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

Optimal. Leaf size=163 \[ \frac {128 c^3 (b+2 c x) (9 b B-10 A c)}{315 b^6 \sqrt {b x+c x^2}}-\frac {32 c^2 (9 b B-10 A c)}{315 b^4 x \sqrt {b x+c x^2}}+\frac {16 c (9 b B-10 A c)}{315 b^3 x^2 \sqrt {b x+c x^2}}-\frac {2 (9 b B-10 A c)}{63 b^2 x^3 \sqrt {b x+c x^2}}-\frac {2 A}{9 b x^4 \sqrt {b x+c x^2}} \]

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Rubi [A]  time = 0.14, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {792, 658, 613} \begin {gather*} \frac {128 c^3 (b+2 c x) (9 b B-10 A c)}{315 b^6 \sqrt {b x+c x^2}}-\frac {32 c^2 (9 b B-10 A c)}{315 b^4 x \sqrt {b x+c x^2}}+\frac {16 c (9 b B-10 A c)}{315 b^3 x^2 \sqrt {b x+c x^2}}-\frac {2 (9 b B-10 A c)}{63 b^2 x^3 \sqrt {b x+c x^2}}-\frac {2 A}{9 b x^4 \sqrt {b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A)/(9*b*x^4*Sqrt[b*x + c*x^2]) - (2*(9*b*B - 10*A*c))/(63*b^2*x^3*Sqrt[b*x + c*x^2]) + (16*c*(9*b*B - 10*A
*c))/(315*b^3*x^2*Sqrt[b*x + c*x^2]) - (32*c^2*(9*b*B - 10*A*c))/(315*b^4*x*Sqrt[b*x + c*x^2]) + (128*c^3*(9*b
*B - 10*A*c)*(b + 2*c*x))/(315*b^6*Sqrt[b*x + c*x^2])

Rule 613

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[(-2*(b + 2*c*x))/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2]), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 658

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

Rule 792

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

Rubi steps

\begin {align*} \int \frac {A+B x}{x^4 \left (b x+c x^2\right )^{3/2}} \, dx &=-\frac {2 A}{9 b x^4 \sqrt {b x+c x^2}}+\frac {\left (2 \left (\frac {1}{2} (b B-2 A c)-4 (-b B+A c)\right )\right ) \int \frac {1}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx}{9 b}\\ &=-\frac {2 A}{9 b x^4 \sqrt {b x+c x^2}}-\frac {2 (9 b B-10 A c)}{63 b^2 x^3 \sqrt {b x+c x^2}}-\frac {(8 c (9 b B-10 A c)) \int \frac {1}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx}{63 b^2}\\ &=-\frac {2 A}{9 b x^4 \sqrt {b x+c x^2}}-\frac {2 (9 b B-10 A c)}{63 b^2 x^3 \sqrt {b x+c x^2}}+\frac {16 c (9 b B-10 A c)}{315 b^3 x^2 \sqrt {b x+c x^2}}+\frac {\left (16 c^2 (9 b B-10 A c)\right ) \int \frac {1}{x \left (b x+c x^2\right )^{3/2}} \, dx}{105 b^3}\\ &=-\frac {2 A}{9 b x^4 \sqrt {b x+c x^2}}-\frac {2 (9 b B-10 A c)}{63 b^2 x^3 \sqrt {b x+c x^2}}+\frac {16 c (9 b B-10 A c)}{315 b^3 x^2 \sqrt {b x+c x^2}}-\frac {32 c^2 (9 b B-10 A c)}{315 b^4 x \sqrt {b x+c x^2}}-\frac {\left (64 c^3 (9 b B-10 A c)\right ) \int \frac {1}{\left (b x+c x^2\right )^{3/2}} \, dx}{315 b^4}\\ &=-\frac {2 A}{9 b x^4 \sqrt {b x+c x^2}}-\frac {2 (9 b B-10 A c)}{63 b^2 x^3 \sqrt {b x+c x^2}}+\frac {16 c (9 b B-10 A c)}{315 b^3 x^2 \sqrt {b x+c x^2}}-\frac {32 c^2 (9 b B-10 A c)}{315 b^4 x \sqrt {b x+c x^2}}+\frac {128 c^3 (9 b B-10 A c) (b+2 c x)}{315 b^6 \sqrt {b x+c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 123, normalized size = 0.75 \begin {gather*} -\frac {2 \left (5 A \left (7 b^5-10 b^4 c x+16 b^3 c^2 x^2-32 b^2 c^3 x^3+128 b c^4 x^4+256 c^5 x^5\right )+9 b B x \left (5 b^4-8 b^3 c x+16 b^2 c^2 x^2-64 b c^3 x^3-128 c^4 x^4\right )\right )}{315 b^6 x^4 \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(9*b*B*x*(5*b^4 - 8*b^3*c*x + 16*b^2*c^2*x^2 - 64*b*c^3*x^3 - 128*c^4*x^4) + 5*A*(7*b^5 - 10*b^4*c*x + 16*
b^3*c^2*x^2 - 32*b^2*c^3*x^3 + 128*b*c^4*x^4 + 256*c^5*x^5)))/(315*b^6*x^4*Sqrt[x*(b + c*x)])

