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

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

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Rubi [A]  time = 0.34, antiderivative size = 292, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {825, 827, 1168, 1162, 617, 204, 1165, 628} \begin {gather*} -\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} \end {gather*}

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 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 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 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]

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 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 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 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)]

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*}

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Mathematica [A]  time = 0.10, size = 290, normalized size = 0.99 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.37, size = 173, normalized size = 0.59 \begin {gather*} -\frac {\left (a^{5/4} B-a^{3/4} A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}}\right )}{\sqrt {2} c^{9/4}}+\frac {\left (a^{3/4} A \sqrt {c}+a^{5/4} B\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}}{\sqrt {a}+\sqrt {c} x}\right )}{\sqrt {2} c^{9/4}}+\frac {2 \left (-15 a B \sqrt {x}+5 A c x^{3/2}+3 B c x^{5/2}\right )}{15 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-15*a*B*Sqrt[x] + 5*A*c*x^(3/2) + 3*B*c*x^(5/2)))/(15*c^2) - ((a^(5/4)*B - a^(3/4)*A*Sqrt[c])*ArcTan[(Sqrt
[a] - Sqrt[c]*x)/(Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x])])/(Sqrt[2]*c^(9/4)) + ((a^(5/4)*B + a^(3/4)*A*Sqrt[c])*ArcT
anh[(Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[c]*x)])/(Sqrt[2]*c^(9/4))

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fricas [B]  time = 0.46, size = 862, normalized size = 2.95 \begin {gather*} -\frac {15 \, c^{2} \sqrt {\frac {c^{4} \sqrt {-\frac {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 \left (-{\left (B^{4} a^{4} - A^{4} a^{2} c^{2}\right )} \sqrt {x} + {\left (A c^{7} \sqrt {-\frac {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}\right )} \sqrt {\frac {c^{4} \sqrt {-\frac {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}}}\right ) - 15 \, c^{2} \sqrt {\frac {c^{4} \sqrt {-\frac {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 \left (-{\left (B^{4} a^{4} - A^{4} a^{2} c^{2}\right )} \sqrt {x} - {\left (A c^{7} \sqrt {-\frac {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}\right )} \sqrt {\frac {c^{4} \sqrt {-\frac {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}}}\right ) - 15 \, c^{2} \sqrt {-\frac {c^{4} \sqrt {-\frac {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 \left (-{\left (B^{4} a^{4} - A^{4} a^{2} c^{2}\right )} \sqrt {x} + {\left (A c^{7} \sqrt {-\frac {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}\right )} \sqrt {-\frac {c^{4} \sqrt {-\frac {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}}}\right ) + 15 \, c^{2} \sqrt {-\frac {c^{4} \sqrt {-\frac {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 \left (-{\left (B^{4} a^{4} - A^{4} a^{2} c^{2}\right )} \sqrt {x} - {\left (A c^{7} \sqrt {-\frac {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}\right )} \sqrt {-\frac {c^{4} \sqrt {-\frac {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}}}\right ) - 4 \, {\left (3 \, B c x^{2} + 5 \, A c x - 15 \, B a\right )} \sqrt {x}}{30 \, c^{2}} \end {gather*}

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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giac [A]  time = 0.20, size = 262, normalized size = 0.90 \begin {gather*} \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 {gather*}

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

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maple [A]  time = 0.06, size = 302, normalized size = 1.03 \begin {gather*} \frac {2 B \,x^{\frac {5}{2}}}{5 c}+\frac {2 A \,x^{\frac {3}{2}}}{3 c}-\frac {\sqrt {2}\, A a \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{2 \left (\frac {a}{c}\right )^{\frac {1}{4}} c^{2}}-\frac {\sqrt {2}\, A a \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{2 \left (\frac {a}{c}\right )^{\frac {1}{4}} c^{2}}-\frac {\sqrt {2}\, A a \ln \left (\frac {x -\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{c}}}{x +\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{c}}}\right )}{4 \left (\frac {a}{c}\right )^{\frac {1}{4}} c^{2}}+\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B a \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{2 c^{2}}+\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B a \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{2 c^{2}}+\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B a \ln \left (\frac {x +\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{c}}}{x -\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{c}}}\right )}{4 c^{2}}-\frac {2 B a \sqrt {x}}{c^{2}} \end {gather*}

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/c*x^(5/2)+2/3*A/c*x^(3/2)-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)+1/4*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)))+1/2*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*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 [A]  time = 1.31, size = 265, normalized size = 0.91 \begin {gather*} \frac {a {\left (\frac {2 \, \sqrt {2} {\left (B a \sqrt {c} - A \sqrt {a} c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} {\left (B a \sqrt {c} - A \sqrt {a} c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {c}} \sqrt {c}} + \frac {\sqrt {2} {\left (B a \sqrt {c} + A \sqrt {a} c\right )} \log \left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {3}{4}}} - \frac {\sqrt {2} {\left (B a \sqrt {c} + A \sqrt {a} c\right )} \log \left (-\sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {3}{4}}}\right )}}{4 \, c^{2}} + \frac {2 \, {\left (3 \, B c x^{\frac {5}{2}} + 5 \, A c x^{\frac {3}{2}} - 15 \, B a \sqrt {x}\right )}}{15 \, c^{2}} \end {gather*}

