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

Optimal. Leaf size=265 \[ -\frac {\left (\sqrt {a} B+A \sqrt {c}\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{2 \sqrt {2} a^{5/4} \sqrt [4]{c}}+\frac {\left (\sqrt {a} B+A \sqrt {c}\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{2 \sqrt {2} a^{5/4} \sqrt [4]{c}}-\frac {\left (\sqrt {a} B-A \sqrt {c}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{5/4} \sqrt [4]{c}}+\frac {\left (\sqrt {a} B-A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} a^{5/4} \sqrt [4]{c}}-\frac {2 A}{a \sqrt {x}} \]

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Rubi [A]  time = 0.22, antiderivative size = 265, 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 = {829, 827, 1168, 1162, 617, 204, 1165, 628} \begin {gather*} -\frac {\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^{5/4} \sqrt [4]{c}}+\frac {\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^{5/4} \sqrt [4]{c}}-\frac {\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^{5/4} \sqrt [4]{c}}+\frac {\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^{5/4} \sqrt [4]{c}}-\frac {2 A}{a \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A)/(a*Sqrt[x]) - ((Sqrt[a]*B - A*Sqrt[c])*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2]*a^(5/4)*
c^(1/4)) + ((Sqrt[a]*B - A*Sqrt[c])*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2]*a^(5/4)*c^(1/4)) -
((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]*a^(5/4)*c^(1/
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]*a^(5/4)*
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 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 )} \, dx &=-\frac {2 A}{a \sqrt {x}}+\frac {\int \frac {a B-A c x}{\sqrt {x} \left (a+c x^2\right )} \, dx}{a}\\ &=-\frac {2 A}{a \sqrt {x}}+\frac {2 \operatorname {Subst}\left (\int \frac {a B-A c x^2}{a+c x^4} \, dx,x,\sqrt {x}\right )}{a}\\ &=-\frac {2 A}{a \sqrt {x}}+\frac {\left (-A+\frac {\sqrt {a} B}{\sqrt {c}}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a} \sqrt {c}+c x^2}{a+c x^4} \, dx,x,\sqrt {x}\right )}{a}+\frac {\left (A+\frac {\sqrt {a} B}{\sqrt {c}}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a} \sqrt {c}-c x^2}{a+c x^4} \, dx,x,\sqrt {x}\right )}{a}\\ &=-\frac {2 A}{a \sqrt {x}}-\frac {\left (A-\frac {\sqrt {a} B}{\sqrt {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}-\frac {\left (A-\frac {\sqrt {a} B}{\sqrt {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}-\frac {\left (\left (A+\frac {\sqrt {a} B}{\sqrt {c}}\right ) \sqrt [4]{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 )}{2 \sqrt {2} a^{5/4}}-\frac {\left (\left (A+\frac {\sqrt {a} B}{\sqrt {c}}\right ) \sqrt [4]{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 )}{2 \sqrt {2} a^{5/4}}\\ &=-\frac {2 A}{a \sqrt {x}}-\frac {\left (A+\frac {\sqrt {a} B}{\sqrt {c}}\right ) \sqrt [4]{c} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} a^{5/4}}+\frac {\left (A+\frac {\sqrt {a} B}{\sqrt {c}}\right ) \sqrt [4]{c} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} a^{5/4}}-\frac {\left (\left (A-\frac {\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 )}{\sqrt {2} a^{5/4}}+\frac {\left (\left (A-\frac {\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 )}{\sqrt {2} a^{5/4}}\\ &=-\frac {2 A}{a \sqrt {x}}+\frac {\left (A-\frac {\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 )}{\sqrt {2} a^{5/4}}-\frac {\left (A-\frac {\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 )}{\sqrt {2} a^{5/4}}-\frac {\left (A+\frac {\sqrt {a} B}{\sqrt {c}}\right ) \sqrt [4]{c} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} a^{5/4}}+\frac {\left (A+\frac {\sqrt {a} B}{\sqrt {c}}\right ) \sqrt [4]{c} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} a^{5/4}}\\ \end {align*}

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Mathematica [C]  time = 0.08, size = 177, normalized size = 0.67 \begin {gather*} \frac {\frac {\sqrt {2} \sqrt [4]{a} B \left (-\log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )+\log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )-2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )+2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )\right )}{\sqrt [4]{c}}-\frac {8 A \, _2F_1\left (-\frac {1}{4},1;\frac {3}{4};-\frac {c x^2}{a}\right )}{\sqrt {x}}}{4 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-8*A*Hypergeometric2F1[-1/4, 1, 3/4, -((c*x^2)/a)])/Sqrt[x] + (Sqrt[2]*a^(1/4)*B*(-2*ArcTan[1 - (Sqrt[2]*c^(
1/4)*Sqrt[x])/a^(1/4)] + 2*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)] - Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/
4)*Sqrt[x] + Sqrt[c]*x] + Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x]))/c^(1/4))/(4*a)

