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

Optimal. Leaf size=289 \[ \frac {\left (3 \sqrt {a} B-A \sqrt {c}\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{8 \sqrt {2} a^{3/4} c^{7/4}}-\frac {\left (3 \sqrt {a} B-A \sqrt {c}\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{8 \sqrt {2} a^{3/4} c^{7/4}}-\frac {\left (3 \sqrt {a} B+A \sqrt {c}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{3/4} c^{7/4}}+\frac {\left (3 \sqrt {a} B+A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} a^{3/4} c^{7/4}}-\frac {\sqrt {x} (A+B x)}{2 c \left (a+c x^2\right )} \]

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Rubi [A]  time = 0.24, antiderivative size = 289, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {819, 827, 1168, 1162, 617, 204, 1165, 628} \begin {gather*} \frac {\left (3 \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 )}{8 \sqrt {2} a^{3/4} c^{7/4}}-\frac {\left (3 \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 )}{8 \sqrt {2} a^{3/4} c^{7/4}}-\frac {\left (3 \sqrt {a} B+A \sqrt {c}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{3/4} c^{7/4}}+\frac {\left (3 \sqrt {a} B+A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} a^{3/4} c^{7/4}}-\frac {\sqrt {x} (A+B x)}{2 c \left (a+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[x]*(A + B*x))/(2*c*(a + c*x^2)) - ((3*Sqrt[a]*B + A*Sqrt[c])*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/
4)])/(4*Sqrt[2]*a^(3/4)*c^(7/4)) + ((3*Sqrt[a]*B + A*Sqrt[c])*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/(
4*Sqrt[2]*a^(3/4)*c^(7/4)) + ((3*Sqrt[a]*B - A*Sqrt[c])*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c
]*x])/(8*Sqrt[2]*a^(3/4)*c^(7/4)) - ((3*Sqrt[a]*B - A*Sqrt[c])*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] +
Sqrt[c]*x])/(8*Sqrt[2]*a^(3/4)*c^(7/4))

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 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 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^{3/2} (A+B x)}{\left (a+c x^2\right )^2} \, dx &=-\frac {\sqrt {x} (A+B x)}{2 c \left (a+c x^2\right )}+\frac {\int \frac {\frac {a A}{2}+\frac {3 a B x}{2}}{\sqrt {x} \left (a+c x^2\right )} \, dx}{2 a c}\\ &=-\frac {\sqrt {x} (A+B x)}{2 c \left (a+c x^2\right )}+\frac {\operatorname {Subst}\left (\int \frac {\frac {a A}{2}+\frac {3}{2} a B x^2}{a+c x^4} \, dx,x,\sqrt {x}\right )}{a c}\\ &=-\frac {\sqrt {x} (A+B x)}{2 c \left (a+c x^2\right )}-\frac {\left (3 B-\frac {A \sqrt {c}}{\sqrt {a}}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a} \sqrt {c}-c x^2}{a+c x^4} \, dx,x,\sqrt {x}\right )}{4 c^2}+\frac {\left (3 B+\frac {A \sqrt {c}}{\sqrt {a}}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a} \sqrt {c}+c x^2}{a+c x^4} \, dx,x,\sqrt {x}\right )}{4 c^2}\\ &=-\frac {\sqrt {x} (A+B x)}{2 