3.21 \(\int \frac {(b x+c x^2)^{3/2}}{x^7} \, dx\)

Optimal. Leaf size=74 \[ -\frac {16 c^2 \left (b x+c x^2\right )^{5/2}}{315 b^3 x^5}+\frac {8 c \left (b x+c x^2\right )^{5/2}}{63 b^2 x^6}-\frac {2 \left (b x+c x^2\right )^{5/2}}{9 b x^7} \]

[Out]

-2/9*(c*x^2+b*x)^(5/2)/b/x^7+8/63*c*(c*x^2+b*x)^(5/2)/b^2/x^6-16/315*c^2*(c*x^2+b*x)^(5/2)/b^3/x^5

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Rubi [A]  time = 0.03, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {658, 650} \[ -\frac {16 c^2 \left (b x+c x^2\right )^{5/2}}{315 b^3 x^5}+\frac {8 c \left (b x+c x^2\right )^{5/2}}{63 b^2 x^6}-\frac {2 \left (b x+c x^2\right )^{5/2}}{9 b x^7} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(3/2)/x^7,x]

[Out]

(-2*(b*x + c*x^2)^(5/2))/(9*b*x^7) + (8*c*(b*x + c*x^2)^(5/2))/(63*b^2*x^6) - (16*c^2*(b*x + c*x^2)^(5/2))/(31
5*b^3*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^7} \, dx &=-\frac {2 \left (b x+c x^2\right )^{5/2}}{9 b x^7}-\frac {(4 c) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^6} \, dx}{9 b}\\ &=-\frac {2 \left (b x+c x^2\right )^{5/2}}{9 b x^7}+\frac {8 c \left (b x+c x^2\right )^{5/2}}{63 b^2 x^6}+\frac {\left (8 c^2\right ) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^5} \, dx}{63 b^2}\\ &=-\frac {2 \left (b x+c x^2\right )^{5/2}}{9 b x^7}+\frac {8 c \left (b x+c x^2\right )^{5/2}}{63 b^2 x^6}-\frac {16 c^2 \left (b x+c x^2\right )^{5/2}}{315 b^3 x^5}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 40, normalized size = 0.54 \[ -\frac {2 (x (b+c x))^{5/2} \left (35 b^2-20 b c x+8 c^2 x^2\right )}{315 b^3 x^7} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(3/2)/x^7,x]

[Out]

(-2*(x*(b + c*x))^(5/2)*(35*b^2 - 20*b*c*x + 8*c^2*x^2))/(315*b^3*x^7)

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fricas [A]  time = 0.92, size = 60, normalized size = 0.81 \[ -\frac {2 \, {\left (8 \, c^{4} x^{4} - 4 \, b c^{3} x^{3} + 3 \, b^{2} c^{2} x^{2} + 50 \, b^{3} c x + 35 \, b^{4}\right )} \sqrt {c x^{2} + b x}}{315 \, b^{3} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^7,x, algorithm="fricas")

[Out]

-2/315*(8*c^4*x^4 - 4*b*c^3*x^3 + 3*b^2*c^2*x^2 + 50*b^3*c*x + 35*b^4)*sqrt(c*x^2 + b*x)/(b^3*x^5)

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giac [B]  time = 0.20, size = 194, normalized size = 2.62 \[ \frac {2 \, {\left (420 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} c^{3} + 1575 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} b c^{\frac {5}{2}} + 2583 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} b^{2} c^{2} + 2310 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} b^{3} c^{\frac {3}{2}} + 1170 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} b^{4} c + 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b^{5} \sqrt {c} + 35 \, b^{6}\right )}}{315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^7,x, algorithm="giac")

[Out]

2/315*(420*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*c^3 + 1575*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*b*c^(5/2) + 2583*(sq
rt(c)*x - sqrt(c*x^2 + b*x))^4*b^2*c^2 + 2310*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*b^3*c^(3/2) + 1170*(sqrt(c)*x
- sqrt(c*x^2 + b*x))^2*b^4*c + 315*(sqrt(c)*x - sqrt(c*x^2 + b*x))*b^5*sqrt(c) + 35*b^6)/(sqrt(c)*x - sqrt(c*x
^2 + b*x))^9

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maple [A]  time = 0.05, size = 44, normalized size = 0.59 \[ -\frac {2 \left (c x +b \right ) \left (8 c^{2} x^{2}-20 b c x +35 b^{2}\right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{315 b^{3} x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x)^(3/2)/x^7,x)

[Out]

-2/315*(c*x+b)*(8*c^2*x^2-20*b*c*x+35*b^2)*(c*x^2+b*x)^(3/2)/b^3/x^6

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maxima [A]  time = 1.36, size = 117, normalized size = 1.58 \[ -\frac {16 \, \sqrt {c x^{2} + b x} c^{4}}{315 \, b^{3} x} + \frac {8 \, \sqrt {c x^{2} + b x} c^{3}}{315 \, b^{2} x^{2}} - \frac {2 \, \sqrt {c x^{2} + b x} c^{2}}{105 \, b x^{3}} + \frac {\sqrt {c x^{2} + b x} c}{63 \, x^{4}} + \frac {\sqrt {c x^{2} + b x} b}{9 \, x^{5}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}}}{3 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^7,x, algorithm="maxima")

[Out]

-16/315*sqrt(c*x^2 + b*x)*c^4/(b^3*x) + 8/315*sqrt(c*x^2 + b*x)*c^3/(b^2*x^2) - 2/105*sqrt(c*x^2 + b*x)*c^2/(b
*x^3) + 1/63*sqrt(c*x^2 + b*x)*c/x^4 + 1/9*sqrt(c*x^2 + b*x)*b/x^5 - 1/3*(c*x^2 + b*x)^(3/2)/x^6

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mupad [B]  time = 0.80, size = 101, normalized size = 1.36 \[ \frac {8\,c^3\,\sqrt {c\,x^2+b\,x}}{315\,b^2\,x^2}-\frac {20\,c\,\sqrt {c\,x^2+b\,x}}{63\,x^4}-\frac {2\,c^2\,\sqrt {c\,x^2+b\,x}}{105\,b\,x^3}-\frac {2\,b\,\sqrt {c\,x^2+b\,x}}{9\,x^5}-\frac {16\,c^4\,\sqrt {c\,x^2+b\,x}}{315\,b^3\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(3/2)/x^7,x)

[Out]

(8*c^3*(b*x + c*x^2)^(1/2))/(315*b^2*x^2) - (20*c*(b*x + c*x^2)^(1/2))/(63*x^4) - (2*c^2*(b*x + c*x^2)^(1/2))/
(105*b*x^3) - (2*b*(b*x + c*x^2)^(1/2))/(9*x^5) - (16*c^4*(b*x + c*x^2)^(1/2))/(315*b^3*x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}{x^{7}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)**(3/2)/x**7,x)

[Out]

Integral((x*(b + c*x))**(3/2)/x**7, x)

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