3.1192 \(\int \frac{(A+B x) (d+e x)^2}{\sqrt{b x+c x^2}} \, dx\)

Optimal. Leaf size=189 \[ \frac{\sqrt{b x+c x^2} \left (2 c e x (6 A c e-5 b B e+4 B c d)+6 A c e (8 c d-3 b e)+B \left (15 b^2 e^2-36 b c d e+16 c^2 d^2\right )\right )}{24 c^3}+\frac{\tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right ) \left (6 b^2 c e (A e+2 B d)-8 b c^2 d (2 A e+B d)+16 A c^3 d^2-5 b^3 B e^2\right )}{8 c^{7/2}}+\frac{B \sqrt{b x+c x^2} (d+e x)^2}{3 c} \]

[Out]

(B*(d + e*x)^2*Sqrt[b*x + c*x^2])/(3*c) + ((6*A*c*e*(8*c*d - 3*b*e) + B*(16*c^2*d^2 - 36*b*c*d*e + 15*b^2*e^2)
+ 2*c*e*(4*B*c*d - 5*b*B*e + 6*A*c*e)*x)*Sqrt[b*x + c*x^2])/(24*c^3) + ((16*A*c^3*d^2 - 5*b^3*B*e^2 + 6*b^2*c
*e*(2*B*d + A*e) - 8*b*c^2*d*(B*d + 2*A*e))*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(8*c^(7/2))

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Rubi [A]  time = 0.190447, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {832, 779, 620, 206} \[ \frac{\sqrt{b x+c x^2} \left (2 c e x (6 A c e-5 b B e+4 B c d)+6 A c e (8 c d-3 b e)+B \left (15 b^2 e^2-36 b c d e+16 c^2 d^2\right )\right )}{24 c^3}+\frac{\tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right ) \left (6 b^2 c e (A e+2 B d)-8 b c^2 d (2 A e+B d)+16 A c^3 d^2-5 b^3 B e^2\right )}{8 c^{7/2}}+\frac{B \sqrt{b x+c x^2} (d+e x)^2}{3 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*(d + e*x)^2*Sqrt[b*x + c*x^2])/(3*c) + ((6*A*c*e*(8*c*d - 3*b*e) + B*(16*c^2*d^2 - 36*b*c*d*e + 15*b^2*e^2)
+ 2*c*e*(4*B*c*d - 5*b*B*e + 6*A*c*e)*x)*Sqrt[b*x + c*x^2])/(24*c^3) + ((16*A*c^3*d^2 - 5*b^3*B*e^2 + 6*b^2*c
*e*(2*B*d + A*e) - 8*b*c^2*d*(B*d + 2*A*e))*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(8*c^(7/2))

Rule 832

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m
- 1)*(a + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m*(c*e*f + c*d*g - b*e*g) + e*(p
+ 1)*(2*c*f - b*g))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
&&  !(IGtQ[m, 0] && EqQ[f, 0])

