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

Optimal. Leaf size=292 \[ -\frac{a^{3/4} \left (\sqrt{a} B+A \sqrt{c}\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{2 \sqrt{2} c^{9/4}}+\frac{a^{3/4} \left (\sqrt{a} B+A \sqrt{c}\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{2 \sqrt{2} c^{9/4}}-\frac{a^{3/4} \left (\sqrt{a} B-A \sqrt{c}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} c^{9/4}}+\frac{a^{3/4} \left (\sqrt{a} B-A \sqrt{c}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{\sqrt{2} c^{9/4}}-\frac{2 a B \sqrt{x}}{c^2}+\frac{2 A x^{3/2}}{3 c}+\frac{2 B x^{5/2}}{5 c} \]

[Out]

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

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Rubi [A]  time = 0.344584, antiderivative size = 292, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {825, 827, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{a^{3/4} \left (\sqrt{a} B+A \sqrt{c}\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{2 \sqrt{2} c^{9/4}}+\frac{a^{3/4} \left (\sqrt{a} B+A \sqrt{c}\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )}{2 \sqrt{2} c^{9/4}}-\frac{a^{3/4} \left (\sqrt{a} B-A \sqrt{c}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} c^{9/4}}+\frac{a^{3/4} \left (\sqrt{a} B-A \sqrt{c}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{\sqrt{2} c^{9/4}}-\frac{2 a B \sqrt{x}}{c^2}+\frac{2 A x^{3/2}}{3 c}+\frac{2 B x^{5/2}}{5 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 825

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

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

Mathematica [A]  time = 0.102424, size = 290, normalized size = 0.99 \[ \frac{-15 \sqrt{2} a^{5/4} B \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )+15 \sqrt{2} a^{5/4} B \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} \sqrt{x}+\sqrt{a}+\sqrt{c} x\right )-30 \sqrt{2} a^{5/4} B \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )+30 \sqrt{2} a^{5/4} B \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}+1\right )+60 (-a)^{3/4} A \sqrt{c} \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-a}}\right )-60 (-a)^{3/4} A \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-a}}\right )-120 a B \sqrt [4]{c} \sqrt{x}+40 A c^{5/4} x^{3/2}+24 B c^{5/4} x^{5/2}}{60 c^{9/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-120*a*B*c^(1/4)*Sqrt[x] + 40*A*c^(5/4)*x^(3/2) + 24*B*c^(5/4)*x^(5/2) - 30*Sqrt[2]*a^(5/4)*B*ArcTan[1 - (Sqr
t[2]*c^(1/4)*Sqrt[x])/a^(1/4)] + 30*Sqrt[2]*a^(5/4)*B*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)] + 60*(-a)^
(3/4)*A*Sqrt[c]*ArcTan[(c^(1/4)*Sqrt[x])/(-a)^(1/4)] - 60*(-a)^(3/4)*A*Sqrt[c]*ArcTanh[(c^(1/4)*Sqrt[x])/(-a)^
(1/4)] - 15*Sqrt[2]*a^(5/4)*B*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] + 15*Sqrt[2]*a^(5/4)*
B*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(60*c^(9/4))

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Maple [A]  time = 0.013, size = 302, normalized size = 1. \begin{align*}{\frac{2\,B}{5\,c}{x}^{{\frac{5}{2}}}}+{\frac{2\,A}{3\,c}{x}^{{\frac{3}{2}}}}-2\,{\frac{aB\sqrt{x}}{{c}^{2}}}+{\frac{aB\sqrt{2}}{2\,{c}^{2}}\sqrt [4]{{\frac{a}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}+1 \right ) }+{\frac{aB\sqrt{2}}{2\,{c}^{2}}\sqrt [4]{{\frac{a}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}-1 \right ) }+{\frac{aB\sqrt{2}}{4\,{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{aA\sqrt{2}}{4\,{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{aA\sqrt{2}}{2\,{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{aA\sqrt{2}}{2\,{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),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.55003, size = 1709, normalized size = 5.85 \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),x, algorithm="fricas")

[Out]

