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

Optimal. Leaf size=132 \[ \frac{b^3 (3 b B-8 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{64 c^{5/2}}-\frac{b (b+2 c x) \sqrt{b x+c x^2} (3 b B-8 A c)}{64 c^2}-\frac{\left (b x+c x^2\right )^{3/2} (3 b B-8 A c)}{24 c}+\frac{B \left (b x+c x^2\right )^{5/2}}{4 c x} \]

[Out]

-(b*(3*b*B - 8*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(64*c^2) - ((3*b*B - 8*A*c)*(b*x + c*x^2)^(3/2))/(24*c) + (
B*(b*x + c*x^2)^(5/2))/(4*c*x) + (b^3*(3*b*B - 8*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(64*c^(5/2))

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Rubi [A]  time = 0.100201, antiderivative size = 132, 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 = {794, 664, 612, 620, 206} \[ \frac{b^3 (3 b B-8 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{64 c^{5/2}}-\frac{b (b+2 c x) \sqrt{b x+c x^2} (3 b B-8 A c)}{64 c^2}-\frac{\left (b x+c x^2\right )^{3/2} (3 b B-8 A c)}{24 c}+\frac{B \left (b x+c x^2\right )^{5/2}}{4 c x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(3*b*B - 8*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(64*c^2) - ((3*b*B - 8*A*c)*(b*x + c*x^2)^(3/2))/(24*c) + (
B*(b*x + c*x^2)^(5/2))/(4*c*x) + (b^3*(3*b*B - 8*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(64*c^(5/2))

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 664

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

Rule 612

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

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{(A+B x) \left (b x+c x^2\right )^{3/2}}{x} \, dx &=\frac{B \left (b x+c x^2\right )^{5/2}}{4 c x}+\frac{\left (b B-A c+\frac{5}{2} (-b B+2 A c)\right ) \int \frac{\left (b x+c x^2\right )^{3/2}}{x} \, dx}{4 c}\\ &=-\frac{(3 b B-8 A c) \left (b x+c x^2\right )^{3/2}}{24 c}+\frac{B \left (b x+c x^2\right )^{5/2}}{4 c x}-\frac{(b (3 b B-8 A c)) \int \sqrt{b x+c x^2} \, dx}{16 c}\\ &=-\frac{b (3 b B-8 A c) (b+2 c x) \sqrt{b x+c x^2}}{64 c^2}-\frac{(3 b B-8 A c) \left (b x+c x^2\right )^{3/2}}{24 c}+\frac{B \left (b x+c x^2\right )^{5/2}}{4 c x}+\frac{\left (b^3 (3 b B-8 A c)\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{128 c^2}\\ &=-\frac{b (3 b B-8 A c) (b+2 c x) \sqrt{b x+c x^2}}{64 c^2}-\frac{(3 b B-8 A c) \left (b x+c x^2\right )^{3/2}}{24 c}+\frac{B \left (b x+c x^2\right )^{5/2}}{4 c x}+\frac{\left (b^3 (3 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^2}\\ &=-\frac{b (3 b B-8 A c) (b+2 c x) \sqrt{b x+c x^2}}{64 c^2}-\frac{(3 b B-8 A c) \left (b x+c x^2\right )^{3/2}}{24 c}+\frac{B \left (b x+c x^2\right )^{5/2}}{4 c x}+\frac{b^3 (3 b B-8 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{64 c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.230145, size = 128, normalized size = 0.97 \[ \frac{\sqrt{x (b+c x)} \left (\sqrt{c} \left (6 b^2 c (4 A+B x)+8 b c^2 x (14 A+9 B x)+16 c^3 x^2 (4 A+3 B x)-9 b^3 B\right )+\frac{3 b^{5/2} (3 b B-8 A c) \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{x} \sqrt{\frac{c x}{b}+1}}\right )}{192 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(-9*b^3*B + 6*b^2*c*(4*A + B*x) + 16*c^3*x^2*(4*A + 3*B*x) + 8*b*c^2*x*(14*A + 9*B
*x)) + (3*b^(5/2)*(3*b*B - 8*A*c)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(192*c^(5/
2))

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Maple [A]  time = 0.007, size = 192, normalized size = 1.5 \begin{align*}{\frac{Bx}{4} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{bB}{8\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{3\,{b}^{2}Bx}{32\,c}\sqrt{c{x}^{2}+bx}}-{\frac{3\,{b}^{3}B}{64\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{3\,{b}^{4}B}{128}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{5}{2}}}}+{\frac{A}{3} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{Abx}{4}\sqrt{c{x}^{2}+bx}}+{\frac{A{b}^{2}}{8\,c}\sqrt{c{x}^{2}+bx}}-{\frac{A{b}^{3}}{16}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*B*x*(c*x^2+b*x)^(3/2)+1/8*B/c*(c*x^2+b*x)^(3/2)*b-3/32*B*b^2/c*(c*x^2+b*x)^(1/2)*x-3/64*B*b^3/c^2*(c*x^2+b
*x)^(1/2)+3/128*B*b^4/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))+1/3*A*(c*x^2+b*x)^(3/2)+1/4*A*b*(c*x^2
+b*x)^(1/2)*x+1/8*A/c*(c*x^2+b*x)^(1/2)*b^2-1/16*A*b^3/c^(3/2)*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((B*x+A)*(c*x^2+b*x)^(3/2)/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.96579, size = 587, normalized size = 4.45 \begin{align*} \left [-\frac{3 \,{\left (3 \, 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} - 9 \, B b^{3} c + 24 \, A b^{2} c^{2} + 8 \,{\left (9 \, B b c^{3} + 8 \, A c^{4}\right )} x^{2} + 2 \,{\left (3 \, B b^{2} c^{2} + 56 \, A b c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{384 \, c^{3}}, -\frac{3 \,{\left (3 \, 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} - 9 \, B b^{3} c + 24 \, A b^{2} c^{2} + 8 \,{\left (9 \, B b c^{3} + 8 \, A c^{4}\right )} x^{2} + 2 \,{\left (3 \, B b^{2} c^{2} + 56 \, A b c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{192 \, c^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/384*(3*(3*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 - 9*B*
b^3*c + 24*A*b^2*c^2 + 8*(9*B*b*c^3 + 8*A*c^4)*x^2 + 2*(3*B*b^2*c^2 + 56*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/c^3, -
1/192*(3*(3*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 - 9*B*b^3*c +
24*A*b^2*c^2 + 8*(9*B*b*c^3 + 8*A*c^4)*x^2 + 2*(3*B*b^2*c^2 + 56*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/c^3]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.21413, size = 186, normalized size = 1.41 \begin{align*} \frac{1}{192} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (4 \,{\left (6 \, B c x + \frac{9 \, B b c^{3} + 8 \, A c^{4}}{c^{3}}\right )} x + \frac{3 \, B b^{2} c^{2} + 56 \, A b c^{3}}{c^{3}}\right )} x - \frac{3 \,{\left (3 \, B b^{3} c - 8 \, A b^{2} c^{2}\right )}}{c^{3}}\right )} - \frac{{\left (3 \, 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{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*sqrt(c*x^2 + b*x)*(2*(4*(6*B*c*x + (9*B*b*c^3 + 8*A*c^4)/c^3)*x + (3*B*b^2*c^2 + 56*A*b*c^3)/c^3)*x - 3*
(3*B*b^3*c - 8*A*b^2*c^2)/c^3) - 1/128*(3*B*b^4 - 8*A*b^3*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c
) - b))/c^(5/2)