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

Optimal. Leaf size=140 \[ \frac {3 c (4 b B-5 A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{4 b^{7/2}}-\frac {3 c \sqrt {x} (4 b B-5 A c)}{4 b^3 \sqrt {b x+c x^2}}-\frac {4 b B-5 A c}{4 b^2 \sqrt {x} \sqrt {b x+c x^2}}-\frac {A}{2 b x^{3/2} \sqrt {b x+c x^2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {792, 672, 666, 660, 207} \begin {gather*} -\frac {3 c \sqrt {x} (4 b B-5 A c)}{4 b^3 \sqrt {b x+c x^2}}-\frac {4 b B-5 A c}{4 b^2 \sqrt {x} \sqrt {b x+c x^2}}+\frac {3 c (4 b B-5 A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{4 b^{7/2}}-\frac {A}{2 b x^{3/2} \sqrt {b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-A/(2*b*x^(3/2)*Sqrt[b*x + c*x^2]) - (4*b*B - 5*A*c)/(4*b^2*Sqrt[x]*Sqrt[b*x + c*x^2]) - (3*c*(4*b*B - 5*A*c)*
Sqrt[x])/(4*b^3*Sqrt[b*x + c*x^2]) + (3*c*(4*b*B - 5*A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(4*b^(
7/2))

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 660

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(
2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^
2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 666

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((2*c*d - b*e)*(d +
e*x)^m*(a + b*x + c*x^2)^(p + 1))/(e*(p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*c*d - b*e)*(m + 2*p + 2))/((p + 1)*
(b^2 - 4*a*c)), Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^
2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && LtQ[0, m, 1] && IntegerQ[2*p]

Rule 672

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*(m + 2*p + 2))/((m + p + 1)*(2*c*d - b*e)), I
nt[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ
[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rule 792

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d*g - e*f)*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/((2*c*d - b*e)*(m + p + 1)), x] + Dist[(m*(g*(c*d - b*e)
+ c*e*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p,
x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((L
tQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]

Rubi steps

\begin {align*} \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^{3/2}} \, dx &=-\frac {A}{2 b x^{3/2} \sqrt {b x+c x^2}}+\frac {\left (\frac {1}{2} (b B-2 A c)-\frac {3}{2} (-b B+A c)\right ) \int \frac {1}{\sqrt {x} \left (b x+c x^2\right )^{3/2}} \, dx}{2 b}\\ &=-\frac {A}{2 b x^{3/2} \sqrt {b x+c x^2}}-\frac {4 b B-5 A c}{4 b^2 \sqrt {x} \sqrt {b x+c x^2}}-\frac {(3 c (4 b B-5 A c)) \int \frac {\sqrt {x}}{\left (b x+c x^2\right )^{3/2}} \, dx}{8 b^2}\\ &=-\frac {A}{2 b x^{3/2} \sqrt {b x+c x^2}}-\frac {4 b B-5 A c}{4 b^2 \sqrt {x} \sqrt {b x+c x^2}}-\frac {3 c (4 b B-5 A c) \sqrt {x}}{4 b^3 \sqrt {b x+c x^2}}-\frac {(3 c (4 b B-5 A c)) \int \frac {1}{\sqrt {x} \sqrt {b x+c x^2}} \, dx}{8 b^3}\\ &=-\frac {A}{2 b x^{3/2} \sqrt {b x+c x^2}}-\frac {4 b B-5 A c}{4 b^2 \sqrt {x} \sqrt {b x+c x^2}}-\frac {3 c (4 b B-5 A c) \sqrt {x}}{4 b^3 \sqrt {b x+c x^2}}-\frac {(3 c (4 b B-5 A c)) \operatorname {Subst}\left (\int \frac {1}{-b+x^2} \, dx,x,\frac {\sqrt {b x+c x^2}}{\sqrt {x}}\right )}{4 b^3}\\ &=-\frac {A}{2 b x^{3/2} \sqrt {b x+c x^2}}-\frac {4 b B-5 A c}{4 b^2 \sqrt {x} \sqrt {b x+c x^2}}-\frac {3 c (4 b B-5 A c) \sqrt {x}}{4 b^3 \sqrt {b x+c x^2}}+\frac {3 c (4 b B-5 A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{4 b^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.02, size = 60, normalized size = 0.43 \begin {gather*} \frac {c x^2 (5 A c-4 b B) \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};\frac {c x}{b}+1\right )-A b^2}{2 b^3 x^{3/2} \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(A*b^2) + c*(-4*b*B + 5*A*c)*x^2*Hypergeometric2F1[-1/2, 2, 1/2, 1 + (c*x)/b])/(2*b^3*x^(3/2)*Sqrt[x*(b + c*
x)])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 1.35, size = 116, normalized size = 0.83 \begin {gather*} \frac {3 \left (4 b B c-5 A c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x+c x^2}}\right )}{4 b^{7/2}}+\frac {\sqrt {b x+c x^2} \left (-2 A b^2+5 A b c x+15 A c^2 x^2-4 b^2 B x-12 b B c x^2\right )}{4 b^3 x^{5/2} (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x + c*x^2]*(-2*A*b^2 - 4*b^2*B*x + 5*A*b*c*x - 12*b*B*c*x^2 + 15*A*c^2*x^2))/(4*b^3*x^(5/2)*(b + c*x))
+ (3*(4*b*B*c - 5*A*c^2)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x + c*x^2]])/(4*b^(7/2))

