3.5.31 \(\int \frac {A+B x}{x^{3/2} (a+c x^2)^3} \, dx\)

Optimal. Leaf size=333 \[ -\frac {3 \left (7 \sqrt {a} B+15 A \sqrt {c}\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{64 \sqrt {2} a^{13/4} \sqrt [4]{c}}+\frac {3 \left (7 \sqrt {a} B+15 A \sqrt {c}\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{64 \sqrt {2} a^{13/4} \sqrt [4]{c}}-\frac {3 \left (7 \sqrt {a} B-15 A \sqrt {c}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{13/4} \sqrt [4]{c}}+\frac {3 \left (7 \sqrt {a} B-15 A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt {2} a^{13/4} \sqrt [4]{c}}-\frac {45 A}{16 a^3 \sqrt {x}}+\frac {9 A+7 B x}{16 a^2 \sqrt {x} \left (a+c x^2\right )}+\frac {A+B x}{4 a \sqrt {x} \left (a+c x^2\right )^2} \]

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Rubi [A]  time = 0.35, antiderivative size = 333, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {823, 829, 827, 1168, 1162, 617, 204, 1165, 628} \begin {gather*} \frac {9 A+7 B x}{16 a^2 \sqrt {x} \left (a+c x^2\right )}-\frac {3 \left (7 \sqrt {a} B+15 A \sqrt {c}\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{64 \sqrt {2} a^{13/4} \sqrt [4]{c}}+\frac {3 \left (7 \sqrt {a} B+15 A \sqrt {c}\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{64 \sqrt {2} a^{13/4} \sqrt [4]{c}}-\frac {3 \left (7 \sqrt {a} B-15 A \sqrt {c}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{13/4} \sqrt [4]{c}}+\frac {3 \left (7 \sqrt {a} B-15 A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt {2} a^{13/4} \sqrt [4]{c}}-\frac {45 A}{16 a^3 \sqrt {x}}+\frac {A+B x}{4 a \sqrt {x} \left (a+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(3/2)*(a + c*x^2)^3),x]

[Out]

(-45*A)/(16*a^3*Sqrt[x]) + (A + B*x)/(4*a*Sqrt[x]*(a + c*x^2)^2) + (9*A + 7*B*x)/(16*a^2*Sqrt[x]*(a + c*x^2))
- (3*(7*Sqrt[a]*B - 15*A*Sqrt[c])*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(13/4)*c^(1/4))
+ (3*(7*Sqrt[a]*B - 15*A*Sqrt[c])*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(13/4)*c^(1/4)
) - (3*(7*Sqrt[a]*B + 15*A*Sqrt[c])*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*a^
(13/4)*c^(1/4)) + (3*(7*Sqrt[a]*B + 15*A*Sqrt[c])*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/
(64*Sqrt[2]*a^(13/4)*c^(1/4))

