3.2.21 \(\int \frac {x^3 (A+B x)}{(b x+c x^2)^{3/2}} \, dx\) [121]

Optimal. Leaf size=135 \[ -\frac {2 (b B-A c) x^3}{b c \sqrt {b x+c x^2}}-\frac {3 (5 b B-4 A c) \sqrt {b x+c x^2}}{4 c^3}+\frac {(5 b B-4 A c) x \sqrt {b x+c x^2}}{2 b c^2}+\frac {3 b (5 b B-4 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{7/2}} \]

[Out]

3/4*b*(-4*A*c+5*B*b)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(7/2)-2*(-A*c+B*b)*x^3/b/c/(c*x^2+b*x)^(1/2)-3/4*(
-4*A*c+5*B*b)*(c*x^2+b*x)^(1/2)/c^3+1/2*(-4*A*c+5*B*b)*x*(c*x^2+b*x)^(1/2)/b/c^2

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Rubi [A]
time = 0.07, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {802, 684, 654, 634, 212} \begin {gather*} \frac {3 b (5 b B-4 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{7/2}}-\frac {3 \sqrt {b x+c x^2} (5 b B-4 A c)}{4 c^3}+\frac {x \sqrt {b x+c x^2} (5 b B-4 A c)}{2 b c^2}-\frac {2 x^3 (b B-A c)}{b c \sqrt {b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 634

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 654

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

Rule 684

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*
((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Dist[(m + p)*((2*c*d - b*e)/(c*(m + 2*p + 1))), Int[(d + e
*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 802

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

Rubi steps

\begin {align*} \int \frac {x^3 (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx &=-\frac {2 (b B-A c) x^3}{b c \sqrt {b x+c x^2}}-\left (\frac {4 A}{b}-\frac {5 B}{c}\right ) \int \frac {x^2}{\sqrt {b x+c x^2}} \, dx\\ &=-\frac {2 (b B-A c) x^3}{b c \sqrt {b x+c x^2}}+\frac {(5 b B-4 A c) x \sqrt {b x+c x^2}}{2 b c^2}-\frac {(3 (5 b B-4 A c)) \int \frac {x}{\sqrt {b x+c x^2}} \, dx}{4 c^2}\\ &=-\frac {2 (b B-A c) x^3}{b c \sqrt {b x+c x^2}}-\frac {3 (5 b B-4 A c) \sqrt {b x+c x^2}}{4 c^3}+\frac {(5 b B-4 A c) x \sqrt {b x+c x^2}}{2 b c^2}+\frac {(3 b (5 b B-4 A c)) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{8 c^3}\\ &=-\frac {2 (b B-A c) x^3}{b c \sqrt {b x+c x^2}}-\frac {3 (5 b B-4 A c) \sqrt {b x+c x^2}}{4 c^3}+\frac {(5 b B-4 A c) x \sqrt {b x+c x^2}}{2 b c^2}+\frac {(3 b (5 b B-4 A c)) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{4 c^3}\\ &=-\frac {2 (b B-A c) x^3}{b c \sqrt {b x+c x^2}}-\frac {3 (5 b B-4 A c) \sqrt {b x+c x^2}}{4 c^3}+\frac {(5 b B-4 A c) x \sqrt {b x+c x^2}}{2 b c^2}+\frac {3 b (5 b B-4 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.18, size = 108, normalized size = 0.80 \begin {gather*} \frac {\sqrt {c} x \left (-15 b^2 B+b c (12 A-5 B x)+2 c^2 x (2 A+B x)\right )-3 b (5 b B-4 A c) \sqrt {x} \sqrt {b+c x} \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )}{4 c^{7/2} \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*x*(-15*b^2*B + b*c*(12*A - 5*B*x) + 2*c^2*x*(2*A + B*x)) - 3*b*(5*b*B - 4*A*c)*Sqrt[x]*Sqrt[b + c*x]*
Log[-(Sqrt[c]*Sqrt[x]) + Sqrt[b + c*x]])/(4*c^(7/2)*Sqrt[x*(b + c*x)])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(267\) vs. \(2(117)=234\).
time = 0.54, size = 268, normalized size = 1.99

method	result	size
risch	\(\frac {\left (2 B c x +4 A c -7 B b \right ) x \left (c x +b \right )}{4 c^{3} \sqrt {x \left (c x +b \right )}}-\frac {3 b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right ) A}{2 c^{\frac {5}{2}}}+\frac {15 b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right ) B}{8 c^{\frac {7}{2}}}+\frac {2 b \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}\, A}{c^{3} \left (\frac {b}{c}+x \right )}-\frac {2 b^{2} \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}\, B}{c^{4} \left (\frac {b}{c}+x \right )}\)	\(182\)
default	\(B \left (\frac {x^{3}}{2 c \sqrt {c \,x^{2}+b x}}-\frac {5 b \left (\frac {x^{2}}{c \sqrt {c \,x^{2}+b x}}-\frac {3 b \left (-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}\right )}{2 c}\right )}{4 c}\right )+A \left (\frac {x^{2}}{c \sqrt {c \,x^{2}+b x}}-\frac {3 b \left (-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}\right )}{2 c}\right )\)	\(268\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*(1/2*x^3/c/(c*x^2+b*x)^(1/2)-5/4*b/c*(x^2/c/(c*x^2+b*x)^(1/2)-3/2*b/c*(-x/c/(c*x^2+b*x)^(1/2)-1/2*b/c*(-1/c/
(c*x^2+b*x)^(1/2)+1/b/c*(2*c*x+b)/(c*x^2+b*x)^(1/2))+1/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2)))))+A*
(x^2/c/(c*x^2+b*x)^(1/2)-3/2*b/c*(-x/c/(c*x^2+b*x)^(1/2)-1/2*b/c*(-1/c/(c*x^2+b*x)^(1/2)+1/b/c*(2*c*x+b)/(c*x^
2+b*x)^(1/2))+1/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))))

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Maxima [A]
time = 0.27, size = 163, normalized size = 1.21 \begin {gather*} \frac {B x^{3}}{2 \, \sqrt {c x^{2} + b x} c} - \frac {5 \, B b x^{2}}{4 \, \sqrt {c x^{2} + b x} c^{2}} + \frac {A x^{2}}{\sqrt {c x^{2} + b x} c} - \frac {15 \, B b^{2} x}{4 \, \sqrt {c x^{2} + b x} c^{3}} + \frac {3 \, A b x}{\sqrt {c x^{2} + b x} c^{2}} + \frac {15 \, B b^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {7}{2}}} - \frac {3 \, A b \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*B*x^3/(sqrt(c*x^2 + b*x)*c) - 5/4*B*b*x^2/(sqrt(c*x^2 + b*x)*c^2) + A*x^2/(sqrt(c*x^2 + b*x)*c) - 15/4*B*b
^2*x/(sqrt(c*x^2 + b*x)*c^3) + 3*A*b*x/(sqrt(c*x^2 + b*x)*c^2) + 15/8*B*b^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x
)*sqrt(c))/c^(7/2) - 3/2*A*b*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2)

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Fricas [A]
time = 3.66, size = 262, normalized size = 1.94 \begin {gather*} \left [-\frac {3 \, {\left (5 \, B b^{3} - 4 \, A b^{2} c + {\left (5 \, B b^{2} c - 4 \, A b c^{2}\right )} x\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (2 \, B c^{3} x^{2} - 15 \, B b^{2} c + 12 \, A b c^{2} - {\left (5 \, B b c^{2} - 4 \, A c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{8 \, {\left (c^{5} x + b c^{4}\right )}}, -\frac {3 \, {\left (5 \, B b^{3} - 4 \, A b^{2} c + {\left (5 \, B b^{2} c - 4 \, A b c^{2}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (2 \, B c^{3} x^{2} - 15 \, B b^{2} c + 12 \, A b c^{2} - {\left (5 \, B b c^{2} - 4 \, A c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{4 \, {\left (c^{5} x + b c^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(3*(5*B*b^3 - 4*A*b^2*c + (5*B*b^2*c - 4*A*b*c^2)*x)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)
) - 2*(2*B*c^3*x^2 - 15*B*b^2*c + 12*A*b*c^2 - (5*B*b*c^2 - 4*A*c^3)*x)*sqrt(c*x^2 + b*x))/(c^5*x + b*c^4), -1
/4*(3*(5*B*b^3 - 4*A*b^2*c + (5*B*b^2*c - 4*A*b*c^2)*x)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (2
*B*c^3*x^2 - 15*B*b^2*c + 12*A*b*c^2 - (5*B*b*c^2 - 4*A*c^3)*x)*sqrt(c*x^2 + b*x))/(c^5*x + b*c^4)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.81, size = 135, normalized size = 1.00 \begin {gather*} \frac {1}{4} \, \sqrt {c x^{2} + b x} {\left (\frac {2 \, B x}{c^{2}} - \frac {7 \, B b c^{5} - 4 \, A c^{6}}{c^{8}}\right )} - \frac {3 \, {\left (5 \, B b^{2} - 4 \, A b c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{8 \, c^{\frac {7}{2}}} - \frac {2 \, {\left (B b^{3} - A b^{2} c\right )}}{{\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} c + b \sqrt {c}\right )} c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(c*x^2 + b*x)*(2*B*x/c^2 - (7*B*b*c^5 - 4*A*c^6)/c^8) - 3/8*(5*B*b^2 - 4*A*b*c)*log(abs(-2*(sqrt(c)*x
- sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(7/2) - 2*(B*b^3 - A*b^2*c)/(((sqrt(c)*x - sqrt(c*x^2 + b*x))*c + b*sqrt(
c))*c^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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