3.1200 \(\int \frac{(A+B x) (d+e x)^2}{(b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=151 \[ \frac{e \sqrt{b x+c x^2} \left (-2 b c (A e+B d)+4 A c^2 d+3 b^2 B e\right )}{b^2 c^2}-\frac{2 (d+e x) \left (x \left (-b c (A e+B d)+2 A c^2 d+b^2 B e\right )+A b c d\right )}{b^2 c \sqrt{b x+c x^2}}+\frac{e \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right ) (2 A c e-3 b B e+4 B c d)}{c^{5/2}} \]

[Out]

(-2*(d + e*x)*(A*b*c*d + (2*A*c^2*d + b^2*B*e - b*c*(B*d + A*e))*x))/(b^2*c*Sqrt[b*x + c*x^2]) + (e*(4*A*c^2*d
+ 3*b^2*B*e - 2*b*c*(B*d + A*e))*Sqrt[b*x + c*x^2])/(b^2*c^2) + (e*(4*B*c*d - 3*b*B*e + 2*A*c*e)*ArcTanh[(Sqr
t[c]*x)/Sqrt[b*x + c*x^2]])/c^(5/2)

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Rubi [A]  time = 0.149722, antiderivative size = 151, 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 = {818, 640, 620, 206} \[ \frac{e \sqrt{b x+c x^2} \left (-2 b c (A e+B d)+4 A c^2 d+3 b^2 B e\right )}{b^2 c^2}-\frac{2 (d+e x) \left (x \left (-b c (A e+B d)+2 A c^2 d+b^2 B e\right )+A b c d\right )}{b^2 c \sqrt{b x+c x^2}}+\frac{e \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right ) (2 A c e-3 b B e+4 B c d)}{c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(d + e*x)*(A*b*c*d + (2*A*c^2*d + b^2*B*e - b*c*(B*d + A*e))*x))/(b^2*c*Sqrt[b*x + c*x^2]) + (e*(4*A*c^2*d
+ 3*b^2*B*e - 2*b*c*(B*d + A*e))*Sqrt[b*x + c*x^2])/(b^2*c^2) + (e*(4*B*c*d - 3*b*B*e + 2*A*c*e)*ArcTanh[(Sqr
t[c]*x)/Sqrt[b*x + c*x^2]])/c^(5/2)

Rule 818

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*(2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g
- c*(b*e*f + b*d*g + 2*a*e*g))*x))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[1/(c*(p + 1)*(b^2 - 4*a*c)), Int[(d +
e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Simp[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a
*e*(e*f*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*(m + p + 1) + 2*c^2*d*f*(m +
2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2*p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && Ne
Q[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] &&
RationalQ[a, b, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

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

Mathematica [A]  time = 0.186197, size = 150, normalized size = 0.99 \[ \frac{\sqrt{c} \left (b B x \left (3 b^2 e^2+b c e (e x-4 d)+2 c^2 d^2\right )-2 A c \left (b^2 e^2 x+b c d (d-2 e x)+2 c^2 d^2 x\right )\right )-b^{5/2} e \sqrt{x} \sqrt{\frac{c x}{b}+1} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right ) (-2 A c e+3 b B e-4 B c d)}{b^2 c^{5/2} \sqrt{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 252, normalized size = 1.7 \begin{align*}{\frac{B{e}^{2}{x}^{2}}{c}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}+3\,{\frac{B{e}^{2}bx}{{c}^{2}\sqrt{c{x}^{2}+bx}}}-{\frac{3\,B{e}^{2}b}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{5}{2}}}}-2\,{\frac{xA{e}^{2}}{c\sqrt{c{x}^{2}+bx}}}-4\,{\frac{Bxde}{c\sqrt{c{x}^{2}+bx}}}+{A{e}^{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{3}{2}}}}+2\,{\frac{Bde}{{c}^{3/2}}\ln \left ({\frac{b/2+cx}{\sqrt{c}}}+\sqrt{c{x}^{2}+bx} \right ) }+4\,{\frac{xAde}{b\sqrt{c{x}^{2}+bx}}}+2\,{\frac{Bx{d}^{2}}{b\sqrt{c{x}^{2}+bx}}}-2\,{\frac{A{d}^{2} \left ( 2\,cx+b \right ) }{{b}^{2}\sqrt{c{x}^{2}+bx}}} \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)^(3/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.6036, size = 918, normalized size = 6.08 \begin{align*} \left [\frac{{\left ({\left (4 \, B b^{2} c^{2} d e -{\left (3 \, B b^{3} c - 2 \, A b^{2} c^{2}\right )} e^{2}\right )} x^{2} +{\left (4 \, B b^{3} c d e -{\left (3 \, B b^{4} - 2 \, A b^{3} c\right )} e^{2}\right )} x\right )} \sqrt{c} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) + 2 \,{\left (B b^{2} c^{2} e^{2} x^{2} - 2 \, A b c^{3} d^{2} +{\left (2 \,{\left (B b c^{3} - 2 \, A c^{4}\right )} d^{2} - 4 \,{\left (B b^{2} c^{2} - A b c^{3}\right )} d e +{\left (3 \, B b^{3} c - 2 \, A b^{2} c^{2}\right )} e^{2}\right )} x\right )} \sqrt{c x^{2} + b x}}{2 \,{\left (b^{2} c^{4} x^{2} + b^{3} c^{3} x\right )}}, -\frac{{\left ({\left (4 \, B b^{2} c^{2} d e -{\left (3 \, B b^{3} c - 2 \, A b^{2} c^{2}\right )} e^{2}\right )} x^{2} +{\left (4 \, B b^{3} c d e -{\left (3 \, B b^{4} - 2 \, A b^{3} c\right )} e^{2}\right )} x\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) -{\left (B b^{2} c^{2} e^{2} x^{2} - 2 \, A b c^{3} d^{2} +{\left (2 \,{\left (B b c^{3} - 2 \, A c^{4}\right )} d^{2} - 4 \,{\left (B b^{2} c^{2} - A b c^{3}\right )} d e +{\left (3 \, B b^{3} c - 2 \, A b^{2} c^{2}\right )} e^{2}\right )} x\right )} \sqrt{c x^{2} + b x}}{b^{2} c^{4} x^{2} + b^{3} c^{3} x}\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)^(3/2),x, algorithm="fricas")