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IntegrateAlgebraic [A]  time = 0.42, size = 139, normalized size = 0.85 \begin {gather*} \frac {2 \sqrt {b x+c x^2} \left (-35 A b^5+50 A b^4 c x-80 A b^3 c^2 x^2+160 A b^2 c^3 x^3-640 A b c^4 x^4-1280 A c^5 x^5-45 b^5 B x+72 b^4 B c x^2-144 b^3 B c^2 x^3+576 b^2 B c^3 x^4+1152 b B c^4 x^5\right )}{315 b^6 x^5 (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[b*x + c*x^2]*(-35*A*b^5 - 45*b^5*B*x + 50*A*b^4*c*x + 72*b^4*B*c*x^2 - 80*A*b^3*c^2*x^2 - 144*b^3*B*c^
2*x^3 + 160*A*b^2*c^3*x^3 + 576*b^2*B*c^3*x^4 - 640*A*b*c^4*x^4 + 1152*b*B*c^4*x^5 - 1280*A*c^5*x^5))/(315*b^6
*x^5*(b + c*x))

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fricas [A]  time = 0.41, size = 142, normalized size = 0.87 \begin {gather*} -\frac {2 \, {\left (35 \, A b^{5} - 128 \, {\left (9 \, B b c^{4} - 10 \, A c^{5}\right )} x^{5} - 64 \, {\left (9 \, B b^{2} c^{3} - 10 \, A b c^{4}\right )} x^{4} + 16 \, {\left (9 \, B b^{3} c^{2} - 10 \, A b^{2} c^{3}\right )} x^{3} - 8 \, {\left (9 \, B b^{4} c - 10 \, A b^{3} c^{2}\right )} x^{2} + 5 \, {\left (9 \, B b^{5} - 10 \, A b^{4} c\right )} x\right )} \sqrt {c x^{2} + b x}}{315 \, {\left (b^{6} c x^{6} + b^{7} x^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(35*A*b^5 - 128*(9*B*b*c^4 - 10*A*c^5)*x^5 - 64*(9*B*b^2*c^3 - 10*A*b*c^4)*x^4 + 16*(9*B*b^3*c^2 - 10*A
*b^2*c^3)*x^3 - 8*(9*B*b^4*c - 10*A*b^3*c^2)*x^2 + 5*(9*B*b^5 - 10*A*b^4*c)*x)*sqrt(c*x^2 + b*x)/(b^6*c*x^6 +
b^7*x^5)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 134, normalized size = 0.82 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (1280 A \,c^{5} x^{5}-1152 B b \,c^{4} x^{5}+640 A b \,c^{4} x^{4}-576 B \,b^{2} c^{3} x^{4}-160 A \,b^{2} c^{3} x^{3}+144 B \,b^{3} c^{2} x^{3}+80 A \,b^{3} c^{2} x^{2}-72 B \,b^{4} c \,x^{2}-50 A \,b^{4} c x +45 B \,b^{5} x +35 A \,b^{5}\right )}{315 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} b^{6} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(c*x+b)*(1280*A*c^5*x^5-1152*B*b*c^4*x^5+640*A*b*c^4*x^4-576*B*b^2*c^3*x^4-160*A*b^2*c^3*x^3+144*B*b^3*
c^2*x^3+80*A*b^3*c^2*x^2-72*B*b^4*c*x^2-50*A*b^4*c*x+45*B*b^5*x+35*A*b^5)/x^3/b^6/(c*x^2+b*x)^(3/2)

