3.23 \(\int \frac {(b x+c x^2)^{3/2}}{x^9} \, dx\)

Optimal. Leaf size=126 \[ -\frac {256 c^4 \left (b x+c x^2\right )^{5/2}}{15015 b^5 x^5}+\frac {128 c^3 \left (b x+c x^2\right )^{5/2}}{3003 b^4 x^6}-\frac {32 c^2 \left (b x+c x^2\right )^{5/2}}{429 b^3 x^7}+\frac {16 c \left (b x+c x^2\right )^{5/2}}{143 b^2 x^8}-\frac {2 \left (b x+c x^2\right )^{5/2}}{13 b x^9} \]

[Out]

-2/13*(c*x^2+b*x)^(5/2)/b/x^9+16/143*c*(c*x^2+b*x)^(5/2)/b^2/x^8-32/429*c^2*(c*x^2+b*x)^(5/2)/b^3/x^7+128/3003
*c^3*(c*x^2+b*x)^(5/2)/b^4/x^6-256/15015*c^4*(c*x^2+b*x)^(5/2)/b^5/x^5

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {658, 650} \[ -\frac {256 c^4 \left (b x+c x^2\right )^{5/2}}{15015 b^5 x^5}+\frac {128 c^3 \left (b x+c x^2\right )^{5/2}}{3003 b^4 x^6}-\frac {32 c^2 \left (b x+c x^2\right )^{5/2}}{429 b^3 x^7}+\frac {16 c \left (b x+c x^2\right )^{5/2}}{143 b^2 x^8}-\frac {2 \left (b x+c x^2\right )^{5/2}}{13 b x^9} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(3/2)/x^9,x]

[Out]

(-2*(b*x + c*x^2)^(5/2))/(13*b*x^9) + (16*c*(b*x + c*x^2)^(5/2))/(143*b^2*x^8) - (32*c^2*(b*x + c*x^2)^(5/2))/
(429*b^3*x^7) + (128*c^3*(b*x + c*x^2)^(5/2))/(3003*b^4*x^6) - (256*c^4*(b*x + c*x^2)^(5/2))/(15015*b^5*x^5)

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]

Rubi steps

\begin {align*} \int \frac {\left (b x+c x^2\right )^{3/2}}{x^9} \, dx &=-\frac {2 \left (b x+c x^2\right )^{5/2}}{13 b x^9}-\frac {(8 c) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^8} \, dx}{13 b}\\ &=-\frac {2 \left (b x+c x^2\right )^{5/2}}{13 b x^9}+\frac {16 c \left (b x+c x^2\right )^{5/2}}{143 b^2 x^8}+\frac {\left (48 c^2\right ) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^7} \, dx}{143 b^2}\\ &=-\frac {2 \left (b x+c x^2\right )^{5/2}}{13 b x^9}+\frac {16 c \left (b x+c x^2\right )^{5/2}}{143 b^2 x^8}-\frac {32 c^2 \left (b x+c x^2\right )^{5/2}}{429 b^3 x^7}-\frac {\left (64 c^3\right ) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^6} \, dx}{429 b^3}\\ &=-\frac {2 \left (b x+c x^2\right )^{5/2}}{13 b x^9}+\frac {16 c \left (b x+c x^2\right )^{5/2}}{143 b^2 x^8}-\frac {32 c^2 \left (b x+c x^2\right )^{5/2}}{429 b^3 x^7}+\frac {128 c^3 \left (b x+c x^2\right )^{5/2}}{3003 b^4 x^6}+\frac {\left (128 c^4\right ) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^5} \, dx}{3003 b^4}\\ &=-\frac {2 \left (b x+c x^2\right )^{5/2}}{13 b x^9}+\frac {16 c \left (b x+c x^2\right )^{5/2}}{143 b^2 x^8}-\frac {32 c^2 \left (b x+c x^2\right )^{5/2}}{429 b^3 x^7}+\frac {128 c^3 \left (b x+c x^2\right )^{5/2}}{3003 b^4 x^6}-\frac {256 c^4 \left (b x+c x^2\right )^{5/2}}{15015 b^5 x^5}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 62, normalized size = 0.49 \[ -\frac {2 (x (b+c x))^{5/2} \left (1155 b^4-840 b^3 c x+560 b^2 c^2 x^2-320 b c^3 x^3+128 c^4 x^4\right )}{15015 b^5 x^9} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(3/2)/x^9,x]

[Out]

(-2*(x*(b + c*x))^(5/2)*(1155*b^4 - 840*b^3*c*x + 560*b^2*c^2*x^2 - 320*b*c^3*x^3 + 128*c^4*x^4))/(15015*b^5*x
^9)

________________________________________________________________________________________

fricas [A]  time = 0.90, size = 82, normalized size = 0.65 \[ -\frac {2 \, {\left (128 \, c^{6} x^{6} - 64 \, b c^{5} x^{5} + 48 \, b^{2} c^{4} x^{4} - 40 \, b^{3} c^{3} x^{3} + 35 \, b^{4} c^{2} x^{2} + 1470 \, b^{5} c x + 1155 \, b^{6}\right )} \sqrt {c x^{2} + b x}}{15015 \, b^{5} x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^9,x, algorithm="fricas")

[Out]

-2/15015*(128*c^6*x^6 - 64*b*c^5*x^5 + 48*b^2*c^4*x^4 - 40*b^3*c^3*x^3 + 35*b^4*c^2*x^2 + 1470*b^5*c*x + 1155*
b^6)*sqrt(c*x^2 + b*x)/(b^5*x^7)