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]

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

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mupad [B]  time = 1.26, size = 665, normalized size = 2.28 \begin {gather*} \frac {2\,A\,x^{3/2}}{3\,c}+\frac {2\,B\,x^{5/2}}{5\,c}-\frac {2\,B\,a\,\sqrt {x}}{c^2}-\mathrm {atan}\left (\frac {A^2\,a^3\,\sqrt {x}\,\sqrt {\frac {A\,B\,a^2}{2\,c^4}-\frac {A^2\,\sqrt {-a^3\,c^9}}{4\,c^8}+\frac {B^2\,a\,\sqrt {-a^3\,c^9}}{4\,c^9}}\,32{}\mathrm {i}}{\frac {16\,A^3\,a^4}{c^2}-\frac {16\,A\,B^2\,a^5}{c^3}-\frac {16\,B^3\,a^4\,\sqrt {-a^3\,c^9}}{c^8}+\frac {16\,A^2\,B\,a^3\,\sqrt {-a^3\,c^9}}{c^7}}-\frac {B^2\,a^4\,\sqrt {x}\,\sqrt {\frac {A\,B\,a^2}{2\,c^4}-\frac {A^2\,\sqrt {-a^3\,c^9}}{4\,c^8}+\frac {B^2\,a\,\sqrt {-a^3\,c^9}}{4\,c^9}}\,32{}\mathrm {i}}{\frac {16\,A^3\,a^4}{c}-\frac {16\,A\,B^2\,a^5}{c^2}-\frac {16\,B^3\,a^4\,\sqrt {-a^3\,c^9}}{c^7}+\frac {16\,A^2\,B\,a^3\,\sqrt {-a^3\,c^9}}{c^6}}\right )\,\sqrt {\frac {B^2\,a\,\sqrt {-a^3\,c^9}-A^2\,c\,\sqrt {-a^3\,c^9}+2\,A\,B\,a^2\,c^5}{4\,c^9}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {A^2\,a^3\,\sqrt {x}\,\sqrt {\frac {A^2\,\sqrt {-a^3\,c^9}}{4\,c^8}+\frac {A\,B\,a^2}{2\,c^4}-\frac {B^2\,a\,\sqrt {-a^3\,c^9}}{4\,c^9}}\,32{}\mathrm {i}}{\frac {16\,A^3\,a^4}{c^2}-\frac {16\,A\,B^2\,a^5}{c^3}+\frac {16\,B^3\,a^4\,\sqrt {-a^3\,c^9}}{c^8}-\frac {16\,A^2\,B\,a^3\,\sqrt {-a^3\,c^9}}{c^7}}-\frac {B^2\,a^4\,\sqrt {x}\,\sqrt {\frac {A^2\,\sqrt {-a^3\,c^9}}{4\,c^8}+\frac {A\,B\,a^2}{2\,c^4}-\frac {B^2\,a\,\sqrt {-a^3\,c^9}}{4\,c^9}}\,32{}\mathrm {i}}{\frac {16\,A^3\,a^4}{c}-\frac {16\,A\,B^2\,a^5}{c^2}+\frac {16\,B^3\,a^4\,\sqrt {-a^3\,c^9}}{c^7}-\frac {16\,A^2\,B\,a^3\,\sqrt {-a^3\,c^9}}{c^6}}\right )\,\sqrt {\frac {A^2\,c\,\sqrt {-a^3\,c^9}-B^2\,a\,\sqrt {-a^3\,c^9}+2\,A\,B\,a^2\,c^5}{4\,c^9}}\,2{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*A*x^(3/2))/(3*c) - atan((A^2*a^3*x^(1/2)*((A^2*(-a^3*c^9)^(1/2))/(4*c^8) + (A*B*a^2)/(2*c^4) - (B^2*a*(-a^3
*c^9)^(1/2))/(4*c^9))^(1/2)*32i)/((16*A^3*a^4)/c^2 - (16*A*B^2*a^5)/c^3 + (16*B^3*a^4*(-a^3*c^9)^(1/2))/c^8 -
(16*A^2*B*a^3*(-a^3*c^9)^(1/2))/c^7) - (B^2*a^4*x^(1/2)*((A^2*(-a^3*c^9)^(1/2))/(4*c^8) + (A*B*a^2)/(2*c^4) -
(B^2*a*(-a^3*c^9)^(1/2))/(4*c^9))^(1/2)*32i)/((16*A^3*a^4)/c - (16*A*B^2*a^5)/c^2 + (16*B^3*a^4*(-a^3*c^9)^(1/
2))/c^7 - (16*A^2*B*a^3*(-a^3*c^9)^(1/2))/c^6))*((A^2*c*(-a^3*c^9)^(1/2) - B^2*a*(-a^3*c^9)^(1/2) + 2*A*B*a^2*
c^5)/(4*c^9))^(1/2)*2i - atan((A^2*a^3*x^(1/2)*((A*B*a^2)/(2*c^4) - (A^2*(-a^3*c^9)^(1/2))/(4*c^8) + (B^2*a*(-
a^3*c^9)^(1/2))/(4*c^9))^(1/2)*32i)/((16*A^3*a^4)/c^2 - (16*A*B^2*a^5)/c^3 - (16*B^3*a^4*(-a^3*c^9)^(1/2))/c^8
+ (16*A^2*B*a^3*(-a^3*c^9)^(1/2))/c^7) - (B^2*a^4*x^(1/2)*((A*B*a^2)/(2*c^4) - (A^2*(-a^3*c^9)^(1/2))/(4*c^8)
+ (B^2*a*(-a^3*c^9)^(1/2))/(4*c^9))^(1/2)*32i)/((16*A^3*a^4)/c - (16*A*B^2*a^5)/c^2 - (16*B^3*a^4*(-a^3*c^9)^
(1/2))/c^7 + (16*A^2*B*a^3*(-a^3*c^9)^(1/2))/c^6))*((B^2*a*(-a^3*c^9)^(1/2) - A^2*c*(-a^3*c^9)^(1/2) + 2*A*B*a
^2*c^5)/(4*c^9))^(1/2)*2i + (2*B*x^(5/2))/(5*c) - (2*B*a*x^(1/2))/c^2

