3.131 \(\int \frac{x^3 (A+B x)}{(b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=84 \[ -\frac{2 x^3 (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac{2 B x}{c^2 \sqrt{b x+c x^2}}+\frac{2 B \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{c^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.084576, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {788, 652, 620, 206} \[ -\frac{2 x^3 (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac{2 B x}{c^2 \sqrt{b x+c x^2}}+\frac{2 B \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{c^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(A + B*x))/(b*x + c*x^2)^(5/2),x]

[Out]

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

Rule 788

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

Rule 652

Int[((d_.) + (e_.)*(x_))^2*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)*(a + b*x +
c*x^2)^(p + 1))/(c*(p + 1)), x] - Dist[(e^2*(p + 2))/(c*(p + 1)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; Fr
eeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && LtQ[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{x^3 (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx &=-\frac{2 (b B-A c) x^3}{3 b c \left (b x+c x^2\right )^{3/2}}+\frac{B \int \frac{x^2}{\left (b x+c x^2\right )^{3/2}} \, dx}{c}\\ &=-\frac{2 (b B-A c) x^3}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac{2 B x}{c^2 \sqrt{b x+c x^2}}+\frac{B \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{c^2}\\ &=-\frac{2 (b B-A c) x^3}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac{2 B x}{c^2 \sqrt{b x+c x^2}}+\frac{(2 B) \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 (b B-A c) x^3}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac{2 B x}{c^2 \sqrt{b x+c x^2}}+\frac{2 B \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.100331, size = 99, normalized size = 1.18 \[ \frac{x \left (2 \sqrt{c} x \left (A c^2 x-3 b^2 B-4 b B c x\right )+6 b^{3/2} B \sqrt{x} (b+c x) \sqrt{\frac{c x}{b}+1} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )\right )}{3 b c^{5/2} (x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(A + B*x))/(b*x + c*x^2)^(5/2),x]

[Out]

(x*(2*Sqrt[c]*x*(-3*b^2*B - 4*b*B*c*x + A*c^2*x) + 6*b^(3/2)*B*Sqrt[x]*(b + c*x)*Sqrt[1 + (c*x)/b]*ArcSinh[(Sq
rt[c]*Sqrt[x])/Sqrt[b]]))/(3*b*c^(5/2)*(x*(b + c*x))^(3/2))

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Maple [B]  time = 0.009, size = 206, normalized size = 2.5 \begin{align*} -{\frac{{x}^{3}B}{3\,c} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}+{\frac{Bb{x}^{2}}{2\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}+{\frac{{b}^{2}Bx}{6\,{c}^{3}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}-{\frac{7\,Bx}{3\,{c}^{2}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}-{\frac{bB}{6\,{c}^{3}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}+{B\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{5}{2}}}}-{\frac{A{x}^{2}}{c} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}-{\frac{Abx}{3\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,Ax}{3\,bc}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}+{\frac{A}{3\,{c}^{2}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(B*x+A)/(c*x^2+b*x)^(5/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.05018, size = 518, normalized size = 6.17 \begin{align*} \left [\frac{3 \,{\left (B b c^{2} x^{2} + 2 \, B b^{2} c x + B b^{3}\right )} \sqrt{c} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) - 2 \,{\left (3 \, B b^{2} c +{\left (4 \, B b c^{2} - A c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{3 \,{\left (b c^{5} x^{2} + 2 \, b^{2} c^{4} x + b^{3} c^{3}\right )}}, -\frac{2 \,{\left (3 \,{\left (B b c^{2} x^{2} + 2 \, B b^{2} c x + B b^{3}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) +{\left (3 \, B b^{2} c +{\left (4 \, B b c^{2} - A c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}\right )}}{3 \,{\left (b c^{5} x^{2} + 2 \, b^{2} c^{4} x + b^{3} c^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="fricas")

[Out]

[1/3*(3*(B*b*c^2*x^2 + 2*B*b^2*c*x + B*b^3)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(3*B*b^2*
c + (4*B*b*c^2 - A*c^3)*x)*sqrt(c*x^2 + b*x))/(b*c^5*x^2 + 2*b^2*c^4*x + b^3*c^3), -2/3*(3*(B*b*c^2*x^2 + 2*B*
b^2*c*x + B*b^3)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (3*B*b^2*c + (4*B*b*c^2 - A*c^3)*x)*sqrt(
c*x^2 + b*x))/(b*c^5*x^2 + 2*b^2*c^4*x + b^3*c^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \left (A + B x\right )}{\left (x \left (b + c x\right )\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(B*x+A)/(c*x**2+b*x)**(5/2),x)

[Out]

Integral(x**3*(A + B*x)/(x*(b + c*x))**(5/2), x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError