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

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

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Rubi [A]  time = 0.37, antiderivative size = 306, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {829, 827, 1168, 1162, 617, 204, 1165, 628} \begin {gather*} \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}} \end {gather*}

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 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 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 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 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 {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*}

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Mathematica [C]  time = 0.01, size = 54, normalized size = 0.18 \begin {gather*} -\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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.41, size = 175, normalized size = 0.57 \begin {gather*} -\frac {\left (\sqrt {a} B c^{5/4}+A c^{7/4}\right ) \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}}\right )}{\sqrt {2} a^{11/4}}-\frac {\left (\sqrt {a} B c^{5/4}-A c^{7/4}\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}}{\sqrt {a}+\sqrt {c} x}\right )}{\sqrt {2} a^{11/4}}-\frac {2 \left (15 a A+21 a B x-35 A c x^2-105 B c x^3\right )}{105 a^2 x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(15*a*A + 21*a*B*x - 35*A*c*x^2 - 105*B*c*x^3))/(105*a^2*x^(7/2)) - ((Sqrt[a]*B*c^(5/4) + A*c^(7/4))*ArcTa
n[(Sqrt[a] - Sqrt[c]*x)/(Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x])])/(Sqrt[2]*a^(11/4)) - ((Sqrt[a]*B*c^(5/4) - A*c^(7/
4))*ArcTanh[(Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[c]*x)])/(Sqrt[2]*a^(11/4))

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fricas [B]  time = 0.44, size = 884, normalized size = 2.89 \begin {gather*} \frac {105 \, a^{2} x^{4} \sqrt {-\frac {a^{5} \sqrt {-\frac {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 \left (-{\left (B^{4} a^{2} c^{4} - A^{4} c^{6}\right )} \sqrt {x} + {\left (B a^{9} \sqrt {-\frac {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}\right )} \sqrt {-\frac {a^{5} \sqrt {-\frac {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}}}\right ) - 105 \, a^{2} x^{4} \sqrt {-\frac {a^{5} \sqrt {-\frac {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 \left (-{\left (B^{4} a^{2} c^{4} - A^{4} c^{6}\right )} \sqrt {x} - {\left (B a^{9} \sqrt {-\frac {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}\right )} \sqrt {-\frac {a^{5} \sqrt {-\frac {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}}}\right ) - 105 \, a^{2} x^{4} \sqrt {\frac {a^{5} \sqrt {-\frac {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 \left (-{\left (B^{4} a^{2} c^{4} - A^{4} c^{6}\right )} \sqrt {x} + {\left (B a^{9} \sqrt {-\frac {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}\right )} \sqrt {\frac {a^{5} \sqrt {-\frac {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}}}\right ) + 105 \, a^{2} x^{4} \sqrt {\frac {a^{5} \sqrt {-\frac {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 \left (-{\left (B^{4} a^{2} c^{4} - A^{4} c^{6}\right )} \sqrt {x} - {\left (B a^{9} \sqrt {-\frac {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}\right )} \sqrt {\frac {a^{5} \sqrt {-\frac {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}}}\right ) + 4 \, {\left (105 \, B c x^{3} + 35 \, A c x^{2} - 21 \, B a x - 15 \, A a\right )} \sqrt {x}}{210 \, a^{2} x^{4}} \end {gather*}

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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giac [A]  time = 0.22, size = 276, normalized size = 0.90 \begin {gather*} \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 {gather*}

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))

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maple [A]  time = 0.07, size = 318, normalized size = 1.04 \begin {gather*} \frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \,c^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{2 a^{3}}+\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \,c^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{2 a^{3}}+\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \,c^{2} \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 a^{3}}+\frac {\sqrt {2}\, B c \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}} a^{2}}+\frac {\sqrt {2}\, B c \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}} a^{2}}+\frac {\sqrt {2}\, B c \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}} a^{2}}+\frac {2 B c}{a^{2} \sqrt {x}}+\frac {2 A c}{3 a^{2} x^{\frac {3}{2}}}-\frac {2 B}{5 a \,x^{\frac {5}{2}}}-\frac {2 A}{7 a \,x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(9/2)/(c*x^2+a),x)

[Out]

