3.82 \(\int (A+B x) \left (b x+c x^2\right )^{3/2} \, dx\)

Optimal. Leaf size=134 \[ -\frac{3 b^4 (b B-2 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{128 c^{7/2}}+\frac{3 b^2 (b+2 c x) \sqrt{b x+c x^2} (b B-2 A c)}{128 c^3}-\frac{(b+2 c x) \left (b x+c x^2\right )^{3/2} (b B-2 A c)}{16 c^2}+\frac{B \left (b x+c x^2\right )^{5/2}}{5 c} \]

[Out]

(3*b^2*(b*B - 2*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(128*c^3) - ((b*B - 2*A*c)*(
b + 2*c*x)*(b*x + c*x^2)^(3/2))/(16*c^2) + (B*(b*x + c*x^2)^(5/2))/(5*c) - (3*b^
4*(b*B - 2*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(128*c^(7/2))

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Rubi [A]  time = 0.126885, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21 \[ -\frac{3 b^4 (b B-2 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{128 c^{7/2}}+\frac{3 b^2 (b+2 c x) \sqrt{b x+c x^2} (b B-2 A c)}{128 c^3}-\frac{(b+2 c x) \left (b x+c x^2\right )^{3/2} (b B-2 A c)}{16 c^2}+\frac{B \left (b x+c x^2\right )^{5/2}}{5 c} \]

Antiderivative was successfully verified.

[In]  Int[(A + B*x)*(b*x + c*x^2)^(3/2),x]

[Out]

(3*b^2*(b*B - 2*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(128*c^3) - ((b*B - 2*A*c)*(
b + 2*c*x)*(b*x + c*x^2)^(3/2))/(16*c^2) + (B*(b*x + c*x^2)^(5/2))/(5*c) - (3*b^
4*(b*B - 2*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(128*c^(7/2))

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Rubi in Sympy [A]  time = 14.9831, size = 126, normalized size = 0.94 \[ \frac{B \left (b x + c x^{2}\right )^{\frac{5}{2}}}{5 c} + \frac{3 b^{4} \left (2 A c - B b\right ) \operatorname{atanh}{\left (\frac{\sqrt{c} x}{\sqrt{b x + c x^{2}}} \right )}}{128 c^{\frac{7}{2}}} - \frac{3 b^{2} \left (b + 2 c x\right ) \left (2 A c - B b\right ) \sqrt{b x + c x^{2}}}{128 c^{3}} + \frac{\left (b + 2 c x\right ) \left (2 A c - B b\right ) \left (b x + c x^{2}\right )^{\frac{3}{2}}}{16 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((B*x+A)*(c*x**2+b*x)**(3/2),x)

[Out]

B*(b*x + c*x**2)**(5/2)/(5*c) + 3*b**4*(2*A*c - B*b)*atanh(sqrt(c)*x/sqrt(b*x +
c*x**2))/(128*c**(7/2)) - 3*b**2*(b + 2*c*x)*(2*A*c - B*b)*sqrt(b*x + c*x**2)/(1
28*c**3) + (b + 2*c*x)*(2*A*c - B*b)*(b*x + c*x**2)**(3/2)/(16*c**2)

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Mathematica [A]  time = 0.269919, size = 148, normalized size = 1.1 \[ \frac{\sqrt{x (b+c x)} \left (\sqrt{c} \left (-10 b^3 c (3 A+B x)+4 b^2 c^2 x (5 A+2 B x)+16 b c^3 x^2 (15 A+11 B x)+32 c^4 x^3 (5 A+4 B x)+15 b^4 B\right )-\frac{15 b^4 (b B-2 A c) \log \left (\sqrt{c} \sqrt{b+c x}+c \sqrt{x}\right )}{\sqrt{x} \sqrt{b+c x}}\right )}{640 c^{7/2}} \]

Antiderivative was successfully verified.

[In]  Integrate[(A + B*x)*(b*x + c*x^2)^(3/2),x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(15*b^4*B - 10*b^3*c*(3*A + B*x) + 4*b^2*c^2*x*(5*A
+ 2*B*x) + 32*c^4*x^3*(5*A + 4*B*x) + 16*b*c^3*x^2*(15*A + 11*B*x)) - (15*b^4*(b
*B - 2*A*c)*Log[c*Sqrt[x] + Sqrt[c]*Sqrt[b + c*x]])/(Sqrt[x]*Sqrt[b + c*x])))/(6
40*c^(7/2))

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Maple [B]  time = 0.01, size = 239, normalized size = 1.8 \[{\frac{Ax}{4} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{Ab}{8\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{3\,Ax{b}^{2}}{32\,c}\sqrt{c{x}^{2}+bx}}-{\frac{3\,A{b}^{3}}{64\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{3\,A{b}^{4}}{128}\ln \left ({1 \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{5}{2}}}}+{\frac{B}{5\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}-{\frac{xBb}{8\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{{b}^{2}B}{16\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{3\,Bx{b}^{3}}{64\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{3\,{b}^{4}B}{128\,{c}^{3}}\sqrt{c{x}^{2}+bx}}-{\frac{3\,B{b}^{5}}{256}\ln \left ({1 \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{7}{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((B*x+A)*(c*x^2+b*x)^(3/2),x)

[Out]

