3.419 \(\int \frac{A+B x}{x^{9/2} (a+c x^2)} \, dx\)

Optimal. Leaf size=306 \[ \frac{c^{5/4} \left (\sqrt{a} B-A \sqrt{c}\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{2 \sqrt{2} a^{11/4}}-\frac{c^{5/4} \left (\sqrt{a} B-A \sqrt{c}\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{2 \sqrt{2} a^{11/4}}-\frac{c^{5/4} \left (\sqrt{a} B+A \sqrt{c}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} a^{11/4}}+\frac{c^{5/4} \left (\sqrt{a} B+A \sqrt{c}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{\sqrt{2} a^{11/4}}+\frac{2 A c}{3 a^2 x^{3/2}}+\frac{2 B c}{a^2 \sqrt{x}}-\frac{2 A}{7 a x^{7/2}}-\frac{2 B}{5 a x^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.368087, antiderivative size = 306, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {829, 827, 1168, 1162, 617, 204, 1165, 628} \[ \frac{c^{5/4} \left (\sqrt{a} B-A \sqrt{c}\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{2 \sqrt{2} a^{11/4}}-\frac{c^{5/4} \left (\sqrt{a} B-A \sqrt{c}\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{2 \sqrt{2} a^{11/4}}-\frac{c^{5/4} \left (\sqrt{a} B+A \sqrt{c}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} a^{11/4}}+\frac{c^{5/4} \left (\sqrt{a} B+A \sqrt{c}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{\sqrt{2} a^{11/4}}+\frac{2 A c}{3 a^2 x^{3/2}}+\frac{2 B c}{a^2 \sqrt{x}}-\frac{2 A}{7 a x^{7/2}}-\frac{2 B}{5 a x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(9/2)*(a + c*x^2)),x]

[Out]

