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

Optimal. Leaf size=153 \[ \frac{5 b^2 (6 A c+b B) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{8 \sqrt{c}}+\frac{2 \left (b x+c x^2\right )^{5/2} (6 A c+b B)}{b x^2}-\frac{5 c \left (b x+c x^2\right )^{3/2} (6 A c+b B)}{3 b}-\frac{5}{8} (b+2 c x) \sqrt{b x+c x^2} (6 A c+b B)-\frac{2 A \left (b x+c x^2\right )^{7/2}}{b x^4} \]

[Out]

(-5*(b*B + 6*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/8 - (5*c*(b*B + 6*A*c)*(b*x + c*x^2)^(3/2))/(3*b) + (2*(b*B +
6*A*c)*(b*x + c*x^2)^(5/2))/(b*x^2) - (2*A*(b*x + c*x^2)^(7/2))/(b*x^4) + (5*b^2*(b*B + 6*A*c)*ArcTanh[(Sqrt[
c]*x)/Sqrt[b*x + c*x^2]])/(8*Sqrt[c])

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Rubi [A]  time = 0.16711, antiderivative size = 153, 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 = {792, 662, 664, 612, 620, 206} \[ \frac{5 b^2 (6 A c+b B) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{8 \sqrt{c}}+\frac{2 \left (b x+c x^2\right )^{5/2} (6 A c+b B)}{b x^2}-\frac{5 c \left (b x+c x^2\right )^{3/2} (6 A c+b B)}{3 b}-\frac{5}{8} (b+2 c x) \sqrt{b x+c x^2} (6 A c+b B)-\frac{2 A \left (b x+c x^2\right )^{7/2}}{b x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*(b*B + 6*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/8 - (5*c*(b*B + 6*A*c)*(b*x + c*x^2)^(3/2))/(3*b) + (2*(b*B +
6*A*c)*(b*x + c*x^2)^(5/2))/(b*x^2) - (2*A*(b*x + c*x^2)^(7/2))/(b*x^4) + (5*b^2*(b*B + 6*A*c)*ArcTanh[(Sqrt[
c]*x)/Sqrt[b*x + c*x^2]])/(8*Sqrt[c])

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 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*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^4} \, dx &=-\frac{2 A \left (b x+c x^2\right )^{7/2}}{b x^4}+\frac{\left (2 \left (-4 (-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^3} \, dx}{b}\\ &=\frac{2 (b B+6 A c) \left (b x+c x^2\right )^{5/2}}{b x^2}-\frac{2 A \left (b x+c x^2\right )^{7/2}}{b x^4}-\frac{(5 c (b B+6 A c)) \int \frac{\left (b x+c x^2\right )^{3/2}}{x} \, dx}{b}\\ &=-\frac{5 c (b B+6 A c) \left (b x+c x^2\right )^{3/2}}{3 b}+\frac{2 (b B+6 A c) \left (b x+c x^2\right )^{5/2}}{b x^2}-\frac{2 A \left (b x+c x^2\right )^{7/2}}{b x^4}-\frac{1}{2} (5 c (b B+6 A c)) \int \sqrt{b x+c x^2} \, dx\\ &=-\frac{5}{8} (b B+6 A c) (b+2 c x) \sqrt{b x+c x^2}-\frac{5 c (b B+6 A c) \left (b x+c x^2\right )^{3/2}}{3 b}+\frac{2 (b B+6 A c) \left (b x+c x^2\right )^{5/2}}{b x^2}-\frac{2 A \left (b x+c x^2\right )^{7/2}}{b x^4}+\frac{1}{16} \left (5 b^2 (b B+6 A c)\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx\\ &=-\frac{5}{8} (b B+6 A c) (b+2 c x) \sqrt{b x+c x^2}-\frac{5 c (b B+6 A c) \left (b x+c x^2\right )^{3/2}}{3 b}+\frac{2 (b B+6 A c) \left (b x+c x^2\right )^{5/2}}{b x^2}-\frac{2 A \left (b x+c x^2\right )^{7/2}}{b x^4}+\frac{1}{8} \left (5 b^2 (b B+6 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{5}{8} (b B+6 A c) (b+2 c x) \sqrt{b x+c x^2}-\frac{5 c (b B+6 A c) \left (b x+c x^2\right )^{3/2}}{3 b}+\frac{2 (b B+6 A c) \left (b x+c x^2\right )^{5/2}}{b x^2}-\frac{2 A \left (b x+c x^2\right )^{7/2}}{b x^4}+\frac{5 b^2 (b B+6 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{8 \sqrt{c}}\\ \end{align*}

Mathematica [A]  time = 0.227675, size = 117, normalized size = 0.76 \[ \frac{\sqrt{x (b+c x)} \left (\frac{15 b^{3/2} \sqrt{x} (6 A c+b B) \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{c} \sqrt{\frac{c x}{b}+1}}-6 A \left (8 b^2-9 b c x-2 c^2 x^2\right )+B x \left (33 b^2+26 b c x+8 c^2 x^2\right )\right )}{24 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(-6*A*(8*b^2 - 9*b*c*x - 2*c^2*x^2) + B*x*(33*b^2 + 26*b*c*x + 8*c^2*x^2) + (15*b^(3/2)*(b*
B + 6*A*c)*Sqrt[x]*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[c]*Sqrt[1 + (c*x)/b])))/(24*x)

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Maple [B]  time = 0.012, size = 358, normalized size = 2.3 \begin{align*} -2\,{\frac{A \left ( c{x}^{2}+bx \right ) ^{7/2}}{b{x}^{4}}}+12\,{\frac{Ac \left ( c{x}^{2}+bx \right ) ^{7/2}}{{b}^{2}{x}^{3}}}-32\,{\frac{A{c}^{2} \left ( c{x}^{2}+bx \right ) ^{7/2}}{{b}^{3}{x}^{2}}}+32\,{\frac{A{c}^{3} \left ( c{x}^{2}+bx \right ) ^{5/2}}{{b}^{3}}}+20\,{\frac{A{c}^{3} \left ( c{x}^{2}+bx \right ) ^{3/2}x}{{b}^{2}}}+10\,{\frac{A{c}^{2} \left ( c{x}^{2}+bx \right ) ^{3/2}}{b}}-{\frac{15\,A{c}^{2}x}{2}\sqrt{c{x}^{2}+bx}}-{\frac{15\,Abc}{4}\sqrt{c{x}^{2}+bx}}+{\frac{15\,A{b}^{2}}{8}\sqrt{c}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ) }+2\,{\frac{B \left ( c{x}^{2}+bx \right ) ^{7/2}}{b{x}^{3}}}-{\frac{16\,Bc}{3\,{b}^{2}{x}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}+{\frac{16\,B{c}^{2}}{3\,{b}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}+{\frac{10\,B{c}^{2}x}{3\,b} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{5\,Bc}{3} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{5\,Bbcx}{4}\sqrt{c{x}^{2}+bx}}-{\frac{5\,{b}^{2}B}{8}\sqrt{c{x}^{2}+bx}}+{\frac{5\,{b}^{3}B}{16}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){\frac{1}{\sqrt{c}}}} \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^4,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.22091, size = 551, normalized size = 3.6 \begin{align*} \left [\frac{15 \,{\left (B b^{3} + 6 \, A b^{2} c\right )} \sqrt{c} x \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) + 2 \,{\left (8 \, B c^{3} x^{3} - 48 \, A b^{2} c + 2 \,{\left (13 \, B b c^{2} + 6 \, A c^{3}\right )} x^{2} + 3 \,{\left (11 \, B b^{2} c + 18 \, A b c^{2}\right )} x\right )} \sqrt{c x^{2} + b x}}{48 \, c x}, -\frac{15 \,{\left (B b^{3} + 6 \, A b^{2} c\right )} \sqrt{-c} x \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) -{\left (8 \, B c^{3} x^{3} - 48 \, A b^{2} c + 2 \,{\left (13 \, B b c^{2} + 6 \, A c^{3}\right )} x^{2} + 3 \,{\left (11 \, B b^{2} c + 18 \, A b c^{2}\right )} x\right )} \sqrt{c x^{2} + b x}}{24 \, c x}\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^4,x, algorithm="fricas")

[Out]

[1/48*(15*(B*b^3 + 6*A*b^2*c)*sqrt(c)*x*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(8*B*c^3*x^3 - 48*A*b
^2*c + 2*(13*B*b*c^2 + 6*A*c^3)*x^2 + 3*(11*B*b^2*c + 18*A*b*c^2)*x)*sqrt(c*x^2 + b*x))/(c*x), -1/24*(15*(B*b^
3 + 6*A*b^2*c)*sqrt(-c)*x*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (8*B*c^3*x^3 - 48*A*b^2*c + 2*(13*B*b*c^2
+ 6*A*c^3)*x^2 + 3*(11*B*b^2*c + 18*A*b*c^2)*x)*sqrt(c*x^2 + b*x))/(c*x)]

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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^{4}}\, 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**4,x)

[Out]

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

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Giac [A]  time = 1.18537, size = 190, normalized size = 1.24 \begin{align*} \frac{2 \, A b^{3}}{\sqrt{c} x - \sqrt{c x^{2} + b x}} + \frac{1}{24} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (4 \, B c^{2} x + \frac{13 \, B b c^{3} + 6 \, A c^{4}}{c^{2}}\right )} x + \frac{3 \,{\left (11 \, B b^{2} c^{2} + 18 \, A b c^{3}\right )}}{c^{2}}\right )} - \frac{5 \,{\left (B b^{3} + 6 \, A b^{2} c\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{16 \, \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^4,x, algorithm="giac")

[Out]

2*A*b^3/(sqrt(c)*x - sqrt(c*x^2 + b*x)) + 1/24*sqrt(c*x^2 + b*x)*(2*(4*B*c^2*x + (13*B*b*c^3 + 6*A*c^4)/c^2)*x
+ 3*(11*B*b^2*c^2 + 18*A*b*c^3)/c^2) - 5/16*(B*b^3 + 6*A*b^2*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sq
rt(c) - b))/sqrt(c)