3.2.18 \(\int \frac {A+B x}{x^5 \sqrt {b x+c x^2}} \, dx\)

Optimal. Leaf size=160 \[ \frac {32 c^3 \sqrt {b x+c x^2} (9 b B-8 A c)}{315 b^5 x}-\frac {16 c^2 \sqrt {b x+c x^2} (9 b B-8 A c)}{315 b^4 x^2}+\frac {4 c \sqrt {b x+c x^2} (9 b B-8 A c)}{105 b^3 x^3}-\frac {2 \sqrt {b x+c x^2} (9 b B-8 A c)}{63 b^2 x^4}-\frac {2 A \sqrt {b x+c x^2}}{9 b x^5} \]

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Rubi [A]  time = 0.14, antiderivative size = 160, 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, 650} \begin {gather*} \frac {32 c^3 \sqrt {b x+c x^2} (9 b B-8 A c)}{315 b^5 x}-\frac {16 c^2 \sqrt {b x+c x^2} (9 b B-8 A c)}{315 b^4 x^2}+\frac {4 c \sqrt {b x+c x^2} (9 b B-8 A c)}{105 b^3 x^3}-\frac {2 \sqrt {b x+c x^2} (9 b B-8 A c)}{63 b^2 x^4}-\frac {2 A \sqrt {b x+c x^2}}{9 b x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A*Sqrt[b*x + c*x^2])/(9*b*x^5) - (2*(9*b*B - 8*A*c)*Sqrt[b*x + c*x^2])/(63*b^2*x^4) + (4*c*(9*b*B - 8*A*c)
*Sqrt[b*x + c*x^2])/(105*b^3*x^3) - (16*c^2*(9*b*B - 8*A*c)*Sqrt[b*x + c*x^2])/(315*b^4*x^2) + (32*c^3*(9*b*B
- 8*A*c)*Sqrt[b*x + c*x^2])/(315*b^5*x)

Rule 650

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))/((p + 1)*(2*c*d - b*e)), 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] && EqQ[m + 2*p + 2, 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^5 \sqrt {b x+c x^2}} \, dx &=-\frac {2 A \sqrt {b x+c x^2}}{9 b x^5}+\frac {\left (2 \left (-5 (-b B+A c)+\frac {1}{2} (-b B+2 A c)\right )\right ) \int \frac {1}{x^4 \sqrt {b x+c x^2}} \, dx}{9 b}\\ &=-\frac {2 A \sqrt {b x+c x^2}}{9 b x^5}-\frac {2 (9 b B-8 A c) \sqrt {b x+c x^2}}{63 b^2 x^4}-\frac {(2 c (9 b B-8 A c)) \int \frac {1}{x^3 \sqrt {b x+c x^2}} \, dx}{21 b^2}\\ &=-\frac {2 A \sqrt {b x+c x^2}}{9 b x^5}-\frac {2 (9 b B-8 A c) \sqrt {b x+c x^2}}{63 b^2 x^4}+\frac {4 c (9 b B-8 A c) \sqrt {b x+c x^2}}{105 b^3 x^3}+\frac {\left (8 c^2 (9 b B-8 A c)\right ) \int \frac {1}{x^2 \sqrt {b x+c x^2}} \, dx}{105 b^3}\\ &=-\frac {2 A \sqrt {b x+c x^2}}{9 b x^5}-\frac {2 (9 b B-8 A c) \sqrt {b x+c x^2}}{63 b^2 x^4}+\frac {4 c (9 b B-8 A c) \sqrt {b x+c x^2}}{105 b^3 x^3}-\frac {16 c^2 (9 b B-8 A c) \sqrt {b x+c x^2}}{315 b^4 x^2}-\frac {\left (16 c^3 (9 b B-8 A c)\right ) \int \frac {1}{x \sqrt {b x+c x^2}} \, dx}{315 b^4}\\ &=-\frac {2 A \sqrt {b x+c x^2}}{9 b x^5}-\frac {2 (9 b B-8 A c) \sqrt {b x+c x^2}}{63 b^2 x^4}+\frac {4 c (9 b B-8 A c) \sqrt {b x+c x^2}}{105 b^3 x^3}-\frac {16 c^2 (9 b B-8 A c) \sqrt {b x+c x^2}}{315 b^4 x^2}+\frac {32 c^3 (9 b B-8 A c) \sqrt {b x+c x^2}}{315 b^5 x}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 100, normalized size = 0.62 \begin {gather*} -\frac {2 \sqrt {x (b+c x)} \left (A \left (35 b^4-40 b^3 c x+48 b^2 c^2 x^2-64 b c^3 x^3+128 c^4 x^4\right )+9 b B x \left (5 b^3-6 b^2 c x+8 b c^2 x^2-16 c^3 x^3\right )\right )}{315 b^5 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[x*(b + c*x)]*(9*b*B*x*(5*b^3 - 6*b^2*c*x + 8*b*c^2*x^2 - 16*c^3*x^3) + A*(35*b^4 - 40*b^3*c*x + 48*b^
2*c^2*x^2 - 64*b*c^3*x^3 + 128*c^4*x^4)))/(315*b^5*x^5)