c \left (a+c x^2\right )}+\frac {\left (3 B+\frac {A \sqrt {c}}{\sqrt {a}}\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 )}{8 c^2}+\frac {\left (3 B+\frac {A \sqrt {c}}{\sqrt {a}}\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 )}{8 c^2}+\frac {\left (3 \sqrt {a} B-A \sqrt {c}\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 )}{8 \sqrt {2} a^{3/4} c^{7/4}}+\frac {\left (3 \sqrt {a} B-A \sqrt {c}\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 )}{8 \sqrt {2} a^{3/4} c^{7/4}}\\ &=-\frac {\sqrt {x} (A+B x)}{2 c \left (a+c x^2\right )}+\frac {\left (3 \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 )}{8 \sqrt {2} a^{3/4} c^{7/4}}-\frac {\left (3 \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 )}{8 \sqrt {2} a^{3/4} c^{7/4}}+\frac {\left (3 \sqrt {a} B+A \sqrt {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 )}{4 \sqrt {2} a^{3/4} c^{7/4}}-\frac {\left (3 \sqrt {a} B+A \sqrt {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 )}{4 \sqrt {2} a^{3/4} c^{7/4}}\\ &=-\frac {\sqrt {x} (A+B x)}{2 c \left (a+c x^2\right )}-\frac {\left (3 \sqrt {a} B+A \sqrt {c}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{3/4} c^{7/4}}+\frac {\left (3 \sqrt {a} B+A \sqrt {c}\right ) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{3/4} c^{7/4}}+\frac {\left (3 \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 )}{8 \sqrt {2} a^{3/4} c^{7/4}}-\frac {\left (3 \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 )}{8 \sqrt {2} a^{3/4} c^{7/4}}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 323, normalized size = 1.12 \begin {gather*} \frac {-\frac {\sqrt {2} \sqrt [4]{a} A \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{c^{5/4}}+\frac {\sqrt {2} \sqrt [4]{a} A \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{c^{5/4}}-\frac {2 \sqrt {2} \sqrt [4]{a} A \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{c^{5/4}}+\frac {2 \sqrt {2} \sqrt [4]{a} A \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{c^{5/4}}+\frac {8 A x^{5/2}}{a+c x^2}-\frac {12 (-a)^{3/4} B \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-a}}\right )}{c^{7/4}}+\frac {12 (-a)^{3/4} B \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-a}}\right )}{c^{7/4}}+\frac {8 B x^{7/2}}{a+c x^2}-\frac {8 A \sqrt {x}}{c}-\frac {8 B x^{3/2}}{c}}{16 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-8*A*Sqrt[x])/c - (8*B*x^(3/2))/c + (8*A*x^(5/2))/(a + c*x^2) + (8*B*x^(7/2))/(a + c*x^2) - (2*Sqrt[2]*a^(1/
4)*A*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/c^(5/4) + (2*Sqrt[2]*a^(1/4)*A*ArcTan[1 + (Sqrt[2]*c^(1/4)
*Sqrt[x])/a^(1/4)])/c^(5/4) - (12*(-a)^(3/4)*B*ArcTan[(c^(1/4)*Sqrt[x])/(-a)^(1/4)])/c^(7/4) + (12*(-a)^(3/4)*
B*ArcTanh[(c^(1/4)*Sqrt[x])/(-a)^(1/4)])/c^(7/4) - (Sqrt[2]*a^(1/4)*A*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*Sq