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 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 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^{3/2} \left (a+c x^2\right )^3} \, dx &=\frac {A+B x}{4 a \sqrt {x} \left (a+c x^2\right )^2}-\frac {\int \frac {-\frac {9}{2} a A c-\frac {7}{2} a B c x}{x^{3/2} \left (a+c x^2\right )^2} \, dx}{4 a^2 c}\\ &=\frac {A+B x}{4 a \sqrt {x} \left (a+c x^2\right )^2}+\frac {9 A+7 B x}{16 a^2 \sqrt {x} \left (a+c x^2\right )}+\frac {\int \frac {\frac {45}{4} a^2 A c^2+\frac {21}{4} a^2 B c^2 x}{x^{3/2} \left (a+c x^2\right )} \, dx}{8 a^4 c^2}\\ &=-\frac {45 A}{16 a^3 \sqrt {x}}+\frac {A+B x}{4 a \sqrt {x} \left (a+c x^2\right )^2}+\frac {9 A+7 B x}{16 a^2 \sqrt {x} \left (a+c x^2\right )}+\frac {\int \frac {\frac {21}{4} a^3 B c^2-\frac {45}{4} a^2 A c^3 x}{\sqrt {x} \left (a+c x^2\right )} \, dx}{8 a^5 c^2}\\ &=-\frac {45 A}{16 a^3 \sqrt {x}}+\frac {A+B x}{4 a \sqrt {x} \left (a+c x^2\right )^2}+\frac {9 A+7 B x}{16 a^2 \sqrt {x} \left (a+c x^2\right )}+\frac {\operatorname {Subst}\left (\int \frac {\frac {21}{4} a^3 B c^2-\frac {45}{4} a^2 A c^3 x^2}{a+c x^4} \, dx,x,\sqrt {x}\right )}{4 a^5 c^2}\\ &=-\frac {45 A}{16 a^3 \sqrt {x}}+\frac {A+B x}{4 a \sqrt {x} \left (a+c x^2\right )^2}+\frac {9 A+7 B x}{16 a^2 \sqrt {x} \left (a+c x^2\right )}-\frac {\left (3 \left (15 A-\frac {7 \sqrt {a} B}{\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 )}{32 a^3}+\frac {\left (3 \left (15 A+\frac {7 \sqrt {a} B}{\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 )}{32 a^3}\\ &=-\frac {45 A}{16 a^3 \sqrt {x}}+\frac {A+B x}{4 a \sqrt {x} \left (a+c x^2\right )^2}+\frac {9 A+7 B x}{16 a^2 \sqrt {x} \left (a+c x^2\right )}-\frac {\left (3 \left (15 A-\frac {7 \sqrt {a} B}{\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 )}{64 a^3}-\frac {\left (3 \left (15 A-\frac {7 \sqrt {a} B}{\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 )}{64 a^3}-\frac {\left (3 \left (7 \sqrt {a} B+15 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 )}{64 \sqrt {2} a^{13/4} \sqrt [4]{c}}-\frac {\left (3 \left (7 \sqrt {a} B+15 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 )}{64 \sqrt {2} a^{13/4} \sqrt [4]{c}}\\ &=-\frac {45 A}{16 a^3 \sqrt {x}}+\frac {A+B x}{4 a \sqrt {x} \left (a+c x^2\right )^2}+\frac {9 A+7 B x}{16 a^2 \sqrt {x} \left (a+c x^2\right )}-\frac {3 \left (7 \sqrt {a} B+15 A \sqrt {c}\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} a^{13/4} \sqrt [4]{c}}+\frac {3 \left (7 \sqrt {a} B+15 A \sqrt {c}\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} a^{13/4} \sqrt [4]{c}}-\frac {\left (3 \left (15 A-\frac {7 \sqrt {a} B}{\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 )}{32 \sqrt {2} a^{13/4}}+\frac {\left (3 \left (15 A-\frac {7 \sqrt {a} B}{\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 )}{32 \sqrt {2} a^{13/4}}\\ &=-\frac {45 A}{16 a^3 \sqrt {x}}+\frac {A+B x}{4 a \sqrt {x} \left (a+c x^2\right )^2}+\frac {9 A+7 B x}{16 a^2 \sqrt {x} \left (a+c x^2\right )}+\frac {3 \left (15 A-\frac {7 \sqrt {a} B}{\sqrt {c}}\right ) \sqrt [4]{c} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{13/4}}-\frac {3 \left (15 A-\frac {7 \sqrt {a} B}{\sqrt {c}}\right ) \sqrt [4]{c} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{13/4}}-\frac {3 \left (7 \sqrt {a} B+15 A \sqrt {c}\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} a^{13/4} \sqrt [4]{c}}+\frac {3 \left (7 \sqrt {a} B+15 A \sqrt {c}\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} a^{13/4} \sqrt [4]{c}}\\ \end {align*}