1/4*c^2/a^3*A*(a/c)^(1/4)*2^(1/2)*ln((x+(a/c)^(1/4)*2^(1/2)*x^(1/2)+(a/c)^(1/2))/(x-(a/c)^(1/4)*2^(1/2)*x^(1/2
)+(a/c)^(1/2)))+1/2*c^2/a^3*A*(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)+1)+1/2*c^2/a^3*A*(a/c)^(1
/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)-1)+1/4*c/a^2*B/(a/c)^(1/4)*2^(1/2)*ln((x-(a/c)^(1/4)*2^(1/2)*x^
(1/2)+(a/c)^(1/2))/(x+(a/c)^(1/4)*2^(1/2)*x^(1/2)+(a/c)^(1/2)))+1/2*c/a^2*B/(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)
/(a/c)^(1/4)*x^(1/2)+1)+1/2*c/a^2*B/(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)-1)-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)

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maxima [A]  time = 1.24, size = 264, normalized size = 0.86 \begin {gather*} \frac {c^{2} {\left (\frac {2 \, \sqrt {2} {\left (B \sqrt {a} + A \sqrt {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 \sqrt {a} + A \sqrt {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 \sqrt {a} - A \sqrt {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 \sqrt {a} - A \sqrt {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 \, a^{2}} + \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 {gather*}

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]

1/4*c^2*(2*sqrt(2)*(B*sqrt(a) + A*sqrt(c))*arctan(1/2*sqrt(2)*(sqrt(2)*a^(1/4)*c^(1/4) + 2*sqrt(c)*sqrt(x))/sq
rt(sqrt(a)*sqrt(c)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(c))*sqrt(c)) + 2*sqrt(2)*(B*sqrt(a) + A*sqrt(c))*arctan(-1/2*s
qrt(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))*sqr
t(c)) - sqrt(2)*(B*sqrt(a) - A*sqrt(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*sqrt(a) - A*sqrt(c))*log(-sqrt(2)*a^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(a))/(a^(3/4)*
c^(3/4)))/a^2 + 2/105*(105*B*c*x^3 + 35*A*c*x^2 - 21*B*a*x - 15*A*a)/(a^2*x^(7/2))

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mupad [B]  time = 1.27, size = 678, normalized size = 2.22 \begin {gather*} 2\,\mathrm {atanh}\left (\frac {32\,A^2\,a^6\,c^7\,\sqrt {x}\,\sqrt {\frac {A^2\,c\,\sqrt {-a^{11}\,c^5}}{4\,a^{11}}-\frac {B^2\,\sqrt {-a^{11}\,c^5}}{4\,a^{10}}-\frac {A\,B\,c^3}{2\,a^5}}}{16\,B^3\,a^5\,c^7+\frac {16\,A^3\,c^6\,\sqrt {-a^{11}\,c^5}}{a^2}-16\,A^2\,B\,a^4\,c^8-\frac {16\,A\,B^2\,c^5\,\sqrt {-a^{11}\,c^5}}{a}}-\frac {32\,B^2\,a^7\,c^6\,\sqrt {x}\,\sqrt {\frac {A^2\,c\,\sqrt {-a^{11}\,c^5}}{4\,a^{11}}-\frac {B^2\,\sqrt {-a^{11}\,c^5}}{4\,a^{10}}-\frac {A\,B\,c^3}{2\,a^5}}}{16\,B^3\,a^5\,c^7+\frac {16\,A^3\,c^6\,\sqrt {-a^{11}\,c^5}}{a^2}-16\,A^2\,B\,a^4\,c^8-\frac {16\,A\,B^2\,c^5\,\sqrt {-a^{11}\,c^5}}{a}}\right )\,\sqrt {-\frac {B^2\,a\,\sqrt {-a^{11}\,c^5}-A^2\,c\,\sqrt {-a^{11}\,c^5}+2\,A\,B\,a^6\,c^3}{4\,a^{11}}}-\frac {\frac {2\,A}{7\,a}+\frac {2\,B\,x}{5\,a}-\frac {2\,A\,c\,x^2}{3\,a^2}-\frac {2\,B\,c\,x^3}{a^2}}{x^{7/2}}+2\,\mathrm {atanh}\left (\frac {32\,A^2\,a^6\,c^7\,\sqrt {x}\,\sqrt {\frac {B^2\,\sqrt {-a^{11}\,c^5}}{4\,a^{10}}-\frac {A^2\,c\,\sqrt {-a^{11}\,c^5}}{4\,a^{11}}-\frac {A\,B\,c^3}{2\,a^5}}}{16\,B^3\,a^5\,c^7-\frac {16\,A^3\,c^6\,\sqrt {-a^{11}\,c^5}}{a^2}-16\,A^2\,B\,a^4\,c^8+\frac {16\,A\,B^2\,c^5\,\sqrt {-a^{11}\,c^5}}{a}}-\frac {32\,B^2\,a^7\,c^6\,\sqrt {x}\,\sqrt {\frac {B^2\,\sqrt {-a^{11}\,c^5}}{4\,a^{10}}-\frac {A^2\,c\,\sqrt {-a^{11}\,c^5}}{4\,a^{11}}-\frac {A\,B\,c^3}{2\,a^5}}}{16\,B^3\,a^5\,c^7-\frac {16\,A^3\,c^6\,\sqrt {-a^{11}\,c^5}}{a^2}-16\,A^2\,B\,a^4\,c^8+\frac {16\,A\,B^2\,c^5\,\sqrt {-a^{11}\,c^5}}{a}}\right )\,\sqrt {-\frac {A^2\,c\,\sqrt {-a^{11}\,c^5}-B^2\,a\,\sqrt {-a^{11}\,c^5}+2\,A\,B\,a^6\,c^3}{4\,a^{11}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(9/2)*(a + c*x^2)),x)

