3.120 \(\int \frac{x^4 (A+B x)}{(b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=170 \[ -\frac{5 b^2 (7 b B-6 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{8 c^{9/2}}+\frac{x^2 \sqrt{b x+c x^2} (7 b B-6 A c)}{3 b c^2}-\frac{5 x \sqrt{b x+c x^2} (7 b B-6 A c)}{12 c^3}+\frac{5 b \sqrt{b x+c x^2} (7 b B-6 A c)}{8 c^4}-\frac{2 x^4 (b B-A c)}{b c \sqrt{b x+c x^2}} \]

[Out]

(-2*(b*B - A*c)*x^4)/(b*c*Sqrt[b*x + c*x^2]) + (5*b*(7*b*B - 6*A*c)*Sqrt[b*x + c*x^2])/(8*c^4) - (5*(7*b*B - 6
*A*c)*x*Sqrt[b*x + c*x^2])/(12*c^3) + ((7*b*B - 6*A*c)*x^2*Sqrt[b*x + c*x^2])/(3*b*c^2) - (5*b^2*(7*b*B - 6*A*
c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(8*c^(9/2))

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Rubi [A]  time = 0.151606, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{5 b^2 (7 b B-6 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{8 c^{9/2}}+\frac{x^2 \sqrt{b x+c x^2} (7 b B-6 A c)}{3 b c^2}-\frac{5 x \sqrt{b x+c x^2} (7 b B-6 A c)}{12 c^3}+\frac{5 b \sqrt{b x+c x^2} (7 b B-6 A c)}{8 c^4}-\frac{2 x^4 (b B-A c)}{b c \sqrt{b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(A + B*x))/(b*x + c*x^2)^(3/2),x]

[Out]

(-2*(b*B - A*c)*x^4)/(b*c*Sqrt[b*x + c*x^2]) + (5*b*(7*b*B - 6*A*c)*Sqrt[b*x + c*x^2])/(8*c^4) - (5*(7*b*B - 6
*A*c)*x*Sqrt[b*x + c*x^2])/(12*c^3) + ((7*b*B - 6*A*c)*x^2*Sqrt[b*x + c*x^2])/(3*b*c^2) - (5*b^2*(7*b*B - 6*A*
c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(8*c^(9/2))

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^4 (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx &=-\frac{2 (b B-A c) x^4}{b c \sqrt{b x+c x^2}}-\left (\frac{6 A}{b}-\frac{7 B}{c}\right ) \int \frac{x^3}{\sqrt{b x+c x^2}} \, dx\\ &=-\frac{2 (b B-A c) x^4}{b c \sqrt{b x+c x^2}}+\frac{(7 b B-6 A c) x^2 \sqrt{b x+c x^2}}{3 b c^2}-\frac{(5 (7 b B-6 A c)) \int \frac{x^2}{\sqrt{b x+c x^2}} \, dx}{6 c^2}\\ &=-\frac{2 (b B-A c) x^4}{b c \sqrt{b x+c x^2}}-\frac{5 (7 b B-6 A c) x \sqrt{b x+c x^2}}{12 c^3}+\frac{(7 b B-6 A c) x^2 \sqrt{b x+c x^2}}{3 b c^2}+\frac{(5 b (7 b B-6 A c)) \int \frac{x}{\sqrt{b x+c x^2}} \, dx}{8 c^3}\\ &=-\frac{2 (b B-A c) x^4}{b c \sqrt{b x+c x^2}}+\frac{5 b (7 b B-6 A c) \sqrt{b x+c x^2}}{8 c^4}-\frac{5 (7 b B-6 A c) x \sqrt{b x+c x^2}}{12 c^3}+\frac{(7 b B-6 A c) x^2 \sqrt{b x+c x^2}}{3 b c^2}-\frac{\left (5 b^2 (7 b B-6 A c)\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{16 c^4}\\ &=-\frac{2 (b B-A c) x^4}{b c \sqrt{b x+c x^2}}+\frac{5 b (7 b B-6 A c) \sqrt{b x+c x^2}}{8 c^4}-\frac{5 (7 b B-6 A c) x \sqrt{b x+c x^2}}{12 c^3}+\frac{(7 b B-6 A c) x^2 \sqrt{b x+c x^2}}{3 b c^2}-\frac{\left (5 b^2 (7 b B-6 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{8 c^4}\\ &=-\frac{2 (b B-A c) x^4}{b c \sqrt{b x+c x^2}}+\frac{5 b (7 b B-6 A c) \sqrt{b x+c x^2}}{8 c^4}-\frac{5 (7 b B-6 A c) x \sqrt{b x+c x^2}}{12 c^3}+\frac{(7 b B-6 A c) x^2 \sqrt{b x+c x^2}}{3 b c^2}-\frac{5 b^2 (7 b B-6 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{8 c^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.223182, size = 137, normalized size = 0.81 \[ \frac{\frac{(b+c x) (7 b B-6 A c) \left (c x \sqrt{\frac{c x}{b}+1} \left (15 b^2-10 b c x+8 c^2 x^2\right )-15 b^{5/2} \sqrt{c} \sqrt{x} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )\right )}{3 \sqrt{\frac{c x}{b}+1}}+16 c^4 x^4 (A c-b B)}{8 b c^5 \sqrt{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(A + B*x))/(b*x + c*x^2)^(3/2),x]

