3.102 \(\int \frac{(A+B x) (b x+c x^2)^{5/2}}{x^6} \, dx\)

Optimal. Leaf size=155 \[ \frac{c^2 \sqrt{b x+c x^2} (2 A c+5 b B)}{b}+c^{3/2} (2 A c+5 b B) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )-\frac{2 \left (b x+c x^2\right )^{5/2} (2 A c+5 b B)}{15 b x^4}-\frac{2 c \left (b x+c x^2\right )^{3/2} (2 A c+5 b B)}{3 b x^2}-\frac{2 A \left (b x+c x^2\right )^{7/2}}{5 b x^6} \]

[Out]

(c^2*(5*b*B + 2*A*c)*Sqrt[b*x + c*x^2])/b - (2*c*(5*b*B + 2*A*c)*(b*x + c*x^2)^(3/2))/(3*b*x^2) - (2*(5*b*B +
2*A*c)*(b*x + c*x^2)^(5/2))/(15*b*x^4) - (2*A*(b*x + c*x^2)^(7/2))/(5*b*x^6) + c^(3/2)*(5*b*B + 2*A*c)*ArcTanh
[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]]

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Rubi [A]  time = 0.162766, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {792, 662, 664, 620, 206} \[ \frac{c^2 \sqrt{b x+c x^2} (2 A c+5 b B)}{b}+c^{3/2} (2 A c+5 b B) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )-\frac{2 \left (b x+c x^2\right )^{5/2} (2 A c+5 b B)}{15 b x^4}-\frac{2 c \left (b x+c x^2\right )^{3/2} (2 A c+5 b B)}{3 b x^2}-\frac{2 A \left (b x+c x^2\right )^{7/2}}{5 b x^6} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(5/2))/x^6,x]

[Out]

(c^2*(5*b*B + 2*A*c)*Sqrt[b*x + c*x^2])/b - (2*c*(5*b*B + 2*A*c)*(b*x + c*x^2)^(3/2))/(3*b*x^2) - (2*(5*b*B +
2*A*c)*(b*x + c*x^2)^(5/2))/(15*b*x^4) - (2*A*(b*x + c*x^2)^(7/2))/(5*b*x^6) + c^(3/2)*(5*b*B + 2*A*c)*ArcTanh
[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]]

Rule 792

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

Rule 662

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

Rule 664

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

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) \left (b x+c x^2\right )^{5/2}}{x^6} \, dx &=-\frac{2 A \left (b x+c x^2\right )^{7/2}}{5 b x^6}+\frac{\left (2 \left (-6 (-b B+A c)+\frac{7}{2} (-b B+2 A c)\right )\right ) \int \frac{\left (b x+c x^2\right )^{5/2}}{x^5} \, dx}{5 b}\\ &=-\frac{2 (5 b B+2 A c) \left (b x+c x^2\right )^{5/2}}{15 b x^4}-\frac{2 A \left (b x+c x^2\right )^{7/2}}{5 b x^6}+\frac{(c (5 b B+2 A c)) \int \frac{\left (b x+c x^2\right )^{3/2}}{x^3} \, dx}{3 b}\\ &=-\frac{2 c (5 b B+2 A c) \left (b x+c x^2\right )^{3/2}}{3 b x^2}-\frac{2 (5 b B+2 A c) \left (b x+c x^2\right )^{5/2}}{15 b x^4}-\frac{2 A \left (b x+c x^2\right )^{7/2}}{5 b x^6}+\frac{\left (c^2 (5 b B+2 A c)\right ) \int \frac{\sqrt{b x+c x^2}}{x} \, dx}{b}\\ &=\frac{c^2 (5 b B+2 A c) \sqrt{b x+c x^2}}{b}-\frac{2 c (5 b B+2 A c) \left (b x+c x^2\right )^{3/2}}{3 b x^2}-\frac{2 (5 b B+2 A c) \left (b x+c x^2\right )^{5/2}}{15 b x^4}-\frac{2 A \left (b x+c x^2\right )^{7/2}}{5 b x^6}+\frac{1}{2} \left (c^2 (5 b B+2 A c)\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx\\ &=\frac{c^2 (5 b B+2 A c) \sqrt{b x+c x^2}}{b}-\frac{2 c (5 b B+2 A c) \left (b x+c x^2\right )^{3/2}}{3 b x^2}-\frac{2 (5 b B+2 A c) \left (b x+c x^2\right )^{5/2}}{15 b x^4}-\frac{2 A \left (b x+c x^2\right )^{7/2}}{5 b x^6}+\left (c^2 (5 b B+2 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )\\ &=\frac{c^2 (5 b B+2 A c) \sqrt{b x+c x^2}}{b}-\frac{2 c (5 b B+2 A c) \left (b x+c x^2\right )^{3/2}}{3 b x^2}-\frac{2 (5 b B+2 A c) \left (b x+c x^2\right )^{5/2}}{15 b x^4}-\frac{2 A \left (b x+c x^2\right )^{7/2}}{5 b x^6}+c^{3/2} (5 b B+2 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0591846, size = 87, normalized size = 0.56 \[ -\frac{2 \sqrt{x (b+c x)} \left (b^2 x (2 A c+5 b B) \, _2F_1\left (-\frac{5}{2},-\frac{3}{2};-\frac{1}{2};-\frac{c x}{b}\right )+3 A \sqrt{\frac{c x}{b}+1} (b+c x)^3\right )}{15 b x^3 \sqrt{\frac{c x}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(5/2))/x^6,x]

[Out]

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

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Maple [B]  time = 0.01, size = 460, normalized size = 3. \begin{align*} -{\frac{2\,A}{5\,b{x}^{6}} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}-{\frac{4\,Ac}{15\,{b}^{2}{x}^{5}} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}-{\frac{16\,A{c}^{2}}{15\,{b}^{3}{x}^{4}} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}+{\frac{32\,A{c}^{3}}{5\,{b}^{4}{x}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}-{\frac{256\,A{c}^{4}}{15\,{b}^{5}{x}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}+{\frac{256\,A{c}^{5}}{15\,{b}^{5}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}+{\frac{32\,A{c}^{5}x}{3\,{b}^{4}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{16\,A{c}^{4}}{3\,{b}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-4\,{\frac{A{c}^{4}\sqrt{c{x}^{2}+bx}x}{{b}^{2}}}-2\,{\frac{A{c}^{3}\sqrt{c{x}^{2}+bx}}{b}}+A{c}^{{\frac{5}{2}}}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ) -{\frac{2\,B}{3\,b{x}^{5}} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}-{\frac{8\,Bc}{3\,{b}^{2}{x}^{4}} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}+16\,{\frac{B{c}^{2} \left ( c{x}^{2}+bx \right ) ^{7/2}}{{b}^{3}{x}^{3}}}-{\frac{128\,B{c}^{3}}{3\,{b}^{4}{x}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}+{\frac{128\,B{c}^{4}}{3\,{b}^{4}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}+{\frac{80\,B{c}^{4}x}{3\,{b}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{40\,B{c}^{3}}{3\,{b}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-10\,{\frac{B{c}^{3}\sqrt{c{x}^{2}+bx}x}{b}}-5\,B{c}^{2}\sqrt{c{x}^{2}+bx}+{\frac{5\,bB}{2}{c}^{{\frac{3}{2}}}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(5/2)/x^6,x)

[Out]