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sympy [A]  time = 33.96, size = 403, normalized size = 1.38 \begin {gather*} \begin {cases} \tilde {\infty } \left (\frac {2 A x^{\frac {3}{2}}}{3} + \frac {2 B x^{\frac {5}{2}}}{5}\right ) & \text {for}\: a = 0 \wedge c = 0 \\\frac {\frac {2 A x^{\frac {7}{2}}}{7} + \frac {2 B x^{\frac {9}{2}}}{9}}{a} & \text {for}\: c = 0 \\\frac {\frac {2 A x^{\frac {3}{2}}}{3} + \frac {2 B x^{\frac {5}{2}}}{5}}{c} & \text {for}\: a = 0 \\\frac {\left (-1\right )^{\frac {3}{4}} A a^{\frac {3}{4}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 c^{2} \sqrt [4]{\frac {1}{c}}} - \frac {\left (-1\right )^{\frac {3}{4}} A a^{\frac {3}{4}} \log {\left (\sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 c^{2} \sqrt [4]{\frac {1}{c}}} - \frac {\left (-1\right )^{\frac {3}{4}} A a^{\frac {3}{4}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{a} \sqrt [4]{\frac {1}{c}}} \right )}}{c^{2} \sqrt [4]{\frac {1}{c}}} + \frac {2 A x^{\frac {3}{2}}}{3 c} - \frac {\sqrt [4]{-1} B a^{\frac {5}{4}} \sqrt [4]{\frac {1}{c}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 c^{2}} + \frac {\sqrt [4]{-1} B a^{\frac {5}{4}} \sqrt [4]{\frac {1}{c}} \log {\left (\sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 c^{2}} - \frac {\sqrt [4]{-1} B a^{\frac {5}{4}} \sqrt [4]{\frac {1}{c}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{a} \sqrt [4]{\frac {1}{c}}} \right )}}{c^{2}} - \frac {2 B a \sqrt {x}}{c^{2}} + \frac {2 B x^{\frac {5}{2}}}{5 c} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(2*A*x**(3/2)/3 + 2*B*x**(5/2)/5), Eq(a, 0) & Eq(c, 0)), ((2*A*x**(7/2)/7 + 2*B*x**(9/2)/9)/a,
Eq(c, 0)), ((2*A*x**(3/2)/3 + 2*B*x**(5/2)/5)/c, Eq(a, 0)), ((-1)**(3/4)*A*a**(3/4)*log(-(-1)**(1/4)*a**(1/4)*
(1/c)**(1/4) + sqrt(x))/(2*c**2*(1/c)**(1/4)) - (-1)**(3/4)*A*a**(3/4)*log((-1)**(1/4)*a**(1/4)*(1/c)**(1/4) +
sqrt(x))/(2*c**2*(1/c)**(1/4)) - (-1)**(3/4)*A*a**(3/4)*atan((-1)**(3/4)*sqrt(x)/(a**(1/4)*(1/c)**(1/4)))/(c*
*2*(1/c)**(1/4)) + 2*A*x**(3/2)/(3*c) - (-1)**(1/4)*B*a**(5/4)*(1/c)**(1/4)*log(-(-1)**(1/4)*a**(1/4)*(1/c)**(
1/4) + sqrt(x))/(2*c**2) + (-1)**(1/4)*B*a**(5/4)*(1/c)**(1/4)*log((-1)**(1/4)*a**(1/4)*(1/c)**(1/4) + sqrt(x)
)/(2*c**2) - (-1)**(1/4)*B*a**(5/4)*(1/c)**(1/4)*atan((-1)**(3/4)*sqrt(x)/(a**(1/4)*(1/c)**(1/4)))/c**2 - 2*B*
a*sqrt(x)/c**2 + 2*B*x**(5/2)/(5*c), True))

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