-1/30*(15*c^2*sqrt((c^4*sqrt(-(B^4*a^5 - 2*A^2*B^2*a^4*c + A^4*a^3*c^2)/c^9) + 2*A*B*a^2)/c^4)*log(-(B^4*a^4 -
A^4*a^2*c^2)*sqrt(x) + (A*c^7*sqrt(-(B^4*a^5 - 2*A^2*B^2*a^4*c + A^4*a^3*c^2)/c^9) + B^3*a^3*c^2 - A^2*B*a^2*
c^3)*sqrt((c^4*sqrt(-(B^4*a^5 - 2*A^2*B^2*a^4*c + A^4*a^3*c^2)/c^9) + 2*A*B*a^2)/c^4)) - 15*c^2*sqrt((c^4*sqrt
(-(B^4*a^5 - 2*A^2*B^2*a^4*c + A^4*a^3*c^2)/c^9) + 2*A*B*a^2)/c^4)*log(-(B^4*a^4 - A^4*a^2*c^2)*sqrt(x) - (A*c
^7*sqrt(-(B^4*a^5 - 2*A^2*B^2*a^4*c + A^4*a^3*c^2)/c^9) + B^3*a^3*c^2 - A^2*B*a^2*c^3)*sqrt((c^4*sqrt(-(B^4*a^
5 - 2*A^2*B^2*a^4*c + A^4*a^3*c^2)/c^9) + 2*A*B*a^2)/c^4)) - 15*c^2*sqrt(-(c^4*sqrt(-(B^4*a^5 - 2*A^2*B^2*a^4*
c + A^4*a^3*c^2)/c^9) - 2*A*B*a^2)/c^4)*log(-(B^4*a^4 - A^4*a^2*c^2)*sqrt(x) + (A*c^7*sqrt(-(B^4*a^5 - 2*A^2*B
^2*a^4*c + A^4*a^3*c^2)/c^9) - B^3*a^3*c^2 + A^2*B*a^2*c^3)*sqrt(-(c^4*sqrt(-(B^4*a^5 - 2*A^2*B^2*a^4*c + A^4*
a^3*c^2)/c^9) - 2*A*B*a^2)/c^4)) + 15*c^2*sqrt(-(c^4*sqrt(-(B^4*a^5 - 2*A^2*B^2*a^4*c + A^4*a^3*c^2)/c^9) - 2*
A*B*a^2)/c^4)*log(-(B^4*a^4 - A^4*a^2*c^2)*sqrt(x) - (A*c^7*sqrt(-(B^4*a^5 - 2*A^2*B^2*a^4*c + A^4*a^3*c^2)/c^
9) - B^3*a^3*c^2 + A^2*B*a^2*c^3)*sqrt(-(c^4*sqrt(-(B^4*a^5 - 2*A^2*B^2*a^4*c + A^4*a^3*c^2)/c^9) - 2*A*B*a^2)
/c^4)) - 4*(3*B*c*x^2 + 5*A*c*x - 15*B*a)*sqrt(x))/c^2

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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Giac [A]  time = 1.27659, size = 354, normalized size = 1.21 \begin{align*} \frac{\sqrt{2}{\left (\left (a c^{3}\right )^{\frac{1}{4}} B a c - \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 )}{2 \, c^{4}} + \frac{\sqrt{2}{\left (\left (a c^{3}\right )^{\frac{1}{4}} B a c - \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 )}{2 \, c^{4}} + \frac{\sqrt{2}{\left (\left (a c^{3}\right )^{\frac{1}{4}} B a c + \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 )}{4 \, c^{4}} - \frac{\sqrt{2}{\left (\left (a c^{3}\right )^{\frac{1}{4}} B a c + \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 )}{4 \, c^{4}} + \frac{2 \,{\left (3 \, B c^{4} x^{\frac{5}{2}} + 5 \, A c^{4} x^{\frac{3}{2}} - 15 \, B a c^{3} \sqrt{x}\right )}}{15 \, c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*((a*c^3)^(1/4)*B*a*c - (a*c^3)^(3/4)*A)*arctan(1/2*sqrt(2)*(sqrt(2)*(a/c)^(1/4) + 2*sqrt(x))/(a/c)
^(1/4))/c^4 + 1/2*sqrt(2)*((a*c^3)^(1/4)*B*a*c - (a*c^3)^(3/4)*A)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a/c)^(1/4) - 2
*sqrt(x))/(a/c)^(1/4))/c^4 + 1/4*sqrt(2)*((a*c^3)^(1/4)*B*a*c + (a*c^3)^(3/4)*A)*log(sqrt(2)*sqrt(x)*(a/c)^(1/
4) + x + sqrt(a/c))/c^4 - 1/4*sqrt(2)*((a*c^3)^(1/4)*B*a*c + (a*c^3)^(3/4)*A)*log(-sqrt(2)*sqrt(x)*(a/c)^(1/4)
+ x + sqrt(a/c))/c^4 + 2/15*(3*B*c^4*x^(5/2) + 5*A*c^4*x^(3/2) - 15*B*a*c^3*sqrt(x))/c^5