-2/5*A*(c*x^2+b*x)^(7/2)/b/x^6-4/15*A/b^2*c/x^5*(c*x^2+b*x)^(7/2)-16/15*A/b^3*c^2/x^4*(c*x^2+b*x)^(7/2)+32/5*A
/b^4*c^3/x^3*(c*x^2+b*x)^(7/2)-256/15*A/b^5*c^4/x^2*(c*x^2+b*x)^(7/2)+256/15*A/b^5*c^5*(c*x^2+b*x)^(5/2)+32/3*
A/b^4*c^5*(c*x^2+b*x)^(3/2)*x+16/3*A/b^3*c^4*(c*x^2+b*x)^(3/2)-4*A/b^2*c^4*(c*x^2+b*x)^(1/2)*x-2*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))-2/3*B/b/x^5*(c*x^2+b*x)^(7/2)-8/3*B/b^2*c/x^4
*(c*x^2+b*x)^(7/2)+16*B/b^3*c^2/x^3*(c*x^2+b*x)^(7/2)-128/3*B/b^4*c^3/x^2*(c*x^2+b*x)^(7/2)+128/3*B/b^4*c^4*(c
*x^2+b*x)^(5/2)+80/3*B/b^3*c^4*(c*x^2+b*x)^(3/2)*x+40/3*B/b^2*c^3*(c*x^2+b*x)^(3/2)-10*B/b*c^3*(c*x^2+b*x)^(1/
2)*x-5*B*c^2*(c*x^2+b*x)^(1/2)+5/2*B*b*c^(3/2)*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((B*x+A)*(c*x^2+b*x)^(5/2)/x^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.9197, size = 529, normalized size = 3.41 \begin{align*} \left [\frac{15 \,{\left (5 \, B b c + 2 \, A c^{2}\right )} \sqrt{c} x^{3} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) + 2 \,{\left (15 \, B c^{2} x^{3} - 6 \, A b^{2} - 2 \,{\left (35 \, B b c + 23 \, A c^{2}\right )} x^{2} - 2 \,{\left (5 \, B b^{2} + 11 \, A b c\right )} x\right )} \sqrt{c x^{2} + b x}}{30 \, x^{3}}, -\frac{15 \,{\left (5 \, B b c + 2 \, A c^{2}\right )} \sqrt{-c} x^{3} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) -{\left (15 \, B c^{2} x^{3} - 6 \, A b^{2} - 2 \,{\left (35 \, B b c + 23 \, A c^{2}\right )} x^{2} - 2 \,{\left (5 \, B b^{2} + 11 \, A b c\right )} x\right )} \sqrt{c x^{2} + b x}}{15 \, x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^6,x, algorithm="fricas")

[Out]

[1/30*(15*(5*B*b*c + 2*A*c^2)*sqrt(c)*x^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(15*B*c^2*x^3 - 6*A
*b^2 - 2*(35*B*b*c + 23*A*c^2)*x^2 - 2*(5*B*b^2 + 11*A*b*c)*x)*sqrt(c*x^2 + b*x))/x^3, -1/15*(15*(5*B*b*c + 2*
A*c^2)*sqrt(-c)*x^3*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (15*B*c^2*x^3 - 6*A*b^2 - 2*(35*B*b*c + 23*A*c^
2)*x^2 - 2*(5*B*b^2 + 11*A*b*c)*x)*sqrt(c*x^2 + b*x))/x^3]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (x \left (b + c x\right )\right )^{\frac{5}{2}} \left (A + B x\right )}{x^{6}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**(5/2)/x**6,x)

[Out]

Integral((x*(b + c*x))**(5/2)*(A + B*x)/x**6, x)

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Giac [B]  time = 1.23469, size = 410, normalized size = 2.65 \begin{align*} \sqrt{c x^{2} + b x} B c^{2} - \frac{{\left (5 \, B b c^{2} + 2 \, A c^{3}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{2 \, \sqrt{c}} + \frac{2 \,{\left (45 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{4} B b^{2} c^{\frac{3}{2}} + 45 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{4} A b c^{\frac{5}{2}} + 15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} B b^{3} c + 45 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} A b^{2} c^{2} + 5 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} B b^{4} \sqrt{c} + 35 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} A b^{3} c^{\frac{3}{2}} + 15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} A b^{4} c + 3 \, A b^{5} \sqrt{c}\right )}}{15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{5} \sqrt{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^6,x, algorithm="giac")

[Out]

sqrt(c*x^2 + b*x)*B*c^2 - 1/2*(5*B*b*c^2 + 2*A*c^3)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/s
qrt(c) + 2/15*(45*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*b^2*c^(3/2) + 45*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*A*b*c
^(5/2) + 15*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^3*c + 45*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*A*b^2*c^2 + 5*(sq
rt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^4*sqrt(c) + 35*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*A*b^3*c^(3/2) + 15*(sqrt(c
)*x - sqrt(c*x^2 + b*x))*A*b^4*c + 3*A*b^5*sqrt(c))/((sqrt(c)*x - sqrt(c*x^2 + b*x))^5*sqrt(c))