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

Optimal. Leaf size=162 \[ -\frac{5 b^2 \sqrt{b x+c x^2} (7 b B-8 A c)}{64 c^4}+\frac{5 b^3 (7 b B-8 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{64 c^{9/2}}+\frac{5 b x \sqrt{b x+c x^2} (7 b B-8 A c)}{96 c^3}-\frac{x^2 \sqrt{b x+c x^2} (7 b B-8 A c)}{24 c^2}+\frac{B x^3 \sqrt{b x+c x^2}}{4 c} \]

[Out]

(-5*b^2*(7*b*B - 8*A*c)*Sqrt[b*x + c*x^2])/(64*c^4) + (5*b*(7*b*B - 8*A*c)*x*Sqrt[b*x + c*x^2])/(96*c^3) - ((7
*b*B - 8*A*c)*x^2*Sqrt[b*x + c*x^2])/(24*c^2) + (B*x^3*Sqrt[b*x + c*x^2])/(4*c) + (5*b^3*(7*b*B - 8*A*c)*ArcTa
nh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(64*c^(9/2))

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Rubi [A]  time = 0.156491, antiderivative size = 162, 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 = {794, 670, 640, 620, 206} \[ -\frac{5 b^2 \sqrt{b x+c x^2} (7 b B-8 A c)}{64 c^4}+\frac{5 b^3 (7 b B-8 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{64 c^{9/2}}+\frac{5 b x \sqrt{b x+c x^2} (7 b B-8 A c)}{96 c^3}-\frac{x^2 \sqrt{b x+c x^2} (7 b B-8 A c)}{24 c^2}+\frac{B x^3 \sqrt{b x+c x^2}}{4 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*b^2*(7*b*B - 8*A*c)*Sqrt[b*x + c*x^2])/(64*c^4) + (5*b*(7*b*B - 8*A*c)*x*Sqrt[b*x + c*x^2])/(96*c^3) - ((7
*b*B - 8*A*c)*x^2*Sqrt[b*x + c*x^2])/(24*c^2) + (B*x^3*Sqrt[b*x + c*x^2])/(4*c) + (5*b^3*(7*b*B - 8*A*c)*ArcTa
nh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(64*c^(9/2))

Rule 794

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), Int[(d + e*x)^m*(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] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 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)}{\sqrt{b x+c x^2}} \, dx &=\frac{B x^3 \sqrt{b x+c x^2}}{4 c}+\frac{\left (3 (-b B+A c)+\frac{1}{2} (-b B+2 A c)\right ) \int \frac{x^3}{\sqrt{b x+c x^2}} \, dx}{4 c}\\ &=-\frac{(7 b B-8 A c) x^2 \sqrt{b x+c x^2}}{24 c^2}+\frac{B x^3 \sqrt{b x+c x^2}}{4 c}+\frac{(5 b (7 b B-8 A c)) \int \frac{x^2}{\sqrt{b x+c x^2}} \, dx}{48 c^2}\\ &=\frac{5 b (7 b B-8 A c) x \sqrt{b x+c x^2}}{96 c^3}-\frac{(7 b B-8 A c) x^2 \sqrt{b x+c x^2}}{24 c^2}+\frac{B x^3 \sqrt{b x+c x^2}}{4 c}-\frac{\left (5 b^2 (7 b B-8 A c)\right ) \int \frac{x}{\sqrt{b x+c x^2}} \, dx}{64 c^3}\\ &=-\frac{5 b^2 (7 b B-8 A c) \sqrt{b x+c x^2}}{64 c^4}+\frac{5 b (7 b B-8 A c) x \sqrt{b x+c x^2}}{96 c^3}-\frac{(7 b B-8 A c) x^2 \sqrt{b x+c x^2}}{24 c^2}+\frac{B x^3 \sqrt{b x+c x^2}}{4 c}+\frac{\left (5 b^3 (7 b B-8 A c)\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{128 c^4}\\ &=-\frac{5 b^2 (7 b B-8 A c) \sqrt{b x+c x^2}}{64 c^4}+\frac{5 b (7 b B-8 A c) x \sqrt{b x+c x^2}}{96 c^3}-\frac{(7 b B-8 A c) x^2 \sqrt{b x+c x^2}}{24 c^2}+\frac{B x^3 \sqrt{b x+c x^2}}{4 c}+\frac{\left (5 b^3 (7 b B-8 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{64 c^4}\\ &=-\frac{5 b^2 (7 b B-8 A c) \sqrt{b x+c x^2}}{64 c^4}+\frac{5 b (7 b B-8 A c) x \sqrt{b x+c x^2}}{96 c^3}-\frac{(7 b B-8 A c) x^2 \sqrt{b x+c x^2}}{24 c^2}+\frac{B x^3 \sqrt{b x+c x^2}}{4 c}+\frac{5 b^3 (7 b B-8 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{64 c^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.342699, size = 129, normalized size = 0.8 \[ \frac{\sqrt{x (b+c x)} \left (\sqrt{c} \left (10 b^2 c (12 A+7 B x)-8 b c^2 x (10 A+7 B x)+16 c^3 x^2 (4 A+3 B x)-105 b^3 B\right )+\frac{15 b^{5/2} (7 b B-8 A c) \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{x} \sqrt{\frac{c x}{b}+1}}\right )}{192 c^{9/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(-105*b^3*B + 16*c^3*x^2*(4*A + 3*B*x) - 8*b*c^2*x*(10*A + 7*B*x) + 10*b^2*c*(12*A
+ 7*B*x)) + (15*b^(5/2)*(7*b*B - 8*A*c)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(19
2*c^(9/2))

