3.2.20 \(\int \frac {x^4 (A+B x)}{(b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=170 \[ -\frac {5 b^2 (7 b B-6 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{9/2}}+\frac {5 b \sqrt {b x+c x^2} (7 b B-6 A c)}{8 c^4}-\frac {5 x \sqrt {b x+c x^2} (7 b B-6 A c)}{12 c^3}+\frac {x^2 \sqrt {b x+c x^2} (7 b B-6 A c)}{3 b c^2}-\frac {2 x^4 (b B-A c)}{b c \sqrt {b x+c x^2}} \]

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Rubi [A]  time = 0.15, antiderivative size = 170, 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 = {788, 670, 640, 620, 206} \begin {gather*} -\frac {5 b^2 (7 b B-6 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{9/2}}+\frac {x^2 \sqrt {b x+c x^2} (7 b B-6 A c)}{3 b c^2}-\frac {5 x \sqrt {b x+c x^2} (7 b B-6 A c)}{12 c^3}+\frac {5 b \sqrt {b x+c x^2} (7 b B-6 A c)}{8 c^4}-\frac {2 x^4 (b B-A c)}{b c \sqrt {b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(A + B*x))/(b*x + c*x^2)^(3/2),x]

[Out]

(-2*(b*B - A*c)*x^4)/(b*c*Sqrt[b*x + c*x^2]) + (5*b*(7*b*B - 6*A*c)*Sqrt[b*x + c*x^2])/(8*c^4) - (5*(7*b*B - 6
*A*c)*x*Sqrt[b*x + c*x^2])/(12*c^3) + ((7*b*B - 6*A*c)*x^2*Sqrt[b*x + c*x^2])/(3*b*c^2) - (5*b^2*(7*b*B - 6*A*
c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(8*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 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 640

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 670

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 788

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^4 (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx &=-\frac {2 (b B-A c) x^4}{b c \sqrt {b x+c x^2}}-\left (\frac {6 A}{b}-\frac {7 B}{c}\right ) \int \frac {x^3}{\sqrt {b x+c x^2}} \, dx\\ &=-\frac {2 (b B-A c) x^4}{b c \sqrt {b x+c x^2}}+\frac {(7 b B-6 A c) x^2 \sqrt {b x+c x^2}}{3 b c^2}-\frac {(5 (7 b B-6 A c)) \int \frac {x^2}{\sqrt {b x+c x^2}} \, dx}{6 c^2}\\ &=-\frac {2 (b B-A c) x^4}{b c \sqrt {b x+c x^2}}-\frac {5 (7 b B-6 A c) x \sqrt {b x+c x^2}}{12 c^3}+\frac {(7 b B-6 A c) x^2 \sqrt {b x+c x^2}}{3 b c^2}+\frac {(5 b (7 b B-6 A c)) \int \frac {x}{\sqrt {b x+c x^2}} \, dx}{8 c^3}\\ &=-\frac {2 (b B-A c) x^4}{b c \sqrt {b x+c x^2}}+\frac {5 b (7 b B-6 A c) \sqrt {b x+c x^2}}{8 c^4}-\frac {5 (7 b B-6 A c) x \sqrt {b x+c x^2}}{12 c^3}+\frac {(7 b B-6 A c) x^2 \sqrt {b x+c x^2}}{3 b c^2}-\frac {\left (5 b^2 (7 b B-6 A c)\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{16 c^4}\\ &=-\frac {2 (b B-A c) x^4}{b c \sqrt {b x+c x^2}}+\frac {5 b (7 b B-6 A c) \sqrt {b x+c x^2}}{8 c^4}-\frac {5 (7 b B-6 A c) x \sqrt {b x+c x^2}}{12 c^3}+\frac {(7 b B-6 A c) x^2 \sqrt {b x+c x^2}}{3 b c^2}-\frac {\left (5 b^2 (7 b B-6 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{8 c^4}\\ &=-\frac {2 (b B-A c) x^4}{b c \sqrt {b x+c x^2}}+\frac {5 b (7 b B-6 A c) \sqrt {b x+c x^2}}{8 c^4}-\frac {5 (7 b B-6 A c) x \sqrt {b x+c x^2}}{12 c^3}+\frac {(7 b B-6 A c) x^2 \sqrt {b x+c x^2}}{3 b c^2}-\frac {5 b^2 (7 b B-6 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 137, normalized size = 0.81 \begin {gather*} \frac {\frac {(b+c x) (7 b B-6 A c) \left (c x \sqrt {\frac {c x}{b}+1} \left (15 b^2-10 b c x+8 c^2 x^2\right )-15 b^{5/2} \sqrt {c} \sqrt {x} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )\right )}{3 \sqrt {\frac {c x}{b}+1}}+16 c^4 x^4 (A c-b B)}{8 b c^5 \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(A + B*x))/(b*x + c*x^2)^(3/2),x]

