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

Optimal. Leaf size=135 \[ \frac{x \sqrt{b x+c x^2} (5 b B-4 A c)}{2 b c^2}-\frac{3 \sqrt{b x+c x^2} (5 b B-4 A c)}{4 c^3}+\frac{3 b (5 b B-4 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{4 c^{7/2}}-\frac{2 x^3 (b B-A c)}{b c \sqrt{b x+c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.116993, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {788, 670, 640, 620, 206} \[ \frac{x \sqrt{b x+c x^2} (5 b B-4 A c)}{2 b c^2}-\frac{3 \sqrt{b x+c x^2} (5 b B-4 A c)}{4 c^3}+\frac{3 b (5 b B-4 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{4 c^{7/2}}-\frac{2 x^3 (b B-A c)}{b c \sqrt{b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 788

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

Rule 670

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

Rule 640

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

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 206

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

Rubi steps

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

Mathematica [A]  time = 0.0993803, size = 109, normalized size = 0.81 \[ \frac{\sqrt{c} x \left (b c (12 A-5 B x)+2 c^2 x (2 A+B x)-15 b^2 B\right )+3 b^{3/2} \sqrt{x} \sqrt{\frac{c x}{b}+1} (5 b B-4 A c) \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{4 c^{7/2} \sqrt{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*x*(-15*b^2*B + b*c*(12*A - 5*B*x) + 2*c^2*x*(2*A + B*x)) + 3*b^(3/2)*(5*b*B - 4*A*c)*Sqrt[x]*Sqrt[1 +
(c*x)/b]*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(4*c^(7/2)*Sqrt[x*(b + c*x)])

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Maple [A]  time = 0.007, size = 166, normalized size = 1.2 \begin{align*}{\frac{{x}^{3}B}{2\,c}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}-{\frac{5\,Bb{x}^{2}}{4\,{c}^{2}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}-{\frac{15\,{b}^{2}Bx}{4\,{c}^{3}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}+{\frac{15\,{b}^{2}B}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{7}{2}}}}+{\frac{A{x}^{2}}{c}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}+3\,{\frac{Abx}{{c}^{2}\sqrt{c{x}^{2}+bx}}}-{\frac{3\,Ab}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*B*x^3/c/(c*x^2+b*x)^(1/2)-5/4*B*b/c^2*x^2/(c*x^2+b*x)^(1/2)-15/4*B*b^2/c^3/(c*x^2+b*x)^(1/2)*x+15/8*B*b^2/
c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))+A*x^2/c/(c*x^2+b*x)^(1/2)+3*A*b/c^2/(c*x^2+b*x)^(1/2)*x-3/2*
A*b/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.94568, size = 585, normalized size = 4.33 \begin{align*} \left [-\frac{3 \,{\left (5 \, B b^{3} - 4 \, A b^{2} c +{\left (5 \, B b^{2} c - 4 \, A b c^{2}\right )} x\right )} \sqrt{c} \log \left (2 \, c x + b - 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) - 2 \,{\left (2 \, B c^{3} x^{2} - 15 \, B b^{2} c + 12 \, A b c^{2} -{\left (5 \, B b c^{2} - 4 \, A c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{8 \,{\left (c^{5} x + b c^{4}\right )}}, -\frac{3 \,{\left (5 \, B b^{3} - 4 \, A b^{2} c +{\left (5 \, B b^{2} c - 4 \, A b c^{2}\right )} x\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) -{\left (2 \, B c^{3} x^{2} - 15 \, B b^{2} c + 12 \, A b c^{2} -{\left (5 \, B b c^{2} - 4 \, A c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{4 \,{\left (c^{5} x + b c^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(3*(5*B*b^3 - 4*A*b^2*c + (5*B*b^2*c - 4*A*b*c^2)*x)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)
) - 2*(2*B*c^3*x^2 - 15*B*b^2*c + 12*A*b*c^2 - (5*B*b*c^2 - 4*A*c^3)*x)*sqrt(c*x^2 + b*x))/(c^5*x + b*c^4), -1
/4*(3*(5*B*b^3 - 4*A*b^2*c + (5*B*b^2*c - 4*A*b*c^2)*x)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (2
*B*c^3*x^2 - 15*B*b^2*c + 12*A*b*c^2 - (5*B*b*c^2 - 4*A*c^3)*x)*sqrt(c*x^2 + b*x))/(c^5*x + b*c^4)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError