\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{3\,A\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\,A\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{c}^{2}}\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{3\,B\sqrt{2}}{64\,a{c}^{2}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}+{\frac{3\,B\sqrt{2}}{64\,a{c}^{2}}\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^(3/2)*(B*x+A)/(c*x^2+a)^3,x)

[Out]

2*(3/32*B/a*x^(7/2)+1/32*A/a*x^(5/2)-1/32*B*x^(3/2)/c-3/32*A*x^(1/2)/c)/(c*x^2+a)^2+3/128/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)))+3/64/a^2/c
*A*(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)+1)+3/64/a^2/c*A*(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(
a/c)^(1/4)*x^(1/2)-1)+3/128/a/c^2*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)))+3/64/a/c^2*B/(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)+1)+3
/64/a/c^2*B/(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^(3/2)*(B*x+A)/(c*x^2+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.89784, size = 2025, normalized size = 6.43 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(3*(a*c^3*x^4 + 2*a^2*c^2*x^2 + a^3*c)*sqrt(-(a^3*c^3*sqrt(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a^7*c^7)
) + 2*A*B)/(a^3*c^3))*log(-27*(B^4*a^2 - A^4*c^2)*sqrt(x) + 27*(B*a^6*c^5*sqrt(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4
*c^2)/(a^7*c^7)) - A*B^2*a^3*c^2 + A^3*a^2*c^3)*sqrt(-(a^3*c^3*sqrt(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a^7*
c^7)) + 2*A*B)/(a^3*c^3))) - 3*(a*c^3*x^4 + 2*a^2*c^2*x^2 + a^3*c)*sqrt(-(a^3*c^3*sqrt(-(B^4*a^2 - 2*A^2*B^2*a
*c + A^4*c^2)/(a^7*c^7)) + 2*A*B)/(a^3*c^3))*log(-27*(B^4*a^2 - A^4*c^2)*sqrt(x) - 27*(B*a^6*c^5*sqrt(-(B^4*a^
2 - 2*A^2*B^2*a*c + A^4*c^2)/(a^7*c^7)) - A*B^2*a^3*c^2 + A^3*a^2*c^3)*sqrt(-(a^3*c^3*sqrt(-(B^4*a^2 - 2*A^2*B
^2*a*c + A^4*c^2)/(a^7*c^7)) + 2*A*B)/(a^3*c^3))) - 3*(a*c^3*x^4 + 2*a^2*c^2*x^2 + a^3*c)*sqrt((a^3*c^3*sqrt(-
(B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a^7*c^7)) - 2*A*B)/(a^3*c^3))*log(-27*(B^4*a^2 - A^4*c^2)*sqrt(x) + 27*(B
*a^6*c^5*sqrt(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a^7*c^7)) + A*B^2*a^3*c^2 - A^3*a^2*c^3)*sqrt((a^3*c^3*sqr
t(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a^7*c^7)) - 2*A*B)/(a^3*c^3))) + 3*(a*c^3*x^4 + 2*a^2*c^2*x^2 + a^3*c)
*sqrt((a^3*c^3*sqrt(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a^7*c^7)) - 2*A*B)/(a^3*c^3))*log(-27*(B^4*a^2 - A^4
*c^2)*sqrt(x) - 27*(B*a^6*c^5*sqrt(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a^7*c^7)) + A*B^2*a^3*c^2 - A^3*a^2*c
^3)*sqrt((a^3*c^3*sqrt(-(B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a^7*c^7)) - 2*A*B)/(a^3*c^3))) + 4*(3*B*c*x^3 + A
*c*x^2 - B*a*x - 3*A*a)*sqrt(x))/(a*c^3*x^4 + 2*a^2*c^2*x^2 + a^3*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**(3/2)*(B*x+A)/(c*x**2+a)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.24995, size = 390, normalized size = 1.24 \begin{align*} \frac{3 \, B c x^{\frac{7}{2}} + A c x^{\frac{5}{2}} - B a x^{\frac{3}{2}} - 3 \, A a \sqrt{x}}{16 \,{\left (c x^{2} + a\right )}^{2} a c} + \frac{3 \, \sqrt{2}{\left (\left (a c^{3}\right )^{\frac{1}{4}} A c^{2} + \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^{2} c^{4}} + \frac{3 \, \sqrt{2}{\left (\left (a c^{3}\right )^{\frac{1}{4}} A c^{2} + \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^{2} c^{4}} + \frac{3 \, \sqrt{2}{\left (\left (a c^{3}\right )^{\frac{1}{4}} A c^{2} - \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^{2} c^{4}} - \frac{3 \, \sqrt{2}{\left (\left (a c^{3}\right )^{\frac{1}{4}} A c^{2} - \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^{2} c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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