rt[x] + Sqrt[c]*x])/c^(5/4) + (Sqrt[2]*a^(1/4)*A*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/c
^(5/4))/(16*a)

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IntegrateAlgebraic [A]  time = 0.81, size = 171, normalized size = 0.59 \begin {gather*} -\frac {\left (3 \sqrt {a} B+A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}}\right )}{4 \sqrt {2} a^{3/4} c^{7/4}}-\frac {\left (3 \sqrt {a} B-A \sqrt {c}\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}}{\sqrt {a}+\sqrt {c} x}\right )}{4 \sqrt {2} a^{3/4} c^{7/4}}-\frac {\sqrt {x} (A+B x)}{2 c \left (a+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.47, size = 888, normalized size = 3.07 \begin {gather*} \frac {{\left (c^{2} x^{2} + a c\right )} \sqrt {-\frac {a c^{3} \sqrt {-\frac {81 \, B^{4} a^{2} - 18 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{3} c^{7}}} + 6 \, A B}{a c^{3}}} \log \left (-{\left (81 \, B^{4} a^{2} - A^{4} c^{2}\right )} \sqrt {x} + {\left (3 \, B a^{3} c^{5} \sqrt {-\frac {81 \, B^{4} a^{2} - 18 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{3} c^{7}}} - 9 \, A B^{2} a^{2} c^{2} + A^{3} a c^{3}\right )} \sqrt {-\frac {a c^{3} \sqrt {-\frac {81 \, B^{4} a^{2} - 18 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{3} c^{7}}} + 6 \, A B}{a c^{3}}}\right ) - {\left (c^{2} x^{2} + a c\right )} \sqrt {-\frac {a c^{3} \sqrt {-\frac {81 \, B^{4} a^{2} - 18 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{3} c^{7}}} + 6 \, A B}{a c^{3}}} \log \left (-{\left (81 \, B^{4} a^{2} - A^{4} c^{2}\right )} \sqrt {x} - {\left (3 \, B a^{3} c^{5} \sqrt {-\frac {81 \, B^{4} a^{2} - 18 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{3} c^{7}}} - 9 \, A B^{2} a^{2} c^{2} + A^{3} a c^{3}\right )} \sqrt {-\frac {a c^{3} \sqrt {-\frac {81 \, B^{4} a^{2} - 18 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{3} c^{7}}} + 6 \, A B}{a c^{3}}}\right ) - {\left (c^{2} x^{2} + a c\right )} \sqrt {\frac {a c^{3} \sqrt {-\frac {81 \, B^{4} a^{2} - 18 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{3} c^{7}}} - 6 \, A B}{a c^{3}}} \log \left (-{\left (81 \, B^{4} a^{2} - A^{4} c^{2}\right )} \sqrt {x} + {\left (3 \, B a^{3} c^{5} \sqrt {-\frac {81 \, B^{4} a^{2} - 18 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{3} c^{7}}} + 9 \, A B^{2} a^{2} c^{2} - A^{3} a c^{3}\right )} \sqrt {\frac {a c^{3} \sqrt {-\frac {81 \, B^{4} a^{2} - 18 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{3} c^{7}}} - 6 \, A B}{a c^{3}}}\right ) + {\left (c^{2} x^{2} + a c\right )} \sqrt {\frac {a c^{3} \sqrt {-\frac {81 \, B^{4} a^{2} - 18 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{3} c^{7}}} - 6 \, A B}{a c^{3}}} \log \left (-{\left (81 \, B^{4} a^{2} - A^{4} c^{2}\right )} \sqrt {x} - {\left (3 \, B a^{3} c^{5} \sqrt {-\frac {81 \, B^{4} a^{2} - 18 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{3} c^{7}}} + 9 \, A B^{2} a^{2} c^{2} - A^{3} a c^{3}\right )} \sqrt {\frac {a c^{3} \sqrt {-\frac {81 \, B^{4} a^{2} - 18 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{3} c^{7}}} - 6 \, A B}{a c^{3}}}\right ) - 4 \, {\left (B x + A\right )} \sqrt {x}}{8 \, {\left (c^{2} x^{2} + a c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((c^2*x^2 + a*c)*sqrt(-(a*c^3*sqrt(-(81*B^4*a^2 - 18*A^2*B^2*a*c + A^4*c^2)/(a^3*c^7)) + 6*A*B)/(a*c^3))*l
og(-(81*B^4*a^2 - A^4*c^2)*sqrt(x) + (3*B*a^3*c^5*sqrt(-(81*B^4*a^2 - 18*A^2*B^2*a*c + A^4*c^2)/(a^3*c^7)) - 9
*A*B^2*a^2*c^2 + A^3*a*c^3)*sqrt(-(a*c^3*sqrt(-(81*B^4*a^2 - 18*A^2*B^2*a*c + A^4*c^2)/(a^3*c^7)) + 6*A*B)/(a*
c^3))) - (c^2*x^2 + a*c)*sqrt(-(a*c^3*sqrt(-(81*B^4*a^2 - 18*A^2*B^2*a*c + A^4*c^2)/(a^3*c^7)) + 6*A*B)/(a*c^3
))*log(-(81*B^4*a^2 - A^4*c^2)*sqrt(x) - (3*B*a^3*c^5*sqrt(-(81*B^4*a^2 - 18*A^2*B^2*a*c + A^4*c^2)/(a^3*c^7))
- 9*A*B^2*a^2*c^2 + A^3*a*c^3)*sqrt(-(a*c^3*sqrt(-(81*B^4*a^2 - 18*A^2*B^2*a*c + A^4*c^2)/(a^3*c^7)) + 6*A*B)
/(a*c^3))) - (c^2*x^2 + a*c)*sqrt((a*c^3*sqrt(-(81*B^4*a^2 - 18*A^2*B^2*a*c + A^4*c^2)/(a^3*c^7)) - 6*A*B)/(a*
c^3))*log(-(81*B^4*a^2 - A^4*c^2)*sqrt(x) + (3*B*a^3*c^5*sqrt(-(81*B^4*a^2 - 18*A^2*B^2*a*c + A^4*c^2)/(a^3*c^
7)) + 9*A*B^2*a^2*c^2 - A^3*a*c^3)*sqrt((a*c^3*sqrt(-(81*B^4*a^2 - 18*A^2*B^2*a*c + A^4*c^2)/(a^3*c^7)) - 6*A*
B)/(a*c^3))) + (c^2*x^2 + a*c)*sqrt((a*c^3*sqrt(-(81*B^4*a^2 - 18*A^2*B^2*a*c + A^4*c^2)/(a^3*c^7)) - 6*A*B)/(
a*c^3))*log(-(81*B^4*a^2 - A^4*c^2)*sqrt(x) - (3*B*a^3*c^5*sqrt(-(81*B^4*a^2 - 18*A^2*B^2*a*c + A^4*c^2)/(a^3*
c^7)) + 9*A*B^2*a^2*c^2 - A^3*a*c^3)*sqrt((a*c^3*sqrt(-(81*B^4*a^2 - 18*A^2*B^2*a*c + A^4*c^2)/(a^3*c^7)) - 6*
A*B)/(a*c^3))) - 4*(B*x + A)*sqrt(x))/(c^2*x^2 + a*c)

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giac [A]  time = 0.20, size = 271, normalized size = 0.94 \begin {gather*} -\frac {B x^{\frac {3}{2}} + A \sqrt {x}}{2 \, {\left (c x^{2} + a\right )} c} + \frac {\sqrt {2} {\left (\left (a c^{3}\right )^{\frac {1}{4}} A c^{2} + 3 \, \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 )}{8 \, a c^{4}} + \frac {\sqrt {2} {\left (\left (a c^{3}\right )^{\frac {1}{4}} A c^{2} + 3 \, \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 )}{8 \, a c^{4}} + \frac {\sqrt {2} {\left (\left (a c^{3}\right )^{\frac {1}{4}} A c^{2} - 3 \, \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 )}{16 \, a c^{4}} - \frac {\sqrt {2} {\left (\left (a c^{3}\right )^{\frac {1}{4}} A c^{2} - 3 \, \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 )}{16 \, a c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(B*x^(3/2) + A*sqrt(x))/((c*x^2 + a)*c) + 1/8*sqrt(2)*((a*c^3)^(1/4)*A*c^2 + 3*(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*c^4) + 1/8*sqrt(2)*((a*c^3)^(1/4)*A*c^2 + 3*(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*c^4) + 1/16*sqrt(2)*((a*c^3)^
(1/4)*A*c^2 - 3*(a*c^3)^(3/4)*B)*log(sqrt(2)*sqrt(x)*(a/c)^(1/4) + x + sqrt(a/c))/(a*c^4) - 1/16*sqrt(2)*((a*c
^3)^(1/4)*A*c^2 - 3*(a*c^3)^(3/4)*B)*log(-sqrt(2)*sqrt(x)*(a/c)^(1/4) + x + sqrt(a/c))/(a*c^4)

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maple [A]  time = 0.08, size = 307, normalized size = 1.06 \begin {gather*} \frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{8 a c}+\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{8 a c}+\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \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 )}{16 a c}+\frac {3 \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {a}{c}\right )^{\frac {1}{4}} c^{2}}+\frac {3 \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {a}{c}\right )^{\frac {1}{4}} c^{2}}+\frac {3 \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 )}{16 \left (\frac {a}{c}\right )^{\frac {1}{4}} c^{2}}+\frac {-\frac {B \,x^{\frac {3}{2}}}{2 c}-\frac {A \sqrt {x}}{2 c}}{c \,x^{2}+a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(-1/4*B/c*x^(3/2)-1/4*A/c*x^(1/2))/(c*x^2+a)+1/16/c*A*(a/c)^(1/4)/a*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)))+1/8/c*A*(a/c)^(1/4)/a*2^(1/2)*arctan(2^(1/2)/(a/c)
^(1/4)*x^(1/2)+1)+1/8/c*A*(a/c)^(1/4)/a*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)-1)+3/16/c^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)))+3/8/c^2*B/(
a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)+1)+3/8/c^2*B/(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1
/4)*x^(1/2)-1)

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maxima [A]  time = 1.21, size = 259, normalized size = 0.90 \begin {gather*} -\frac {B x^{\frac {3}{2}} + A \sqrt {x}}{2 \, {\left (c^{2} x^{2} + a c\right )}} + \frac {\frac {2 \, \sqrt {2} {\left (3 \, 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 (3 \, 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 (3 \, 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 (3 \, 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}}}}{16 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(B*x^(3/2) + A*sqrt(x))/(c^2*x^2 + a*c) + 1/16*(2*sqrt(2)*(3*B*sqrt(a) + A*sqrt(c))*arctan(1/2*sqrt(2)*(s
qrt(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)*(3*B*sqrt(a) + A*sqrt(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)*(3*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)*(3*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)))/c

