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

Optimal. Leaf size=151 \[ \frac {b^5 (7 b B-12 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{512 c^{9/2}}-\frac {b^3 (b+2 c x) \sqrt {b x+c x^2} (7 b B-12 A c)}{512 c^4}+\frac {b (b+2 c x) \left (b x+c x^2\right )^{3/2} (7 b B-12 A c)}{192 c^3}-\frac {\left (b x+c x^2\right )^{5/2} (-12 A c+7 b B-10 B c x)}{60 c^2} \]

[Out]

1/192*b*(-12*A*c+7*B*b)*(2*c*x+b)*(c*x^2+b*x)^(3/2)/c^3-1/60*(-10*B*c*x-12*A*c+7*B*b)*(c*x^2+b*x)^(5/2)/c^2+1/
512*b^5*(-12*A*c+7*B*b)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(9/2)-1/512*b^3*(-12*A*c+7*B*b)*(2*c*x+b)*(c*x^
2+b*x)^(1/2)/c^4

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Rubi [A]  time = 0.07, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {779, 612, 620, 206} \[ -\frac {b^3 (b+2 c x) \sqrt {b x+c x^2} (7 b B-12 A c)}{512 c^4}+\frac {b^5 (7 b B-12 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{512 c^{9/2}}+\frac {b (b+2 c x) \left (b x+c x^2\right )^{3/2} (7 b B-12 A c)}{192 c^3}-\frac {\left (b x+c x^2\right )^{5/2} (-12 A c+7 b B-10 B c x)}{60 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b^3*(7*b*B - 12*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(512*c^4) + (b*(7*b*B - 12*A*c)*(b + 2*c*x)*(b*x + c*x^2
)^(3/2))/(192*c^3) - ((7*b*B - 12*A*c - 10*B*c*x)*(b*x + c*x^2)^(5/2))/(60*c^2) + (b^5*(7*b*B - 12*A*c)*ArcTan
h[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(512*c^(9/2))

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])

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 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int x (A+B x) \left (b x+c x^2\right )^{3/2} \, dx &=-\frac {(7 b B-12 A c-10 B c x) \left (b x+c x^2\right )^{5/2}}{60 c^2}+\frac {(b (7 b B-12 A c)) \int \left (b x+c x^2\right )^{3/2} \, dx}{24 c^2}\\ &=\frac {b (7 b B-12 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^3}-\frac {(7 b B-12 A c-10 B c x) \left (b x+c x^2\right )^{5/2}}{60 c^2}-\frac {\left (b^3 (7 b B-12 A c)\right ) \int \sqrt {b x+c x^2} \, dx}{128 c^3}\\ &=-\frac {b^3 (7 b B-12 A c) (b+2 c x) \sqrt {b x+c x^2}}{512 c^4}+\frac {b (7 b B-12 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^3}-\frac {(7 b B-12 A c-10 B c x) \left (b x+c x^2\right )^{5/2}}{60 c^2}+\frac {\left (b^5 (7 b B-12 A c)\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{1024 c^4}\\ &=-\frac {b^3 (7 b B-12 A c) (b+2 c x) \sqrt {b x+c x^2}}{512 c^4}+\frac {b (7 b B-12 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^3}-\frac {(7 b B-12 A c-10 B c x) \left (b x+c x^2\right )^{5/2}}{60 c^2}+\frac {\left (b^5 (7 b B-12 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{512 c^4}\\ &=-\frac {b^3 (7 b B-12 A c) (b+2 c x) \sqrt {b x+c x^2}}{512 c^4}+\frac {b (7 b B-12 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^3}-\frac {(7 b B-12 A c-10 B c x) \left (b x+c x^2\right )^{5/2}}{60 c^2}+\frac {b^5 (7 b B-12 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{512 c^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 166, normalized size = 1.10 \[ \frac {\sqrt {x (b+c x)} \left (\frac {15 b^{9/2} (7 b B-12 A c) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {x} \sqrt {\frac {c x}{b}+1}}+\sqrt {c} \left (10 b^4 c (18 A+7 B x)-8 b^3 c^2 x (15 A+7 B x)+48 b^2 c^3 x^2 (2 A+B x)+64 b c^4 x^3 (33 A+26 B x)+256 c^5 x^4 (6 A+5 B x)-105 b^5 B\right )\right )}{7680 c^{9/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(-105*b^5*B + 48*b^2*c^3*x^2*(2*A + B*x) + 256*c^5*x^4*(6*A + 5*B*x) - 8*b^3*c^2*x
*(15*A + 7*B*x) + 10*b^4*c*(18*A + 7*B*x) + 64*b*c^4*x^3*(33*A + 26*B*x)) + (15*b^(9/2)*(7*b*B - 12*A*c)*ArcSi
nh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(7680*c^(9/2))