[Out]

[1/2*(((4*B*b^2*c^2*d*e - (3*B*b^3*c - 2*A*b^2*c^2)*e^2)*x^2 + (4*B*b^3*c*d*e - (3*B*b^4 - 2*A*b^3*c)*e^2)*x)*
sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(B*b^2*c^2*e^2*x^2 - 2*A*b*c^3*d^2 + (2*(B*b*c^3 - 2*
A*c^4)*d^2 - 4*(B*b^2*c^2 - A*b*c^3)*d*e + (3*B*b^3*c - 2*A*b^2*c^2)*e^2)*x)*sqrt(c*x^2 + b*x))/(b^2*c^4*x^2 +
b^3*c^3*x), -(((4*B*b^2*c^2*d*e - (3*B*b^3*c - 2*A*b^2*c^2)*e^2)*x^2 + (4*B*b^3*c*d*e - (3*B*b^4 - 2*A*b^3*c)
*e^2)*x)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (B*b^2*c^2*e^2*x^2 - 2*A*b*c^3*d^2 + (2*(B*b*c^3
- 2*A*c^4)*d^2 - 4*(B*b^2*c^2 - A*b*c^3)*d*e + (3*B*b^3*c - 2*A*b^2*c^2)*e^2)*x)*sqrt(c*x^2 + b*x))/(b^2*c^4*x
^2 + b^3*c^3*x)]

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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}}{\left (x \left (b + c x\right )\right )^{\frac{3}{2}}}\, 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)**(3/2),x)

[Out]

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

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Giac [A]  time = 1.34234, size = 209, normalized size = 1.38 \begin{align*} -\frac{\frac{2 \, A d^{2}}{b} -{\left (\frac{B x e^{2}}{c} + \frac{2 \, B b c^{2} d^{2} - 4 \, A c^{3} d^{2} - 4 \, B b^{2} c d e + 4 \, A b c^{2} d e + 3 \, B b^{3} e^{2} - 2 \, A b^{2} c e^{2}}{b^{2} c^{2}}\right )} x}{\sqrt{c x^{2} + b x}} - \frac{{\left (4 \, B c d e - 3 \, B b e^{2} + 2 \, A c e^{2}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{2 \, c^{\frac{5}{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)^(3/2),x, algorithm="giac")

[Out]

-(2*A*d^2/b - (B*x*e^2/c + (2*B*b*c^2*d^2 - 4*A*c^3*d^2 - 4*B*b^2*c*d*e + 4*A*b*c^2*d*e + 3*B*b^3*e^2 - 2*A*b^
2*c*e^2)/(b^2*c^2))*x)/sqrt(c*x^2 + b*x) - 1/2*(4*B*c*d*e - 3*B*b*e^2 + 2*A*c*e^2)*log(abs(-2*(sqrt(c)*x - sqr
t(c*x^2 + b*x))*sqrt(c) - b))/c^(5/2)