3.2.30 \(\int \frac {x^4 (A+B x)}{(b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=136 \[ -\frac {(5 b B-2 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{7/2}}+\frac {\sqrt {b x+c x^2} (5 b B-2 A c)}{b c^3}-\frac {2 x^2 (5 b B-2 A c)}{3 b c^2 \sqrt {b x+c x^2}}-\frac {2 x^4 (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.12, antiderivative size = 136, 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 = {788, 668, 640, 620, 206} \begin {gather*} -\frac {2 x^2 (5 b B-2 A c)}{3 b c^2 \sqrt {b x+c x^2}}+\frac {\sqrt {b x+c x^2} (5 b B-2 A c)}{b c^3}-\frac {(5 b B-2 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{7/2}}-\frac {2 x^4 (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(A + B*x))/(b*x + c*x^2)^(5/2),x]

[Out]

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

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 668

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*(p + 1)), x] - Dist[(e^2*(m + p))/(c*(p + 1)), Int[(d + e*x)^(m - 2)*(a + b*x +
c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &
& LtQ[p, -1] && GtQ[m, 1] && 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 )^{5/2}} \, dx &=-\frac {2 (b B-A c) x^4}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac {1}{3} \left (\frac {2 A}{b}-\frac {5 B}{c}\right ) \int \frac {x^3}{\left (b x+c x^2\right )^{3/2}} \, dx\\ &=-\frac {2 (b B-A c) x^4}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac {2 (5 b B-2 A c) x^2}{3 b c^2 \sqrt {b x+c x^2}}+\frac {(5 b B-2 A c) \int \frac {x}{\sqrt {b x+c x^2}} \, dx}{b c^2}\\ &=-\frac {2 (b B-A c) x^4}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac {2 (5 b B-2 A c) x^2}{3 b c^2 \sqrt {b x+c x^2}}+\frac {(5 b B-2 A c) \sqrt {b x+c x^2}}{b c^3}-\frac {(5 b B-2 A c) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{2 c^3}\\ &=-\frac {2 (b B-A c) x^4}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac {2 (5 b B-2 A c) x^2}{3 b c^2 \sqrt {b x+c x^2}}+\frac {(5 b B-2 A c) \sqrt {b x+c x^2}}{b c^3}-\frac {(5 b B-2 A c) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{c^3}\\ &=-\frac {2 (b B-A c) x^4}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac {2 (5 b B-2 A c) x^2}{3 b c^2 \sqrt {b x+c x^2}}+\frac {(5 b B-2 A c) \sqrt {b x+c x^2}}{b c^3}-\frac {(5 b B-2 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 80, normalized size = 0.59 \begin {gather*} \frac {2 x^4 \left ((b+c x) \sqrt {\frac {c x}{b}+1} (5 b B-2 A c) \, _2F_1\left (\frac {3}{2},\frac {5}{2};\frac {7}{2};-\frac {c x}{b}\right )+5 b (A c-b B)\right )}{15 b^2 c (x (b+c x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(A + B*x))/(b*x + c*x^2)^(5/2),x]

[Out]

(2*x^4*(5*b*(-(b*B) + A*c) + (5*b*B - 2*A*c)*(b + c*x)*Sqrt[1 + (c*x)/b]*Hypergeometric2F1[3/2, 5/2, 7/2, -((c
*x)/b)]))/(15*b^2*c*(x*(b + c*x))^(3/2))

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IntegrateAlgebraic [A]  time = 0.51, size = 107, normalized size = 0.79 \begin {gather*} \frac {\sqrt {b x+c x^2} \left (-6 A b c-8 A c^2 x+15 b^2 B+20 b B c x+3 B c^2 x^2\right )}{3 c^3 (b+c x)^2}+\frac {(5 b B-2 A c) \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right )}{2 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^4*(A + B*x))/(b*x + c*x^2)^(5/2),x]

[Out]