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maxima [A]  time = 0.52, size = 234, normalized size = 1.44 \begin {gather*} \frac {256 \, B c^{4} x}{35 \, \sqrt {c x^{2} + b x} b^{5}} - \frac {512 \, A c^{5} x}{63 \, \sqrt {c x^{2} + b x} b^{6}} + \frac {128 \, B c^{3}}{35 \, \sqrt {c x^{2} + b x} b^{4}} - \frac {256 \, A c^{4}}{63 \, \sqrt {c x^{2} + b x} b^{5}} - \frac {32 \, B c^{2}}{35 \, \sqrt {c x^{2} + b x} b^{3} x} + \frac {64 \, A c^{3}}{63 \, \sqrt {c x^{2} + b x} b^{4} x} + \frac {16 \, B c}{35 \, \sqrt {c x^{2} + b x} b^{2} x^{2}} - \frac {32 \, A c^{2}}{63 \, \sqrt {c x^{2} + b x} b^{3} x^{2}} - \frac {2 \, B}{7 \, \sqrt {c x^{2} + b x} b x^{3}} + \frac {20 \, A c}{63 \, \sqrt {c x^{2} + b x} b^{2} x^{3}} - \frac {2 \, A}{9 \, \sqrt {c x^{2} + b x} b x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

256/35*B*c^4*x/(sqrt(c*x^2 + b*x)*b^5) - 512/63*A*c^5*x/(sqrt(c*x^2 + b*x)*b^6) + 128/35*B*c^3/(sqrt(c*x^2 + b
*x)*b^4) - 256/63*A*c^4/(sqrt(c*x^2 + b*x)*b^5) - 32/35*B*c^2/(sqrt(c*x^2 + b*x)*b^3*x) + 64/63*A*c^3/(sqrt(c*
x^2 + b*x)*b^4*x) + 16/35*B*c/(sqrt(c*x^2 + b*x)*b^2*x^2) - 32/63*A*c^2/(sqrt(c*x^2 + b*x)*b^3*x^2) - 2/7*B/(s
qrt(c*x^2 + b*x)*b*x^3) + 20/63*A*c/(sqrt(c*x^2 + b*x)*b^2*x^3) - 2/9*A/(sqrt(c*x^2 + b*x)*b*x^4)

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mupad [B]  time = 1.36, size = 191, normalized size = 1.17 \begin {gather*} \frac {\sqrt {c\,x^2+b\,x}\,\left (x\,\left (\frac {1300\,A\,c^5-1044\,B\,b\,c^4}{315\,b^6}-\frac {4\,c^4\,\left (965\,A\,c-837\,B\,b\right )}{315\,b^6}\right )-\frac {2\,c^3\,\left (965\,A\,c-837\,B\,b\right )}{315\,b^5}\right )}{x\,\left (b+c\,x\right )}-\frac {2\,A\,\sqrt {c\,x^2+b\,x}}{9\,b^2\,x^5}-\frac {\left (18\,B\,b^2-34\,A\,b\,c\right )\,\sqrt {c\,x^2+b\,x}}{63\,b^4\,x^4}-\frac {2\,c\,\sqrt {c\,x^2+b\,x}\,\left (55\,A\,c-39\,B\,b\right )}{105\,b^4\,x^3}+\frac {2\,c^2\,\sqrt {c\,x^2+b\,x}\,\left (325\,A\,c-261\,B\,b\right )}{315\,b^5\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*x + c*x^2)^(1/2)*(x*((1300*A*c^5 - 1044*B*b*c^4)/(315*b^6) - (4*c^4*(965*A*c - 837*B*b))/(315*b^6)) - (2*c
^3*(965*A*c - 837*B*b))/(315*b^5)))/(x*(b + c*x)) - (2*A*(b*x + c*x^2)^(1/2))/(9*b^2*x^5) - ((18*B*b^2 - 34*A*
b*c)*(b*x + c*x^2)^(1/2))/(63*b^4*x^4) - (2*c*(b*x + c*x^2)^(1/2)*(55*A*c - 39*B*b))/(105*b^4*x^3) + (2*c^2*(b
*x + c*x^2)^(1/2)*(325*A*c - 261*B*b))/(315*b^5*x^2)

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

[Out]

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

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