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

Optimal. Leaf size=317 \[ -\frac{\sqrt [4]{c} \left (5 \sqrt{a} B-7 A \sqrt{c}\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{8 \sqrt{2} a^{11/4}}+\frac{\sqrt [4]{c} \left (5 \sqrt{a} B-7 A \sqrt{c}\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{8 \sqrt{2} a^{11/4}}+\frac{\sqrt [4]{c} \left (5 \sqrt{a} B+7 A \sqrt{c}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} a^{11/4}}-\frac{\sqrt [4]{c} \left (5 \sqrt{a} B+7 A \sqrt{c}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt{2} a^{11/4}}-\frac{7 A}{6 a^2 x^{3/2}}-\frac{5 B}{2 a^2 \sqrt{x}}+\frac{A+B x}{2 a x^{3/2} \left (a+c x^2\right )} \]

[Out]

(-7*A)/(6*a^2*x^(3/2)) - (5*B)/(2*a^2*Sqrt[x]) + (A + B*x)/(2*a*x^(3/2)*(a + c*x^2)) + ((5*Sqrt[a]*B + 7*A*Sqr
t[c])*c^(1/4)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(11/4)) - ((5*Sqrt[a]*B + 7*A*Sqrt[c
])*c^(1/4)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(11/4)) - ((5*Sqrt[a]*B - 7*A*Sqrt[c])*
c^(1/4)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*a^(11/4)) + ((5*Sqrt[a]*B - 7*A
*Sqrt[c])*c^(1/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*a^(11/4))

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Rubi [A]  time = 0.349584, antiderivative size = 317, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.45, Rules used = {823, 829, 827, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{\sqrt [4]{c} \left (5 \sqrt{a} B-7 A \sqrt{c}\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{8 \sqrt{2} a^{11/4}}+\frac{\sqrt [4]{c} \left (5 \sqrt{a} B-7 A \sqrt{c}\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{8 \sqrt{2} a^{11/4}}+\frac{\sqrt [4]{c} \left (5 \sqrt{a} B+7 A \sqrt{c}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} a^{11/4}}-\frac{\sqrt [4]{c} \left (5 \sqrt{a} B+7 A \sqrt{c}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt{2} a^{11/4}}-\frac{7 A}{6 a^2 x^{3/2}}-\frac{5 B}{2 a^2 \sqrt{x}}+\frac{A+B x}{2 a x^{3/2} \left (a+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*A)/(6*a^2*x^(3/2)) - (5*B)/(2*a^2*Sqrt[x]) + (A + B*x)/(2*a*x^(3/2)*(a + c*x^2)) + ((5*Sqrt[a]*B + 7*A*Sqr
t[c])*c^(1/4)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(11/4)) - ((5*Sqrt[a]*B + 7*A*Sqrt[c
])*c^(1/4)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(11/4)) - ((5*Sqrt[a]*B - 7*A*Sqrt[c])*
c^(1/4)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*a^(11/4)) + ((5*Sqrt[a]*B - 7*A
*Sqrt[c])*c^(1/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*a^(11/4))

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 829

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

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}{x^{5/2} \left (a+c x^2\right )^2} \, dx &=\frac{A+B x}{2 a x^{3/2} \left (a+c x^2\right )}-\frac{\int \frac{-\frac{7}{2} a A c-\frac{5}{2} a B c x}{x^{5/2} \left (a+c x^2\right )} \, dx}{2 a^2 c}\\ &=-\frac{7 A}{6 a^2 x^{3/2}}+\frac{A+B x}{2 a x^{3/2} \left (a+c x^2\right )}-\frac{\int \frac{-\frac{5}{2} a^2 B c+\frac{7}{2} a A c^2 x}{x^{3/2} \left (a+c x^2\right )} \, dx}{2 a^3 c}\\ &=-\frac{7 A}{6 a^2 x^{3/2}}-\frac{5 B}{2 a^2 \sqrt{x}}+\frac{A+B x}{2 a x^{3/2} \left (a+c x^2\right )}-\frac{\int \frac{\frac{7}{2} a^2 A c^2+\frac{5}{2} a^2 B c^2 x}{\sqrt{x} \left (a+c x^2\right )} \, dx}{2 a^4 c}\\ &=-\frac{7 A}{6 a^2 x^{3/2}}-\frac{5 B}{2 a^2 \sqrt{x}}+\frac{A+B x}{2 a x^{3/2} \left (a+c x^2\right )}-\frac{\operatorname{Subst}\left (\int \frac{\frac{7}{2} a^2 A c^2+\frac{5}{2} a^2 B c^2 x^2}{a+c x^4} \, dx,x,\sqrt{x}\right )}{a^4 c}\\ &=-\frac{7 A}{6 a^2 x^{3/2}}-\frac{5 B}{2 a^2 \sqrt{x}}+\frac{A+B x}{2 a x^{3/2} \left (a+c x^2\right )}+\frac{\left (5 \sqrt{a} B-7 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 )}{4 a^{5/2}}-\frac{\left (5 \sqrt{a} B+7 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 )}{4 a^{5/2}}\\ &=-\frac{7 A}{6 a^2 x^{3/2}}-\frac{5 B}{2 a^2 \sqrt{x}}+\frac{A+B x}{2 a x^{3/2} \left (a+c x^2\right )}-\frac{\left (5 \sqrt{a} B+7 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 )}{8 a^{5/2}}-\frac{\left (5 \sqrt{a} B+7 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 )}{8 a^{5/2}}-\frac{\left (\left (5 \sqrt{a} B-7 A \sqrt{c}\right ) \sqrt [4]{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 )}{8 \sqrt{2} a^{11/4}}-\frac{\left (\left (5 \sqrt{a} B-7 A \sqrt{c}\right ) \sqrt [4]{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 )}{8 \sqrt{2} a^{11/4}}\\ &=-\frac{7 A}{6 a^2 x^{3/2}}-\frac{5 B}{2 a^2 \sqrt{x}}+\frac{A+B x}{2 a x^{3/2} \left (a+c x^2\right )}-\frac{\left (5 \sqrt{a} B-7 A \sqrt{c}\right ) \sqrt [4]{c} \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} a^{11/4}}+\frac{\left (5 \sqrt{a} B-7 A \sqrt{c}\right ) \sqrt [4]{c} \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} a^{11/4}}-\frac{\left (\left (5 \sqrt{a} B+7 A \sqrt{c}\right ) \sqrt [4]{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 )}{4 \sqrt{2} a^{11/4}}+\frac{\left (\left (5 \sqrt{a} B+7 A \sqrt{c}\right ) \sqrt [4]{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 )}{4 \sqrt{2} a^{11/4}}\\ &=-\frac{7 A}{6 a^2 x^{3/2}}-\frac{5 B}{2 a^2 \sqrt{x}}+\frac{A+B x}{2 a x^{3/2} \left (a+c x^2\right )}+\frac{\left (5 \sqrt{a} B+7 A \sqrt{c}\right ) \sqrt [4]{c} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} a^{11/4}}-\frac{\left (5 \sqrt{a} B+7 A \sqrt{c}\right ) \sqrt [4]{c} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} a^{11/4}}-\frac{\left (5 \sqrt{a} B-7 A \sqrt{c}\right ) \sqrt [4]{c} \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} a^{11/4}}+\frac{\left (5 \sqrt{a} B-7 A \sqrt{c}\right ) \sqrt [4]{c} \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} a^{11/4}}\\ \end{align*}

Mathematica [C]  time = 0.0340898, size = 85, normalized size = 0.27 \[ \frac{3 a (A+B x)-7 A \left (a+c x^2\right ) \, _2F_1\left (-\frac{3}{4},1;\frac{1}{4};-\frac{c x^2}{a}\right )-15 B x \left (a+c x^2\right ) \, _2F_1\left (-\frac{1}{4},1;\frac{3}{4};-\frac{c x^2}{a}\right )}{6 a^2 x^{3/2} \left (a+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*(A + B*x) - 7*A*(a + c*x^2)*Hypergeometric2F1[-3/4, 1, 1/4, -((c*x^2)/a)] - 15*B*x*(a + c*x^2)*Hypergeome
tric2F1[-1/4, 1, 3/4, -((c*x^2)/a)])/(6*a^2*x^(3/2)*(a + c*x^2))

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Maple [A]  time = 0.016, size = 327, normalized size = 1. \begin{align*} -{\frac{2\,A}{3\,{a}^{2}}{x}^{-{\frac{3}{2}}}}-2\,{\frac{B}{{a}^{2}\sqrt{x}}}-{\frac{Bc}{2\,{a}^{2} \left ( c{x}^{2}+a \right ) }{x}^{{\frac{3}{2}}}}-{\frac{Ac}{2\,{a}^{2} \left ( c{x}^{2}+a \right ) }\sqrt{x}}-{\frac{7\,Ac\sqrt{2}}{16\,{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{7\,Ac\sqrt{2}}{8\,{a}^{3}}\sqrt [4]{{\frac{a}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}+1 \right ) }-{\frac{7\,Ac\sqrt{2}}{8\,{a}^{3}}\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}}{16\,{a}^{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{5\,B\sqrt{2}}{8\,{a}^{2}}\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}}{8\,{a}^{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((B*x+A)/x^(5/2)/(c*x^2+a)^2,x)

[Out]

-2/3*A/a^2/x^(3/2)-2*B/a^2/x^(1/2)-1/2/a^2*c/(c*x^2+a)*B*x^(3/2)-1/2/a^2*c/(c*x^2+a)*A*x^(1/2)-7/16/a^3*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)))-
7/8/a^3*c*A*(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)+1)-7/8/a^3*c*A*(a/c)^(1/4)*2^(1/2)*arctan(2
^(1/2)/(a/c)^(1/4)*x^(1/2)-1)-5/16/a^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)))-5/8/a^2*B/(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)+1)
-5/8/a^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((B*x+A)/x^(5/2)/(c*x^2+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.61056, size = 2091, normalized size = 6.6 \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^(5/2)/(c*x^2+a)^2,x, algorithm="fricas")

