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

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 = 0.345084, 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 \[ \frac{5 b^2 (6 A c+b B) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{8 \sqrt{c}}+\frac{2 \left (b x+c x^2\right )^{5/2} (6 A c+b B)}{b x^2}-\frac{5 c \left (b x+c x^2\right )^{3/2} (6 A c+b B)}{3 b}-\frac{5}{8} (b+2 c x) \sqrt{b x+c x^2} (6 A c+b B)-\frac{2 A \left (b x+c x^2\right )^{7/2}}{b x^4} \]

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])

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Rubi in Sympy [A]  time = 19.755, size = 146, normalized size = 0.95 \[ - \frac{2 A \left (b x + c x^{2}\right )^{\frac{7}{2}}}{b x^{4}} + \frac{5 b^{2} \left (6 A c + B b\right ) \operatorname{atanh}{\left (\frac{\sqrt{c} x}{\sqrt{b x + c x^{2}}} \right )}}{8 \sqrt{c}} - \left (b + 2 c x\right ) \left (\frac{15 A c}{4} + \frac{5 B b}{8}\right ) \sqrt{b x + c x^{2}} - \frac{5 c \left (6 A c + B b\right ) \left (b x + c x^{2}\right )^{\frac{3}{2}}}{3 b} + \frac{2 \left (6 A c + B b\right ) \left (b x + c x^{2}\right )^{\frac{5}{2}}}{b x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-2*A*(b*x + c*x**2)**(7/2)/(b*x**4) + 5*b**2*(6*A*c + B*b)*atanh(sqrt(c)*x/sqrt(
b*x + c*x**2))/(8*sqrt(c)) - (b + 2*c*x)*(15*A*c/4 + 5*B*b/8)*sqrt(b*x + c*x**2)
- 5*c*(6*A*c + B*b)*(b*x + c*x**2)**(3/2)/(3*b) + 2*(6*A*c + B*b)*(b*x + c*x**2
)**(5/2)/(b*x**2)

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Mathematica [A]  time = 0.220306, size = 119, normalized size = 0.78 \[ \frac{\sqrt{x (b+c x)} \left (\frac{15 b^2 \sqrt{x} (6 A c+b B) \log \left (\sqrt{c} \sqrt{b+c x}+c \sqrt{x}\right )}{\sqrt{c} \sqrt{b+c x}}-6 A \left (8 b^2-9 b c x-2 c^2 x^2\right )+B x \left (33 b^2+26 b c x+8 c^2 x^2\right )\right )}{24 x} \]

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^2*(b*B + 6*A*c)*Sqrt[x]*Log[c*Sqrt[x] + Sqrt[c]*Sqrt[b + c*
x]])/(Sqrt[c]*Sqrt[b + c*x])))/(24*x)

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Maple [B]  time = 0.015, size = 358, normalized size = 2.3 \[ -2\,{\frac{A \left ( c{x}^{2}+bx \right ) ^{7/2}}{b{x}^{4}}}+12\,{\frac{Ac \left ( c{x}^{2}+bx \right ) ^{7/2}}{{b}^{2}{x}^{3}}}-32\,{\frac{A{c}^{2} \left ( c{x}^{2}+bx \right ) ^{7/2}}{{b}^{3}{x}^{2}}}+32\,{\frac{A{c}^{3} \left ( c{x}^{2}+bx \right ) ^{5/2}}{{b}^{3}}}+20\,{\frac{A{c}^{3} \left ( c{x}^{2}+bx \right ) ^{3/2}x}{{b}^{2}}}+10\,{\frac{A{c}^{2} \left ( c{x}^{2}+bx \right ) ^{3/2}}{b}}-{\frac{15\,Ax{c}^{2}}{2}\sqrt{c{x}^{2}+bx}}-{\frac{15\,Abc}{4}\sqrt{c{x}^{2}+bx}}+{\frac{15\,{b}^{2}A}{8}\sqrt{c}\ln \left ({1 \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ) }+2\,{\frac{B \left ( c{x}^{2}+bx \right ) ^{7/2}}{b{x}^{3}}}-{\frac{16\,Bc}{3\,{b}^{2}{x}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}+{\frac{16\,B{c}^{2}}{3\,{b}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}+{\frac{10\,B{c}^{2}x}{3\,b} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{5\,Bc}{3} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{5\,Bxbc}{4}\sqrt{c{x}^{2}+bx}}-{\frac{5\,{b}^{2}B}{8}\sqrt{c{x}^{2}+bx}}+{\frac{5\,B{b}^{3}}{16}\ln \left ({1 \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){\frac{1}{\sqrt{c}}}} \]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x^2 + b*x)^(5/2)*(B*x + A)/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.302443, size = 1, normalized size = 0.01 \[ \left [\frac{15 \,{\left (B b^{3} + 6 \, A b^{2} c\right )} x \log \left ({\left (2 \, c x + b\right )} \sqrt{c} + 2 \, \sqrt{c x^{2} + b x} c\right ) + 2 \,{\left (8 \, B c^{2} x^{3} - 48 \, A b^{2} + 2 \,{\left (13 \, B b c + 6 \, A c^{2}\right )} x^{2} + 3 \,{\left (11 \, B b^{2} + 18 \, A b c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{c}}{48 \, \sqrt{c} x}, \frac{15 \,{\left (B b^{3} + 6 \, A b^{2} c\right )} x \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) +{\left (8 \, B c^{2} x^{3} - 48 \, A b^{2} + 2 \,{\left (13 \, B b c + 6 \, A c^{2}\right )} x^{2} + 3 \,{\left (11 \, B b^{2} + 18 \, A b c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{-c}}{24 \, \sqrt{-c} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x^2 + b*x)^(5/2)*(B*x + A)/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{\left (x \left (b + c x\right )\right )^{\frac{5}{2}} \left (A + B x\right )}{x^{4}}\, dx \]

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/XCAS [A]  time = 0.28738, size = 190, normalized size = 1.24 \[ \frac{2 \, A b^{3}}{\sqrt{c} x - \sqrt{c x^{2} + b x}} + \frac{1}{24} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (4 \, B c^{2} x + \frac{13 \, B b c^{3} + 6 \, A c^{4}}{c^{2}}\right )} x + \frac{3 \,{\left (11 \, B b^{2} c^{2} + 18 \, A b c^{3}\right )}}{c^{2}}\right )} - \frac{5 \,{\left (B b^{3} + 6 \, A b^{2} c\right )}{\rm ln}\left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{16 \, \sqrt{c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x^2 + b*x)^(5/2)*(B*x + A)/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)*ln(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/sqrt(c)