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fricas [A]  time = 0.90, size = 350, normalized size = 2.32 \[ \left [-\frac {15 \, {\left (7 \, B b^{6} - 12 \, A b^{5} c\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (1280 \, B c^{6} x^{5} - 105 \, B b^{5} c + 180 \, A b^{4} c^{2} + 128 \, {\left (13 \, B b c^{5} + 12 \, A c^{6}\right )} x^{4} + 48 \, {\left (B b^{2} c^{4} + 44 \, A b c^{5}\right )} x^{3} - 8 \, {\left (7 \, B b^{3} c^{3} - 12 \, A b^{2} c^{4}\right )} x^{2} + 10 \, {\left (7 \, B b^{4} c^{2} - 12 \, A b^{3} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{15360 \, c^{5}}, -\frac {15 \, {\left (7 \, B b^{6} - 12 \, A b^{5} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (1280 \, B c^{6} x^{5} - 105 \, B b^{5} c + 180 \, A b^{4} c^{2} + 128 \, {\left (13 \, B b c^{5} + 12 \, A c^{6}\right )} x^{4} + 48 \, {\left (B b^{2} c^{4} + 44 \, A b c^{5}\right )} x^{3} - 8 \, {\left (7 \, B b^{3} c^{3} - 12 \, A b^{2} c^{4}\right )} x^{2} + 10 \, {\left (7 \, B b^{4} c^{2} - 12 \, A b^{3} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{7680 \, c^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/15360*(15*(7*B*b^6 - 12*A*b^5*c)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(1280*B*c^6*x^5
- 105*B*b^5*c + 180*A*b^4*c^2 + 128*(13*B*b*c^5 + 12*A*c^6)*x^4 + 48*(B*b^2*c^4 + 44*A*b*c^5)*x^3 - 8*(7*B*b^3
*c^3 - 12*A*b^2*c^4)*x^2 + 10*(7*B*b^4*c^2 - 12*A*b^3*c^3)*x)*sqrt(c*x^2 + b*x))/c^5, -1/7680*(15*(7*B*b^6 - 1
2*A*b^5*c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (1280*B*c^6*x^5 - 105*B*b^5*c + 180*A*b^4*c^2 +
128*(13*B*b*c^5 + 12*A*c^6)*x^4 + 48*(B*b^2*c^4 + 44*A*b*c^5)*x^3 - 8*(7*B*b^3*c^3 - 12*A*b^2*c^4)*x^2 + 10*(
7*B*b^4*c^2 - 12*A*b^3*c^3)*x)*sqrt(c*x^2 + b*x))/c^5]

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giac [A]  time = 0.22, size = 194, normalized size = 1.28 \[ \frac {1}{7680} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (10 \, B c x + \frac {13 \, B b c^{5} + 12 \, A c^{6}}{c^{5}}\right )} x + \frac {3 \, {\left (B b^{2} c^{4} + 44 \, A b c^{5}\right )}}{c^{5}}\right )} x - \frac {7 \, B b^{3} c^{3} - 12 \, A b^{2} c^{4}}{c^{5}}\right )} x + \frac {5 \, {\left (7 \, B b^{4} c^{2} - 12 \, A b^{3} c^{3}\right )}}{c^{5}}\right )} x - \frac {15 \, {\left (7 \, B b^{5} c - 12 \, A b^{4} c^{2}\right )}}{c^{5}}\right )} - \frac {{\left (7 \, B b^{6} - 12 \, A b^{5} c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{1024 \, c^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7680*sqrt(c*x^2 + b*x)*(2*(4*(2*(8*(10*B*c*x + (13*B*b*c^5 + 12*A*c^6)/c^5)*x + 3*(B*b^2*c^4 + 44*A*b*c^5)/c
^5)*x - (7*B*b^3*c^3 - 12*A*b^2*c^4)/c^5)*x + 5*(7*B*b^4*c^2 - 12*A*b^3*c^3)/c^5)*x - 15*(7*B*b^5*c - 12*A*b^4
*c^2)/c^5) - 1/1024*(7*B*b^6 - 12*A*b^5*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(9/2)