(Sqrt[b*x + c*x^2]*(15*b^2*B - 6*A*b*c + 20*b*B*c*x - 8*A*c^2*x + 3*B*c^2*x^2))/(3*c^3*(b + c*x)^2) + ((5*b*B
- 2*A*c)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[b*x + c*x^2]])/(2*c^(7/2))

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fricas [A]  time = 0.43, size = 321, normalized size = 2.36 \begin {gather*} \left [-\frac {3 \, {\left (5 \, B b^{3} - 2 \, A b^{2} c + {\left (5 \, B b c^{2} - 2 \, A c^{3}\right )} x^{2} + 2 \, {\left (5 \, B b^{2} c - 2 \, 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 (3 \, B c^{3} x^{2} + 15 \, B b^{2} c - 6 \, A b c^{2} + 4 \, {\left (5 \, B b c^{2} - 2 \, A c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{6 \, {\left (c^{6} x^{2} + 2 \, b c^{5} x + b^{2} c^{4}\right )}}, \frac {3 \, {\left (5 \, B b^{3} - 2 \, A b^{2} c + {\left (5 \, B b c^{2} - 2 \, A c^{3}\right )} x^{2} + 2 \, {\left (5 \, B b^{2} c - 2 \, A b c^{2}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (3 \, B c^{3} x^{2} + 15 \, B b^{2} c - 6 \, A b c^{2} + 4 \, {\left (5 \, B b c^{2} - 2 \, A c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{3 \, {\left (c^{6} x^{2} + 2 \, b c^{5} x + b^{2} c^{4}\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)^(5/2),x, algorithm="fricas")

[Out]

[-1/6*(3*(5*B*b^3 - 2*A*b^2*c + (5*B*b*c^2 - 2*A*c^3)*x^2 + 2*(5*B*b^2*c - 2*A*b*c^2)*x)*sqrt(c)*log(2*c*x + b
+ 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(3*B*c^3*x^2 + 15*B*b^2*c - 6*A*b*c^2 + 4*(5*B*b*c^2 - 2*A*c^3)*x)*sqrt(c*
x^2 + b*x))/(c^6*x^2 + 2*b*c^5*x + b^2*c^4), 1/3*(3*(5*B*b^3 - 2*A*b^2*c + (5*B*b*c^2 - 2*A*c^3)*x^2 + 2*(5*B*
b^2*c - 2*A*b*c^2)*x)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (3*B*c^3*x^2 + 15*B*b^2*c - 6*A*b*c^
2 + 4*(5*B*b*c^2 - 2*A*c^3)*x)*sqrt(c*x^2 + b*x))/(c^6*x^2 + 2*b*c^5*x + b^2*c^4)]

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giac [A]  time = 0.29, size = 224, normalized size = 1.65 \begin {gather*} \frac {\sqrt {c x^{2} + b x} B}{c^{3}} + \frac {{\left (5 \, B b - 2 \, A c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{2 \, c^{\frac {7}{2}}} + \frac {2 \, {\left (9 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{2} c^{\frac {3}{2}} - 6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b c^{\frac {5}{2}} + 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} B b^{3} c - 9 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{2} c^{2} + 7 \, B b^{4} \sqrt {c} - 4 \, A b^{3} c^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b\right )}^{3} 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)^(5/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.05, size = 283, normalized size = 2.08 \begin {gather*} \frac {B \,x^{4}}{\left (c \,x^{2}+b x \right )^{\frac {3}{2}} c}-\frac {A \,x^{3}}{3 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c}+\frac {5 B b \,x^{3}}{6 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{2}}+\frac {A b \,x^{2}}{2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{2}}-\frac {5 B \,b^{2} x^{2}}{4 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{3}}+\frac {A \,b^{2} x}{6 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{3}}-\frac {5 B \,b^{3} x}{12 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{4}}-\frac {7 A x}{3 \sqrt {c \,x^{2}+b x}\, c^{2}}+\frac {35 B b x}{6 \sqrt {c \,x^{2}+b x}\, c^{3}}+\frac {A \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {5}{2}}}-\frac {5 B b \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {7}{2}}}-\frac {A b}{6 \sqrt {c \,x^{2}+b x}\, c^{3}}+\frac {5 B \,b^{2}}{12 \sqrt {c \,x^{2}+b x}\, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(B*x+A)/(c*x^2+b*x)^(5/2),x)

