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

Optimal. Leaf size=172 \[ -\frac{2 x^3 (7 b B-4 A c)}{3 b c^2 \sqrt{b x+c x^2}}+\frac{5 x \sqrt{b x+c x^2} (7 b B-4 A c)}{6 b c^3}-\frac{5 \sqrt{b x+c x^2} (7 b B-4 A c)}{4 c^4}+\frac{5 b (7 b B-4 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{4 c^{9/2}}-\frac{2 x^5 (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}} \]

[Out]

(-2*(b*B - A*c)*x^5)/(3*b*c*(b*x + c*x^2)^(3/2)) - (2*(7*b*B - 4*A*c)*x^3)/(3*b*c^2*Sqrt[b*x + c*x^2]) - (5*(7
*b*B - 4*A*c)*Sqrt[b*x + c*x^2])/(4*c^4) + (5*(7*b*B - 4*A*c)*x*Sqrt[b*x + c*x^2])/(6*b*c^3) + (5*b*(7*b*B - 4
*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(4*c^(9/2))

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Rubi [A]  time = 0.168229, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {788, 668, 670, 640, 620, 206} \[ -\frac{2 x^3 (7 b B-4 A c)}{3 b c^2 \sqrt{b x+c x^2}}+\frac{5 x \sqrt{b x+c x^2} (7 b B-4 A c)}{6 b c^3}-\frac{5 \sqrt{b x+c x^2} (7 b B-4 A c)}{4 c^4}+\frac{5 b (7 b B-4 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{4 c^{9/2}}-\frac{2 x^5 (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*B - A*c)*x^5)/(3*b*c*(b*x + c*x^2)^(3/2)) - (2*(7*b*B - 4*A*c)*x^3)/(3*b*c^2*Sqrt[b*x + c*x^2]) - (5*(7
*b*B - 4*A*c)*Sqrt[b*x + c*x^2])/(4*c^4) + (5*(7*b*B - 4*A*c)*x*Sqrt[b*x + c*x^2])/(6*b*c^3) + (5*b*(7*b*B - 4
*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(4*c^(9/2))

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 670

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*(m + 2*p + 1)), x] + Dist[((m + p)*(2*c*d - b*e))/(c*(m + 2*p + 1)), Int[(d + e
*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{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] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && 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^5 (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx &=-\frac{2 (b B-A c) x^5}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac{1}{3} \left (\frac{4 A}{b}-\frac{7 B}{c}\right ) \int \frac{x^4}{\left (b x+c x^2\right )^{3/2}} \, dx\\ &=-\frac{2 (b B-A c) x^5}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac{2 (7 b B-4 A c) x^3}{3 b c^2 \sqrt{b x+c x^2}}+\frac{(5 (7 b B-4 A c)) \int \frac{x^2}{\sqrt{b x+c x^2}} \, dx}{3 b c^2}\\ &=-\frac{2 (b B-A c) x^5}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac{2 (7 b B-4 A c) x^3}{3 b c^2 \sqrt{b x+c x^2}}+\frac{5 (7 b B-4 A c) x \sqrt{b x+c x^2}}{6 b c^3}-\frac{(5 (7 b B-4 A c)) \int \frac{x}{\sqrt{b x+c x^2}} \, dx}{4 c^3}\\ &=-\frac{2 (b B-A c) x^5}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac{2 (7 b B-4 A c) x^3}{3 b c^2 \sqrt{b x+c x^2}}-\frac{5 (7 b B-4 A c) \sqrt{b x+c x^2}}{4 c^4}+\frac{5 (7 b B-4 A c) x \sqrt{b x+c x^2}}{6 b c^3}+\frac{(5 b (7 b B-4 A c)) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{8 c^4}\\ &=-\frac{2 (b B-A c) x^5}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac{2 (7 b B-4 A c) x^3}{3 b c^2 \sqrt{b x+c x^2}}-\frac{5 (7 b B-4 A c) \sqrt{b x+c x^2}}{4 c^4}+\frac{5 (7 b B-4 A c) x \sqrt{b x+c x^2}}{6 b c^3}+\frac{(5 b (7 b B-4 A c)) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{4 c^4}\\ &=-\frac{2 (b B-A c) x^5}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac{2 (7 b B-4 A c) x^3}{3 b c^2 \sqrt{b x+c x^2}}-\frac{5 (7 b B-4 A c) \sqrt{b x+c x^2}}{4 c^4}+\frac{5 (7 b B-4 A c) x \sqrt{b x+c x^2}}{6 b c^3}+\frac{5 b (7 b B-4 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{4 c^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0585966, size = 80, normalized size = 0.47 \[ \frac{2 x^5 \left ((b+c x) \sqrt{\frac{c x}{b}+1} (7 b B-4 A c) \, _2F_1\left (\frac{3}{2},\frac{7}{2};\frac{9}{2};-\frac{c x}{b}\right )+7 b (A c-b B)\right )}{21 b^2 c (x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^5*(7*b*(-(b*B) + A*c) + (7*b*B - 4*A*c)*(b + c*x)*Sqrt[1 + (c*x)/b]*Hypergeometric2F1[3/2, 7/2, 9/2, -((c
*x)/b)]))/(21*b^2*c*(x*(b + c*x))^(3/2))