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 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) (d+e x)^2}{\sqrt{b x+c x^2}} \, dx &=\frac{B (d+e x)^2 \sqrt{b x+c x^2}}{3 c}+\frac{\int \frac{(d+e x) \left (-\frac{1}{2} (b B-6 A c) d+\frac{1}{2} (4 B c d-5 b B e+6 A c e) x\right )}{\sqrt{b x+c x^2}} \, dx}{3 c}\\ &=\frac{B (d+e x)^2 \sqrt{b x+c x^2}}{3 c}+\frac{\left (6 A c e (8 c d-3 b e)+B \left (16 c^2 d^2-36 b c d e+15 b^2 e^2\right )+2 c e (4 B c d-5 b B e+6 A c e) x\right ) \sqrt{b x+c x^2}}{24 c^3}+\frac{\left (16 A c^3 d^2-5 b^3 B e^2+6 b^2 c e (2 B d+A e)-8 b c^2 d (B d+2 A e)\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{16 c^3}\\ &=\frac{B (d+e x)^2 \sqrt{b x+c x^2}}{3 c}+\frac{\left (6 A c e (8 c d-3 b e)+B \left (16 c^2 d^2-36 b c d e+15 b^2 e^2\right )+2 c e (4 B c d-5 b B e+6 A c e) x\right ) \sqrt{b x+c x^2}}{24 c^3}+\frac{\left (16 A c^3 d^2-5 b^3 B e^2+6 b^2 c e (2 B d+A e)-8 b c^2 d (B d+2 A e)\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{8 c^3}\\ &=\frac{B (d+e x)^2 \sqrt{b x+c x^2}}{3 c}+\frac{\left (6 A c e (8 c d-3 b e)+B \left (16 c^2 d^2-36 b c d e+15 b^2 e^2\right )+2 c e (4 B c d-5 b B e+6 A c e) x\right ) \sqrt{b x+c x^2}}{24 c^3}+\frac{\left (16 A c^3 d^2-5 b^3 B e^2+6 b^2 c e (2 B d+A e)-8 b c^2 d (B d+2 A e)\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{8 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.230248, size = 190, normalized size = 1.01 \[ \frac{\sqrt{c} x (b+c x) \left (6 A c e (-3 b e+8 c d+2 c e x)+B \left (15 b^2 e^2-2 b c e (18 d+5 e x)+8 c^2 \left (3 d^2+3 d e x+e^2 x^2\right )\right )\right )-3 \sqrt{b} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right ) \left (-6 b^2 c e (A e+2 B d)+8 b c^2 d (2 A e+B d)-16 A c^3 d^2+5 b^3 B e^2\right )}{24 c^{7/2} \sqrt{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*x*(b + c*x)*(6*A*c*e*(8*c*d - 3*b*e + 2*c*e*x) + B*(15*b^2*e^2 - 2*b*c*e*(18*d + 5*e*x) + 8*c^2*(3*d^
2 + 3*d*e*x + e^2*x^2))) - 3*Sqrt[b]*(-16*A*c^3*d^2 + 5*b^3*B*e^2 - 6*b^2*c*e*(2*B*d + A*e) + 8*b*c^2*d*(B*d +
2*A*e))*Sqrt[x]*Sqrt[1 + (c*x)/b]*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(24*c^(7/2)*Sqrt[x*(b + c*x)])

