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

Optimal. Leaf size=97 \[ \frac{b^2 (b B-2 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{8 c^{5/2}}-\frac{(b+2 c x) \sqrt{b x+c x^2} (b B-2 A c)}{8 c^2}+\frac{B \left (b x+c x^2\right )^{3/2}}{3 c} \]

[Out]

-((b*B - 2*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(8*c^2) + (B*(b*x + c*x^2)^(3/2))/(3*c) + (b^2*(b*B - 2*A*c)*Ar
cTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(8*c^(5/2))

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Rubi [A]  time = 0.0376421, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {640, 612, 620, 206} \[ \frac{b^2 (b B-2 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{8 c^{5/2}}-\frac{(b+2 c x) \sqrt{b x+c x^2} (b B-2 A c)}{8 c^2}+\frac{B \left (b x+c x^2\right )^{3/2}}{3 c} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)*Sqrt[b*x + c*x^2],x]

[Out]

-((b*B - 2*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(8*c^2) + (B*(b*x + c*x^2)^(3/2))/(3*c) + (b^2*(b*B - 2*A*c)*Ar
cTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(8*c^(5/2))

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

Mathematica [A]  time = 0.187183, size = 108, normalized size = 1.11 \[ \frac{\sqrt{x (b+c x)} \left (\sqrt{c} \left (2 b c (3 A+B x)+4 c^2 x (3 A+2 B x)-3 b^2 B\right )+\frac{3 b^{3/2} (b B-2 A c) \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{x} \sqrt{\frac{c x}{b}+1}}\right )}{24 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(-3*b^2*B + 2*b*c*(3*A + B*x) + 4*c^2*x*(3*A + 2*B*x)) + (3*b^(3/2)*(b*B - 2*A*c)*
ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(24*c^(5/2))

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Maple [A]  time = 0.007, size = 157, normalized size = 1.6 \begin{align*}{\frac{B}{3\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{bBx}{4\,c}\sqrt{c{x}^{2}+bx}}-{\frac{{b}^{2}B}{8\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{{b}^{3}B}{16}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{5}{2}}}}+{\frac{Ax}{2}\sqrt{c{x}^{2}+bx}}+{\frac{Ab}{4\,c}\sqrt{c{x}^{2}+bx}}-{\frac{A{b}^{2}}{8}\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)^(1/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.02513, size = 468, normalized size = 4.82 \begin{align*} \left [-\frac{3 \,{\left (B b^{3} - 2 \, A b^{2} c\right )} \sqrt{c} \log \left (2 \, c x + b - 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) - 2 \,{\left (8 \, B c^{3} x^{2} - 3 \, B b^{2} c + 6 \, A b c^{2} + 2 \,{\left (B b c^{2} + 6 \, A c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{48 \, c^{3}}, -\frac{3 \,{\left (B b^{3} - 2 \, A b^{2} c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) -{\left (8 \, B c^{3} x^{2} - 3 \, B b^{2} c + 6 \, A b c^{2} + 2 \,{\left (B b c^{2} + 6 \, A c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{24 \, 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)^(1/2),x, algorithm="fricas")

[Out]

[-1/48*(3*(B*b^3 - 2*A*b^2*c)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(8*B*c^3*x^2 - 3*B*b^2*
c + 6*A*b*c^2 + 2*(B*b*c^2 + 6*A*c^3)*x)*sqrt(c*x^2 + b*x))/c^3, -1/24*(3*(B*b^3 - 2*A*b^2*c)*sqrt(-c)*arctan(
sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (8*B*c^3*x^2 - 3*B*b^2*c + 6*A*b*c^2 + 2*(B*b*c^2 + 6*A*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 \sqrt{x \left (b + c x\right )} \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)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.18063, size = 138, normalized size = 1.42 \begin{align*} \frac{1}{24} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (4 \, B x + \frac{B b c + 6 \, A c^{2}}{c^{2}}\right )} x - \frac{3 \,{\left (B b^{2} - 2 \, A b c\right )}}{c^{2}}\right )} - \frac{{\left (B b^{3} - 2 \, A b^{2} c\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{16 \, 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)^(1/2),x, algorithm="giac")

[Out]

1/24*sqrt(c*x^2 + b*x)*(2*(4*B*x + (B*b*c + 6*A*c^2)/c^2)*x - 3*(B*b^2 - 2*A*b*c)/c^2) - 1/16*(B*b^3 - 2*A*b^2
*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(5/2)