3.427 \(\int \frac{x^{5/2} (A+B x)}{(a+c x^2)^3} \, dx\)

Optimal. Leaf size=325 \[ -\frac{\left (5 \sqrt{a} B-3 A \sqrt{c}\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{64 \sqrt{2} a^{5/4} c^{9/4}}+\frac{\left (5 \sqrt{a} B-3 A \sqrt{c}\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{64 \sqrt{2} a^{5/4} c^{9/4}}-\frac{\left (5 \sqrt{a} B+3 A \sqrt{c}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{5/4} c^{9/4}}+\frac{\left (5 \sqrt{a} B+3 A \sqrt{c}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt{2} a^{5/4} c^{9/4}}-\frac{\sqrt{x} (5 a B-3 A c x)}{16 a c^2 \left (a+c x^2\right )}-\frac{x^{3/2} (A+B x)}{4 c \left (a+c x^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.282118, antiderivative size = 325, 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, 821, 827, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{\left (5 \sqrt{a} B-3 A \sqrt{c}\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{64 \sqrt{2} a^{5/4} c^{9/4}}+\frac{\left (5 \sqrt{a} B-3 A \sqrt{c}\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{64 \sqrt{2} a^{5/4} c^{9/4}}-\frac{\left (5 \sqrt{a} B+3 A \sqrt{c}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{5/4} c^{9/4}}+\frac{\left (5 \sqrt{a} B+3 A \sqrt{c}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt{2} a^{5/4} c^{9/4}}-\frac{\sqrt{x} (5 a B-3 A c x)}{16 a c^2 \left (a+c x^2\right )}-\frac{x^{3/2} (A+B x)}{4 c \left (a+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^(5/2)*(A + B*x))/(a + c*x^2)^3,x]

[Out]

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

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 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 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^{5/2} (A+B x)}{\left (a+c x^2\right )^3} \, dx &=-\frac{x^{3/2} (A+B x)}{4 c \left (a+c x^2\right )^2}+\frac{\int \frac{\sqrt{x} \left (\frac{3 a A}{2}+\frac{5 a B x}{2}\right )}{\left (a+c x^2\right )^2} \, dx}{4 a c}\\ &=-\frac{x^{3/2} (A+B x)}{4 c \left (a+c x^2\right )^2}-\frac{\sqrt{x} (5 a B-3 A c x)}{16 a c^2 \left (a+c x^2\right )}+\frac{\int \frac{\frac{5 a^2 B}{4}+\frac{3}{4} a A c x}{\sqrt{x} \left (a+c x^2\right )} \, dx}{8 a^2 c^2}\\ &=-\frac{x^{3/2} (A+B x)}{4 c \left (a+c x^2\right )^2}-\frac{\sqrt{x} (5 a B-3 A c x)}{16 a c^2 \left (a+c x^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{\frac{5 a^2 B}{4}+\frac{3}{4} a A c x^2}{a+c x^4} \, dx,x,\sqrt{x}\right )}{4 a^2 c^2}\\ &=-\frac{x^{3/2} (A+B x)}{4 c \left (a+c x^2\right )^2}-\frac{\sqrt{x} (5 a B-3 A c x)}{16 a c^2 \left (a+c x^2\right )}-\frac{\left (3 A-\frac{5 \sqrt{a} B}{\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 c^2}+\frac{\left (3 A+\frac{5 \sqrt{a} B}{\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 c^2}\\ &=-\frac{x^{3/2} (A+B x)}{4 c \left (a+c x^2\right )^2}-\frac{\sqrt{x} (5 a B-3 A c x)}{16 a c^2 \left (a+c x^2\right )}+\frac{\left (3 A+\frac{5 \sqrt{a} B}{\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 c^2}+\frac{\left (3 A+\frac{5 \sqrt{a} B}{\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 c^2}+\frac{\left (3 A-\frac{5 \sqrt{a} B}{\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^{5/4} c^{7/4}}+\frac{\left (3 A-\frac{5 \sqrt{a} B}{\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^{5/4} c^{7/4}}\\ &=-\frac{x^{3/2} (A+B x)}{4 c \left (a+c x^2\right )^2}-\frac{\sqrt{x} (5 a B-3 A c x)}{16 a c^2 \left (a+c x^2\right )}+\frac{\left (3 A-\frac{5 \sqrt{a} B}{\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^{5/4} c^{7/4}}-\frac{\left (3 A-\frac{5 \sqrt{a} B}{\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^{5/4} c^{7/4}}+\frac{\left (5 \sqrt{a} B+3 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^{5/4} c^{9/4}}-\frac{\left (5 \sqrt{a} B+3 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^{5/4} c^{9/4}}\\ &=-\frac{x^{3/2} (A+B x)}{4 c \left (a+c x^2\right )^2}-\frac{\sqrt{x} (5 a B-3 A c x)}{16 a c^2 \left (a+c x^2\right )}-\frac{\left (5 \sqrt{a} B+3 A \sqrt{c}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{5/4} c^{9/4}}+\frac{\left (5 \sqrt{a} B+3 A \sqrt{c}\right ) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{5/4} c^{9/4}}+\frac{\left (3 A-\frac{5 \sqrt{a} B}{\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^{5/4} c^{7/4}}-\frac{\left (3 A-\frac{5 \sqrt{a} B}{\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^{5/4} c^{7/4}}\\ \end{align*}

Mathematica [A]  time = 0.401082, size = 372, normalized size = 1.14 \[ \frac{-\frac{5 \sqrt{2} a^{5/4} B \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{c^{9/4}}+\frac{5 \sqrt{2} a^{5/4} B \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{c^{9/4}}-\frac{10 \sqrt{2} a^{5/4} B \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )}{c^{9/4}}+\frac{10 \sqrt{2} a^{5/4} B \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{c^{9/4}}-\frac{12 (-a)^{3/4} A \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-a}}\right )}{c^{7/4}}+\frac{12 (-a)^{3/4} A \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-a}}\right )}{c^{7/4}}+\frac{8 A x^{7/2}}{a+c x^2}+\frac{32 a A x^{7/2}}{\left (a+c x^2\right )^2}-\frac{40 a B \sqrt{x}}{c^2}-\frac{8 B x^{9/2}}{a+c x^2}+\frac{32 a B x^{9/2}}{\left (a+c x^2\right )^2}-\frac{8 A x^{3/2}}{c}+\frac{8 B x^{5/2}}{c}}{128 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(5/2)*(A + B*x))/(a + c*x^2)^3,x]

[Out]

((-40*a*B*Sqrt[x])/c^2 - (8*A*x^(3/2))/c + (8*B*x^(5/2))/c + (32*a*A*x^(7/2))/(a + c*x^2)^2 + (32*a*B*x^(9/2))
/(a + c*x^2)^2 + (8*A*x^(7/2))/(a + c*x^2) - (8*B*x^(9/2))/(a + c*x^2) - (10*Sqrt[2]*a^(5/4)*B*ArcTan[1 - (Sqr
t[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/c^(9/4) + (10*Sqrt[2]*a^(5/4)*B*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])
/c^(9/4) - (12*(-a)^(3/4)*A*ArcTan[(c^(1/4)*Sqrt[x])/(-a)^(1/4)])/c^(7/4) + (12*(-a)^(3/4)*A*ArcTanh[(c^(1/4)*
Sqrt[x])/(-a)^(1/4)])/c^(7/4) - (5*Sqrt[2]*a^(5/4)*B*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x
])/c^(9/4) + (5*Sqrt[2]*a^(5/4)*B*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/c^(9/4))/(128*a^
2)

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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{3\,A{x}^{7/2}}{32\,a}}-{\frac{9\,B{x}^{5/2}}{32\,c}}-1/32\,{\frac{A{x}^{3/2}}{c}}-{\frac{5\,aB\sqrt{x}}{32\,{c}^{2}}} \right ) }+{\frac{5\,B\sqrt{2}}{64\,a{c}^{2}}\sqrt [4]{{\frac{a}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}-1 \right ) }+{\frac{5\,B\sqrt{2}}{128\,a{c}^{2}}\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\,B\sqrt{2}}{64\,a{c}^{2}}\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}}{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\,A\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\,A\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^(5/2)*(B*x+A)/(c*x^2+a)^3,x)

[Out]

2*(3/32*A/a*x^(7/2)-9/32*B*x^(5/2)/c-1/32*A*x^(3/2)/c-5/32*a*B*x^(1/2)/c^2)/(c*x^2+a)^2+5/64/a/c^2*B*(a/c)^(1/
4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)-1)+5/128/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)))+5/64/a/c^2*B*(a/c)^(1/4)*2^(1/2)*arctan(2^(1/
2)/(a/c)^(1/4)*x^(1/2)+1)+3/128/a/c^2*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/c^2*A/(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)+
1)+3/64/a/c^2*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^(5/2)*(B*x+A)/(c*x^2+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.76797, size = 2253, normalized size = 6.93 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)/(c*x^2+a)^3,x, algorithm="fricas")

[Out]

