3.2.10 \(\int \frac {x^3 (A+B x)}{\sqrt {b x+c x^2}} \, dx\) [110]

Optimal. Leaf size=162 \[ -\frac {5 b^2 (7 b B-8 A c) \sqrt {b x+c x^2}}{64 c^4}+\frac {5 b (7 b B-8 A c) x \sqrt {b x+c x^2}}{96 c^3}-\frac {(7 b B-8 A c) x^2 \sqrt {b x+c x^2}}{24 c^2}+\frac {B x^3 \sqrt {b x+c x^2}}{4 c}+\frac {5 b^3 (7 b B-8 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{9/2}} \]

[Out]

5/64*b^3*(-8*A*c+7*B*b)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(9/2)-5/64*b^2*(-8*A*c+7*B*b)*(c*x^2+b*x)^(1/2)
/c^4+5/96*b*(-8*A*c+7*B*b)*x*(c*x^2+b*x)^(1/2)/c^3-1/24*(-8*A*c+7*B*b)*x^2*(c*x^2+b*x)^(1/2)/c^2+1/4*B*x^3*(c*
x^2+b*x)^(1/2)/c

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Rubi [A]
time = 0.10, antiderivative size = 162, normalized size of antiderivative = 1.00, 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 = {808, 684, 654, 634, 212} \begin {gather*} \frac {5 b^3 (7 b B-8 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{9/2}}-\frac {5 b^2 \sqrt {b x+c x^2} (7 b B-8 A c)}{64 c^4}+\frac {5 b x \sqrt {b x+c x^2} (7 b B-8 A c)}{96 c^3}-\frac {x^2 \sqrt {b x+c x^2} (7 b B-8 A c)}{24 c^2}+\frac {B x^3 \sqrt {b x+c x^2}}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*b^2*(7*b*B - 8*A*c)*Sqrt[b*x + c*x^2])/(64*c^4) + (5*b*(7*b*B - 8*A*c)*x*Sqrt[b*x + c*x^2])/(96*c^3) - ((7
*b*B - 8*A*c)*x^2*Sqrt[b*x + c*x^2])/(24*c^2) + (B*x^3*Sqrt[b*x + c*x^2])/(4*c) + (5*b^3*(7*b*B - 8*A*c)*ArcTa
nh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(64*c^(9/2))

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 808

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), Int[(d + e*x)^m*(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] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 0])

Rubi steps

\begin {align*} \int \frac {x^3 (A+B x)}{\sqrt {b x+c x^2}} \, dx &=\frac {B x^3 \sqrt {b x+c x^2}}{4 c}+\frac {\left (3 (-b B+A c)+\frac {1}{2} (-b B+2 A c)\right ) \int \frac {x^3}{\sqrt {b x+c x^2}} \, dx}{4 c}\\ &=-\frac {(7 b B-8 A c) x^2 \sqrt {b x+c x^2}}{24 c^2}+\frac {B x^3 \sqrt {b x+c x^2}}{4 c}+\frac {(5 b (7 b B-8 A c)) \int \frac {x^2}{\sqrt {b x+c x^2}} \, dx}{48 c^2}\\ &=\frac {5 b (7 b B-8 A c) x \sqrt {b x+c x^2}}{96 c^3}-\frac {(7 b B-8 A c) x^2 \sqrt {b x+c x^2}}{24 c^2}+\frac {B x^3 \sqrt {b x+c x^2}}{4 c}-\frac {\left (5 b^2 (7 b B-8 A c)\right ) \int \frac {x}{\sqrt {b x+c x^2}} \, dx}{64 c^3}\\ &=-\frac {5 b^2 (7 b B-8 A c) \sqrt {b x+c x^2}}{64 c^4}+\frac {5 b (7 b B-8 A c) x \sqrt {b x+c x^2}}{96 c^3}-\frac {(7 b B-8 A c) x^2 \sqrt {b x+c x^2}}{24 c^2}+\frac {B x^3 \sqrt {b x+c x^2}}{4 c}+\frac {\left (5 b^3 (7 b B-8 A c)\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{128 c^4}\\ &=-\frac {5 b^2 (7 b B-8 A c) \sqrt {b x+c x^2}}{64 c^4}+\frac {5 b (7 b B-8 A c) x \sqrt {b x+c x^2}}{96 c^3}-\frac {(7 b B-8 A c) x^2 \sqrt {b x+c x^2}}{24 c^2}+\frac {B x^3 \sqrt {b x+c x^2}}{4 c}+\frac {\left (5 b^3 (7 b B-8 A c)\right ) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{64 c^4}\\ &=-\frac {5 b^2 (7 b B-8 A c) \sqrt {b x+c x^2}}{64 c^4}+\frac {5 b (7 b B-8 A c) x \sqrt {b x+c x^2}}{96 c^3}-\frac {(7 b B-8 A c) x^2 \sqrt {b x+c x^2}}{24 c^2}+\frac {B x^3 \sqrt {b x+c x^2}}{4 c}+\frac {5 b^3 (7 b B-8 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{9/2}}\\ \end {align*}