[Out]

2*atanh((32*A^2*a^6*c^7*x^(1/2)*((A^2*c*(-a^11*c^5)^(1/2))/(4*a^11) - (B^2*(-a^11*c^5)^(1/2))/(4*a^10) - (A*B*
c^3)/(2*a^5))^(1/2))/(16*B^3*a^5*c^7 + (16*A^3*c^6*(-a^11*c^5)^(1/2))/a^2 - 16*A^2*B*a^4*c^8 - (16*A*B^2*c^5*(
-a^11*c^5)^(1/2))/a) - (32*B^2*a^7*c^6*x^(1/2)*((A^2*c*(-a^11*c^5)^(1/2))/(4*a^11) - (B^2*(-a^11*c^5)^(1/2))/(
4*a^10) - (A*B*c^3)/(2*a^5))^(1/2))/(16*B^3*a^5*c^7 + (16*A^3*c^6*(-a^11*c^5)^(1/2))/a^2 - 16*A^2*B*a^4*c^8 -
(16*A*B^2*c^5*(-a^11*c^5)^(1/2))/a))*(-(B^2*a*(-a^11*c^5)^(1/2) - A^2*c*(-a^11*c^5)^(1/2) + 2*A*B*a^6*c^3)/(4*
a^11))^(1/2) - ((2*A)/(7*a) + (2*B*x)/(5*a) - (2*A*c*x^2)/(3*a^2) - (2*B*c*x^3)/a^2)/x^(7/2) + 2*atanh((32*A^2
*a^6*c^7*x^(1/2)*((B^2*(-a^11*c^5)^(1/2))/(4*a^10) - (A^2*c*(-a^11*c^5)^(1/2))/(4*a^11) - (A*B*c^3)/(2*a^5))^(
1/2))/(16*B^3*a^5*c^7 - (16*A^3*c^6*(-a^11*c^5)^(1/2))/a^2 - 16*A^2*B*a^4*c^8 + (16*A*B^2*c^5*(-a^11*c^5)^(1/2
))/a) - (32*B^2*a^7*c^6*x^(1/2)*((B^2*(-a^11*c^5)^(1/2))/(4*a^10) - (A^2*c*(-a^11*c^5)^(1/2))/(4*a^11) - (A*B*
c^3)/(2*a^5))^(1/2))/(16*B^3*a^5*c^7 - (16*A^3*c^6*(-a^11*c^5)^(1/2))/a^2 - 16*A^2*B*a^4*c^8 + (16*A*B^2*c^5*(
-a^11*c^5)^(1/2))/a))*(-(A^2*c*(-a^11*c^5)^(1/2) - B^2*a*(-a^11*c^5)^(1/2) + 2*A*B*a^6*c^3)/(4*a^11))^(1/2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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