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mupad [B]  time = 1.29, size = 656, normalized size = 2.27 \begin {gather*} 2\,\mathrm {atanh}\left (\frac {18\,B^2\,a\,\sqrt {x}\,\sqrt {\frac {9\,B^2\,\sqrt {-a^3\,c^7}}{64\,a^2\,c^7}-\frac {A^2\,\sqrt {-a^3\,c^7}}{64\,a^3\,c^6}-\frac {3\,A\,B}{32\,a\,c^3}}}{\frac {3\,A^2\,B}{4\,c}-\frac {27\,B^3\,a}{4\,c^2}+\frac {A^3\,\sqrt {-a^3\,c^7}}{4\,a^2\,c^4}-\frac {9\,A\,B^2\,\sqrt {-a^3\,c^7}}{4\,a\,c^5}}-\frac {2\,A^2\,c\,\sqrt {x}\,\sqrt {\frac {9\,B^2\,\sqrt {-a^3\,c^7}}{64\,a^2\,c^7}-\frac {A^2\,\sqrt {-a^3\,c^7}}{64\,a^3\,c^6}-\frac {3\,A\,B}{32\,a\,c^3}}}{\frac {3\,A^2\,B}{4\,c}-\frac {27\,B^3\,a}{4\,c^2}+\frac {A^3\,\sqrt {-a^3\,c^7}}{4\,a^2\,c^4}-\frac {9\,A\,B^2\,\sqrt {-a^3\,c^7}}{4\,a\,c^5}}\right )\,\sqrt {-\frac {A^2\,c\,\sqrt {-a^3\,c^7}-9\,B^2\,a\,\sqrt {-a^3\,c^7}+6\,A\,B\,a^2\,c^4}{64\,a^3\,c^7}}+2\,\mathrm {atanh}\left (\frac {18\,B^2\,a\,\sqrt {x}\,\sqrt {\frac {A^2\,\sqrt {-a^3\,c^7}}{64\,a^3\,c^6}-\frac {3\,A\,B}{32\,a\,c^3}-\frac {9\,B^2\,\sqrt {-a^3\,c^7}}{64\,a^2\,c^7}}}{\frac {3\,A^2\,B}{4\,c}-\frac {27\,B^3\,a}{4\,c^2}-\frac {A^3\,\sqrt {-a^3\,c^7}}{4\,a^2\,c^4}+\frac {9\,A\,B^2\,\sqrt {-a^3\,c^7}}{4\,a\,c^5}}-\frac {2\,A^2\,c\,\sqrt {x}\,\sqrt {\frac {A^2\,\sqrt {-a^3\,c^7}}{64\,a^3\,c^6}-\frac {3\,A\,B}{32\,a\,c^3}-\frac {9\,B^2\,\sqrt {-a^3\,c^7}}{64\,a^2\,c^7}}}{\frac {3\,A^2\,B}{4\,c}-\frac {27\,B^3\,a}{4\,c^2}-\frac {A^3\,\sqrt {-a^3\,c^7}}{4\,a^2\,c^4}+\frac {9\,A\,B^2\,\sqrt {-a^3\,c^7}}{4\,a\,c^5}}\right )\,\sqrt {-\frac {9\,B^2\,a\,\sqrt {-a^3\,c^7}-A^2\,c\,\sqrt {-a^3\,c^7}+6\,A\,B\,a^2\,c^4}{64\,a^3\,c^7}}-\frac {\frac {A\,\sqrt {x}}{2\,c}+\frac {B\,x^{3/2}}{2\,c}}{c\,x^2+a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*atanh((18*B^2*a*x^(1/2)*((9*B^2*(-a^3*c^7)^(1/2))/(64*a^2*c^7) - (A^2*(-a^3*c^7)^(1/2))/(64*a^3*c^6) - (3*A*
B)/(32*a*c^3))^(1/2))/((3*A^2*B)/(4*c) - (27*B^3*a)/(4*c^2) + (A^3*(-a^3*c^7)^(1/2))/(4*a^2*c^4) - (9*A*B^2*(-
a^3*c^7)^(1/2))/(4*a*c^5)) - (2*A^2*c*x^(1/2)*((9*B^2*(-a^3*c^7)^(1/2))/(64*a^2*c^7) - (A^2*(-a^3*c^7)^(1/2))/
(64*a^3*c^6) - (3*A*B)/(32*a*c^3))^(1/2))/((3*A^2*B)/(4*c) - (27*B^3*a)/(4*c^2) + (A^3*(-a^3*c^7)^(1/2))/(4*a^
2*c^4) - (9*A*B^2*(-a^3*c^7)^(1/2))/(4*a*c^5)))*(-(A^2*c*(-a^3*c^7)^(1/2) - 9*B^2*a*(-a^3*c^7)^(1/2) + 6*A*B*a
^2*c^4)/(64*a^3*c^7))^(1/2) + 2*atanh((18*B^2*a*x^(1/2)*((A^2*(-a^3*c^7)^(1/2))/(64*a^3*c^6) - (3*A*B)/(32*a*c
^3) - (9*B^2*(-a^3*c^7)^(1/2))/(64*a^2*c^7))^(1/2))/((3*A^2*B)/(4*c) - (27*B^3*a)/(4*c^2) - (A^3*(-a^3*c^7)^(1
/2))/(4*a^2*c^4) + (9*A*B^2*(-a^3*c^7)^(1/2))/(4*a*c^5)) - (2*A^2*c*x^(1/2)*((A^2*(-a^3*c^7)^(1/2))/(64*a^3*c^
6) - (3*A*B)/(32*a*c^3) - (9*B^2*(-a^3*c^7)^(1/2))/(64*a^2*c^7))^(1/2))/((3*A^2*B)/(4*c) - (27*B^3*a)/(4*c^2)
- (A^3*(-a^3*c^7)^(1/2))/(4*a^2*c^4) + (9*A*B^2*(-a^3*c^7)^(1/2))/(4*a*c^5)))*(-(9*B^2*a*(-a^3*c^7)^(1/2) - A^
2*c*(-a^3*c^7)^(1/2) + 6*A*B*a^2*c^4)/(64*a^3*c^7))^(1/2) - ((A*x^(1/2))/(2*c) + (B*x^(3/2))/(2*c))/(a + c*x^2
)

