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

Optimal. Leaf size=139 \[ \frac{e \sqrt{b x+c x^2} \left (3 b^2 e^2+2 c e x (2 c d-b e)-6 b c d e+8 c^2 d^2\right )}{b^2 c^2}-\frac{2 (d+e x)^2 (x (2 c d-b e)+b d)}{b^2 \sqrt{b x+c x^2}}+\frac{3 e^2 (2 c d-b e) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{c^{5/2}} \]

[Out]

(-2*(d + e*x)^2*(b*d + (2*c*d - b*e)*x))/(b^2*Sqrt[b*x + c*x^2]) + (e*(8*c^2*d^2 - 6*b*c*d*e + 3*b^2*e^2 + 2*c
*e*(2*c*d - b*e)*x)*Sqrt[b*x + c*x^2])/(b^2*c^2) + (3*e^2*(2*c*d - b*e)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]]
)/c^(5/2)

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Rubi [A]  time = 0.0931574, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {738, 779, 620, 206} \[ \frac{e \sqrt{b x+c x^2} \left (3 b^2 e^2+2 c e x (2 c d-b e)-6 b c d e+8 c^2 d^2\right )}{b^2 c^2}-\frac{2 (d+e x)^2 (x (2 c d-b e)+b d)}{b^2 \sqrt{b x+c x^2}}+\frac{3 e^2 (2 c d-b e) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(d + e*x)^2*(b*d + (2*c*d - b*e)*x))/(b^2*Sqrt[b*x + c*x^2]) + (e*(8*c^2*d^2 - 6*b*c*d*e + 3*b^2*e^2 + 2*c
*e*(2*c*d - b*e)*x)*Sqrt[b*x + c*x^2])/(b^2*c^2) + (3*e^2*(2*c*d - b*e)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]]
)/c^(5/2)

Rule 738

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] &
& NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
e, m, p, x]

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{(d+e x)^3}{\left (b x+c x^2\right )^{3/2}} \, dx &=-\frac{2 (d+e x)^2 (b d+(2 c d-b e) x)}{b^2 \sqrt{b x+c x^2}}-\frac{2 \int \frac{(d+e x) (-2 b d e-2 e (2 c d-b e) x)}{\sqrt{b x+c x^2}} \, dx}{b^2}\\ &=-\frac{2 (d+e x)^2 (b d+(2 c d-b e) x)}{b^2 \sqrt{b x+c x^2}}+\frac{e \left (8 c^2 d^2-6 b c d e+3 b^2 e^2+2 c e (2 c d-b e) x\right ) \sqrt{b x+c x^2}}{b^2 c^2}+\frac{\left (3 e^2 (2 c d-b e)\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{2 c^2}\\ &=-\frac{2 (d+e x)^2 (b d+(2 c d-b e) x)}{b^2 \sqrt{b x+c x^2}}+\frac{e \left (8 c^2 d^2-6 b c d e+3 b^2 e^2+2 c e (2 c d-b e) x\right ) \sqrt{b x+c x^2}}{b^2 c^2}+\frac{\left (3 e^2 (2 c d-b e)\right ) \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)^2 (b d+(2 c d-b e) x)}{b^2 \sqrt{b x+c x^2}}+\frac{e \left (8 c^2 d^2-6 b c d e+3 b^2 e^2+2 c e (2 c d-b e) x\right ) \sqrt{b x+c x^2}}{b^2 c^2}+\frac{3 e^2 (2 c d-b e) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.112178, size = 129, normalized size = 0.93 \[ \frac{\sqrt{c} \left (b^2 c e^2 x (e x-6 d)+3 b^3 e^3 x-2 b c^2 d^2 (d-3 e x)-4 c^3 d^3 x\right )-3 b^{5/2} e^2 \sqrt{x} \sqrt{\frac{c x}{b}+1} (b e-2 c d) \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{b^2 c^{5/2} \sqrt{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*(-4*c^3*d^3*x + 3*b^3*e^3*x - 2*b*c^2*d^2*(d - 3*e*x) + b^2*c*e^2*x*(-6*d + e*x)) - 3*b^(5/2)*e^2*(-2
*c*d + b*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 = 0.054, size = 177, normalized size = 1.3 \begin{align*}{\frac{{e}^{3}{x}^{2}}{c}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}+3\,{\frac{b{e}^{3}x}{{c}^{2}\sqrt{c{x}^{2}+bx}}}-{\frac{3\,b{e}^{3}}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{5}{2}}}}-6\,{\frac{d{e}^{2}x}{c\sqrt{c{x}^{2}+bx}}}+3\,{\frac{d{e}^{2}}{{c}^{3/2}}\ln \left ({\frac{b/2+cx}{\sqrt{c}}}+\sqrt{c{x}^{2}+bx} \right ) }+6\,{\frac{{d}^{2}ex}{b\sqrt{c{x}^{2}+bx}}}-2\,{\frac{{d}^{3} \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((e*x+d)^3/(c*x^2+b*x)^(3/2),x)