[Out]

-1/24*(3*(a^2*c*x^4 + a^3*x^2)*sqrt(-(a^5*sqrt(-(625*B^4*a^2*c - 2450*A^2*B^2*a*c^2 + 2401*A^4*c^3)/a^11) + 70
*A*B*c)/a^5)*log(-(625*B^4*a^2*c - 2401*A^4*c^3)*sqrt(x) + (5*B*a^9*sqrt(-(625*B^4*a^2*c - 2450*A^2*B^2*a*c^2
+ 2401*A^4*c^3)/a^11) - 175*A*B^2*a^4*c + 343*A^3*a^3*c^2)*sqrt(-(a^5*sqrt(-(625*B^4*a^2*c - 2450*A^2*B^2*a*c^
2 + 2401*A^4*c^3)/a^11) + 70*A*B*c)/a^5)) - 3*(a^2*c*x^4 + a^3*x^2)*sqrt(-(a^5*sqrt(-(625*B^4*a^2*c - 2450*A^2
*B^2*a*c^2 + 2401*A^4*c^3)/a^11) + 70*A*B*c)/a^5)*log(-(625*B^4*a^2*c - 2401*A^4*c^3)*sqrt(x) - (5*B*a^9*sqrt(
-(625*B^4*a^2*c - 2450*A^2*B^2*a*c^2 + 2401*A^4*c^3)/a^11) - 175*A*B^2*a^4*c + 343*A^3*a^3*c^2)*sqrt(-(a^5*sqr
t(-(625*B^4*a^2*c - 2450*A^2*B^2*a*c^2 + 2401*A^4*c^3)/a^11) + 70*A*B*c)/a^5)) - 3*(a^2*c*x^4 + a^3*x^2)*sqrt(
(a^5*sqrt(-(625*B^4*a^2*c - 2450*A^2*B^2*a*c^2 + 2401*A^4*c^3)/a^11) - 70*A*B*c)/a^5)*log(-(625*B^4*a^2*c - 24
01*A^4*c^3)*sqrt(x) + (5*B*a^9*sqrt(-(625*B^4*a^2*c - 2450*A^2*B^2*a*c^2 + 2401*A^4*c^3)/a^11) + 175*A*B^2*a^4
*c - 343*A^3*a^3*c^2)*sqrt((a^5*sqrt(-(625*B^4*a^2*c - 2450*A^2*B^2*a*c^2 + 2401*A^4*c^3)/a^11) - 70*A*B*c)/a^
5)) + 3*(a^2*c*x^4 + a^3*x^2)*sqrt((a^5*sqrt(-(625*B^4*a^2*c - 2450*A^2*B^2*a*c^2 + 2401*A^4*c^3)/a^11) - 70*A
*B*c)/a^5)*log(-(625*B^4*a^2*c - 2401*A^4*c^3)*sqrt(x) - (5*B*a^9*sqrt(-(625*B^4*a^2*c - 2450*A^2*B^2*a*c^2 +
2401*A^4*c^3)/a^11) + 175*A*B^2*a^4*c - 343*A^3*a^3*c^2)*sqrt((a^5*sqrt(-(625*B^4*a^2*c - 2450*A^2*B^2*a*c^2 +
2401*A^4*c^3)/a^11) - 70*A*B*c)/a^5)) + 4*(15*B*c*x^3 + 7*A*c*x^2 + 12*B*a*x + 4*A*a)*sqrt(x))/(a^2*c*x^4 + a
^3*x^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((B*x+A)/x**(5/2)/(c*x**2+a)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.39735, size = 393, normalized size = 1.24 \begin{align*} -\frac{B c x^{\frac{3}{2}} + A c \sqrt{x}}{2 \,{\left (c x^{2} + a\right )} a^{2}} - \frac{2 \,{\left (3 \, B x + A\right )}}{3 \, a^{2} x^{\frac{3}{2}}} - \frac{\sqrt{2}{\left (7 \, \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 )}{8 \, a^{3} c^{2}} - \frac{\sqrt{2}{\left (7 \, \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 )}{8 \, a^{3} c^{2}} - \frac{\sqrt{2}{\left (7 \, \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 )}{16 \, a^{3} c^{2}} + \frac{\sqrt{2}{\left (7 \, \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 )}{16 \, a^{3} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(B*c*x^(3/2) + A*c*sqrt(x))/((c*x^2 + a)*a^2) - 2/3*(3*B*x + A)/(a^2*x^(3/2)) - 1/8*sqrt(2)*(7*(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^2) - 1
/8*sqrt(2)*(7*(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^2) - 1/16*sqrt(2)*(7*(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^2) + 1/16*sqrt(2)*(7*(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^2)