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Mathematica [A]
time = 0.21, size = 137, normalized size = 0.85 \begin {gather*} \frac {-\sqrt {c} x (b+c x) \left (105 b^3 B-16 c^3 x^2 (4 A+3 B x)+8 b c^2 x (10 A+7 B x)-10 b^2 c (12 A+7 B x)\right )-15 b^3 (7 b B-8 A c) \sqrt {x} \sqrt {b+c x} \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )}{192 c^{9/2} \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[c]*x*(b + c*x)*(105*b^3*B - 16*c^3*x^2*(4*A + 3*B*x) + 8*b*c^2*x*(10*A + 7*B*x) - 10*b^2*c*(12*A + 7*B
*x))) - 15*b^3*(7*b*B - 8*A*c)*Sqrt[x]*Sqrt[b + c*x]*Log[-(Sqrt[c]*Sqrt[x]) + Sqrt[b + c*x]])/(192*c^(9/2)*Sqr
t[x*(b + c*x)])

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Maple [A]
time = 0.54, size = 224, normalized size = 1.38

method	result	size
risch	\(\frac {\left (48 B \,c^{3} x^{3}+64 A \,c^{3} x^{2}-56 b B \,x^{2} c^{2}-80 A b \,c^{2} x +70 B \,b^{2} c x +120 A \,b^{2} c -105 B \,b^{3}\right ) x \left (c x +b \right )}{192 c^{4} \sqrt {x \left (c x +b \right )}}-\frac {5 b^{3} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right ) A}{16 c^{\frac {7}{2}}}+\frac {35 b^{4} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right ) B}{128 c^{\frac {9}{2}}}\)	\(146\)
default	\(B \left (\frac {x^{3} \sqrt {c \,x^{2}+b x}}{4 c}-\frac {7 b \left (\frac {x^{2} \sqrt {c \,x^{2}+b x}}{3 c}-\frac {5 b \left (\frac {x \sqrt {c \,x^{2}+b x}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\right )}{6 c}\right )}{8 c}\right )+A \left (\frac {x^{2} \sqrt {c \,x^{2}+b x}}{3 c}-\frac {5 b \left (\frac {x \sqrt {c \,x^{2}+b x}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\right )}{6 c}\right )\)	\(224\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(B*x+A)/(c*x^2+b*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

B*(1/4*x^3*(c*x^2+b*x)^(1/2)/c-7/8*b/c*(1/3*x^2*(c*x^2+b*x)^(1/2)/c-5/6*b/c*(1/2*x*(c*x^2+b*x)^(1/2)/c-3/4*b/c
*((c*x^2+b*x)^(1/2)/c-1/2*b/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))))))+A*(1/3*x^2*(c*x^2+b*x)^(1/2)
/c-5/6*b/c*(1/2*x*(c*x^2+b*x)^(1/2)/c-3/4*b/c*((c*x^2+b*x)^(1/2)/c-1/2*b/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 = 0.26, size = 206, normalized size = 1.27 \begin {gather*} \frac {\sqrt {c x^{2} + b x} B x^{3}}{4 \, c} - \frac {7 \, \sqrt {c x^{2} + b x} B b x^{2}}{24 \, c^{2}} + \frac {\sqrt {c x^{2} + b x} A x^{2}}{3 \, c} + \frac {35 \, \sqrt {c x^{2} + b x} B b^{2} x}{96 \, c^{3}} - \frac {5 \, \sqrt {c x^{2} + b x} A b x}{12 \, c^{2}} + \frac {35 \, B b^{4} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {9}{2}}} - \frac {5 \, A b^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, c^{\frac {7}{2}}} - \frac {35 \, \sqrt {c x^{2} + b x} B b^{3}}{64 \, c^{4}} + \frac {5 \, \sqrt {c x^{2} + b x} A b^{2}}{8 \, 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)^(1/2),x, algorithm="maxima")