1/4*A*x*(c*x^2+b*x)^(3/2)+1/8*A/c*(c*x^2+b*x)^(3/2)*b-3/32*A*b^2/c*(c*x^2+b*x)^(
1/2)*x-3/64*A*b^3/c^2*(c*x^2+b*x)^(1/2)+3/128*A*b^4/c^(5/2)*ln((1/2*b+c*x)/c^(1/
2)+(c*x^2+b*x)^(1/2))+1/5*B*(c*x^2+b*x)^(5/2)/c-1/8*B*b/c*(c*x^2+b*x)^(3/2)*x-1/
16*B*b^2/c^2*(c*x^2+b*x)^(3/2)+3/64*B*b^3/c^2*(c*x^2+b*x)^(1/2)*x+3/128*B*b^4/c^
3*(c*x^2+b*x)^(1/2)-3/256*B*b^5/c^(7/2)*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)^(3/2)*(B*x + A),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.281195, size = 1, normalized size = 0.01 \[ \left [\frac{2 \,{\left (128 \, B c^{4} x^{4} + 15 \, B b^{4} - 30 \, A b^{3} c + 16 \,{\left (11 \, B b c^{3} + 10 \, A c^{4}\right )} x^{3} + 8 \,{\left (B b^{2} c^{2} + 30 \, A b c^{3}\right )} x^{2} - 10 \,{\left (B b^{3} c - 2 \, A b^{2} c^{2}\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{c} - 15 \,{\left (B b^{5} - 2 \, A b^{4} c\right )} \log \left ({\left (2 \, c x + b\right )} \sqrt{c} + 2 \, \sqrt{c x^{2} + b x} c\right )}{1280 \, c^{\frac{7}{2}}}, \frac{{\left (128 \, B c^{4} x^{4} + 15 \, B b^{4} - 30 \, A b^{3} c + 16 \,{\left (11 \, B b c^{3} + 10 \, A c^{4}\right )} x^{3} + 8 \,{\left (B b^{2} c^{2} + 30 \, A b c^{3}\right )} x^{2} - 10 \,{\left (B b^{3} c - 2 \, A b^{2} c^{2}\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{-c} - 15 \,{\left (B b^{5} - 2 \, A b^{4} c\right )} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right )}{640 \, \sqrt{-c} c^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x^2 + b*x)^(3/2)*(B*x + A),x, algorithm="fricas")

[Out]

[1/1280*(2*(128*B*c^4*x^4 + 15*B*b^4 - 30*A*b^3*c + 16*(11*B*b*c^3 + 10*A*c^4)*x
^3 + 8*(B*b^2*c^2 + 30*A*b*c^3)*x^2 - 10*(B*b^3*c - 2*A*b^2*c^2)*x)*sqrt(c*x^2 +
b*x)*sqrt(c) - 15*(B*b^5 - 2*A*b^4*c)*log((2*c*x + b)*sqrt(c) + 2*sqrt(c*x^2 +
b*x)*c))/c^(7/2), 1/640*((128*B*c^4*x^4 + 15*B*b^4 - 30*A*b^3*c + 16*(11*B*b*c^3
+ 10*A*c^4)*x^3 + 8*(B*b^2*c^2 + 30*A*b*c^3)*x^2 - 10*(B*b^3*c - 2*A*b^2*c^2)*x
)*sqrt(c*x^2 + b*x)*sqrt(-c) - 15*(B*b^5 - 2*A*b^4*c)*arctan(sqrt(c*x^2 + b*x)*s
qrt(-c)/(c*x)))/(sqrt(-c)*c^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \int \left (x \left (b + c x\right )\right )^{\frac{3}{2}} \left (A + B x\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x+A)*(c*x**2+b*x)**(3/2),x)

[Out]

Integral((x*(b + c*x))**(3/2)*(A + B*x), x)

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GIAC/XCAS [A]  time = 0.282473, size = 219, normalized size = 1.63 \[ \frac{1}{640} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \, B c x + \frac{11 \, B b c^{4} + 10 \, A c^{5}}{c^{4}}\right )} x + \frac{B b^{2} c^{3} + 30 \, A b c^{4}}{c^{4}}\right )} x - \frac{5 \,{\left (B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )}}{c^{4}}\right )} x + \frac{15 \,{\left (B b^{4} c - 2 \, A b^{3} c^{2}\right )}}{c^{4}}\right )} + \frac{3 \,{\left (B b^{5} - 2 \, A b^{4} c\right )}{\rm ln}\left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{256 \, c^{\frac{7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x^2 + b*x)^(3/2)*(B*x + A),x, algorithm="giac")

[Out]

1/640*sqrt(c*x^2 + b*x)*(2*(4*(2*(8*B*c*x + (11*B*b*c^4 + 10*A*c^5)/c^4)*x + (B*
b^2*c^3 + 30*A*b*c^4)/c^4)*x - 5*(B*b^3*c^2 - 2*A*b^2*c^3)/c^4)*x + 15*(B*b^4*c
- 2*A*b^3*c^2)/c^4) + 3/256*(B*b^5 - 2*A*b^4*c)*ln(abs(-2*(sqrt(c)*x - sqrt(c*x^
2 + b*x))*sqrt(c) - b))/c^(7/2)