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Mathematica [C]  time = 0.29, size = 300, normalized size = 0.90 \begin {gather*} \frac {\sqrt [4]{a} \left (\frac {32 a^{7/4} A}{\sqrt {x} \left (a+c x^2\right )^2}+\frac {72 a^{3/4} A}{\sqrt {x} \left (a+c x^2\right )}+\frac {32 a^{7/4} B \sqrt {x}}{\left (a+c x^2\right )^2}+\frac {56 a^{3/4} B \sqrt {x}}{a+c x^2}-\frac {21 \sqrt {2} B \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{\sqrt [4]{c}}+\frac {21 \sqrt {2} B \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{\sqrt [4]{c}}-\frac {42 \sqrt {2} B \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt [4]{c}}+\frac {42 \sqrt {2} B \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt [4]{c}}\right )-\frac {360 A \, _2F_1\left (-\frac {1}{4},1;\frac {3}{4};-\frac {c x^2}{a}\right )}{\sqrt {x}}}{128 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(3/2)*(a + c*x^2)^3),x]

[Out]

((-360*A*Hypergeometric2F1[-1/4, 1, 3/4, -((c*x^2)/a)])/Sqrt[x] + a^(1/4)*((32*a^(7/4)*A)/(Sqrt[x]*(a + c*x^2)
^2) + (32*a^(7/4)*B*Sqrt[x])/(a + c*x^2)^2 + (72*a^(3/4)*A)/(Sqrt[x]*(a + c*x^2)) + (56*a^(3/4)*B*Sqrt[x])/(a
+ c*x^2) - (42*Sqrt[2]*B*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/c^(1/4) + (42*Sqrt[2]*B*ArcTan[1 + (Sq
rt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/c^(1/4) - (21*Sqrt[2]*B*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[
c]*x])/c^(1/4) + (21*Sqrt[2]*B*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/c^(1/4)))/(128*a^3)

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IntegrateAlgebraic [A]  time = 0.92, size = 206, normalized size = 0.62 \begin {gather*} -\frac {3 \left (7 \sqrt {a} B-15 A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}}\right )}{32 \sqrt {2} a^{13/4} \sqrt [4]{c}}+\frac {3 \left (7 \sqrt {a} B+15 A \sqrt {c}\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}}{\sqrt {a}+\sqrt {c} x}\right )}{32 \sqrt {2} a^{13/4} \sqrt [4]{c}}+\frac {-32 a^2 A+11 a^2 B x-81 a A c x^2+7 a B c x^3-45 A c^2 x^4}{16 a^3 \sqrt {x} \left (a+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/(x^(3/2)*(a + c*x^2)^3),x]

[Out]

(-32*a^2*A + 11*a^2*B*x - 81*a*A*c*x^2 + 7*a*B*c*x^3 - 45*A*c^2*x^4)/(16*a^3*Sqrt[x]*(a + c*x^2)^2) - (3*(7*Sq
rt[a]*B - 15*A*Sqrt[c])*ArcTan[(Sqrt[a] - Sqrt[c]*x)/(Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x])])/(32*Sqrt[2]*a^(13/4)*
c^(1/4)) + (3*(7*Sqrt[a]*B + 15*A*Sqrt[c])*ArcTanh[(Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[c]*x)])/(
32*Sqrt[2]*a^(13/4)*c^(1/4))