[Out]

(16*c^4*(-(b*B) + A*c)*x^4 + ((7*b*B - 6*A*c)*(b + c*x)*(c*x*Sqrt[1 + (c*x)/b]*(15*b^2 - 10*b*c*x + 8*c^2*x^2)
- 15*b^(5/2)*Sqrt[c]*Sqrt[x]*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]]))/(3*Sqrt[1 + (c*x)/b]))/(8*b*c^5*Sqrt[x*(b +
c*x)])

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IntegrateAlgebraic [A]  time = 0.62, size = 136, normalized size = 0.80 \begin {gather*} \frac {5 \left (7 b^3 B-6 A b^2 c\right ) \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right )}{16 c^{9/2}}+\frac {\sqrt {b x+c x^2} \left (-90 A b^2 c-30 A b c^2 x+12 A c^3 x^2+105 b^3 B+35 b^2 B c x-14 b B c^2 x^2+8 B c^3 x^3\right )}{24 c^4 (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^4*(A + B*x))/(b*x + c*x^2)^(3/2),x]

[Out]

(Sqrt[b*x + c*x^2]*(105*b^3*B - 90*A*b^2*c + 35*b^2*B*c*x - 30*A*b*c^2*x - 14*b*B*c^2*x^2 + 12*A*c^3*x^2 + 8*B
*c^3*x^3))/(24*c^4*(b + c*x)) + (5*(7*b^3*B - 6*A*b^2*c)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[b*x + c*x^2]])/(16*c^(
9/2))

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fricas [A]  time = 0.42, size = 313, normalized size = 1.84 \begin {gather*} \left [-\frac {15 \, {\left (7 \, B b^{4} - 6 \, A b^{3} c + {\left (7 \, B b^{3} c - 6 \, A b^{2} 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 (8 \, B c^{4} x^{3} + 105 \, B b^{3} c - 90 \, A b^{2} c^{2} - 2 \, {\left (7 \, B b c^{3} - 6 \, A c^{4}\right )} x^{2} + 5 \, {\left (7 \, B b^{2} c^{2} - 6 \, A b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{48 \, {\left (c^{6} x + b c^{5}\right )}}, \frac {15 \, {\left (7 \, B b^{4} - 6 \, A b^{3} c + {\left (7 \, B b^{3} c - 6 \, A b^{2} c^{2}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (8 \, B c^{4} x^{3} + 105 \, B b^{3} c - 90 \, A b^{2} c^{2} - 2 \, {\left (7 \, B b c^{3} - 6 \, A c^{4}\right )} x^{2} + 5 \, {\left (7 \, B b^{2} c^{2} - 6 \, A b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{24 \, {\left (c^{6} x + b c^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x+A)/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

[-1/48*(15*(7*B*b^4 - 6*A*b^3*c + (7*B*b^3*c - 6*A*b^2*c^2)*x)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqr
t(c)) - 2*(8*B*c^4*x^3 + 105*B*b^3*c - 90*A*b^2*c^2 - 2*(7*B*b*c^3 - 6*A*c^4)*x^2 + 5*(7*B*b^2*c^2 - 6*A*b*c^3
)*x)*sqrt(c*x^2 + b*x))/(c^6*x + b*c^5), 1/24*(15*(7*B*b^4 - 6*A*b^3*c + (7*B*b^3*c - 6*A*b^2*c^2)*x)*sqrt(-c)
*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (8*B*c^4*x^3 + 105*B*b^3*c - 90*A*b^2*c^2 - 2*(7*B*b*c^3 - 6*A*c^4
)*x^2 + 5*(7*B*b^2*c^2 - 6*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/(c^6*x + b*c^5)]

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giac [A]  time = 0.26, size = 163, normalized size = 0.96 \begin {gather*} \frac {1}{24} \, \sqrt {c x^{2} + b x} {\left (2 \, x {\left (\frac {4 \, B x}{c^{2}} - \frac {11 \, B b c^{10} - 6 \, A c^{11}}{c^{13}}\right )} + \frac {3 \, {\left (19 \, B b^{2} c^{9} - 14 \, A b c^{10}\right )}}{c^{13}}\right )} + \frac {5 \, {\left (7 \, B b^{3} - 6 \, A b^{2} c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{16 \, c^{\frac {9}{2}}} + \frac {2 \, {\left (B b^{4} - A b^{3} c\right )}}{{\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} c + b \sqrt {c}\right )} c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x+A)/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

