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

Optimal. Leaf size=134 \[ \frac{3 b^2 (b+2 c x) \sqrt{b x+c x^2} (b B-2 A c)}{128 c^3}-\frac{3 b^4 (b B-2 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{128 c^{7/2}}-\frac{(b+2 c x) \left (b x+c x^2\right )^{3/2} (b B-2 A c)}{16 c^2}+\frac{B \left (b x+c x^2\right )^{5/2}}{5 c} \]

[Out]

(3*b^2*(b*B - 2*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(128*c^3) - ((b*B - 2*A*c)*(b + 2*c*x)*(b*x + c*x^2)^(3/2)
)/(16*c^2) + (B*(b*x + c*x^2)^(5/2))/(5*c) - (3*b^4*(b*B - 2*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(128
*c^(7/2))

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Rubi [A]  time = 0.0511963, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {640, 612, 620, 206} \[ \frac{3 b^2 (b+2 c x) \sqrt{b x+c x^2} (b B-2 A c)}{128 c^3}-\frac{3 b^4 (b B-2 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{128 c^{7/2}}-\frac{(b+2 c x) \left (b x+c x^2\right )^{3/2} (b B-2 A c)}{16 c^2}+\frac{B \left (b x+c x^2\right )^{5/2}}{5 c} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)*(b*x + c*x^2)^(3/2),x]

[Out]

(3*b^2*(b*B - 2*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(128*c^3) - ((b*B - 2*A*c)*(b + 2*c*x)*(b*x + c*x^2)^(3/2)
)/(16*c^2) + (B*(b*x + c*x^2)^(5/2))/(5*c) - (3*b^4*(b*B - 2*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(128
*c^(7/2))

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 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 (A+B x) \left (b x+c x^2\right )^{3/2} \, dx &=\frac{B \left (b x+c x^2\right )^{5/2}}{5 c}+\frac{(-b B+2 A c) \int \left (b x+c x^2\right )^{3/2} \, dx}{2 c}\\ &=-\frac{(b B-2 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac{B \left (b x+c x^2\right )^{5/2}}{5 c}+\frac{\left (3 b^2 (b B-2 A c)\right ) \int \sqrt{b x+c x^2} \, dx}{32 c^2}\\ &=\frac{3 b^2 (b B-2 A c) (b+2 c x) \sqrt{b x+c x^2}}{128 c^3}-\frac{(b B-2 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac{B \left (b x+c x^2\right )^{5/2}}{5 c}-\frac{\left (3 b^4 (b B-2 A c)\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{256 c^3}\\ &=\frac{3 b^2 (b B-2 A c) (b+2 c x) \sqrt{b x+c x^2}}{128 c^3}-\frac{(b B-2 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac{B \left (b x+c x^2\right )^{5/2}}{5 c}-\frac{\left (3 b^4 (b B-2 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{128 c^3}\\ &=\frac{3 b^2 (b B-2 A c) (b+2 c x) \sqrt{b x+c x^2}}{128 c^3}-\frac{(b B-2 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac{B \left (b x+c x^2\right )^{5/2}}{5 c}-\frac{3 b^4 (b B-2 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{128 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.262827, size = 146, normalized size = 1.09 \[ \frac{\sqrt{x (b+c x)} \left (\sqrt{c} \left (4 b^2 c^2 x (5 A+2 B x)-10 b^3 c (3 A+B x)+16 b c^3 x^2 (15 A+11 B x)+32 c^4 x^3 (5 A+4 B x)+15 b^4 B\right )-\frac{15 b^{7/2} (b B-2 A c) \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{x} \sqrt{\frac{c x}{b}+1}}\right )}{640 c^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(15*b^4*B - 10*b^3*c*(3*A + B*x) + 4*b^2*c^2*x*(5*A + 2*B*x) + 32*c^4*x^3*(5*A + 4
*B*x) + 16*b*c^3*x^2*(15*A + 11*B*x)) - (15*b^(7/2)*(b*B - 2*A*c)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]
*Sqrt[1 + (c*x)/b])))/(640*c^(7/2))

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Maple [B]  time = 0.006, size = 239, normalized size = 1.8 \begin{align*}{\frac{B}{5\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}-{\frac{bBx}{8\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{{b}^{2}B}{16\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{3\,{b}^{3}Bx}{64\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{3\,{b}^{4}B}{128\,{c}^{3}}\sqrt{c{x}^{2}+bx}}-{\frac{3\,B{b}^{5}}{256}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{7}{2}}}}+{\frac{Ax}{4} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{Ab}{8\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{3\,A{b}^{2}x}{32\,c}\sqrt{c{x}^{2}+bx}}-{\frac{3\,A{b}^{3}}{64\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{3\,A{b}^{4}}{128}\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((B*x+A)*(c*x^2+b*x)^(3/2),x)

