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

Optimal. Leaf size=157 \[ -\frac{2 \left (b x+c x^2\right )^{5/2} (4 A c+3 b B)}{3 b x^3}+\frac{5 c \left (b x+c x^2\right )^{3/2} (4 A c+3 b B)}{6 b x}+\frac{5}{4} c \sqrt{b x+c x^2} (4 A c+3 b B)+\frac{5}{4} b \sqrt{c} (4 A c+3 b B) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )-\frac{2 A \left (b x+c x^2\right )^{7/2}}{3 b x^5} \]

[Out]

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

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Rubi [A]  time = 0.163561, antiderivative size = 157, 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} \[ -\frac{2 \left (b x+c x^2\right )^{5/2} (4 A c+3 b B)}{3 b x^3}+\frac{5 c \left (b x+c x^2\right )^{3/2} (4 A c+3 b B)}{6 b x}+\frac{5}{4} c \sqrt{b x+c x^2} (4 A c+3 b B)+\frac{5}{4} b \sqrt{c} (4 A c+3 b B) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )-\frac{2 A \left (b x+c x^2\right )^{7/2}}{3 b x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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^5} \, dx &=-\frac{2 A \left (b x+c x^2\right )^{7/2}}{3 b x^5}+\frac{\left (2 \left (-5 (-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^4} \, dx}{3 b}\\ &=-\frac{2 (3 b B+4 A c) \left (b x+c x^2\right )^{5/2}}{3 b x^3}-\frac{2 A \left (b x+c x^2\right )^{7/2}}{3 b x^5}+\frac{(5 c (3 b B+4 A c)) \int \frac{\left (b x+c x^2\right )^{3/2}}{x^2} \, dx}{3 b}\\ &=\frac{5 c (3 b B+4 A c) \left (b x+c x^2\right )^{3/2}}{6 b x}-\frac{2 (3 b B+4 A c) \left (b x+c x^2\right )^{5/2}}{3 b x^3}-\frac{2 A \left (b x+c x^2\right )^{7/2}}{3 b x^5}+\frac{1}{4} (5 c (3 b B+4 A c)) \int \frac{\sqrt{b x+c x^2}}{x} \, dx\\ &=\frac{5}{4} c (3 b B+4 A c) \sqrt{b x+c x^2}+\frac{5 c (3 b B+4 A c) \left (b x+c x^2\right )^{3/2}}{6 b x}-\frac{2 (3 b B+4 A c) \left (b x+c x^2\right )^{5/2}}{3 b x^3}-\frac{2 A \left (b x+c x^2\right )^{7/2}}{3 b x^5}+\frac{1}{8} (5 b c (3 b B+4 A c)) \int \frac{1}{\sqrt{b x+c x^2}} \, dx\\ &=\frac{5}{4} c (3 b B+4 A c) \sqrt{b x+c x^2}+\frac{5 c (3 b B+4 A c) \left (b x+c x^2\right )^{3/2}}{6 b x}-\frac{2 (3 b B+4 A c) \left (b x+c x^2\right )^{5/2}}{3 b x^3}-\frac{2 A \left (b x+c x^2\right )^{7/2}}{3 b x^5}+\frac{1}{4} (5 b c (3 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 )\\ &=\frac{5}{4} c (3 b B+4 A c) \sqrt{b x+c x^2}+\frac{5 c (3 b B+4 A c) \left (b x+c x^2\right )^{3/2}}{6 b x}-\frac{2 (3 b B+4 A c) \left (b x+c x^2\right )^{5/2}}{3 b x^3}-\frac{2 A \left (b x+c x^2\right )^{7/2}}{3 b x^5}+\frac{5}{4} b \sqrt{c} (3 b B+4 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0530951, size = 86, normalized size = 0.55 \[ -\frac{2 \sqrt{x (b+c x)} \left (b^2 x (4 A c+3 b B) \, _2F_1\left (-\frac{5}{2},-\frac{1}{2};\frac{1}{2};-\frac{c x}{b}\right )+A \sqrt{\frac{c x}{b}+1} (b+c x)^3\right )}{3 b x^2 \sqrt{\frac{c x}{b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.012, size = 411, normalized size = 2.6 \begin{align*} -2\,{\frac{B \left ( c{x}^{2}+bx \right ) ^{7/2}}{b{x}^{4}}}+12\,{\frac{Bc \left ( c{x}^{2}+bx \right ) ^{7/2}}{{b}^{2}{x}^{3}}}-32\,{\frac{B{c}^{2} \left ( c{x}^{2}+bx \right ) ^{7/2}}{{b}^{3}{x}^{2}}}+32\,{\frac{B{c}^{3} \left ( c{x}^{2}+bx \right ) ^{5/2}}{{b}^{3}}}+20\,{\frac{B{c}^{3} \left ( c{x}^{2}+bx \right ) ^{3/2}x}{{b}^{2}}}+10\,{\frac{B{c}^{2} \left ( c{x}^{2}+bx \right ) ^{3/2}}{b}}-{\frac{15\,B{c}^{2}x}{2}\sqrt{c{x}^{2}+bx}}-{\frac{15\,bBc}{4}\sqrt{c{x}^{2}+bx}}+{\frac{15\,{b}^{2}B}{8}\sqrt{c}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ) }-{\frac{2\,A}{3\,b{x}^{5}} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}-{\frac{8\,Ac}{3\,{b}^{2}{x}^{4}} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}+16\,{\frac{A{c}^{2} \left ( c{x}^{2}+bx \right ) ^{7/2}}{{b}^{3}{x}^{3}}}-{\frac{128\,A{c}^{3}}{3\,{b}^{4}{x}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}+{\frac{128\,A{c}^{4}}{3\,{b}^{4}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}+{\frac{80\,A{c}^{4}x}{3\,{b}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{40\,A{c}^{3}}{3\,{b}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-10\,{\frac{A{c}^{3}\sqrt{c{x}^{2}+bx}x}{b}}-5\,A{c}^{2}\sqrt{c{x}^{2}+bx}+{\frac{5\,Ab}{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^5,x)

