3.9.49 \(\int \frac {(a+b x)^2}{x^2 (c x^2)^{5/2}} \, dx\) [849]

Optimal. Leaf size=66 \[ -\frac {a^2}{6 c^2 x^5 \sqrt {c x^2}}-\frac {2 a b}{5 c^2 x^4 \sqrt {c x^2}}-\frac {b^2}{4 c^2 x^3 \sqrt {c x^2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 45} \begin {gather*} -\frac {a^2}{6 c^2 x^5 \sqrt {c x^2}}-\frac {2 a b}{5 c^2 x^4 \sqrt {c x^2}}-\frac {b^2}{4 c^2 x^3 \sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*a^2/(c^2*x^5*Sqrt[c*x^2]) - (2*a*b)/(5*c^2*x^4*Sqrt[c*x^2]) - b^2/(4*c^2*x^3*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*} \int \frac {(a+b x)^2}{x^2 \left (c x^2\right )^{5/2}} \, dx &=\frac {x \int \frac {(a+b x)^2}{x^7} \, dx}{c^2 \sqrt {c x^2}}\\ &=\frac {x \int \left (\frac {a^2}{x^7}+\frac {2 a b}{x^6}+\frac {b^2}{x^5}\right ) \, dx}{c^2 \sqrt {c x^2}}\\ &=-\frac {a^2}{6 c^2 x^5 \sqrt {c x^2}}-\frac {2 a b}{5 c^2 x^4 \sqrt {c x^2}}-\frac {b^2}{4 c^2 x^3 \sqrt {c x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 38, normalized size = 0.58 \begin {gather*} -\frac {\sqrt {c x^2} \left (10 a^2+24 a b x+15 b^2 x^2\right )}{60 c^3 x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/60*(Sqrt[c*x^2]*(10*a^2 + 24*a*b*x + 15*b^2*x^2))/(c^3*x^7)

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Mathics [A]
time = 2.00, size = 32, normalized size = 0.48 \begin {gather*} \frac {\left (-\frac {a^2}{6}+\frac {b x \left (-8 a-5 b x\right )}{20}\right ) {\left (c x^2\right )}^{\frac {5}{2}}}{c^5 x^{11}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-a ^ 2 / 6 + b x (-8 a - 5 b x) / 20) (c x ^ 2) ^ (5 / 2) / (c ^ 5 x ^ 11)

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Maple [A]
time = 0.11, size = 32, normalized size = 0.48

method	result	size
gosper	\(-\frac {15 x^{2} b^{2}+24 a b x +10 a^{2}}{60 x \left (c \,x^{2}\right )^{\frac {5}{2}}}\)	\(32\)
default	\(-\frac {15 x^{2} b^{2}+24 a b x +10 a^{2}}{60 x \left (c \,x^{2}\right )^{\frac {5}{2}}}\)	\(32\)
risch	\(\frac {-\frac {1}{4} x^{2} b^{2}-\frac {2}{5} a b x -\frac {1}{6} a^{2}}{c^{2} x^{5} \sqrt {c \,x^{2}}}\)	\(34\)
trager	\(\frac {\left (-1+x \right ) \left (10 a^{2} x^{5}+24 a b \,x^{5}+15 b^{2} x^{5}+10 a^{2} x^{4}+24 a b \,x^{4}+15 b^{2} x^{4}+10 a^{2} x^{3}+24 a b \,x^{3}+15 b^{2} x^{3}+10 a^{2} x^{2}+24 a b \,x^{2}+15 x^{2} b^{2}+10 a^{2} x +24 a b x +10 a^{2}\right ) \sqrt {c \,x^{2}}}{60 c^{3} x^{7}}\)	\(128\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(15*b^2*x^2+24*a*b*x+10*a^2)/x/(c*x^2)^(5/2)

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Maxima [A]
time = 0.26, size = 33, normalized size = 0.50 \begin {gather*} -\frac {b^{2}}{4 \, c^{\frac {5}{2}} x^{4}} - \frac {2 \, a b}{5 \, c^{\frac {5}{2}} x^{5}} - \frac {a^{2}}{6 \, c^{\frac {5}{2}} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.29, size = 34, normalized size = 0.52 \begin {gather*} -\frac {{\left (15 \, b^{2} x^{2} + 24 \, a b x + 10 \, a^{2}\right )} \sqrt {c x^{2}}}{60 \, c^{3} x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(15*b^2*x^2 + 24*a*b*x + 10*a^2)*sqrt(c*x^2)/(c^3*x^7)

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Sympy [A]
time = 0.38, size = 46, normalized size = 0.70 \begin {gather*} - \frac {a^{2}}{6 x \left (c x^{2}\right )^{\frac {5}{2}}} - \frac {2 a b}{5 \left (c x^{2}\right )^{\frac {5}{2}}} - \frac {b^{2} x}{4 \left (c x^{2}\right )^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**2/(6*x*(c*x**2)**(5/2)) - 2*a*b/(5*(c*x**2)**(5/2)) - b**2*x/(4*(c*x**2)**(5/2))

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Giac [A]
time = 0.00, size = 41, normalized size = 0.62 \begin {gather*} \frac {-15 b^{2} x^{2}-24 a b x-10 a^{2}}{\sqrt {c}\cdot 60 \left (c^{2} x^{6} \mathrm {sign}\left (x\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(15*b^2*x^2 + 24*a*b*x + 10*a^2)/(c^(5/2)*x^6*sgn(x))

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Mupad [B]
time = 0.18, size = 42, normalized size = 0.64 \begin {gather*} -\frac {10\,a^2\,\sqrt {x^2}+15\,b^2\,x^2\,\sqrt {x^2}+24\,a\,b\,x\,\sqrt {x^2}}{60\,c^{5/2}\,x^7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(10*a^2*(x^2)^(1/2) + 15*b^2*x^2*(x^2)^(1/2) + 24*a*b*x*(x^2)^(1/2))/(60*c^(5/2)*x^7)

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