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fricas [B]  time = 0.45, size = 958, normalized size = 2.88 \begin {gather*} -\frac {3 \, {\left (a^{3} c^{2} x^{5} + 2 \, a^{4} c x^{3} + a^{5} x\right )} \sqrt {\frac {a^{6} \sqrt {-\frac {2401 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 50625 \, A^{4} c^{2}}{a^{13} c}} + 210 \, A B}{a^{6}}} \log \left (-9 \, {\left (2401 \, B^{4} a^{2} - 50625 \, A^{4} c^{2}\right )} \sqrt {x} + 9 \, {\left (15 \, A a^{10} c \sqrt {-\frac {2401 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 50625 \, A^{4} c^{2}}{a^{13} c}} + 343 \, B^{3} a^{5} - 1575 \, A^{2} B a^{4} c\right )} \sqrt {\frac {a^{6} \sqrt {-\frac {2401 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 50625 \, A^{4} c^{2}}{a^{13} c}} + 210 \, A B}{a^{6}}}\right ) - 3 \, {\left (a^{3} c^{2} x^{5} + 2 \, a^{4} c x^{3} + a^{5} x\right )} \sqrt {\frac {a^{6} \sqrt {-\frac {2401 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 50625 \, A^{4} c^{2}}{a^{13} c}} + 210 \, A B}{a^{6}}} \log \left (-9 \, {\left (2401 \, B^{4} a^{2} - 50625 \, A^{4} c^{2}\right )} \sqrt {x} - 9 \, {\left (15 \, A a^{10} c \sqrt {-\frac {2401 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 50625 \, A^{4} c^{2}}{a^{13} c}} + 343 \, B^{3} a^{5} - 1575 \, A^{2} B a^{4} c\right )} \sqrt {\frac {a^{6} \sqrt {-\frac {2401 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 50625 \, A^{4} c^{2}}{a^{13} c}} + 210 \, A B}{a^{6}}}\right ) - 3 \, {\left (a^{3} c^{2} x^{5} + 2 \, a^{4} c x^{3} + a^{5} x\right )} \sqrt {-\frac {a^{6} \sqrt {-\frac {2401 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 50625 \, A^{4} c^{2}}{a^{13} c}} - 210 \, A B}{a^{6}}} \log \left (-9 \, {\left (2401 \, B^{4} a^{2} - 50625 \, A^{4} c^{2}\right )} \sqrt {x} + 9 \, {\left (15 \, A a^{10} c \sqrt {-\frac {2401 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 50625 \, A^{4} c^{2}}{a^{13} c}} - 343 \, B^{3} a^{5} + 1575 \, A^{2} B a^{4} c\right )} \sqrt {-\frac {a^{6} \sqrt {-\frac {2401 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 50625 \, A^{4} c^{2}}{a^{13} c}} - 210 \, A B}{a^{6}}}\right ) + 3 \, {\left (a^{3} c^{2} x^{5} + 2 \, a^{4} c x^{3} + a^{5} x\right )} \sqrt {-\frac {a^{6} \sqrt {-\frac {2401 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 50625 \, A^{4} c^{2}}{a^{13} c}} - 210 \, A B}{a^{6}}} \log \left (-9 \, {\left (2401 \, B^{4} a^{2} - 50625 \, A^{4} c^{2}\right )} \sqrt {x} - 9 \, {\left (15 \, A a^{10} c \sqrt {-\frac {2401 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 50625 \, A^{4} c^{2}}{a^{13} c}} - 343 \, B^{3} a^{5} + 1575 \, A^{2} B a^{4} c\right )} \sqrt {-\frac {a^{6} \sqrt {-\frac {2401 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 50625 \, A^{4} c^{2}}{a^{13} c}} - 210 \, A B}{a^{6}}}\right ) + 4 \, {\left (45 \, A c^{2} x^{4} - 7 \, B a c x^{3} + 81 \, A a c x^{2} - 11 \, B a^{2} x + 32 \, A a^{2}\right )} \sqrt {x}}{64 \, {\left (a^{3} c^{2} x^{5} + 2 \, a^{4} c x^{3} + a^{5} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(3/2)/(c*x^2+a)^3,x, algorithm="fricas")

[Out]