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IntegrateAlgebraic [A]  time = 0.38, size = 108, normalized size = 0.68 \begin {gather*} -\frac {2 \sqrt {b x+c x^2} \left (35 A b^4-40 A b^3 c x+48 A b^2 c^2 x^2-64 A b c^3 x^3+128 A c^4 x^4+45 b^4 B x-54 b^3 B c x^2+72 b^2 B c^2 x^3-144 b B c^3 x^4\right )}{315 b^5 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[b*x + c*x^2]*(35*A*b^4 + 45*b^4*B*x - 40*A*b^3*c*x - 54*b^3*B*c*x^2 + 48*A*b^2*c^2*x^2 + 72*b^2*B*c^2
*x^3 - 64*A*b*c^3*x^3 - 144*b*B*c^3*x^4 + 128*A*c^4*x^4))/(315*b^5*x^5)

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fricas [A]  time = 0.41, size = 106, normalized size = 0.66 \begin {gather*} -\frac {2 \, {\left (35 \, A b^{4} - 16 \, {\left (9 \, B b c^{3} - 8 \, A c^{4}\right )} x^{4} + 8 \, {\left (9 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x^{3} - 6 \, {\left (9 \, B b^{3} c - 8 \, A b^{2} c^{2}\right )} x^{2} + 5 \, {\left (9 \, B b^{4} - 8 \, A b^{3} c\right )} x\right )} \sqrt {c x^{2} + b x}}{315 \, b^{5} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^5/(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.22, size = 251, normalized size = 1.57 \begin {gather*} \frac {2 \, {\left (630 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} B c^{\frac {3}{2}} + 756 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} B b c + 1008 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} A c^{2} + 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} B b^{2} \sqrt {c} + 1680 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} A b c^{\frac {3}{2}} + 45 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{3} + 1080 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b^{2} c + 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{3} \sqrt {c} + 35 \, A b^{4}\right )}}{315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^5/(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

2/315*(630*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*B*c^(3/2) + 756*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*b*c + 1008*(s
qrt(c)*x - sqrt(c*x^2 + b*x))^4*A*c^2 + 315*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^2*sqrt(c) + 1680*(sqrt(c)*x
- sqrt(c*x^2 + b*x))^3*A*b*c^(3/2) + 45*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^3 + 1080*(sqrt(c)*x - sqrt(c*x^2
+ b*x))^2*A*b^2*c + 315*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^3*sqrt(c) + 35*A*b^4)/(sqrt(c)*x - sqrt(c*x^2 + b
*x))^9

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maple [A]  time = 0.05, size = 110, normalized size = 0.69 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (128 A \,c^{4} x^{4}-144 B b \,c^{3} x^{4}-64 A b \,c^{3} x^{3}+72 B \,b^{2} c^{2} x^{3}+48 A \,b^{2} c^{2} x^{2}-54 B \,b^{3} c \,x^{2}-40 A \,b^{3} c x +45 b^{4} B x +35 A \,b^{4}\right )}{315 \sqrt {c \,x^{2}+b x}\, b^{5} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^5/(c*x^2+b*x)^(1/2),x)