[Out]

-2*B/b/x^4*(c*x^2+b*x)^(7/2)+12*B/b^2*c/x^3*(c*x^2+b*x)^(7/2)-32*B/b^3*c^2/x^2*(c*x^2+b*x)^(7/2)+32*B/b^3*c^3*
(c*x^2+b*x)^(5/2)+20*B/b^2*c^3*(c*x^2+b*x)^(3/2)*x+10*B/b*c^2*(c*x^2+b*x)^(3/2)-15/2*B*c^2*(c*x^2+b*x)^(1/2)*x
-15/4*B*b*c*(c*x^2+b*x)^(1/2)+15/8*B*b^2*c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))-2/3*A*(c*x^2+b*x)^(
7/2)/b/x^5-8/3*A/b^2*c/x^4*(c*x^2+b*x)^(7/2)+16*A/b^3*c^2/x^3*(c*x^2+b*x)^(7/2)-128/3*A/b^4*c^3/x^2*(c*x^2+b*x
)^(7/2)+128/3*A/b^4*c^4*(c*x^2+b*x)^(5/2)+80/3*A/b^3*c^4*(c*x^2+b*x)^(3/2)*x+40/3*A/b^2*c^3*(c*x^2+b*x)^(3/2)-
10*A/b*c^3*(c*x^2+b*x)^(1/2)*x-5*A*c^2*(c*x^2+b*x)^(1/2)+5/2*A*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^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.00727, size = 518, normalized size = 3.3 \begin{align*} \left [\frac{15 \,{\left (3 \, B b^{2} + 4 \, A b c\right )} \sqrt{c} x^{2} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) + 2 \,{\left (6 \, B c^{2} x^{3} - 8 \, A b^{2} + 3 \,{\left (9 \, B b c + 4 \, A c^{2}\right )} x^{2} - 8 \,{\left (3 \, B b^{2} + 7 \, A b c\right )} x\right )} \sqrt{c x^{2} + b x}}{24 \, x^{2}}, -\frac{15 \,{\left (3 \, B b^{2} + 4 \, A b c\right )} \sqrt{-c} x^{2} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) -{\left (6 \, B c^{2} x^{3} - 8 \, A b^{2} + 3 \,{\left (9 \, B b c + 4 \, A c^{2}\right )} x^{2} - 8 \,{\left (3 \, B b^{2} + 7 \, A b c\right )} x\right )} \sqrt{c x^{2} + b x}}{12 \, x^{2}}\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^5,x, algorithm="fricas")

[Out]

[1/24*(15*(3*B*b^2 + 4*A*b*c)*sqrt(c)*x^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(6*B*c^2*x^3 - 8*A*
b^2 + 3*(9*B*b*c + 4*A*c^2)*x^2 - 8*(3*B*b^2 + 7*A*b*c)*x)*sqrt(c*x^2 + b*x))/x^2, -1/12*(15*(3*B*b^2 + 4*A*b*
c)*sqrt(-c)*x^2*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (6*B*c^2*x^3 - 8*A*b^2 + 3*(9*B*b*c + 4*A*c^2)*x^2
- 8*(3*B*b^2 + 7*A*b*c)*x)*sqrt(c*x^2 + b*x))/x^2]

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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^{5}}\, 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**5,x)

[Out]

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

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Giac [A]  time = 1.18783, size = 285, normalized size = 1.82 \begin{align*} \frac{1}{4} \,{\left (2 \, B c^{2} x + \frac{9 \, B b c^{2} + 4 \, A c^{3}}{c}\right )} \sqrt{c x^{2} + b x} - \frac{5 \,{\left (3 \, B b^{2} c + 4 \, A b c^{2}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{8 \, \sqrt{c}} + \frac{2 \,{\left (3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} B b^{3} \sqrt{c} + 9 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} A b^{2} c^{\frac{3}{2}} + 3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} A b^{3} c + A b^{4} \sqrt{c}\right )}}{3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} \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^5,x, algorithm="giac")

[Out]

1/4*(2*B*c^2*x + (9*B*b*c^2 + 4*A*c^3)/c)*sqrt(c*x^2 + b*x) - 5/8*(3*B*b^2*c + 4*A*b*c^2)*log(abs(-2*(sqrt(c)*
x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/sqrt(c) + 2/3*(3*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^3*sqrt(c) + 9*(sqr
t(c)*x - sqrt(c*x^2 + b*x))^2*A*b^2*c^(3/2) + 3*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^3*c + A*b^4*sqrt(c))/((sqr
t(c)*x - sqrt(c*x^2 + b*x))^3*sqrt(c))