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IntegrateAlgebraic [A]  time = 0.32, size = 149, normalized size = 0.56 \begin {gather*} -\frac {\left (\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 )}{\sqrt {2} a^{5/4} \sqrt [4]{c}}+\frac {\left (\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 )}{\sqrt {2} a^{5/4} \sqrt [4]{c}}-\frac {2 A}{a \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A)/(a*Sqrt[x]) - ((Sqrt[a]*B - A*Sqrt[c])*ArcTan[(Sqrt[a] - Sqrt[c]*x)/(Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x])])
/(Sqrt[2]*a^(5/4)*c^(1/4)) + ((Sqrt[a]*B + A*Sqrt[c])*ArcTanh[(Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[a] + Sqr
t[c]*x)])/(Sqrt[2]*a^(5/4)*c^(1/4))

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fricas [B]  time = 0.44, size = 767, normalized size = 2.89 \begin {gather*} -\frac {a x \sqrt {\frac {a^{2} \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{5} c}} + 2 \, A B}{a^{2}}} \log \left (-{\left (B^{4} a^{2} - A^{4} c^{2}\right )} \sqrt {x} + {\left (A a^{4} c \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{5} c}} + B^{3} a^{3} - A^{2} B a^{2} c\right )} \sqrt {\frac {a^{2} \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{5} c}} + 2 \, A B}{a^{2}}}\right ) - a x \sqrt {\frac {a^{2} \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{5} c}} + 2 \, A B}{a^{2}}} \log \left (-{\left (B^{4} a^{2} - A^{4} c^{2}\right )} \sqrt {x} - {\left (A a^{4} c \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{5} c}} + B^{3} a^{3} - A^{2} B a^{2} c\right )} \sqrt {\frac {a^{2} \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{5} c}} + 2 \, A B}{a^{2}}}\right ) - a x \sqrt {-\frac {a^{2} \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{5} c}} - 2 \, A B}{a^{2}}} \log \left (-{\left (B^{4} a^{2} - A^{4} c^{2}\right )} \sqrt {x} + {\left (A a^{4} c \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{5} c}} - B^{3} a^{3} + A^{2} B a^{2} c\right )} \sqrt {-\frac {a^{2} \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{5} c}} - 2 \, A B}{a^{2}}}\right ) + a x \sqrt {-\frac {a^{2} \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{5} c}} - 2 \, A B}{a^{2}}} \log \left (-{\left (B^{4} a^{2} - A^{4} c^{2}\right )} \sqrt {x} - {\left (A a^{4} c \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{5} c}} - B^{3} a^{3} + A^{2} B a^{2} c\right )} \sqrt {-\frac {a^{2} \sqrt {-\frac {B^{4} a^{2} - 2 \, A^{2} B^{2} a c + A^{4} c^{2}}{a^{5} c}} - 2 \, A B}{a^{2}}}\right ) + 4 \, A \sqrt {x}}{2 \, a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 249, normalized size = 0.94 \begin {gather*} -\frac {2 \, A}{a \sqrt {x}} + \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 \, a^{2} c^{2}} + \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 \, a^{2} c^{2}} + \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 \, a^{2} c^{2}} - \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 \, a^{2} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*A/(a*sqrt(x)) + 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))/(a^2*c^2) + 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))/(a^2*c^2) + 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))/(a^2*c^2) - 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))/(a^2*c^2)

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maple [A]  time = 0.06, size = 277, normalized size = 1.05 \begin {gather*} -\frac {\sqrt {2}\, 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}} a}-\frac {\sqrt {2}\, 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}} a}-\frac {\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 )}{4 \left (\frac {a}{c}\right )^{\frac {1}{4}} a}+\frac {\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 )}{2 a}+\frac {\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 )}{2 a}+\frac {\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 )}{4 a}-\frac {2 A}{a \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),x)

[Out]

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

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maxima [A]  time = 1.19, size = 246, normalized size = 0.93 \begin {gather*} \frac {\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}}}}{4 \, a} - \frac {2 \, A}{a \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(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))/sq
rt(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)))/a - 2*A/(a*sqrt(x))