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Maple [A]  time = 0.008, size = 209, normalized size = 1.3 \begin{align*}{\frac{{x}^{3}B}{4\,c}\sqrt{c{x}^{2}+bx}}-{\frac{7\,Bb{x}^{2}}{24\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{35\,{b}^{2}Bx}{96\,{c}^{3}}\sqrt{c{x}^{2}+bx}}-{\frac{35\,{b}^{3}B}{64\,{c}^{4}}\sqrt{c{x}^{2}+bx}}+{\frac{35\,{b}^{4}B}{128}\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}^{2}}{3\,c}\sqrt{c{x}^{2}+bx}}-{\frac{5\,Abx}{12\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{5\,A{b}^{2}}{8\,{c}^{3}}\sqrt{c{x}^{2}+bx}}-{\frac{5\,A{b}^{3}}{16}\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^3*(B*x+A)/(c*x^2+b*x)^(1/2),x)

[Out]

1/4*B*x^3*(c*x^2+b*x)^(1/2)/c-7/24*B*b/c^2*x^2*(c*x^2+b*x)^(1/2)+35/96*B*b^2/c^3*x*(c*x^2+b*x)^(1/2)-35/64*B*b
^3/c^4*(c*x^2+b*x)^(1/2)+35/128*B*b^4/c^(9/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))+1/3*A*x^2/c*(c*x^2+b*x
)^(1/2)-5/12*A*b/c^2*x*(c*x^2+b*x)^(1/2)+5/8*A*b^2/c^3*(c*x^2+b*x)^(1/2)-5/16*A*b^3/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^3*(B*x+A)/(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.01187, size = 598, normalized size = 3.69 \begin{align*} \left [-\frac{15 \,{\left (7 \, B b^{4} - 8 \, A b^{3} c\right )} \sqrt{c} \log \left (2 \, c x + b - 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) - 2 \,{\left (48 \, B c^{4} x^{3} - 105 \, B b^{3} c + 120 \, A b^{2} c^{2} - 8 \,{\left (7 \, B b c^{3} - 8 \, A c^{4}\right )} x^{2} + 10 \,{\left (7 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{384 \, c^{5}}, -\frac{15 \,{\left (7 \, B b^{4} - 8 \, A b^{3} c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) -{\left (48 \, B c^{4} x^{3} - 105 \, B b^{3} c + 120 \, A b^{2} c^{2} - 8 \,{\left (7 \, B b c^{3} - 8 \, A c^{4}\right )} x^{2} + 10 \,{\left (7 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{192 \, c^{5}}\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)^(1/2),x, algorithm="fricas")

[Out]

[-1/384*(15*(7*B*b^4 - 8*A*b^3*c)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(48*B*c^4*x^3 - 105
*B*b^3*c + 120*A*b^2*c^2 - 8*(7*B*b*c^3 - 8*A*c^4)*x^2 + 10*(7*B*b^2*c^2 - 8*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/c^
5, -1/192*(15*(7*B*b^4 - 8*A*b^3*c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (48*B*c^4*x^3 - 105*B*
b^3*c + 120*A*b^2*c^2 - 8*(7*B*b*c^3 - 8*A*c^4)*x^2 + 10*(7*B*b^2*c^2 - 8*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/c^5]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \left (A + B x\right )}{\sqrt{x \left (b + c x\right )}}\, 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)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.20911, size = 185, normalized size = 1.14 \begin{align*} \frac{1}{192} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (4 \,{\left (\frac{6 \, B x}{c} - \frac{7 \, B b c^{2} - 8 \, A c^{3}}{c^{4}}\right )} x + \frac{5 \,{\left (7 \, B b^{2} c - 8 \, A b c^{2}\right )}}{c^{4}}\right )} x - \frac{15 \,{\left (7 \, B b^{3} - 8 \, A b^{2} c\right )}}{c^{4}}\right )} - \frac{5 \,{\left (7 \, B b^{4} - 8 \, A b^{3} c\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{128 \, c^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)/(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

1/192*sqrt(c*x^2 + b*x)*(2*(4*(6*B*x/c - (7*B*b*c^2 - 8*A*c^3)/c^4)*x + 5*(7*B*b^2*c - 8*A*b*c^2)/c^4)*x - 15*
(7*B*b^3 - 8*A*b^2*c)/c^4) - 5/128*(7*B*b^4 - 8*A*b^3*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) -
b))/c^(9/2)