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

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^{9/2} \left (a+c x^2\right )} \, dx &=-\frac{2 A}{7 a x^{7/2}}+\frac{\int \frac{a B-A c x}{x^{7/2} \left (a+c x^2\right )} \, dx}{a}\\ &=-\frac{2 A}{7 a x^{7/2}}-\frac{2 B}{5 a x^{5/2}}+\frac{\int \frac{-a A c-a B c x}{x^{5/2} \left (a+c x^2\right )} \, dx}{a^2}\\ &=-\frac{2 A}{7 a x^{7/2}}-\frac{2 B}{5 a x^{5/2}}+\frac{2 A c}{3 a^2 x^{3/2}}+\frac{\int \frac{-a^2 B c+a A c^2 x}{x^{3/2} \left (a+c x^2\right )} \, dx}{a^3}\\ &=-\frac{2 A}{7 a x^{7/2}}-\frac{2 B}{5 a x^{5/2}}+\frac{2 A c}{3 a^2 x^{3/2}}+\frac{2 B c}{a^2 \sqrt{x}}+\frac{\int \frac{a^2 A c^2+a^2 B c^2 x}{\sqrt{x} \left (a+c x^2\right )} \, dx}{a^4}\\ &=-\frac{2 A}{7 a x^{7/2}}-\frac{2 B}{5 a x^{5/2}}+\frac{2 A c}{3 a^2 x^{3/2}}+\frac{2 B c}{a^2 \sqrt{x}}+\frac{2 \operatorname{Subst}\left (\int \frac{a^2 A c^2+a^2 B c^2 x^2}{a+c x^4} \, dx,x,\sqrt{x}\right )}{a^4}\\ &=-\frac{2 A}{7 a x^{7/2}}-\frac{2 B}{5 a x^{5/2}}+\frac{2 A c}{3 a^2 x^{3/2}}+\frac{2 B c}{a^2 \sqrt{x}}-\frac{\left (\left (\sqrt{a} B-A \sqrt{c}\right ) c\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a} \sqrt{c}-c x^2}{a+c x^4} \, dx,x,\sqrt{x}\right )}{a^{5/2}}+\frac{\left (\left (\sqrt{a} B+A \sqrt{c}\right ) c\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a} \sqrt{c}+c x^2}{a+c x^4} \, dx,x,\sqrt{x}\right )}{a^{5/2}}\\ &=-\frac{2 A}{7 a x^{7/2}}-\frac{2 B}{5 a x^{5/2}}+\frac{2 A c}{3 a^2 x^{3/2}}+\frac{2 B c}{a^2 \sqrt{x}}+\frac{\left (\left (\sqrt{a} B+A \sqrt{c}\right ) 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 )}{2 a^{5/2}}+\frac{\left (\left (\sqrt{a} B+A \sqrt{c}\right ) 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 )}{2 a^{5/2}}+\frac{\left (\left (\sqrt{a} B-A \sqrt{c}\right ) c^{5/4}\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 )}{2 \sqrt{2} a^{11/4}}+\frac{\left (\left (\sqrt{a} B-A \sqrt{c}\right ) c^{5/4}\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 )}{2 \sqrt{2} a^{11/4}}\\ &=-\frac{2 A}{7 a x^{7/2}}-\frac{2 B}{5 a x^{5/2}}+\frac{2 A c}{3 a^2 x^{3/2}}+\frac{2 B c}{a^2 \sqrt{x}}+\frac{\left (\sqrt{a} B-A \sqrt{c}\right ) c^{5/4} \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{2 \sqrt{2} a^{11/4}}-\frac{\left (\sqrt{a} B-A \sqrt{c}\right ) c^{5/4} \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{2 \sqrt{2} a^{11/4}}+\frac{\left (\left (\sqrt{a} B+A \sqrt{c}\right ) c^{5/4}\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 )}{\sqrt{2} a^{11/4}}-\frac{\left (\left (\sqrt{a} B+A \sqrt{c}\right ) c^{5/4}\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 )}{\sqrt{2} a^{11/4}}\\ &=-\frac{2 A}{7 a x^{7/2}}-\frac{2 B}{5 a x^{5/2}}+\frac{2 A c}{3 a^2 x^{3/2}}+\frac{2 B c}{a^2 \sqrt{x}}-\frac{\left (\sqrt{a} B+A \sqrt{c}\right ) c^{5/4} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} a^{11/4}}+\frac{\left (\sqrt{a} B+A \sqrt{c}\right ) c^{5/4} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} a^{11/4}}+\frac{\left (\sqrt{a} B-A \sqrt{c}\right ) c^{5/4} \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{2 \sqrt{2} a^{11/4}}-\frac{\left (\sqrt{a} B-A \sqrt{c}\right ) c^{5/4} \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{2 \sqrt{2} a^{11/4}}\\ \end{align*}

Mathematica [C]  time = 0.0141569, size = 54, normalized size = 0.18 \[ -\frac{2 \left (5 A \, _2F_1\left (-\frac{7}{4},1;-\frac{3}{4};-\frac{c x^2}{a}\right )+7 B x \, _2F_1\left (-\frac{5}{4},1;-\frac{1}{4};-\frac{c x^2}{a}\right )\right )}{35 a x^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(9/2)*(a + c*x^2)),x]

[Out]

(-2*(5*A*Hypergeometric2F1[-7/4, 1, -3/4, -((c*x^2)/a)] + 7*B*x*Hypergeometric2F1[-5/4, 1, -1/4, -((c*x^2)/a)]
))/(35*a*x^(7/2))

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Maple [A]  time = 0.011, size = 318, normalized size = 1. \begin{align*} -{\frac{2\,A}{7\,a}{x}^{-{\frac{7}{2}}}}-{\frac{2\,B}{5\,a}{x}^{-{\frac{5}{2}}}}+{\frac{2\,Ac}{3\,{a}^{2}}{x}^{-{\frac{3}{2}}}}+2\,{\frac{Bc}{{a}^{2}\sqrt{x}}}+{\frac{A{c}^{2}\sqrt{2}}{4\,{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{A{c}^{2}\sqrt{2}}{2\,{a}^{3}}\sqrt [4]{{\frac{a}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}+1 \right ) }+{\frac{A{c}^{2}\sqrt{2}}{2\,{a}^{3}}\sqrt [4]{{\frac{a}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}-1 \right ) }+{\frac{Bc\sqrt{2}}{4\,{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{Bc\sqrt{2}}{2\,{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{Bc\sqrt{2}}{2\,{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^(9/2)/(c*x^2+a),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.63339, size = 1782, normalized size = 5.82 \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^(9/2)/(c*x^2+a),x, algorithm="fricas")