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maple [B]  time = 0.05, size = 283, normalized size = 1.87 \[ -\frac {3 A \,b^{5} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{256 c^{\frac {7}{2}}}+\frac {7 B \,b^{6} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{1024 c^{\frac {9}{2}}}+\frac {3 \sqrt {c \,x^{2}+b x}\, A \,b^{3} x}{64 c^{2}}-\frac {7 \sqrt {c \,x^{2}+b x}\, B \,b^{4} x}{256 c^{3}}+\frac {3 \sqrt {c \,x^{2}+b x}\, A \,b^{4}}{128 c^{3}}-\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} A b x}{8 c}-\frac {7 \sqrt {c \,x^{2}+b x}\, B \,b^{5}}{512 c^{4}}+\frac {7 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,b^{2} x}{96 c^{2}}-\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} A \,b^{2}}{16 c^{2}}+\frac {7 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,b^{3}}{192 c^{3}}+\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}} B x}{6 c}+\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}} A}{5 c}-\frac {7 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} B b}{60 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*B*x*(c*x^2+b*x)^(5/2)/c-7/60*B*b/c^2*(c*x^2+b*x)^(5/2)+7/96*B*b^2/c^2*x*(c*x^2+b*x)^(3/2)+7/192*B*b^3/c^3*
(c*x^2+b*x)^(3/2)-7/256*B*b^4/c^3*(c*x^2+b*x)^(1/2)*x-7/512*B*b^5/c^4*(c*x^2+b*x)^(1/2)+7/1024*B*b^6/c^(9/2)*l
n((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))+1/5*A*(c*x^2+b*x)^(5/2)/c-1/8*A*b/c*x*(c*x^2+b*x)^(3/2)-1/16*A*b^2/c^
2*(c*x^2+b*x)^(3/2)+3/64*A*b^3/c^2*(c*x^2+b*x)^(1/2)*x+3/128*A*b^4/c^3*(c*x^2+b*x)^(1/2)-3/256*A*b^5/c^(7/2)*l
n((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))

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maxima [B]  time = 0.95, size = 280, normalized size = 1.85 \[ -\frac {7 \, \sqrt {c x^{2} + b x} B b^{4} x}{256 \, c^{3}} + \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{2} x}{96 \, c^{2}} + \frac {3 \, \sqrt {c x^{2} + b x} A b^{3} x}{64 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} B x}{6 \, c} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b x}{8 \, c} + \frac {7 \, B b^{6} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{1024 \, c^{\frac {9}{2}}} - \frac {3 \, A b^{5} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{256 \, c^{\frac {7}{2}}} - \frac {7 \, \sqrt {c x^{2} + b x} B b^{5}}{512 \, c^{4}} + \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{3}}{192 \, c^{3}} + \frac {3 \, \sqrt {c x^{2} + b x} A b^{4}}{128 \, c^{3}} - \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} B b}{60 \, c^{2}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b^{2}}{16 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} A}{5 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+A)*(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

-7/256*sqrt(c*x^2 + b*x)*B*b^4*x/c^3 + 7/96*(c*x^2 + b*x)^(3/2)*B*b^2*x/c^2 + 3/64*sqrt(c*x^2 + b*x)*A*b^3*x/c
^2 + 1/6*(c*x^2 + b*x)^(5/2)*B*x/c - 1/8*(c*x^2 + b*x)^(3/2)*A*b*x/c + 7/1024*B*b^6*log(2*c*x + b + 2*sqrt(c*x
^2 + b*x)*sqrt(c))/c^(9/2) - 3/256*A*b^5*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/2) - 7/512*sqrt(c*x
^2 + b*x)*B*b^5/c^4 + 7/192*(c*x^2 + b*x)^(3/2)*B*b^3/c^3 + 3/128*sqrt(c*x^2 + b*x)*A*b^4/c^3 - 7/60*(c*x^2 +
b*x)^(5/2)*B*b/c^2 - 1/16*(c*x^2 + b*x)^(3/2)*A*b^2/c^2 + 1/5*(c*x^2 + b*x)^(5/2)*A/c

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,{\left (c\,x^2+b\,x\right )}^{3/2}\,\left (A+B\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \left (x \left (b + c x\right )\right )^{\frac {3}{2}} \left (A + B x\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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