\(\int \frac {(a+b x)^2}{x^3 (c x^2)^{5/2}} \, dx\) [850]

Optimal result

Integrand size = 20, antiderivative size = 66 \[ \int \frac {(a+b x)^2}{x^3 \left (c x^2\right )^{5/2}} \, dx=-\frac {a^2}{7 c^2 x^6 \sqrt {c x^2}}-\frac {a b}{3 c^2 x^5 \sqrt {c x^2}}-\frac {b^2}{5 c^2 x^4 \sqrt {c x^2}} \]

[Out]

-1/7*a^2/c^2/x^6/(c*x^2)^(1/2)-1/3*a*b/c^2/x^5/(c*x^2)^(1/2)-1/5*b^2/c^2/x^4/(c*x^2)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 45} \[ \int \frac {(a+b x)^2}{x^3 \left (c x^2\right )^{5/2}} \, dx=-\frac {a^2}{7 c^2 x^6 \sqrt {c x^2}}-\frac {a b}{3 c^2 x^5 \sqrt {c x^2}}-\frac {b^2}{5 c^2 x^4 \sqrt {c x^2}} \]

[In]

Int[(a + b*x)^2/(x^3*(c*x^2)^(5/2)),x]

[Out]

-1/7*a^2/(c^2*x^6*Sqrt[c*x^2]) - (a*b)/(3*c^2*x^5*Sqrt[c*x^2]) - b^2/(5*c^2*x^4*Sqrt[c*x^2])

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {x \int \frac {(a+b x)^2}{x^8} \, dx}{c^2 \sqrt {c x^2}} \\ & = \frac {x \int \left (\frac {a^2}{x^8}+\frac {2 a b}{x^7}+\frac {b^2}{x^6}\right ) \, dx}{c^2 \sqrt {c x^2}} \\ & = -\frac {a^2}{7 c^2 x^6 \sqrt {c x^2}}-\frac {a b}{3 c^2 x^5 \sqrt {c x^2}}-\frac {b^2}{5 c^2 x^4 \sqrt {c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.53 \[ \int \frac {(a+b x)^2}{x^3 \left (c x^2\right )^{5/2}} \, dx=\frac {-15 a^2-35 a b x-21 b^2 x^2}{105 x^2 \left (c x^2\right )^{5/2}} \]

[In]

Integrate[(a + b*x)^2/(x^3*(c*x^2)^(5/2)),x]

[Out]

(-15*a^2 - 35*a*b*x - 21*b^2*x^2)/(105*x^2*(c*x^2)^(5/2))

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.48

method	result	size
gosper	\(-\frac {21 b^{2} x^{2}+35 a b x +15 a^{2}}{105 x^{2} \left (c \,x^{2}\right )^{\frac {5}{2}}}\)	\(32\)
default	\(-\frac {21 b^{2} x^{2}+35 a b x +15 a^{2}}{105 x^{2} \left (c \,x^{2}\right )^{\frac {5}{2}}}\)	\(32\)
risch	\(\frac {-\frac {1}{5} b^{2} x^{2}-\frac {1}{3} a b x -\frac {1}{7} a^{2}}{c^{2} x^{6} \sqrt {c \,x^{2}}}\)	\(34\)
trager	\(\frac {\left (-1+x \right ) \left (15 a^{2} x^{6}+35 a b \,x^{6}+21 b^{2} x^{6}+15 a^{2} x^{5}+35 a b \,x^{5}+21 b^{2} x^{5}+15 a^{2} x^{4}+35 a b \,x^{4}+21 b^{2} x^{4}+15 a^{2} x^{3}+35 a b \,x^{3}+21 b^{2} x^{3}+15 a^{2} x^{2}+35 a b \,x^{2}+21 b^{2} x^{2}+15 a^{2} x +35 a b x +15 a^{2}\right ) \sqrt {c \,x^{2}}}{105 c^{3} x^{8}}\)	\(151\)

[In]

int((b*x+a)^2/x^3/(c*x^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/105*(21*b^2*x^2+35*a*b*x+15*a^2)/x^2/(c*x^2)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.52 \[ \int \frac {(a+b x)^2}{x^3 \left (c x^2\right )^{5/2}} \, dx=-\frac {{\left (21 \, b^{2} x^{2} + 35 \, a b x + 15 \, a^{2}\right )} \sqrt {c x^{2}}}{105 \, c^{3} x^{8}} \]

[In]

integrate((b*x+a)^2/x^3/(c*x^2)^(5/2),x, algorithm="fricas")

[Out]

-1/105*(21*b^2*x^2 + 35*a*b*x + 15*a^2)*sqrt(c*x^2)/(c^3*x^8)

Sympy [A] (verification not implemented)

Time = 0.65 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.70 \[ \int \frac {(a+b x)^2}{x^3 \left (c x^2\right )^{5/2}} \, dx=- \frac {a^{2}}{7 x^{2} \left (c x^{2}\right )^{\frac {5}{2}}} - \frac {a b}{3 x \left (c x^{2}\right )^{\frac {5}{2}}} - \frac {b^{2}}{5 \left (c x^{2}\right )^{\frac {5}{2}}} \]

[In]

integrate((b*x+a)**2/x**3/(c*x**2)**(5/2),x)

[Out]

-a**2/(7*x**2*(c*x**2)**(5/2)) - a*b/(3*x*(c*x**2)**(5/2)) - b**2/(5*(c*x**2)**(5/2))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.50 \[ \int \frac {(a+b x)^2}{x^3 \left (c x^2\right )^{5/2}} \, dx=-\frac {b^{2}}{5 \, c^{\frac {5}{2}} x^{5}} - \frac {a b}{3 \, c^{\frac {5}{2}} x^{6}} - \frac {a^{2}}{7 \, c^{\frac {5}{2}} x^{7}} \]

[In]

integrate((b*x+a)^2/x^3/(c*x^2)^(5/2),x, algorithm="maxima")

[Out]

-1/5*b^2/(c^(5/2)*x^5) - 1/3*a*b/(c^(5/2)*x^6) - 1/7*a^2/(c^(5/2)*x^7)

Giac [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.47 \[ \int \frac {(a+b x)^2}{x^3 \left (c x^2\right )^{5/2}} \, dx=-\frac {21 \, b^{2} x^{2} + 35 \, a b x + 15 \, a^{2}}{105 \, c^{\frac {5}{2}} x^{7} \mathrm {sgn}\left (x\right )} \]

[In]

integrate((b*x+a)^2/x^3/(c*x^2)^(5/2),x, algorithm="giac")

[Out]

-1/105*(21*b^2*x^2 + 35*a*b*x + 15*a^2)/(c^(5/2)*x^7*sgn(x))

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.64 \[ \int \frac {(a+b x)^2}{x^3 \left (c x^2\right )^{5/2}} \, dx=-\frac {15\,a^2\,\sqrt {x^2}+21\,b^2\,x^2\,\sqrt {x^2}+35\,a\,b\,x\,\sqrt {x^2}}{105\,c^{5/2}\,x^8} \]

[In]

int((a + b*x)^2/(x^3*(c*x^2)^(5/2)),x)

[Out]

-(15*a^2*(x^2)^(1/2) + 21*b^2*x^2*(x^2)^(1/2) + 35*a*b*x*(x^2)^(1/2))/(105*c^(5/2)*x^8)