### 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*(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)

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Rubi [A]  time = 0.0270773, antiderivative size = 74, normalized size of antiderivative = 1., 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 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 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]

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*}

Mathematica [A]  time = 0.0157896, 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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Maple [A]  time = 0.046, size = 44, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( 8\,{c}^{2}{x}^{2}-20\,bcx+35\,{b}^{2} \right ) }{315\,{x}^{6}{b}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}} \end{align*}

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)/x^6/b^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.93022, size = 135, normalized size = 1.82 \begin{align*} -\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}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (x \left (b + c x\right )\right )^{\frac{3}{2}}}{x^{7}}\, dx \end{align*}

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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Giac [B]  time = 1.23412, size = 262, normalized size = 3.54 \begin{align*} \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}} \end{align*}

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