-1/64*(3*(a^3*c^2*x^5 + 2*a^4*c*x^3 + a^5*x)*sqrt((a^6*sqrt(-(2401*B^4*a^2 - 22050*A^2*B^2*a*c + 50625*A^4*c^2
)/(a^13*c)) + 210*A*B)/a^6)*log(-9*(2401*B^4*a^2 - 50625*A^4*c^2)*sqrt(x) + 9*(15*A*a^10*c*sqrt(-(2401*B^4*a^2
- 22050*A^2*B^2*a*c + 50625*A^4*c^2)/(a^13*c)) + 343*B^3*a^5 - 1575*A^2*B*a^4*c)*sqrt((a^6*sqrt(-(2401*B^4*a^
2 - 22050*A^2*B^2*a*c + 50625*A^4*c^2)/(a^13*c)) + 210*A*B)/a^6)) - 3*(a^3*c^2*x^5 + 2*a^4*c*x^3 + a^5*x)*sqrt
((a^6*sqrt(-(2401*B^4*a^2 - 22050*A^2*B^2*a*c + 50625*A^4*c^2)/(a^13*c)) + 210*A*B)/a^6)*log(-9*(2401*B^4*a^2
- 50625*A^4*c^2)*sqrt(x) - 9*(15*A*a^10*c*sqrt(-(2401*B^4*a^2 - 22050*A^2*B^2*a*c + 50625*A^4*c^2)/(a^13*c)) +
343*B^3*a^5 - 1575*A^2*B*a^4*c)*sqrt((a^6*sqrt(-(2401*B^4*a^2 - 22050*A^2*B^2*a*c + 50625*A^4*c^2)/(a^13*c))
+ 210*A*B)/a^6)) - 3*(a^3*c^2*x^5 + 2*a^4*c*x^3 + a^5*x)*sqrt(-(a^6*sqrt(-(2401*B^4*a^2 - 22050*A^2*B^2*a*c +
50625*A^4*c^2)/(a^13*c)) - 210*A*B)/a^6)*log(-9*(2401*B^4*a^2 - 50625*A^4*c^2)*sqrt(x) + 9*(15*A*a^10*c*sqrt(-
(2401*B^4*a^2 - 22050*A^2*B^2*a*c + 50625*A^4*c^2)/(a^13*c)) - 343*B^3*a^5 + 1575*A^2*B*a^4*c)*sqrt(-(a^6*sqrt
(-(2401*B^4*a^2 - 22050*A^2*B^2*a*c + 50625*A^4*c^2)/(a^13*c)) - 210*A*B)/a^6)) + 3*(a^3*c^2*x^5 + 2*a^4*c*x^3
+ a^5*x)*sqrt(-(a^6*sqrt(-(2401*B^4*a^2 - 22050*A^2*B^2*a*c + 50625*A^4*c^2)/(a^13*c)) - 210*A*B)/a^6)*log(-9
*(2401*B^4*a^2 - 50625*A^4*c^2)*sqrt(x) - 9*(15*A*a^10*c*sqrt(-(2401*B^4*a^2 - 22050*A^2*B^2*a*c + 50625*A^4*c
^2)/(a^13*c)) - 343*B^3*a^5 + 1575*A^2*B*a^4*c)*sqrt(-(a^6*sqrt(-(2401*B^4*a^2 - 22050*A^2*B^2*a*c + 50625*A^4
*c^2)/(a^13*c)) - 210*A*B)/a^6)) + 4*(45*A*c^2*x^4 - 7*B*a*c*x^3 + 81*A*a*c*x^2 - 11*B*a^2*x + 32*A*a^2)*sqrt(
x))/(a^3*c^2*x^5 + 2*a^4*c*x^3 + a^5*x)

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giac [A]  time = 0.21, size = 304, normalized size = 0.91 \begin {gather*} -\frac {2 \, A}{a^{3} \sqrt {x}} + \frac {3 \, \sqrt {2} {\left (7 \, \left (a c^{3}\right )^{\frac {1}{4}} B a c - 15 \, \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 )}{64 \, a^{4} c^{2}} + \frac {3 \, \sqrt {2} {\left (7 \, \left (a c^{3}\right )^{\frac {1}{4}} B a c - 15 \, \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 )}{64 \, a^{4} c^{2}} + \frac {3 \, \sqrt {2} {\left (7 \, \left (a c^{3}\right )^{\frac {1}{4}} B a c + 15 \, \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 )}{128 \, a^{4} c^{2}} - \frac {3 \, \sqrt {2} {\left (7 \, \left (a c^{3}\right )^{\frac {1}{4}} B a c + 15 \, \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 )}{128 \, a^{4} c^{2}} - \frac {13 \, A c^{2} x^{\frac {7}{2}} - 7 \, B a c x^{\frac {5}{2}} + 17 \, A a c x^{\frac {3}{2}} - 11 \, B a^{2} \sqrt {x}}{16 \, {\left (c x^{2} + a\right )}^{2} a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(3/2)/(c*x^2+a)^3,x, algorithm="giac")

