3.130 \(\int \frac{x^4 (A+B x)}{(b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=136 \[ -\frac{2 x^2 (5 b B-2 A c)}{3 b c^2 \sqrt{b x+c x^2}}+\frac{\sqrt{b x+c x^2} (5 b B-2 A c)}{b c^3}-\frac{(5 b B-2 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{c^{7/2}}-\frac{2 x^4 (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}} \]

[Out]

(-2*(b*B - A*c)*x^4)/(3*b*c*(b*x + c*x^2)^(3/2)) - (2*(5*b*B - 2*A*c)*x^2)/(3*b*c^2*Sqrt[b*x + c*x^2]) + ((5*b
*B - 2*A*c)*Sqrt[b*x + c*x^2])/(b*c^3) - ((5*b*B - 2*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/c^(7/2)

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Rubi [A]  time = 0.124706, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {788, 668, 640, 620, 206} \[ -\frac{2 x^2 (5 b B-2 A c)}{3 b c^2 \sqrt{b x+c x^2}}+\frac{\sqrt{b x+c x^2} (5 b B-2 A c)}{b c^3}-\frac{(5 b B-2 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{c^{7/2}}-\frac{2 x^4 (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(A + B*x))/(b*x + c*x^2)^(5/2),x]

[Out]

(-2*(b*B - A*c)*x^4)/(3*b*c*(b*x + c*x^2)^(3/2)) - (2*(5*b*B - 2*A*c)*x^2)/(3*b*c^2*Sqrt[b*x + c*x^2]) + ((5*b
*B - 2*A*c)*Sqrt[b*x + c*x^2])/(b*c^3) - ((5*b*B - 2*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/c^(7/2)

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 668

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

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

Mathematica [C]  time = 0.0561486, size = 80, normalized size = 0.59 \[ \frac{2 x^4 \left ((b+c x) \sqrt{\frac{c x}{b}+1} (5 b B-2 A c) \, _2F_1\left (\frac{3}{2},\frac{5}{2};\frac{7}{2};-\frac{c x}{b}\right )+5 b (A c-b B)\right )}{15 b^2 c (x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(A + B*x))/(b*x + c*x^2)^(5/2),x]

[Out]

(2*x^4*(5*b*(-(b*B) + A*c) + (5*b*B - 2*A*c)*(b + c*x)*Sqrt[1 + (c*x)/b]*Hypergeometric2F1[3/2, 5/2, 7/2, -((c
*x)/b)]))/(15*b^2*c*(x*(b + c*x))^(3/2))

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Maple [B]  time = 0.009, size = 283, normalized size = 2.1 \begin{align*}{\frac{{x}^{4}B}{c} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}+{\frac{5\,bB{x}^{3}}{6\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}-{\frac{5\,{b}^{2}B{x}^{2}}{4\,{c}^{3}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}-{\frac{5\,{b}^{3}Bx}{12\,{c}^{4}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}+{\frac{35\,bBx}{6\,{c}^{3}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}+{\frac{5\,{b}^{2}B}{12\,{c}^{4}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}-{\frac{5\,bB}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{7}{2}}}}-{\frac{A{x}^{3}}{3\,c} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}+{\frac{Ab{x}^{2}}{2\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}+{\frac{A{b}^{2}x}{6\,{c}^{3}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}-{\frac{7\,Ax}{3\,{c}^{2}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}-{\frac{Ab}{6\,{c}^{3}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}+{A\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(B*x+A)/(c*x^2+b*x)^(5/2),x)

[Out]

B*x^4/c/(c*x^2+b*x)^(3/2)+5/6*B*b/c^2*x^3/(c*x^2+b*x)^(3/2)-5/4*B*b^2/c^3*x^2/(c*x^2+b*x)^(3/2)-5/12*B*b^3/c^4
/(c*x^2+b*x)^(3/2)*x+35/6*B*b/c^3/(c*x^2+b*x)^(1/2)*x+5/12*B*b^2/c^4/(c*x^2+b*x)^(1/2)-5/2*B*b/c^(7/2)*ln((1/2
*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))-1/3*A*x^3/c/(c*x^2+b*x)^(3/2)+1/2*A*b/c^2*x^2/(c*x^2+b*x)^(3/2)+1/6*A*b^2/c
^3/(c*x^2+b*x)^(3/2)*x-7/3*A/c^2/(c*x^2+b*x)^(1/2)*x-1/6*A*b/c^3/(c*x^2+b*x)^(1/2)+A/c^(5/2)*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(x^4*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.03374, size = 710, normalized size = 5.22 \begin{align*} \left [-\frac{3 \,{\left (5 \, B b^{3} - 2 \, A b^{2} c +{\left (5 \, B b c^{2} - 2 \, A c^{3}\right )} x^{2} + 2 \,{\left (5 \, B b^{2} c - 2 \, A b c^{2}\right )} x\right )} \sqrt{c} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) - 2 \,{\left (3 \, B c^{3} x^{2} + 15 \, B b^{2} c - 6 \, A b c^{2} + 4 \,{\left (5 \, B b c^{2} - 2 \, A c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{6 \,{\left (c^{6} x^{2} + 2 \, b c^{5} x + b^{2} c^{4}\right )}}, \frac{3 \,{\left (5 \, B b^{3} - 2 \, A b^{2} c +{\left (5 \, B b c^{2} - 2 \, A c^{3}\right )} x^{2} + 2 \,{\left (5 \, B b^{2} c - 2 \, A b c^{2}\right )} x\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) +{\left (3 \, B c^{3} x^{2} + 15 \, B b^{2} c - 6 \, A b c^{2} + 4 \,{\left (5 \, B b c^{2} - 2 \, A c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{3 \,{\left (c^{6} x^{2} + 2 \, b c^{5} x + b^{2} c^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="fricas")

[Out]

[-1/6*(3*(5*B*b^3 - 2*A*b^2*c + (5*B*b*c^2 - 2*A*c^3)*x^2 + 2*(5*B*b^2*c - 2*A*b*c^2)*x)*sqrt(c)*log(2*c*x + b
+ 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(3*B*c^3*x^2 + 15*B*b^2*c - 6*A*b*c^2 + 4*(5*B*b*c^2 - 2*A*c^3)*x)*sqrt(c*
x^2 + b*x))/(c^6*x^2 + 2*b*c^5*x + b^2*c^4), 1/3*(3*(5*B*b^3 - 2*A*b^2*c + (5*B*b*c^2 - 2*A*c^3)*x^2 + 2*(5*B*
b^2*c - 2*A*b*c^2)*x)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (3*B*c^3*x^2 + 15*B*b^2*c - 6*A*b*c^
2 + 4*(5*B*b*c^2 - 2*A*c^3)*x)*sqrt(c*x^2 + b*x))/(c^6*x^2 + 2*b*c^5*x + b^2*c^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4} \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**4*(B*x+A)/(c*x**2+b*x)**(5/2),x)

[Out]

Integral(x**4*(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^4*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError