3.9.48 \(\int \frac {(a+b x)^2}{x (c x^2)^{5/2}} \, dx\) [848]

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

[Out]

-1/5*a^2/c^2/x^4/(c*x^2)^(1/2)-1/2*a*b/c^2/x^3/(c*x^2)^(1/2)-1/3*b^2/c^2/x^2/(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}{5 c^2 x^4 \sqrt {c x^2}}-\frac {a b}{2 c^2 x^3 \sqrt {c x^2}}-\frac {b^2}{3 c^2 x^2 \sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/(x*(c*x^2)^(5/2)),x]

[Out]

-1/5*a^2/(c^2*x^4*Sqrt[c*x^2]) - (a*b)/(2*c^2*x^3*Sqrt[c*x^2]) - b^2/(3*c^2*x^2*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 \left (c x^2\right )^{5/2}} \, dx &=\frac {x \int \frac {(a+b x)^2}{x^6} \, dx}{c^2 \sqrt {c x^2}}\\ &=\frac {x \int \left (\frac {a^2}{x^6}+\frac {2 a b}{x^5}+\frac {b^2}{x^4}\right ) \, dx}{c^2 \sqrt {c x^2}}\\ &=-\frac {a^2}{5 c^2 x^4 \sqrt {c x^2}}-\frac {a b}{2 c^2 x^3 \sqrt {c x^2}}-\frac {b^2}{3 c^2 x^2 \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 (6 a^2+15 a b x+10 b^2 x^2\right )}{30 c^3 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/30*(Sqrt[c*x^2]*(6*a^2 + 15*a*b*x + 10*b^2*x^2))/(c^3*x^6)

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Mathics [A]
time = 1.93, size = 27, normalized size = 0.41 \begin {gather*} \frac {-\frac {a^2}{5}-\frac {a b x}{2}-\frac {b^2 x^2}{3}}{{\left (c x^2\right )}^{\frac {5}{2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-a ^ 2 / 5 - a b x / 2 - b ^ 2 x ^ 2 / 3) / (c x ^ 2) ^ (5 / 2)

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Maple [A]
time = 0.12, size = 29, normalized size = 0.44

method	result	size
gosper	\(-\frac {10 x^{2} b^{2}+15 a b x +6 a^{2}}{30 \left (c \,x^{2}\right )^{\frac {5}{2}}}\)	\(29\)
default	\(-\frac {10 x^{2} b^{2}+15 a b x +6 a^{2}}{30 \left (c \,x^{2}\right )^{\frac {5}{2}}}\)	\(29\)
risch	\(\frac {-\frac {1}{3} x^{2} b^{2}-\frac {1}{2} a b x -\frac {1}{5} a^{2}}{c^{2} x^{4} \sqrt {c \,x^{2}}}\)	\(34\)
trager	\(\frac {\left (-1+x \right ) \left (6 a^{2} x^{4}+15 a b \,x^{4}+10 b^{2} x^{4}+6 a^{2} x^{3}+15 a b \,x^{3}+10 b^{2} x^{3}+6 a^{2} x^{2}+15 a b \,x^{2}+10 x^{2} b^{2}+6 a^{2} x +15 a b x +6 a^{2}\right ) \sqrt {c \,x^{2}}}{30 c^{3} x^{6}}\)	\(105\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(10*b^2*x^2+15*a*b*x+6*a^2)/(c*x^2)^(5/2)

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Maxima [A]
time = 0.26, size = 37, normalized size = 0.56 \begin {gather*} -\frac {b^{2}}{3 \, \left (c x^{2}\right )^{\frac {3}{2}} c} - \frac {a b}{2 \, c^{\frac {5}{2}} x^{4}} - \frac {a^{2}}{5 \, c^{\frac {5}{2}} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.29, size = 34, normalized size = 0.52 \begin {gather*} -\frac {{\left (10 \, b^{2} x^{2} + 15 \, a b x + 6 \, a^{2}\right )} \sqrt {c x^{2}}}{30 \, c^{3} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(10*b^2*x^2 + 15*a*b*x + 6*a^2)*sqrt(c*x^2)/(c^3*x^6)

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Sympy [A]
time = 0.34, size = 46, normalized size = 0.70 \begin {gather*} - \frac {a^{2}}{5 \left (c x^{2}\right )^{\frac {5}{2}}} - \frac {a b x}{2 \left (c x^{2}\right )^{\frac {5}{2}}} - \frac {b^{2} x^{2}}{3 \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/(c*x**2)**(5/2),x)

[Out]

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

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Giac [A]
time = 0.00, size = 41, normalized size = 0.62 \begin {gather*} \frac {-10 b^{2} x^{2}-15 a b x-6 a^{2}}{\sqrt {c}\cdot 30 \left (c^{2} x^{5} \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/(c*x^2)^(5/2),x)

[Out]

-1/30*(10*b^2*x^2 + 15*a*b*x + 6*a^2)/(c^(5/2)*x^5*sgn(x))

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Mupad [B]
time = 0.18, size = 42, normalized size = 0.64 \begin {gather*} -\frac {6\,a^2\,\sqrt {x^2}+10\,b^2\,x^2\,\sqrt {x^2}+15\,a\,b\,x\,\sqrt {x^2}}{30\,c^{5/2}\,x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^2/(x*(c*x^2)^(5/2)),x)

[Out]

-(6*a^2*(x^2)^(1/2) + 10*b^2*x^2*(x^2)^(1/2) + 15*a*b*x*(x^2)^(1/2))/(30*c^(5/2)*x^6)

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