[Out]

-2*A/(a^3*sqrt(x)) + 3/64*sqrt(2)*(7*(a*c^3)^(1/4)*B*a*c - 15*(a*c^3)^(3/4)*A)*arctan(1/2*sqrt(2)*(sqrt(2)*(a/
c)^(1/4) + 2*sqrt(x))/(a/c)^(1/4))/(a^4*c^2) + 3/64*sqrt(2)*(7*(a*c^3)^(1/4)*B*a*c - 15*(a*c^3)^(3/4)*A)*arcta
n(-1/2*sqrt(2)*(sqrt(2)*(a/c)^(1/4) - 2*sqrt(x))/(a/c)^(1/4))/(a^4*c^2) + 3/128*sqrt(2)*(7*(a*c^3)^(1/4)*B*a*c
+ 15*(a*c^3)^(3/4)*A)*log(sqrt(2)*sqrt(x)*(a/c)^(1/4) + x + sqrt(a/c))/(a^4*c^2) - 3/128*sqrt(2)*(7*(a*c^3)^(
1/4)*B*a*c + 15*(a*c^3)^(3/4)*A)*log(-sqrt(2)*sqrt(x)*(a/c)^(1/4) + x + sqrt(a/c))/(a^4*c^2) - 1/16*(13*A*c^2*
x^(7/2) - 7*B*a*c*x^(5/2) + 17*A*a*c*x^(3/2) - 11*B*a^2*sqrt(x))/((c*x^2 + a)^2*a^3)

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maple [A]  time = 0.06, size = 354, normalized size = 1.06 \begin {gather*} -\frac {13 A \,c^{2} x^{\frac {7}{2}}}{16 \left (c \,x^{2}+a \right )^{2} a^{3}}+\frac {7 B c \,x^{\frac {5}{2}}}{16 \left (c \,x^{2}+a \right )^{2} a^{2}}-\frac {17 A c \,x^{\frac {3}{2}}}{16 \left (c \,x^{2}+a \right )^{2} a^{2}}+\frac {11 B \sqrt {x}}{16 \left (c \,x^{2}+a \right )^{2} a}-\frac {45 \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{64 \left (\frac {a}{c}\right )^{\frac {1}{4}} a^{3}}-\frac {45 \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{64 \left (\frac {a}{c}\right )^{\frac {1}{4}} a^{3}}-\frac {45 \sqrt {2}\, 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 )}{128 \left (\frac {a}{c}\right )^{\frac {1}{4}} a^{3}}+\frac {21 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{64 a^{3}}+\frac {21 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{64 a^{3}}+\frac {21 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \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 )}{128 a^{3}}-\frac {2 A}{a^{3} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(3/2)/(c*x^2+a)^3,x)

[Out]

-13/16/a^3/(c*x^2+a)^2*A*x^(7/2)*c^2+7/16/a^2/(c*x^2+a)^2*B*x^(5/2)*c-17/16/a^2/(c*x^2+a)^2*A*x^(3/2)*c+11/16/
a/(c*x^2+a)^2*B*x^(1/2)+21/64/a^3*B*(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)+1)+21/64/a^3*B*(a/c
)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)-1)+21/128/a^3*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)))-45/128/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)))-45/64/a^3*A/(a/c)^(1/4)*2^
(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)+1)-45/64/a^3*A/(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)
-1)-2*A/a^3/x^(1/2)

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maxima [A]  time = 1.38, size = 313, normalized size = 0.94 \begin {gather*} -\frac {45 \, A c^{2} x^{4} - 7 \, B a c x^{3} + 81 \, A a c x^{2} - 11 \, B a^{2} x + 32 \, A a^{2}}{16 \, {\left (a^{3} c^{2} x^{\frac {9}{2}} + 2 \, a^{4} c x^{\frac {5}{2}} + a^{5} \sqrt {x}\right )}} + \frac {3 \, {\left (\frac {2 \, \sqrt {2} {\left (7 \, B a \sqrt {c} - 15 \, 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 (7 \, B a \sqrt {c} - 15 \, 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 (7 \, B a \sqrt {c} + 15 \, 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 (7 \, B a \sqrt {c} + 15 \, 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 )}}{128 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(3/2)/(c*x^2+a)^3,x, algorithm="maxima")

