3.1164 \(\int (A+B x) (d+e x) \sqrt{b x+c x^2} \, dx\)

Optimal. Leaf size=154 \[ \frac{(b+2 c x) \sqrt{b x+c x^2} \left (-8 b c (A e+B d)+16 A c^2 d+5 b^2 B e\right )}{64 c^3}-\frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right ) \left (-8 b c (A e+B d)+16 A c^2 d+5 b^2 B e\right )}{64 c^{7/2}}-\frac{\left (b x+c x^2\right )^{3/2} (-8 c (A e+B d)+5 b B e-6 B c e x)}{24 c^2} \]

[Out]

((16*A*c^2*d + 5*b^2*B*e - 8*b*c*(B*d + A*e))*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(64*c^3) - ((5*b*B*e - 8*c*(B*d +
A*e) - 6*B*c*e*x)*(b*x + c*x^2)^(3/2))/(24*c^2) - (b^2*(16*A*c^2*d + 5*b^2*B*e - 8*b*c*(B*d + A*e))*ArcTanh[(
Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(64*c^(7/2))

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Rubi [A]  time = 0.14077, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {779, 612, 620, 206} \[ \frac{(b+2 c x) \sqrt{b x+c x^2} \left (-8 b c (A e+B d)+16 A c^2 d+5 b^2 B e\right )}{64 c^3}-\frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right ) \left (-8 b c (A e+B d)+16 A c^2 d+5 b^2 B e\right )}{64 c^{7/2}}-\frac{\left (b x+c x^2\right )^{3/2} (-8 c (A e+B d)+5 b B e-6 B c e x)}{24 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((16*A*c^2*d + 5*b^2*B*e - 8*b*c*(B*d + A*e))*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(64*c^3) - ((5*b*B*e - 8*c*(B*d +
A*e) - 6*B*c*e*x)*(b*x + c*x^2)^(3/2))/(24*c^2) - (b^2*(16*A*c^2*d + 5*b^2*B*e - 8*b*c*(B*d + A*e))*ArcTanh[(
Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(64*c^(7/2))

Rule 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[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) (d+e x) \sqrt{b x+c x^2} \, dx &=-\frac{(5 b B e-8 c (B d+A e)-6 B c e x) \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac{\left (\frac{5}{2} b^2 B e+4 c (2 A c d-b (B d+A e))\right ) \int \sqrt{b x+c x^2} \, dx}{8 c^2}\\ &=\frac{\left (16 A c^2 d+5 b^2 B e-8 b c (B d+A e)\right ) (b+2 c x) \sqrt{b x+c x^2}}{64 c^3}-\frac{(5 b B e-8 c (B d+A e)-6 B c e x) \left (b x+c x^2\right )^{3/2}}{24 c^2}-\frac{\left (b^2 \left (16 A c^2 d+5 b^2 B e-8 b c (B d+A e)\right )\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{128 c^3}\\ &=\frac{\left (16 A c^2 d+5 b^2 B e-8 b c (B d+A e)\right ) (b+2 c x) \sqrt{b x+c x^2}}{64 c^3}-\frac{(5 b B e-8 c (B d+A e)-6 B c e x) \left (b x+c x^2\right )^{3/2}}{24 c^2}-\frac{\left (b^2 \left (16 A c^2 d+5 b^2 B e-8 b c (B d+A e)\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{64 c^3}\\ &=\frac{\left (16 A c^2 d+5 b^2 B e-8 b c (B d+A e)\right ) (b+2 c x) \sqrt{b x+c x^2}}{64 c^3}-\frac{(5 b B e-8 c (B d+A e)-6 B c e x) \left (b x+c x^2\right )^{3/2}}{24 c^2}-\frac{b^2 \left (16 A c^2 d+5 b^2 B e-8 b c (B d+A e)\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{64 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.464507, size = 177, normalized size = 1.15 \[ \frac{\sqrt{x (b+c x)} \left (\sqrt{c} \left (-2 b^2 c (12 A e+12 B d+5 B e x)+8 b c^2 (2 A (3 d+e x)+B x (2 d+e x))+16 c^3 x (A (6 d+4 e x)+B x (4 d+3 e x))+15 b^3 B e\right )-\frac{3 b^{3/2} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right ) \left (-8 b c (A e+B d)+16 A c^2 d+5 b^2 B e\right )}{\sqrt{x} \sqrt{\frac{c x}{b}+1}}\right )}{192 c^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(15*b^3*B*e - 2*b^2*c*(12*B*d + 12*A*e + 5*B*e*x) + 8*b*c^2*(B*x*(2*d + e*x) + 2*A
*(3*d + e*x)) + 16*c^3*x*(B*x*(4*d + 3*e*x) + A*(6*d + 4*e*x))) - (3*b^(3/2)*(16*A*c^2*d + 5*b^2*B*e - 8*b*c*(
B*d + A*e))*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(192*c^(7/2))

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Maple [B]  time = 0.007, size = 372, normalized size = 2.4 \begin{align*}{\frac{Bex}{4\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{5\,bBe}{24\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{b}^{2}Bex}{32\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{5\,{b}^{3}Be}{64\,{c}^{3}}\sqrt{c{x}^{2}+bx}}-{\frac{5\,Be{b}^{4}}{128}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{7}{2}}}}+{\frac{Ae}{3\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{Bd}{3\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{Abex}{4\,c}\sqrt{c{x}^{2}+bx}}-{\frac{Bbdx}{4\,c}\sqrt{c{x}^{2}+bx}}-{\frac{A{b}^{2}e}{8\,{c}^{2}}\sqrt{c{x}^{2}+bx}}-{\frac{{b}^{2}Bd}{8\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{A{b}^{3}e}{16}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{5}{2}}}}+{\frac{{b}^{3}Bd}{16}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{5}{2}}}}+{\frac{Adx}{2}\sqrt{c{x}^{2}+bx}}+{\frac{Abd}{4\,c}\sqrt{c{x}^{2}+bx}}-{\frac{Ad{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)*(e*x+d)*(c*x^2+b*x)^(1/2),x)

[Out]

1/4*B*e*x*(c*x^2+b*x)^(3/2)/c-5/24*B*e*b/c^2*(c*x^2+b*x)^(3/2)+5/32*B*e*b^2/c^2*(c*x^2+b*x)^(1/2)*x+5/64*B*e*b
^3/c^3*(c*x^2+b*x)^(1/2)-5/128*B*e*b^4/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))+1/3*(c*x^2+b*x)^(3/2)
/c*A*e+1/3*(c*x^2+b*x)^(3/2)/c*B*d-1/4*b/c*(c*x^2+b*x)^(1/2)*x*A*e-1/4*b/c*(c*x^2+b*x)^(1/2)*x*B*d-1/8*b^2/c^2
*(c*x^2+b*x)^(1/2)*A*e-1/8*b^2/c^2*(c*x^2+b*x)^(1/2)*B*d+1/16*b^3/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(
1/2))*A*e+1/16*b^3/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))*B*d+1/2*A*d*(c*x^2+b*x)^(1/2)*x+1/4*A*d/c
*(c*x^2+b*x)^(1/2)*b-1/8*A*d*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)*(e*x+d)*(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.55701, size = 896, normalized size = 5.82 \begin{align*} \left [\frac{3 \,{\left (8 \,{\left (B b^{3} c - 2 \, A b^{2} c^{2}\right )} d -{\left (5 \, B b^{4} - 8 \, A b^{3} c\right )} e\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} e x^{3} + 8 \,{\left (8 \, B c^{4} d +{\left (B b c^{3} + 8 \, A c^{4}\right )} e\right )} x^{2} - 24 \,{\left (B b^{2} c^{2} - 2 \, A b c^{3}\right )} d + 3 \,{\left (5 \, B b^{3} c - 8 \, A b^{2} c^{2}\right )} e + 2 \,{\left (8 \,{\left (B b c^{3} + 6 \, A c^{4}\right )} d -{\left (5 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} e\right )} x\right )} \sqrt{c x^{2} + b x}}{384 \, c^{4}}, -\frac{3 \,{\left (8 \,{\left (B b^{3} c - 2 \, A b^{2} c^{2}\right )} d -{\left (5 \, B b^{4} - 8 \, A b^{3} c\right )} e\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) -{\left (48 \, B c^{4} e x^{3} + 8 \,{\left (8 \, B c^{4} d +{\left (B b c^{3} + 8 \, A c^{4}\right )} e\right )} x^{2} - 24 \,{\left (B b^{2} c^{2} - 2 \, A b c^{3}\right )} d + 3 \,{\left (5 \, B b^{3} c - 8 \, A b^{2} c^{2}\right )} e + 2 \,{\left (8 \,{\left (B b c^{3} + 6 \, A c^{4}\right )} d -{\left (5 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} e\right )} x\right )} \sqrt{c x^{2} + b x}}{192 \, c^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)*(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