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Maple [B]  time = 0.007, size = 395, normalized size = 2.1 \begin{align*}{\frac{B{e}^{2}{x}^{2}}{3\,c}\sqrt{c{x}^{2}+bx}}-{\frac{5\,B{e}^{2}bx}{12\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{5\,B{e}^{2}{b}^{2}}{8\,{c}^{3}}\sqrt{c{x}^{2}+bx}}-{\frac{5\,{b}^{3}B{e}^{2}}{16}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{7}{2}}}}+{\frac{xA{e}^{2}}{2\,c}\sqrt{c{x}^{2}+bx}}+{\frac{Bxde}{c}\sqrt{c{x}^{2}+bx}}-{\frac{3\,Ab{e}^{2}}{4\,{c}^{2}}\sqrt{c{x}^{2}+bx}}-{\frac{3\,bBde}{2\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{3\,A{b}^{2}{e}^{2}}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{5}{2}}}}+{\frac{3\,{b}^{2}Bde}{4}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{5}{2}}}}+2\,{\frac{\sqrt{c{x}^{2}+bx}Ade}{c}}+{\frac{B{d}^{2}}{c}\sqrt{c{x}^{2}+bx}}-{Abde\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{3}{2}}}}-{\frac{bB{d}^{2}}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{3}{2}}}}+{A{d}^{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){\frac{1}{\sqrt{c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*B*e^2*x^2/c*(c*x^2+b*x)^(1/2)-5/12*B*e^2*b/c^2*x*(c*x^2+b*x)^(1/2)+5/8*B*e^2*b^2/c^3*(c*x^2+b*x)^(1/2)-5/1
6*B*e^2*b^3/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))+1/2*x/c*(c*x^2+b*x)^(1/2)*A*e^2+x/c*(c*x^2+b*x)^
(1/2)*B*d*e-3/4*b/c^2*(c*x^2+b*x)^(1/2)*A*e^2-3/2*b/c^2*(c*x^2+b*x)^(1/2)*B*d*e+3/8*b^2/c^(5/2)*ln((1/2*b+c*x)
/c^(1/2)+(c*x^2+b*x)^(1/2))*A*e^2+3/4*b^2/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))*B*d*e+2/c*(c*x^2+b
*x)^(1/2)*A*d*e+1/c*(c*x^2+b*x)^(1/2)*B*d^2-b/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))*A*d*e-1/2*b/c^
(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))*B*d^2+A*d^2*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))/c^(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)^2/(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.84927, size = 869, normalized size = 4.6 \begin{align*} \left [-\frac{3 \,{\left (8 \,{\left (B b c^{2} - 2 \, A c^{3}\right )} d^{2} - 4 \,{\left (3 \, B b^{2} c - 4 \, A b c^{2}\right )} d e +{\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} e^{2}\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} e^{2} x^{2} + 24 \, B c^{3} d^{2} - 12 \,{\left (3 \, B b c^{2} - 4 \, A c^{3}\right )} d e + 3 \,{\left (5 \, B b^{2} c - 6 \, A b c^{2}\right )} e^{2} + 2 \,{\left (12 \, B c^{3} d e -{\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} e^{2}\right )} x\right )} \sqrt{c x^{2} + b x}}{48 \, c^{4}}, \frac{3 \,{\left (8 \,{\left (B b c^{2} - 2 \, A c^{3}\right )} d^{2} - 4 \,{\left (3 \, B b^{2} c - 4 \, A b c^{2}\right )} d e +{\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} e^{2}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) +{\left (8 \, B c^{3} e^{2} x^{2} + 24 \, B c^{3} d^{2} - 12 \,{\left (3 \, B b c^{2} - 4 \, A c^{3}\right )} d e + 3 \,{\left (5 \, B b^{2} c - 6 \, A b c^{2}\right )} e^{2} + 2 \,{\left (12 \, B c^{3} d e -{\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} e^{2}\right )} x\right )} \sqrt{c x^{2} + b x}}{24 \, c^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(3*(8*(B*b*c^2 - 2*A*c^3)*d^2 - 4*(3*B*b^2*c - 4*A*b*c^2)*d*e + (5*B*b^3 - 6*A*b^2*c)*e^2)*sqrt(c)*log(
2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(8*B*c^3*e^2*x^2 + 24*B*c^3*d^2 - 12*(3*B*b*c^2 - 4*A*c^3)*d*e +
3*(5*B*b^2*c - 6*A*b*c^2)*e^2 + 2*(12*B*c^3*d*e - (5*B*b*c^2 - 6*A*c^3)*e^2)*x)*sqrt(c*x^2 + b*x))/c^4, 1/24*(
3*(8*(B*b*c^2 - 2*A*c^3)*d^2 - 4*(3*B*b^2*c - 4*A*b*c^2)*d*e + (5*B*b^3 - 6*A*b^2*c)*e^2)*sqrt(-c)*arctan(sqrt
(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (8*B*c^3*e^2*x^2 + 24*B*c^3*d^2 - 12*(3*B*b*c^2 - 4*A*c^3)*d*e + 3*(5*B*b^2*c
- 6*A*b*c^2)*e^2 + 2*(12*B*c^3*d*e - (5*B*b*c^2 - 6*A*c^3)*e^2)*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 \frac{\left (A + B x\right ) \left (d + e x\right )^{2}}{\sqrt{x \left (b + c x\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*x)*(d + e*x)**2/sqrt(x*(b + c*x)), x)

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Giac [A]  time = 1.45027, size = 265, normalized size = 1.4 \begin{align*} \frac{1}{24} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (\frac{4 \, B x e^{2}}{c} + \frac{12 \, B c^{2} d e - 5 \, B b c e^{2} + 6 \, A c^{2} e^{2}}{c^{3}}\right )} x + \frac{3 \,{\left (8 \, B c^{2} d^{2} - 12 \, B b c d e + 16 \, A c^{2} d e + 5 \, B b^{2} e^{2} - 6 \, A b c e^{2}\right )}}{c^{3}}\right )} + \frac{{\left (8 \, B b c^{2} d^{2} - 16 \, A c^{3} d^{2} - 12 \, B b^{2} c d e + 16 \, A b c^{2} d e + 5 \, B b^{3} e^{2} - 6 \, A b^{2} c e^{2}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{16 \, c^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*sqrt(c*x^2 + b*x)*(2*(4*B*x*e^2/c + (12*B*c^2*d*e - 5*B*b*c*e^2 + 6*A*c^2*e^2)/c^3)*x + 3*(8*B*c^2*d^2 -
12*B*b*c*d*e + 16*A*c^2*d*e + 5*B*b^2*e^2 - 6*A*b*c*e^2)/c^3) + 1/16*(8*B*b*c^2*d^2 - 16*A*c^3*d^2 - 12*B*b^2*
c*d*e + 16*A*b*c^2*d*e + 5*B*b^3*e^2 - 6*A*b^2*c*e^2)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))
/c^(7/2)