[Out]

-1/16*(45*A*c^2*x^4 - 7*B*a*c*x^3 + 81*A*a*c*x^2 - 11*B*a^2*x + 32*A*a^2)/(a^3*c^2*x^(9/2) + 2*a^4*c*x^(5/2) +
a^5*sqrt(x)) + 3/128*(2*sqrt(2)*(7*B*a*sqrt(c) - 15*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)*(7*B*a*sqrt(c)
- 15*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)))/(s
qrt(a)*sqrt(sqrt(a)*sqrt(c))*sqrt(c)) + sqrt(2)*(7*B*a*sqrt(c) + 15*A*sqrt(a)*c)*log(sqrt(2)*a^(1/4)*c^(1/4)*s
qrt(x) + sqrt(c)*x + sqrt(a))/(a^(3/4)*c^(3/4)) - sqrt(2)*(7*B*a*sqrt(c) + 15*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)))/a^3

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mupad [B]  time = 0.28, size = 673, normalized size = 2.02 \begin {gather*} 2\,\mathrm {atanh}\left (\frac {66355200\,A^2\,a^{10}\,c^4\,\sqrt {x}\,\sqrt {\frac {945\,A\,B}{2048\,a^6}-\frac {2025\,A^2\,\sqrt {-a^{13}\,c}}{4096\,a^{13}}+\frac {441\,B^2\,\sqrt {-a^{13}\,c}}{4096\,a^{12}\,c}}}{46656000\,A^3\,a^7\,c^4-4741632\,B^3\,a^2\,c^2\,\sqrt {-a^{13}\,c}-10160640\,A\,B^2\,a^8\,c^3+21772800\,A^2\,B\,a\,c^3\,\sqrt {-a^{13}\,c}}-\frac {14450688\,B^2\,a^{11}\,c^3\,\sqrt {x}\,\sqrt {\frac {945\,A\,B}{2048\,a^6}-\frac {2025\,A^2\,\sqrt {-a^{13}\,c}}{4096\,a^{13}}+\frac {441\,B^2\,\sqrt {-a^{13}\,c}}{4096\,a^{12}\,c}}}{46656000\,A^3\,a^7\,c^4-4741632\,B^3\,a^2\,c^2\,\sqrt {-a^{13}\,c}-10160640\,A\,B^2\,a^8\,c^3+21772800\,A^2\,B\,a\,c^3\,\sqrt {-a^{13}\,c}}\right )\,\sqrt {\frac {9\,\left (49\,B^2\,a\,\sqrt {-a^{13}\,c}-225\,A^2\,c\,\sqrt {-a^{13}\,c}+210\,A\,B\,a^7\,c\right )}{4096\,a^{13}\,c}}+2\,\mathrm {atanh}\left (\frac {66355200\,A^2\,a^{10}\,c^4\,\sqrt {x}\,\sqrt {\frac {2025\,A^2\,\sqrt {-a^{13}\,c}}{4096\,a^{13}}+\frac {945\,A\,B}{2048\,a^6}-\frac {441\,B^2\,\sqrt {-a^{13}\,c}}{4096\,a^{12}\,c}}}{46656000\,A^3\,a^7\,c^4+4741632\,B^3\,a^2\,c^2\,\sqrt {-a^{13}\,c}-10160640\,A\,B^2\,a^8\,c^3-21772800\,A^2\,B\,a\,c^3\,\sqrt {-a^{13}\,c}}-\frac {14450688\,B^2\,a^{11}\,c^3\,\sqrt {x}\,\sqrt {\frac {2025\,A^2\,\sqrt {-a^{13}\,c}}{4096\,a^{13}}+\frac {945\,A\,B}{2048\,a^6}-\frac {441\,B^2\,\sqrt {-a^{13}\,c}}{4096\,a^{12}\,c}}}{46656000\,A^3\,a^7\,c^4+4741632\,B^3\,a^2\,c^2\,\sqrt {-a^{13}\,c}-10160640\,A\,B^2\,a^8\,c^3-21772800\,A^2\,B\,a\,c^3\,\sqrt {-a^{13}\,c}}\right )\,\sqrt {\frac {9\,\left (225\,A^2\,c\,\sqrt {-a^{13}\,c}-49\,B^2\,a\,\sqrt {-a^{13}\,c}+210\,A\,B\,a^7\,c\right )}{4096\,a^{13}\,c}}-\frac {\frac {2\,A}{a}-\frac {11\,B\,x}{16\,a}+\frac {81\,A\,c\,x^2}{16\,a^2}-\frac {7\,B\,c\,x^3}{16\,a^2}+\frac {45\,A\,c^2\,x^4}{16\,a^3}}{a^2\,\sqrt {x}+c^2\,x^{9/2}+2\,a\,c\,x^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(3/2)*(a + c*x^2)^3),x)