[Out]

1/210*(105*a^2*x^4*sqrt(-(a^5*sqrt(-(B^4*a^2*c^5 - 2*A^2*B^2*a*c^6 + A^4*c^7)/a^11) + 2*A*B*c^3)/a^5)*log(-(B^
4*a^2*c^4 - A^4*c^6)*sqrt(x) + (B*a^9*sqrt(-(B^4*a^2*c^5 - 2*A^2*B^2*a*c^6 + A^4*c^7)/a^11) - A*B^2*a^4*c^3 +
A^3*a^3*c^4)*sqrt(-(a^5*sqrt(-(B^4*a^2*c^5 - 2*A^2*B^2*a*c^6 + A^4*c^7)/a^11) + 2*A*B*c^3)/a^5)) - 105*a^2*x^4
*sqrt(-(a^5*sqrt(-(B^4*a^2*c^5 - 2*A^2*B^2*a*c^6 + A^4*c^7)/a^11) + 2*A*B*c^3)/a^5)*log(-(B^4*a^2*c^4 - A^4*c^
6)*sqrt(x) - (B*a^9*sqrt(-(B^4*a^2*c^5 - 2*A^2*B^2*a*c^6 + A^4*c^7)/a^11) - A*B^2*a^4*c^3 + A^3*a^3*c^4)*sqrt(
-(a^5*sqrt(-(B^4*a^2*c^5 - 2*A^2*B^2*a*c^6 + A^4*c^7)/a^11) + 2*A*B*c^3)/a^5)) - 105*a^2*x^4*sqrt((a^5*sqrt(-(
B^4*a^2*c^5 - 2*A^2*B^2*a*c^6 + A^4*c^7)/a^11) - 2*A*B*c^3)/a^5)*log(-(B^4*a^2*c^4 - A^4*c^6)*sqrt(x) + (B*a^9
*sqrt(-(B^4*a^2*c^5 - 2*A^2*B^2*a*c^6 + A^4*c^7)/a^11) + A*B^2*a^4*c^3 - A^3*a^3*c^4)*sqrt((a^5*sqrt(-(B^4*a^2
*c^5 - 2*A^2*B^2*a*c^6 + A^4*c^7)/a^11) - 2*A*B*c^3)/a^5)) + 105*a^2*x^4*sqrt((a^5*sqrt(-(B^4*a^2*c^5 - 2*A^2*
B^2*a*c^6 + A^4*c^7)/a^11) - 2*A*B*c^3)/a^5)*log(-(B^4*a^2*c^4 - A^4*c^6)*sqrt(x) - (B*a^9*sqrt(-(B^4*a^2*c^5
- 2*A^2*B^2*a*c^6 + A^4*c^7)/a^11) + A*B^2*a^4*c^3 - A^3*a^3*c^4)*sqrt((a^5*sqrt(-(B^4*a^2*c^5 - 2*A^2*B^2*a*c
^6 + A^4*c^7)/a^11) - 2*A*B*c^3)/a^5)) + 4*(105*B*c*x^3 + 35*A*c*x^2 - 21*B*a*x - 15*A*a)*sqrt(x))/(a^2*x^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**(9/2)/(c*x**2+a),x)

[Out]

Timed out

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Giac [A]  time = 1.3126, size = 373, normalized size = 1.22 \begin{align*} \frac{\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 )}{2 \, a^{3} c} + \frac{\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 )}{2 \, a^{3} c} + \frac{\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 )}{4 \, a^{3} c} - \frac{\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 )}{4 \, a^{3} c} + \frac{2 \,{\left (105 \, B c x^{3} + 35 \, A c x^{2} - 21 \, B a x - 15 \, A a\right )}}{105 \, a^{2} x^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(9/2)/(c*x^2+a),x, algorithm="giac")

[Out]

1/2*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^3*c) + 1/2*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^3*c) + 1/4*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^3*c) - 1/4*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^3*c) + 2/105*(105*B*c*x^3 + 35*A*c*x^2 - 21*B*a*x - 15*A*a)/(a^2*x^(7/2))