[Out]

(16*c^4*(-(b*B) + A*c)*x^4 + ((7*b*B - 6*A*c)*(b + c*x)*(c*x*Sqrt[1 + (c*x)/b]*(15*b^2 - 10*b*c*x + 8*c^2*x^2)
- 15*b^(5/2)*Sqrt[c]*Sqrt[x]*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]]))/(3*Sqrt[1 + (c*x)/b]))/(8*b*c^5*Sqrt[x*(b +
c*x)])

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Maple [A]  time = 0.009, size = 215, normalized size = 1.3 \begin{align*}{\frac{{x}^{4}B}{3\,c}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}-{\frac{7\,bB{x}^{3}}{12\,{c}^{2}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}+{\frac{35\,{b}^{2}B{x}^{2}}{24\,{c}^{3}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}+{\frac{35\,{b}^{3}Bx}{8\,{c}^{4}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}-{\frac{35\,{b}^{3}B}{16}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{9}{2}}}}+{\frac{A{x}^{3}}{2\,c}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}-{\frac{5\,Ab{x}^{2}}{4\,{c}^{2}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}-{\frac{15\,A{b}^{2}x}{4\,{c}^{3}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}+{\frac{15\,A{b}^{2}}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(B*x+A)/(c*x^2+b*x)^(3/2),x)

[Out]

1/3*B*x^4/c/(c*x^2+b*x)^(1/2)-7/12*B*b/c^2*x^3/(c*x^2+b*x)^(1/2)+35/24*B*b^2/c^3*x^2/(c*x^2+b*x)^(1/2)+35/8*B*
b^3/c^4/(c*x^2+b*x)^(1/2)*x-35/16*B*b^3/c^(9/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))+1/2*A*x^3/c/(c*x^2+b
*x)^(1/2)-5/4*A*b/c^2*x^2/(c*x^2+b*x)^(1/2)-15/4*A*b^2/c^3/(c*x^2+b*x)^(1/2)*x+15/8*A*b^2/c^(7/2)*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^4*(B*x+A)/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.01869, size = 699, normalized size = 4.11 \begin{align*} \left [-\frac{15 \,{\left (7 \, B b^{4} - 6 \, A b^{3} c +{\left (7 \, B b^{3} c - 6 \, A b^{2} 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 (8 \, B c^{4} x^{3} + 105 \, B b^{3} c - 90 \, A b^{2} c^{2} - 2 \,{\left (7 \, B b c^{3} - 6 \, A c^{4}\right )} x^{2} + 5 \,{\left (7 \, B b^{2} c^{2} - 6 \, A b c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{48 \,{\left (c^{6} x + b c^{5}\right )}}, \frac{15 \,{\left (7 \, B b^{4} - 6 \, A b^{3} c +{\left (7 \, B b^{3} c - 6 \, A b^{2} c^{2}\right )} x\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) +{\left (8 \, B c^{4} x^{3} + 105 \, B b^{3} c - 90 \, A b^{2} c^{2} - 2 \,{\left (7 \, B b c^{3} - 6 \, A c^{4}\right )} x^{2} + 5 \,{\left (7 \, B b^{2} c^{2} - 6 \, A b c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{24 \,{\left (c^{6} x + b c^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x+A)/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

[-1/48*(15*(7*B*b^4 - 6*A*b^3*c + (7*B*b^3*c - 6*A*b^2*c^2)*x)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqr
t(c)) - 2*(8*B*c^4*x^3 + 105*B*b^3*c - 90*A*b^2*c^2 - 2*(7*B*b*c^3 - 6*A*c^4)*x^2 + 5*(7*B*b^2*c^2 - 6*A*b*c^3
)*x)*sqrt(c*x^2 + b*x))/(c^6*x + b*c^5), 1/24*(15*(7*B*b^4 - 6*A*b^3*c + (7*B*b^3*c - 6*A*b^2*c^2)*x)*sqrt(-c)
*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (8*B*c^4*x^3 + 105*B*b^3*c - 90*A*b^2*c^2 - 2*(7*B*b*c^3 - 6*A*c^4
)*x^2 + 5*(7*B*b^2*c^2 - 6*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/(c^6*x + b*c^5)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4} \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**4*(B*x+A)/(c*x**2+b*x)**(3/2),x)

[Out]

Integral(x**4*(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^4*(B*x+A)/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError