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

Optimal. Leaf size=169 \[ -\frac{b^2 (b+2 c x) \sqrt{b x+c x^2} (3 b B-10 A c)}{128 c^2}+\frac{b^4 (3 b B-10 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{128 c^{5/2}}+\frac{(b+2 c x) \left (b x+c x^2\right )^{3/2} (3 b B-10 A c)}{48 c}+\frac{\left (b x+c x^2\right )^{5/2} (3 b B-10 A c)}{15 b}+\frac{2 A \left (b x+c x^2\right )^{7/2}}{3 b x^2} \]

[Out]

-(b^2*(3*b*B - 10*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(128*c^2) + ((3*b*B - 10*A*c)*(b + 2*c*x)*(b*x + c*x^2)^
(3/2))/(48*c) + ((3*b*B - 10*A*c)*(b*x + c*x^2)^(5/2))/(15*b) + (2*A*(b*x + c*x^2)^(7/2))/(3*b*x^2) + (b^4*(3*
b*B - 10*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(128*c^(5/2))

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Rubi [A]  time = 0.169981, antiderivative size = 169, 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, 664, 612, 620, 206} \[ -\frac{b^2 (b+2 c x) \sqrt{b x+c x^2} (3 b B-10 A c)}{128 c^2}+\frac{b^4 (3 b B-10 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{128 c^{5/2}}+\frac{(b+2 c x) \left (b x+c x^2\right )^{3/2} (3 b B-10 A c)}{48 c}+\frac{\left (b x+c x^2\right )^{5/2} (3 b B-10 A c)}{15 b}+\frac{2 A \left (b x+c x^2\right )^{7/2}}{3 b x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b^2*(3*b*B - 10*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(128*c^2) + ((3*b*B - 10*A*c)*(b + 2*c*x)*(b*x + c*x^2)^
(3/2))/(48*c) + ((3*b*B - 10*A*c)*(b*x + c*x^2)^(5/2))/(15*b) + (2*A*(b*x + c*x^2)^(7/2))/(3*b*x^2) + (b^4*(3*
b*B - 10*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(128*c^(5/2))

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 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^2} \, dx &=\frac{2 A \left (b x+c x^2\right )^{7/2}}{3 b x^2}-\frac{\left (2 \left (-2 (-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} \, dx}{3 b}\\ &=\frac{(3 b B-10 A c) \left (b x+c x^2\right )^{5/2}}{15 b}+\frac{2 A \left (b x+c x^2\right )^{7/2}}{3 b x^2}-\frac{1}{6} (-3 b B+10 A c) \int \left (b x+c x^2\right )^{3/2} \, dx\\ &=\frac{(3 b B-10 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{48 c}+\frac{(3 b B-10 A c) \left (b x+c x^2\right )^{5/2}}{15 b}+\frac{2 A \left (b x+c x^2\right )^{7/2}}{3 b x^2}-\frac{\left (b^2 (3 b B-10 A c)\right ) \int \sqrt{b x+c x^2} \, dx}{32 c}\\ &=-\frac{b^2 (3 b B-10 A c) (b+2 c x) \sqrt{b x+c x^2}}{128 c^2}+\frac{(3 b B-10 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{48 c}+\frac{(3 b B-10 A c) \left (b x+c x^2\right )^{5/2}}{15 b}+\frac{2 A \left (b x+c x^2\right )^{7/2}}{3 b x^2}+\frac{\left (b^4 (3 b B-10 A c)\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{256 c^2}\\ &=-\frac{b^2 (3 b B-10 A c) (b+2 c x) \sqrt{b x+c x^2}}{128 c^2}+\frac{(3 b B-10 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{48 c}+\frac{(3 b B-10 A c) \left (b x+c x^2\right )^{5/2}}{15 b}+\frac{2 A \left (b x+c x^2\right )^{7/2}}{3 b x^2}+\frac{\left (b^4 (3 b B-10 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{128 c^2}\\ &=-\frac{b^2 (3 b B-10 A c) (b+2 c x) \sqrt{b x+c x^2}}{128 c^2}+\frac{(3 b B-10 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{48 c}+\frac{(3 b B-10 A c) \left (b x+c x^2\right )^{5/2}}{15 b}+\frac{2 A \left (b x+c x^2\right )^{7/2}}{3 b x^2}+\frac{b^4 (3 b B-10 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{128 c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.272028, size = 147, normalized size = 0.87 \[ \frac{\sqrt{x (b+c x)} \left (\sqrt{c} \left (4 b^2 c^2 x (295 A+186 B x)+30 b^3 c (5 A+B x)+16 b c^3 x^2 (85 A+63 B x)+96 c^4 x^3 (5 A+4 B x)-45 b^4 B\right )+\frac{15 b^{7/2} (3 b B-10 A c) \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{x} \sqrt{\frac{c x}{b}+1}}\right )}{1920 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(-45*b^4*B + 30*b^3*c*(5*A + B*x) + 96*c^4*x^3*(5*A + 4*B*x) + 16*b*c^3*x^2*(85*A
+ 63*B*x) + 4*b^2*c^2*x*(295*A + 186*B*x)) + (15*b^(7/2)*(3*b*B - 10*A*c)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/
(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(1920*c^(5/2))

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Maple [A]  time = 0.011, size = 266, normalized size = 1.6 \begin{align*}{\frac{B}{5} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}+{\frac{bBx}{8} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{{b}^{2}B}{16\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{3\,{b}^{3}Bx}{64\,c}\sqrt{c{x}^{2}+bx}}-{\frac{3\,{b}^{4}B}{128\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{3\,B{b}^{5}}{256}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{5}{2}}}}+{\frac{2\,A}{3\,b{x}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}-{\frac{2\,Ac}{3\,b} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}-{\frac{5\,Acx}{12} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{5\,Ab}{24} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{5\,A{b}^{2}x}{32}\sqrt{c{x}^{2}+bx}}+{\frac{5\,A{b}^{3}}{64\,c}\sqrt{c{x}^{2}+bx}}-{\frac{5\,A{b}^{4}}{128}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{3}{2}}}} \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^2,x)

[Out]

1/5*B*(c*x^2+b*x)^(5/2)+1/8*B*b*(c*x^2+b*x)^(3/2)*x+1/16*B/c*(c*x^2+b*x)^(3/2)*b^2-3/64*B*b^3/c*(c*x^2+b*x)^(1
/2)*x-3/128*B*b^4/c^2*(c*x^2+b*x)^(1/2)+3/256*B*b^5/c^(5/2)*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^2-2/3*A/b*c*(c*x^2+b*x)^(5/2)-5/12*A*c*(c*x^2+b*x)^(3/2)*x-5/24*A*b*(c*x^2+b*x)^(3/2)+5/32
*A*b^2*(c*x^2+b*x)^(1/2)*x+5/64*A*b^3/c*(c*x^2+b*x)^(1/2)-5/128*A*b^4/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^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.0194, size = 722, normalized size = 4.27 \begin{align*} \left [-\frac{15 \,{\left (3 \, B b^{5} - 10 \, A b^{4} c\right )} \sqrt{c} \log \left (2 \, c x + b - 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) - 2 \,{\left (384 \, B c^{5} x^{4} - 45 \, B b^{4} c + 150 \, A b^{3} c^{2} + 48 \,{\left (21 \, B b c^{4} + 10 \, A c^{5}\right )} x^{3} + 8 \,{\left (93 \, B b^{2} c^{3} + 170 \, A b c^{4}\right )} x^{2} + 10 \,{\left (3 \, B b^{3} c^{2} + 118 \, A b^{2} c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{3840 \, c^{3}}, -\frac{15 \,{\left (3 \, B b^{5} - 10 \, A b^{4} c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) -{\left (384 \, B c^{5} x^{4} - 45 \, B b^{4} c + 150 \, A b^{3} c^{2} + 48 \,{\left (21 \, B b c^{4} + 10 \, A c^{5}\right )} x^{3} + 8 \,{\left (93 \, B b^{2} c^{3} + 170 \, A b c^{4}\right )} x^{2} + 10 \,{\left (3 \, B b^{3} c^{2} + 118 \, A b^{2} c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{1920 \, c^{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^2,x, algorithm="fricas")

[Out]

[-1/3840*(15*(3*B*b^5 - 10*A*b^4*c)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(384*B*c^5*x^4 -
45*B*b^4*c + 150*A*b^3*c^2 + 48*(21*B*b*c^4 + 10*A*c^5)*x^3 + 8*(93*B*b^2*c^3 + 170*A*b*c^4)*x^2 + 10*(3*B*b^3
*c^2 + 118*A*b^2*c^3)*x)*sqrt(c*x^2 + b*x))/c^3, -1/1920*(15*(3*B*b^5 - 10*A*b^4*c)*sqrt(-c)*arctan(sqrt(c*x^2
+ b*x)*sqrt(-c)/(c*x)) - (384*B*c^5*x^4 - 45*B*b^4*c + 150*A*b^3*c^2 + 48*(21*B*b*c^4 + 10*A*c^5)*x^3 + 8*(93
*B*b^2*c^3 + 170*A*b*c^4)*x^2 + 10*(3*B*b^3*c^2 + 118*A*b^2*c^3)*x)*sqrt(c*x^2 + b*x))/c^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^{2}}\, 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**2,x)

[Out]

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

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Giac [A]  time = 1.17782, size = 230, normalized size = 1.36 \begin{align*} \frac{1}{1920} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (4 \,{\left (6 \,{\left (8 \, B c^{2} x + \frac{21 \, B b c^{5} + 10 \, A c^{6}}{c^{4}}\right )} x + \frac{93 \, B b^{2} c^{4} + 170 \, A b c^{5}}{c^{4}}\right )} x + \frac{5 \,{\left (3 \, B b^{3} c^{3} + 118 \, A b^{2} c^{4}\right )}}{c^{4}}\right )} x - \frac{15 \,{\left (3 \, B b^{4} c^{2} - 10 \, A b^{3} c^{3}\right )}}{c^{4}}\right )} - \frac{{\left (3 \, B b^{5} - 10 \, A b^{4} c\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{256 \, c^{\frac{5}{2}}} \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^2,x, algorithm="giac")

[Out]

1/1920*sqrt(c*x^2 + b*x)*(2*(4*(6*(8*B*c^2*x + (21*B*b*c^5 + 10*A*c^6)/c^4)*x + (93*B*b^2*c^4 + 170*A*b*c^5)/c
^4)*x + 5*(3*B*b^3*c^3 + 118*A*b^2*c^4)/c^4)*x - 15*(3*B*b^4*c^2 - 10*A*b^3*c^3)/c^4) - 1/256*(3*B*b^5 - 10*A*
b^4*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(5/2)