[Out]

e^3*x^2/c/(c*x^2+b*x)^(1/2)+3*e^3*b/c^2/(c*x^2+b*x)^(1/2)*x-3/2*e^3*b/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*
x)^(1/2))-6*d*e^2/c/(c*x^2+b*x)^(1/2)*x+3*d*e^2/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))+6*d^2*e/b/(c
*x^2+b*x)^(1/2)*x-2*d^3*(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((e*x+d)^3/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.11231, size = 730, normalized size = 5.25 \begin{align*} \left [-\frac{3 \,{\left ({\left (2 \, b^{2} c^{2} d e^{2} - b^{3} c e^{3}\right )} x^{2} +{\left (2 \, b^{3} c d e^{2} - b^{4} e^{3}\right )} x\right )} \sqrt{c} \log \left (2 \, c x + b - 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) - 2 \,{\left (b^{2} c^{2} e^{3} x^{2} - 2 \, b c^{3} d^{3} -{\left (4 \, c^{4} d^{3} - 6 \, b c^{3} d^{2} e + 6 \, b^{2} c^{2} d e^{2} - 3 \, b^{3} c e^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{2 \,{\left (b^{2} c^{4} x^{2} + b^{3} c^{3} x\right )}}, -\frac{3 \,{\left ({\left (2 \, b^{2} c^{2} d e^{2} - b^{3} c e^{3}\right )} x^{2} +{\left (2 \, b^{3} c d e^{2} - b^{4} e^{3}\right )} x\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) -{\left (b^{2} c^{2} e^{3} x^{2} - 2 \, b c^{3} d^{3} -{\left (4 \, c^{4} d^{3} - 6 \, b c^{3} d^{2} e + 6 \, b^{2} c^{2} d e^{2} - 3 \, b^{3} c e^{3}\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((e*x+d)^3/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

[-1/2*(3*((2*b^2*c^2*d*e^2 - b^3*c*e^3)*x^2 + (2*b^3*c*d*e^2 - b^4*e^3)*x)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^
2 + b*x)*sqrt(c)) - 2*(b^2*c^2*e^3*x^2 - 2*b*c^3*d^3 - (4*c^4*d^3 - 6*b*c^3*d^2*e + 6*b^2*c^2*d*e^2 - 3*b^3*c*
e^3)*x)*sqrt(c*x^2 + b*x))/(b^2*c^4*x^2 + b^3*c^3*x), -(3*((2*b^2*c^2*d*e^2 - b^3*c*e^3)*x^2 + (2*b^3*c*d*e^2
- b^4*e^3)*x)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (b^2*c^2*e^3*x^2 - 2*b*c^3*d^3 - (4*c^4*d^3
- 6*b*c^3*d^2*e + 6*b^2*c^2*d*e^2 - 3*b^3*c*e^3)*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 (d + e x\right )^{3}}{\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((e*x+d)**3/(c*x**2+b*x)**(3/2),x)

[Out]

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

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Giac [A]  time = 1.32247, size = 169, normalized size = 1.22 \begin{align*} -\frac{\frac{2 \, d^{3}}{b} - x{\left (\frac{x e^{3}}{c} - \frac{4 \, c^{3} d^{3} - 6 \, b c^{2} d^{2} e + 6 \, b^{2} c d e^{2} - 3 \, b^{3} e^{3}}{b^{2} c^{2}}\right )}}{\sqrt{c x^{2} + b x}} - \frac{3 \,{\left (2 \, c d e^{2} - b e^{3}\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((e*x+d)^3/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

-(2*d^3/b - x*(x*e^3/c - (4*c^3*d^3 - 6*b*c^2*d^2*e + 6*b^2*c*d*e^2 - 3*b^3*e^3)/(b^2*c^2)))/sqrt(c*x^2 + b*x)
- 3/2*(2*c*d*e^2 - b*e^3)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(5/2)