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Maple [B]  time = 0.01, size = 338, normalized size = 2. \begin{align*}{\frac{{x}^{5}B}{2\,c} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}-{\frac{7\,bB{x}^{4}}{4\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}-{\frac{35\,{b}^{2}B{x}^{3}}{24\,{c}^{3}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}+{\frac{35\,{b}^{3}B{x}^{2}}{16\,{c}^{4}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}+{\frac{35\,{b}^{4}Bx}{48\,{c}^{5}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}-{\frac{245\,{b}^{2}Bx}{24\,{c}^{4}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}-{\frac{35\,{b}^{3}B}{48\,{c}^{5}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}+{\frac{35\,{b}^{2}B}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{9}{2}}}}+{\frac{A{x}^{4}}{c} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}+{\frac{5\,Ab{x}^{3}}{6\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}-{\frac{5\,A{b}^{2}{x}^{2}}{4\,{c}^{3}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}-{\frac{5\,A{b}^{3}x}{12\,{c}^{4}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}+{\frac{35\,Abx}{6\,{c}^{3}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}+{\frac{5\,A{b}^{2}}{12\,{c}^{4}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}-{\frac{5\,Ab}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*B*x^5/c/(c*x^2+b*x)^(3/2)-7/4*B*b/c^2*x^4/(c*x^2+b*x)^(3/2)-35/24*B*b^2/c^3*x^3/(c*x^2+b*x)^(3/2)+35/16*B*
b^3/c^4*x^2/(c*x^2+b*x)^(3/2)+35/48*B*b^4/c^5/(c*x^2+b*x)^(3/2)*x-245/24*B*b^2/c^4/(c*x^2+b*x)^(1/2)*x-35/48*B
*b^3/c^5/(c*x^2+b*x)^(1/2)+35/8*B*b^2/c^(9/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))+A*x^4/c/(c*x^2+b*x)^(3
/2)+5/6*A*b/c^2*x^3/(c*x^2+b*x)^(3/2)-5/4*A*b^2/c^3*x^2/(c*x^2+b*x)^(3/2)-5/12*A*b^3/c^4/(c*x^2+b*x)^(3/2)*x+3
5/6*A*b/c^3/(c*x^2+b*x)^(1/2)*x+5/12*A*b^2/c^4/(c*x^2+b*x)^(1/2)-5/2*A*b/c^(7/2)*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^5*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.133, size = 838, normalized size = 4.87 \begin{align*} \left [-\frac{15 \,{\left (7 \, B b^{4} - 4 \, A b^{3} c +{\left (7 \, B b^{2} c^{2} - 4 \, A b c^{3}\right )} x^{2} + 2 \,{\left (7 \, B b^{3} c - 4 \, A b^{2} 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 (6 \, B c^{4} x^{3} - 105 \, B b^{3} c + 60 \, A b^{2} c^{2} - 3 \,{\left (7 \, B b c^{3} - 4 \, A c^{4}\right )} x^{2} - 20 \,{\left (7 \, B b^{2} c^{2} - 4 \, A b c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{24 \,{\left (c^{7} x^{2} + 2 \, b c^{6} x + b^{2} c^{5}\right )}}, -\frac{15 \,{\left (7 \, B b^{4} - 4 \, A b^{3} c +{\left (7 \, B b^{2} c^{2} - 4 \, A b c^{3}\right )} x^{2} + 2 \,{\left (7 \, B b^{3} c - 4 \, A b^{2} c^{2}\right )} x\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) -{\left (6 \, B c^{4} x^{3} - 105 \, B b^{3} c + 60 \, A b^{2} c^{2} - 3 \,{\left (7 \, B b c^{3} - 4 \, A c^{4}\right )} x^{2} - 20 \,{\left (7 \, B b^{2} c^{2} - 4 \, A b c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{12 \,{\left (c^{7} x^{2} + 2 \, b c^{6} x + b^{2} c^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/24*(15*(7*B*b^4 - 4*A*b^3*c + (7*B*b^2*c^2 - 4*A*b*c^3)*x^2 + 2*(7*B*b^3*c - 4*A*b^2*c^2)*x)*sqrt(c)*log(2
*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(6*B*c^4*x^3 - 105*B*b^3*c + 60*A*b^2*c^2 - 3*(7*B*b*c^3 - 4*A*c^4
)*x^2 - 20*(7*B*b^2*c^2 - 4*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/(c^7*x^2 + 2*b*c^6*x + b^2*c^5), -1/12*(15*(7*B*b^4
- 4*A*b^3*c + (7*B*b^2*c^2 - 4*A*b*c^3)*x^2 + 2*(7*B*b^3*c - 4*A*b^2*c^2)*x)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x
)*sqrt(-c)/(c*x)) - (6*B*c^4*x^3 - 105*B*b^3*c + 60*A*b^2*c^2 - 3*(7*B*b*c^3 - 4*A*c^4)*x^2 - 20*(7*B*b^2*c^2
- 4*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/(c^7*x^2 + 2*b*c^6*x + b^2*c^5)]

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

[Out]

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

[Out]

Exception raised: TypeError