[Out]

B*x^4/c/(c*x^2+b*x)^(3/2)+5/6*B*b/c^2*x^3/(c*x^2+b*x)^(3/2)-5/4*B*b^2/c^3*x^2/(c*x^2+b*x)^(3/2)-5/12*B*b^3/c^4
/(c*x^2+b*x)^(3/2)*x+35/6*B*b/c^3/(c*x^2+b*x)^(1/2)*x+5/12*B*b^2/c^4/(c*x^2+b*x)^(1/2)-5/2*B*b/c^(7/2)*ln((c*x
+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))-1/3*A*x^3/c/(c*x^2+b*x)^(3/2)+1/2*A*b/c^2*x^2/(c*x^2+b*x)^(3/2)+1/6*A*b^2/c
^3/(c*x^2+b*x)^(3/2)*x-7/3*A/c^2/(c*x^2+b*x)^(1/2)*x-1/6*A*b/c^3/(c*x^2+b*x)^(1/2)+A/c^(5/2)*ln((c*x+1/2*b)/c^
(1/2)+(c*x^2+b*x)^(1/2))

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maxima [B]  time = 0.60, size = 310, normalized size = 2.28 \begin {gather*} -\frac {1}{3} \, A x {\left (\frac {3 \, x^{2}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} + \frac {b x}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c^{2}} - \frac {2 \, x}{\sqrt {c x^{2} + b x} b c} - \frac {1}{\sqrt {c x^{2} + b x} c^{2}}\right )} + \frac {5 \, B b x {\left (\frac {3 \, x^{2}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} + \frac {b x}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c^{2}} - \frac {2 \, x}{\sqrt {c x^{2} + b x} b c} - \frac {1}{\sqrt {c x^{2} + b x} c^{2}}\right )}}{6 \, c} + \frac {B x^{4}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} + \frac {10 \, B b x}{3 \, \sqrt {c x^{2} + b x} c^{3}} - \frac {4 \, A x}{3 \, \sqrt {c x^{2} + b x} c^{2}} - \frac {5 \, B b \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, c^{\frac {7}{2}}} + \frac {A \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{c^{\frac {5}{2}}} + \frac {5 \, \sqrt {c x^{2} + b x} B}{3 \, c^{3}} - \frac {2 \, \sqrt {c x^{2} + b x} A}{3 \, b c^{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)^(5/2),x, algorithm="maxima")

[Out]

-1/3*A*x*(3*x^2/((c*x^2 + b*x)^(3/2)*c) + b*x/((c*x^2 + b*x)^(3/2)*c^2) - 2*x/(sqrt(c*x^2 + b*x)*b*c) - 1/(sqr
t(c*x^2 + b*x)*c^2)) + 5/6*B*b*x*(3*x^2/((c*x^2 + b*x)^(3/2)*c) + b*x/((c*x^2 + b*x)^(3/2)*c^2) - 2*x/(sqrt(c*
x^2 + b*x)*b*c) - 1/(sqrt(c*x^2 + b*x)*c^2))/c + B*x^4/((c*x^2 + b*x)^(3/2)*c) + 10/3*B*b*x/(sqrt(c*x^2 + b*x)
*c^3) - 4/3*A*x/(sqrt(c*x^2 + b*x)*c^2) - 5/2*B*b*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/2) + A*log
(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2) + 5/3*sqrt(c*x^2 + b*x)*B/c^3 - 2/3*sqrt(c*x^2 + b*x)*A/(b*c
^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 )}^{5/2}} \,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)^(5/2),x)

[Out]

int((x^4*(A + B*x))/(b*x + c*x^2)^(5/2), 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 {5}{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)**(5/2),x)

[Out]

Integral(x**4*(A + B*x)/(x*(b + c*x))**(5/2), x)

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