\(\int \frac {(b x+c x^2)^{3/2}}{x^7} \, dx\) [21]

Optimal result

Integrand size = 17, antiderivative size = 74 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^7} \, dx=-\frac {2 \left (b x+c x^2\right )^{5/2}}{9 b x^7}+\frac {8 c \left (b x+c x^2\right )^{5/2}}{63 b^2 x^6}-\frac {16 c^2 \left (b x+c x^2\right )^{5/2}}{315 b^3 x^5} \]

[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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {672, 664} \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^7} \, dx=-\frac {16 c^2 \left (b x+c x^2\right )^{5/2}}{315 b^3 x^5}+\frac {8 c \left (b x+c x^2\right )^{5/2}}{63 b^2 x^6}-\frac {2 \left (b x+c x^2\right )^{5/2}}{9 b x^7} \]

[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 664

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 672

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*} \text {integral}& = -\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] (verified)

Time = 0.17 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.54 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^7} \, dx=-\frac {2 (x (b+c x))^{5/2} \left (35 b^2-20 b c x+8 c^2 x^2\right )}{315 b^3 x^7} \]

[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)

Maple [A] (verified)

Time = 2.48 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.57

method	result	size
pseudoelliptic	\(-\frac {2 \left (\frac {8}{35} c^{2} x^{2}-\frac {4}{7} b c x +b^{2}\right ) \left (c x +b \right )^{2} \sqrt {x \left (c x +b \right )}}{9 x^{5} b^{3}}\)	\(42\)
gosper	\(-\frac {2 \left (c x +b \right ) \left (8 c^{2} x^{2}-20 b c x +35 b^{2}\right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{315 b^{3} x^{6}}\)	\(44\)
trager	\(-\frac {2 \left (8 c^{4} x^{4}-4 b \,c^{3} x^{3}+3 b^{2} c^{2} x^{2}+50 b^{3} c x +35 b^{4}\right ) \sqrt {c \,x^{2}+b x}}{315 b^{3} x^{5}}\)	\(61\)
risch	\(-\frac {2 \left (c x +b \right ) \left (8 c^{4} x^{4}-4 b \,c^{3} x^{3}+3 b^{2} c^{2} x^{2}+50 b^{3} c x +35 b^{4}\right )}{315 x^{4} \sqrt {x \left (c x +b \right )}\, b^{3}}\)	\(64\)
default	\(-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{9 b \,x^{7}}-\frac {4 c \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{7 b \,x^{6}}+\frac {4 c \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{35 b^{2} x^{5}}\right )}{9 b}\)	\(67\)

[In]

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

[Out]

-2/9*(8/35*c^2*x^2-4/7*b*c*x+b^2)*(c*x+b)^2*(x*(c*x+b))^(1/2)/x^5/b^3

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.81 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^7} \, dx=-\frac {2 \, {\left (8 \, c^{4} x^{4} - 4 \, b c^{3} x^{3} + 3 \, b^{2} c^{2} x^{2} + 50 \, b^{3} c x + 35 \, b^{4}\right )} \sqrt {c x^{2} + b x}}{315 \, b^{3} x^{5}} \]

[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)

Sympy [F]

\[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^7} \, dx=\int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}{x^{7}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.58 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^7} \, dx=-\frac {16 \, \sqrt {c x^{2} + b x} c^{4}}{315 \, b^{3} x} + \frac {8 \, \sqrt {c x^{2} + b x} c^{3}}{315 \, b^{2} x^{2}} - \frac {2 \, \sqrt {c x^{2} + b x} c^{2}}{105 \, b x^{3}} + \frac {\sqrt {c x^{2} + b x} c}{63 \, x^{4}} + \frac {\sqrt {c x^{2} + b x} b}{9 \, x^{5}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}}}{3 \, x^{6}} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (62) = 124\).

Time = 0.28 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.62 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^7} \, dx=\frac {2 \, {\left (420 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} c^{3} + 1575 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} b c^{\frac {5}{2}} + 2583 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} b^{2} c^{2} + 2310 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} b^{3} c^{\frac {3}{2}} + 1170 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} b^{4} c + 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b^{5} \sqrt {c} + 35 \, b^{6}\right )}}{315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{9}} \]

[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

Mupad [B] (verification not implemented)

Time = 9.68 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.36 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^7} \, dx=\frac {8\,c^3\,\sqrt {c\,x^2+b\,x}}{315\,b^2\,x^2}-\frac {20\,c\,\sqrt {c\,x^2+b\,x}}{63\,x^4}-\frac {2\,c^2\,\sqrt {c\,x^2+b\,x}}{105\,b\,x^3}-\frac {2\,b\,\sqrt {c\,x^2+b\,x}}{9\,x^5}-\frac {16\,c^4\,\sqrt {c\,x^2+b\,x}}{315\,b^3\,x} \]

[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)