________________________________________________________________________________________

giac [B]  time = 0.20, size = 252, normalized size = 2.00 \[ \frac {2 \, {\left (48048 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{8} c^{4} + 240240 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{7} b c^{\frac {7}{2}} + 531960 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} b^{2} c^{3} + 675675 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} b^{3} c^{\frac {5}{2}} + 535535 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} b^{4} c^{2} + 270270 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} b^{5} c^{\frac {3}{2}} + 84630 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} b^{6} c + 15015 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b^{7} \sqrt {c} + 1155 \, b^{8}\right )}}{15015 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{13}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^9,x, algorithm="giac")

[Out]

2/15015*(48048*(sqrt(c)*x - sqrt(c*x^2 + b*x))^8*c^4 + 240240*(sqrt(c)*x - sqrt(c*x^2 + b*x))^7*b*c^(7/2) + 53
1960*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*b^2*c^3 + 675675*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*b^3*c^(5/2) + 535535
*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*b^4*c^2 + 270270*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*b^5*c^(3/2) + 84630*(sqr
t(c)*x - sqrt(c*x^2 + b*x))^2*b^6*c + 15015*(sqrt(c)*x - sqrt(c*x^2 + b*x))*b^7*sqrt(c) + 1155*b^8)/(sqrt(c)*x
- sqrt(c*x^2 + b*x))^13

________________________________________________________________________________________

maple [A]  time = 0.04, size = 66, normalized size = 0.52 \[ -\frac {2 \left (c x +b \right ) \left (128 c^{4} x^{4}-320 x^{3} c^{3} b +560 c^{2} x^{2} b^{2}-840 c x \,b^{3}+1155 b^{4}\right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{15015 b^{5} x^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x)^(3/2)/x^9,x)

[Out]

-2/15015*(c*x+b)*(128*c^4*x^4-320*b*c^3*x^3+560*b^2*c^2*x^2-840*b^3*c*x+1155*b^4)*(c*x^2+b*x)^(3/2)/b^5/x^8

________________________________________________________________________________________

maxima [A]  time = 1.41, size = 161, normalized size = 1.28 \[ -\frac {256 \, \sqrt {c x^{2} + b x} c^{6}}{15015 \, b^{5} x} + \frac {128 \, \sqrt {c x^{2} + b x} c^{5}}{15015 \, b^{4} x^{2}} - \frac {32 \, \sqrt {c x^{2} + b x} c^{4}}{5005 \, b^{3} x^{3}} + \frac {16 \, \sqrt {c x^{2} + b x} c^{3}}{3003 \, b^{2} x^{4}} - \frac {2 \, \sqrt {c x^{2} + b x} c^{2}}{429 \, b x^{5}} + \frac {3 \, \sqrt {c x^{2} + b x} c}{715 \, x^{6}} + \frac {3 \, \sqrt {c x^{2} + b x} b}{65 \, x^{7}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}}}{5 \, x^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^9,x, algorithm="maxima")

[Out]

-256/15015*sqrt(c*x^2 + b*x)*c^6/(b^5*x) + 128/15015*sqrt(c*x^2 + b*x)*c^5/(b^4*x^2) - 32/5005*sqrt(c*x^2 + b*
x)*c^4/(b^3*x^3) + 16/3003*sqrt(c*x^2 + b*x)*c^3/(b^2*x^4) - 2/429*sqrt(c*x^2 + b*x)*c^2/(b*x^5) + 3/715*sqrt(
c*x^2 + b*x)*c/x^6 + 3/65*sqrt(c*x^2 + b*x)*b/x^7 - 1/5*(c*x^2 + b*x)^(3/2)/x^8

________________________________________________________________________________________

mupad [B]  time = 1.21, size = 145, normalized size = 1.15 \[ \frac {16\,c^3\,\sqrt {c\,x^2+b\,x}}{3003\,b^2\,x^4}-\frac {28\,c\,\sqrt {c\,x^2+b\,x}}{143\,x^6}-\frac {2\,c^2\,\sqrt {c\,x^2+b\,x}}{429\,b\,x^5}-\frac {2\,b\,\sqrt {c\,x^2+b\,x}}{13\,x^7}-\frac {32\,c^4\,\sqrt {c\,x^2+b\,x}}{5005\,b^3\,x^3}+\frac {128\,c^5\,\sqrt {c\,x^2+b\,x}}{15015\,b^4\,x^2}-\frac {256\,c^6\,\sqrt {c\,x^2+b\,x}}{15015\,b^5\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(3/2)/x^9,x)

[Out]

(16*c^3*(b*x + c*x^2)^(1/2))/(3003*b^2*x^4) - (28*c*(b*x + c*x^2)^(1/2))/(143*x^6) - (2*c^2*(b*x + c*x^2)^(1/2
))/(429*b*x^5) - (2*b*(b*x + c*x^2)^(1/2))/(13*x^7) - (32*c^4*(b*x + c*x^2)^(1/2))/(5005*b^3*x^3) + (128*c^5*(
b*x + c*x^2)^(1/2))/(15015*b^4*x^2) - (256*c^6*(b*x + c*x^2)^(1/2))/(15015*b^5*x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}{x^{9}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)**(3/2)/x**9,x)

[Out]

Integral((x*(b + c*x))**(3/2)/x**9, x)

________________________________________________________________________________________