[Out]

1/5*B*(c*x^2+b*x)^(5/2)/c-1/8*B*b/c*(c*x^2+b*x)^(3/2)*x-1/16*B*b^2/c^2*(c*x^2+b*x)^(3/2)+3/64*B*b^3/c^2*(c*x^2
+b*x)^(1/2)*x+3/128*B*b^4/c^3*(c*x^2+b*x)^(1/2)-3/256*B*b^5/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))+
1/4*A*x*(c*x^2+b*x)^(3/2)+1/8*A/c*(c*x^2+b*x)^(3/2)*b-3/32*A*b^2/c*(c*x^2+b*x)^(1/2)*x-3/64*A*b^3/c^2*(c*x^2+b
*x)^(1/2)+3/128*A*b^4/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((B*x+A)*(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.94128, size = 687, normalized size = 5.13 \begin{align*} \left [-\frac{15 \,{\left (B b^{5} - 2 \, A b^{4} c\right )} \sqrt{c} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) - 2 \,{\left (128 \, B c^{5} x^{4} + 15 \, B b^{4} c - 30 \, A b^{3} c^{2} + 16 \,{\left (11 \, B b c^{4} + 10 \, A c^{5}\right )} x^{3} + 8 \,{\left (B b^{2} c^{3} + 30 \, A b c^{4}\right )} x^{2} - 10 \,{\left (B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{1280 \, c^{4}}, \frac{15 \,{\left (B b^{5} - 2 \, A b^{4} c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) +{\left (128 \, B c^{5} x^{4} + 15 \, B b^{4} c - 30 \, A b^{3} c^{2} + 16 \,{\left (11 \, B b c^{4} + 10 \, A c^{5}\right )} x^{3} + 8 \,{\left (B b^{2} c^{3} + 30 \, A b c^{4}\right )} x^{2} - 10 \,{\left (B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{640 \, c^{4}}\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, algorithm="fricas")

[Out]

[-1/1280*(15*(B*b^5 - 2*A*b^4*c)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(128*B*c^5*x^4 + 15*
B*b^4*c - 30*A*b^3*c^2 + 16*(11*B*b*c^4 + 10*A*c^5)*x^3 + 8*(B*b^2*c^3 + 30*A*b*c^4)*x^2 - 10*(B*b^3*c^2 - 2*A
*b^2*c^3)*x)*sqrt(c*x^2 + b*x))/c^4, 1/640*(15*(B*b^5 - 2*A*b^4*c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/
(c*x)) + (128*B*c^5*x^4 + 15*B*b^4*c - 30*A*b^3*c^2 + 16*(11*B*b*c^4 + 10*A*c^5)*x^3 + 8*(B*b^2*c^3 + 30*A*b*c
^4)*x^2 - 10*(B*b^3*c^2 - 2*A*b^2*c^3)*x)*sqrt(c*x^2 + b*x))/c^4]

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

[Out]

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

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Giac [A]  time = 1.17154, size = 219, normalized size = 1.63 \begin{align*} \frac{1}{640} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \, B c x + \frac{11 \, B b c^{4} + 10 \, A c^{5}}{c^{4}}\right )} x + \frac{B b^{2} c^{3} + 30 \, A b c^{4}}{c^{4}}\right )} x - \frac{5 \,{\left (B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )}}{c^{4}}\right )} x + \frac{15 \,{\left (B b^{4} c - 2 \, A b^{3} c^{2}\right )}}{c^{4}}\right )} + \frac{3 \,{\left (B b^{5} - 2 \, A b^{4} c\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{256 \, c^{\frac{7}{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, algorithm="giac")

[Out]

1/640*sqrt(c*x^2 + b*x)*(2*(4*(2*(8*B*c*x + (11*B*b*c^4 + 10*A*c^5)/c^4)*x + (B*b^2*c^3 + 30*A*b*c^4)/c^4)*x -
5*(B*b^3*c^2 - 2*A*b^2*c^3)/c^4)*x + 15*(B*b^4*c - 2*A*b^3*c^2)/c^4) + 3/256*(B*b^5 - 2*A*b^4*c)*log(abs(-2*(
sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(7/2)