________________________________________________________________________________________

fricas [A]  time = 0.42, size = 304, normalized size = 2.17 \begin {gather*} \left [-\frac {3 \, {\left ({\left (4 \, B b c^{2} - 5 \, A c^{3}\right )} x^{4} + {\left (4 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{3}\right )} \sqrt {b} \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (2 \, A b^{3} + 3 \, {\left (4 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{2} + {\left (4 \, B b^{3} - 5 \, A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{8 \, {\left (b^{4} c x^{4} + b^{5} x^{3}\right )}}, -\frac {3 \, {\left ({\left (4 \, B b c^{2} - 5 \, A c^{3}\right )} x^{4} + {\left (4 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (2 \, A b^{3} + 3 \, {\left (4 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{2} + {\left (4 \, B b^{3} - 5 \, A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{4 \, {\left (b^{4} c x^{4} + b^{5} x^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(3*((4*B*b*c^2 - 5*A*c^3)*x^4 + (4*B*b^2*c - 5*A*b*c^2)*x^3)*sqrt(b)*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2
+ b*x)*sqrt(b)*sqrt(x))/x^2) + 2*(2*A*b^3 + 3*(4*B*b^2*c - 5*A*b*c^2)*x^2 + (4*B*b^3 - 5*A*b^2*c)*x)*sqrt(c*x^
2 + b*x)*sqrt(x))/(b^4*c*x^4 + b^5*x^3), -1/4*(3*((4*B*b*c^2 - 5*A*c^3)*x^4 + (4*B*b^2*c - 5*A*b*c^2)*x^3)*sqr
t(-b)*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (2*A*b^3 + 3*(4*B*b^2*c - 5*A*b*c^2)*x^2 + (4*B*b^3 - 5*A*b
^2*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^4*c*x^4 + b^5*x^3)]

________________________________________________________________________________________

giac [A]  time = 0.28, size = 125, normalized size = 0.89 \begin {gather*} -\frac {3 \, {\left (4 \, B b c - 5 \, A c^{2}\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{4 \, \sqrt {-b} b^{3}} - \frac {2 \, {\left (B b c - A c^{2}\right )}}{\sqrt {c x + b} b^{3}} - \frac {4 \, {\left (c x + b\right )}^{\frac {3}{2}} B b c - 4 \, \sqrt {c x + b} B b^{2} c - 7 \, {\left (c x + b\right )}^{\frac {3}{2}} A c^{2} + 9 \, \sqrt {c x + b} A b c^{2}}{4 \, b^{3} c^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/4*(4*B*b*c - 5*A*c^2)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^3) - 2*(B*b*c - A*c^2)/(sqrt(c*x + b)*b^3)
- 1/4*(4*(c*x + b)^(3/2)*B*b*c - 4*sqrt(c*x + b)*B*b^2*c - 7*(c*x + b)^(3/2)*A*c^2 + 9*sqrt(c*x + b)*A*b*c^2)
/(b^3*c^2*x^2)

________________________________________________________________________________________

maple [A]  time = 0.09, size = 124, normalized size = 0.89 \begin {gather*} -\frac {\sqrt {\left (c x +b \right ) x}\, \left (15 \sqrt {c x +b}\, A \,c^{2} x^{2} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-12 \sqrt {c x +b}\, B b c \,x^{2} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-15 A \sqrt {b}\, c^{2} x^{2}+12 B \,b^{\frac {3}{2}} c \,x^{2}-5 A \,b^{\frac {3}{2}} c x +4 B \,b^{\frac {5}{2}} x +2 A \,b^{\frac {5}{2}}\right )}{4 \left (c x +b \right ) b^{\frac {7}{2}} x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/x^(5/2)*((c*x+b)*x)^(1/2)*(15*A*(c*x+b)^(1/2)*arctanh((c*x+b)^(1/2)/b^(1/2))*x^2*c^2-12*B*(c*x+b)^(1/2)*a
rctanh((c*x+b)^(1/2)/b^(1/2))*x^2*b*c+4*B*b^(5/2)*x+12*B*b^(3/2)*x^2*c+2*A*b^(5/2)-5*A*b^(3/2)*x*c-15*A*b^(1/2
)*x^2*c^2)/(c*x+b)/b^(7/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {B x + A}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} x^{\frac {3}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {A+B\,x}{x^{3/2}\,{\left (c\,x^2+b\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B x}{x^{\frac {3}{2}} \left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*x)/(x**(3/2)*(x*(b + c*x))**(3/2)), x)

________________________________________________________________________________________