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

Optimal. Leaf size=131 \[ -\frac {128 c^2 (b+2 c x) (7 b B-10 A c)}{105 b^6 \sqrt {b x+c x^2}}+\frac {16 c (b+2 c x) (7 b B-10 A c)}{105 b^4 \left (b x+c x^2\right )^{3/2}}-\frac {2 (7 b B-10 A c)}{35 b^2 x \left (b x+c x^2\right )^{3/2}}-\frac {2 A}{7 b x^2 \left (b x+c x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.12, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {792, 658, 614, 613} \begin {gather*} -\frac {128 c^2 (b+2 c x) (7 b B-10 A c)}{105 b^6 \sqrt {b x+c x^2}}+\frac {16 c (b+2 c x) (7 b B-10 A c)}{105 b^4 \left (b x+c x^2\right )^{3/2}}-\frac {2 (7 b B-10 A c)}{35 b^2 x \left (b x+c x^2\right )^{3/2}}-\frac {2 A}{7 b x^2 \left (b x+c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A)/(7*b*x^2*(b*x + c*x^2)^(3/2)) - (2*(7*b*B - 10*A*c))/(35*b^2*x*(b*x + c*x^2)^(3/2)) + (16*c*(7*b*B - 10
*A*c)*(b + 2*c*x))/(105*b^4*(b*x + c*x^2)^(3/2)) - (128*c^2*(7*b*B - 10*A*c)*(b + 2*c*x))/(105*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 614

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^(p + 1))/((p +
1)*(b^2 - 4*a*c)), x] - Dist[(2*c*(2*p + 3))/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

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^2 \left (b x+c x^2\right )^{5/2}} \, dx &=-\frac {2 A}{7 b x^2 \left (b x+c x^2\right )^{3/2}}+\frac {\left (2 \left (-2 (-b B+A c)-\frac {3}{2} (-b B+2 A c)\right )\right ) \int \frac {1}{x \left (b x+c x^2\right )^{5/2}} \, dx}{7 b}\\ &=-\frac {2 A}{7 b x^2 \left (b x+c x^2\right )^{3/2}}-\frac {2 (7 b B-10 A c)}{35 b^2 x \left (b x+c x^2\right )^{3/2}}-\frac {(8 c (7 b B-10 A c)) \int \frac {1}{\left (b x+c x^2\right )^{5/2}} \, dx}{35 b^2}\\ &=-\frac {2 A}{7 b x^2 \left (b x+c x^2\right )^{3/2}}-\frac {2 (7 b B-10 A c)}{35 b^2 x \left (b x+c x^2\right )^{3/2}}+\frac {16 c (7 b B-10 A c) (b+2 c x)}{105 b^4 \left (b x+c x^2\right )^{3/2}}+\frac {\left (64 c^2 (7 b B-10 A c)\right ) \int \frac {1}{\left (b x+c x^2\right )^{3/2}} \, dx}{105 b^4}\\ &=-\frac {2 A}{7 b x^2 \left (b x+c x^2\right )^{3/2}}-\frac {2 (7 b B-10 A c)}{35 b^2 x \left (b x+c x^2\right )^{3/2}}+\frac {16 c (7 b B-10 A c) (b+2 c x)}{105 b^4 \left (b x+c x^2\right )^{3/2}}-\frac {128 c^2 (7 b B-10 A c) (b+2 c x)}{105 b^6 \sqrt {b x+c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 123, normalized size = 0.94 \begin {gather*} -\frac {2 \left (5 A \left (3 b^5-6 b^4 c x+16 b^3 c^2 x^2-96 b^2 c^3 x^3-384 b c^4 x^4-256 c^5 x^5\right )+7 b B x \left (3 b^4-8 b^3 c x+48 b^2 c^2 x^2+192 b c^3 x^3+128 c^4 x^4\right )\right )}{105 b^6 x^2 (x (b+c x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(7*b*B*x*(3*b^4 - 8*b^3*c*x + 48*b^2*c^2*x^2 + 192*b*c^3*x^3 + 128*c^4*x^4) + 5*A*(3*b^5 - 6*b^4*c*x + 16*
b^3*c^2*x^2 - 96*b^2*c^3*x^3 - 384*b*c^4*x^4 - 256*c^5*x^5)))/(105*b^6*x^2*(x*(b + c*x))^(3/2))

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IntegrateAlgebraic [A]  time = 0.44, size = 139, normalized size = 1.06 \begin {gather*} -\frac {2 \sqrt {b x+c x^2} \left (15 A b^5-30 A b^4 c x+80 A b^3 c^2 x^2-480 A b^2 c^3 x^3-1920 A b c^4 x^4-1280 A c^5 x^5+21 b^5 B x-56 b^4 B c x^2+336 b^3 B c^2 x^3+1344 b^2 B c^3 x^4+896 b B c^4 x^5\right )}{105 b^6 x^4 (b+c x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[b*x + c*x^2]*(15*A*b^5 + 21*b^5*B*x - 30*A*b^4*c*x - 56*b^4*B*c*x^2 + 80*A*b^3*c^2*x^2 + 336*b^3*B*c^
2*x^3 - 480*A*b^2*c^3*x^3 + 1344*b^2*B*c^3*x^4 - 1920*A*b*c^4*x^4 + 896*b*B*c^4*x^5 - 1280*A*c^5*x^5))/(105*b^
6*x^4*(b + c*x)^2)

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fricas [A]  time = 0.40, size = 153, normalized size = 1.17 \begin {gather*} -\frac {2 \, {\left (15 \, A b^{5} + 128 \, {\left (7 \, B b c^{4} - 10 \, A c^{5}\right )} x^{5} + 192 \, {\left (7 \, B b^{2} c^{3} - 10 \, A b c^{4}\right )} x^{4} + 48 \, {\left (7 \, B b^{3} c^{2} - 10 \, A b^{2} c^{3}\right )} x^{3} - 8 \, {\left (7 \, B b^{4} c - 10 \, A b^{3} c^{2}\right )} x^{2} + 3 \, {\left (7 \, B b^{5} - 10 \, A b^{4} c\right )} x\right )} \sqrt {c x^{2} + b x}}{105 \, {\left (b^{6} c^{2} x^{6} + 2 \, b^{7} c x^{5} + b^{8} x^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(15*A*b^5 + 128*(7*B*b*c^4 - 10*A*c^5)*x^5 + 192*(7*B*b^2*c^3 - 10*A*b*c^4)*x^4 + 48*(7*B*b^3*c^2 - 10*
A*b^2*c^3)*x^3 - 8*(7*B*b^4*c - 10*A*b^3*c^2)*x^2 + 3*(7*B*b^5 - 10*A*b^4*c)*x)*sqrt(c*x^2 + b*x)/(b^6*c^2*x^6
+ 2*b^7*c*x^5 + b^8*x^4)

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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 {5}{2}} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 134, normalized size = 1.02 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (-1280 A \,c^{5} x^{5}+896 B b \,c^{4} x^{5}-1920 A b \,c^{4} x^{4}+1344 B \,b^{2} c^{3} x^{4}-480 A \,b^{2} c^{3} x^{3}+336 B \,b^{3} c^{2} x^{3}+80 A \,b^{3} c^{2} x^{2}-56 B \,b^{4} c \,x^{2}-30 A \,b^{4} c x +21 B \,b^{5} x +15 A \,b^{5}\right )}{105 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} b^{6} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(c*x+b)*(-1280*A*c^5*x^5+896*B*b*c^4*x^5-1920*A*b*c^4*x^4+1344*B*b^2*c^3*x^4-480*A*b^2*c^3*x^3+336*B*b^
3*c^2*x^3+80*A*b^3*c^2*x^2-56*B*b^4*c*x^2-30*A*b^4*c*x+21*B*b^5*x+15*A*b^5)/x/b^6/(c*x^2+b*x)^(5/2)

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maxima [A]  time = 0.59, size = 224, normalized size = 1.71 \begin {gather*} \frac {32 \, B c^{2} x}{15 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{3}} - \frac {256 \, B c^{3} x}{15 \, \sqrt {c x^{2} + b x} b^{5}} - \frac {64 \, A c^{3} x}{21 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{4}} + \frac {512 \, A c^{4} x}{21 \, \sqrt {c x^{2} + b x} b^{6}} + \frac {16 \, B c}{15 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2}} - \frac {128 \, B c^{2}}{15 \, \sqrt {c x^{2} + b x} b^{4}} - \frac {32 \, A c^{2}}{21 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{3}} + \frac {256 \, A c^{3}}{21 \, \sqrt {c x^{2} + b x} b^{5}} - \frac {2 \, B}{5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b x} + \frac {4 \, A c}{7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2} x} - \frac {2 \, A}{7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32/15*B*c^2*x/((c*x^2 + b*x)^(3/2)*b^3) - 256/15*B*c^3*x/(sqrt(c*x^2 + b*x)*b^5) - 64/21*A*c^3*x/((c*x^2 + b*x
)^(3/2)*b^4) + 512/21*A*c^4*x/(sqrt(c*x^2 + b*x)*b^6) + 16/15*B*c/((c*x^2 + b*x)^(3/2)*b^2) - 128/15*B*c^2/(sq
rt(c*x^2 + b*x)*b^4) - 32/21*A*c^2/((c*x^2 + b*x)^(3/2)*b^3) + 256/21*A*c^3/(sqrt(c*x^2 + b*x)*b^5) - 2/5*B/((
c*x^2 + b*x)^(3/2)*b*x) + 4/7*A*c/((c*x^2 + b*x)^(3/2)*b^2*x) - 2/7*A/((c*x^2 + b*x)^(3/2)*b*x^2)

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mupad [B]  time = 1.33, size = 235, normalized size = 1.79 \begin {gather*} \frac {\sqrt {c\,x^2+b\,x}\,\left (\frac {1280\,A\,c^3-896\,B\,b\,c^2}{105\,b^5}+\frac {2\,c\,x\,\left (1280\,A\,c^3-896\,B\,b\,c^2\right )}{105\,b^6}\right )}{x\,\left (b+c\,x\right )}-\frac {\sqrt {c\,x^2+b\,x}\,\left (14\,B\,b^3-40\,A\,b^2\,c\right )}{35\,b^6\,x^3}-\frac {\sqrt {c\,x^2+b\,x}\,\left (x\,\left (\frac {4\,c^2\,\left (185\,A\,c-98\,B\,b\right )}{105\,b^4}+\frac {2\,c^2\,\left (230\,A\,c-91\,B\,b\right )}{105\,b^4}+\frac {b\,\left (\frac {160\,A\,c^4-56\,B\,b\,c^3}{105\,b^5}-\frac {4\,c^3\,\left (230\,A\,c-91\,B\,b\right )}{105\,b^5}\right )}{c}\right )+\frac {2\,c\,\left (185\,A\,c-98\,B\,b\right )}{105\,b^3}\right )}{x^2\,{\left (b+c\,x\right )}^2}-\frac {2\,A\,\sqrt {c\,x^2+b\,x}}{7\,b^3\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*x + c*x^2)^(1/2)*((1280*A*c^3 - 896*B*b*c^2)/(105*b^5) + (2*c*x*(1280*A*c^3 - 896*B*b*c^2))/(105*b^6)))/(x
*(b + c*x)) - ((b*x + c*x^2)^(1/2)*(14*B*b^3 - 40*A*b^2*c))/(35*b^6*x^3) - ((b*x + c*x^2)^(1/2)*(x*((4*c^2*(18
5*A*c - 98*B*b))/(105*b^4) + (2*c^2*(230*A*c - 91*B*b))/(105*b^4) + (b*((160*A*c^4 - 56*B*b*c^3)/(105*b^5) - (
4*c^3*(230*A*c - 91*B*b))/(105*b^5)))/c) + (2*c*(185*A*c - 98*B*b))/(105*b^3)))/(x^2*(b + c*x)^2) - (2*A*(b*x
+ c*x^2)^(1/2))/(7*b^3*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^{2} \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((B*x+A)/x**2/(c*x**2+b*x)**(5/2),x)

[Out]

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

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