[Out]

1/4*sqrt(c*x^2 + b*x)*B*x^3/c - 7/24*sqrt(c*x^2 + b*x)*B*b*x^2/c^2 + 1/3*sqrt(c*x^2 + b*x)*A*x^2/c + 35/96*sqr
t(c*x^2 + b*x)*B*b^2*x/c^3 - 5/12*sqrt(c*x^2 + b*x)*A*b*x/c^2 + 35/128*B*b^4*log(2*c*x + b + 2*sqrt(c*x^2 + b*
x)*sqrt(c))/c^(9/2) - 5/16*A*b^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/2) - 35/64*sqrt(c*x^2 + b*x
)*B*b^3/c^4 + 5/8*sqrt(c*x^2 + b*x)*A*b^2/c^3

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Fricas [A]
time = 3.04, size = 256, normalized size = 1.58 \begin {gather*} \left [-\frac {15 \, {\left (7 \, B b^{4} - 8 \, A b^{3} c\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (48 \, B c^{4} x^{3} - 105 \, B b^{3} c + 120 \, A b^{2} c^{2} - 8 \, {\left (7 \, B b c^{3} - 8 \, A c^{4}\right )} x^{2} + 10 \, {\left (7 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{384 \, c^{5}}, -\frac {15 \, {\left (7 \, B b^{4} - 8 \, A b^{3} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (48 \, B c^{4} x^{3} - 105 \, B b^{3} c + 120 \, A b^{2} c^{2} - 8 \, {\left (7 \, B b c^{3} - 8 \, A c^{4}\right )} x^{2} + 10 \, {\left (7 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{192 \, c^{5}}\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)^(1/2),x, algorithm="fricas")

[Out]

[-1/384*(15*(7*B*b^4 - 8*A*b^3*c)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(48*B*c^4*x^3 - 105
*B*b^3*c + 120*A*b^2*c^2 - 8*(7*B*b*c^3 - 8*A*c^4)*x^2 + 10*(7*B*b^2*c^2 - 8*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/c^
5, -1/192*(15*(7*B*b^4 - 8*A*b^3*c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (48*B*c^4*x^3 - 105*B*
b^3*c + 120*A*b^2*c^2 - 8*(7*B*b*c^3 - 8*A*c^4)*x^2 + 10*(7*B*b^2*c^2 - 8*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/c^5]

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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 )}{\sqrt {x \left (b + c x\right )}}\, 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)**(1/2),x)

[Out]

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

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Giac [A]
time = 0.81, size = 137, normalized size = 0.85 \begin {gather*} \frac {1}{192} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (\frac {6 \, B x}{c} - \frac {7 \, B b c^{2} - 8 \, A c^{3}}{c^{4}}\right )} x + \frac {5 \, {\left (7 \, B b^{2} c - 8 \, A b c^{2}\right )}}{c^{4}}\right )} x - \frac {15 \, {\left (7 \, B b^{3} - 8 \, A b^{2} c\right )}}{c^{4}}\right )} - \frac {5 \, {\left (7 \, B b^{4} - 8 \, A b^{3} c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{128 \, c^{\frac {9}{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)^(1/2),x, algorithm="giac")

[Out]

1/192*sqrt(c*x^2 + b*x)*(2*(4*(6*B*x/c - (7*B*b*c^2 - 8*A*c^3)/c^4)*x + 5*(7*B*b^2*c - 8*A*b*c^2)/c^4)*x - 15*
(7*B*b^3 - 8*A*b^2*c)/c^4) - 5/128*(7*B*b^4 - 8*A*b^3*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) -
b))/c^(9/2)

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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 )}{\sqrt {c\,x^2+b\,x}} \,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)^(1/2),x)

[Out]

int((x^3*(A + B*x))/(b*x + c*x^2)^(1/2), x)

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