[Out]

2*atanh((66355200*A^2*a^10*c^4*x^(1/2)*((945*A*B)/(2048*a^6) - (2025*A^2*(-a^13*c)^(1/2))/(4096*a^13) + (441*B
^2*(-a^13*c)^(1/2))/(4096*a^12*c))^(1/2))/(46656000*A^3*a^7*c^4 - 4741632*B^3*a^2*c^2*(-a^13*c)^(1/2) - 101606
40*A*B^2*a^8*c^3 + 21772800*A^2*B*a*c^3*(-a^13*c)^(1/2)) - (14450688*B^2*a^11*c^3*x^(1/2)*((945*A*B)/(2048*a^6
) - (2025*A^2*(-a^13*c)^(1/2))/(4096*a^13) + (441*B^2*(-a^13*c)^(1/2))/(4096*a^12*c))^(1/2))/(46656000*A^3*a^7
*c^4 - 4741632*B^3*a^2*c^2*(-a^13*c)^(1/2) - 10160640*A*B^2*a^8*c^3 + 21772800*A^2*B*a*c^3*(-a^13*c)^(1/2)))*(
(9*(49*B^2*a*(-a^13*c)^(1/2) - 225*A^2*c*(-a^13*c)^(1/2) + 210*A*B*a^7*c))/(4096*a^13*c))^(1/2) + 2*atanh((663
55200*A^2*a^10*c^4*x^(1/2)*((2025*A^2*(-a^13*c)^(1/2))/(4096*a^13) + (945*A*B)/(2048*a^6) - (441*B^2*(-a^13*c)
^(1/2))/(4096*a^12*c))^(1/2))/(46656000*A^3*a^7*c^4 + 4741632*B^3*a^2*c^2*(-a^13*c)^(1/2) - 10160640*A*B^2*a^8
*c^3 - 21772800*A^2*B*a*c^3*(-a^13*c)^(1/2)) - (14450688*B^2*a^11*c^3*x^(1/2)*((2025*A^2*(-a^13*c)^(1/2))/(409
6*a^13) + (945*A*B)/(2048*a^6) - (441*B^2*(-a^13*c)^(1/2))/(4096*a^12*c))^(1/2))/(46656000*A^3*a^7*c^4 + 47416
32*B^3*a^2*c^2*(-a^13*c)^(1/2) - 10160640*A*B^2*a^8*c^3 - 21772800*A^2*B*a*c^3*(-a^13*c)^(1/2)))*((9*(225*A^2*
c*(-a^13*c)^(1/2) - 49*B^2*a*(-a^13*c)^(1/2) + 210*A*B*a^7*c))/(4096*a^13*c))^(1/2) - ((2*A)/a - (11*B*x)/(16*
a) + (81*A*c*x^2)/(16*a^2) - (7*B*c*x^3)/(16*a^2) + (45*A*c^2*x^4)/(16*a^3))/(a^2*x^(1/2) + c^2*x^(9/2) + 2*a*
c*x^(5/2))

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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**(3/2)/(c*x**2+a)**3,x)

[Out]

Timed out

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