1/24*sqrt(c*x^2 + b*x)*(2*x*(4*B*x/c^2 - (11*B*b*c^10 - 6*A*c^11)/c^13) + 3*(19*B*b^2*c^9 - 14*A*b*c^10)/c^13)
+ 5/16*(7*B*b^3 - 6*A*b^2*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(9/2) + 2*(B*b^4 - A*
b^3*c)/(((sqrt(c)*x - sqrt(c*x^2 + b*x))*c + b*sqrt(c))*c^4)

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maple [A]  time = 0.06, size = 215, normalized size = 1.26 \begin {gather*} \frac {B \,x^{4}}{3 \sqrt {c \,x^{2}+b x}\, c}+\frac {A \,x^{3}}{2 \sqrt {c \,x^{2}+b x}\, c}-\frac {7 B b \,x^{3}}{12 \sqrt {c \,x^{2}+b x}\, c^{2}}-\frac {5 A b \,x^{2}}{4 \sqrt {c \,x^{2}+b x}\, c^{2}}+\frac {35 B \,b^{2} x^{2}}{24 \sqrt {c \,x^{2}+b x}\, c^{3}}-\frac {15 A \,b^{2} x}{4 \sqrt {c \,x^{2}+b x}\, c^{3}}+\frac {35 B \,b^{3} x}{8 \sqrt {c \,x^{2}+b x}\, c^{4}}+\frac {15 A \,b^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {7}{2}}}-\frac {35 B \,b^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{16 c^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(B*x+A)/(c*x^2+b*x)^(3/2),x)

[Out]

1/3*B*x^4/c/(c*x^2+b*x)^(1/2)-7/12*B*b/c^2*x^3/(c*x^2+b*x)^(1/2)+35/24*B*b^2/c^3*x^2/(c*x^2+b*x)^(1/2)+35/8*B*
b^3/c^4/(c*x^2+b*x)^(1/2)*x-35/16*B*b^3/c^(9/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))+1/2*A*x^3/c/(c*x^2+b
*x)^(1/2)-5/4*A*b/c^2*x^2/(c*x^2+b*x)^(1/2)-15/4*A*b^2/c^3/(c*x^2+b*x)^(1/2)*x+15/8*A*b^2/c^(7/2)*ln((c*x+1/2*
b)/c^(1/2)+(c*x^2+b*x)^(1/2))

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maxima [A]  time = 0.94, size = 212, normalized size = 1.25 \begin {gather*} \frac {B x^{4}}{3 \, \sqrt {c x^{2} + b x} c} - \frac {7 \, B b x^{3}}{12 \, \sqrt {c x^{2} + b x} c^{2}} + \frac {A x^{3}}{2 \, \sqrt {c x^{2} + b x} c} + \frac {35 \, B b^{2} x^{2}}{24 \, \sqrt {c x^{2} + b x} c^{3}} - \frac {5 \, A b x^{2}}{4 \, \sqrt {c x^{2} + b x} c^{2}} + \frac {35 \, B b^{3} x}{8 \, \sqrt {c x^{2} + b x} c^{4}} - \frac {15 \, A b^{2} x}{4 \, \sqrt {c x^{2} + b x} c^{3}} - \frac {35 \, B b^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, c^{\frac {9}{2}}} + \frac {15 \, A b^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x+A)/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

1/3*B*x^4/(sqrt(c*x^2 + b*x)*c) - 7/12*B*b*x^3/(sqrt(c*x^2 + b*x)*c^2) + 1/2*A*x^3/(sqrt(c*x^2 + b*x)*c) + 35/
24*B*b^2*x^2/(sqrt(c*x^2 + b*x)*c^3) - 5/4*A*b*x^2/(sqrt(c*x^2 + b*x)*c^2) + 35/8*B*b^3*x/(sqrt(c*x^2 + b*x)*c
^4) - 15/4*A*b^2*x/(sqrt(c*x^2 + b*x)*c^3) - 35/16*B*b^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(9/2)
+ 15/8*A*b^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^4\,\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^4*(A + B*x))/(b*x + c*x^2)^(3/2),x)

[Out]

int((x^4*(A + B*x))/(b*x + c*x^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4} \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**4*(B*x+A)/(c*x**2+b*x)**(3/2),x)

[Out]

Integral(x**4*(A + B*x)/(x*(b + c*x))**(3/2), x)

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