[Out]

-2/315*(c*x+b)*(128*A*c^4*x^4-144*B*b*c^3*x^4-64*A*b*c^3*x^3+72*B*b^2*c^2*x^3+48*A*b^2*c^2*x^2-54*B*b^3*c*x^2-
40*A*b^3*c*x+45*B*b^4*x+35*A*b^4)/x^4/b^5/(c*x^2+b*x)^(1/2)

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maxima [A]  time = 0.89, size = 198, normalized size = 1.24 \begin {gather*} \frac {32 \, \sqrt {c x^{2} + b x} B c^{3}}{35 \, b^{4} x} - \frac {256 \, \sqrt {c x^{2} + b x} A c^{4}}{315 \, b^{5} x} - \frac {16 \, \sqrt {c x^{2} + b x} B c^{2}}{35 \, b^{3} x^{2}} + \frac {128 \, \sqrt {c x^{2} + b x} A c^{3}}{315 \, b^{4} x^{2}} + \frac {12 \, \sqrt {c x^{2} + b x} B c}{35 \, b^{2} x^{3}} - \frac {32 \, \sqrt {c x^{2} + b x} A c^{2}}{105 \, b^{3} x^{3}} - \frac {2 \, \sqrt {c x^{2} + b x} B}{7 \, b x^{4}} + \frac {16 \, \sqrt {c x^{2} + b x} A c}{63 \, b^{2} x^{4}} - \frac {2 \, \sqrt {c x^{2} + b x} A}{9 \, b x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^5/(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

32/35*sqrt(c*x^2 + b*x)*B*c^3/(b^4*x) - 256/315*sqrt(c*x^2 + b*x)*A*c^4/(b^5*x) - 16/35*sqrt(c*x^2 + b*x)*B*c^
2/(b^3*x^2) + 128/315*sqrt(c*x^2 + b*x)*A*c^3/(b^4*x^2) + 12/35*sqrt(c*x^2 + b*x)*B*c/(b^2*x^3) - 32/105*sqrt(
c*x^2 + b*x)*A*c^2/(b^3*x^3) - 2/7*sqrt(c*x^2 + b*x)*B/(b*x^4) + 16/63*sqrt(c*x^2 + b*x)*A*c/(b^2*x^4) - 2/9*s
qrt(c*x^2 + b*x)*A/(b*x^5)

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mupad [B]  time = 1.10, size = 146, normalized size = 0.91 \begin {gather*} \frac {\sqrt {c\,x^2+b\,x}\,\left (128\,A\,c^3-144\,B\,b\,c^2\right )}{315\,b^4\,x^2}-\frac {\sqrt {c\,x^2+b\,x}\,\left (256\,A\,c^4-288\,B\,b\,c^3\right )}{315\,b^5\,x}-\frac {\left (32\,A\,c^2-36\,B\,b\,c\right )\,\sqrt {c\,x^2+b\,x}}{105\,b^3\,x^3}-\frac {2\,A\,\sqrt {c\,x^2+b\,x}}{9\,b\,x^5}+\frac {\sqrt {c\,x^2+b\,x}\,\left (16\,A\,c-18\,B\,b\right )}{63\,b^2\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^5*(b*x + c*x^2)^(1/2)),x)

[Out]

((b*x + c*x^2)^(1/2)*(128*A*c^3 - 144*B*b*c^2))/(315*b^4*x^2) - ((b*x + c*x^2)^(1/2)*(256*A*c^4 - 288*B*b*c^3)
)/(315*b^5*x) - ((32*A*c^2 - 36*B*b*c)*(b*x + c*x^2)^(1/2))/(105*b^3*x^3) - (2*A*(b*x + c*x^2)^(1/2))/(9*b*x^5
) + ((b*x + c*x^2)^(1/2)*(16*A*c - 18*B*b))/(63*b^2*x^4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B x}{x^{5} \sqrt {x \left (b + c x\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**5/(c*x**2+b*x)**(1/2),x)

[Out]

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

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