[1/384*(3*(8*(B*b^3*c - 2*A*b^2*c^2)*d - (5*B*b^4 - 8*A*b^3*c)*e)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*
sqrt(c)) + 2*(48*B*c^4*e*x^3 + 8*(8*B*c^4*d + (B*b*c^3 + 8*A*c^4)*e)*x^2 - 24*(B*b^2*c^2 - 2*A*b*c^3)*d + 3*(5
*B*b^3*c - 8*A*b^2*c^2)*e + 2*(8*(B*b*c^3 + 6*A*c^4)*d - (5*B*b^2*c^2 - 8*A*b*c^3)*e)*x)*sqrt(c*x^2 + b*x))/c^
4, -1/192*(3*(8*(B*b^3*c - 2*A*b^2*c^2)*d - (5*B*b^4 - 8*A*b^3*c)*e)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c
)/(c*x)) - (48*B*c^4*e*x^3 + 8*(8*B*c^4*d + (B*b*c^3 + 8*A*c^4)*e)*x^2 - 24*(B*b^2*c^2 - 2*A*b*c^3)*d + 3*(5*B
*b^3*c - 8*A*b^2*c^2)*e + 2*(8*(B*b*c^3 + 6*A*c^4)*d - (5*B*b^2*c^2 - 8*A*b*c^3)*e)*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 \sqrt{x \left (b + c x\right )} \left (A + B x\right ) \left (d + e x\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)*(c*x**2+b*x)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.36562, size = 277, normalized size = 1.8 \begin{align*} \frac{1}{192} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (4 \,{\left (6 \, B x e + \frac{8 \, B c^{3} d + B b c^{2} e + 8 \, A c^{3} e}{c^{3}}\right )} x + \frac{8 \, B b c^{2} d + 48 \, A c^{3} d - 5 \, B b^{2} c e + 8 \, A b c^{2} e}{c^{3}}\right )} x - \frac{3 \,{\left (8 \, B b^{2} c d - 16 \, A b c^{2} d - 5 \, B b^{3} e + 8 \, A b^{2} c e\right )}}{c^{3}}\right )} - \frac{{\left (8 \, B b^{3} c d - 16 \, A b^{2} c^{2} d - 5 \, B b^{4} e + 8 \, A b^{3} c e\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{128 \, c^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*sqrt(c*x^2 + b*x)*(2*(4*(6*B*x*e + (8*B*c^3*d + B*b*c^2*e + 8*A*c^3*e)/c^3)*x + (8*B*b*c^2*d + 48*A*c^3*
d - 5*B*b^2*c*e + 8*A*b*c^2*e)/c^3)*x - 3*(8*B*b^2*c*d - 16*A*b*c^2*d - 5*B*b^3*e + 8*A*b^2*c*e)/c^3) - 1/128*
(8*B*b^3*c*d - 16*A*b^2*c^2*d - 5*B*b^4*e + 8*A*b^3*c*e)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) -
b))/c^(7/2)