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sympy [A]  time = 90.99, size = 1316, normalized size = 4.55

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(-2*A/(3*x**(3/2)) - 2*B/sqrt(x)), Eq(a, 0) & Eq(c, 0)), ((-2*A/(3*x**(3/2)) - 2*B/sqrt(x))/c**
2, Eq(a, 0)), ((2*A*x**(5/2)/5 + 2*B*x**(7/2)/7)/a**2, Eq(c, 0)), (-4*(-1)**(1/4)*A*a**(5/4)*c*sqrt(x)*(1/c)**
(1/4)/(8*(-1)**(1/4)*a**(9/4)*c**2*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(5/4)*c**3*x**2*(1/c)**(1/4)) - I*A*a**(3/2
)*c*sqrt(1/c)*log(-(-1)**(1/4)*a**(1/4)*(1/c)**(1/4) + sqrt(x))/(8*(-1)**(1/4)*a**(9/4)*c**2*(1/c)**(1/4) + 8*
(-1)**(1/4)*a**(5/4)*c**3*x**2*(1/c)**(1/4)) + I*A*a**(3/2)*c*sqrt(1/c)*log((-1)**(1/4)*a**(1/4)*(1/c)**(1/4)
+ sqrt(x))/(8*(-1)**(1/4)*a**(9/4)*c**2*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(5/4)*c**3*x**2*(1/c)**(1/4)) - 2*I*A*
a**(3/2)*c*sqrt(1/c)*atan((-1)**(3/4)*sqrt(x)/(a**(1/4)*(1/c)**(1/4)))/(8*(-1)**(1/4)*a**(9/4)*c**2*(1/c)**(1/
4) + 8*(-1)**(1/4)*a**(5/4)*c**3*x**2*(1/c)**(1/4)) - I*A*sqrt(a)*c**2*x**2*sqrt(1/c)*log(-(-1)**(1/4)*a**(1/4
)*(1/c)**(1/4) + sqrt(x))/(8*(-1)**(1/4)*a**(9/4)*c**2*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(5/4)*c**3*x**2*(1/c)**
(1/4)) + I*A*sqrt(a)*c**2*x**2*sqrt(1/c)*log((-1)**(1/4)*a**(1/4)*(1/c)**(1/4) + sqrt(x))/(8*(-1)**(1/4)*a**(9
/4)*c**2*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(5/4)*c**3*x**2*(1/c)**(1/4)) - 2*I*A*sqrt(a)*c**2*x**2*sqrt(1/c)*ata
n((-1)**(3/4)*sqrt(x)/(a**(1/4)*(1/c)**(1/4)))/(8*(-1)**(1/4)*a**(9/4)*c**2*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(5
/4)*c**3*x**2*(1/c)**(1/4)) - 4*(-1)**(1/4)*B*a**(5/4)*c*x**(3/2)*(1/c)**(1/4)/(8*(-1)**(1/4)*a**(9/4)*c**2*(1
/c)**(1/4) + 8*(-1)**(1/4)*a**(5/4)*c**3*x**2*(1/c)**(1/4)) + 3*B*a**2*log(-(-1)**(1/4)*a**(1/4)*(1/c)**(1/4)
+ sqrt(x))/(8*(-1)**(1/4)*a**(9/4)*c**2*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(5/4)*c**3*x**2*(1/c)**(1/4)) - 3*B*a*
*2*log((-1)**(1/4)*a**(1/4)*(1/c)**(1/4) + sqrt(x))/(8*(-1)**(1/4)*a**(9/4)*c**2*(1/c)**(1/4) + 8*(-1)**(1/4)*
a**(5/4)*c**3*x**2*(1/c)**(1/4)) - 6*B*a**2*atan((-1)**(3/4)*sqrt(x)/(a**(1/4)*(1/c)**(1/4)))/(8*(-1)**(1/4)*a
**(9/4)*c**2*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(5/4)*c**3*x**2*(1/c)**(1/4)) + 3*B*a*c*x**2*log(-(-1)**(1/4)*a**
(1/4)*(1/c)**(1/4) + sqrt(x))/(8*(-1)**(1/4)*a**(9/4)*c**2*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(5/4)*c**3*x**2*(1/
c)**(1/4)) - 3*B*a*c*x**2*log((-1)**(1/4)*a**(1/4)*(1/c)**(1/4) + sqrt(x))/(8*(-1)**(1/4)*a**(9/4)*c**2*(1/c)*
*(1/4) + 8*(-1)**(1/4)*a**(5/4)*c**3*x**2*(1/c)**(1/4)) - 6*B*a*c*x**2*atan((-1)**(3/4)*sqrt(x)/(a**(1/4)*(1/c
)**(1/4)))/(8*(-1)**(1/4)*a**(9/4)*c**2*(1/c)**(1/4) + 8*(-1)**(1/4)*a**(5/4)*c**3*x**2*(1/c)**(1/4)), True))

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