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mupad [B]  time = 0.24, size = 602, normalized size = 2.27 \begin {gather*} 2\,\mathrm {atanh}\left (\frac {32\,A^2\,a^4\,c^4\,\sqrt {x}\,\sqrt {\frac {A\,B}{2\,a^2}-\frac {A^2\,\sqrt {-a^5\,c}}{4\,a^5}+\frac {B^2\,\sqrt {-a^5\,c}}{4\,a^4\,c}}}{16\,A^3\,a^3\,c^4-16\,B^3\,a^2\,c^2\,\sqrt {-a^5\,c}-16\,A\,B^2\,a^4\,c^3+16\,A^2\,B\,a\,c^3\,\sqrt {-a^5\,c}}-\frac {32\,B^2\,a^5\,c^3\,\sqrt {x}\,\sqrt {\frac {A\,B}{2\,a^2}-\frac {A^2\,\sqrt {-a^5\,c}}{4\,a^5}+\frac {B^2\,\sqrt {-a^5\,c}}{4\,a^4\,c}}}{16\,A^3\,a^3\,c^4-16\,B^3\,a^2\,c^2\,\sqrt {-a^5\,c}-16\,A\,B^2\,a^4\,c^3+16\,A^2\,B\,a\,c^3\,\sqrt {-a^5\,c}}\right )\,\sqrt {\frac {B^2\,a\,\sqrt {-a^5\,c}-A^2\,c\,\sqrt {-a^5\,c}+2\,A\,B\,a^3\,c}{4\,a^5\,c}}+2\,\mathrm {atanh}\left (\frac {32\,A^2\,a^4\,c^4\,\sqrt {x}\,\sqrt {\frac {A^2\,\sqrt {-a^5\,c}}{4\,a^5}+\frac {A\,B}{2\,a^2}-\frac {B^2\,\sqrt {-a^5\,c}}{4\,a^4\,c}}}{16\,A^3\,a^3\,c^4+16\,B^3\,a^2\,c^2\,\sqrt {-a^5\,c}-16\,A\,B^2\,a^4\,c^3-16\,A^2\,B\,a\,c^3\,\sqrt {-a^5\,c}}-\frac {32\,B^2\,a^5\,c^3\,\sqrt {x}\,\sqrt {\frac {A^2\,\sqrt {-a^5\,c}}{4\,a^5}+\frac {A\,B}{2\,a^2}-\frac {B^2\,\sqrt {-a^5\,c}}{4\,a^4\,c}}}{16\,A^3\,a^3\,c^4+16\,B^3\,a^2\,c^2\,\sqrt {-a^5\,c}-16\,A\,B^2\,a^4\,c^3-16\,A^2\,B\,a\,c^3\,\sqrt {-a^5\,c}}\right )\,\sqrt {\frac {A^2\,c\,\sqrt {-a^5\,c}-B^2\,a\,\sqrt {-a^5\,c}+2\,A\,B\,a^3\,c}{4\,a^5\,c}}-\frac {2\,A}{a\,\sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 14.43, size = 355, normalized size = 1.34 \begin {gather*} \begin {cases} \tilde {\infty } \left (- \frac {2 A}{5 x^{\frac {5}{2}}} - \frac {2 B}{3 x^{\frac {3}{2}}}\right ) & \text {for}\: a = 0 \wedge c = 0 \\\frac {- \frac {2 A}{5 x^{\frac {5}{2}}} - \frac {2 B}{3 x^{\frac {3}{2}}}}{c} & \text {for}\: a = 0 \\\frac {- \frac {2 A}{\sqrt {x}} + 2 B \sqrt {x}}{a} & \text {for}\: c = 0 \\- \frac {2 A}{a \sqrt {x}} + \frac {\left (-1\right )^{\frac {3}{4}} A \log {\left (- \sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 a^{\frac {5}{4}} \sqrt [4]{\frac {1}{c}}} - \frac {\left (-1\right )^{\frac {3}{4}} A \log {\left (\sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 a^{\frac {5}{4}} \sqrt [4]{\frac {1}{c}}} - \frac {\left (-1\right )^{\frac {3}{4}} A \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{a} \sqrt [4]{\frac {1}{c}}} \right )}}{a^{\frac {5}{4}} \sqrt [4]{\frac {1}{c}}} - \frac {\sqrt [4]{-1} B \sqrt [4]{\frac {1}{c}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 a^{\frac {3}{4}}} + \frac {\sqrt [4]{-1} B \sqrt [4]{\frac {1}{c}} \log {\left (\sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 a^{\frac {3}{4}}} - \frac {\sqrt [4]{-1} B \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 )}}{a^{\frac {3}{4}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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