-1/64*((a*c^4*x^4 + 2*a^2*c^3*x^2 + a^3*c^2)*sqrt(-(a^2*c^4*sqrt(-(625*B^4*a^2 - 450*A^2*B^2*a*c + 81*A^4*c^2)
/(a^5*c^9)) + 30*A*B)/(a^2*c^4))*log(-(625*B^4*a^2 - 81*A^4*c^2)*sqrt(x) + (3*A*a^4*c^7*sqrt(-(625*B^4*a^2 - 4
50*A^2*B^2*a*c + 81*A^4*c^2)/(a^5*c^9)) + 125*B^3*a^3*c^2 - 45*A^2*B*a^2*c^3)*sqrt(-(a^2*c^4*sqrt(-(625*B^4*a^
2 - 450*A^2*B^2*a*c + 81*A^4*c^2)/(a^5*c^9)) + 30*A*B)/(a^2*c^4))) - (a*c^4*x^4 + 2*a^2*c^3*x^2 + a^3*c^2)*sqr
t(-(a^2*c^4*sqrt(-(625*B^4*a^2 - 450*A^2*B^2*a*c + 81*A^4*c^2)/(a^5*c^9)) + 30*A*B)/(a^2*c^4))*log(-(625*B^4*a
^2 - 81*A^4*c^2)*sqrt(x) - (3*A*a^4*c^7*sqrt(-(625*B^4*a^2 - 450*A^2*B^2*a*c + 81*A^4*c^2)/(a^5*c^9)) + 125*B^
3*a^3*c^2 - 45*A^2*B*a^2*c^3)*sqrt(-(a^2*c^4*sqrt(-(625*B^4*a^2 - 450*A^2*B^2*a*c + 81*A^4*c^2)/(a^5*c^9)) + 3
0*A*B)/(a^2*c^4))) - (a*c^4*x^4 + 2*a^2*c^3*x^2 + a^3*c^2)*sqrt((a^2*c^4*sqrt(-(625*B^4*a^2 - 450*A^2*B^2*a*c
+ 81*A^4*c^2)/(a^5*c^9)) - 30*A*B)/(a^2*c^4))*log(-(625*B^4*a^2 - 81*A^4*c^2)*sqrt(x) + (3*A*a^4*c^7*sqrt(-(62
5*B^4*a^2 - 450*A^2*B^2*a*c + 81*A^4*c^2)/(a^5*c^9)) - 125*B^3*a^3*c^2 + 45*A^2*B*a^2*c^3)*sqrt((a^2*c^4*sqrt(
-(625*B^4*a^2 - 450*A^2*B^2*a*c + 81*A^4*c^2)/(a^5*c^9)) - 30*A*B)/(a^2*c^4))) + (a*c^4*x^4 + 2*a^2*c^3*x^2 +
a^3*c^2)*sqrt((a^2*c^4*sqrt(-(625*B^4*a^2 - 450*A^2*B^2*a*c + 81*A^4*c^2)/(a^5*c^9)) - 30*A*B)/(a^2*c^4))*log(
-(625*B^4*a^2 - 81*A^4*c^2)*sqrt(x) - (3*A*a^4*c^7*sqrt(-(625*B^4*a^2 - 450*A^2*B^2*a*c + 81*A^4*c^2)/(a^5*c^9
)) - 125*B^3*a^3*c^2 + 45*A^2*B*a^2*c^3)*sqrt((a^2*c^4*sqrt(-(625*B^4*a^2 - 450*A^2*B^2*a*c + 81*A^4*c^2)/(a^5
*c^9)) - 30*A*B)/(a^2*c^4))) - 4*(3*A*c^2*x^3 - 9*B*a*c*x^2 - A*a*c*x - 5*B*a^2)*sqrt(x))/(a*c^4*x^4 + 2*a^2*c
^3*x^2 + a^3*c^2)

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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**(5/2)*(B*x+A)/(c*x**2+a)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.36526, size = 416, normalized size = 1.28 \begin{align*} \frac{\sqrt{2}{\left (5 \, \left (a c^{3}\right )^{\frac{1}{4}} B a c + 3 \, \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^{2} c^{4}} - \frac{\sqrt{2}{\left (5 \, \left (a c^{3}\right )^{\frac{1}{4}} B a c - 3 \, \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^{2} c^{4}} + \frac{3 \, A c^{2} x^{\frac{7}{2}} - 9 \, B a c x^{\frac{5}{2}} - A a c x^{\frac{3}{2}} - 5 \, B a^{2} \sqrt{x}}{16 \,{\left (c x^{2} + a\right )}^{2} a c^{2}} + \frac{\sqrt{2}{\left (5 \, \left (a c^{3}\right )^{\frac{1}{4}} B a c^{3} + 3 \, \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^{2} c^{6}} + \frac{\sqrt{2}{\left (5 \, \left (a c^{3}\right )^{\frac{1}{4}} B a c^{3} - 3 \, \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^{2} c^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)/(c*x^2+a)^3,x, algorithm="giac")

[Out]

1/64*sqrt(2)*(5*(a*c^3)^(1/4)*B*a*c + 3*(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^2*c^4) - 1/128*sqrt(2)*(5*(a*c^3)^(1/4)*B*a*c - 3*(a*c^3)^(3/4)*A)*log(-sqrt(2)*sqrt(x)*(a/c)
^(1/4) + x + sqrt(a/c))/(a^2*c^4) + 1/16*(3*A*c^2*x^(7/2) - 9*B*a*c*x^(5/2) - A*a*c*x^(3/2) - 5*B*a^2*sqrt(x))
/((c*x^2 + a)^2*a*c^2) + 1/64*sqrt(2)*(5*(a*c^3)^(1/4)*B*a*c^3 + 3*(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^2*c^6) + 1/128*sqrt(2)*(5*(a*c^3)^(1/4)*B*a*c^3 - 3*(a*c^3)^(3/
4)*A*c^2)*log(sqrt(2)*sqrt(x)*(a/c)^(1/4) + x + sqrt(a/c))/(a^2*c^6)